the official guide companion 12th [2010.11.17]

Patrick Siewe - GMATFix.com OG Companion 12th Table of Contents

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Table of Contents Table of Contents ....................................................................................................................................................... 1

About The Author ...................................................................................................................................................... 2

Why You Need This Book ......................................................................................................................................... 3

How to Use this Book ................................................................................................................................................ 6

GMAT Interviews ...................................................................................................................................................... 7

Diagnostic Test Solutions ........................................................................................................................................ 16

Problem Solving Solutions ...................................................................................................................................... 65

Data Sufficiency Solutions .................................................................................................................................... 252

Advanced Speed Drills .......................................................................................................................................... 382

Advanced Drills Solutions ..................................................................................................................................... 403

Take Aways List .................................................................................................................................................... 539

Reference Tables .................................................................................................................................................... 543

Questions By Topic ............................................................................................................................................ 544

Topic Frequency ................................................................................................................................................ 546

Scores to Percentiles Conversion Scales ............................................................................................................ 547

Patrick Siewe – GMATFix.com OG Companion 12th About the Author

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About The Author

Since 2004, Patrick has worked as an exam content writer and professional tutor in New York City. After graduating from Cornell University with a Bachelor’s in Political Science, he worked for the Princeton Review where he taught GMAT, GRE and SAT classes, and where he was subcontracted to write math and verbal questions for the state of New York’s Regents examination. He later moved to ManhattanGMAT, a company that specializes in GMAT tutoring.

Over the course of his early career, Patrick noticed that the best prepared test takers were those who successfully mastered the Official questions published by GMAC™ (the owner of the GMAT). It became clear that the most effective program of study would be one that aimed to achieve this mastery. In 2007, Patrick funded GMATFix, a specialist in providing detailed explanations and solving strategies to questions from GMAC’s Official Guides and GMATPrep Test software. In his private practice, Patrick couples GMATFix tools with the holistic tutoring approach outlined below, yielding typical score increases of 50 to 180 points for 1-on-1 clients who follow his full course.

• Technical Training, examining each question type and sub-type in detail and teaching the student the skills to solve all question types. One can think of this training as a classic grammar or math academic course.

• Tactical Training, teaching the student ways to eliminate wrong answers when facing a question too difficult to properly solve. Problem solving techniques such as “Variable Substitution”, and “Reverse Engineering” can drastically improve a student’s odds on a difficult or unfamiliar question. Tactical Training also includes teaching the student when to cut his losses and move on from a question.

• Strategic Planning, helping the student understand when to take the test, whether to retake the test, how the test is scored, how much time to allot to each question, and how to study in order to transfer hard work into concrete score improvement. Good study skills are lost on most people after high school and are a critical aspect of preparation that is too often ignored by standard prep courses.

• Psychological Preparation. Patrick is trained in test anxiety management and makes sure to address this dimension in his tutoring. Too many students simply shut down on test day because fear or anxiety can be overwhelming. Unfortunately, very few GMAT prep programs make it a point to address this issue. It is not possible to completely get rid of anxiety, but there are several useful techniques that can make it manageable. These techniques, like every other, need to be practiced before the day of the test and they are integral to Patrick’s GMAT prep curriculum.

Patrick is a native French speaker who learned English in high school. This gives him a particularly good understanding of students who struggle with the verbal section of the GMAT. He can be reached via GMATFix.com

Patrick Siewe - GMATFix.com OG Companion 12th Is This Book for You?

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Why You Need This Book

The best GMAT practice comes from actual GMAT questions. The most extensive resource of past GMAT questions is the “Official Guide for GMAT Review, 12th Edition” also known as the OG12. The importance of understanding every single question in that book cannot be overstated. The OG12, along with the GMATPrep Test software, should be the cornerstone of any GMAT prep program.

Unfortunately, the OG’s flaws prevent many students from getting the most out of it. Below I have outlined complaints from my students about the OG, starting with the most common complaint. This Companion book has been designed to complement the OG by addressing these grievances. For each deficiency, I have provided a solution.1

“The Official Guide has lots of practice questions but is very thin on theory review. The math section in particular has skimpy explanations insufficient to fully master some questions.”

The greatest weakness of The Official Guide is the lack of details in its math solutions. The primary aim of the OG Companion is to provide a thorough review of every single math question from The Official Guide. One of the beauties of math is that the same problem can often be solved via different approaches. Knowing more about a problem than the bare minimum is necessary to master the problem type and will improve your odds of doing a similar problem better. In this book, you will find two, three, or sometimes four thorough explanations with illustrations for every question. In The Official Guide, most solutions are much shorter than this paragraph whereas in the OG Companion, solutions are much more thoroughly discussed and are often 2 or 3 pages long. Spend time studying the different ways to solve the same question, because in doing so you will be improving various skills. While writing my solutions, I’ve made sure to point out patterns I’ve noticed in how the question writers attempt to trick students into picking wrong answers.

“The Official Guide doesn’t provide ways to eliminate wrong answers and perform educated guesses when facing very difficult questions. It doesn’t help me improve my test-taking strategy.”

Unless you are in the top 0.3% of test takers in your math ability, you will run into questions that are either too difficult to do or that would take too long to do. Because you must answer 37 questions in 75 minutes, you have about 2 minutes per question on average on the real exam. This means that you will sometimes need to cut your losses, make an educated guess and move on. Thankfully many GMAT wrong answers can be eliminated with a few clever observations or approximations. Where applicable, my solutions feature a “30 seconds Hack”, a guide

1 To read more about what people liked or didn’t like about the Official Guide, look up “Official Guide for GMAT Review, 11th Edition” on Amazon.com and read its user reviews.

Patrick Siewe – GMATFix.com OG Companion 12th Is This Book for You?

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to eliminating one or more wrong answers in 30 seconds without actually solving the problem. The Hacks are meant to teach you how to significantly improve your guessing odds when you don’t have enough time to solve.

“Many of the math questions in The Official Guide are way easier than those I saw on my real GMAT. Only the last 50 to 75 questions in each math section were on par with the more difficult questions on the actual exam”

The Official Guide questions are organized by difficulty level, but are geared toward a mainstream audience. If your target score is anything above 650, you may not find enough challenging practice problems in the OG.2

“I know the Official Guide questions are roughly ordered from easier to harder, but how can I know exactly how difficult a question is so I can better gauge how I’m doing and also know which questions to spend my study time on?”

The OG Companion includes a set of 10 advanced speed drills containing 150 questions. These questions were carefully drafted to test the same concepts in the same ways as the GMAT tests. Almost all the questions from these drills are of the 600 – 700 and the 700+ difficulty level. Of course, I have provided explanations to every question for your reviewing fancy. Before you get to these drills however, be sure that you fully complete and understand all the questions from the Official Guide. Doing a lot of difficult questions without having a good grasp of easier and moderate questions can not only be frustrating, but also be counter-productive.

In this book, the solution to every Official Guide question includes that question’s difficulty level. There are five levels loosely corresponding to GMAT scores. The difficulty levels are 200-400, 400-500, 500-600, 600-700, and 700+. Along with the difficulty level, each solution specifies which math topics are tested by the question. There are about 13 distinct topics of arithmetic, algebra and geometry, and one question may test multiple topics.

“If I want to practice a particular question type, it takes too long to read through the pages of the Official Guide looking for questions of that type. There is no easy way to get a listing of questions by topic”

I have put together a “Reference” section that organizes question numbers by topic and difficulty. If you need to practice “Rates & Work” questions for instance, lookup that topic in the Reference section, select the difficulty level you want to practice and you will see a list of all Official Guide questions that test this concept.

“I would like a way to know which percentiles different scores correspond to.”

The “Reference” section of this book offers statistical data on how sub-scores and percentiles correlate with GMAT scores. From this section you can answer questions such as, “What GMAT score should I aim for if I want to be in the top 10% of test-takers?” and “Which essay score would put me in the top 20%?”

“The Official Guide doesn’t tell me which question types are most common, so I don’t know what topics are most important to master.”

Not all topics are equal on the GMAT. There are a lot more questions about Fractions & Percentages than questions about Combinations & Permutations. The “Reference” section of this book provides an analysis of the frequency of topics in The Official Guide, so you can devise a study plan ensuring that you have mastered the topics that come up the most.

2 According to my calculations, only about 13.7% of questions in The Official Guide are of the 700+ difficulty level

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“I have general questions about the GMAT and the B-school application process that are not addressed in the Official Guide”

The GMAT Interviews section of this book is comprised of four interviews. Ryan Childs, a GMAT Tutor, goes over how much score improvement you can expect, how to decide between classes and tutoring, how to recognize a good teacher, and how much you can expect to spend on a GMAT prep program. Mia Tallico, a GMAT Program Developer, talks about the registration and rescheduling procedures of the GMAT, what you can expect on test day, how much time you should spend preparing for your exam, and which GMAT prep books are the best. Steven Martins, an admissions consultant, discusses the application essay, work experience, recommendations, and other important aspects of the business school application. Lastly, Christina Chang, my former 740 GMAT student, shares her views on online courses vs. live classes vs. 1-on-1 tutoring, how she organized her study program, and how her score progressed throughout her preparation.

Head to www.GMATFix.com to learn more about this as well as other resources that I’ve developed for GMAT takers.

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Patrick Siewe – GMATFix.com OG Companion 12th How to Use this Book

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How to Use this Book

The OG Companion 12th provides thorough solutions to every single math problem from the “Official Guide for GMAT Review, 12th Edition” (OG).

Each solution is headed by an information panel that shows where you can find the corresponding question in the OG.

The panel also shows how difficult the question is. There are five difficulty levels: 200-400, 400-500, 500-600, 600-700 and 700+. Finally, the “Topics” line shows which of 13 math topics are tested by that particular question.

Where applicable, I have added a Hack to the standard solutions. Each hack is a guide to eliminating one or more wrong answers in 30 seconds without actually solving the problem. The Hacks are meant to teach you how to increase your

guessing odds when you cannot solve a problem.

The Official Guide solutions are split into three sections in this book: “Diagnostic Test” solutions review the questions from pg 20-26; “Problem Solving” solutions review the questions from pg 152-185; and “Data Sufficiency” solutions review the questions from pg 273-288. Page numbers are not from this book, but are from The Official Guide.

A “GMAT Interviews” section includes interviews with four people who have very different perspectives of the GMAT – a tutor, an admission consultant, a GMAT program developer, and a 740 GMAT student. They answer questions about the exam, the study process, the business school applications, and GMAC policies

An “Advanced Speed Drills” section has 10 drills geared for those students who target quant scores above the 90th percentile. I have included this section because The Official Guide doesn’t include enough practice with challenging questions. Most quant questions from that book are of easy and moderate difficulty. The speed drills are of course accompanied by solutions with explanations.

A “Take-Aways” section lists over 60 of the most important lessons you should have learned while going through the Official Guide problems and solutions. These Take-Aways address GMAT specific concepts that you wouldn’t find in standard math manuals such as patterns in the way the question writers attempt to trick test-takers.

Finally, a “Reference” section organizes every question in the Official Guide by topic and difficulty, so if you need to practice a particular math topic you can quickly devise a study plan. The Reference section also includes information about how GMAT scores and sub-scores translate to percentiles. Finally, this section measures the frequency with which every math topic is tested on the GMAT, so you can determine which topics are most important to master

#1, pg. 20 Difficulty Level: 200-400 Topics: Translations & Manipulations

30 Seconds Hack

Patrick Siewe - GMATFix.com OG Companion 12th GMAT Interviews

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GMAT Interviews

• Ryan Childs, GMAT Tutor • Mia Tallico, GMAT Program Developer • Steven Martins, Admissions Consultant • Christina Chang, GMAT Student

Patrick Siewe – GMATFix.com OG Companion 12th GMAT Interviews

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Ryan Childs, Tutor - 770 Raw Score Percentile

Analytical Writing 6.0 95 Quantitative 51 99 Verbal 44 97 OVERALL 770 99

Ryan Childs is a GMAT, GRE, and LSAT tutor in New York City. He specializes in large corporate classes but occasionally takes on individual students. Ryan has been teaching the GMAT for over 5 years. To date, he has helped countless students improve their GMAT performance and enroll in Business school.

What is the first thing you would say to someone who is considering taking the GMAT?

Too many students claim that they want to do what it takes to get a great GMAT score, but 3 or 4 weeks later, they’ve already fallen behind assignments. This exam requires a lot of work. Before starting the GMAT preparation process, find out why. Ask yourself “Why take the GMAT?” “Why go to business school?” Why get an MBA in __?” “Why make all these sacrifices?” You will find that ultimately, you may want to take the GMAT because you want more freedom with your time, more respect from peers, more comfort for yourself and family, etc… Now consider what effort is necessary to fully prepare for the test. Are you prepared to spend 10-15hrs every single week of dedicated, focused study? If you find the gains to be worth the sacrifices, great. Whenever you catch yourself wanting to give up or cut corners, visualize the goal and remember why you first chose to make these sacrifices.

How much improvement can students expect from a GMAT preparation program?

First, it’s important to realize that the higher your score, the harder it is to improve. Thus, the increase in ability required to go from a 550 to a 600 is a lot smaller than that required to go from a 700 to a 750. Another consideration is that one really shouldn’t compare test scores from different sources. A Princeton Review or Kaplan GMAT score isn’t derived using the same algorithm as an official GMAT score. So… your initial level should be based either on an official test or on a GMATPrep3

I’d also like to add that some companies offer “score improvement guarantees”, but these require that the student attend all classes and complete all work assigned. This is as it should be, for without doing all the work, you won’t see all the benefits.

exam, because both are from GMAC and have similar questions and identical scoring algorithms. From my experience, students who take a class can expect to see 30-70 points improvements while those who hire a good tutor can expect anywhere from 50-120 points.

Which is better: Tutoring or Classes?

The short answer: good tutoring is the best option. In all educational environments, a lower student-to-teacher ratio is usually indicative of higher quality of education thanks to the increased individual attention that the students receive. This advantage is reflected in my students’ score improvements: My tutoring clients tend to score 30 or 40 points more than my classroom clients. However, as with most things, this issue is a bit more complex. A more appropriate question may be “Which is better for me?” Among other things, students should 3 GMATPrep is a free test software that can be obtained from MBA.com after registration


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consider financial considerations. Are the extra 30 or 40 points you would get from tutoring worth the extra money it would cost to get a tutor? Another important consideration is the ability of the tutor. Most courses are the result of established prep programs that the teacher simply needs to follow. The teacher’s ability is important, but an average teacher can deliver pretty good results with a good prep program. On the other hand, the ability of the tutor is second in importance only to the student’s work ethic as far as improvement is concerned. If your tutor is unproven, be wary.

How can a student recognize a good tutor from their first meeting?

Ask your tutor what his GMAT score is. At a minimum, a good tutor must have a 720 GMAT score, meaning that he or she must be in the top 5% of all scorers. If your tutor hesitates to tell you his score, this should raise a red flag. Even the best teacher can only teach what he knows.

Did the tutor set clear expectations either before or at the start of your meeting? Did the tutor show up on time and prepared? The answer to these questions will inform you as to whether your tutor is well organized, organization being an important trait of any GMAT prep program.

Is your tutor a good student? The best tutors are great students. They will patiently listen to what you have to say, and try to learn as much as possible from you, because knowing the student’s learning process is crucial in helping him improve his score. A tutor cannot figure out your learning process unless he allows enough time for questions, listens closely to what you have to say, and responds to you rather than spew out a memorized teaching speech. Generally, a 1hr meeting is sufficient to determine whether your instructor is likely to be a good one.

How much should a good GMAT prep program cost?

For classroom courses, most of the major companies such as Veritas, Kaplan, the Princeton Review, Manhattan GMAT… charge from $36 to $60 per hour. In general, the smaller the class size, the more expensive the course. Private, 1-on-1 tutoring costs more: from $125 to over $350 per hour.

If you choose to opt for a lesser known option, quality can vary widely, so be careful and don’t be too shy to ask for references. A good independent tutor may offer the best value, but may also be a more risky choice.


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Mia Tallico, GMAT Program Developer - 770

Raw Score Percentile Analytical Writing 6.0 95 Quantitative 49 88 Verbal 47 99 OVERALL 770 99

Mia Tallico has several years of experience designing prep courses for the SAT and the GMAT. She has developed a few hundred questions for two major test preparation companies, and has also worked with the state of New York to develop items for the Regents Examination. Mia has also worked as a teacher evaluator, recommending teacher and tutor candidates for hire.

What is the registration process for the GMAT?

To register for the exam, head to www.mba.com with a credit or debit card at hand. You will need to create an account with your name exactly as it would appear on your government ID. You will then be taken through a series of steps to select your test date and location – the GMAT is offered most weekdays and because some location are heavily booked, I would suggest registering 4 weeks ahead of your test. Once you’ve made your selection, you will pay a $250 registration fee and receive confirmation by email.

Can I cancel or reschedule my appointment?

Yes. For a $50 rescheduling fee, you can return to www.mba.com, login and reschedule your appointment. Note that you will have to pay the full $250 again if you try to reschedule within 7 days of your original appointment. From your mba.com account, you can also cancel your GMAT appointment and receive a partial $80 refund. However, by cancelling within 7 days of your original appointment you will forfeit the entire $250

What should I expect on test day from the testing center?

First time GMAT testers tend to be surprised by the attention given to security at GMAT test centers. You’re never allowed to forget that there is someone watching over your shoulder. All you need is a non-expired government picture ID with a name matching the name you used to book your appointment. Arrive at the test center 15 minutes early. Once you present proper identification, your fingerprints and your photograph will be taken. Starting in the fall of 2008, many test centers will start using palm scans instead of fingerprints. The palm scan reads the vein patterns inside your hand, and is much more difficult to defraud than a fingerprint reader. This new technology however will only be used on test takers who were not previously fingerprinted.

You will then be handed a set of test center rules to review, and access to a locker in which to store all your belongings, ID excepted. You will receive a 10 page dry-erase pad with one or two markers to work with. You will be asked not to erase any of your work, but to ask for another pad if you run out of space.

There are two optional breaks: one between the second essay and the Math, and the other between the Math and the Verbal sections. Each break lasts 10 minutes, and any time you spend beyond the 10 will be deducted from your next section. Every time you go in or come out of the test room, you will have to present your ID and be

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fingerprinted (or palm scanned). One last important note is that the use of study materials as well as cell phones is not allowed, even during breaks.4

How much time should I allot to study for the GMAT?

Serious test takers must allocate 12 to 15hrs or more per week on GMAT preparation. It’s quite important to figure out exactly how you will accommodate this significant burden into your daily schedule. As soon as possible, spend a full 5 minutes identifying what obstacles and distractions are likely to come up: Television show? Late working hours? Kids? Friends? Your lack of patience? Be specific. Your list should have at least 7-10 specific items. Then take 60 seconds to think about each item and write down 3 concrete measures you can take to stay on course for a great GMAT score.

Although some people can prepare for this test in 2 weeks, I strongly recommend that you start preparing seriously 8 to 12 weeks before you actually take your test. Whatever you do, don’t go in cold.

There are tons of GMAT books available. Where can I find the best prep material to work from?

Of the many, many GMAT materials available for sale, the official GMAT guides are the best. There are three current guides: The Official Guide for GMAT Review (11th Ed.) aka the Big Orange, the smaller green Official Guide for GMAT Quantitative Review, and the smaller purple Official Guide for GMAT Verbal Review. Because these books contain retired GMAT questions, they are great source of practice content. The downside however is that because these books are aimed at the majority of test takers, they do not contain enough of the hardest questions to satisfy those who want a 700+ score.

Don’t forget about the old official guide either: the 10th Edition contains a lot more questions than the 11th and many, many great questions indeed. One common shortcoming of these guides is their explanations which can be less than thorough at times. Many GMAT math questions can be “cracked” in a few seconds and with a few simple steps, and many questions can be solved even when the test taker really doesn’t know how to fully do the question. The official guides however will not show any of that. They will show one solution, and not always the most efficient. This is where a good tutor or teacher can be very helpful.

4 You can find a presentation on the test taking experience at http://www.mba.com/mbasite/resources/globalgmat/

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Steven Martins, Admissions Consultant Steven Martins runs his own admission consultancy firm in Houston Texas. He specializes in business schools and law schools. His experience and background in education has given him great insight into how schools view the GMAT and GMAT scores. He has successfully placed students in business schools worldwide, and was kind enough to give me a few minutes of his time.

What key things should I do or avoid in my application essay?

Although this may sound cliché, make sure you answer the question asked! Many students desperately want to share a particular part of their experiences with the admissions officer, and almost ignore the assignment. Before you begin writing, really think about the question and make an outline of what you want to get across so long as it is on topic. Be sure to cover all items specified. In the spirit of following instructions, respect the length limits and other instructions set. Whatever you do, avoid “one size fits all” essays.

Use vivid language and demonstrate your points with concrete examples. It isn’t enough to say that you’re a quick learner if you don’t go on to show how you’ve had to quickly adapt to an unfamiliar environment to survive or thrive. Vividness can be achieved by using modifiers – adjectives, adverbs and other descriptive phrases that add color to your picture. For instance, instead of “In my three years at Widget Corp I rose from salesman to regional director of marketing,“ you might write “In three short years, the relentless drive and high performance that mark my work have propelled me from salesman to director of marketing for the Atlanta metropolitan area.”

Don’t submit an unfinished product. Make sure to proofread. Nothing says “I’m careless” like mistakenly quoting the wrong school name. If you’re applying to Columbia, an essay that reads “I would relish the challenge of Stern’s MBA program” isn’t exactly smooth. Ask a friend you trust to read your essay and give you feedback. A fresh pair of eyes is something you just can’t provide yourself.

How important is the GMAT relative to undergraduate GPA?

The more work experience you’ve accumulated since graduation, the less weight your GPA will hold. If you graduated within the last 2 years, admissions officers won’t have much to evaluate in terms of work accomplishments. In this case, your GPA will naturally grow in importance relative to the GMAT.

The GMAT score is important to most schools regardless of what else is in your application package. The fact that average GMAT score for the top 20 MBA programs is 697 while the next 20 average 642 is an indication of this importance. To be competitive, get the highest score you can.

Another point to consider is that generally, the larger the school, the more importance the GMAT holds whereas the smaller – or more resourceful – the school, the less importance it holds. An admissions office that has to review 6000 applications is bound to rely heavily on GMAT score as a sort of blunt tool to hack away at applications. Conversely an office that has to review 200 applications may have the luxury of carefully inspecting each app.

The bottom line is: don’t take the GMAT lightly.

How can I help my letters of recommendations stand out?


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Simple: in all your endeavors, go above and beyond what is expected of you, and produce results! The best letters of recommendations tend to come from professionals you have worked for. Avoid asking for letters from people who are equal to or below you in the chain of command. Also avoid executives who haven’t had a chance to observe you directly. A recommendation from your company’s CEO may not be as relevant as one from your direct supervisor.

Request recommendations early, and when making the initial request, ask when it would be ok to check on its progress. Your recommenders are busy, so check back a week or two later. Do not write your recommendations yourself. Any experienced admissions official will likely spot this farce and it will immediately destroy your chances of admission. Rather, hand each of your recommenders distinct “talking points” highlighting those aspects you would like them to stress. Give them some material to work with. The talking points should highlight your strengths with clear examples of demonstrated leadership or work ethic.

Bottom line: what are admissions officials looking for in the ideal application?

Admissions officials want to be convinced that an MBA from their particular school is the next logical step linking your current position to your stated goal(s). The different parts of your application should work in sync to paint a clear picture of an individual who knows exactly where he wants to go, and why.

Secondly, the application should demonstrate your ability to get where you’re headed. By highlighting your past successes as well as your reactions to failure, the application can convince admissions officials that if given the opportunity, you will achieve. Ultimately, a body of alumni who are leaders in their fields means everything to an MBA program, and this body is what school officials want to build in the long term via selective admissions. Get them to believe that they’d be sorry to miss out on you.


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Christina Chang, GMAT Student - 740

Raw Score Percentile Analytical Writing 5.0 75 Quantitative 48 86 Verbal 41 93 OVERALL 740 98

Christina Chang came to me on her friend’s recommendation. She needed.1-on-1 GMAT tutoring because she believed that a high GMAT score would enhance her chances of admission to a top 20 MBA program. We worked together for 8 weeks, meeting twice a week for 1.5hrs per meeting.

How did you decide between self-study, online course, a class, and 1-on-1 tutoring?

My sister had taken the GMAT after studying by herself and she had done very well. However, she had 4 months to prepare and she took the test twice. I realized that almost anyone who is motivated enough could get great results through self-study. However, this can be risky, and take more time than is necessary. Jenny studied for about 15hrs a week every week for 4 months. In addition, she spent a couple of hours each week scouring the web for information on how to best prepare. She was never quite sure what the most effective thing to do was. Because she wasn’t in a course, she had to figure everything out including which materials are best, how to use them, how to build a good study plan etc… If I had 4 months to prepare and her relentlessness, I might’ve tried self-study. However, I knew that I had only 2 months and that I needed to get it right the first time. From my days trying to follow a strict gym workout regimen, I also knew that I could benefit from a motivator. So I decided to consider a GMAT course or tutoring option.

It made sense to me to go with tutoring because I wanted the prep program to be personalized to my weaknesses and learning style, not mass produced for the thousands who take that course. Of course tutoring costs a lot more, but I figured that a higher score would get me into a better school. So I made my decision thinking about the long term.

How did you structure your study?

I think consistency is the most important thing. I met with Patrick – my tutor – twice a week, usually on Tuesdays and Friday evenings. Roughly, the first week was spent pinpointing my weaknesses and figuring my learning style; the next four weeks were spent reviewing all topics with particular emphasis on those most heavily tested; the last three weeks were spent refining my skills, perfecting my pacing, dealing with test anxiety, and cutting down careless errors. I took practice tests on weeks 1, 3, 5, 6, 7 and 8.

Along with maintaining a consistent study schedule, I would say that it’s important to put yourself in an environment free of distractions when you study, and even more so when you take a practice exam. I used an empty conference room at work after hours where I knew I wouldn’t be interrupted. Patrick assigned about 10-12hrs of homework each week, mostly from the official guides. One thing he helped me realize is that I should spend as much time reviewing a question as I spent doing it. At first it seemed like a waste of time, and I have to admit that sometimes I just quickly checked the answers and moved on. However, when I worked with my tutor, he forced me to verbalize my thinking, and run through each step in the solution process. It was a bit annoying when he would ask me things I thought were obvious, but my responses sometimes revealed either a flaw or


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inefficiency in my thinking. By pointing out these flaws, he helped me reduce mistakes that I had been unaware of.

I studied 5 days a week, and took Fridays and Sundays off. During my 40 minute commute to and from work on the train, I would review lessons and go over my notes. Sometimes, I would also do some of the exercises that had been assigned for homework. The hard core study would happen after work in the evenings. On days I didn’t have tutoring, I would study for 2.5hrs before heading home from work. Tuesdays and Thursdays I would study for only 1.5hrs after my tutoring sessions. It was difficult to be consistent, but with Patrick constantly checking up on me, I managed it for the most part.

How long did it take for your score to increase?

When we started, Patrick asked me to take a diagnostic test right away, so we could figure out my starting level and my weaknesses. I scored a 600 on that test. I felt like I needed more help in verbal, because math had always been my strong point in school. I quickly realized though that I would need to work on both areas. I remember that the second week, we reviewed verb tense errors and pronoun errors in sentence corrections, and we did some work with rate problems in math. My second test was on week 3, and my tutor had warned me that I may not see any improvement from test #1 to test #2, because there is typically a lag between when a student learns a concept and when she becomes good at applying it in a test. In fact, my score dropped to 590!

I was discouraged because of course I thought I would be the exception and I was eager to see my score go up. One great thing about having a tutor though is that you get the insight of someone who has seen many people go through the journey you’re on. Even though I didn’t like it, I understood that as long as I did the work, this wasn’t really a step back. It was more like a dip in the road to the hilltop. As tacky as it sounds, my tutor helped me see the forest, and not focus on the trees.

One major mistake that many students make is that they do not spend enough time reviewing their test performance. Just looking at your score isn’t even the tip of the iceberg. We worked together to identify exactly what I’d done well, where I had been inefficient, not just what I did wrong, but why. From week 5 onward, we spent one full tutoring session studying my test performance, and setting specific goals for the next practice test. Getting a high score is the ultimate objective of the tutoring, but there is so much that goes into it, and it was helpful to have smaller concrete objectives for each test such as improving pacing, increasing my efficiency rating, and sketching all reading comprehension passages. I took a test every week from week 5 on, and my score started increasing but not at regular intervals. My scores from week 5 to week 8 were 660, 690, 710, and 730. I then took the actual GMAT and scored a 740

Patrick Siewe – GMATFix.com OG Companion 12th Diagnostic Test

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Diagnostic Test Solutions

Solutions to Questions from pages 20-26 in The Official Guide for GMAT Review, 12th Edition

Patrick Siewe - GMATFix.com OG Companion 12th Diagnostic Test

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DIAGNOSTIC TEST Quantitative Answer Key

1. A 25. E

2. D 26. E

3. E 27. E

4. B 28. E

5. B 29. E

6. A 30. A

7. E 31. D

8. E 32. C

9. D 33. D

10. C 34. C

11. C 35. D

12. C 36. B

13. E 37. A

14. B 38. B

15. C 39. E

16. E 40. D

17. D 41. C

18. A 42. C

19. A 43. B

20. B 44. A

21. D 45. D

22. E 46. E

23. B 47. D

24. C 48. C


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#1, pg.20 Difficulty Level: 200-400 Topics: Translations & Manipulations

Solution

The first CD costs $15.95, and each of the other five costs $3.99, so the expenditure can be expressed as . A quick look at the answer choices reveals that this question has a trick up its sleeve, as none

of the answer choices offer an exact match. One thing we could do to distinguish the correct answer is to compute our expression as well as the expressions in the answer choices, and find out where there is a match. This would get us to the right answer in a few minutes, but would this be the most efficient approach?

A better alternative is to take a closer look at the answer choices. Note that the writers have replaced 3.99 with 4.00. We ought to do the same to our expression before we compare it to the answers:

What must we put in place of x in the equation above to offset our modification? By replacing 3.99 with 4.00, we have added 0.01. Since this new amount is multiplied by 5, we have actually added 0.01 five times, hence an addition of 0.05 to the first term of the addition. To offset this extra 0.05, we must remove that amount from the second part. Thus 15.95 should be replaced with 15.90. In one quick step, we have changed our expression to.

.

The correct answer is A

Take-Aways • Do not rush to do long calculations. Give yourself a few seconds to think about whether there is a faster

way or whether you can approximate. Because the GMAT only grants you 2 minutes per math question, you will seldom be required to complete these operations.

• Always glance at the answer choices before you start solving. In many cases, this will help you avoid

unnecessary work. For instance, you might notice that the answers are expressed as fractions and thus discover that there is no need to engage in long division.

#2, pg.20 Difficulty Level: 500-600 Topics: Statistics

Solution

To find the average of an evenly distributed set (an arithmetic sequence), average the highest and the lowest values. In an evenly distributed set, the difference between each value and the next higher value is always the same. Some examples include:

95.15)99.3(5 +

x+=+ )00.4(595.15)99.3(5

90.15)00.4(5 +


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• A set of consecutive integers: 12, 13, 14… Difference is 1 • A set of consecutive even integers: 2, 4, 6… Difference is 2 • A set of multiples of seven: 7, 14, 21… Difference is 7.

In our case, we are dealing with consecutive integers from 200 to 400, and from 50 to 75. In each case, we need to average the highest and lowest value to find the average of the set.

and .

Now that we know the averages, our question has become easier: Instead of “The average of the integers from 200 to 400 is how much greater than the average of the integers from 50 to 100?”, we can now answer to: “300 is how much greater than 75?” The answer is

The correct answer is D

Take-Aways • A set of numbers in which the difference between each value and the next higher value is constant is called

an arithmetic sequence. • The average of an arithmetic sequence is the average of its highest and lowest values. This average is

always equal to the median of the sequence. • The sum of an arithmetic sequence is its average multiplied by the number of numbers in the sequence

#3, pg.20 Difficulty Level: 600-700 Topics: Functions & Sequences

30 Seconds Hack

In fewer than 60 seconds, you can dramatically increase your guessing odds. Here is how:

A sequence formula describes how to solve for each term in the sequence by using the terms before or after it.

The term described is given the subscript “n” (eg. ). The previous term is and the term before that is

3002

400200=

+ 75210050

=+

22575300 =−

na 1−na

2−na


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From the formula given, , we can deduce that each term ( ) is equal to the average of the

two terms that precede it ( and ). In other words, the value of each term will be between the two

preceding values. If we were to graph these terms, we would see something like:

. A quick look at the graph tells us that , so the correct answer must be greater than 20 and we

can eliminate answers A, B, and C.

Guess D or E.

Solution

We are given the data below:

4 20

The formula given, , tells us that to find a term, we need to know the two previous terms in

the sequence. Therefore, to find , we must first find . Thankfully this can be done very quickly.

From the formula, we know that each term ( ) is equal to the average of the two terms that precede it ( and

). Specifically, we know that is the average of and . Since their average must be 20,

, and thus .

221 −− +

= nnn

aaa na

1−na 2−na

205 =a 56 aa >

3a 4a 5a 6a

221 −− +

= nnn

aaa

6a 4a

na 1−na

2−na 5a 3a 4a

3 4 40a a+ = 4 36a =

a3

a4

a5

a6 a3 must be between a1 and a2. Note that this graph agrees with the data we have. Namely that a5 is bigger than a3


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Now we have and . Average them to find and solve the problem:

The correct answer is E

#4, pg.20 Difficulty Level: 500-600 Topics: Combinatorics

30 Seconds Hack

In fewer than 60 seconds, you can eliminate most answers with certainty. Wonder how? :

Forget about the total number of people. This is a distraction aimed at taking up precious seconds. Because we are asked to find a probability, the correct answer would be the same whether we had 2500 people, 25000 people, or 5218 people. It is the percentages that matter.

We are asked to find the probability of selecting someone who invests in bonds but NOT in stocks. This group is a subset of all people who invest in bonds; in other words, it is a subset of 35% of the people. For this reason, the right answer will be smaller than 35%. Take a look at the answer choices.

To easily compare the answer choices to 35%, adjust each fraction so that its denominator is 100

(A) 9 1850 100

=

(B)

(C)

(D)

(E)

Since our probability must be smaller than 35%, we can safely eliminate C, D, and E

Guess A or B

4 36a = 5 20a = 6a

5 46

20 36 282 2

a aa + += = =

7 2825 100

=

7 3520 100

=

21 4250 100

=

27 5450 100

=


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Solution A – Venn Diagrams

Most importantly, do not spend any time trying to work out 35%, 18% or 7% of 2500 because it is irrelevant. If we had been asked to find “How many people…” these numbers would have been relevant. To find the probability, all we need are percentages (or ratios). Realizing this fact alone could easily shave 90 to 120 seconds off your solution time.

One relatively easy way to represent overlapping sets is to use the Venn diagram. In this diagram, each circle represents one group, and the overlap represents the group of people who belong to both circles

From the Venn diagrams, and without any involved computation, we can quickly see that the probability that someone invest in bonds but NOT in stocks is 28%. A quick look at the answer choices informs us that we must

express our solution as a fraction:

The correct answer is B

Solution B – Group Formula

The group formula is a useful tool for handling questions of two overlapping sets. You’ve probably seen in before: . In this specific question, the formula would therefore be:

. With this formula in mind, this question can be solved very quickly.

What we are looking for is Bonds Only. Bonds Only = All Bonds – both . So Bonds Only .

28 14 728%100 50 25

= = =

1 2Total Group Group neither both= + + −

Total Stocks Bonds neither both= + + −

35% 7% 28%= − =

Stocks Bonds

7%

Always start with the center. Use a variable if you don’t know the value. In this case, 7% invest in both bonds and stocks.

Stocks Bonds

7%

The 35% who invest in bonds include 7% who invest in both, and 35 7 28%− = who invest in Bonds only. Similarly, 18 7 11%− = invest in Stocks only.

28% 11%


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A quick look at the answer choices informs us that we must express our solution as a fraction:


Take-Aways • When solving a probability question, you can often use your intuition to eliminate answers that are way off

• When solving probability, percentages, or ratios, consider ignoring the total given and using your own to

make the math easy. 100 makes a great total in percentage questions. • The group formula, , is a useful way to solve questions about

2 overlapping groups.

#5, pg.20 Difficulty Level: 600-700 Topics: Geometry

Solution

It is a good idea to start geometry questions by writing down formulas relevant to the question, and filling in what

data we find. In this case, the relevant formula is volume of a cylinder:

To better understand what we’re asked (the elevation of the water when the cylinder is on its side), let us draw the cylinder on its side. The tank is half full, so the water level will be exactly half way.

To find out the height of the water, we must solve for the radius. Which unknowns stand between us and the radius? In the volume formula above, we can see that to find the radius, we must find V, and h.

Given that it takes to fill half the cylinder, its volume is twice that: . Update your equation:

We know that the cylinder is half full, so whether it stands on its base, or lies on its side (its height), the water will reach half way up. When the cylinder is upright, we are told that the water reaches 4 feet. If 4 feet is half way up,

28 14 728%100 50 25

= = =


2V r hπ=

36π 72V π=272 r hπ π=

r

The height of the water level is the length of the radius


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then what is the full height? . Update your equation: . Now we have all the elements needed to solve for r.

Divide both sides by

Square root both sides


Take-Aways • Drawing figures reveals connections between data and often makes the problem easier to understand and

handle. Always make a drawing of your own for geometry questions, even if one is already provided • In 3-d Geometry, you are expected to know the volume of a cylinder, , and the volume of a box,

. Just remember (area of base)(height).

#6, pg.21 Difficulty Level: 600-700 Topics: Sets & Groups

Solution A – Group formula

The group formula is a useful tool to handle questions of two overlapping sets. You’ve probably seen in before: . It is important to note that in this formula, each group has two

parts: items that belong to that group alone, and items that belong to both groups. Algebraically, we can say: , and . Take a second to really grasp this

relationship…

In this specific question, our groups are Brand A and Brand B, so the relevant formulas are:

(Eq. 1)

(Eq. 2)

(Eq. 3)

Again, make sure you understand these relationships before continuing.

First, identify what you need to solve for. The question is “How many…used both brands?” To solve for both, we need to get rid of the other four variables in (Eq. 1) above. To get rid of a variable, replace it with its value, or express it in terms of the variable you want to solve for (in this case, in terms of both).

Express A in terms of both, using (Eq. 2) above.

“60 used Only A”

8h = 272 8rπ π=

8π 29 r=

3 r=

2V r hπ=V l w h= × ×


1 1Group OnlyGroup both= + 2 2Group OnlyGroup both= +

Total A B neither both= + + −

A OnlyA both= +

B OnlyB both= +

60OnlyA =


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In (Eq. 2) replace OnlyA with 60 (Eq. 4)

Now, Express B in terms of both. This is a bit more difficult

“For each…that uses both, 3 use OnlyB”

Cross multiply

In (Eq. 3) replace OnlyB with

Simplify (Eq. 5)

Next, get rid of Total by replacing it with its value

“200 households surveyed” (Eq. 6)

Finally, get rid of neither by replacing it with its value

“80 used neither” (Eq. 7)

We are nearly done. Using Equations 4-7, replace all the variables from (Eq. 1) with their new values

Original formula (Eq. 1)

Plug-in values

Simplify

Solve for both


Solution B – Venn diagrams

First, identify what you need to solve for. The question is “How many…used both brands?”

Draw a Venn diagram.

60A both= +

13

bothOnlyB

=

3 ( )both OnlyB⋅ =

(3 )both⋅ (3 )B both both= ⋅ +

4B both= ⋅

200Total =

80neither =

Total A B neither both= + + −

200 (60 ) (4 ) 80both both both= + + ⋅ + −

200 140 (4 )both= + ⋅

15both =


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Since of the 200 homes, 80 use neither A nor B, by adding all diagram regions we would only get the remaining 120 homes.

Add all regions

Simplify

Solve for both


Take-Aways • When drawing a Venn diagram, always start with the center

• The group formula, , is a useful way to solve questions about



60 (3 ) 120b b+ + ⋅ =

60 4 120b+ =

15b =


B A

b

Always start with the center. Use a variable if you don’t know the value. In this case, we used b for both.

B A

b

“60 used OnlyA”, so we place 60 inside the A bubble, but outside the both region 60

B A

b

“For each that used both, 3 used OnlyB”. This ratio tells us that 3OnlyB b= ⋅ . So we place (3 )both⋅ inside B, but outside the both region

60 (3 )b⋅


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30 Seconds Hack

In about 30 seconds, you can eliminate 3 wrong answers and guess form the remaining two. With minimum additional effort, you can get to the right answer in a few more seconds.

There are 10 people, and two of them will be chosen to be secretary and treasurer. The probability that Harry is chosen for one or the other is not easy to find, but intuitively, it is higher than the probability that he be chosen to just one specific position. In other words, if you had to bet, which of these odds would be higher?

• Odds that Harry is chosen, out of 10 people, to be secretary. • Odds that Harry is chosen, out of 10 people, to be secretary or treasurer

The 2nd scenario is what we are asked to find, and it is the more likely scenario (higher probability). The first

scenario has easy to find odds, because Harry is one of ten possible choices, this probability is . Since the

probability we are looking for is higher, we can eliminate all answers that are or smaller. Cross off answers

A, B and C

Consider answers D and E. Their values are respectively and . Ask yourself; if instead of just the secretary’s position, Harry can occupy anyone of 2 positions, will the probability go up only a little, or will it go up significantly? It will go up significantly, because Harry’s options doubled; so the likely answer is E.

Guess E.

Solution

To solve this one quickly, consider the following law of probability:

In our case, the probability that Harry is picked as secretary, out of 10 people, is

The probability that Harry is picked as treasurer, out of 10 people is also

The probability that Harry is picked as both secretary and treasurer is 0

So


110

110

19

15

( ) ( ) ( ) ( )P AorB P A P B P both= + −1

10

110

1 1 2 1(Secretary or Treasurer) 010 10 10 5

P = + − = =


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#8, pg.21 Difficulty Level: 500-600 Topics: FDPs & Ratios

Solution – Plug In

One way to do this question is to plug in a value for one of the revenues. Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem. If you work correctly, whatever you plug in will reduce to the same answer.

To decide what to plug in, look at the fractions in the problem: and . You want to pick a number that will easily divide into all the denominators. Let us use 20 as a starting number.

Now we need to decide where to put the 20. The first sentence in the question says “…revenue in November was 2/5 of revenue in December”. This tells us that our first operation will be to take 2/5 of December, so December will be the starting point.

Plug in your chosen value in your starting point:

Then simply follow the question, step by step.

“November was 2/5 of December”

“January was 1/4 of November”

To answer the question, find the average of Nov and Jan:

“December is how many times the average of Nov and Jan?” With our newfound data, the question becomes “20 is how many times 5?” The answer, of course, is 4 times.

I encourage you to try the same exercise with a different plug-in value (eg. 60 or 100) to see that you will still get the same answer


25

14

2 2 20 85 5

Nov Dec= = ⋅ =

1 1 8 24 4

Jan Nov= = ⋅ =

8 2 52 2

Nov Jan+ += =

Our plug-in 20December =


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Take-Aways • Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete

numbers, you can plug in a value that will be easy to deal with and use it to solve the problem.


Solution

This question is rather straightforward and simply looks to test your understanding of statistical terms

• Mean – The average of all values:

• Median – The middle value, or the average of the two middle values (when there is an even number of values)

• Standard Deviation – a measure of how tightly/loosely values are grouped around the average. Standard Deviation must be at least zero.

If we increase each value by 5, the Mean and Median will each increase by 5, but the Standard Deviation will remain constant, since the new set of values will have the same distribution (spread from the new average).


Take-Aways • For the purposes of the GMAT, you do not need to memorize the formula for standard deviation. However,

you need to have a good understanding of what standard deviation means. Broadly, standard deviation is a measure of how tightly/loosely values are grouped around the average. The more dispersed the values, the higher the standard deviation.

#10, pg.21 Difficulty Level: 700+ Topics: Geometry

30 Seconds Hack

In about 30 seconds, you can get the right answer with very little work.

Most geometric figures on the GMAT are drawn to scale, unless otherwise specified. This is an extremely useful factoid because it means that if you are clueless about how to solve a geometric problem, you can sometimes estimate lengths, angles, etc...

Sum of valuesNumber of values


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Take a look at the angles, and estimate them. They are certainly smaller than . When estimating, if you feel too nervous about picking one value, pick a range. When I first looked at these angles, I felt that they were all roughly in the to range. Since there are 5 angles, the total will be in the range. Once you have an estimated answer, look at the answer choices and pick the answer that falls closest to the center of your range.

Guess C

Solution

To find the solution, we need to figure out a way to isolate:” ” in an equation. The key to building this equation is to look at different triangles in isolation. Because each triangle has , we can express the 3rd angle in terms of the other 2. For example, below I have highlighted the triangle with angles

, the 3rd angle will be .

Now, consider the five angles in the middle of the figure. These angles make up a pentagon. As for any other polygon on the GMAT, you can find the total degree measure of a pentagon using the formula below:

All the angles in the middle add up to 540, so we now have our looong equation:

Combine like terms

Divide by 2

90°

30° 45° 150 to 225° °

v w x y z+ + + +180°

and v y° ° (180 )v y− − °

180(Number of sides 2) 180(5 2) 540Total = − = − =

(180 ) (180 ) (180 ) (180 ) (180 ) 540v z w x v y x z w y− − + − − + − − + − − + − − =

900 2 2 2 2 2 540v w x y z− − − − − =

450 270v w x y z− − − − − =

v w

x

y

z

180-w-y

180-v-z

180-x-z

180-v-y

180-w-x


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Isolate and solve

The correct answer is C

Take-Aways • If you find yourself needing an equation that involves angles in a triangle, remember to use the “180° sum”

equation. • Most geometric figures are drawn to scale. When stuck, you can approximate lengths, angles, etc…

#11, pg.22 Difficulty Level: 700+ Topics: Combinatorics

Solution

Three different pairs of digits can be matched:

Tens & Ones:

Hundreds & Tens:

Hundreds & Ones:

To find all possible combinations, we will need to find the number of combinations for each matched pair, and then add all three results.

To find these combinations, we will use a concept of arithmetic that is similar to probability. To make it simple to understand, let’s suppose you have to pick a cell in a 5 by 6 matrix. How many possible options (cells) are there? How do you find the number of options? There are 30 options, which you found by multiplying 5 by 6. This works because selecting a cell is actually a succession of two actions:

1. Picking a row – 5 options 2. Then picking a column in that row – 6 options available, for each of the 5 rows.

The formula here is simple. Let be the number of options/outcomes possible for event A, and the number of options/outcomes possible for event B. Then, if (A and B)n are the number of options when A, then B happen: In other words, the number of way to choose a row and choose a column is equal to the number of ways to choose a row (5) multiplied by the number of ways to choose a column (6).

Let’s return to our question, and find all combinations of matched pairs.

Combinations of matching Tens & Ones:

450 270 180v w x y z+ + + + = − =

_

_

_

nA nB

(A and B)n n nA B= ⋅

_


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(3 options) x (9 options) x (1 option) = 27 combinations.

Combinations of matching Hundreds & Tens:

(3 options) x (1 options) x (9 options) = 27 combinations

Combinations of matching Hundreds & Ones:

(3 options) x (9 options) x (1 option) = 27 combinations

Add up all the combinations, and remember to remove (700) from the mix since we only want integers greater than 700



Solution A

To do this question quickly, pay close attention to the first bit of info: “y is 50% of 50% of x”. This means that y

is 1/2 of 1/2 of x, so (Eq.1)

Now, let’s move onto the 2nd piece of information: “y percent of x equals 100”. In algebra, percent literally means divided by 100. So we can build our 2nd equation:

“y percent of x equals 100”

From (Eq. 1), replace y with

Multiply by 100 and combine x’s

_

_

Total Options 27 27 27 1 80= + + − =

14

y x=

100100

y x =

14

x14( ) 100

100x x =

2 21 1004

x =

{7, 8, or 9} since the number is greater than 700

All digits, less what is already picked for hundreds

Must be same as Tens

Use logic above to figure out how many options there are for each digit.


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Multiply by 4 and square root

Simplify and solve


Solution B – Reverse Engineering

When a question gives you non-variable answers along with a word problem, you can often use reverse engineering.

Reverse Engineering is working from the answer choices, to find which choice can perfectly accommodate all the given data in the question. As a very simple example, consider the following question:

“Five years ago, Jack was 3. How old is Jack today?”

(A) 7yrs (B) 8yrs (C) 9yrs (D)10yrs (E) 11yrs

Suppose you had no idea how to solve the question (humor me). Through reverse engineering, you can try each answer choice until you find the one that fits the data given (“Five years ago, Jack was 3”). The answer cannot be “9yrs” because it doesn’t match the info we already know, whereas the right answer (B) is a perfect match: 8 yrs old today is the only answer that fits the given data.

When using reverse engineering, start with answer choice (C). Why? Well answer choices are typically ordered; you will often know whether the choice you picked is too large or too small, and you will thus be able to eliminate multiple answers at once and save yourself some time. In the example above, “9yrs” is too large, so there is no need to even check answers D and E.

Ok, back to our original question. Let’s reverse engineer it and find out which answer choice makes “y% of x equals 100” true.

What is value of x?

y is ½ of ½ of x. So y =

y% of x (should be 100)

Notes

(A) 50 12.5 ? Don’t start long calculations unless you absolutely have to

(B) 100 25 25 y% of x is too small (25 instead of 100)

(C) 200 50 100 Yes! x=200 makes everything else fall into place as expected. y% of x is 100

(D) 1000 -- -- Don’t bother, we’ve found a match (E) 2000 -- -- Same as above


2 2100 4x = ⋅

200x =


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Take-Aways • When a question gives you non-variable answers choices along with a word problem, you can often use

reverse engineering. Start with answer C, especially if it will take more than a few seconds to check each answer.

#13, pg.22 Difficulty Level: 700+ Topics: Number Properties

Solution A

.By cross-multiplying the equation given, we can see that . Note that this can also be written as . In other words, t goes into s sixty-four times, and what remains is 0.12t. This is the remainder.

We now have a new equation: . Since t is an integer, it must be true that

(Eq. 1)

To find out which answer satisfies this constraint, change 0.12 into a fraction:

Replace 0.12 with its fraction form in (Eq. 1)

Simplify

For this fraction to equal an integer, remainder must be divisible by 3. Only answer choice (E) fulfills this constraint.



Solution A – Group Formula

.The easiest way to solve this problem is to use the group formula (see #6, pg 21 for more)

(Eq. 1)

64.12s t=64 0.12s t t= +

0.12remainder t= int0.12

remainder=

12 30.12100 25

= =

int325

remainder=

25 int3

remainder⋅=

Total Supervise Refresh neither both= + + −


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As always, we must first know what we are looking for; in this case, we want Refresh. Therefore, we must get rid of all other variables either by replacing them with their numerical values or by expressing them in terms of Refresh

“Of 84 parents…” Replace Total with 84 in (Eq. 1)

“35 volunteered to supervise…” Replace Supervise with 35 in (Eq. 1)

“11 volunteered both…” Replace both with 11 in (Eq.1)

“parents who bring refreshment was 1.5 times…neither”

Replace neither with in (Eq.1)

Once you complete all four replacements, you can solve quickly:

Original equation (Eq. 1)

Make all 4 replacements

Combine like terms

Solve for Refresh


Solution B – Venn diagrams

Before drawing the diagram, know exactly what you’re solving for. We want Refresh. Therefore, we must get rid of all other variables either by replacing them with their numerical values or by expressing them in terms of Refresh

1.5Refresh neither=

32

Refresh neither=

23

Refresh neither=

23

Refresh

Total Supervise Refresh neither both= + + −

284 35 113

Refresh Refresh= + + −

5603

Refresh=

36Refresh =


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e36

Now, we are left only with the variable we want to solve for. We know that all regions must add up to the total of 84:

Replace 1.5 with 3/2

Divide by the fraction

Combine like terms

Solve





24 11 ( 11) 841.5RefRef+ + − + =

24 11 ( 11) 8432

RefRef+ + − + =

224 11 ( 11) 843

RefRef+ + − + =

524 843

Ref+ =

36Ref =


Ref Sup

11

Always start with the center.

“Supervise only” is all people who supervise minus the 11 who both supervise and serve refreshments. This is why the left-most region is (Sup-11). The same logic applies to “Ref only”

Ref Sup

11

• Sup-11 = 24. • “parents who bring refreshment was

1.5 times…neither”:

1.51.5RefRef neither neither= → =

24 (Ref-11)

(Sup-11) (Ref-11)

neither

1.5Ref


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#15, pg.22 Difficulty Level: 700+ Topics: Exponents & Roots

30 Seconds Hack

The product of all primes less than 20 is . This question is particularly challenging because you just do not have enough time to do the required multiplication. Instead, you ought to approximate.

Group the numbers roughly in products of 10 as below

Original #s 2 x 5 3 x 7 11 19 13 17 Product estimate = 10 2 x 10 (10) 2 x 10 10 2 x 10

Of course, these are just estimates, but they should be good enough. The GMAT will never force you to go through computations that would take 5 minutes to complete. If you find yourself in such an operation, you’ve missed something.

Let’s rearrange the 2nd row of the table above


Take-Aways • Do not rush to do long calculations. Give yourself a few seconds to think about whether there is a faster

way or whether you can approximate. Because the GMAT only grants you 2 minutes per math question, you will seldom be required to complete these operations

#16, pg.22 Difficulty Level: 700+ Topics: Exponents & Roots

Solution

This question is perhaps more difficult than it first seems. Since we are asked to find , we know that we’ll need to manipulate the equation to isolate this expression. A close look at the answer choices gives us one important clue. None of the answers contains a square root; so in manipulating our equation, we will need to square each side intelligently, so as to get rid of all the roots.

2 3 5 7 11 13 17 19⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

≈ ≈ ≈ ≈ ≈

2 3 5 7 11 13 17 19 10 2 10 10 2 10 10 2 10⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ≈ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

610 8≈ ⋅

710≈

24x


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Original equation

Square each side

To get rid of the remaining square root, we must isolate it before squaring again:

Isolate square root

Divide by 2

Square each side

Solve for


Take-Aways Always glance at the answer choices before you start solving. In many cases, this will help you avoid

unnecessary work. For instance, you might notice that the answers are expressed as fractions and thus discover that there is no need to engage in long division.

#17, pg.22 Difficulty Level: 200-400 Topics: Exponents & Roots

Solution A

Solve for root of n

Original equation

Break up roots

Calculate roots

3 2 2 1x x− = +

23 2 ( 2 1)x x− = +

2 23 2 ( 2 ) 2( 2 1) 1x x x− = + ⋅ +

3 2 2 2 2 1x x x− = + +

2 4 2 2x x− =

1 2 2x x− =

2 2(1 2 ) ( 2 )x x− =

21 4 4 2x x x− + =

24x 24 6 1x x= −

1681

n =

1681

n =

49

n =


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Finally, find root of n

The only curveball is the inclusion of n in the answer choices (C). Make sure you pick , and not n


Take-Aways • In the answer choices, the GMAT will often include values that you are likely to come across as you work

the problem. For instance, you may be asked to solve for y, and one of the answer choices will be the value for x. It is generally a good idea to take an extra 3 or 5 seconds to re-read the end of each problem (what you’re asked to solve for) before making your final selection.

#18, pg.22 Difficulty Level: 400-500 Topics: Number Properties

Solution

When solving a Number Properties question, express all values in their prime factorizations. In this format, it is easier to deal with very large numbers, see patterns, and make connections.

Express n in its prime factorization format:

In this format, we can quickly see that the prime factors of n are 2, 3, 5, and 7. So n has four prime factors


Take-Aways • When solving a Number Properties question, express all values in their prime factorizations (eg.

). In this format, it is easier to deal with very large numbers, see patterns, and make connections.


Solution

4 29 3

n = =

n

1 2 3 4 5 6 7 8n = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

2 32 3 2 5 (2 3) 7 2n = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

212 2 3x x= → = ⋅


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This question tests your knowledge of the 3rd side rule of triangles. In any triangle, the length of the third side must be greater than the difference, but smaller than the sum of the other two sides

In our case, the triangle has . According to the 3rd side rule, .

Since the possible values of k are 3, 4, 5, and 6, there is only one value of k that could be the length of .

That value is 6.


Take-Aways • 3rd side rule of triangles: In any triangle, the length of the third side must be greater than the difference, but

smaller than the sum of the other two sides.


Solution

This is an exercise in visualizing a cone inside a hemisphere. You’re asked to compare the height of the cone to the radius of the hemisphere.

• The height is the distance from the center to the top, which touches the surface of the hemisphere • The radius is the distance from the center to the surface of the hemisphere.

These distances are the same, so the ration is 1:1



30 Seconds Hack

This question can be done with little to no computation

1 2 3 1 2side side side side side− < < +

1 22 and 7side side= = 35 9side< <

3side

1 2 3 1 2side side side side side− < < +


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Compound interest accumulates faster than simple interest. If the annual interest is 4%, the quarterly interest is 1%. When adding interest, be careful to only add it twice, since we want the amount “6 months (2 quarters) after the account was opened”.

Our starting amount is $10,000. If we added 1% simple interest two times, we would be adding for a total of $10,200. Since we are compounding interest, the correct total will be slightly higher.

Guess D

Solution

One way to increase an amount by x% is to multiply it by . Conversely, to decrease an amount by y%,

you can multiply it by

Since we want the amount after two quarters (6 months), the $10,000 principal will grow twice. Because the annual interest is 4%, each growth is actually a 1% increase.

Apply the 1% increase twice 100 1 100 110,000

100 100+ +

Simplify by crossing off common factors 100 1 100 110,000

100 100+ +

Multiply


Take-Aways • Generally, you should simplify fractions before you perform multiplications. This will keep your numbers

from growing needlessly large. For instance, suppose you’re faced with . You could multiply

across top and bottom, and then divide . Solving this way might take you 90 seconds. On the other

hand, you can simplify by crossing off common factors in 21 & 7, as well as in 27 & 9. In this case you’ll

be left with to divide. Solving this way might take you 15-20 seconds, and you would’ve saved over

60 seconds to use on a more difficult problem.


$100 2 $200⋅ =

100100

x+

100100

y−

(101) (101) 10201⋅ =

21 27 19 7⋅

⋅

56763

3 31⋅


Pag

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Solution

As always, with geometry it’s important to write down and fill out formulas relevant to the question.

Volume of cylinder

“height is 9 inches”

“container is 1/2 full of water…” and “volume of water is 36” means that total volume of cylinder is

Now, we can plug-in 72 for the volume in the formula, and solve for the radius

Plug-in 72 for volume

Divide by

Square root

So the radius is , so why is this not in the answer choices?

Because we’re asked to solve for the Diameter: 2 22 2 2 4diameter rπ π

= → ⋅ →



Solution

Because this is a number properties question, it’s a good idea to express each variable in its prime factorization form. In the equations below, “int” is a place-holder for an unknown integer.

“x is multiple of 4”

“y is multiple of 6”

It follows from the above that . Now, check the solutions offered:

2Volume r hπ=

29Volume rπ=

2 36 72⋅ =

272 9rπ=

9π 2 8rπ

=

8 22rπ π

= =

22π

22 intx = ⋅

2 3 inty = ⋅ ⋅

32 3 int 24 intxy = ⋅ ⋅ = ⋅


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I. . The answer MUST BE an integer

II. . The answer MUST BE an integer

III. . The answer COULD be an integer, depending on the value of the unknown

int, but it doesn’t have to be, so we cannot include III in the right answer.


#24, pg.23 Difficulty Level: 700+ Topics: Rates & Work

Solution

We’re asked to find the distance of the jog. Let’s call it d. This is the same as the distance of the walk. To solve complex rate questions, try to draw out what is happening to help you visualize the problem. The key to solving these questions is to spot the relationships between the three parts of all rate/work equations: Distance, Rate, and Time.

There are two trips: a jog, and a walk.

Notice that the distance we’re solving for is the same for either trip, so I used the same variable (d) to express it on either trip. It is generally a good idea to use as few variables as you have to, and avoid needlessly introducing

new variables (such as if they are the same distance; just use d).

We know that the roundtrip took t hours, so this will be the sum of the time spent on the two trips. Write out this sum, and solve for the distance of each trip (d).

Express the total time in terms of t

24 int 3 int8⋅

= ⋅

24 int 2 int12⋅

= ⋅

24 int 4 int18 3⋅ ⋅

=

1 2 and d d

d dtx y

= +

Jog Distance = d Rate = x Time = d/x

Distance = d Rate = y Time = d/y

DistanceTimeRate

=

Walk


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Multiply equation by xy

Cross out common factors

Factor out d

Divide (x+y)


Solution B – Plug-in

Even a 700+ problem can be solved without full knowledge of the underlying algebra. When variables appear in the answer choices, you can often plug in easy values for those variables within the question, solve and then find out which answer choice agrees with your solution.

Let’s pick easy values:

Our plug-ins

Picking easy values here takes some careful planning. I picked Aaron’s jog rate that is twice the walk

rate, so he will spend twice as much time walking. The time spent will be in a ratio of . This is

important because I should then pick a total time that is a multiple of 3 to make my work easy.

With the values we picked the question becomes: “Aaron jogs at 4mph then walks back at 2mph. How many miles from home can he jog so that he spends a total of 3hrs jogging and walking?”

Since with our plug-ins , we know that Aaron will spend 2hrs walking and 1hr jogging. Thus we

can simplify the question further: “How many miles does Aaron jog, at 4mph, for 1hr of jogging?”

Our solution 4 miles

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variables in each answer with our plug-ins until we find the answer that equals 4 miles.

(A)

(B)

dxy dxytxyx y

= +

txy dy dx= +

( )txy d x y= +

txydx y

=+

4mphx = 2mphy = 3hrst =

(4mph)jog time 1

walk time 2=

( )3hrst =

jog time 1walk time 2

=

4 6 43

xyt

⋅= ≠

4 3 44 2

x txy+ +

= ≠⋅


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(C) This matches our solution!

(D)

(E)


Take-Aways • To solve complex rate questions, try to draw out what is happening to help you visualize the problem. The

key to solving these questions is to spot the relationships between the three parts of all rate/work equations: Distance, Rate, and Time

• It is generally a good idea to use as few variables as you have to, and avoid needlessly introducing new

variables (such as if they are the same distance; just use d). • When variables appear in the answer choices, you can often plug in easy values for those variables within

the question, solve and then find out which answer choice agrees with your solution. Avoid plugging-in 0, 1, or the same value for different variables (unless the variables are equal). By coincidence, two answers may agree with your solution. In that case, just plug in another set of easy numbers and try again


The question doesn’t need to be rephrased. What is the units digit of n?

(1) If n and have the same units digit, n could be 5, or 6.

Statement (1) is NOT SUFFICIENT

(2) If n and have the same units digit, n could be 5, or 6.


MERGE STATEMENTS

Even merging the two statements does give us a unique solution, as 5 or 6 are still possible values for n. Together, the two statements are NOT SUFFICIENT.


4 2 3 24 44 2 6

xytx y

⋅ ⋅= = =

+ +

4 2 3 9 44 2 8

x y txy+ + + +

= = ≠⋅

2 3 3 44 2

y t tx y+ +

− = − ≠

1 2 and d d

2n

3n


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Take-Aways Don’t spend more than a few seconds trying to anticipate what data would be sufficient. If you have a gut feeling,

make a mental note. If you don’t, move on.


p = ?

(1) p is divisible by 2, 3, and 5. Does that limit us to a unique value of p? No, it only means that where int can be any integer.

Statement (1) is NOT SUFFICIENT.

(2) p is divisible by 2, 5, and 7. Does that limit us to a unique value of p? No, it means that where int can be any integer


MERGE STATEMENTS

By merging the two statements, we learn that p is a multiple of 2, 3, 5, and 7. But even this doesn’t limit us to a unique possible p.

Together, the two statements are NOT SUFFICIENT.



How many minutes was Wanda charged for?

. By manipulating this equation to isolate (# of mins charged), we get the our rephrase:

Our rephrase “What is ?”

(1) Knowing how much the call cost, alone, does not tell us how long it lasted.


2 3 5 intp = ⋅ ⋅ ⋅

2 5 7 intp = ⋅ ⋅ ⋅

(per minute cost)(# of mins charged) Total Cost=

Total Cost(per minute cost)


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(2) We still have no information about the length of the call.


MERGE STATEMENTS

By merging the two statements, it may seem like we have a lot more data, but is it sufficient? Consider the following two scenarios:

a. 1st minute costs $2.50, and each additional minute costs $2.00. Total length of call would be 3 minutes b. 1st minute costs $3.50, and each additional minute costs $3.00. Total length of call would be 2 minutes

This demonstrates that we still do not have a unique value for the number of minutes that Wanda was charged for.



Take-Aways • One way to check whether a statement or two statements together are sufficient is to plug-in a couple of

values that fit the data given in those statements. If the values yield more than one unique answer to the question, then the statement or statements together are NOT sufficient. On the other hand, if all values that you plug-in yield a unique answer to the question, then the statement or statements together may (not must) be sufficient.



(1) The length of MN alone doesn’t help me find the sum of all three sides.


(2) The length of NP alone doesn’t help me find the sum of all three sides


MERGE STATEMENTS

By merging the two statements, I have 2 pieces of the puzzle from my rephrase (in step I). The fact that we’re dealing with an isosceles triangle is somewhat helpful because it tells us that the unknown side shares its length with one of the known sides. So the 3rd side, MP, has length 16 or 20. However, we cannot narrow the perimeter value to a unique solution

Together, the statements are NOT SUFFICIENT

MN NP MP+ +


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(1) This statement tells us nothing about retailers who did not own their own store, so we cannot get the full picture.


(2) The percentage who owned their store tells me nothing about purchasing computers.


MERGE STATEMENTS

By merging the two statements, we know that 40% owned their own stores, and that 85% of these retailers purchased computers. However, we cannot tell anything about retailers who did not own their own stores. Once again, we only have a partial picture



#30, pg.25 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values

Let be the number of $10 certificates. Because there are 20 certificates total, the number of $100 tickets will be ( ).

Our rephrase “What is t?”

(1) The formula for total revenue is “revenue from $10 certificates + revenue from $100 certificates”. Because the total revenue is between $1650 and $1800, we can build an inequality:

.

If we were dealing with an equation instead of an inequality, we could probably safely conclude without additional calculation, that we have sufficient data to find t. However, when dealing with inequalities, always do the work before you decide whether you have sufficient data. Let’s

purchased computersTotal

t20 t−

1650 10 100(20 ) 1800t t< + − <


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simplify the equation to find out what it tells us about t. Remember to operate on all three sides of the inequality.

Original

Get rid of parentheses

Merge like terms

-2000 on all sides

Divide by -90 (remember to switch signs)

Simplify

Does this inequality limit t to a unique value? Yes! the number of $10 certificates (t) must be an integer. Only one integer falls in this range:

Statement (1) is SUFFICIENT.

(2) This statement tells us that the number of $10 certificates (t) is less than 5. However, we do not know the exact value of t.




Our rephrase “Is standard deviation less than 3?”

(1) By definition, the standard deviation is the positive square root of the variance. So if variance is 4, standard deviation must be 2


(2) Standard deviation is the average amount by which individual values in a set differ from the mean of the set. This statement gives us this average amount, since the difference between each value and the mean is the same for all values. Standard deviation is 2.



1650 10 100(20 ) 1800t t< + − <

1650 10 2000 100 1800t t< + − <

1650 90 2000 1800t< − + <

350 90 200t− < − < −

35 209 9

t> >

8 29 93 2t> >

3t =


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you need to have a good understanding of what standard deviation means. Broadly, standard deviation is a measure of how tightly/loosely values are grouped around the average. The more dispersed the values, the higher the standard deviation.


The range is the positive difference between the highest and the lowest values. The range would be greater than 9 in the following cases. Each case leads to a rephrased question:

if x or y is greater than 12. “Is x or y greater than 12?”

if x or y is smaller than -3 “Is x or y smaller than -3?”

if “Is ?”

The answer to any of these rephrased questions would be sufficient information.

(1) Let’s test alternate scenarios to evaluate this statement. Remember that when using this method, only use values that fit the data from the statement you’re checking. In this case,

a. . In this case, the range would be 5, and the answer to the question (or rephrase) would be NO

b. . In this case, the range would be 12, and the answer to the question (or rephrase) would be YES

The two scenarios have yielded two different answers, so we do not have a unique solution.


(2) To determine whether this statement limits us to a unique solution—and is therefore sufficient—consider the following two scenarios:

a. . In this case, the range would be 3, and the answer to the question (or rephrase) would be NO

b. . In this case, the range would be 12, and the answer to the question (or rephrase) would be YES

The two scenarios have yielded two different answers, so we do not have a unique solution.


9range >

9range >

9range > 9x y− > 9x y− >

3y x>

1 and 4x y= =

4 and 15x y= =

4 and 5x y= =

4 and 15x y= =


Pag

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MERGE STATEMENTS

We know from statement (2) that x and y are greater than 3, and we know from the original question that they are integers. This means that . Since (statement 1), we can also deduce that

Look at our 1st rephrase in step I above. “Is x or y bigger than 12?” Now, we have enough info to definitively answer the question.

Together, the statements are SUFFICIENT.




#33, pg.25 Difficulty Level: 600-700 Topics: Exponents & Roots; Inequalities & Absolute Values Whenever you can, simplify given equations and isolate variables:

Original

Cross multiply

Our Rephrase “is ”

(1) The only way for to be a fraction (smaller than 1) is if the exponent is negative. For example,

. So we know that 0x <

Statement (1) is SUFFICIENT

(2) This statement directly gives the answer to our rephrase.



4x ≥ 3y x> 12y >

25 1?25

x+

<

2 25 5 ?x+ <

0?x <

5x

1 155

− =


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In a question about 2 overlapping sets, you can typically use either Venn diagrams or the Group Formula. Let’s use the latter. The Group formula is:

(Eq. 1)

Plug in (Eq. 2)

Isolate what we’re looking for (Eq. 3)

We are asked to find neither. From (Eq. 3) above, we can rephrase to:

Rephrase “What is ”

(1) Statement (1) tells us that ½ of those who required computer required writing. In an equation, this would

be: 12

writ comp= .

Although this gives us a ratio of writ to comp, it gives us no way to find and thus no way to answer our rephrase.


(2) 45% required writing only. Companies that require writing only is found by taking all companies that require writing, and subtracting companies that also require computer skills.

Algebraically, the statement is

Plug in

This statement will allow us to find out that , but still gives us no way to find and thus no way to answer our rephrase.


MERGE STATEMENTS

Let’s combine the statements. Take a look at the equations from each statement side by side:

(1) (2)

Plug in (2) into (1)

Solve for comp

Total comp writ neither both= + + −

20%both = 100 20comp writ neither= + + −

120 ( )neither comp writ= − +

?comp writ+

( )comp writ+

45writ both− =

20%both = 20 45writ − =

65writ = ( )comp writ+

12

writ comp= 65writ =

1652

comp=

130comp =


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By merging the two statements, we now have . This data is sufficient to answer our

rephrase “What is ”

In Data Sufficiency, if your rephrase is a good one, data sufficient to answer your rephrase is necessarily sufficient to answer the original question. Consequently, while taking your test, there is no need to check whether you have sufficient info to answer the original question if you’ve already checked whether you can answer your rephrase.

If you are still not convinced, you can plug into (Eq. 2).



Take-Aways • In Data Sufficiency, if your rephrase is a good one, data sufficient to answer your rephrase is necessarily

sufficient to answer the original question. Consequently, while taking your test, there is no need to check whether you have sufficient info to answer the original question if you’ve already checked whether you can answer your rephrase.




Our rephrase “What is ”

(1) Before combining, or evaluating data, try to write all your equations in a similar format. For example, note that the question asks for , but in this statement, w is on one side while q is on the other side of the equation. Restructure your equations so they look similar before you combine or compare them.

Original statement

Divide by 3

Add q


(2) Divide by 5


65 and 130writ comp= =?comp writ+

65 and 130writ comp= =


?w q+

w q+

3 3 3w q= −

1w q= −

1w q+ =

1w q+ =


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Take-Aways • Before combining, or evaluating data, try to write all your equations in a similar format.


Circle C has radius of 2. Does Y lie inside the circle can be rephrased as:

Our Rephrase “is ”

(1) Let’s use alternate scenarios to evaluate this statement.

According to Scenario A, the answer to the original question (does Y lie inside the circle?) is YES. According to Scenario B, the answer to the original question is NO. We do not have a unique answer.


(2) This statement directly answers our rephrase.



#37, pg.25 Difficulty Level: 500-600 Topics: Inequalities

Our rephrase “Is ”

(1) This tells us that x must be greater than y, since it is

2?OY <

?x y>

2x y= +

O

2

Y

X

O

2

Y

X

Scenario A Scenario B


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(2) Manipulate & simplify this equation first:

Original statement (2)

Multiply by 2

Now that we have x in terms of y, we can simplify the original question one step further:

Original question “Is ”

Plug in “Is ”

Simplify the question “Is ”

We have simplified the question further, but this statement alone still doesn’t give us the info needed to answer the question.



#38, pg.26 Difficulty Level: 600-700 Topics: Rates & Work; Inequalities & Absolute Values

The Distance, Rate & Time equation . This question is a bit tricky. We know that the rate is

more than 70, so we can write

The only way for the distance to be smaller than 210 kilometers is for the time to be less than 3hrs

Our rephrase “Is ”

(1) If , I have no way to determine whether it is less or greater than 210.


(2) If , we can definitely conclude that . Thus we can answer our rephrase (and thus the original question).



12x y= −

2 2x y= −

?x y>

2 2x y= − 2 2 ?y y− >

2?y >

( )( )d rate time=(more than 70)( )d time=

210?d <

200d >

205d < 210d <


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#39, pg.26 Difficulty Level: 600-700 Topics: Coordinate Geometry

Let’s look at all possible configurations of line k

As you can see, if r is positive, we cannot answer the question without more information. On the other hand, if r < 0, then the x-intercept must be negative.

(1) The slope of the line gives us no additional information because we have already been told that the slope is negative. You could draw a line of slope = -5 that matches all three scenarios drawn in the figures above.


(2) We know that r >0, but as Scenarios A and B demonstrate above, this isn’t enough to tell us whether the x-intercept is positive or negative


MERGE STATEMENTS

When we consider the statements together, we are limited to the first two scenarios drawn (since r > 0). However, this is no more helpful than statement (1) alone was.

Together, the statements are NOT SUFFICIENT.


(-5, r)

x-int

(-5, r)

x-int

x-int

(-5, r)

Scenario A

r > 0

Scenario B

r > 0

Scenario C

r < 0


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To find the interest yielded by p percent, we can use the formula . In this case,

, so . Thus we know the value of p

The question is “What amount…at k%...yields $500”. Let X be this amount. Algebraically, . To

find X, all we need is the value of k.

Our rephrase “What is k?”

(1) We already know that from the original question. So this statement gives us a way to find the value of k and thus an answer to our rephrase


(2) Statement (2) clearly gives us a value for k and thus an answer to our rephrase.






For a fraction to be positive, the top and bottom must have the same sign. Therefore has the same sign as z. Is x negative?

(1) x, y, and z could all be positive, or they could all be negative. We cannot answer our rephrase.


(2) If , then . x itself could still be positive or negative.

100pPrincipal interest⋅ =

5000 and 500Principal interest= = 10p =

500100

kX ⋅ =

10p =

( )x y+

0z < ( ) 0x y+ <


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MERGE STATEMENTS

From statement (2), we deduced that 0x y+ < , meaning that one or both of these is negative. Since x y< (from statement (1)), we can conclude that x must be negative.




When solving a Number Properties question, express all values in their prime factorizations. In this format, it is easier to deal with very large numbers, see patterns, and make connections.

Our rephrase “If we write k in its prime factorization form, are there at least 3 different numbers present?”

(1) We are told:

Multiply by 15

We can tell that k has at least 2 prime factors. Depending on the value of the unknown “int”, k could have no other prime factor (eg. if int = 1), or 2 other prime factors (eg. if )


(2) We are told:

Multiply by 10

We can only tell that k has at least 2 prime factors. This isn’t sufficient to tell us whether k has at least three different prime factors.


MERGE STATEMENTS

By combining both statements, we can tell that . We do not know how many different prime factors k has, but we do know that it has at least 3 different prime factors, so we can definitively answer the original question.


int15k=

3 5 intk = ⋅ ⋅

int 2 7= ⋅

int10k=

2 5 intk = ⋅ ⋅

2 3 5 intk = ⋅ ⋅ ⋅


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The question is already simplified. Is average median> ?

(1) With the total and the number of items (30 days), we can find the average (do not waste time calculating it). However, there is no way to compare it to the median.


(2) The bottom 60 percent of all values must include the median, since when you order the values the median is in the middle. In other words, if 60% of all temperatures are less than the average, the median must be less than the average.



#44, pg.26 Difficulty Level: 500-600 Topics: Exponents & Roots

Our rephrase “Is ” (Eq. 1)

(1) We now know that is an integer, so we can rephrase our question further:

Plug in (Eq. 1) to simplify the question. “Is ”. Another way to say the

same thing is “If you raise an integer to a positive integer power, do you get an integer result?”

The answer is a definitive YES. The only way to get a non-integer result would be to use a negative power, or a decimal power, like 7/3. We cannot do this because we are told in the beginning that m and n are positive integers.

Statement (1) is SUFFICIENT. (2) This statement tells us that n is a perfect square such as 1, 4, 9, 16, 25…. To see whether it is sufficient

info, let’s try alternate scenarios

212 2 3x x= → = ⋅

( ) int?nm =

m

1intm = 1(int ) int?n =


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(a) . In this scenario, and the answer to our rephrase (Eq. 1) is YES

(b) . In this scenario, and the answer to our rephrase (Eq. 1) is NO.

This statement does not limit us to a unique answer. Depending on the values we use, we get either YES or NO






Notice that there are 66 people, and that at most 6 people can have same-month birthdays. There is only one way that nobody has a January birthday. That is, if all the other 11 months are maxed out at 6 birthdays.

If even one of the other months has fewer than 6 birthdays, then January must have a birthday.

Our Rephrase “Do all months from Feb to Dec have 6 birthdays?”

(1) This statement tells us that Feb and Mar don’t have the same value, so only one of those two months could have 6 birthdays. We can definitively answer our rephrase. February through December cannot all have 6 birthdays.


1 and 4n m= = ( ) 2nm =

1 and 7n m= = ( ) 7nm =

0

1

2

3

4

5

6

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Births


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(2) This statement tells us that March has 5 birthdays. Not all months from Feb to Dec have 6 birthdays. We can definitively answer our rephrase





#46, pg.26 Difficulty Level: 700+ Topics: FDPs & Ratios; Statistics

. We already know that there are 10 employees. To find the average, we need the total of all

salaries.

Our rephrase “What is the total of all salaries this year?”

(1) Statement (1) only gives us a partial picture. We do not know what happened to the other 2 employees. Did their salaries triple? Did it decrease?


(2) Alone, this statement is no better than Statement (1) alone. We do not know what happened to the other 8 employees.


MERGE STATEMENTS

Together, it seems as though the two statements give all the info we need. Be careful however, and consider the following:

If the 8 lowest paid employees are the ones to receive a 15% increase, the new total salaries amount would be lower than if the 8 highest paid employees are the ones to receive the 15% increase. This is because a 15% increase of a larger number is a greater dollar increase. Thus we still cannot derive the new total from the information given.



total# of items

avg =


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#47, pg.26 Difficulty Level: 700+ Topics: Sets & Groups

This is an overlapping sets question. My preferred methods for these questions are Group Formula and Venn Diagrams. Either method would work here. Let’s use Venn Diagrams.

We know that and . We are looking for both (Fiction AND written in Spanish).

Our rephrase “What is the value of both?”

(1) Statement (1) allows us to break up the Fiction books into “Fiction but NOT in Spanish” (far right of diagram), and “both” (Fiction AND written in Spanish). We already know that there are 24 total Fiction books. So if “Fiction but NOT in Spanish” is 6 more than “both”, the breakdown must be 15 and 9. On the diagram, it would look like this:

We have the information needed to answer our rephrase.


(2) Statement (2) allows us to break up the Spanish books into “Spanish but Nonfiction” (far left of diagram), and “both” (written in Spanish AND Fiction). We already know that there are 23 total Spanish books. So if “Spanish but Nonfiction” is 5 more than “both”, the breakdown must be 14 and 9. On the diagram, it

23spanish = 24fiction =

Fiction Spanish

9 Note that Fiction correctly adds up to 24. “Of the fiction books, 6 more are NOT in Spanish…”

(23 - both) 15

neither

Fiction Spanish

both

Always start with the center.

“Spanish & Nonfiction” is every book in Spanish minus books that belong to both. This is the far left is (23 – both). Similarly, “Fiction but NOT in Spanish” is all of Fiction books minus books that belong to both. Thus the far right is (24 – both).

(23 - both) (24 - both)

neither


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would look like this:

We have the information needed to answer our rephrase.






#48, pg.26 Difficulty Level: 600-700 Topics: Translations & Manipulations; Geometry

. To find the perimeter, all we need is .


(1) With the diagonal, we can build the Pythagorean theorem

We have one equation, but it is not possible to isolate from this equation.


(2) From this statement, we know that , but as the two scenarios below show, we still don’t know . Consider the two possibilities below:


Perimeter 2( )l w= + ( )l w+

( )l w+

( )l w+

48l w⋅ =( )l w+

Fiction Spanish

9 Note that Spanish correctly adds up to 23. “Of the books in Spanish, there are 5 more Nonfiction…”

14 (24-both)

neither

w

l

2 2 210l w+ = 10


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(a) (b)

We do not have enough info to find the perimeter.


MERGE STATEMENTS

Together, we now have two independent equations, and two variables. If the equations were linear, we could conclude without any additional work that we have enough information. However, because we have a non-linear equation ( ), we must go further.

Statement (1) told us that 2 2 100l w+ = and statement (2) added that 48lw = . Because l and w are lengths, they must be positive. We could try a few numbers that multiply to 48 and see whether the sum of their squares is 100.

• 1 48 48⋅ = , but 2 21 48+ is too large (bigger than 100)

• 2 24 48⋅ = , but 224 is already too large. We need two numbers smaller than 10.

• 6 8 48⋅ = , and 2 26 8 100+ = .

The length and width are 6 and 8. This is sufficient to find the perimeter.

Together, the statements are SUFFICIENT


8 and 6l w= = 48 and 1l w= =

48l w⋅ =

Patrick Siewe - GMATFix.com OG Companion 12th Problem Solving

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Problem Solving Solutions

Solutions to Questions from pg. 152-185 in the Official Guide for GMAT Review, 12th Edition

Patrick Siewe – GMATFix.com OG Companion 12th Problem Solving

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PROBLEM SOLVING Answer Key

1. A

2. B

3. C

4. E

5. C

6. C

7. B

8. E

9. E

10. D

11. E

12. A

13. C

14. A

15. B

16. D

17. D

18. E

19. E

20. A

21. B

22. A

23. B

24. D

25. E

26. B

27. E

28. B

29. C

30. B

31. B

32. B

33. C

34. A

35. B

36. A

37. A

38. B

39. B

40. C

41. A

42. E

43. B

44. D

45. B

46. C

47. C

48. C

49. B

50. B

51. C

52. E

53. E

54. A

55. E

56. B

57. A

58. B

59. D

60. D

61. E

62. A

63. A

64. B

65. B

66. E

67. B

68. D

69. C

70. C

71. A

72. B

73. D

74. D

75. E

76. D

77. D

78. C

79. A

80. B

81. A

82. A

83. C

84. A

85. A

86. A

87. E

88. D

89. D

90. E

91. B

92. C

93. B

94. B

95. E

96. D

97. B

98. C

99. B

100. D

101. D

102. B

103. E

104. B

105. C

106. B

107. D

108. B

109. C

110. C

111. B

112. D

113. D

114. B

115. C

116. B

117. E

118. C

119. D

120. C


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PROBLEM SOLVING Answer Key

121. C

122. E

123. B

124. B

125. E

126. C

127. A

128. A

129. D

130. E

131. E

132. D

133. E

134. B

135. C

136. A

137. E

138. E

139. B

140. D

141. C

142. E

143. A

144. C

145. C

146. C

147. A

148. D

149. E

150. A

151. B

152. E

153. D

154. B

155. D

156. D

157. B

158. E

159. D

160. D

161. E

162. B

163. E

164. D

165. A

166. D

167. E

168. E

169. C

170. A

171. E

172. C

173. E

174. D

175. D

176. C

177. A

178. C

179. E

180. B

181. A

182. A

183. C

184. D

185. C

186. C

187. A

188. E

189. B

190. E

191. C

192. E

193. D

194. D

195. C

196. D

197. B

198. C

199. D

200. A

201. B

202. E

203. C

204. E

205. B

206. E

207. E

208. A

209. D

210. B

211. A

212. B

213. D

214. E

215. C

216. B

217. D

218. D

219. E

220. B

221. D

222. C

223. B

224. E

225. D

226. A

227. A

228. D

229. C

230. C


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#1, pg.152 Difficulty Level: 200-400 Topics: Translations & Manipulations, FDPs & Ratios

30 Seconds Hack

A glance at the answer choices tells us that they are a few hundreds apart. We can approximate. An annual budget of $12,600 means that the monthly budget is a bit more than $1000. Say…$1050.

At the end of the fourth month, the budget should be about $4200. The actual expenditure is about $4600, so we are roughly $400 over-budget. Pick the answer closest to $400

Guess A.

Solution

The annual budget is $12600, so the monthly budget is . Don’t do the math yet.

After four months, the budget should be .

Cross off common factors .

Divide by 3 4-month budget should be $4200.

The actual expenditure is $4580, so the excess is


Take-Aways • Do not rush to “do the math”. Instead, build your equations and see whether you can cross off common

factors. On the GMAT, things tend to simplify to easy numbers.


Solution A – Algebraic Solution

Set up an equation: the sum of 5, 8, 12 and 15 5 + 8 + 12 + 15

…is equal to the sum of 3, 4, x and x + 3 = 3 + 4 + x + x + 3.

1260012

12600 412

⋅

12600 13

⋅

$4580 $4200 $380− =


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Solve: 5 8 12 15 3 4 3 40 2 10 30 2 15x x x x x+ + + = + + + + → = + → = → =



Since the right answer gives us the only value of x that will work with the rest of the problem, we can work backward. Start with C and find the answer choice fits.

Value of x Sum of 5, 8, 12, 15 Sum of 3, 4, x, x+3 Are sums equal? (A) 14

(B) 15 40 ( ) ( )3 4 15 15 3 40+ + + + = YES. When x = 15, the sums equal each other

(C) 16 5+8+12+15 = 40 ( ) ( )3 4 16 16 3 42+ + + + = 2nd sum is too large. we need a smaller x. Eliminate C, D & E

(D) 17 (E) 18


Take-Aways • When a question gives you non-variable answer choices along with a word problem, you can often use


#3, pg.152 Difficulty Level: 500-600 Topics: Statistics; Functions & Sequences

Solution

Simplify the given fraction by breaking it apart.

The question becomes “For which value of n is NOT an integer?” In other words, we want a value of n that

does NOT divide into 100.


100 100 100 1n nn n n n+

= + = +

100n


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30 Seconds Hack

If the area is the same, is constant. So if we change one number, we must adjust with the other:

Floor X is 12 by 18.

Floor Y is length by 9

The way I set it up above, I’ve lined up 18 with 9 so that we can easily see that one dimension is cut in ½. The other must be doubled to offset the change. So, length goes from 12 to 24.

Guess E.

Solution

If the area is the same, is constant. Set up an equation to equate the two areas:

Plug-in values

Isolate length

Cross off common factors in 9 & 18



Solution

( )length width⋅

24length =

( )length width⋅

AreaX AreaY=

12 18 9 length⋅ = ⋅

12 189

length ⋅=

12 21

length ⋅=

24length =


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This question gives us a frequency distribution table. Frequency distribution data tells you which values occur in the set as well as how many times each of them occurs. In our case, 5 salaries equal $20,000, 4 salaries equal

$22,000… The average is by definition: total

# of valuesavg = .

• The number of values (total number of employees) is 5 + 4 + 8 + 3 = 20. • The first group of 5 employees earn a total of 5 20,000 $100,000⋅ =

• The next group of 4 employees earn a total of 4 22,000 $88,000⋅ =

• The next group of 8 employees earn a total of 8 25,000 $200,000⋅ =

• The last group of 3 employees earn a total of 3 30,000 $90,000⋅ =

• The total earned by all employees is 100 88 200 90 $478+ + + = $478,000

Now that we have the number of values and the total, we can find the average:

478,000 $23,90020

total# of values

avg = ==


#6, pg.153 Difficulty Level: 200-400 Topics: Translations & Manipulations; FDPs & Ratios

Solution A

Take in the info, one bit at a time.

Each box has 100 paperclips, so 3 boxes would have paperclips. There are b boxes in a carton, so there

are paperclips in a carton.

Since there are paperclips in a carton, 5 cartons would have paperclips. There are c cartons

in a case, so there are paperclips in a case.

Since there are paperclips in a case, 2 cases must have paperclips.


Solution B – Plug In

When variables appear in the answer choices, you can often plug in easy values for those variables within the question, solve and then find out which answer choice agrees with your solution.

(100 3)⋅(100 )b⋅

(100 )b⋅ (100 5)b⋅ ⋅

( )100 b c⋅ ⋅

( )100 b c⋅ ⋅ ( )200 b c⋅ ⋅


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Our plug-ins

With these values plugged in, re-consider the question: “A case has 2 cartons. Each carton contains 10 boxes, and each box contains 100 paperclips…”

There will be 20 boxes in each case. In two cases, there are 40 boxes. Since each box has 100 paperclips, there are a total of 4000 paperclips.

Our solution 4000

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variables in each answer with our plug-ins until we find the answer that equals 4000.

Only answer choice C matches our solution.

(C)


Take-Aways • When variables appear in the answer choices, you can often plug in easy values for those variables within

the question, solve and then find out which answer choice agrees with your solution. Avoid plugging-in 0, 1, or the same value for different variables (unless the variables are equal). By coincidence, two answers may agree with your solution. In that case, just plug in another set of easy numbers and try again.

#7, pg. 153 Difficulty Level: 200-400 Topics: Number Properties

Solution

Memorize your multiplication tables up to 12’s before taking the GMAT. This will allow you to save a lot of time on laborious calculations.

Between 60 and 70, the only prime numbers are 61 and 67. Their sum,


Take-Aways • Memorize your multiplication tables up to 12 times 12 before taking the GMAT. You should also know

your perfect squares up to and how to recognize at a glance whether a number is divisible by 2, 3, 4, 5, 6, 8, and 9.

10b = 2c =

200 200 10 2 4000bc = ⋅ ⋅ =

61 67 128+ =

220


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#8, pg. 153 Difficulty Level: 600-700 Topics: FDPs & Ratios; Translations & Manipulations

Solution - Approximate

To the answer, we need the total capacity (t). We’re told that the new volume of water, 138 billion gallons, is 82% of total capacity. This can be expressed in the equation . To avoid dealing with decimals, I recommend using fractions instead.

Replace 0.82 with 82/100

Don’t do the math yet. Isolate t

It’s critical that you not rush into calculations. Instead look to simplify. In this case, dividing 138 by 82 would be quite ugly. Notice however that even the question says “Approximately how many gallons…” This is a very clear invitation to approximate. Instead of 138 and 82, use two easier integers. I chose 136 and 80 because these share a lot more common factors (than 138 and 82) and thus will allow me to simplify a lot more.

Approximate

Simplify by a factor of 4

Simplify by a factor of 2

Approximate t

Note that we simplified before multiplying. This is a useful technique to keep numbers low. “ ” is a reminder that our numbers are just estimates.

Now that we have an approximate value for t, we can answer the question. How much was the reservoir short of capacity before the storm?

Approximate a solution


138 0.82t=

82138100

t=

138 10082

t ⋅=

138 100 136 10082 80

t ⋅ ⋅= ≈

34 10020

t ⋅≈

17 10010

t ⋅≈

170t ≈

≈

Answer 170 124 46

≈ −≈


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Take-Aways • When the question itself asks you to approximate, do so because the math will likely be unpleasant. • Do not rush to “do the math”. Instead, build your equations and see whether you can cross off common

factors. On the GMAT, things tend to simplify to easy numbers. • Consider using fractions instead of decimals in your calculations because the former are easier to simplify.

#9, pg. 153 Difficulty Level: 200-400 Topics: Coordinate Geometry Solution

Because the center of symmetry is at , the value of y will be the same when x is 1 as it is when x is 3.

When


#10, pg. 153 Difficulty Level: 200-400 Topics: FDPs & Ratios

Solution

“Percent” literally means “divided by 100”. So .

Subtract of 5000 from of 5000

Simplify


#11, pg. 153 Difficulty Level: 400-500 Topics: Exponents & Roots

Solution A – Intuitive solution

The square root of a cube root is the 6th root (ie ), so we are looking for a number that when multiplied by itself 6 times, will give us 0.000064.

2x =

1, 1 so when 3, 1x y x y= = = =

( )1101

101 1 1%

100 10 100 1000= = ⋅ =

110 % 1

101 15000 5000

10 1000 ⋅ − ⋅

500 5 495= − =

3 6x x=


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Because 64 is , you ought to suspect that the correct answer is either 0.2 or 0.02.

Because 0.02 has two decimal digits, will have twelve digits past the decimal point. On the other hand,

will have six digits past the decimal point. Given that (0.000064) has six decimal digits, 0.2 is the correct

answer.


Solution B

When manipulating complex roots, consider:

a. Changing roots to exponents: b. Rewriting numbers in their prime factorization

Original

Merge exponents

Change decimals into fractions

Use Prime factorization



Solution

“Favorable outcomes” is the number of tickets numbered 200 to 299. There are 100 favorable outcomes.

“All possible outcomes” is the number of tickets numbered 101 to 350. There are 250 possible outcomes

62

6(0.02)

( )60.2

1n nx x=

( )( )1/21/33 0.000064 0.000064=

( )1/60.000064=

1/664

1,000,000

=

1/66

6

2 2 0.210 10

= = =

favorable outcomesProbabilityAll possible outcomes

=


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30 Seconds Hack

We know that the tax paid is 7% of the amount in excess of $1,000. So the total value of the item is the non-taxable $1,000 plus the taxable amount. The 7% tax on this taxable amount is $87.50.

$87.50 is 10% of $875. This same $87.50 is a smaller portion (7%) of the taxable amount. So the taxable amount must be greater than $875. Therefore the total value must be greater than $1,000 (the non-taxable portion) plus $875. The total value must be greater than $1,875.

Guess C, D or E


Since the first $1,000 are not taxed, the tax ($87.50) must be 7% of the taxable amount only. Set this up as an equation and find the taxable amount, x:

Tax equals 7% of x 7 100 8,75087.50 87.50 $1,250

100 7 7x x x x= ⋅ → = ⋅ → = → =

Now, we’ve found x, the taxable portion of the value. Since the first $1,000 are not taxed, the total value is 1,000 + 1,250 = $2,250



The answer choices represent the total value of the item. Working backwards, we can find out which value fits into the problem. We know that a 7% tax on the portion of the value in excess of $1,000 should equal $87.50. Only the right answer can meet this constraint. When doing reverse engineering, start with answer C

Total Value Taxable amount (amount above 1,000)

7% tax (should equal $87.50) Are sums equal?

(A) 1,600

100 10 2Probability250 25 5

= = =


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(B) 1,850

(C) 2,250 1,250 71250 12.5 7 $87.50

100⋅ = ⋅ =

YES. If the total value is $2,250, the tax is exactly $87.50. If this answer had been too large, we would have been able to eliminate C, D and E. That’s why it’s smart to start with C.

(D) 2,400 (E) 2,750


Take-Aways • When a question gives you non-variable answer choices along with a word problem, you can often use


#14, pg. 154 Difficulty Level: 500-600 Topics: FDPs & Ratios; Statistics

30 Seconds Hack

The overall average must be between the two numbers given. So the answer must be greater than but less

than . Eliminate C, D, and E.

We can go further. Because there are twice as many packages at the lower weight, the overall average should be much closer to

Guess A.

Solution A

To find the overall average, use the average formula

Factor out a 4 to simplify

3812

1415

3812

total# of items

avg =( ) ( )3 1

8 48 12 4 1512+

( ) ( )3 18 44 2 12 1512

+


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Simplify by factor of 4

Change mixed numbers to fractions

Simplify 2 and 8 in first fraction

Merge fractions


#15, pg. 154 Difficulty Level: 200-400 Topics: Exponents & Roots; FDPs & Ratios

Solution


#16, pg. 154 Difficulty Level: 500-600 Topics: FDPs & Ratios; Geometry

Solution

. So we know that

If we double, each dimension,


( ) ( )3 18 42 12 15

3 +

99 6128 4

3

+

99 614 4

3

+

13

160404 13

3 3= =

( ) ( )2 30.1 0.1 0.1 0.1 0.01 0.001 0.111+ + = + + =

Volume l w h= ⋅ ⋅ 10l w h⋅ ⋅ =

( )New 2 2 2 8 8 (10) 80V l w h l w h= ⋅ ⋅ = ⋅ ⋅ = ⋅ =


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#17, pg. 154 Difficulty Level: 200-400 Topics: Translations & Manipulations; FDPs & Ratios

Solution

Starting amount 40

1/2 sold by noon, so half is left 20 is left at noon

80% of remaining sold means 20% of remaining left at closing time.


#18, pg. 154 Difficulty Level: 200-400 Topics: Geometry

30 Seconds Hack

The square and rectangle have the same area. The width of the rectangle (2.5) is obtained by dividing the side of the square (6) by almost 3. To balance this out (so that areas be equal), the length of the rectangle must obtained by multiplying the side of the square (6) by almost 3. Thus the length of the rectangle can be obtained my multiplying the side of the square (6) by a number that is almost 3. That number is between 2 and 3, so the length must be between 6 2⋅ and 6 3⋅ . The length must be between 12 and 18.

Guess E

Solution A

We know the dimensions of the square (all sides are 6). We know the width of the rectangle (2.5). Draw:

Solve for the length: 5 2 722.5 36 36 36 14.42 5 5

l l l l l= → = → = ⋅ → = → =

Notice that I converted the decimal 2.5 into a fraction. Doing so allows me to work with integers, because I don’t like having decimals in my division.

20 20 4 left100

⋅ =

Since the two figures have the same area, we can write: 6 6 2.5 l⋅ = ⋅ where l is the length.


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Because the answers give us specific values for the length of the rectangle, we can work backwards. We know

that the rectangle and square have the same area (the area of the square is 2 26 36side = = ). Only the right answer will match this constraint. When doing reverse engineering, start with answer answer C.

Length Area (length x width) – width = 2.5

Does the area of rectangle equal the area of square? (36)

(A) 8.5

(B) 9.5

(C) 9.6 ( ) ( )9.6 2.5 too small⋅ = ( )10 2.5⋅ would equal 25, so this is smaller than 25

and definitely smaller than 36. We need a longer length. Eliminate A, B and C

(D) 10.5 ( ) ( )10.5 2.5 25⋅ ≈ This is about 25 ( a bit bigger), but definitely not as large as 36. STOP HERE. E must be the answer because it’s the only choice remaining.

(E) 14.4 Must be the right answer. Area must equal 36.


Take-Aways

• When a question gives you non-variable answer choices along with a word problem, you can often use reverse engineering. Start with answer C, especially if it will take more than a few seconds to check each answer.

• Drawing figures reveals connections between data and often makes the problem easier to understand and handle. Always make a drawing of your own for geometry questions, even if one is already provided


Solution A – Intuitive Solution

150 is five times 30, so it must be 500% of 30.


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Solution B

Percent literally means “divided by 100”. Consequently, “what percent” is the same as .

Let’s express the question algebraically:

“150 is what percent of 30?”

Isolate x

Simplify, then multiply



Solution

The answer choices express the ratio using only integers. To change 2 to into integers, multiply each number

by three. The ratio can thus be written as 6 to 1.


#21, pg. 155 Difficulty Level: 500-600 Topics: Rates & Work

30 Seconds Hack

In fewer than 60 seconds, we can solve this problem. If we don’t have the time, or the skill to quickly find how many bottles 10 machines can produce in 4 minutes, we can come up with a range.

Six machines make 270 bottles a minute, so in four minutes, they will make bottles.

100x

150 30100

x= ⋅

150 10030

x⋅=

500x =

13

270 4 1080⋅ =


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Twelve machines would make twice as many bottles; so in four minutes, they will make 2160 bottles.

Ten machines would thus have to make more than 1080 but fewer than 2160 bottles.

Guess B.

Solution

If six machines make 270 bottles per minutes, each machine makes bottles per minute.

Ten machines would make bottles per minutes.

In four minutes, these ten machines would make bottles



Solution

On some copies of the Official Guide 11th Ed, the graph in the explanation does not match the graph in the question.

Absolute value of a number is distance from 0. The greatest distance from 0 – and thus the greatest absolute value – leads to point A.



Solution – Plug In

The only thing we know about n is that it’s a prime number bigger than 3. Thus this question is asking us about the property not of a unique value, but of a group of numbers (primes greater than 3). Because only 1 answer can be correct, whatever prime number we use must give us the same result.

To solve, select an a value for n that fits the restriction given (prime greater than 3).

• If n = 5, 2 25n = and the remainder when 2n is divided by 12 equals 1.

270 456

=

10 45 450⋅ =

4 (450) 1800⋅ =


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Because there can only be one answer that equals the remainder, B must be correct. It doesn’t matter what n we use.

• If n = 7, 2 49n = and the remainder when 2n is divided by 12 equals 1.



30 Seconds Hack

1 1 1 11 1 1.33 1.501 13 2

→− −+ +

. Since the denominator of the 2nd fraction is greater, the 2nd fraction as a

whole is the smaller fraction. Thus we’re subtracting a smaller value from a larger value. The result must be positive. Eliminate A, B and C.

Guess D or E

Solution

Start away from the main fraction bar: 1 14 33 2

1 1 1 11 13 2

−− →+ +

When fractions are involved in a division, be sure to express both the top and bottom as fractions:

1 11 1 1 3 1 2 3 21 1 4 3 4 3 1 4 1 3 4 33 2 3 2

→

− − → ⋅ − ⋅ → −

Find a common denominator (12) and solve: 3 2 9 8 1 4 3 12 12 12

→− − →



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#25, pg. 155 Difficulty Level: 200-400 Topics: Coordinate Geometry

Solution

V has a positive x-coordinate and a negative y-coordinate. Only answer choice E offers the correct combination




The total length is 40 feet. If the short piece is 18 feet shorter, the short piece must be 9 less than the average and the long piece must be 9 more than the average length. So the short piece must be .


Solution B – Build Equations

We can build two equations:

Total length is 40 feet (Eq. 1)

long is 18 more than short (Eq. 2)

By merging these two equations, we can solve for short:

Replace long with in (Eq. 1)

Solve for short

Be careful to answer the question asked. We have the length of the short piece and we should stop there. The answer choices include the length of the long piece, in an attempt to distract unfocused test-takers.


Solution C – Reverse Engineering


20 9 11− =

40short long+ =

18long short= +

( 18)short + ( ) 40short long+ =

( 18) 40short short+ + =

11short =


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Reverse Engineering is working from the answer choices, to find which choice can perfectly accommodate all the given data in the question. Let’s reverse engineer the question and figure out which answer choice makes “rope length equals 40” true. When reverse engineering, start with answer C

Length of the short piece

Long piece (18 longer than short)

Total length (should be 40)

Notes

(A) 9

(B) 11 29 40 Yes! A short piece of length 11 is needed to make total length 40

(C) 18 36 54 The lengths are too long. Eliminate C, D and E

(D) 22 -- (E) 29 --




#27, pg. 156 Difficulty Level: 400-500 Topics: Statistics

30 Seconds Hack

4 scores average 78 points. At the end of the day the average should be 80 points. The 5th score, by itself will need to balance the other 4. Since the 4 scores are 2 points less than the final average, there is total “deficit”, between the 4 scores and the average, of 8 points. To balance this out, the 5th score will need to have an “excess” of 8 points. Thus the 5th score will be 8 points more than the overall average. The 5th score will be 80 + 8 = 88

Guess E


Questions about averages that tell you the number of items are really asking you about the total. In this case, we

know that the average of 4 tests is 78. The average formula is: total

# of valuesavg = . We know the average (78)

and we know the number of values (4), so we can find the total: ( )( )total # of values 78 4 312avg= = ⋅ =


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At the end of the day, 5 scores should have an average of 80. So the total of the 5 scores should be: 80 5 400⋅ = . To take his total from 312 to 400, the student must score the difference on his last test. His final score must be 400 – 312 = 88



Because the answers give us exact values for the 5th score, we could reverse engineer. Work backwards and find the answer choice that will result in an overall average of 80 points. Only the right answer will fit all the constraints of the problem (1st 4 scores must average 78). Start with answer C.

5th score Average of all 5 scores: 5

sum Is overall average equal to 80?

(A) 80

(B) 82

(C) 84 ( )4 78 84 396 79.2

5 5+

= = Average is close to 80, but too small. The final score should be greater. Eliminate A, B and C

(D) 86 ( )4 78 86 398 79.6

5 5+

= = Overall average is still too small. STOP. Pick E. It’s the only possible right answer.

(E) 88 No need to do the math. Nothing else worked, so this must be correct.



Solution

Each kilometer is 43.9 10⋅ inches. The number of kilometers in 142.3 10⋅ inches can be found by setting up a

proportion: 4 141

3.9 10 2.3 10km x

in in=

⋅ ⋅ . Solve:

( )( )( )( )

1414 4

4

2.3 10 2.3 103.93.9 10

x −= → ⋅


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Because the answer choices are far apart (different exponents means each answer choice is about 10 times the previous), it’s safe to round 2.3 and 3.9 to integers 2 and 4. Thus the distance we’re looking for becomes:

10 9 92 2010 10 5 104 4

x x≈ ⋅ → → ≈ ⋅ . Pick the closest answer



Solution

If , a and b must have the same sign; either they are both negative, or both positive

(A) This would be false if a and b were both negative.

(B) This would be false if a and b were both negative.

(C) Since a and b have the same sign, their product must be positive. This MUST BE true.

(D) This would be false if a were smaller than b

(E) This would be false if a and b were both negative.



Solution

By observing the numbers printed on the grid, we can tell the scale. Along the horizontal axis, the x-interval is 1. On the vertical axis, the y-interval is 2.

Only account for points that are on the right of and above . There are five such points

Never draw or write directly on the images and text in a GMAT book, because you will be unable to do that on test day.


0ab>

25x = 22y =


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Solution

This is a rate/work question. We know the amount of work that needs doing – to type y words – as well as the rate – x words per minute. What we need to solve for is time. So let’s start by plugging what we know into the rate formula:

Rate Formula

Plug-in work (y) and rate (x)

Solve for the time


Solution B – Plug-In



Our plug-ins

With these chosen values, the question becomes: “How many minutes does it take John to type 10 words if he types at the rate of 2 words per minute”

At 2 words each minutes, John will take 5 minutes to complete the task.

Our solution 5


(A)

(B) This answer matches our solution!

(C)

work rate time= ⋅

y x time= ⋅

ytimex

=

10 wordsy = 2 words/minutex =

2 510

xy= ≠

10 52

yx= =

2 10 5xy = ⋅ ≠


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(D)

(E)





Solution

Instead of doing all the multiplication and addition, express the numbers in their prime factorizations so that you can take out common factors.

Original expression

Write prime factorizations ( )( ) ( )( )4 2 3 52 2 5 2 2⋅ +

Combine like factors 6 82 5 2⋅ +

Factor ( )6 22 5 2+

Take 62 out and do the addition 32 9 24=



212 2 3x x= → = ⋅ ). In this format, it is easier to deal with very large numbers, see patterns, and make connections.

60 60 2 55

xy

⋅= ≠

10 560 60 2

yx= ≠

⋅

(16)(20) (8)(32)+


Pag

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Solution



Solution

This is a rate/work question. We know the amount of work that needs doing – to run x yards – as well as the rate

– y yards in 11 seconds, or . What we need to solve for is time. So let’s start by plugging

what we know into the rate formula:

Rate Formula

Plug-in work (x) and rate

Solve for the time



yards per seconds11y

work rate time= ⋅

11y

11

yx time= ⋅

11xtimey

=

A straight line is 180° , so the angles immediately on either side of 150 must be 30° each. Thus, the fraction shaded is 2 30 60 1360 360 6⋅ ° °

= =° °

150


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Our plug-ins

With these chosen values, the question becomes: “If Juan takes 11 seconds to run 2 yards, how many seconds will it take to run 4 yards?”

Since he must run twice as far, it will take twice as long. Juan will take 22 seconds.

Our solution 22 seconds


(A) This answer matches our solution!

(B)

(C)

(D)

(E)




#35, pg. 157 Difficulty Level: 200-400 Topics: Combinatorics

Solution

2 yardsy = 4 yardsx =

11 11 4 11 2 222

xy

⋅= = ⋅ =

11 11 2 224

yx

⋅= ≠

4 2211 11 2

xy= ≠

⋅

11 11 224 2xy

= ≠⋅

4 2 2211 11xy ⋅

= ≠


Pag

e92

To leave John with the greatest number of pairs intact, make sure that the socks he loses affect the minimum pairs possible. The figure below shows 6 pairs of matched socks left.



Solution A – Reverse Engineering

When a question gives you non-variable answers along with a word problem, you can often use reverse engineering. Reverse Engineering is working from the answer choices, to find which choice can perfectly accommodate all the given data in the question.

Let’s reverse engineer the question and find out the smallest answer choice that is divisible by 1, 2, 3, 4, 5, 6, and 7. In this case, because we want the lowest integer, we should start with the smallest answer (A).

by 2? by 3? by 4? by 5? by 6? by 7?

(A) 420 Yes, it is even

Yes, sum of digits is a multiple of 3

Yes, last two digits make up a multiple of four (20)

Yes, ends in a 0 or 5

Yes, even and a multiple of 3

Yes,

There is no need to look further than (A), since it fulfills all requirements.

The rules of divisibility by…

2 – The number is even

3 – The sum of the digits add up to a multiple of 3

4 – The last two digits make up a multiple of 4. (ie. 1248240 works, but 78849 doesn’t)

5 – The number ends with a 5 or a 0

6 – The number is even and divisible by 3

8 – The last three digits make up a multiple of 8 (ie. 1248240 works but 1248243 doesn’t)

9 – The sum of the digits add up to a multiple of 9

420 607

=

Not lost

Lost


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Solution B – Least Common Multiple

The lowest positive integer divisible by 1 through 7 is the least common multiple of 1 through 7. 1 can be ignored since it is a factor of all integers. Let’s find the least common multiple of 2 – 7.

To find the least common multiple:

1. Write out all your values in their prime factorizations

2. Write down the product of all primes that appear in the different factorizations (without duplicates)

3. Assign to each prime, the highest exponent that appeared in any single factorization. This new product is the least common multiple





30 Seconds Hack

The bottom of the fraction, 0.75 – 1, is a negative number (-0.25) while the top is positive. Because the top and bottom of the fraction have opposite signs, the result must be negative. Eliminate C, D and E.

Guess A or B

Solution

1 1 100 40.75 1 0.25 25

→ → → −− − −

1

1

2 23 3=

=

2

1

4 25 5=

=

1 1

1

6 2 37 7= ⋅

=

2 3 5 7⋅ ⋅ ⋅

2LCM 2 3 5 7 420= ⋅ ⋅ ⋅ =


Pag

e94



Solution

The fraction is equal to 5, so the top must be five times the bottom. Since the top is 1.5, the bottom must add up to 0.3. Consequently,


Solution B

Original

Multiply by 0.2+x

Isolate x



Solution

Approximate the coordinates of the point that is twice as far from P as from Q. On the y-axis, P and Q are 3 units apart. On the x-axis, they are also 3 units apart. The point we seek should be 2 units away from P on both axes. This is the point (2, 1)



0.1x =

1.5 50.2 x

=+

1.5 1 5x= +

1.5 1 0.5 0.15 5

x −= = =


Pag

e95

Solution

Rely on your number property rules of odd and even (or plug in an even n and an odd n – the right answer must be even whatever n is).

(A) n + 1 This would be odd if n = 2.

(B) n + 2 This would be odd if n = 1

(C) 2n Any integer multiplied by an even is even. This answer must be even.

(D) 2n + 1 This will always be odd because it is 2n (which is even) + odd.

(E) 2n 2n has the same property as n. This would be odd if n were odd.


Take-Aways

• To find the answer that must be something, you can eliminate all the other answers by showing that they don’t have to be that thing.


Solution

A solid understanding of quadratics would allow you to solve this question in 10 to 15 seconds. Let’s quickly review how to solve a quadratic.

To solve the quadratic , It is necessary to find two numbers m and n such that:

1. m and n add up to b 2. m and n multiply to c

The equation could then be factored into and the solutions will be . It is important here to note that the two solutions will add up to since

.

Let’s apply this to our question. In this question, the quadratic is . The two solutions of x will

add up to -3. We are already told that one solution is 4, so the other must be .


2 0x bx c+ + =

m n b+ =m n c⋅ =

2 0x bx c+ + = ( )( ) 0x m x n+ + = or x m x n= − = − b−

( ) ( ) ( )m n m n b− + − = − + = −

2 3 10x x k+ + =

7−


Pag

e96

#42, pg. 158 Difficulty Level: 400-500 Topics: Functions & Sequences

Solution

Function questions basically have 2 parts: (1) definition and (2) application. The definition tells you how to use the function by showing you an example (typically with variables). The application asks you to repeat the process with other variables or with numbers. To solve, just replace the variables in the definition with the values in the application.

Definition: a b

ad bcc d

= − . To apply, replace a, b, c, and d with the values that take their place.

Application: ( )( ) ( )( ) ( )3 53 4 5 2 12 10 22

2 4

= − − → − − →−



30 Seconds Hack

We want 7 18 9+ .

78

is smaller than 89

because the latter is closer to 1. Since 8 1 19 9+ = ,

7 18 9+ must be a bit

smaller than 1.

Guess B

Solution

7 1 7 9 1 8 63 8 71 8 9 8 9 9 8 72 72 72

⋅ ⋅+ → + → + →⋅ ⋅

. The sum is slightly smaller than 1



Pag

e97


Solution A

Set . We are told that

Isolate t

Solve




Let’s reverse engineer the question and figure out which answer choice makes “ ” true.

What is value of t? , so x is… , so y is… Notes: At the correct t, x should equal y

(A) , this answer is wrong

(B) , this answer is wrong

(C) , this answer is wrong

(D) Yes! At this value of t,

(E) --- Don’t bother. We’ve found a match


x y= 1 3 and 2 1x t y t= − = −

x y= 1 3 2 1t t− = −

2 5t=

25

t =

x y=

1 3x t= − 2 1y t= −52

5 131 32 2

− = −

52 1 42

− =

x y≠

32

3 71 32 2

− = −

32 1 22

− =

x y≠

23

21 3 13

− = −

2 12 13 3

− =

x y≠

25

2 11 35 5

− = −

2 12 15 5

− = −

x y=

0 ( )1 3 0 1− =


Pag

e98



• Reverse Engineering is a great technique when the algebra is unclear or too difficult. Sometimes however, it takes longer than one might like. If you can conventionally solve a problem quickly, you should. If not, consider techniques such as approximation, plug-in, and reverse engineering.


Start with the parentheses: 1 2 3 4 1 11 1 1 12 3 6 6 6 6

− − → − − → − − → +

Change 1 to a fraction & add: 1 6 1 71 6 6 6 6

+ → + →



Solution

According to exponent rules, m

m nn

x xx

−= . Let’s solve our problem:

Before solving, glance at the answer choices to determine whether the answer should be in fractions or decimals. This will determine how we simplify our expression.

( )( )

( )5

23

0.30.3 0.09

0.3= →




Pag

e99

30 Seconds Hack

This is a weighted average question. In these questions, the larger group will influence the average more than the smaller group will. Here are our two groups:

I- 200 seeds at 57% germination II- 300 seeds at 42% germination

If the groups had equal size, the overall average would be right in the middle, at 49.5%. However, because Plot II has more seeds larger, the overall average will be closer to 42 than to 57. Cross off D and E. You may be able to narrow down the answers further, by realizing that for the overall average to be really close to 42 (like answer choice A), the size of Plot II would need to be several times larger than Plot I. Given that Plot II is only 1.5 times Plot I, it is more reasonable to suspect an overall average of 46-48%

Guess B or C

Solution

To find what percentage of all seeds germinated, we need

Multiply, then add

Divide by 5



Solution

all germinated seedsall seeds planted

57 42200 300all germinated seeds 100 100

all seeds planted 200 300

⋅ + ⋅ =

+

( ) ( )57 2 42 3500

⋅ + ⋅=

114 126 240500 500+

= =

48 48%100

= =


Pag

e100


Take-Aways

• Drawing figures reveals connections between data and often makes the problem easier to understand and handle. Always make a drawing of your own for geometry questions, even if one is already provided.

#49, pg. 159 Difficulty Level: 200-400 Topics: Inequalities & Absolute Values

Since side CE = DE, the angles they face (angle D and angle C) must be equal.

We know that angle C = x, so angle D is also equal to x

Put in the value of y (45). To find x, look at triangle ECD. Because its angles must add up to 180, we can write:

45 180 67.5x x x+ + = → =

A line that crosses 2 parallel lines creates corresponding angles.

AB and EC are parallel, so angle B corresponds to angle C. They are both x. Next, angle A corresponds to angle E. They are both y.


Pag

e101

30 Seconds Hack

Note that the middle of the inequality will always be a multiple of 5, starting with 5 itself (when ).

The possible values for the middle are all multiples of 5 between 1 and 25, exclusive. These are {5, 10, 15, and 20}. Thus there are four values of n possible.

Guess B.

Solution

Manipulate the inequality to isolate n. Remember perform each operation on all 3 sides.

Original Equation

Subtract 5 4 5 20n− < <

Divide by 5 4 45

n− < <

Since n must be an integer, the only possible values within that range are {0, 1, 2, 3}



Solution A

Because absolute values cannot be negative, to minimize them we must make their values as close to 0 as possible. Thus to find the smallest possible value of 23 5y− , we need to find the value of y that will 23 – 5y as

close to zero as possible. Because we’re working with small numbers, we can use trial and error

• If y = 4, 23 – 5y will be 3 • If y = 5, 23 – 5y will be -2. This value is closest to zero.

23 5y− is smallest (closest to 0) when y = 5. Be careful not to pick 5!! The question is not asking for the value

of y, but the smallest value of 23 5y− .

When y = 5, 23 5 2 2y− = − =

0n =

1 5 5 25n< + <


Pag

e102



Solution A


Solution B

Apply the following property to this question:


Take-Aways

• Know the following quadratic factorizations by heart: ( )22 22x xy y x y+ + = + ;

( )22 22x xy y x y− + = − and ( )( )2 2x y x y x y− = + − .


Solution A

Think of ratios as relationships between different parts of the same whole. That there are 3 times as many people under-21, than there are people over-21 can be expressed as:

The ratio has four total parts, and three of them are under-21. So the ratio

( ) ( )2 27 7 2 7 4 7 28+ = = ⋅ =

( )2 2 22x y x xy y+ = + +

( ) ( ) ( )2 2 27 7 7 2 7 7 7 14 7 28+ = + ⋅ + = + + =

under-21 : over-21 : total 3 : 1 : 4

under-21 : total 3 : 4=


Pag

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One way to do this question is to plug in values for the numbers of people in each group. Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem. If you work correctly, whatever you plug in will reduce to the same answer.

Only plug in values that agree with the information you are given. “There are 3 times as many people aged 21 or under as there are people over 21”

Plug in your chosen values:

These plug-ins mean that the total population is 8. The ration we want,





Solution

The sum of all angles that form a straight line is 180. It is also true that vertical angles (angles that face each other) must be equal. So we can build two equations:

(Eq. 1)

(Eq. 2)

Plug-in the value of x from (Eq. 1) into (Eq. 2)


under-21 6 3: 4total 8

= =

2 3 180 36x x x+ = → =

2 30 2 30x y y x= + → = −

2 36 30 42y = ⋅ − =

Under-21 6= Over-21 2=


Pag

e104


30 Seconds Hack

29 81= and 211 121= , so 80 is about 9, while 125 is slightly bigger than 11. Therefore,

80 125 20+ ≈ . Have a look at the answer choices:

(A) 9 5 is a little bigger than 9 2⋅ . This is very close to 20.

(B) 20 5 is a little bigger than 20 2⋅ . This is close to 40. It’s way too large.

The answer choices only get bigger. We can safely eliminate B through E

Guess A

Solution

To simplify a square root, break it into a product of two square roots, one of which can be solved for:

• 80 16 5 4 5= ⋅ →

• 125 25 5 5 5= ⋅ →

Find the sum: 80 125 4 5 5 5 9 5+ = + →


Take-Aways

• To simplify a square root, break it into a product of two square roots, one of which has an integer solution. Example: 63 9 7 3 7= = .


Solution

Chris packed 60% of boxes, so Kelly packed 40%. Kelly to Chris is 40 to 60. This reduces to 2 to 3



Pag

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Solution

Approximate


Take-Aways • When the question itself asks you to approximate, do so because the math will likely be unpleasant.


Solution A

. The question can be expressed as the following equation:

Let’s solve for x, the missing part of the 2nd average.

Multiply by 3

Add

Solve



15050.2 0.49 2199.8 200

⋅×≈

25 1200 8

≈ ≈

totalaverage# of items

=

10 30 50 15 20 40 x+ + = + + +

90 75 x= +

15x =

Average of 10, 30, and 50

...is 5 more than average of 20, 40, and x

10 30 50 20 4053 3

x+ + + += +


Pag

e106


Let’s reverse engineer the question and figure out which answer choice makes “the average of 10, 30, and 50 is 5 more than the average of 20, 40, and x” true. Start with answer C

What is x? The average of 10, 30, and 50

The average of 20, 40, and x (should be 5 less)

Notes

(A) 15 30 Yes! When , the average of 10, 30, and 50 is 5 more.

(B) 25 30

(C) 35 30 20 40 35 31.7

3+ + = The 2nd average is way too big. We

need a smaller x. Eliminate C, D & E

(D) 45 30

(E) 55 30






Solution

In the equation 3y kx= + , there are 3 variables. Knowing that when y = 17, x = 2 gives us the value of 2 out of

3. We can solve for k. ( )2 3 2 14 73 17 k ky kx k + → = → == + → =

We’re told that k is a constant therefore its value never changes. k is always 7.

We need to find y when x = 4 and k = 7. Plug the known values in the equation and find y:

( )( )3 7 4 3 31y kx y y= + → = + → =


20 40 15 253

+ += 15x =

20 40 253

+ +=

20 40 453

+ +=

20 40 553

+ +=


Pag

e107



Harry gets paid x dollars per hour for the first 30 hours, so after 30 hrs he has earned 30x. For each extra hour, Harry gets paid 1.5x. If we call his “extra” hours h, Harry receives (1.5x)(h) for his overtime. Therefore the total Harry earns is ( )( )30 1.5x x h+ where h is the number of overtime hours.

James gets paid x dollars per hour for the first 40 hours, so after 40 hrs he has earned 40x. For his overtime, James gets paid 2x for each hour. Since he only worked 1hr of overtime (his total time is 41hrs), his overtime pay is exactly 2x. Therefore James’ total pay is exactly 40 2 42x x x+ =

Because Harry and James received the same pay, we can set up their pay as equal:

Harry’s pay equals James’ pay ( )( )30 1.5 42x x h x+ =

To find how long Harry worked, we need to find h, his overtime, and add it to the first 30 hrs.

Solve for h ( )( ) 12 121.5 42 30 1.5 1.5

xx h x x h hx

= − → = → =

Multiply by 2 to get rid of decimals 12 24 8hrs1.5 3

h = → →

Harry worked 8 overtime hours, so his total number of hours is 30 + 8 = 38hrs



When a question gives you non-variable answers along with a word problem, you can often use reverse engineering. Reverse Engineering is working from the answer choices, to find which choice can perfectly accommodate all the given data in the question. We already know that James worked 41hrs, and that he was paid x for each of the first 40 hrs (a total of 40x) and 2x for his single overtime hour. Thus we know that Harry was paid 42x dollars last week.

Let’s reverse engineer the question and figure out which answer choice makes “Harry and James were paid the same amount” true. Start with answer C

Harry’s hours Harry’s pay (x for each of 1st 30 hours + Is Harry equal to James (42x)?


Pag

e108

1.5x for each overtime hour)

(A) 35

(B) 36

(C) 37 ( )30 7 1.5 40.5x x x+ = Harry’s pay is too small. He needs to work more hours. Eliminate A, B & C

(D) 38 ( )30 8 1.5 42x x x+ = YES. If Harry works 38hrs, his pay is equal to Jame’s

(E) 39


Take-Aways



Solution

We start with 10 ounces in the glass. 1% of 10 is 0.1 ounce. This is also the amount that evaporates after 10 days (0.01 ounce each day 0.1 ounce in 10 days). Thus after 10 days 1% of the original amount is gone.

It’s safe to conclude that after 20 days, 2% of the original amount will have evaporated because the same quantity evaporates each day.



30 Seconds Hack


Pag

e109

The amount of solution poured is a bit less than ½ of the original amount (45 poured out of 100 total). Consequently, the amount of glucose poured must be a bit less than ½ of the original 15 grams. Look for an answer a bit less than 7.5

Guess E, or possibly D.

Solution

Set up a proportion:

Isolate x

Simplify fraction

Before the next step, glance at the answer choices to find out whether to keep the answer as a fraction or as a decimal.

Solve



Solution

In a parallelogram, adjacent angles add up to (Eq. 1)

PQ and RS are parallel (opposite sides of a parallelogram are parallel), so , and .

First, let’s find x

Plug 70 for y in (Eq. 1)

Isolate x

Now, we know and . So,


15100 45

x=

15 45100

x ⋅=

3 94

x ⋅=

6.75x =

180° 2 180x y+ =

2 140y = 70y =

2 70 180x + ⋅ =

40x =

70y = 40x = 30y x− =


Pag

e110


30 Seconds Hack

We’re told that 0.6 miles equal 1 kilometer. 2 miles is a bit more than three times as long as 0.6 miles, so it will equal a bit more than 3 kilometers.

Guess A.

Solution

This is a direct proportion. A glance at the answer choices tells us that we should change decimals to fractions.

Set up the proportion

Isolate x and change 0.6 to



30 Seconds Hack

As a result, the overall change will likely be an increase of less than 5% of the original amount. Thus, the final amount will be less than $10,500

1 kilometer kilometers0.6 miles 2 miles

x=

35

1 2 2 5 103 3 35

x ⋅ ⋅= = =

$10,000

Successively, the changes are +10%, +5%, -10%. The last decrease is a bigger change than the first increase, because the last change is 10% of a larger amount. Graphically, the changes might look like the graph on the right:


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Guess A or B

Solution

To increase a value by y%, multiply it by . To decrease a value by y%, multiply it by .

Let’s apply all changes to our initial amount: 10,000 +10%, +5%, -10%

Set up the changes

Simplify 10,000 with the denominators

Simplify first, then multiply


Take-Aways • Generally, you should simplify fractions before you perform multiplications. This will keep your numbers

from growing needlessly large. For instance, suppose you’re faced with . You could multiply

across top and bottom, and then divide . Solving this way might take you 90 seconds. On the other

hand, you can simplify by crossing off common factors in 21 & 7, as well as in 27 & 9. In this case you’ll

be left with to divide. Solving this way might take you 15-20 seconds, and you would’ve saved over

60 seconds to use on a more difficult problem.


Solution A

To avoid dealing with decimals, it would make sense to work with cents rather than dollars. Each apple was sold for 70 cents and each banana for 50. If we call a and b the numbers of apples and bananas respectively, the cost of all apples is 70a and the cost of all bananas is 50b. Since the total cost of all fruits is 630 cents ($6.30), we can write: 70 50 630 7 5 63a b a b+ = → + = (Eq.1)

100100

x+ 100100

y−

110 105 9010,000100 100 100⋅ ⋅ ⋅

110 105 901 1 100

= ⋅ ⋅

11 105 9 10,3951 1 1

= ⋅ ⋅ =

21 27 19 7⋅

⋅

56763

3 31⋅


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e112

We’re asked to find the total number of fruits, a + b. Because (Eq.1) has two variables, it may seem impossible to solve. However, a and b are restricted to positive integers (there is no 0.3 apples), so by isolating one variable, we may learn a bit about its possible values. Let’s find a, the number of apples:

63 5 57 5 63 7 63 5 97 7 7

b ba b a b a a+ = → = − → = − → = −

Since a must an integer, 57b

must also be an integer. For this to happen, b must be a multiple of 7.

• If b = 7, then 5 79 47

a a⋅= − → = .

• On the other hand, if b = 14, then 5 149 1

7a ⋅= − = − . This cannot be since a, the number of apples,

must be a counting number (positive integer)

b must equal 7. Any higher value of b such as 14 or 21 will make a less than zero. Thus there are 7 bananas and 4 apples. The total number of fruits sold was 7 + 4 = 11.

Notes: We could have solved for b instead of a and we would have reached the same values. This alternative path is shown in the Official Guide’s solution (page 206)



Solution A

To find the ratio of 1st graders to 3rd graders, we need both of these quantities in the same ratio. Unfortunately the question gives us 2nd-to-4th, 1st-to-2nd, and 3rd-to-4th only. To synchronize several overlapping ratios into a unique ratio, make sure that the overlapping values are represented by the same number in all their ratios. Below is the information we start with.

Ratios 1st 2nd 3rd 4th R1 2nd to 4th is 8 to 5 --- 8 --- 5 R2 1st to 2nd is 3 to 4 3 4 --- --- R3 3rd to 4th is 3 to 2 --- --- 3 2

Notice that 2nd graders are part of the first 2 ratios (the ratios overlap). We could merge these ratios into one by multiplying the second ratio by 2 (multiplying or dividing all terms of a ratio by the same factor doesn’t change the value of the ratio). As a result of this multiplication the 2nd graders will be represented by the same number in both ratios. The first 2 ratios would then be synchronized:

Ratios 1st 2nd 3rd 4th R1 2nd to 4th is 8 to 5 --- 8 --- 5


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R2 1st to 2nd is 3 to 4 6 8 --- --- R3 3rd to 4th is 3 to 2 --- --- 3 2

Because the ratios R1 and R2 are synchronized, we can write them as one ratio. We now know that the ratio of 1st to 2nd to 4th graders is 6:8:5

Ratios 1st 2nd 3rd 4th R1 2nd to 4th is 8 to 5 6 8 --- 5 R2 1st to 2nd is 3 to 4 6 8 --- 5 R3 3rd to 4th is 3 to 2 --- --- 3 2

To be able to answer the question, we need to incorporate 3rd graders into the overall ratio. This can be done easily because 4th graders overlap R2 and R3. By manipulating both ratios so that the overlapping quantity is represented by the same number, we will synchronize R2 and R3. The easiest way to accomplish this is to multiply R2 by 2 and multiply R3 by 5.

Ratios 1st 2nd 3rd 4th R1 2nd to 4th is 8 to 5 6 8 --- 5 R2 1st to 2nd is 3 to 4 12 16 --- 10 R3 3rd to 4th is 3 to 2 --- --- 15 10

Now that all ratios are synchronized, we can write them as one:

1st 2nd 3rd 4th 12 16 15 10

The ratio we’re asked to find, 1st graders to 3rd graders, is 12 to 15 which simplifies to 4 to 5



Solution

.

Find the number of pairs that will add up to 9

There are 4 such pairs: {2+7, 3+6, 4+5, and 5+4}

Find the number of pairs possible

# of favorable outcomes # of pairs that add to 9Probabilityall possible outcomes all possible pairs

= =


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e114

The formula here is simple. Let be the number of options/outcomes possible for event A, and the number of options/outcomes possible for event B. Then, if (A and B)n is the number of options/outcomes possible when A, then B happen: In this question, the total number of ways that we can make a selection from A, then a selection from B is

We can now find probability



Solution


Definition: 220

600LdN

s=

+ where L is # of lanes, d is distance in feet, and s is speed in miles per hour.

Application: There are 2 lanes (L = 2), the distance is 1/2 mile, and the speed is 40 mph (s = 40). In the definition, the distance was given in feet, so we must convert our distance from 1/2 miles into feet before we apply the formula.

We’re told that 5,280 feet = 1 mile, so half the distance (1/2 mile) must be 5,280 2,640

2= feet. So we have d =

2640. Now that we have all 3 values, we can apply the formula to find N, the number of cars:

( )( )( )22

20 2 2,64020 600 600 40

LdNs

= →+ +

Factor and simplify as much as you can before you even think about

multiplying: ( )( )

( )( )( )( )( )

( )2

20 2 2,640 20 2 2640 40 2640 2640 600 40 40 40 15 40 55600 40

N

= → → →+ ++

. Factor

out a 5 and simplify: 2640 528 4855 11

N = → →


nA nB


4 5 20⋅ =

# of pairs that add to 9 4Probability 0.2all possible pairs 20

= = =


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e115


30 Seconds Hack

Guess C

Solution

To find the median of a list, we must first order the list then select the middle value (or the average of the 2 middle values if there is no single middle value because the number of values is even). In this case, because there are 11 values, the median is the 6th value. We’re not given exact values, so we’re forced to approximate them:

Year 1990 1991 1992 1993 1994 1995 Shipments 190 180 210 270 310 350 Year 1996 1997 1998 1999 2000 Shipments 380 370 390 360 270

In increasing order, the bottom values are about 180, 190, 210, 270, 270, and 310. STOP. We’re only interested in the middle, or 6th, value: 310,000


The median is the middle value in an ordered list. A quick look tells us that there are 11 values. 5 values are below 300 and 6values are above. The middle value (the 6th value) must be the smallest value above 300. This is 1994, or just above 300


Pag

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Solution

Original equation

Divide by y

Multiply by 2

Solve for x



Solution

Solve for x in each inequality:

and

Before you attempt to combine inequalities, make the symbols face the same way. In this case, change the first inequality to . Now, merge it with to get


#72, pg. 162 Difficulty Level: 500-600 Topics: Translations & Manipulations; Number Properties

Solution A – Least Common Multiple

3 52

xy y− =

3 5 12

x −=

3 5 2x − =

2 5 73 3

x += =

5 23

xx+ >> −

3 710

xx− <<

3 x− < 10x < 3 10x− < <


Pag

e117

If the students can be divided into 8 or into 12 teams, the number of students must be a multiple of 8 and 12. To find the smallest number than can be evenly divided by 8 and by 12, find the Least Common Multiple (LCM) of 8 and 12. To find the least common multiple:

1. Write out all your values in their prime factorizations

2. Write down the product of all primes that appear in the different factorizations (without duplicates)

3. Assign to each prime, the highest exponent that appeared in any single factorization. This new product is the least common multiple




Reverse Engineering is working from the answer choices, to find which choice can perfectly accommodate all the given data in the question.

Let’s reverse engineer the question and figure out the smallest answer choice that can be evenly divided by 8 and by 12. We should start with the smallest answer.

What is lowest possible number of students?

Divisible by 8?

Divisible by 12?

Notes

(A) 20 No -- No need to check 12

(B) 24 Yes Yes Yes! 24 is the smallest answer that is divisible by both 8 and 12

(C) 36 Don’t bother, we’ve found a match (D) 48 Same as above (E) 96 Same as above




#73, pg. 162 Difficulty Level: 200-400 Topics: Exponents & Roots; Number Properties

3 28 2 and 12 2 3= = ⋅

2 3⋅

3LCM 2 3 24= ⋅ =


Pag

e118

Solution A

Unlike “regular number” proper fractions (values between-1 and 1) get smaller as they are raised to higher exponents. In this case, the base is the decimal 0.345. The highest value will be the one with the smallest exponent.

To make comparisons between roots and exponents, or to simplify expressions that include both, rewrite roots as

exponents. Remember that roots are just fractional exponents: 12x x= ,

13 3x x= , etc… In our case, 0.345

is an exponent of 1/2. Aligned from smallest to greatest, the values are: ( )20.345 , 0.345 and 0.345 . In

terms of r, s, and t, s r t< < .


Solution B

It’s clear that we’re not expected to find the square or the root of 0.345. This question actually tests the property of positive fractions. One approach to solving the question is to use a more friendly value within the same “number group”. A number group is a group of numbers that behave in a similar fashion in roots and exponents. The groups are all values smaller than -1, values between -1 and 0, values between 0 and 1, and values larger than 1.

In this case, we need a number whose root and square can be found easily. Let’s replace 0.345 with 14

.

• 10.345 4

r = →

• ( )( )

22

1 10.345 164

s = = →

• 1 10.345 4 2

t = = → .




Because 1/4 and 0.345 are both between 0 and 1, they have similar properties in roots and exponents. Using 1/4 helps us get through the math quickly. s is the smallest value and t is the greatest.


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“The number of cars is 1/4 of the number of trucks” means that out of every 5 vehicles there are 1 car and 4

trucks. Therefore the number of trucks is four-fifths of all vehicles: 4trucks5

n=

“Two thirds of the trucks are pickups” 2 2 4 8pickups trucks 3 3 5 15

n n

= → =



When variables appear in the answer choices, you can often plug in easy values for those variables within the question, solve and then find out which answer choice agrees with your solution. In this case, we can plug in a value for n, the total number of vehicles.

Because the ratio of cars to trucks is 1:4, the value of n must be a multiple of 5. Since 2/3 of trucks are pickups, we must also be able to break n into thirds. Let n be 15.

The ratio of cars to trucks is 1:4, so there are 4 trucks for every 5 vehicles. Out of 15 vehicles, there must be 12 trucks.

2/3 of trucks are pickups, so there must be 2 12 83⋅ = pickups. Our solution is 8. The final step is to find out

which answer is equal to our solution, replacing n with our plug-in value (15)

(A) ( )1 1 156 6

n = . This is not an integer.

(B) ( )5 5 1512 12


(C) ( )1 1 152 2


(D) ( )8 8 15 815 15

n = → . Yes! This matches our solution

(E) ( )11 11 1512 12

n = . This is not an integer


Take-Aways

• When variables appear in the answer choices, you can often plug in easy values for those variables within the question, solve and then find out which answer choice agrees with your solution. Avoid plugging-in 0, 1, or the same value for different variables (unless the variables are equal). By coincidence, two answers may agree with your solution. In that case, just plug in another set of easy numbers and try again.


Pag

e120


Solution

. This means at least 27 people need to vote in favor for a resolution to pass. This also means that

the maximum who can vote against a passing resolution is



Solution – Intuitive Solution

A glance at the ratio (household: food: misc. 5:2:1) tells us that food makes up 2 out of 8 total parts, or a fourth of expenses. Find a fourth of the total budget.




There are 10 members. If there are four more women than men, women must be 2 more than the average, and men must be 2 fewer than the average. So there must be 7 women.


Solution B – Build Equations

We can build two equations from the info given.

“4 more women than men” (Eq 1)

23

2 40 263⋅ =

40 27 13 people− =

1800 4504

=

4w m= +


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e121

“10 members total”

Replace m with in the first equation and solve for w:

Rewrite (Eq. 1)

Solve for w




Let’s reverse engineer the question and figure out which answer choice makes “there are 10 members in total” true.

How many women?

There are 4 fewer men. How many men?

Total members (should be 10)

Notes

(A) 3 -- -- (B) 4 0 4 (C) 5 1 6 (D) 6 2 8

(E) 7 3 10 Yes! 7 women give us 10 total members.





30 Seconds Hack

Since interest is compounded at 4% every 6 months, it will grow faster than a simple 8% interest. Because 8% simple interest on $10,000 would be $800, we should pick an answer that is a bit more than $800

10 10w m m w+ = → = −

10 w−

(10 ) 4w w= − +

2 14 7w w= → =


Pag

e122

Guess C or possibly B.

Solution

To find the interest that accumulates after 1-year, find out what the total amount is after applying two 4% increases to $10,000 (because interest is compounded semiannually).

To increase an amount by p%, multiply it by . To decrease an amount by n%, multiply it by .

Increase $10,000 by 4% twice


The final amount is 10,816 so the interest accumulated is



Solution A

This question tests your ability to manipulate the position of decimal points and adjust powers of 10. It’s a bit counter-intuitive:

• If you move the decimal point one spot to the right, you must decrease the power of 10 by 1. For

example, 1 43.4567 34.567 10 34567 10− −= ⋅ = ⋅ • If you move the decimal point one spot to the left, you must increase the power of 10 by 1. For example,

2 31234.5 12.345 10 1.2345 10= ⋅ = ⋅

To simplify expressions that contain many decimal spaces, it’s typically useful to write them all as powers of 10.

Remember that according to rules of exponents, a b a bx x x +⋅ =

( )( )( )( )( )

( )( )( )( )( )( )( )( )( )( )

( )( )( )( )( )

4 1 5

62 1 3

36 10 28 100.0036 2.8 36 28 10

0.04 0.1 0.003 4 1 3 104 10 1 10 3 10

− − −

−− − −

⋅→ →

⋅

Complete the simplification, and remember the rule a

a bb

x xx

−= :

100100

p+ 100100

n−

104 10410,000100 100 ⋅

104 104 10,816= ⋅ =

10,816 10,000 816− =


Pag

e123

( )( )( )( )( )

( ) ( )55 6 1

636 28 10

3 28 10 84 10 8404 1 3 10

−− − −

−

⋅→ ⋅ ⋅ → ⋅ →

⋅



Solution

The rate/work formula is . We are asked to solve for time, so we should rewrite the formula

as . We know how much work needs to get done (200 bolts), but we first need to figure out the

combined rates since the machines will be doing the work together.

To find the combined rate, simply add individual rates

Armed with the combined rates, we have all the ingredients needed for the rate/work formula:


#81, pg. 163 Difficulty Level: 600-700 Topics: FDPs & Ratios; Translations & Manipulations

30 Seconds Hack

Time Amount 1pm 10.0 4pm x 7pm 14.4

work rate time= ⋅worktimerate

=

120 bolts 100 bolts 3 5 8 bolts per second40 seconds 20 secondsA Brate rate+ = + = + =

200 258

worktimerate

= = =

Since the amount increases by the same factor (not the same quantity), the amount is multiplied by some value every 3 hours. This is the definition of exponential growth (for example, a value that increases by 10% each term is multiplied by 1.1)


Pag

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A property of exponential growth is that the growth gets faster over time (it curves up). This means that in this case, since there has been two increases, the 2nd increase must be greater than the 1st. In short, x must be closer to 10.0 than to 14.4. Since the midpoint of 10.0 and 14.4 is 12.2, we can safely conclude that x is smaller than 12.2

Guess A or B

Solution

It’s important to be able to distinguish a constant quantity increase (in which the same amount is added each term) and a constant factor increase (in which the amount is multiplied by the same factor each term). A constant factor increase means that each term equals the product of previous term with the factor. If we call this factor f, the table data can be rewritten as:

Time Amount Amount 1pm 10.0 10.0 4pm x 10 f 7pm 14.4 210 f

( )10 10 1.2 12x f x= → → =


Take-Aways

• Memorize your multiplication tables up to 12 times 12 before taking the GMAT. You should also know your perfect squares up to 220 and how to recognize at a glance whether a number is divisible by 2, 3, 4, 5, 6, 8, and 9.


Solution A – Algebraic

Multiples of 3 occur every 3rd term. This means that every integer is either a multiple of 3 (ex: 3, 6, 9), one more than a multiple of 3 (ex: 4, 7, 10), or two more than a multiple of 3 (ex: 5, 8, 11). If an integer is three more than a multiple of 3, the integer will equal another multiple of 3. Understanding this is the basis of solving this question algebraically.

There are 3 possibilities:

• if n is a multiple of 3, then 3n x= where x is an unknown integer • if n is one more than a multiple of 3, then 3 1n x= + • if n is two more than a multiple of 3, then 3 2n x= +

To find x, we need the value of f. The 3rd row gives us a useable equation: 2 214.4 10 1.44f f= → = . 2144 12= , so 21.44 1.2= . The value of f is 1.2. Now we can find the amount at 4pm


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We can go through the answer choices and find out which answer is always divisible by 3, regardless of which of the 3 possibilities above is actually true.

If 3n x= If 3 1n x= + If 3 2n x= + (A) ( )( )1 4n n n+ − ( )( )( )3 3 1 3 4x x x+ − ( )( )( )3 1 3 2 3 3x x x+ + − ( )( )( )3 2 3 3 3 2x x x+ + −

Answer A is always divisible by 3 because each possibility has a factor that is divisible by 3:

( ) ( )and3 , 3 3 , 3 3x x x− + . No need to check other answers.



One way to find out which answer must be a divisible by 3 is to plug in a couple of values of n and eliminate any answer that doesn’t have to be divisible by 3

If n = 7 If n = 8 If n = 9 Notes

(A) ( )( )1 4n n n+ − ( )( )7 8 3 ( )( )8 9 4 --- Must be the right answer b/c the others fail to be div by 3

(B) ( )( )2 1n n n+ − ( )( )7 9 8 ( )( )8 10 7 --- Not div by 3 if n is 8

(C) ( )( )3 5n n n+ − ( )( )7 10 2 --- --- Not div by 3 if n is 7

(D) ( )( )4 2n n n+ − ( )( )7 11 5 --- --- Not div by 3 if n is 7

(E) ( )( )5 6n n n+ − ( )( )7 12 1 ( )( )8 13 2 --- Not div by 3 if n is 8

Note: there is no need to actually do the multiplications. Just check whether 3 is one of the factors of the products


Take-Aways

• To find the answer that must be something, you can eliminate all the other answers by showing that they don’t have to be that thing.


Solution

Profit = Revenue – Cost. If we can find the gross profit for all 20,000 tools, we will be able to find the profit per tool by dividing the gross profit by 20,000.


Pag

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Find Total Revenue: There are 20,000 tools, and each tool is sold for $8, so the total revenue is 20,000 8 $160,000⋅ =

Find Total Cost: There is a fixed cost of $10,000, plus an extra cost of $3 per tool. For 20,000 tools, the extra cost will be 20,000 3 $60,000⋅ = . Once the fixed cost is added, the total cost is $70,000

Find Total Profit: 160 70 $90,000Revenue Cost k k Profit− = − → =

Find Profit per tool: The total profit for 20,000 tools is $90,000, so the profit per tool is 90,000 9 $4.5020,000 2

= =



Solution A

If q dollars bought the entire lot of 100 batteries, then each battery cost dollars. To find the resale price of

each battery, we need to increase the purchase price by 50%.



Increase by 50%

Simplify to find the resale price





100q

100100

x+

100100

y−

100q 100 50 150

100 100 100 100q q+

⋅ = ⋅

3 3100 2 200

q q= ⋅ =


Pag

e127

Our plug-ins

Most of the time, “easy values” would be 2, 3, 5 or 10. In this case, 400 is a good number to use because I know that q is the price of 100 batteries, so I’ll need to divide it by 100 to find the price of each battery.

Let’s solve the question with our plug in.

100 batteries cost a total of $400, so each battery cost $4.

Each battery was sold at 50% above purchase price, so each battery was sold for

We’ve solved the question: Our solution $6

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variable in each answer with our plug-in until we find the answer that equals 6.

(A) This matches our solution!

(B)

(C)

(D)

(E)





Solution A – Intuitive solution

The 10 consecutive integers have been split into two groups. Imagine writing each group in ascending order, and stacking the two groups as in the figure below

400q =

$4 $2 $6+ =

3 3 400 6200 200

q ⋅= =

3 3 400 62 2q ⋅= ≠

150 150 400 6q = ⋅ ≠

40050 50 6100 100

q+ = + ≠

150 150 6400q

= ≠


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Note that the difference between each stacked pair is 5. Since there are 5 pairs, the 2nd group sum will be 25 more than the 1st group sum. We are told that the 1st group sum is 560, so the 2nd group must be


Solution B

In a list of consecutive integers, the middle integer is equal to the average of the list. The average of the first five

integers is 560 112

5= , so 112 must be the middle of the first five integers, so we now know our list of ten

integers:

110, 111, 112, 113, 114 and 115, 116, 117, 118, 119.

Find the sum of the last 5:



Solution

Since Machine A works twice as fast, A will produce 100 parts in 20 minutes. Thus

. The rate/work formula is . We are looking for

work (“How many parts…”) done in 6 minutes, so we simply need to solve



Solution

560 25 585+ =

115 116 117 118 119 585+ + + + =

100 parts 5 parts per minute20 minutesArate = = work rate time= ⋅

5 6 30work = ⋅ =

int1 int2 int3 int4 int5


Difference between each pair is 5. Difference between the two groups is 25


Pag

e129

This is a pattern problem. To solve patterns, it is always best to draw out what is happening as this will increases your chances of spotting the pattern. Below, I have reproduced each cycle on its own line.

Note that because each cycle has 5 beads, at the end of a cycle, the total number of beads must be a multiple of 5. The necklace ends with a white bead, which means that the total length (N) must be a multiple of 5 (end of a cycle) plus 3 additional beads. where n is an integer. We need to find the answer choice that can be expressed in this fashion:

(A)

(B)

(C)

(D)

(E) This is the only answer in the form where n is an integer


Take-Aways • To solve pattern problems, it is always best to draw out what is happening as this will increases your

chances of spotting the pattern.

5 3N n= +

( )16 5 3 1= ⋅ +

( )32 5 6 2= ⋅ +

( )41 5 8 1= ⋅ +

( )54 5 10 4= ⋅ +

( )68 5 13 3= ⋅ + 5 3n +

Red Green White Blue Yellow Total

5

10

15

20

5 3n +

Start


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When working with equations of a line, it is advisable to write them in the standard format where m is the slope and b is the y-intercept (y value when x is 0).

Our equation is . Rewrite as . The slope is .

One definition of slope is . Look at the two points given. From (a,b) to (a+3, b+k), we can

determine that the slope is . Set this expression of slope to equal what we know the slope to be, and find have

k:



As above, the first step is to rewrite the equation of the line in its standard format. In this case, we change

to .

Although there are no variables in the answers, we can plug in an easy value for a (or b), solve and then find out which answer choice agrees with our solution.

Let’s pick an easy value:

Our plug-ins

6 is a good value because it is divisible by the denominator that appears in the problem (3). It’s very important to realize that we should not arbitrarily pick a second value for b, because b depends on a. Instead, we solve for b, using our plug in for a.

Now that we’ve picked a, let’s figure out what (a, b) and (a+3, b+k) really are by using the equation of the line

. Since a is 6, (a, b) is now (6, b) while (a+3, b+k) is now (9, b+k)

y mx b= +

3 7x y= − 1 73 3

y x= +13

change in ychange in x

3k++

1 13 3

k k= → =

3 7x y= − 1 73 3

y x= +

6a =

1 73 3

y x= +


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e131

Point (6, b): In the equation of the line, if , then b (the y-value) will be . So the first point is

.

Point (9, b+k): In the equation of the line, if , then b+k (the y-value) will be . So the 2nd

point is .

Since k is the only difference in the y value of the two points, and



Solution

“s is the product from 100 to 200” ( )( ) ( )( )100 101 ... 199 200s =

“t is the product from 100 to 201” ( )( ) ( )( )( )100 101 ... 199 200 201t =

Note that s and t are the same, except that t has 201 as an extra factor. In other words, 201t s= ⋅ (Eq.1)

We’re asked to express 1 1s t+ in terms of t. Because the answers are in terms of t, we need to find a way to get

rid of s.

Isolate s in (Eq.1): 201 201

tt s s= ⋅ → =

Replace s with its equivalent:

11 1 1 1 1 201 1 2021

201 201t ts t t t t t t

+ = + = + → + →


6x = 13

1 76 43 3

+ =

( )136, 4

9x = 13

1 79 53 3

+ =

( )139,5

1 13 34 5k+ = 1k =


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Take-Aways

• In general, if you want to get rid of a variable in a system of equations, isolate that variable in one equation and replace it with its equivalent in the other.



Let j and s be jack’s and his sister’s weights. Since they weight 278 together we can write 278j s+ = (Eq.1). Once Jack loses 8lbs, his weight will be j – 8. We’re told that this would be “twice as much as his sister”. Since the sister’s weight doesn’t change we can write: 8 2j s− = (Eq.2)

To find j, we could isolate s in one equation and replace s with its equivalent in the 2nd equation.

Isolate s in (Eq.1) 278 278j s s j+ = → = −

Replace s in (Eq.2)

( )8 2 8 2 278 556 2 8 3 564 188j s j j j j j j− = → − = − → = − + → = → =




Reverse Engineering is working from the answer choices, to find which choice can perfectly accommodate all the given data in the question. Let’s reverse engineer the question and figure out which answer choice makes “If Jack loses 8 he will weight 2x his sister” true. When reverse engineering, start with answer C

Jack now Sister now (278 total) Jack loses 8 Notes (A) 131 -- -- (B) 135 -- --

(C) 139 278 139 139− = 131 After losing 8lbs, Jack < Sister. Jack needs to be much heavier. Eliminate A, B & C

(D) 147

(E) 188 278 188 90− = 180 Yes! After losing 8lbs, Jack will weight 180, twice his sister’s weight of 90.



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Take-Aways



Solution

Let m and b be the price of maps and books respectively. We’re given two relationships:

• 12 maps and 10 books cost $38 12 10 38 6 5 19m b m b+ = → + = (Eq.1) • 20 maps and 15 books cost $60 20 15 60m b+ = (Eq.2)

To find the difference between maps and books, we could solve for the prices of each item first. One of the quickest ways to find a variable in a system of 2 equations is to manipulate the equations so that the variable you don’t want has the same coefficient (number in front of it) in both equations.

• triple (Eq.1) 18 15 57m b+ = • (Eq.2) 20 15 60m b+ =

We can get rid of b by doing (Eq.2) – (Eq.1). The result is 2 3 1.5m m= → =

To find b, replace m with 1.5 in either equation. I chose (Eq.1):

( )6 5 19 6 1.5 5 19 9 5 19 2m b b b b+ = → + = → + = → =

The difference between the two prices is 0.50



Solution A

Percentage change is always difference 100original

⋅ . In a percentage increase, the original value is the smaller value.

In our case, the sales go from 320 to 385. The percentage increase is


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385 320 65 65100 100 10320 320 32− ⋅ → ⋅ → ⋅ . This value will not simplify to an integer. This is why

we’re asked to approximate. To approximate, think about how you could wiggle the values a bit to further simplify. I like to leave 10 alone because it will make the multiplication at the end very easy: 65 6410 10 2 10 20%32 32

⋅ ≈ ⋅ → ⋅ =



Because the answer choices give us specific percentage increases, we could work backwards and find out which increase gets us from 320 to 385. Notice that because the question asks us to approximate, no answer will be exactly right. We will need to pick the answer that comes closest to being the percentage increase from 320 to 385. When reverse engineering, start with C

% inc. Start Increase Final value (should be 385)

Notes

(A) 2% 320 -- (B) 17% 320 --

(C) 20% 320 ( )0.20 320 64= 384

Yes! There is no reason to look elsewhere. 20% increase gets us to 384. The exact % must be tiny bit larger than 20%. D & E are too large.

(D) 65% 320 (E) 83% 320


Take-Aways



Solution A

In an ordered list of even values, the median is the average of the middle two values. The median of List I is 7. List II has an odd number of numbers, so the median will be the middle value. Thus the middle value must be 7. x must be 7.


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Let’s reverse engineer the question and figure out which answer choice would correctly yield the same median for both lists.

What is x? What is median of List I (average of the 2 middle values)

With this x, what is median of List II (should be equal to median of List I)

Notes

(A) 6 7 6 Medians are not equal

(B) 7 7 7 Yes! gives us equal medians

(C) 8 7 -- Don’t bother, we’ve found a match

(D) 9 7 -- Same as above (E) 10 7 -- Same as above





Solution

We aren’t told how many voters there are, but we can use any friendly number. Let’s suppose we had 100 voters

Our plug in

Let’s breakdown the problem:

“60% are democrats and 40% republicans” There are 60 democrats and 40 republicans

“75% of democrats vote for A” democrats vote for A

7x =

Total 100 voters=

75 60 45100

⋅ =


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“20% of republicans vote for A” republicans vote for A

53 out of 100 people vote for A. This is 53%





Solution

Respect the order of operations; work inside out:

Original expression

Simplify the inner parentheses

Change

Use common denominators to solve



30 Seconds Hack

The ratio of hydrogen to oxygen given is 2:16, so the mixture (water) is almost entirely oxygen. In 144 grams of water, we should have a lot more oxygen than hydrogen. Cross off A and B. Answer E is too extreme because it basically leaves no room for hydrogen.

20 40 8100

⋅ =

1 2 3 942 3 8 16

+ × ÷ −

1 1 942 4 16

= + ÷ −

144

÷ →×1 1 1 92 4 4 16

= + × −

8 1 9 016 16 16

= + − =


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Guess C or D.

Solution

First, simplify the ratio, and rewrite it including the total mixture:

We are told hydrogen to oxygen is 2:16

Rewrite as hydrogen to oxygen to water is 1:8:9

Looking at the entire ratio including the total mixture (water) allows us to see that oxygen to water is 8:9. To find out how much oxygen in 144 grams of water, set up a proportion:

Isolate x

Simplify first, and then multiply



Solution

Both equations are already factored and set to 0. x could equal whatever value makes each factor zero.

Equation1: . What value of x would make each factor equal to 0?

1st factor x. This equals 0 when

2nd factor when

From the first equation, we know that

Equation2: . What value of x would make each factor equal to 0?

89 144

oxygen xwater

= =

8 1449

x ⋅=

8 16 1281

x ⋅= =

( )2 1 0x x + =

0x =

( )2 1 0x + =12

x = −

10 or 2

x = −

( )1 2 3 02

x x + − =


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1st factor when

2nd factor when

From the second equation, we know that

The only value of x that agrees with both equations is




“a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n”. In simpler term, that means “Each time the reading goes up by 1, the intensity is multiplied by 10.”

If the reading increases from 3 to 8, it has gone up by 1 five times, and each time the intensity is multiplied by 10. The table below tracks each increase in reading and the effect on intensity.

Reading Intensity 3 original 34 45 56 67 78



Solution A – Algebraic

The arithmetic mean is the same thing as the average.

1 02

x + =

12

x = −

( )2 3 0x − =32

x =

1 3 or 2 2

x = −

12

−

10original ⋅10 10original ⋅ ⋅10 10 10original ⋅ ⋅ ⋅10 10 10 10original ⋅ ⋅ ⋅ ⋅

510 10 10 10 10 10original original⋅ ⋅ ⋅ ⋅ ⋅ = ⋅


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Mean is sum

# of values. In this case:

( ) ( ) ( ) ( ) ( )1 2 4 8 5 15 35 5

n n n n n n n+ + + + + + + + +→ = +

The median is the middle value in the ordered list. Since we have 5 values, the median is the 3rd: 2n+

The difference between mean and median is ( ) ( )3 2 1n n+ − + =




To find the answer we could simply pick an easy value for n. Let n = 1.

With this value, the question becomes “what is the difference between the mean and median of the 5 numbers: 1,

2, 3, 5, and 9”? The mean (average) is 1 2 3 5 9 4

5+ + + + = and the median (middle value) is 3. the difference

between the two is 1. Our solution is 1

The last step of the plug-in is to pick the answer that matches your solution.


Take-Aways



Solution

Plug what you’re given into the equation and solve for the other variable (K)

Plug 290 in for T

( )290T =

( )5290 329

K= −


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Multiply by

Simplify fraction, and Isolate K



30 Seconds Hack

The 2nd outlet can fill the pool in 5hrs. If the 1st outlet could work just as fast, then together they would take half the time, or 2.5hrs. Because the 1st outlet is slower, they will take more than 2.5hrs together. Cross off A, B, and C.

On the other hand, if both outlets worked at Outlet 1’s speed (9hrs to fill the pool), together they would take 4.5hrs. Since Outlet 2 is faster, they will take fewer than 4.5hrs. Cross off E.

Guess. D

Solution


as . We know how much work needs to get done (fill 1 pool), but we first need to figure out the

combined rates since the outlets will be doing the work together.



No need to go into decimal division. Only one answer begins with 3.


95

290 9 325

K⋅= −

(58 9) 32 554K K⋅ + = → =


=

1 pool 1 pool 14 pool per hour9 hours 5 hours 45A Brate rate+ = + =

1 45 314 1445

worktimerate

= = = ≈


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30 Seconds Hack

The side must be shorter than 20 (length of diagonal), so the perimeter must be less than 80. Cross off C, D and E.

Approximate. Look at the picture. If the diagonal is 20, pick a likely range for the length of the side. I picked 15 to 17. Multiply your estimate by 4 to estimate the perimeter. The perimeter should be about 60 to 68. Pick the closest value to the middle of your range

Guess B.

Solution

The diagonal of a square is also the hypotenuse (longest side) of a 45-45-90 triangle. The ratio of side to diagonal

is . To find the side, set up a proportion:

Now that we have one side, find the perimeter by multiplying the side by 4:

Perimeter equals side times 4

A glance at the answer choices tells us that we should approximate to the nearest whole number.

Multiply top & bottom by & approximate


Take-Aways Most geometric figures are drawn to scale. When stuck, you can approximate lengths, angles, etc…

#103, pg. 166 Difficulty Level: 700+ Topics: Translations & Manipulations; FDPs & Ratios

: 2x x

20 side 20 20side2 2 2

xxx x

= → = =

20 8042 2

p = ⋅ =

2 80 2 40 2 40 1.4 562

p = = ≈ ⋅ ≈


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Solution A

There are two ratios mentioned. The initial ratio is 30:1 and the new ratio is 25:1 after we add 50 students and 5 teachers. One way to do this question is to build two equations one for each ratio:

“Students to teachers is 30:1”

“+50 students, +5 teachers and ratio becomes 25:1” (Eq. 1)

We want t, the number of teachers. Combine the equations:

Replace s with 30t in (Eq. 1)

Cross multiply

Solve for t


Solution B

There is a much faster way. You can combine the two given ratios into one equation directly. Initial ratio of

students to teachers can be expressed as where 30t is the number of students and t is the number of teachers.

If we add 50 students and 5 teachers to these numbers, the new ratio will be 25 to 1. This change can be

expressed as . Now you have an equation; solve it for the number of teachers (t)

Cross multiply

Distribute 25 then isolate t

Solve for t



30 301

s t st= → =

50 255 1

st+

=+

( )30 50 255 1

tt

+=

+

30 50 25 125t t+ = +

5 75 15t t= → =

30tt

30 50 255 1

tt+

=+

( )30 50 25 5t t+ = +

30 25 125 50t t− = −

75 155

t = =


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Let’s reverse engineer the question and figure out which answer choice gives us the correct ratio of 25 to 1 after we add 50 students and 5 teachers.

How many teachers now?

How many students now? (30 times teachers)

What would be ratio if we added 50 students & 5 teachers? (should be 25 to 1)

Notes

(A) 5 150 200:10 20:1 Final ratio is not 25:1 (B) 8 240 290:13 Same as above (C) 10 300 350:15 70:3 Same as above (D) 12 360 410:17 Same as above

(E) 15 450 500:2025:1 Yes! Starting with 15 teachers gives us the correct final ratio


Take-Aways When a question gives you non-variable answers choices along with a word problem, you can often use reverse

engineering. Start with answer C, especially if it will take more than a few seconds to check each answer.

#104, pg. 166 Difficulty Level: 500-600 Topics: Exponents & Roots; Inequalities & Absolute Values

Solution

To manipulate huge numbers, change them into their prime factorizations

Original inequality

Factorize 25

Equate the exponents

So n must be greater than 6. The smallest integer n could be is 7.


#105, pg. 166 Difficulty Level: 500-600 Topics: Combinatorics; FDPs & Ratios

1225 5n >

( )2 12 2 125 5 5 5n n> → >

2 12 6n n> → >


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The probability that the person selected is a woman and also a lawyer is equal to the percentage of women lawyers.

60% of the people are women, and 45% of them (a little under half) are also lawyers. So the percentage of all people who are women and lawyers is a little under half of 60%. The little under 30% of the people in the room are women lawyer.


Solution B – Hidden Plug In

The question asks about a ratio (probability is a ratio of success/all possibilities) without giving us any information about specific concrete number of people. We can plug in any value for the total and our math will be reduced to the same probability.

Let there be 100 people in the room. 60%, or 60 people are women. 45% of these 60 women are lawyers. So the number of women lawyers is ( )0.45 60 27= .

The probability in general equals successful outcomesall possible outcomes

. The probability that a woman lawyer is picked at

random is 27 27%

100# of women lawyers

all people= =


Take-Aways

• Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem.

#106, pg.167 Difficulty Level: 700+ Topics: Number Properties

Solution A - Algebraic

Since the remainder when x is divided by y is 9, x must be 9 more than a multiple of y. For example, if y is 10, then x could be 19, 29, 39… Whatever y is, x is 9 more than a multiple of y. A multiple of y is the product of y and another integer. Algebraically, a multiple of y can be expressed as yq


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Since x is 9 more than a multiple of y, we can write: 9x yq= +

Because we’re interested in the value of y, we can replace x with yq+9 in the equation given: 9 996.12 96.12 96.12x yq q

y y y+= → = → + = (Eq.1)

Because q is the integer part of the result when x is divided by y, q must equal 96. If we can get the right side of

(Eq.1) in a format similar to the left, we will easily spot the value of y: 12 3 996.12 96 96 96100 25 75

= + = + = +

Rewrite (Eq.1) 9 9 996.12 96 75

75q q y

y y+ = → + = + → =




Reverse Engineering is working from the answer choices, to find which choice can perfectly accommodate all the given data in the question. Let’s reverse engineer the question and figure out which answer choice makes “When x is divided by y the remainder is 9” true. Typically, when reverse engineering we should start with C. In this case, it won’t really matter because if C is incorrect, I won’t know whether to eliminate all smaller or all greater answers.

y x Remainder when x is divided by y (should be 9)

Notes

(A) 96 96.12 96x x= → = non-

integer

x is not an integer. GMAT numbers tend to work evenly. Move on.

(B) 75 96.12 720975x x= → = 9 Yes. If y = 75, x will be 7,209

and the remainder of x/y is 9 (C) 48 --- --- (D) 25 --- --- (E) 12 --- ---

The math in this case is ugly. Reverse Engineering wouldn’t be my tactic of choice for this question



Pag

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Take-Aways



Solution

We’re given 2 expressions for x. (1) x is the product of integers 1 – 8 and (2) 2 3 5 7 pi k mx = ⋅ ⋅ ⋅ . To find out more about the exponents, we need to find out more about the product of integers 1 – 8:

( )2 3 7 21 2 3 4 5 6 7 8 2 3 2 5 2 3 7 2 2 3 5 7x x x= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ → = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ → = ⋅ ⋅ ⋅ . Now that we have both

representations of x in similar formats, we can equate them: 7 22 3 5 7 2 3 5 7pi k mx = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

7 2 1 1 11i k m p+ + + = + + + =


Take-Aways

• When solving a Number Properties question, express all values in their prime factorizations (eg. 212 2 3x x= → = ⋅ ). In this format, it is easier to deal with very large numbers, see patterns, and make

connections.

#108, pg. 167 Difficulty Level: 700+ Topics: Exponents & Roots

Solution

9 31

2 5t =

⋅. We need to find out how many zeroes there are between the decimal point and the first nonzero

digit. Because the placement of the decimal point has to do with powers of 10 (the decimal point of 10x is one spot to the right of the decimal point of x) it would be smart to factor out as many powers of 10 as possible to see what type of number we’re dealing with.

( ) ( )9 3 6 3 6 66 3 31 1 1 1 1 1 1 0.001

10002 5 2 10 2 22 2 5t t t t= = → = ⋅ → = ⋅ → = ⋅

⋅ ⋅ ⋅


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62 64= , so the value of t is actually 0.001

64. Without a calculator, this is a real pain to compute. one thing we

could do however, is “temporarily” multiply the ratio by 100,000 in order to move the decimal place to the right far enough that we can actually perform the calculation, then at the end dividing the result by 100,000

0.001 100100,000 100,000 100,000 100,000 1._64 64

t t t

= → = → =

This allows to find that 100,000t is “one point something” (doesn’t matter what). Dividing this value be 100,000 (this will bring the decimal point 5 spaces to the left) will give us t:

100,000 1._ 0.00001_t t= → =

Thus through some gymnastic we find out that there are four zeroes before the first non-zero digit.



Solution

First, notice that we’re asked to approximate, which means that the numbers won’t work evenly. The ratio of

royalties to sales for the first part is 3 0.1520

= (you can find this quickly if you think about what 3/2 equals).

The ratio for the second part is 9 1 0.08

108 12= ≈ (either do the math or like me, memorize that 1/6 = 16.6%, so

1/12 must be 8.3%)

The decrease from 0.15 to 0.08 is a drop of a little less than half (exactly half would have dropped all the way to 0.075), so the percentage decrease should be a bit smaller than 50%


Take-Aways

• I recommend memorizing the decimal values of 1 1 1, ,...,2 3 10

to save time on the exam.

#110, pg. 167 Difficulty Level: 700+ Topics: Exponents & Roots; Number Properties


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Solution

For 3k to be a factor of p, 3kp

must simplify to an integer. The only way this can happen is if p has 3 among its

factors at least k times. If p has fewer 3’s among its factors, there will not be enough 3’s in the top of the fraction for it to simplify into an integer. Consequently, the greatest possible value of k will be the exact number of 3’s in p. Although p is the product of all integers from 1 to 30, factors of 3 can only occur in integers that are divisible by 3, so for the sake of this exercise, we can ignore the other integers that make up p. We can write:

...3...6...9...12...15...18...21...24...27...303 3k kp = .

How many 3’s occur in p? to find out we need to break out the multiples of 3:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 33... 2 3 ... 3 ... 3 4 ... 3 5 ... 2 3 ... 3 7 ... 3 8 ... 3 ... 3 10

3 3k kp ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= . The largest possible value

of k that will allow this monster to be simplified is equal to the number of 3’s on top: that’s 143 . If k is larger than 14, there will be a 3 at the bottom that cannot be simplified because the top will have run out of 3’s.


Take-Aways

• When solving a Number Properties question, express all values in their prime factorizations (ex: 212 2 3x x= → = ⋅ ). In this format, it is easier to deal with very large numbers, see patterns, and make

connections.


Solution

Instead of $0.80, the pair of candy bars costs $0.75. The percent reduction is the same as the percent from 80 to 75. It is a reduction of 5 out of 80. So a simple way to rephrase the question is “5 is what percent of 80”?

5 is what percent of 80?

Isolate x

Simplify and solve


5 80100

x= ⋅

5or 80 100

x =

5 100 5 1080 8

x ⋅ ⋅= =

14

25 64

x = =


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Solution A

We are looking for r. Let’s isolate it

Original equation

Square both sides and find r





Our plug-in

Because r depends on s, we can only plug in for one of the two, and then solve the other. In this case, with our

plug in, the equation becomes . What must be under the square root for it to simplify to 2? The fraction

under the square root must equal 4, since . Thus with this plug-in, .

Our Solution


(A)

(B)

(C)

r ss=

2 3 r s r ss= → =

2s =

22r=

4 2= 4 and 82r r= =

8r =

1 1 82s

= ≠

2 8s = ≠

2 2 8s s = ≠


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(D) This answer matches our solution!

(E)


Take-Aways When variables appear in the answer choices, you can often plug in easy values for those variables within the

question, solve and then find out which answer choice agrees with your solution. Avoid plugging-in 0, 1, or the same value for different variables (unless the variables are equal). By coincidence, two answers may agree with your solution. In that case, just plug in another set of easy numbers and try again.


30 Seconds Hack

Glance at the picture and approximate. The shade takes up all sides and cuts a large swath through the middle of the figure. I estimated that the shade is about ½ of the whole figure.

Guess C or D

Solution

The easiest way to find the shaded portion is to find the unshaded area and subtract it from the total. In this case the total area is


3 32 8s = =

2 22 2 8s s− = − ≠

6 8 48⋅ =

4

2.5

Total height is 8. The three horizontal bands leave a combined 5 feet of height for the white rectangles. Each rectangle is 2.5 feet high.

Total width is 6. The two vertical bands leave 4 feet for the width of each white rectangle.

The unshaded area equals ( )2 2.5 4 5 4 20⋅ = ⋅ = and the shaded

area is the remainder ( )48 20 28− =

The shaded fraction is 28 76 8 12

shadedtotal

= =⋅


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handle. Always make a drawing of your own for geometry questions, even if one is already provided.



We start with a negative proper fraction. When you square it, the result will be positive (and thus bigger). Raise that number to the third power and it becomes negative again (and thus smaller than the square). So must be the biggest value. Cross off A, C, and D.

When a proper fraction is raised to an exponent, its absolute value becomes smaller. Since we are dealing with

negative numbers, a smaller absolute value actually means a bigger number.


Solution B

Just as quickly, you can just find out what the values are because the numbers are small.

So



Solution A

“Mary is 60% more than Tim” is the same as “Mary is 160% of Tim” (Eq. 1)

“Tim is 40% less than Juan” is the same as “Tim is 60% of Juan” (Eq. 2)

2a

3a a<

0.3a = − ( )22 0.3 0.09a = − = 3 0.09 ( 0.3) 0.027a = ⋅ − = −

3 2a a a< <

160 8100 5

m t t= =

60 3100 5

t j j= =


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To compare Mary to Juan, we need an equation that gets Tim out of the way.

Combine equations 1 & 2

Multiply and put over 100



One way to do this question is to plug in a value for one of the incomes. Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem. If you work correctly, whatever you plug in will reduce to the same answer.

We need to decide which variable to plug in for. Anyone would work, but the right decision would make the question easier to solve. The question is “What percentage of Juan is Mary?” Because at the end I will need to find a percentage of Juan and because the number that is easiest to find a percentage of is 100, I chose to plug in 100 for Juan

Plug in your chosen value to start off:


“Tim is 40% less than Juan” means “Tim is 40% less than 100”

“Mary is 60% more than Tim” means “Mary is 60% more than 60”

To quickly find 60% of any number, find 10% and multiply it by 6. 10% of 60 is 6, so 60% of 60 is . I just added this 36 to t in order to get

We can now easily answer the question: “Mary is what percent of Juan” means “96 is what percent of 100?” 96%!


Take-Aways Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you

can plug in a value that will be easy to deal with and use it to solve the problem.

8 8 3 5 5 5

m t m j = → =

24 96 96% of 25 100

m j j m j= = → =

60t =

96m =

6 6 36⋅ =96m =

Our plug-in 100j =


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#116, pg. 168 Difficulty Level: 700+ Topics: Combinatorics

30 Seconds Hack

Because “each distance is represented by only one entry”, we have only one dot between each city-pair. For example, there is no dot at row B, column A because we already have a dot at row A, column B. In addition, there are no dots down the diagonal, so the total number of dots will be a bit fewer than half the number of cells in

the table. A 30 cities table would have cells, and a bit fewer than dots

Guess B.

Solution A - Combinations

This question can be solved in fewer than 30 seconds with a good understanding of combinations. We are basically asked “how many ways can we pick combinations of 2 cities from a lot of 30 cities?” Because we do not care whether we are going from city AB or BA (we are told “each distance is to be represented by only one entry”), these are not permutation (pairs are not ordered).

How many combinations of 2 are possible out of 30 cities?


Solution B – Intuitive Solution

A 30 cities table has 900 cells. Ignoring the 30 diagonal cells (we only indicate distance between two distinct cities) leaves us with 870 cells. Since we are told that “each distance is represented by only one entry”, we do not care whether we are going from city AB or BA we only need entries for half of the remaining cells:


#117, pg. 168 Difficulty Level: 700+ Topics: Exponents & Roots

30 30 900⋅ =900 450

2=

30! 30 29 15 29 4352!28! 2 1

⋅= = ⋅ =

⋅

870 435 entries2

=


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30 Seconds Hack

One option is to approximate. As n gets very large, n and 1n+ get closer and closer. As n gets huge, these

two roots are almost equal to each other, so the denominator of 11n n+ −

approaches zero. As the bottom of

a fraction gets very small, the fraction gets very large. In other words, as n gets bigger and bigger, the fraction gets very big as well. Because the right answer is a value equivalent to the fraction for all positive values of n, we can be confident that as n gets large, the biggest answer has the best chance of being right.

The easiest choices to eliminate are A and C because they are both about 1 as n gets very large. We can also

safely eliminate D because as n gets large, n and 1n+ get closer and answer D approaches zero. The answer must be B or E

As n gets really large, E is about 2n n n+ = whereas B approaches 2 2n n= . Thus as n gets large E is the biggest of all answer choices. E has a better chance than B at being correct

Guess E

Solution

This question is really testing whether you know how to “rationalize” a denominator. Rationalization is the process via which we can get rid of roots in denominators. This is done because by mathematical conventions, a

fraction should be written without a root in its denominator. For example, a solution 32

should be rationalized

to 3 2 3 2

22 2⋅ → . What most people don’t know how to do is rationalize a denominator that contains a

sum or difference of incompatible terms such as 11n n+ −

. To rationalize such a denominator, you must

multiply it by its “complement”, the same terms linked by the opposite arithmetic operation. In this case, because

the terms in the denominator are linked by subtraction, you will use addition: 1 11 1

n nn n n n

+ +⋅+ − + +

.

The reason this will make your roots disappear has to do with the common quadratic: ( )( ) 2 2x y x y x y+ − = + .

Each term will be squared, and the square root will disappear! Let’s do it:

( ) ( )2 21 1 1 1 1

11 1 1

n n n n n n n nn nn n n n n n

+ + + + + +⋅ → → → + ++ −+ − + + + −



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Take-Aways


( )22 22x xy y x y− + = − and ( )( )2 2x y x y x y− = + −


30 Seconds Hack

Ratio of length to width is 3.3 to 2, so the length is a bit more than 1.5 times the width ( ). Since the width equals 8, the length must be a bit more than .

Guess C.

Solution

You can set up a proportion and solve for l

Cross multiply & divide 3.3 8 3.3 4 13.2

2l ⋅= = ⋅ =



Solution

To learn more about x itself, isolate it in each inequality

• 6 10 4x x+ > → > • 3 5 8x x− ≤ → ≤

Merging the two inequalities gives us 4 8x< ≤ .


1.5 2 3× =1.5 8 12× =

3.32 8

length lwidth

= =


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Solution A

Let d, j and p be the number of books of David, Jeff and Paula. Set up your equations:

“David has d books”

“David has 3 times as many as Jeff”

“David has ½ as many as Paula”

Together, David + Jeff + Paula have





Our plug-in

6 is a great number because it is small but divisible by all the potential denominators that we could encounter (3 and 2).

If David has 6 books:

“David has 3 times as many as Jeff” Jeff has 2 books.

“David has ½ as many as Paula” Paula has 12 books

Together, they must have

Our solution 20


3 3dd j j= → =

1 22

d p p d= → =

1023 3d dd d+ + =

6d =

6 2 12 20 books+ + =


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(A)

(B)

(C) This answer matches our solution!

(D)

(E)





Solution A

This is a combination question. It can be rephrased as “how many combinations of 2 can we select from 8 available teams”? Each combination of 2 represents a game played between the two teams. It is NOT a permutation, because the order in which the two teams are picked does not matter.

Find combination of 2 out of 8 items


Solution B

To find out how many ways there are of accomplishing a succession of tasks is straightforward. Let be the number of options/outcomes possible for event A, and the number of options/outcomes possible for event B. Then, if (A and B)n is the number of possibilities for A, then B to happen:

5 5 6 206 6

d = ⋅ ≠

7 7 6 203 3

d = ⋅ ≠

10 10 6 203 3

d = ⋅ =

7 7 6 202 2

d = ⋅ ≠

9 9 6 202 2

d = ⋅ ≠

8! 8 7 282!6! 2

⋅= =

nA

nB


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In other words, to find out how many ways you can do two things, multiply the number of ways you can do each. You can view this problem as a succession of two tasks.

1. Select one team out of eight 8 options 2. Select another team out of seven remaining 7 options

To do both tasks (to select two teams that will then play each other), there are ways to pick. Now,

because this methodology accounts for picking team A and then B as well as for picking team B and then A, we

need to get rid of half those ways (we’re told that teams play each other only once). The final number of games is

games.




and . The only way for a fraction to equal 0 is if the top equals 0. (a – c) is 0 only if c

equals a.


#123, pg. 169 Difficulty Level: 600-700 Topics: Translations & Manipulations ; FDPs & Ratios;

Solution

A quick glance at the answers tells us that not only is the math likely to get messy, but we shouldn’t venture to approximate because the answers are very close to one another.

When I suspect difficult computation, I try to set up the entire equation before I do the math rather than compute a little bit at a time.

$207 is the total cost of 15 meals + 15% in tip. In other words, $207 is 115% of the total cost of 15 meals. Let’s call the cost of each meal c. The price of lunch for all people (excluding tax) is 15c


8 7 56⋅ =

56 282=

a ca ca c

θ −=

+0a cθ =


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“$207 is 115% of 15c”

Cross multiply & divide to isolate c

Divide 207 & 15 by 3. Divide 100 & 115 by 5

Divide 69 & 23 by 23. Divide 20 & 5 by 5

Note that we simplify as much as we can before multiplying. This keeps numbers manageable. The cost of each meal, c, is $12

Note: Technically, you could solve this question through Reverse Engineering, but I would not recommend it as it would take too long.


#124, pg. 169 Difficulty Level: 500-600 Topics: FDPs & Ratios; Sets & Groups

Solution

One efficient way to deal with groups that don’t overlap (no single person can belong to both men & women, or to both employed & unemployed groups) is to draw a table and fill it in with what we know. Since the question is asking us about a percentage without giving us any specific number of people, we can plug in anything we want for the total. Since we’re dealing with percentages,

• Let there be 100 people. • 64 people are employed • 48 people are employed males

Put this in a table and figure out as much as you can. When building a table in this context, it’s a good idea to identify what you’re looking for from the start. “what percent of employed people are females?”

Men Women Total Employed 48 64 48 16− = 64 Unemployed Total 100

We’ve now found that 16 people, or 16% of the total, are employed females. Don’t rush to pick 16% however. This question has one last trick up its sleeve. It doesn’t ask what percent of people are females (that would be

115207 15100

c= ⋅

207 100115 15

c⋅=

⋅

69 2023 5

c⋅=

⋅

3 4 121 1

c ⋅= =

⋅


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16%), but it asks “what percent of employed people are females?” To find this we must calculate employed females 100

all employed⋅ . The solution is

16 1100 100 25%64 4

⋅ = ⋅ →


Take-Aways


#125, pg. 169 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values; Number Properties


Since we are dealing with positive integers, the fraction only if p<q. Let’s go through the answer choices

and try to spot which MUST be greater than 1. If the fraction is positive, then the top MUST be greater than the bottom for the fraction to be greater than 1

(A) The root of a positive fraction less than 1 is another positive fraction less than 1.

(B) Since p<q, the top of this fraction is smaller than the bottom. The fraction is smaller than 1.

(C) 2p q q< < , so the top of this fraction is smaller than the bottom.

(D) If 3p = and 4q = , this fraction would be smaller than 1

(E) We already know that p<q, so in this fraction, the top is bigger than the bottom. This fraction MUST BE greater than 1.


Take-Aways When evaluating one answer choice at a time, skip choices that would take long to work with, because in many

cases, you will know the right answer almost as soon as you see it


30 Seconds Hack

1pq<


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If each machine took 3hrs, then together, they would take 1.5hrs. On the other hand, if each machine took 4hrs, then together, they would take 2hrs. Since we have a 3hr and a 4hr machine, working together will take between 1.5 and 2hrs. Only one answer choice falls in this interval.

Guess C.

Solution


as . We know how much work needs to get done (1 production order), but we first need to figure

out the combined rates since the machines will be doing the work together.






By combining shipments, we will pay the higher 1st pound rate only once. Thus combining shipments is the cheaper option. Cross off D and E.

When we combine shipments, instead of costing the higher rate the first pound of the 2nd package will be charged at the lower rate. What we save is the difference between the higher and lower rate. We save


Solution B


=

1 order 1 order 7 order per hour4 hours 3 hours 12A Brate rate+ = + =

57

1 12 17 712

worktimerate

= = = =

x y−


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Find the shipping cost for each scenario:

Separate Shipping costs:

Package #1: 1st pound will cost x, the other 2 pounds will cost 2y. Total cost will be

Package #2: 1st pound will cost x, the other 4 pounds will cost 4y. Total cost will be

“Separate Shipping” costs will be

Combined Shipping costs:

1st pound will cost x, the other 7 pounds will cost 7y. Total cost will be

Separate shipping costs an extra x and combined shipping an extra y. Since , separate is more expensive.

To find out how much would be saved in combined shipping, find the difference.


Solution C – Plug In



Our plug-ins

Whatever you plug in must respect any given restriction. In this case, . With these values plugged in, solve the question:

Separate shipping costs would be 18cents as below:

3lb package would cost

5lb package would cost

Combined shipping costs would be .

Our solution Combined, with a saving of 1cent

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variables in each answer with our plug-ins. Answer Choice A matches our solution.


2x y+

4x y+

2 6x y+

7x y+

x y>

( ) ( )2 6 7x y x y x y+ − + = −

3x = 2y =

x y>

3 2 2 7 cents+ ⋅ =

3 2 4 11 cents+ ⋅ =

3 2 7 17 cents+ ⋅ =


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Solution - Approximation

Doing the math to this question would be nightmarish and include raising a decimal to the 18th power. You’ll never be expected to do such a feat on the GMAT. Instead, we are given a shortcut and told that the amount

doubles every years. In this case, r is 8%. So the amount will double every . In 18years the

investment will double, and double again. So 5,000 will become 10,000 after 9 years, and 20,000 after 18years.




Solution

The actual distance traveled must be between 285 and 295 miles

The actual amount of gas used must be between 11.5 and 12.5 gallons

To find the smallest fraction, make the top small and the bottom big the smallest is

To find the greatest fraction, make the top big and the bottom small the greatest is

In the answers, find the range closest to:

70r

70 9 years8≈

28512.5

29511.5

285 295 and 12.5 11.5


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Notes: I’ve yet to see a GMAT question in which deciding whether to round 0.5 up or down is crucial in getting the question right. Even if your exact answer isn’t there, only one answer will be close



Solution A – Check the limits

A generally simple way to check whether an inequality corresponds to a number line graph is to see whether the edges of the graph correspond to the limits of the inequality. For instance, in this case, the edges of the graph are

. If we plug these values in the correct inequality, the resulting expression should match the limit of the inequality. In answer choices A & C, this limit is 3. In answer choices D & E, this limit is 4. In answer choice B, it is 5.

Plug in each answer choice:

Put in Put in Notes: (A) -- 5 does not equal the limit (3)

(B) The first plug in matches the limit (5), but the second plug in does not.

(C) -- 7 does not equal the limit (3).

(D) -- 6 does not equal the limit (4)

(E) Yes, we have a match in both cases


Solution B

To quickly derive the absolute value equation that corresponds to a number line graph, follow the following steps:

1. Find the center of the shaded graph. will be the absolute value component of the equation, where c

is the center of the shaded graph. For example, if the center of the graph is at +10, then will be

the absolute value component. In our case, the center of the graph is at -1, so will be the absolute value component. Only Answer E has this component, so we could stop here and pick E

2. Figure out the distance from the center to the limits of the shaded graph. Let’s call this distance d. If the shaded points are between the center and the limits, the equation will be .

5 and 3x x= − =

5 and 3x x= − =

5x = − 3x =5 3 5 3− ≤ → ≤

5 5 5 5− ≤ → ≤ 3 5 3 5≤ → ≤

( 5) 2 3 7 3− − ≤ → ≤

( 5) 1 4 6 4− − ≤ → ≤

( 5) 1 4 4 4− + ≤ → ≤ (3) 1 4 4 4+ ≤ → ≤

x c−

( 10)x − +

( 1) 1x x− − = +

or x c d x c d− < − ≤


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Otherwise, if the shaded points are away from the center, the equation will be . In our case, the distance from the center to the limits is 4, and the shaded points are between the center and the limits. As a result, the correct expression is



Solution A

The number of workers will go from 500 to 550. Let’s examine the women population:

15% of 500 workers are women There are currently

We want 20% of 550 workers to be women We want

To get from 75 to 110 women, we would need to hire 35 women.




The current population of women is . Let’s reverse engineer the question

and figure out which answer choice makes the population 20% women after hiring 50 workers. When reverse engineering, start with answer C.

or x c d x c d− > − ≥

1 4x + ≤

15 500 75 women100

⋅ =

20 550 2 55 110 women100

⋅ = ⋅ =

1515% of 500 500 75 women100

= ⋅ =


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How many women hired?

The women population would go from 75 to…

Women will be what percentage of 550? (should be 20% or 1/5)

Notes

(A) 3 (B) 10

(C) 25 100 100550

is less than 20% % is too low. Nee more. Eliminate A, B and C

(D) 30 105 105 21 20%550 110

= ≠ Since this doesn’t give us 20%, E must be the right answer. STOP

(E) 35 110 Yes!






Total average is $400. The first 6 average $360 so together the have created a deficit of $240 (each of the 6 fell short of the mean by $40). The next 4 must together make up this deficit by having a combined surplus of $240.

So the average surplus will be . This surplus makes the average revenue for the days $460.


Solution B

Set up the weighted average equation in which x is the average revenue of the remaining 4 days:

Cross multiply & drop parentheses

Solve for x


240 $604

=

( ) ( )6 360 4400

10x+

=

400 10 2160 4x⋅ = +

1840 4604

x = =


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Solution

To handle this question, express all numbers in scientific notation before you start working. Per Capita

means “divided by the population”



Solution A

“Window is twice as long as it is wide” means

“Perimeter is 10ft” means

No need to spend time looking for l. Only one answer has the correct w




Let’s reverse engineer the question and figure out which answer choice makes the perimeter equal 10 feet. We can ignore A, since the length must be twice the width. As always, start with C

( )10nx ⋅

12 124

8

1.2 10 1.2 10 1 110 10,000 $5,000240,000,000 2.4 10 2 2

⋅ ⋅= = ⋅ = ⋅ =

⋅

2l w=

( ) 52 10 5 2 5 3

l w l w w w w+ = → + = → + = → =


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What are dimensions?

Perimeter should

be 10

Notes

(A) -- --

(B) Yes! These dimensions give us the correct perimeter.

(C) 2 by 4 12 Perimeter is too long. Ignore greater dimensions and check smaller ones

(D) 3 by 6 -- --

(E) -- --





Solution

To find out how many ways there are of accomplishing a succession of tasks is straightforward. Let be the number of options/outcomes possible for event A, and the number of options/outcomes possible for event B. Then, if (A and B)n is the number of ways for A, then B to happen: In other words, to find out how many total ways you can do several things, multiply the number of ways you can do each. In our particular question, each time we reach a fork in the road, we have new options. The number of paths from X to Y will be the product of the number of options at each fork.

( )( )2 l w+

3 7 by 2 25 10 by 3 3

152 103

=

10 20 by 3 3

nA

nB



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Solution

Work from the inside the parentheses first.

Simplify the root

Now, you can include 60

Simplify the root



Solution

Expand

( )5 45 60 5 45 60= ⋅

5 5 3 3 60 15 60= ⋅ ⋅ ⋅ =

15 60= ⋅

( )(3 5) 2 2 3 5 2 3 5 30= ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ =

( ) ( ) ( )4 2 4 210 10 0.0012 10 0.0012 10 0.0012− = −

X

The 3 bold points represent the locations at which there are multiple options. The number of paths from X to Y is the product of the options at each point. 2 2 3 12 options⋅ ⋅ =

Y


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Shift the decimal point according to the exponent of 10


#138, pg. 171 Difficulty Level: 700+ Topics: FDPs & Ratios


Each night worker loads of boxes of each day worker. At the same time, there are as many night workers as

day workers. Since to find the total number of boxes loaded, we multiply number of workers by the number of boxes per worker, these ratios will be multiplied and the number of boxes loaded by the night crew will be

of the number of boxes loaded by the day crew. So we have a ratio of night to day to total boxes equal

to 3:5:8. The fraction loaded by the day crew is 5 out of 8.



One way to do this question is to plug in values that respect the ratios given. Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem. If you work correctly, whatever you plug in will reduce to the same answer.

“Each worker on night crew loaded as many boxes as each worker on the day crew.”

Our plug ins Each night worker loads 3 boxes; Each day worker loads 4 boxes

“The night crew has as many workers as the day crew”

Our plug ins Night crew has 4 workers; day crew has 5 workers

Since to find the number of boxes loaded, we multiply number of workers by the number of boxes per worker, we can tell that the day crew loaded 20 boxes, and the night crew 12 boxes. So the fraction loaded by the day crew

equals


12.12 0.12 12= − =

34

45

3 4 34 5 5⋅ =

34

45

20 520 12 8

=+


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Solution C

We will use the following variables:

- Number of boxes loaded by each day worker

- Number of boxes loaded by each night worker

- Number of workers on the day crew

- Number of workers on the night crew

Using the variables above, we can express the number of boxes loaded by each crew as the number of workers on the crew times the number of boxes loaded by each worker.

The day crew loaded

The night crew loaded

Express each given relationship as an equation.

“Each worker on night crew loaded as many boxes as each worker on the day crew.” (Eq.

1)

“The night crew has as many workers as the day crew” (Eq. 2)

Now let’s answer the question:

“What fraction of all boxes did day crew load?”

Rewrite & include (Eq. 1) & (Eq. 2)

Factor out & Simplify by

dbox

nbox

dworkers

nworkers

( )( )d dbox workers

( )( )n nbox workers

34

34n dbox box=

45

45n dworkers workers=

( )( )( )( ) ( )( )

d d

d d n n

box workersbox workers box workers+

( )( )

( )( ) 3 44 5

d d

d d d d

box workers

box workers box workers +

( )( )d dbox workers( )( )( )

( )( )

1 1 583 4 8154 5

d d

d d

box workers

box workers= =

+ ⋅


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Take-Aways



Solution - Approximate

10% of $35.50 is $3.55, so a 10% tip will raise the minimum amount paid to $39.05.

To quickly calculate a 15% tip, take 10% ($3.55) and add half of 10% to it (add about $1.75). This extra $1.75 will raise the maximum amount paid to about $40.80.

The total amount paid is approximately between $39.05 and $40.80



Solution A

“Total weight is 750” (Eq. 1)

“Twice the first left was 300 more than the 2nd lift” (Eq. 2)

Since we want to find the first lift, we should replace the 2nd lift with its equivalent from (Eq. 1)

Rewrite (Eq. 2)

Add first to both sides



750 750first second second first+ = → = −

( )2 300first second= +

( ) ( ) ( )2 300 2 750 300first second first first= + → = − +

3 1050 350first first= → =


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Let’s reverse engineer the question and figure out which answer choice makes the two lifts add up to 750lbs. When reverse engineering, start with answer C

First lift? Twice the first lift

Twice 1st is 300 more than 2nd. 2nd lift is…

Total of two lifts (should be 750)

Notes

(A) 225 -- -- -- -- (B) 275 -- -- -- --

(C) 325 650 350 675 Total is too small. Try a higher answer

(D) 350 700 400 750 Yes! A first lift of 350 gives us a total of 750.

(E) 400 -- Don’t bother, we’ve found a match.






To find the maximum membership, we must assume that each member contributed the minimum allowed ($12). The club collected almost $599. If there were 50 people, the total collection would have to be at least $600. Since the actual amount collected is smaller, there must be 49 or fewer people.


Solution B

Since each member contributed at least $12, we can find the number of members by doing . If the

bottom is exactly 12, the value of this fraction is between 49 and 50. As the bottom gets bigger, the fraction gets smaller. So the maximum population is 49.


( )599

12 or more


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Solution

A perfect square (a number such as 4, 9, 16 or 25 with an integer square root) is a number that can be written as two groups of identical factors. For example, the perfect square 100 is ( ) ( )2 5 2 5⋅ ⋅ ⋅ . We’re told that when y

and 3,150 are multiplied, the result is a perfect square. So 3,150y is the product of two groups of identical factors. Because this question is about factors, it would be smart to break down 3,150 into its prime factors. We would then be able to see what y would have to be for 3,150y to be a perfect square.

23,150 315 10 9 35 10 3 5 7 2 5y y y y= ⋅ ⋅ = ⋅ ⋅ ⋅ → ⋅ ⋅ ⋅ ⋅ ⋅

Let’s organize 3,150y into two groups of factors to see what y has to be:

( )( )2 2 23 5 7 2 5 5 7 2 3 5 7 3 5 2 3y y y⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ → ⋅ ⋅ ⋅ ⋅ ⋅→ ⋅

For this to be a perfect square, y must be at least 2 7⋅ so that the two groups of factors are identical.


Notes: There are actually an infinite number of values that y can have, as long as the final string of factors can be broken into two identical groups. We’re only interested in the smallest possible y though. Secondly, This question could be solved through reverse engineering: work backwards starting with the smallest answer choice to see the smallest value of y that will make 3150y a perfect square. I wouldn’t recommend this technique for this problem however because it would be very time consuming.

Take-Aways


connections. • When a question gives you non-variable answers choices along with a word problem, you can often use



Solution

Function questions basically have 2 parts: (1) definition and (2) application. The definition tells you how to use the function by showing you an example (typically with variables). The application asks you to repeat the process


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with other variables or with numbers. To solve, just replace the variables in the definition with the values in the application.

Definition: x greatest integer less than or equal to x. The function rounds down to the next integer.

Application: ( ) ( ) ( )1.6 3.4 2.7 2 3 2 2 − + + = − + + →

The only tricky bit is 1.6 − . Because the function rounds down, we must replace 1.6 − with the next

smaller integer: -2.



Solution

To simplify algebraic expressions, it is often important to get rid of denominators and roots (if present).

Cross multiply

Distribute x and set to 0



30 Seconds Hack

The area of a trapezoid is equal to the average of the two parallel sides times the perpendicular side (typically

expressed as ). Approximate the length of the base by comparing it to the 5ft side. The base

is at least twice as long as the height. Let’s estimate 10-13. This estimate yields an area of the trapezoid between

( )4 2 42

x x x x xx

−= → + = −

+

2 22 4 3 4 0x x x x x+ = − → + − =

1 2

2base base h+

⋅


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and . The middle of our range is just above 40. This would be the best

guess based on the estimate above

Guess B or C.


Be familiar with the most used Pythagorean Triples. These are right triangles of which all sides are whole numbers. On the GMAT, you will most often see the 3:4:5 and the 5:12:13 Pythagorean triples, or their multiples (for example the 6:8:10 or 30:40:50 are multiples of the 3:4:5).

A quick glance at the figure provided tells us that since AB is 13 feet, a 5:12:13 triangle is formed, and the base is 12.

The area of a trapezoid is equal to the average of the two parallel sides times the perpendicular side (typically

expressed as ). In this case, the area is


Take-Aways • You should be able to recognize a Pythagorean triple when you see it. 3:4:5, and 5:12:13 along with their

multiples are often used on the GMAT.



Solution

5 2 (10) 352+

⋅ =5 2 (13) 45.5

2+

⋅ =

1 2

2base base h+

⋅ 2 5 12 422+

⋅ =

5

12

13

B

A

We are given 5, and 13. Knowing the Pythagorean triple 5:12:13, we can deduce the 3rd side without any effort.

Alternatively, we could have used the Pythagorean theorem to find the base:

2 2 25 13x + = 2


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In sequences, the subscript (small number below the main line) is an index that identifies each term. For example,

1x is the 1st term and 3x is the 3rd term. In this notation, notation nx is the nth term, 1nx − is 1 term before the

nth, and 2nx + is 2 terms after the nth.

We’re told that ( )1 2122n n nx x x− −= − . This means that to find the nth term, start with twice the previous term

(n-1) and subtract half of the term before the previous (n-2). Thus to find a term we need to know the two terms that come before it. We’re given the values of the 0th and 1st terms: 0 13, 2x x= = , and asked to find the 3rd

term.

According to the formula, the 3rd term is: ( ) ( )3 2 1 2 1 12 2 22 2

x x x x→= − − (Eq.1)

We need 2x to find the 3rd term: ( ) ( ) ( )2 1 0 1 12 2 2 3 2.52 2

x x x → →= − −

Now that we have 2x , plug it into (Eq.1): ( ) ( ) ( )3 2 3 1 12 2 2 2.5 2 42 2

x x x→ →= − − =




The knowledge that VR is 10 feet yields the triangle below:

Note: You can accomplish the same solution by using the Pythagorean Theorem to find s. The proper setup

would be . 2 2 25 10s + =

V

R S

s

5 10

A right triangle with a side of 5 and a hypotenuse of 10 must be a 30 60 90° − °− ° triangle with side

lengths of the ratio : 3 : 2x x x . Thus

the length of side s must be 5 3 .

10


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Take-Aways • Memorize the ratios of the lengths of the sides of 30-60-90 degree triangles as well as 45-45-

90 degree triangles also called isosceles right triangle

#148, pg. 173 Difficulty Level: 700+ Topics: Number Properties

Solution

If we factor out 10, we end of with 210 2 10x y x ykx y x y x y

+= + →+ + +

Since we don’t know the exact values of x and y, we must rely on the clues we’re given to find out what sort of

number the fraction 2x y

x y++

is.

• Because x and y are both positive, k will also be positive. The top of the fraction, 2x y x y y+ = + + , is bigger than the bottom, x y+ , so the fraction is greater than 1. k must be greater than 10 1⋅ . Eliminate A.

• The fraction however is smaller than 2, because the top is less than twice the bottom (twice the bottom would be 2 2x y+ ). k must be smaller than 10 2⋅ . Eliminate E.

We still have choice B, C and D. What do we do? Well one property we haven’t used yet is x y<

• If x and y were equal, the fraction 2x y

x y++

would equal 3 1.52

xx= . If x were really large (1 million) and

y were insignificantly small (0.00001) the fraction 2x y

x y++

would be really close to 1xx= . On the other

hand, if y were really large and x were insignificantly small, the fraction 2x y

x y++

would be really close to

2 2yy= . Because x y< , the real value of the fraction must be closer to 2 than it is to 1. In other

words, the real value of the fraction must be greater than 1.5. Since k is 210 x yx y

++

, k must be greater

than ( )10 1.5 15⋅ = . Eliminate B and C

( ): 3 : 2x x x

( ): : 2x x x


Pag

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Notes: The Official Guide explanation shows another excellent way to solve. Both ways come down to realizing that you must use the properties given (positive #s, x < y) to learn about what k could be.

#149, pg. 173 Difficulty Level: 700+ Topics: Rates & Work

30 Seconds Hack

We know that x% of the distance was traveled at 40mph and the remainder at 60mph. If x is 100 (meaning if the entire trip was traveled at 40mph), the average speed would have to be 40. On the other hand, if x is 0 (the entire trip traveled at 60mph), the average speed would have to be 60. We can quickly check which answers give us the correct average speed for both of these values of x. The right answer must do that.

Unfortunately, if we plug in 100 for x, all the answer choices will give us the expected average speed of 40. However, if we plug in 0 for x, only answers C and E give us the expected average speed of 60mph so one of these two answers must be correct.

Guess C or E

Solution A – Plug In



Our plug-in x = 50

With these values plugged in, re-consider the question: “Francine traveled 50% of the distance at 40mph and 50% at 60mph. What was the average speed?” To find this, we also need to pick an easy distance. Let the total distance be 240 miles. 240 is a great distance because it’s divisible by the two speeds.

The average speed over the whole trip is total distance

total time. We’ve used 240 as the distance. Find time.

• Francine travels 50% of the distance, or 120 miles at 40mph. The first part will take 120 3hrs40

=

• She travels the last 120 miles at 60mph. The 2nd part will take 120 2hrs60

=


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Francine’s total distance is 240 miles and her total time is 5hrs. Her average speed is 240 48mph5

=

Our solution 48

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variable in each answer with our plug-ins until we find the answer that equals 48.

(A) 180 180 50 48

2 2x− −= ≠

(B) 60 50 60 48

4 4x+ += ≠

(C) 300 300 50 48

5 5x− −= ≠

(D) 600 600 48

115 115 50x= ≠

− −. On the test I would stop here. E must be the right answer.

(E) 12,000 12,000 12,000 48

200 50 200 250x= = →

+ +

Notes: As soon as you realize an answer will not give you 48, move on. Don’t do the full math.


Solution B – Algebraic

Because the trip is divided into two legs, we have two separate distance-rate-time relationships. We can express these in a table. Let d be the total distance, and 1t and 2t be the times. The first leg is x% of the distance, or

100x d , so the other leg must be the remainder, or (100-x)% of the distance, or

100100

x d−

Leg 1 Leg 2 Total Time

1t 2t 1 2t t+ Rate 40 60 --- Distance

100x d

100100

x d−

d

Leg 1 Leg 2 Total ( 1 2t t+ )

Since we can express the time in terms of x and d, it doesn’t make sense to use the variables 1t and 2t (in general, the fewer variables you introduce into the problem, the better).

distancetimerate

= , so we can rewrite our table as:


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Time

10040 100 40

x d x d=⋅

100100100

60 100 60

x d x d−

−=⋅

100100 40 100 60

100 3 200 24,000 6,000 12,000 12,000

20012,000

x xd d

xd d xd xd d xd

xd d

=

−+⋅ ⋅

− −+ = +

+=

Rate 40 60 --- Distance

100x d

100100

x d−

d

You can see why I would plug in . Ok the final step is to find out the average speed over the entire trip. This is total distance

total time, or

12,000200 200

12,000

d dxd d xd d

=+ +. Factor out and simplify top and bottom by d, and you’re left

with 12,000

200x+


Take-Aways


• It is generally a good idea to use as few variables as you have to, and avoid needlessly introducing new variables (such as 1 2 and d d if they are the same distance; just use d).


Solution

To solve, substitute -1 for x and do the math carefully


( ) ( ) ( )( )

4 3 24 3 2 1 1 11 1 1

x x xx

− − − + −− +=

− − −

( ) ( ) ( )1 1 1 3 32 2 2

− − += = = −

− −


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e182




To find the lowest price, apply the highest discount. We start with 100% of the original price. A 40% discount leaves us with 60% of the price. An additional 25% discount leaves 75% of 60%. This equals 45% of the original price.

Find 45% of $16


Solution B

To find the lowest price, apply the highest discounts: 40%, then 25%. One way to increase an amount by x% is to

multiply it by . Conversely, to decrease an amount by y%, you can multiply it by

Decrease by 40%, then by 25%

Simplify first, then divide



30 Seconds Hack

The area of the triangle is 24, so . An observation of the figure tells us that x is only a little

bit longer than y. To estimate, find two numbers close to each other that multiply to 48. It makes most sense to

45 45 916 4 4 $7.20100 25 5

⋅ = ⋅ = ⋅ =

100100

x+ 100100

y−

60 7516100 100⋅ ⋅

3 3 3 316 4 $7.205 4 5 1

= ⋅ ⋅ = ⋅ ⋅ =

24 and 482xy xy= =


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use because the GMAT tends to use easy numbers. Based on the Pythagorean Theorem (or knowing the 6:8:10 Pythagorean triple), solve

Guess E.

Solution

Based on the information given , we should replace x with y+2 in the area formula. Since we know

that the area of the triangle is 24, we can write:

Set to 0



The equation could then be factored into and the solutions will be .

In this case, will factor to with solution . We ignore the negative

solution because y represents a distance. We now have two sides of the triangle and can quickly solve for z



30 Seconds Hack

6 and 8y x= =10z =

( )2x y= +

( ) ( )224 48 2

2 2y yb h y y

+⋅= = → = +

2 248 2 y 2 48 0y y y= + → + − =

2 0y by c+ + =

m n b+ =m n c⋅ =

2 0y by c+ + = ( )( ) 0y m y n+ + = or yy m n= − = −

2 2 48 0y y+ − = ( )( )8 6 0y y+ − = 6y =

6

8 z You should recognize this as the Pythagorean triple 6:8:10, with 10z =

Alternatively, set up 2 2 26 8 z+ = to solve for z.


Pag

e184

Jack is 14yrs older than Bill now, which means Jack is at least 15yrs old. So in 5 years, Jack will be at least 20yrs old.

Guess C, D, or E

Solution A

“Jack is 14 yrs older than Bill” (Eq. 1)

“In 10yrs, Jack will be twice as old as Bill” means Jack plus 10 is twice Bill plus 10 (Eq. 2)

Combine the (Eq. 1) and (Eq. 2)

Find j

Since Jack is now 18, in five years he will be 23.




Let’s reverse engineer the question and figure out which answer choice makes Jack twice as old as Bill in 10 years. When reverse engineering, start with answer C

Jack in 5yrs?

Jack now

Bill now (14yrs younger than Jack)

Ages in 10yrs (Jack should be twice Bill)

Notes

(A) 9 4 -- (B) 19 14 --

(C) 21 16 2 26 and 12 Jack not old enough. Eliminate A, B and C

(D) 23 18 4 28 and 14 Yes! In this scenario, in 10yrs, Jack is twice as old as Bill

(E) 33 28 -- -- Don’t bother, we’ve found a match


Take-Aways When a question gives you non-variable answers choices along with a word problem, you can often use reverse

engineering. Start with answer C, especially if it will take more than a few seconds to check each answer.

14 14j b b j= + → = −

( )10 2 10 2 10j b j b+ = + → = +

( ) ( )2 10 2 14 10j b j j= + → = − +

2 28 10 18j j j= − + → =


Pag

e185


30 Seconds Hack

In 8hrs, 3/5 (or 60%) of the pool is filled. 40% is left, so this will take more than half as much time as the first part did

Guess A, B, or C


The question divides the pool in five; 3 parts are filled, and 2 parts are left. The quantity left is thus of the

quantity filled. To fill the rest of the pool will take of the 8hrs it took to fill the first 3 parts.


Solution B – Proportions

The portion filled is proportional to the time spent. 60% of the pool has been filled in 8hrs. How long will it take to fill the remaining 40%? Let x be the time it will take to fill the remaining 40%:

We can set up a proportion


1/3 of an hour is 20 minutes, so



23

23

13

2 168 5 5hrs. 20mins.3 3

x = ⋅ = = =

8 8 40 60 40 60

x x ⋅= → =

13

16 53

x = =

5hrs. 20minsx =


Pag

e186

Solution

Build the equation one step at a time:

“Positive number x is multiplied by 2” 2x

“The product is divided by 3”

“Positive square root of the result equals x”

This is the proper equation. Now solve for x

Square both sides and set to 0

Factor and find the roots

Solving for the 2nd root gives us



30 Seconds Hack

Since only about 25% of the water is removed, the concentration of sodium chloride will increase from 5%, but not by much.

Guess C or D

Solution

Sodium chloride makes up 5% of the 10,000gallons, or 500 gallons. After 2500 gallons of water evaporate, sodium chloride will be 500 of the new 7,500 gallons total. The amount of sodium chloride is the same since it did not evaporate. Find out what percentage 500 is of 7,500.

23x

23x x=

2 22 3 2 03x x x x= → − =

( )3 2 0 0 or 3 2 0x x x x− = → = − =

23

x =


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e187

“500 is what percent of 7500?”

Isolate x and simplify



30 Seconds Hack

From 99 to 301, there are about 200 numbers, about 100 of which are even. The average of these evens (halfway between 99 and 301) is roughly 200.

( )( ) ( )( )sumaverage sum average # of values sum 200 100 20,000# of values

= → = → ≈ =

Guess B

Solution A

The formula gives us a way to find the sum of the first n positive integers, but we’re asked to find the sum of even integers between 99 and 301. Using the formula, we can find the sum of all integers from 100 to 300 by subtracting the sum the sum from 1 to 99 from the sum from 1 to 300.

• Sum from 1 to 300: 300 301 45,150

2⋅ =

• Sum from 1 to 99: 99 100 4,950

2⋅ =

• The difference is the sum from 100 to 300: 45,150 4,950 40,200− = . This number encompasses both even and odd integers. About half of this sum, or 20,100 is the sum of the evens in the range.


Solution B

If we can find the average as well as the number of values involved, we’ll be able to solve for the sum.

( )( )sumaverage sum average # of values# of values

= → =

500 7500100

x= ⋅

500 100 500 20 6.677500 75 3

x ⋅= = = =


Pag

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A sequence in which there is a constant difference between adjacent values is an arithmetic sequence. Some examples include:

• A set of consecutive integers: 12, 13, 14… Difference is 1 • A set of consecutive even integers: 2, 4, 6… Difference is 2 • A set of consecutive multiples of seven: 7, 14, 21… Difference is 7.

The average of an arithmetic sequence is the average of its highest and lowest values. In our case, we’re dealing with consecutive even integers between 99 and 301. The lowest value is 100 and the highest is 300. The average

of the sequence is 100 300 200

2+ = .

The sum of the sequence is ( )( ) ( )sum average # of values sum 200 # of values= → = . To find the

sum, all we need now is the number of values. Because we are only interested in even integers, we can ignore 99 and 301. From 100 to 300, there are 300 100 1 201− + = integers. There is 1 more even integer than odd integer in that list because it begins and ends with an even. Therefore the number of values in the sequence is 101. the sum is ( ) ( )101 20,200sum 200 # of values sum 200 == → =


Take-Aways • A set of numbers in which the difference between each value and the next higher value is constant is

called an arithmetic sequence. • The average of an arithmetic sequence is the average of its highest and lowest values. The average is

always equal to the median of this sequence.

• The sum of an arithmetic sequence is its average multiplied by the number of numbers in the sequence

#158, pg. 174 Difficulty Level: 400-500 Topics: Translations & Manipulations; Combinatorics

Solution A

After 3 women and 2 men are added, we have . The probability of selecting a

woman is


3 women and m 2 menw+ +

( ) ( )women 3 3

total 3 2 5w w

w m w m+ +

= =+ + + + +


Pag

e189




Our plug-ins

With this plug-in, after 3 women and 2 men are added, we will have 6 women and 7 men. The probability of

selecting a woman is

Our solution

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the

variables in each answer with our plug-ins until we find the answer that equals

(A)

(B)

(C)

(D)

(E) 3 6

5 13w

w m+

=+ +

This answer matches our solution!





3 womenw = 5 menx =

womentotal

613

613

3 65 13

wm= ≠

3 68 13

ww m

= ≠+

3 6 62 7 13

wm+

= ≠+

3 6 63 11 13

ww m

+= ≠

+ +


Pag

e190

To find out the prime factors of 7,150, we must break it down into its prime factors: 7,150 50 143= ⋅ . Is 143 prime? Well to check whether a number is prime, test it against all prime factors whose square is smaller than the number. For instance, to determine whether 143 is prime, you only need to check whether it’s divisible by 2, 3, 5, 7 and 11. There is no need to check the next prime number (13) because the square of 13 is bigger than 143. It

turns out that 143 is divisible by 11: 27,150 50 143 50 11 13 11 13 2 5= ⋅ = ⋅ ⋅ ⋅ ⋅→ ⋅

The numbers that are prime factors of 7,150 are 2, 5, 11 and 13.


Take-Aways • When solving a Number Properties question, express all values in their prime factorizations (ex:


• Memorize your multiplication tables up to 12 times 12 before taking the GMAT. You should also know your perfect squares up to 220 and how to recognize at a glance whether a number is divisible by 2, 3, 4, 5, 6, 8, and 9.


Solution

A common way to find shaded areas is to find the unshaded area and subtract it from the total area. Since the shaded path is 3 feet wide, the total radius is 11, and the total area is . The unshaded area, of

radius 8 has area . Finally, the shaded area is .



Solution

Because we are asked about , we should attempt to manipulate the equation we are given to isolate this

fraction.

We are given Since both values are positive, squaring will not flip the sign.

2 211 121rπ π π= =2 28 64rπ π π= = 121 64 57π π π− =

25n

25n >


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e191

Square both sides

Divide by 25

Only one answer is bigger than 25



Solution

The weight distribution of apples to peaches to grapes is 6x to 5x to 2x, and the total weight is 13x. Since the total weight is also 39lbs, . The difference between apples and grapes is




To solve this algebraically, we need to setup some variables. Let I be Henry’s income and s be the amount he saves this year. Since for each dollar saved, Henry will have 1+r dollars to spend next year, after saving s dollars this year Henry will have ( )1s r+ to spend next year. Finally because I is the total income and s is the amount

saved this year, the amount Henry spends this year is the remainder, I s−

“Next year the amount Henry has available to spend will be half the amount that he spends this year”. Translating

this gives us ( ) ( )112

s r I s+ = − (Eq.1)

We’re asked to find “what fraction of his income should Henry save this year”. Since s is the amount saved this

year and I is the income, we need to solve for sI

25 25n > ⋅

2525n>

13 39 lbs 3 lbsx x= → =6 2 4 12 lbsx x x− = =


Pag

e192

Let’s manipulate (Eq.1) to find sI

:

( ) ( ) ( )11 2 1 2 2 3 22

s r I s s r I s s rs I s s rs I+ = − → + = − → + = − → + =

Factor out s and solve ( ) ( )3 2 13 2 1 3 2

r I ss r Is r I

++ = → = → =

+

The fraction of income Henry needs to save, sI

, is equal to 1

3 2r+





Our plug-in r = 2

With this the question becomes. “For each dollar Henry saves this year, he will have 1 $3r+ = next year. What fraction should Henry save this year so that the amount he spends next year is half what he spends this year?”

We can use whatever amount we want as Henry’s income, because the question asks about fractions without limiting us to specific concrete values. Instead of starting with Henry’s income however, it’s much easier to start with how much he saves.

Let Henry save $100. Since he will have $3 to spend next year for each $1 saved now, Henry will have $300 to spend next year. We know that next year’s amount should be “equal to half the amount that he spends this year”. Therefore $300 to spend next year means $600 are spent this year. Now we can answer the question: “what fraction of his income should Henry save this year?”

saves now saves now 100 1 income saves now + spends now 100 600 7

= = →+

Our solution 17

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variable

in each answer with our plug-in ( r = 2) until we find the answer that equals 17

. Only answer E matches our

solution


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e193

E) ( )

1 1 1 2 3 2 2 3 7r

= →+ +


Take-Aways

• Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem. In this case, we chose to make Henry save $100. Whether we had chosen $10, $25 or $35.12, the final fraction would be the same.

• When variables appear in the answer choices, you can often plug in easy values for those variables within the question, solve and then find out which answer choice agrees with your solution. Avoid plugging-in 0, 1, or the same value for different variables (unless the variables are equal). By coincidence, two answers may agree with your solution. In that case, just plug in another set of easy numbers and try again


Solution

According to rules of exponents, ( )bab ax x= . Notice that what we’re asked to find, 2m− , is just the square of

what we’re given. ( )2

22 1 1 1= 3 9

m m− −

= − →




If Lois has x more than Jim, Lois must have more than the average, while Jim must have less than the

average. The total is y, so the average is . Therefore, Jim must have


2x

2x

2y

2 2 2y x y x−− =


Pag

e194

Solution B

Let j be the amount Jim has, and l be the amount Lois has

“Lois has x more than Jim” (Eq. 1)

“Together, they have y” (Eq. 2)

We want to solve for j, in terms of x and y, so we need a way to get rid of l. One way to do this is to stack the equations and subtract them.

(Eq. 1) – (Eq. 2)





Our plug-ins

With these values plugged in, re-read the question as: “Lois has $2 more than Jim has, and together they have a total of $10. How much does Jim have?”

In this scenario, Lois would have $6, and Jim would have $4..

Our solution $4

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variables in each answer with our plug-ins until we find the answer that equals $4.


(B) 210 4

2 2xy − = − ≠

(C) 10 2 4

2 2y x− = − ≠

(D) 2 2 10 2 4y x− = ⋅ − ≠

(E) 2 10 2 2 4y x− = − ⋅ ≠

l j x= +

l j y+ =

2 2

y xj j x y j y x j −− = + − → = − → =

2x = 10y =

10 2 42 2

y x− −= =


Pag

e195






Intuitively this can be solved as a weighted average question. To do this effectively, instead of talking in terms of winning percentage for a group of games, we assign a value to each game. The overall average value is 70 (corresponding to the winning percentage). The first 100 games have a value of 80, and the remaining games have a value of 50. The first 100 games create a surplus of 1000 (each of the 100 games exceeds the average value by 10). The remaining games must together create a deficit of 1000 to offset this surplus. Since each of these remaining games has a deficit of 20, we need only 50 such games to accumulate the needed deficit of 1000.

100 initial games, plus the 50 needed to offset the surplus gives us 150 total games


Solution B

To solve algebraically, we need to build an equation that uses only the variable t, to represent the total number of games. Express the following data as an equation: “The team won 80% of the first 100 games plus 50% of the remaining games which is equivalent to 70% of the total games”

I use common denominator to make math easy

Multiply by 10

Solve for t



8 5 7100 ( 100)10 10 10

t t⋅ + ⋅ − =

800 5 500 7t t+ − =

300 2 150t t= → =


Pag

e196


Let’s reverse engineer the question and figure out which answer choice makes “the team won 70% of games for the season” true. Keep in mind that the team has already won 80 games (80% of the first 100 games).

How many total games?

How many total wins (80 plus 50% of games over 100)

What is total winning%? (should be 70% or 7/10)

Notes

(A) 180

(B) 170

(C) 156

(D) 150 Yes! 150 total games yields a 70% overall winning percentage

(E) 105 -- Don’t bother, we’ve found a match





#167, pg. 176 Difficulty Level: 600-700 Topics: Sets & Groups

Solution A

The group formula is a useful tool to handle questions of two overlapping sets. You’ve probably seen in before: . In this specific question, the formula would therefore be:


We want to find both. Let’s input the data we have:

80 50% 80 120+ ⋅ =120 2180 3

=

80 50% 70 115+ ⋅ =115 23170 34

=

80 50% 56 108+ ⋅ =108 9156 13

=

80 50% 50 105+ ⋅ =105 7150 10

=

80 50% 5 82.5+ ⋅ =


Total Experience Degree neither both= + + −

Total Experience Degree neither both= + + − 30 14 18 3 both= + + −


Pag

e197

Solve for both


Solution B

Another option is to represent the given data in a table. Each axis of the table will focus on one group. Before solving, circle the cell that you want:

4+ yrs of Exp. Less Exp. Total Degree 18 No Degree 3 Total 14 30

Solve for the missing pieces until you arrive at the circled cell

4+ yrs of Exp. Less Exp. Total Degree 5 13 18 No Degree 3 Total 14 16 30


Take-Aways • The group formula, , is a useful way to solve questions about



Solution

Original equation

Multiply by x



30 35 5both both= − → =


1 21 2x x

+ = −

1 2 2 3x x x+ = − → =


Pag

e198

Solution

Set up a proportion

Simplify

x is slightly less than 100 times 30 which would be 3000


#170, pg. 176 Difficulty Level: 700+ Topics: Translations & Manipulations; Sets & Groups

Solution A

One option is to represent the given data in a table. Each axis of the table will focus on one group. Before solving, circle the cell that you want:

Passed. Not Passed Total Course 12 No Course 30 Total 0.3T T

Add up the “Not Passed” column

4+ yrs of Exp. Less Exp. Total Degree 12 No Degree 30 Total 0.3T 42 T

The bottom row yields a simple equation for T

Isolate T and solve


Solution B – Group Formula

Let T be the total, Passed be the number who have passed the course, and Course be the number of students who have taken the course. The group formula is a useful tool to handle questions of two overlapping sets. You’ve

96 96 3,000,000,000 100,000,000 3,000,000,000 100,000,000

x x ⋅= → =

96 30x = ⋅

0.3 42T T+ =

7 1042 42 6010 7

T T= = ⋅ =


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e199

probably seen in before: . In this case,

. Let’s input the data given

“30% have passed the test”

“Among those who haven’t passed test, 12 have the course”

“Among those who haven’t passed test, 30 don’t have the course”

Putting all this data into the group formula yields the following equation:




From “Among those who have not passed the test, 12 have taken the course and 30 have not”, we can tell that a total of 42 people have not passed the test. Let’s reverse engineer the question and figure out which answer choice agrees with this information.

What is the total?

How many have passed the test? (30% of total)

How many have not passed the test? (should be 42)

Notes

(A) 60 Yes! A total of 60 makes everything else work

(B) 80 -- Don’t bother, we’ve found a match

(C) 100 -- Same as above (D) 120 -- Same as above (E) 140 -- Same as above




• The group formula, , is a useful way to solve questions about 2 overlapping groups.


T Passed Course neither both= + + −

0.3Passed T=

12Course both− =

30neither =

100.3 12 30 0.7 42 42 607

T T T T= + + → = → = ⋅ =

total passed−

30% 60 18⋅ = 60 18 42− =

30% 80 24⋅ =

30% 100 30⋅ =30% 120 36⋅ =30% 140 42⋅ =



Pag

e200


Solution

We’re told how to calculate the nth term: 12nn −+ . To find any term, replace n with the term’s position.

• The 6th term is 6 16 2 6 32 38−+ = + =

• The 5th term is 5 15 2 5 16 21−+ = + =

The difference between the 6th and 5th terms is 38 21 17− =



Solution

Think of as one number, which when squared results in 400. What is that number? . Let’s

solve for both values of x

or

Careful, we are not asked to find x, but rather . This has two corresponding values:

or

Pick whichever solution is presented in the answers




Manipulate the equation given to obtain . Which numbers when squared will fit this range? All fractions, positive or negative, as well as 1 and -1 themselves would work.

( )1x − ( )1 20x − = ±

1 20 21x x− = → = 1 20 19x x− = − → = −

( )5x −

21 5 16x x= → − = 19 5 24x x= − → − = −

2 1x ≤


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Solution B – Find the Limits

An alternative is to factor the expression into and find the solutions 1 and -1. The solutions

give the limits of the final inequality. To find out how x compares to the limits, plug values into the original

inequality

Try a number less than -1. If , then . This is illogical, so x cannot be less

than -1

Try a number between -1 and 1. If , then . This is logical, so x is between -1 and

1.

Try a number above 1. If , then . This is illogical, so x cannot be greater than 1

Because 1 and -1 themselves work, we must include them in the inequality. Together these plug-ins give us the correct inequality. .



30 Seconds Hack

Approximate. On any given toss, there is a 50% probability of tails. On three tosses, it is very likely that at least one tail will show. The answer should be well above 50%, though the probability isn’t a near certainty

Guess C or D

Solution

Keep two rules of probability in mind:

1 1x− ≤ ≤

( )( )1 1 0x x− + =

( )21 0x− ≥

2x = − ( )21 2 0 3 0− − ≥ → − ≥

0x = ( )21 0 0 1 0− ≥ → ≥

2x = ( )21 2 0 3 0− ≥ → − ≥

1 1x− ≤ ≤


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Probability of two successive independent outcomes is the probability of one times the probability of the other

Probability of an outcome is 1 minus the probability that the outcome does not happen

Whenever you are asked to find the probability that something happens “at least one time”, it is easier to find the probability that it doesn’t happen at all, and subtract it from 1. In this case, the probability that we get at least one tail equals 1 minus the probability of no tail (or all heads)


Take-Aways • When solving a probability question, you can often use your intuition to eliminate answers that are way off

• Whenever you are asked to find the probability that something happens “at least one time”, it is easier to

find the probability that it doesn’t happen at all, and subtract it from 1


Solution A

To quickly solve, convert the fractions into percentages:

are A’s 20% are A’s are B’s 25% are B’s are C’s 50% are C’s

95% are accounted for, so the 10 D’s must be 5% of all students. To find 100%, we need only multiply 5% by 20. Multiplying the number of D’s by 20 gives us




Let’s reverse engineer the question and figure out which answer choice correctly leaves 10 students with D’s

( ) ( ) ( )&P A B P A P B= ⋅

( ) ( )1P A P notA= −

( ) ( ) 1 1 1 7at least 1 tail 1 all heads 12 2 2 8

P P= − = − ⋅ ⋅ =

15

14

12

10 20 200 students⋅ =


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How many total?

A’s (a fifth of total)

B’s (a fourth of total)

C’s (half of total)

D’s (remainder, should be 10)

Notes

(A) 80 -- -- -- -- (B) 110 -- -- -- --

(C) 160 Too few D’s. Start with a higher number

(D) 200 Yes! Starting with 200 students leaves 10 D’s

(E) 400 -- -- -- --





Solution A

Let’s think about what happens to each expression as a positive x gets larger

I. As x gets bigger, 2x gets bigger. Since we are subtracting the same amount (5) from both, the expression as a whole will get larger. must increase

II. As x gets bigger, the denominator gets bigger, and the fraction gets smaller. So as x gets bigger, we subtract

a smaller amount from 1. The expression as a whole gets larger as we subtract less and less. must

increase III. As x gets bigger, the difference between gets larger, so the denominator gets larger and the

fraction as a whole gets smaller. must decrease


Solution B

It would be difficult to do this question by plugging values between 165 and 166 because the math would not be friendly. However, we can use any other positive number bigger than 1 to check. Let’s make x grow from 10 to 100 and see which expressions get larger.

1160 325

=1160 404

=1160 802

= 160 32 40 80 8− − − =

1 200 405

=1 200 504

=1 200 1002

= 200 40 50 100 10− − − =

2 5x −

11x

−

2 and x x

2

1x x−


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I. When x is 10, . When x is 100, . As x gets larger, this expression increases.

II. When x is 10, . When x is 100, . As x gets larger, this

expression increases.

III. When x is 10, . When x is 100, . As x gets larger, this expression decreases.



Solution

To find the diagonal of a box, use the following formula, an extension of the Pythagorean Theorem: where l, w, and h are the length, width and height.

In this case, our measurements are 10, 10, and 5. Put it in the formula to get:


Take-Aways • To find the diagonal of a box, use the following formula, an extension of the Pythagorean Theorem:

• Memorize your multiplication tables up to 12 times 12 before taking the GMAT. You should also know your perfect squares up to and how to recognize at a glance whether a number is divisible by 2, 3, 4, 5, 6, 8, and 9.

#178, pg. 177 Difficulty Level: 700+ Topics: Translations & Manipulations; Sets & Groups


The following solution would take fewer than 20 seconds. In the table provided, the students who are part of two groups have been counted twice. For example, suppose John is part of the Chess and Drama clubs. He would be counted once as one of the 40 Chess club members, but he would be counted again as one of the 30 Drama club members. To get rid of the redundant count, add the membership of all clubs and subtract those who belong to two clubs (so that they will be counted once instead of twice).

2 5 15x − = 2 5 195x − =11 1 0.1 0.9x

− = − =11 1 0.01 0.99x

− = − =

2

1 190x x

=− 2

1 19900x x

=−

2 2 2 2d l w h= + +

2 2 2 210 10 5 225 15d d= + + = → =

2 2 2 2d l w h= + +

220

40 30 25 10 5 6 74total = + + − − − =


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Solution B – Venn Diagrams

A more lengthy solution involves building Venn Diagrams.



Chess Drama

Math

0 5

10

6

40 – 5 – 10 - 0 =

30 – 10 – 6 – 0 =

25 – 5 – 6 – 0 = 14

By putting in the given data, we can find all regions of the diagram. The total number of students is the sum of all regions. 25 14 14 10 5 6 0 74+ + + + + + =

Chess Drama

Math

x a

b

c

Chess – a – b – x

Drama – b – c – x

Math – a – c – x

Always start with the innermost circle. To find the regions that do not overlap, you must subtract the overlapping regions from the circle total. For instance to find students who belong only to the Chess club, take the entire Chess population and subtract those who also belong to Math (a), those who also belong to Drama (b), and those who belong to all three clubs (x). This is why Chess only is equal to Chess a b x− − −


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Solution A

The initial ratio can be expressed . Add 5 to each quantity and the new ratio is . There is no way to

reduce this ratio and cancel out the x’s.



An alternative approach is to plug in two sets of numbers that befit the 3 to 4 ratio and see what the resulting ratio would be if we added 5 to each quantity:

If the initial quantities were 3, and 4, the initial ratio would be . Adding 5 to each quantity would result in a

ratio of .

On the other hand, initial quantities of 9 and 12 give an initial ratio of . Adding 5 to each quantity would result

in a ratio of .

We can conclude that the final ratio cannot be deduced without knowing the actual values used in the initial 3:4 ratio.




The total of y and z ( )80 2⋅ is 40 more than the total of x and y ( )60 2⋅ . Since y is the same in both groups, this

40 difference must be the difference between z and x.


34

xx

3 54 5

xx++

34

89

34

1417


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Solution B – Average Formula

“The average of x and y is 60” (Eq. 1)

“The average of y and z is 80” (Eq. 2)

To find , do (Eq. 2) – (Eq. 1) to get



30 Seconds Hack

Half is removed after the first stroke. Half of what’s left after the 2nd stroke. So after 2 strokes, 75% of the air is already removed. After 4 strokes, almost all the air should be out.

Guess A, or possibly B.

Solution A

Figure out what fraction is left in the tank and subtract it from 1 to find out how much has been removed. Each stroke of the pump takes out half of the air. Additional strokes have the cumulative effect of cutting the amount

left by ½. After four strokes, is left. The rest, or , has been removed



One way to do this question is to plug in a value for the amount of air in the tank. Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem. If you work correctly, whatever you plug in will reduce to the same answer.

Since we’ll be cutting our quantity in half several times, let’s take a power of two:

60 1202

x y x y+= → + =

80 1602

y z y z+= → + =

z x− 160 120 40z x− = − =

1 1 1 1 12 2 2 2 16⋅ ⋅ ⋅ =

1516


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Our plug-in 32 units of air total

After the first stroke, 16 units are left. After the 2nd stroke, 8 units left. 4 units after the 3rd, and only 2 units are left after the last stroke.

If 2 units are left, then 30 out of 32 have been removed.

Our solution




#182, pg. 178 Difficulty Level: 700+ Topics: Translations & Manipulations; Number Properties


You should be able to quickly eliminate answers D and E, because it is relatively straightforward to find that .

To find whether 121 or 181 can be obtained, we must realize that the sum of the two digits must be 11 (because the units digit of 121 and of 181 is 1). This gives us a very limited set of digit pairs check: (5, 6) (4, 7) (3, 8) and (2, 9). In fact, they all give us 121. , , , and

We can stop here because we’ve tested all digit pairs that add up to 11, and none of the resulting numbers will add to 181 (they all give us 121). Thus it would be impossible to get 181 from this addition of numbers with reversed digits

If the conclusion above isn’t clear, you can look for numbers that add up to 165 by using two digits that add up to 15.


Solution B

All two digit numbers can be expressed as the sum of the tens digit times 10 plus the units digit. For example, , and . This becomes important when we’re asked to manipulate unknown digits.

In this question, M and N have reversed digits. Let’s say where X and Y are digits. We could write that . Now, we can add the two:

30 1532 16

=

44 13 31, and 99 36 63= + = +

56 65 121+ = 47 74 121+ = 38 83 121+ = 29 92 121+ =

24 2 10 4= ⋅ + 93 9 10 3= ⋅ +XY and YXM N= =

10 and 10M X Y N Y X= + = +


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.

The final expression tells us that the sum must be a multiple of 11. All answers, with the exception of A, are multiples of 11.




Car X traveled 80 miles in 2hrs, so its speed is 40mph. Since Y goes 50% faster than X, the speed of Y will be .

Traveling 80 miles at 60mph will take between 1 and 2hrs, but closer to 1hr.


Solution B

The rate/work formula is . We are asked to solve for time, so we should rewrite the

formula as . We already know the distance (80 miles), but we need to find Y’s rate by adding

50% to X’s rate.

X covers 80 miles in 2hrs, so X’s rate is 40mph. To add 50%, multiply by 1.5 (or 3/2).

. Now that we have Y’s rate, plug it into the rate/work formula to find the

time:


Solution C

( ) ( ) ( )10 10 11 11 11M N X Y Y X X Y X Y+ = + + + = + = +

40 20 60mph+ =

distance rate time= ⋅distancetime

rate=

3 340 60mph2 2y xrate rate= ⋅ = ⋅ =

13

80 160

distancetimerate

= = =


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Let r be X’s rate. Let T be the time it took car Y and 32

r (a 50% increase in rate) be Y’s rate. distancetime

rate= ,

so can write 802hrsr

= , and ( ) 13

80 80 2 22hrs 1 hrs3 3 32

Trr

= = ⋅ = ⋅ =


#184, pg. 178 Difficulty Level: 400-500 Topics: Translations & Manipulations; Statistics

Solution

Write the average formula

Multiply by 4 and remove parentheses

Combine and isolate K’s



Solution A

Examine each answer in turn

(A) We don’t even know whether q is a factor of p, so we can’t even tell whether this answer is an integer, let

alone even or odd. Note that the certain prints of The Official Guide are wrong in their explanation of this

answer.

(B) even odd even⋅ =

(C) 2p must be even. Adding q gives us which must be odd. STOP here


Solution B – Eliminate by trial and error

( ) ( ) ( )2 3 3 5 5 163

4K K K K+ + + − + +

=

2 3 3 5 5 1 63 4K K K K+ + + − + + = ⋅

11 252 1 23K K= + → =

even odd+


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An alternative to solving these “MUST BE” questions is to plug in numbers that show that each answer doesn’t have to be what you seek. Using this method, you must eliminate all four wrong answers before you select the one left.

(A) If p is 2 and q is 3, then this is not odd.

(B) If p is 2 and q is 3, then this is not odd

(C) This is always odd.

(D) If p is 2 and q is 3, then this is not odd

(E) If p is 2 and q is 3, then this is not odd




Since Y is twice the size of X, a full X would be half of Y, but X is only half-full, so it contains just a quarter of Y.

Y is already 2/3 full, adding the oil in X would take the amount in Y to



One way to do this question is to plug in a value for one of the capacities. Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem. If you work correctly, whatever you plug in will reduce to the same answer.

To decide what to plug in, look at the fractions in the problem: and . You want to pick a number that will easily divide into all the denominators.

Plug in your chosen value in your starting point:


“X is half full” X contains 3

2 1 8 3 113 4 12 12 12

Y Y Y+ = + =

12

23

Our plug-in 6X =


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“Y has twice the capacity of X”

“Y is 2/3 full” Y contains 2 12 83⋅ =

If all the oil from X is poured into Y, Y will contain 11 out of 12.





Solution A

. Together they equal



You can substitute an easy value for x, based on the fractions given.

Our Plug in

. Thus the question would become “12 is what percent of 200?”

Our solution



Solution

12Y =

2 4, and 50 100 25 100x xx x= =

6 or 6% of 100

x x

200x =

200 200 1250 25 50 25x x+ = + =

12 6 6%200 100

= =


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Simply replace a and b with their respective values in the function.

Start working inside the parentheses .

Replace with – 3 in main equation


#189, pg. 179 Difficulty Level: 700+ Topics: Geometry

30 Seconds Hack

Since the circular base of the cylinder will be inscribed against a rectangular side of the box, the diameter of the base will be equal to the width of the side. For instance, if the circular base rests against a 6 by 8 side, the diameter must be 6. Any bigger and it would not fit. There are only two possible values for the diameter:

If the base is inscribed in a 6 by 8, or a 6 by 10 side, the diameter will be 6 (radius is 3)

If the base is inscribed in a 8 by 10 side, the diameter will be 8 (radius is 4).

The only possible values for the radius are 3 and 4. Eliminate C, D, and E

Guess A or B


The question can be solved in 15 seconds, but this solution requires grasping some abstract concepts. To maximize the size of the cylinder, we need to pick the configuration that leaves as little unoccupied space in the box as possible. The height of the cylinder will equal the height of the box, so space will only be wasted at the base. More specifically, space will be wasted by the corners that the circular base cannot reach.

( ) ( )22 3 1 3 1 3

3 3a ba b

−= → − = = −

( )3 1− ( ) ( )22 2 3 2 3 4

3 3a ba b

−= → − = = −


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The figures above show that less space is lost if the rectangle is almost square. To solve this question in 15 seconds, we need find out which side of our box is closest to a square (width to length ratio must be closest to 1:1). It is on this almost-square side that the circular base of the cylinder must be so that the least volume is wasted. The side in question is either the 6 by 8 or the 8 by 10 side. Compare the sides:

6 by 8 side has a ratio of 6:8 30:40

8 by 10 side has a ratio of 8:10 32:40. This is closest to a square (1:1).

To minimize lost space (and thus maximize volume), the circular base must be inscribed in the 8 by 10 side, so the diameter will be 8, and the radius 4


Solution B

A simpler, though longer, alternative is to try all possible configurations of the cylinder (there are only 3) and find the radius of the configuration with the highest volume. The height of the cylinder in each case will equal to the height of the box. For example, if the cylinder is standing on a 6 by 8 side, the height will be 10. For the reason I pointed out in the previous solutions, the diameter of the base will be equal to the width of the side

A cylinder standing on the 6 by 8 side would have radius 3 and volume



As the data shows, the highest volume is from a cylinder of radius 4.


#190, pg. 179 Difficulty Level: 600-700 Topics: Functions & Sequences; Number Properties

Solution

Note that the units digit of is the same as that of since in multiplication, the units digit of the result

depends only on the units digit of each factor. has units digit of 1.

Likewise, the units digit of is the same as that of and is 9.

The units digit of is the same as that of and is 9.

Finally, the units digit of is equal to the units digit of , which is 1.

23 10 90π π⋅ =

23 8 72π π⋅ =

24 6 96π π⋅ =

413 4343

217 27

329 39

4 2 313 17 29⋅ ⋅ 1 9 9⋅ ⋅


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Solution

To get to point Y, Pat must go up 3 times, and right 2 times. Each path can thus be seen as an arrangement (permutation) of three U’s and two R’s where U means “go up” and R means “go right”. With this understanding, the question becomes one of straightforward arrangements:

“How many arrangements of three R’s and two U’s are possible?” An easy way to remember the permutations/combinations formulas is to build a fraction as follows:

• In the numerator: the factorial of the number of items in the experiment • In the denominator: the product of factorials. Each factorial indicates how many items are grouped

together



Solution

To double or half a ratio, double or half its first part (the numerator).

The initial ratio is S:A:W 2:50:100

Let’s change the ratios according to instructions. We will replace the amount that is not changed with a variable:

5! 5 4 10 arrangements3!2! 2

⋅= =

The number of arrangements is5!

5 total items to be arranged: 3 U’s, 2 R’s

The R’s form a group of 2 items

The U’s form a group of 3 items


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Ratio of soap to alcohol is doubled. Initial ratio of 2:50:100 becomes 4:50:W

Ratio of soap to water is halved. Initial ratio of 2:50:100 becomes 1:A:100

To combine these ratios into one, multiply the second ratio by 4, so that the quantities of Soap match. This will ensure that the other quantities also match. The new, unified ratio of S:A:W is 4:50:400, or 8:100:800. Thus if there are 100 cubic centimeters of alcohol, there will be 800 cubic centimeters of water.



Solution A – Group Formula



Input the given data:




1 2Total Q Q neither both= + + −

100 75 55 20 50both both= + + − → =

Q2 Q1

x

Always start with the center. Use a variable if you don’t know the value.

neither goes outside both circles

Q2 Q1

x

Students who answer only Q1 correctly equals all students who answered it correctly minus those who answered both correctly. A similar logic is used to find 55 – x, the fraction who answer only Q2 correctly.

75 - x 55 - x

neither

20


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We know that all portions must add up to 100%, so we can set up an equation as follows: .

Solve for x







Draw and estimate.


Solution B

( ) ( )75 55 20 100x x x+ + + − + =

150 100 50x x+ = → =


(2,3) B will be somewhere in this region

(2,3)

C will be somewhere in this region

The final estimate of C has positive x and negative y coordinates. Only answers C and D satisfy this restriction.

Because points B and C are farther to the right than A is, an additional restriction is that the x coordinate of C be greater than the x coordinate of A. Only answer choice D satisfies both requirements


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“ is perpendicular bisector of AB” means that B is the reflection of A across that line. To reflect a point across the line , simply exchange their coordinates. A is at (2, 3), so B is at (3, 2)

“The x-axis is the perpendicular bisector of BC” means that C is the reflection of B across the axis. To reflect a point across the x-axis, simply negate its y-coordinate. B is at (3, 2), so C is at (3, -2).


#195, pg. 180 Difficulty Level: 700+ Topics: Translations & Manipulations

Solution A

Let p be the current price, and n be the number of towels that can be bought for $120. Since to find out how many towels you can buy, you would need to divide how much money you spend by the price of each towel, one

equation we can set up is (Eq. 1).

If the price were increased by $1, to a new price of , the number of towels that could be bought for $120 would be 10 fewer, or . This allows us to setup a 2nd equation:

(Eq. 2)

Combine (Eq. 1) and (Eq. 2) by replacing n with and by replacing np with 120 in (Eq. 2)

Rewrite (Eq. 2):

Subtract 120, and then multiply by p

Divide by -10

Factor and find p

We ignore the other solution (-4) because the price must be positive, so p could not equal -4.



y x=y x=

120 or 120n npp

= =

1p +10n −

( ) ( ) ( )( )12010 10 1 120 10 10 1201

n n p np n pp

− = → − + = → + − − =+

120p

10 10 120np n p+ − − =120120 10 10 120p

p+ − − =

2120 10 10 0p p− − =

2 12 0p p+ − =

( )( )4 3 0 3 p p p+ − = → =


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Let’s reverse engineer the question and figure out which answer choice makes a $1 increase in price result in 10 fewer towels for $120.

What is current price?

# of towels for

$120

# of towels for $120 after price goes up by $1 (should be 10 fewer towels than at lower price)

Notes

(A) $1 120 60 60 fewer towels (B) $2 60 40 20 fewer towels

(C) $3 40 30 Yes! # of towels decreased by 10 when price went up by $1

(D) $4 30 -- Don’t bother, we have a match (E) $12 10 -- Same as above





Solution

The Official Guide’s solution is a very good one. Below is an alternative. First identify what you want: “green marbles in jar R” is z. Write out the 3 equations:

(Eq. 1) (Eq. 2) (Eq. 3)

Do (Eq. 3) – (Eq. 1) to get rid of x (Eq. 4)

Do (Eq. 4) + (Eq. 2) to get rid of y


#197, pg. 180 Difficulty Level: 700+ Topics: Geometry Solution

120price

80x y+ = 120y z+ = 160x z+ =

80z y− =

2 80 120 100z z= + → =


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We know that 300 revolutions are completed in 1 minute (60 seconds). In 15 seconds (a quarter of 1 minute), there will be one fourth as many revolutions. Therefore in 15 seconds, the point will complete 300 75 revolutions

4= .

Thus in 15 seconds, the point completes 75 revolutions and each revolution is 20π centimeters. The total distance covered in 15 seconds is 75 20 1,500 centimetersπ π⋅ =


Take-Aways



Solution A

Find the prime factorization of n

List out all the factors of n by finding all combinations of the prime factors, first using only 1 number from the prime factorization, then using two numbers, and finally using all three.

• Using only 1 number, the factors are 1, 2, and p. Of these, only 2 is even. • Using 2 numbers, the factors are . Both of these are even. • Using 3 numbers, the only factor is . This factor is even.

There are 4 even factors:


2 2n p= ⋅ ⋅

2 2, and 2 p⋅ ⋅2 2 p⋅ ⋅

( ) ( ) ( )2, 2 2 , 2 , and 2 2p p⋅ ⋅ ⋅ ⋅

The point is rotating 10 centimeters away from the center, so the path covered is a circle of radius 10. Each revolution, the point covers a circumference, C.

( )2 2 10 20C rπ π π= = →

We know that the point covers 20π centimeters each revolution. To find out the distance traveled in 15 seconds, we need to find the number of revolutions covered in this timeframe.


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Solution B

Alternatively, we could pick a number that has the property defined ( , where p is a prime greater than 2) and find its even factors.

Our plug-in

The even factors of 12 are 2, 4, 6, and 12.

Our solution There are 4 even factors





Solution

Standard deviation is a measure of how tightly/loosely values are grouped around the average. The more dispersed the values, the higher the standard deviation. Data set II has no dispersion at all (standard deviation of 0), while data set I’s values are more tightly grouped than III’s values are.

Even if you know how to do it, do NOT calculate standard deviation. This takes more time than you have on the GMAT.



you need to have a good understanding of what standard deviation means. Broadly, standard deviation is a measure of how tightly/loosely values are grouped around the average. The more dispersed the values, the higher the standard deviation


4n p=

4 3 12n = ⋅ =

212 2 3x x= → = ⋅


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Solution

Imagine that you are the manager, and you want to please as many people as possible

Team A has that need to be filled.

want to be on Team A

Team B has that need to be filled.

want to be on Team B

All 15 people who want to be on B will get their wish, and so would the first 20 who want to be on A. The remaining 15 people who also wanted to be on A will be disappointed however, because there is no more space in A, and they will have to go to B.


Take-Aways To quickly find a percentage, find 10% and adjust. For example, to quickly find 40% of 50, find 10% – which is

5 – and multiply it by 4 to get 40%.



“The first 10 multiples of 5” is an arithmetic sequence – the difference between any value and the next greater value is the same. As a rule, the average of an arithmetic sequence is always equal to the median of the sequence.

There is no difference between m and M.


Solution B

To find the average of an evenly distributed set (an arithmetic sequence), average the highest and the lowest values. In an evenly distributed set, the difference between each value and the next higher value is always the same. Some examples include:

• A set of consecutive integers: 12, 13, 14… Difference is 1 • A set of consecutive even integers: 2, 4, 6… Difference is 2 • A set of multiples of five: 5, 15, 20… Difference is 5.

In this case, the average of our set is .

40% 50 20 slots⋅ =

70% 50 35 researchers⋅ =

60% 50 30 slots⋅ =

30% 50 15 reasearchers⋅ =

5 50 27.52

m += =


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Since there are 10 numbers, the median will be the average of the 5th and 6th numbers:



an arithmetic sequence. • The average of an arithmetic sequence is the average of its highest and lowest values. This average is

always equal to the median of the sequence. • The sum of an arithmetic sequence is its average multiplied by the number of numbers in the sequence


Solution A

Percent literally means “divided by 100”, “x is m percent of y” means (Eq. 1)

Percent is also , where is the fraction (for example ¼), and is multiplied by 100 to get the

percentage (for example 25%). To find “y is what percent of x”, we need to manipulate (Eq. 1) to get .

(Eq. 1)

Divide by y

Multiply by 100



( ) ( )5 5 6 527.5

2M

+= =

27.5 27.5 0M m− = − =

100mx y=

part 100whole

⋅part

whole

100yx⋅

100mx y=

100 100

x m yy x m= → =

10,000100yx m⋅ =


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Our plug-ins

Picking easy values here takes some careful planning. The values must agree with the information “x is m percent of y” In our case, this becomes “2 is 50% of 4”.

Now that we’ve plugged in, let’s find our solution. “y is what percent of x” becomes “4 is what percent of 2”?

Since 4 is twice 2, it is 200% of 2. You can find this by setting up and solving

Our solution 200

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variables in each answer with our plug-in until we find the answer that equals 200.

(A)

(B)

(C)

(D)

(E) This matches our solution!



the question, solve and then find out which answer choice agrees with your solution. Avoid plugging-in 0, 1, or the same value for different variables (unless the variables are equal). By coincidence, two answers may agree with your solution. In that case, just plug in another set of easy numbers and try again


2x = 50m = 4y =

4004 2 200100 2

p p= ⋅ → = =

( )100 100 50 200m = ≠

( )1 1 200

100 100 50m= ≠

( )1 1 200

50m= ≠

( )10 10 200

50m= ≠

( )10,000 10,000 200

50m= =


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Solution

This is a pattern problem. To solve patterns, it is always best to draw out what is happening as this will increases your chances of spotting the pattern. Do the division until the decimals repeat themselves in sequence:

After the decimal point, the odd-numbered digits (1st, 3rd, 5th digits…) are 5, so the 25th digit must be a 5.





Solution A

At the start, they each receive x dollars. However, because John worked 10hrs vs Mary’s 8hrs, Mary gave him y dollars so that they would have the same hourly pay. At the end of the day, John will have dollars for 10hrs of work while Mary will have dollars for 8hrs of work.

Because their hourly pay is the same, must be equal for Mary and John. This allows us to write

the following equation:

John’s hourly wage equal Mary’s

We want the dollar amount John was paid in advance – this is x dollars. Let’s solve for it.

Cross multiply

Isolate x




6 0.545454...11

=

x y+x y−

amount earned# of hours

10 8x y x y+ −

=

8 8 10 10x y x y+ = −

18 2 9y x x y= → =


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In this case, the plug in process is quite tricky. If we tried to plug in directly for x (how much they were each paid in advance), we would have to find out how much Mary gave John so that their hourly wages are equal. The math could become ugly quickly. A safer policy is to plug in for their common hourly wages, and thus indirectly find out how much money they were paid in advance.


Our plug-in Hourly wage is $10 per hour

Let’s solve the question with our plug in.

Since they work a total of 18hrs, the total amount they must have been paid is . Because John

and Mary were given the same amount at the start, they each received half of this lump sum: .

However, Mary only worked 8hrs, and at her $10 per hour rate, she should have received $80. On the other hand, John worked 10hrs, and at the same rate, he should have received $100. The amount that Mary must give John so that they each receive what is fair is $10.

We’ve solved the question. “How much was John paid in advance?”

Our solution

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variable in each answer with our value of y ($10) until we find the answer that equals $90. Only answer E agrees with our solution.

(E) This matches our solution!





Solution

The y coordinate of R corresponds to the height of triangle ORP.

( )18 $10 $180⋅ =

180 $902

x = =

$10y =

$90x =

( )9 9 10 90y = ⋅ =


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Since the area (12) is , . We already know that the base is 4 (distance from

O to P), so the height must be 6.


#206, pg. 181 Difficulty Level: 700+ Topics: Translations & Manipulations; Rates & Work


Draw out what is happening to help you solve. Because A is going 8 miles per hour faster than B, A will reduce the gap between them by 8 miles each hour.

To get 8miles ahead of B, A will take between 3 and 4 hrs.


Solution B – Intuitive Solution

The rate/work formula is . We are asked to solve for time, so we should rewrite the

formula as .

( )( )2

base height( )( ) 24base height =

distance rate time= ⋅distancetime

rate=

At the start, A is 20 miles behind B

One hour later, A is 12 miles behind B

2hrs after the start, A is just 4 miles behind B

3hrs after the start, A is now 4 miles ahead of B

4hrs after the start, A is 12 miles ahead of B

A

A

A

A

A

B

B

B

B

B


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To find out how quickly A can get 8 miles ahead of B, only the difference in their speeds matters. Whether A and B are traveling at 58mph and 50mph, or traveling at 108mph and 100mph, it is the 8mph difference that will determine how quickly A catches up. Understanding this greatly simplifies the problem.

In the equation avove, distance is the gap that A must close (20 miles) plus how far ahead of B it must go (8 miles). .

In the equation above, rate is much faster than B, A is traveling. .

Solve


Solution C

The key to solving complex rate is to spot the relationships between the three parts of all rate/work equations: Distance, Rate, and Time. The table below compares these parts for both cars:

Car A Car B Notes Rate 58 50 Time T T A & B will travel for the same period Distance 58t 50t Distance is Rate times Time

Since A starts 20 miles back and must finish 8 miles ahead, we can also say that A’s distance must be 28 more than B’s distance


Take-Aways • To solve complex rate questions, try to draw out what is happening to help you visualize the problem.

The key to solving these questions is to spot the relationships between the three parts of all rate/work equations: Distance, Rate, and Time



If today’s production had been 50 units, the average would remain at 50. However, the extra 40 units (today’s

production is 90 instead of 50) raised the average by 5. Since average is , these extra 40 units must be

28distance =

8rate =

28 3.58

distancetimerate

= = =

58 50 28 8 28 3.5t t t t= + → = → =

total# of days


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divided into 8 days because . However we’re not looking for the total number of days; we’re looking

for the number of days before today.


Solution B

Use the weighted average formula. The overall average is 55. The average of the first n days is 50, and the value of the last day is 90:

Weighed avg formula

Cross multiply

Isolate n and solve




Let’s reverse engineer the question and figure out how many days there must be in order to raise the average to 55 with the addition of a 90 units production day.

What’s n?

Units in n days

New average after a 90 units day

should be 55

Notes

(A) 30 -- -- (B) 18 -- --

(C) 10 500 Average did not go up enough. Use fewer days

(D) 9 450 Same as above. On the test,stop here. Only E can work.

(E) 7 350 Yes! Adding a 90 units day raises the average to 55.


4058

=

( ) ( )50 1 9055

1n

n+

=+

50 90 55 55n n+ = +

35 5 7n n= → =

( )50 n⋅ total unitstotal days

500 90 5311+

≈

450 90 5410+

=

350 90 558+

=


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#208, pg. 182 Difficulty Level: 700+ Topics: Translations & Manipulations; Exponents & Roots

Solution A

Replace x with 1/x

Multiply top and bottom by x

The problem becomes difficult because the expression above isn’t in the answer choices, so we must manipulate it to look like the right answer. Because we are dealing with a value squared, the expression above is the same as



30 Seconds Hack

Approximate

21 1

1 1

x

x

+ −

211

xx

+ −

22 21 1 1 1 (1 ) 1

x x xx x x

+ + + − → → − − − −


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Guess D

Solution

In any given triangle, the sum of two adjacent angles is always equal to the supplementary of the 3rd angle. For example, in the 3 triangles below,

In our question, the measure of angle z is actually irrelevant. The sum of the two adjacent angles inside the smaller triangle is equal to x

a b c+ =

y

z

x

180 – y

90

( )180 90y x− + =

y

z

x

Each of these angles is about 45 degrees. Since a straight line is 180 degrees, y and x are each about 180 – 45 = 135.

a

b c a

b

c

a

c

b


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Take-Aways • In any given triangle, the sum of two adjacent angles is always equal to the supplementary of the 3rd angle.

• Most geometric figures are drawn to scale. When stuck, you can approximate lengths, angles, etc…


Solution A – Check Points

The drawing gives us two points on the line: (0, 2) and (3, 0). Let’s evaluate each answer choice; for each point, we could plug in one coordinate and see whether the equation gives us the 2nd coordinate.

When x is 0, y is… (should be 2)

When y is 0, x is… (should be 3)

Notes

(A) -- Stop. This answer failed the first check. (B) This equation fits the two known points

(0,2) and (3,0) (C) Don’t bother, we have a match (D) Same as above (E) Same as above


Solution B

The standard format for the equation of the line is where m is the slope and b is the y-intercept. To build the equation of the line, we need to know the y-intercept (point where x = 0), and we need to know the slope of the line

• When x is 0, y = 2, so the y-intercept is (0,2). With this info, we can refine our equation from to

• The slope is as the line travels from one point to the next. As our line goes from the two

given points (0,2) to (3,0), y decreases by 2, while x increases by 3. As a result, our slope is . With

this info, we can refine our equation from to

( )180 90 270 270y x y x x y− + = → − = → = +

( )2 0 3 6 2y y− = → = −

( )2 0 3 6 2y y+ = → = ( )2 3 0 6 x 3x + = → =

y mx b= +

y mx b= +2y mx= +change in ychange in x

23−

2y mx= + 2 23

y x= − +


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To compare our equation to the answers, we must put x and y on the left side, and get rid of the fraction:


Take-Aways • The standard format for the equation of a line is where m is the slope and b is the y-intercept.


Solution

All two digit numbers can be expressed as the sum of the tens digit times 10 plus the units digit. For example,, and . This becomes important when we’re asked to manipulate unknown digits.

Our unknown integer is where T is the tens digit, and U is the units digit. Reversing the digits results in. Because the two integers differ by 27, we can setup an equation and find how much the digits differ

(find T – U or U – T):

Difference is 27




30 Seconds Hack

2 22 2 2 3 63 3

y x x y x y= − + → + = → + =

y mx b= +

24 2 10 4= ⋅ + 93 9 10 3= ⋅ +

10T U+10U T+

( ) ( )10 10 27T U U T+ − + =

9 9 27 3T U T U− = → − =

Approximate:

k is the distance from the center to O. The radius stops at the edge of the circle. The radius is more than ½ of k, but it is shorter than k

O


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The answers are ordered from greatest to least. To pick an answer between k/2 and k, cross off A, D, and E

Guess B or C

Solution

Because the circle is tangent to the axes, we can build a right triangle for which segment OC is the hypotenuse and for which the base and height are the radii of the circle





Solution

The trick here is to not interpret “r is the combined resistance” as “r is the sum”. In this problem, you are told specifically how the resistances are combined: “the reciprocal or r is the sum of the reciprocals of x and y.” Set up the corresponding equation and solve for r

Reciprocal or r is sum of reciprocals of x and y

Find a common denominator (xy)

Invert the fractions to find r

1 1 1r x y= +

1 1 y x x yr xy xy r xy

+= + → =

xyrx y

=+

O

k r

r

C

According to the Pythagorean Theorem, 2 2 2 2 2 2r r k r k+ = → = .

Divide by 2, and take the square root to find 2

2 2 2k kr r= → =


Pag

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30 Seconds Hack

The probability that Xavier alone solves the problem is only 1/4. So the chances that all three people do exactly as predicted will be rather small. Cross off A and B

Guess C, D, or E

Solution

Keep two rules of probability in mind:

Probability of two successive independent outcomes is the probability of one times the probability of the other, and this can extend to any number of outcomes.

Probability of an outcome is 1 minus the probability that the outcome does not happen , so

the probability that Zelda does not solve is 1 – the probability that she does

The probability that Xavier and Yvonne get it right but not Zelda is


Take-Aways • When solving a probability question, you can often use your intuition to eliminate answers that are way off.


30 Seconds Hack

( ) ( ) ( )&P A B P A P B= ⋅

( ) ( )1P A P notA= −

5 318 8

− =

1 1 3 34 2 8 64

=


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Because 0 is not allowed at the bottom of a fraction, x cannot equal any value which would make a denominator 0. x cannot be 0, -1, or -4. Cross off A, B, and E

Guess C or D

Solution A

Multiply by to get rid of denominators

Factor out (x+4) from the left side

Simplify & remove the brackets





Let’s reverse engineer the question and figure out which answer choice gives a value of x that fits in the equation given.

What’s x

Is

Notes

(A) 0 -- x cannot be 0, otherwise 1/x would be undefined

(B) -1 -- x cannot be -1, otherwise, 1/(x+1) would be undefined

(C) -2 Yes! This value of x makes the equation true.

(D) -3 Don’t bother, we have a match (E) -4 Same as above




( )( 1) 4x x x+ + ( )( ) ( ) ( )1 4 4 1x x x x x x+ + − + = +

( ) ( ) ( )4 1 1x x x x x + + − = +

( )( ) ( )4 1 1x x x+ = +

2 24 4 2x x x x x+ = + → = → = ±

1 1 1 ?1 4x x x

− =+ +

( ) ( ) ( )1 1 1 1 1 12 2 1 2 4 2 2

− = → − + =− − + − +


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Solution

Take a look at the answer choices to see the format you should aim for. We need to put all fractions over a common denominator before we merge them. Two rules of exponents are important to remember:

• . For example,

• . For example,

Original expression

Use common denominators

Simplify exponents by multiplication

Combine terms


#217, pg. 183 Difficulty Level: 700+ Topics: Translations & Manipulations; Number Properties

Solution

The key phrase is “the product of the point values”. To find this product, each time a bead was removed, we multiplied its points by the points from the other beads. For example, suppose one green and two blue beads are removed; the green is worth 3 points and each blue bead is worth 2 points. The product of the point values of these three beads is . In other words, to find how many red beads were removed, we need to find out how many times the number “7” appears in the prime factorization of 147,000

m n m nx x x +⋅ = ( )2 62 6 45 5 5 5+ −− −⋅ = =

( )nm mnx x= ( ) ( )6 2 62 125 5 5− ⋅ − −= =

3 2 11 1 12 4 16

− − −

2 13 2 41 1 12 2 2

− −− =

3 4 41 1 12 2 2

− − − =

( ) ( ) ( )3 4 4 111 12 2

− + − + − − = =

3 2 2 12⋅ ⋅ =

147,000 1,000 7 21 1,000 3 7 7= ⋅ ⋅ = ⋅ ⋅ ⋅


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We can actually stop here because there is no factor of 7 inside 1000. The number “7” appears only twice in the prime factorization of 147,000, so two red beads were removed



Solution A

Since the fraction is equal to 1, the top must equal the bottom:

Numerator equals denominator



Solution A

Before going to the choices, consider what the information means about your integers. They are consecutive, so if a is the smallest, and . In addition, at least one of the three is even.

I. . I MUST be true. II. abc is either (even)(odd)(even) or (odd)(even)(odd). Either way, the result is even. II MUST be true

III. is the average of the 3 consecutive integers. This will always be the middle value and so will

always be an integer. III MUST be true


Solution B

Another option is to plug in values for your consecutive integers and test the options. This technique isn’t as good as the previous solution because we cannot show by trial and error that something MUST be either true or false. We can only demonstrate that it can be true or false. However, when the algebra isn’t clear, this demonstration is better than blind guessing. To improve your odds, try two or three sets of values. We’ll try

and

2 22 1 1 2yy y

= + → = → =

1b a= + 2c a= +

( )2 2c a a a− = + − =

3a b c+ +

1, 2, 3a b c= = =4, 5, 6a b c= = =


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I. For both sets, c – a = 2. II. abc is 6 for the 1st set, and 120 for the 2nd set. Either way, it is even.

III. is 2 for the 1st set, and 5 for the 2nd. Either way, it is an integer.



30 Seconds Hack

If we are to guess blindly here, we should definitely not pick C, because it would be too straightforward and too easy to say that a 25% increase in wage would be offset by a 25% drop in number of hours. Cross off C

In addition, you may be able to tell that if the wage had doubled, the employee would need to cut his hours in half (50% reduction) to keep his income steady. Since the wage hasn’t increased that much, he will need a smaller reduction in the hours worked. Cross off D and E.

Guess A or B

Solution A



Income is . If hourlyWage goes up by 25%, the new income would be

. To cancel off this percentage increase, we will need to multiply hours by

as this will bring income back to the original:

. So to adjust the hours, we multiplied them by 80%, which is the same as reducing them by

20%


3a b c+ +

100100

x+

100100

y−

( )( )hours hourlyWage

( )( ) 125100

hours hourlyWage

100125

( )( ) ( )( )125 100100 125

hours hourlyWage hours hourlyWage =

100 4 80%125 5

= =


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e240

Solution B

Original income is . After the 25% increase in hourly wage, the new income is

. To find out the percentage by which we must reduce hours to get back to

the original income, solve for x in the following:

Divide by

x is 0.8, so we had to multiply hours by 0.8, which is the same as reduce hours by 20%



Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem. If you work correctly, whatever you plug in will reduce to the same answer.

Our plug-ins

With these values, the starting income is . Once hourly wage goes up by 25% (to $10), the higher income would be . To keep the income at the $800 level, the hours must be reduced to 80hrs, because .

The reduction of hours from 100hrs to 80hrs is a 20% reduction.

Our solution 20%




#221, pg. 184 Difficulty Level: 700+ Topics: Translations & Manipulations; Sets & Groups;

Solution – Group Formula

( )( )I hours hourlyWage=

( )( )1.25newI hours hourlyWage=

( )( ) ( )( )1.25x hours hourlyWage hours hourlyWage=

( )( )hours hourlyWage 1 41.25 1 0.81.25 5

x x= = = =

100hours = 8hourlyWage =

100 8 $800⋅ =100 10 $1000⋅ =

80 10 $800⋅ =


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e241


.

Plug in the given data: 200 total, 130 in Chem, 150 in Bio, at least 30 in neither

Put in data

Add both, subtract 200

Add terms

Both answers D and E have the correct minimum both, but what is the maximum? Well, there cannot be more students majoring in both Chem and Bio than there are majoring in Chem alone, so the maximum number majoring in both is 130.



Put the info given into the Venn Diagrams

All regions must add up to the total of 200 students:

Sum of regions is 200


Total Chem Bio neither both= + + −

( )200 130 150 at least 30 both= + + −

( )130 150 200 at least 30both = + − +

( )80 at least 30 110both both= + → ≥

110 130both≤ ≤

( ) ( ) ( )130 150 at least 30 200x x x− + + − + =

Bio Chem

x

Always start with the center. Use a variable if you don’t know the value. People belonging to neither are outside both circles.

Bio Chem

x

The left-most region is for Chem majors only. This includes everyone who majors in Chem (130) minus those who major in both, hence 130-x. The same logic yields 150-x in the right-most region. Finally, “at least 30” do not major in either.

130-

150-x

neither

At least 30


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Combine numbers

Isolate x

Both answers D and E have the correct minimum of 110, but what is the maximum? Well, there cannot be more students majoring in both Chem and Bio than there are majoring in Chem alone, so the maximum number majoring in both is 130.


Take-Aways • The group formula, , is a useful way to solve questions about



30 Seconds Hack

We want to determine how many values of x will fit into the equation 65 xx

− = . Notice that when we cross

multiply to get rid of the fraction, we will have an 2x . All quadratics (equation with 2 as the highest exponent) can have at most 2 solutions. Eliminate D or E.

Since this equation is not of the form 2 negativex = , there will be at least 1 solution. Eliminate A.

Guess B or C

Solution

Solve for x. Get rid of fractions by multiplying both sides by x:

2 26 65 5 5 6x x x x x xx x

− = → − ⋅ = → − = .

To solve quadratics, set them to zero and factor: 2 25 6 0 5 6x x x x− = → = − +

Let’s quickly review how to solve a quadratic.



( )280 at least 30 200x− + =

( )80 at least 30 110x x+ = → ≥

110 130x≤ ≤


2 0x bx c+ + =

m n b+ =m n c⋅ =


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The equation could then be factored into and the solutions will be .

Let’s apply this to our question. To solve 2 5 6 0x x− + = . The two numbers we want must add up to -5 and multiply to 6. These numbers must be -2 and -3.

( )( )2 5 6 0 2 3 0 2 or 3x x x x x− + = → − − = → = . There are two possible values of x.



30 Seconds Hack

X is 40% ryegrass, Y is 25% ryegrass, but the mixture is only 30% ryegrass. We can conclude that because the overall mixture is much closer to Y in its concentration of ryegrass, this mixture is mostly Y and contains only a little X. Cross off D and E, and possibly C

Guess A or B, or possibly C

Solution A - Intuitive Solution

If the amounts of X and Y were equal in the mixture, the overall concentration of ryegrass would have been right in the middle of X and Y’s concentrations of ryegrass. Because the concentration of ryegrass in the mix (30%) is twice as close to its concentration in Y (25%, 5 away) as it is to its concentration in X (40%, 10 away), we can conclude that there is twice as much Y in the mixture as there is X.

The mixture is 1 part X, and 2 parts Y. So X makes up a third, or percent of the whole mixture.

Note: Whatever else X and Y may contain is irrelevant because it is only what they have in common that allows us to find out their relative quantity once they are mixed.


Solution B

The formula that would allow us to find the concentration of ryegrass in a mixture of X and Y is:

. The total quantity of ryegrass is the ryegrass in X (40% of X) plus the ryegrass in Y

2 0x bx c+ + = ( )( ) 0x m x n+ + = or x m x n= − = −

1333

total quantity of ryetotal quantity of mix


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(25% of Y). The total quantity of the mix is all of X plus all of Y. Finally, we are told that the concentration of ryegrass in the mixture is 30%. So we can build the equation below:

Concentration of rye in mixture is 30%

Cross Multiply

Isolate variables

Multiply by 20 to eliminate decimals

If X is ½ of Y, the mixture must be 1 part X, and 2 parts Y. So X makes up a third, or percent of the whole

mixture



Solution A

Before going to the choices, consider what the information means about our integers. Each integer is one more than the one before it, so we are dealing with a product of three consecutive integers.

(A) Our product is either (even)(odd)(even) or (odd)(even)(odd). Either way, the product must ALWAYS be even, so this answer is not correct.

(B) The product is always even, so this answer is not correct

(C) The product is always even, so this answer is not correct

(D) Multiples of 3 occur every third integer, so among three consecutive integer, one must be a multiple of 3. Since one of the factors is a multiple of three, the product must ALWAYS be divisible by three. This answer is not correct

(E) To be divisible by 4, a number must be divisible by 2 twice. If n is even, then n+2 is also even (divisible by 2). Thus, the product of all three would be divisible by 2 twice. If n is even, the product is divisible by 4. This answer is correct.

Notes: Note that answer E would be incorrect if the product was always divisible by 4, regardless of whether n is even. If n is odd however, the product’s divisibility by 4 depends on whether n+1 is divisible by 4.

0.4 0.25 0.3X YX Y+

=+

0.4 0.25 0.3 0.3X Y X Y+ = +

0.1 0.05X Y=

12 2

X Y X Y= → =

1333


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Solution B

Another option is to plug in values for your consecutive integers and test the choices.

(A) If n was 1, the product would still be even. n doesn’t have to be even. This answer is incorrect (B) If n was 2, the product would still be even. n doesn’t have to be odd. This answer is incorrect (C) As the previous examples show, the product is even whether n is even or odd. This answer is incorrect. (D) If n was 2, the product would still be divisible by 3. n doesn’t have to be odd. This answer is

incorrect. (E) If n was 1, the product would not be divisible by 4. On the other hand, if n were 2, or 4, the product

would still be divisible by 4. This answer is correct.

The drawback to this technique is that because a single counterexample could show that a statement doesn’t HAVE to be true, you can never prove by plug-in that the statement MUST BE anything. The best you can do is try 2 or 3 different plug-ins, eliminate answers that you’ve disproved, and hope that what is left is the right answer.



Solution

To easily compare different fractions, we ought to use a common denominator. Instead of marking the pipe in

fourths and thirds, we will mark it in and

321 ⋅⋅432 ⋅⋅

432 ⋅⋅

321 ⋅⋅

3 's12

4 's12

3

12

6

12

9

12

1 0

0 4

12

8

12

1

Yard is marked in fourths

Yard is marked in thirds

3

12

6

12

9

12

1 0

4

12

4

12

By merging the lines, we can see all distances between points: 3 1 2 2 1 3, , , , , and

12 12 12 12 12 12


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If we simplify the distances and remove duplicates, we’d be left with


Solution B

An alternative solution is to plug in for the length of the pipe, and find out which fractions of the pipe make up the distances between markings. Since the pipe will be marked in thirds and fourths, we need a length that will be easily divisible by 3 and 4.

Our plug-in length is 24

Marking the pipe of length 24 in fourths would mean markings at 6, at 12, at 18, and at 24. Marking the pipe in thirds would mean adding markings at 8, 16, and 24. Thus all markings would be at 6, 8, 12, 16, 18, and 24. The

lengths of different pieces (distances between markings) will be . If we simplify

these fractions and remove the duplicates, we’d be left with .



30 Seconds Hack

The two sides of the equation are equal. Since is smaller than 5, the power of 10 on the left side must be

greater than the power of 10 on the right side of the equation to makeup this difference. So

. Cross off C, D, and E

Guess A or B

Solution

It would be easier to divide the left side if 0.0015 was changed to 0.15. Well, we can do that by multiplying the left side by . Since we are dealing with an equation, we must do the same to the right side:

1 1 1, , and 12 6 4

6 2 4 4 2 6, , , , , and 24 24 24 24 24 24

1 1 1, , and 12 6 4

0.00150.03

710 10 1010

mm k

k−= >

210


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Original equation

Multiply both sides by

Simplify both sides



Solution A

The Official Guide shows a good algebraic solution. Below is another good solution.

The two given equations are: (Eq. 1) (Eq. 2)

(Eq. 1) + (Eq. 2) 2x a b= + (Eq. 3)

We can use this value of x to solve for y in either (Eq. 1) or (Eq. 2). Let’s use (Eq. 2)

Replace x with in (Eq. 2)

Simplify

We know from (Eq. 3) that 2x is a+b, and we’ve now found y, so we can solve for 2xy

Setup 2xy

Multiply



70.0015 10 5 100.03 10

m

k⋅

= ⋅⋅

210 7 20.15 10 5 10 100.03 10

m

k⋅

= ⋅ ⋅⋅

95 10 5 10 9m k m k−⋅ = ⋅ → − =

x y a+ = x y b− =

2a b+

2 2a b a by b y b+ +

− = → = −

2 2 2 2

a b b a by y+ −= − → =

( ) ( )( )2 22 2

a b a ba bxy a b xy+ −− = + → =

2 2

22

a bxy −=


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Our plug-ins

Since we can derive that and .

We are asked to find

Our solution

The final step is to find out which answer choice agrees with our solution. To do this, we will replace the variables in each answer with our values until we find the answer that equals 12.


(B)

(C)

(D)

(E)





Solution A

3x = 2y =

,x y a x y b+ = − = 5a = 1b =

( )( )2 2 3 2 12xy = =

12

2 2 2 25 1 122 2

a b− −= =

2 2 2 21 5 122 2

b a− −= ≠

5 1 122 2

a b− −= ≠

( )( )5 112

2 2ab

= ≠

2 2 2 25 1 122 2

a b+ += ≠


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A simpler way to think of an arithmetic sequence is a set of numbers for which the difference between each value and the next higher value is always the same. Some examples include:

• A set of consecutive integers: 12, 13, 14… Difference is 1 • A set of consecutive even integers: 2, 4, 6… Difference is 2 • A set of multiples of seven: 7, 14, 21… Difference is 7.

To simplify the question, let’s consider only the first 3 terms: p, r, and s. If the difference between terms is d, our sequence becomes . By using the same variable to describe all terms, we can manipulate the sequence more easily. Let’s evaluate the options

I. Using p, and d, the sequence is now . This is still an arithmetic sequence with a difference of 2d

II. Subtracting 3 from each term does not change the difference between the terms. There will still be a constant difference, and the sequence is still arithmetic.

III. Using p, and d, the sequence is now . Expanding the terms gives

. Note that the difference between the 1st and 2nd terms is

while the difference between the 2nd and 3rd terms is . Since the difference is not constant, the sequence is NOT arithmetic.


Solution B

Another option is to plug in easy numbers and test each answer choice. Let’s use 1, 2, 3, 4, 5. This is an arithmetic sequence with a constant difference of 1 between terms.

I. 2, 4, 6, 8, 10. This is also an arithmetic sequence II. -2, -1, 0, 1, 2. This is also an arithmetic sequence III. 1, 4, 9, 16, 25. The difference between two adjacent terms is no longer constant, so this is NOT an

arithmetic sequence.


#229, pg. 185 Difficulty Level: 700+ Topics: Combinatorics; Coordinate Geometry;

30 Seconds Hack

, , and 2p r p d s p d= + = +

( ) ( )2 ,2 ,2 2 2 , 2 2 ,2 4p p d p d p p d p d+ + → + +

( ) ( )2 22 , , 2p p d p d+ +

( ) ( )2 2 2 2 2, 2 , 4 4p p pd d p pd d+ + + +22 pd d+ + 22 3pd d+ +


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If we ignore all other restrictions, we can tell that there are 10 choices for x-coordinates of P, Q, or R. These choices are integers from -4 to 5. There are 11 choices for the y-coordinates, from 6 to 16. It is safe to say therefore that the first point drawn can occupy any of positions. We can also safely say that each of the other two points can occupy “several” positions each. Thus without doing the work, we know that the number of triangles possible (the number of ways to draw the 3 points) is where a and b are unknown, but “several”. The correct answer should be a fairly large multiple of 110. Cross off A, B, and D

Guess C or E

Solution

The number of triangles is equal to the number of ways we can draw P, Q, and R. Generally, the number of ways to do 2 things is the product of the number of ways to do each. Ways to draw P, Q and R equals

.

But how do we find the ways to draw each point? Thankfully this is straightforward. Drawing a point is akin to selecting an x and then a y coordinate. So the number of ways to draw a point is the product of the number of possible x and the number of possible y coordinates.

1. Find , and , so there are 10 possible x, and 11 possible y. Hence the number of ways to

draw P,

2. Find PQ must be vertical (parallel to the y axis), so the x coordinate of Q is the same as P’s. Thus, there is only one possible x for Q. On the other hand, the y coordinate of Q can be any of the 11 y-values within the range, with the exception of P’s (since they share x, if they had the same y, they would be the same point). This means that Q has 10 possible y values. Hence the number of ways to draw Q,

3. Find PR must be horizontal (parallel to the x axis), so the y coordinate of R is the same as P’s. Thus, there is only one possible y for R. On the other hand, the x coordinate of R can be any of the 10 x-values within the range, with the exception of P’s (since they share y, if they had the same x, they would be the same point). This means that R has 9 possible x values. Hence the number of ways to draw R,

The number of triangles possible is the number of ways to draw P, Q, and R. This is


10 11 110⋅ =

110 a b⋅ ⋅

P Q Rways ways ways⋅ ⋅

Pways4 5x− ≤ ≤ 6 16y≤ ≤

10 11 110Pways = ⋅ =

Qways

1 10 10Qways = ⋅ =

Rways

1 9 9Rways = ⋅ =

110 10 9 9900P Q Rways ways ways⋅ ⋅ = ⋅ ⋅ =


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Solution

Let’s call 14 15 16 172 2 2 2

5

− − − −+ + + “The Thing”. The question is asking: “The Thing” is how many times the

value of 172− ? To answer the question, it would make sense to try to simplify “The Thing”. We can get rid of

the negative exponents by multiplying the top and bottom by 172

14 15 16 17 17 3 2 0

17 17 17 17"The Thing" 2 2 2 2 2 2 2 2 2 15 3 5 2 5 2 5 2 2

− − − −=

+ + + + + +⋅ → → →⋅ ⋅

Now that we’ve simplified “The Thing”, we can rewrite 172− as 171

2. So the question is now much simpler:

173

2 is how many times the value of 17

12

? Without doing any more work, we can be sure that the first is 3 times

the value of the second.


Patrick Siewe - GMATFix.com OG Companion 12th Data Sufficiency

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Data Sufficiency Solutions

Solutions to Questions from pg. 273-288 in the Official Guide for GMAT Review, 12th Edition


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DATA SUFFICIENCY Answer Key

1. B 2. E 3. A 4. E 5. E 6. D 7. A 8. C 9. A 10. B 11. A 12. B 13. D 14. C 15. B 16. A 17. D 18. B 19. C 20. A 21. E 22. E 23. B 24. D 25. D 26. D 27. D 28. E 29. E 30. B 31. B 32. E 33. C 34. D 35. D

36. B 37. A 38. C 39. D 40. B 41. E 42. A 43. E 44. C 45. D 46. B 47. A 48. C 49. B 50. C 51. B 52. B 53. A 54. C 55. E 56. A 57. D 58. E 59. E 60. D 61. C 62. D 63. C 64. B 65. D 66. C 67. D 68. A 69. A 70. D

71. C 72. A 73. B 74. A 75. A 76. E 77. D 78. C 79. E 80. D 81. C 82. B 83. C 84. D 85. A 86. A 87. E 88. B 89. C 90. A 91. E 92. A 93. D 94. A 95. E 96. A 97. A 98. B 99. D 100. D 101. B 102. D 103. D 104. E 105. B

106. D 107. B 108. D 109. A 110. A 111. E 112. B 113. D 114. D 115. D 116. A 117. D 118. E 119. E 120. B 121. E 122. C 123. A 124. A 125. D 126. E 127. D 128. B 129. C 130. C 131. D 132. C 133. B 134. B 135. D 136. D 137. A 138. E 139. B 140. E

141. C 142. E 143. D 144. D 145. D 146. B 147. C 148. B 149. A 150. A 151. D 152. B 153. D 154. A 155. D 156. B 157. D 158. D 159. A 160. D 161. C 162. A 163. B 164. C 165. A 166. D 167. A 168. A 169. B 170. A 171. A 172. C 173. D 174. B


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The question doesn’t need to be rephrased. What is the absolute value of x?

(3) Absolute value is always 0 or positive, so if x is the negative of an absolute value, either x is 0, or x is negative such as -1 or -10. The statement tells us that 0x ≤ , but we cannot get a unique value for the absolute value of x


(4) 2 4 2 or 2x x= → = − . Whether x is 2 or -2, we know that 2x = . This statement gives us a

unique answer.




Our rephrase “what is ?”

(1) If the group is all women, then 5% of women have red hair would mean that 5% of the whole group have red hair. But if the group is mostly men, then 5% of women is a much smaller percentage of the whole group.


(2) Tells us nothing about the women.


MERGE STATEMENTS

By merging the two statements, it may seem like we have a lot more data, but is it sufficient? Consider the following two scenarios:

a. There are 100 men & 100 women. 5% of women have red hair, so 5 women would have red hair. Out of 200 people, that is 2.5% are women with red hair

red hair womenall men all women+


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b. There are 400 men & 100 women. 5% of women have red hair, so 5 women would have red hair. Out of 500 people, that is 1% are women with red hair.

This demonstrates that we still do not have a unique value for the percentage of people who are women with red hair






The probability that a boy be selected is

Our rephrase “What percentage of the students are boys?”

(1) Two-thirds are boys means 66.6% are boys. This is also the probability that a boy is selected at random. This answers our rephrase


(2) This tells us nothing about the percentage of the students that are boys. If this statement also told us how many students in total were present, we could find the percentage of girls and then solve for the percentage of boys.




# of boystotal students


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I recommend always using the group formula to deal with 2 overlapping groups. The group formula is 1 2Total Group Group neither both= + + − . In this case, T C B n b= + + − where C and B are the

number of chemistry and biology students. The question asks “both…is how much less than…neither”?

Our rephrase “What is n – b ?”

(1) There are 60 chemistry students. This gives us no way to determine how many students are enrolled in neither course or how many students are enrolled in both courses. So we have no way to find the difference between “neither” and “both”. 60C = . By putting this into the group formula we can rewrite it as 60T B n b= + + − . There is no way to find the value of n – b.


(2) There are 85 biology students. This statement is the mirror image of (1). There is no way to find out how many students are enrolled in neither course or how many are enrolled in both courses, so the difference cannot be found.


MERGE STATEMENTS

Together, the statements tell us that 60C = and 85B = . However, we don’t know how many of these students belong to both courses. In addition, we’re never given any information about other students, who are enrolled in neither course. Putting all we know into the group formula gives us 60 85T n b= + + − . Without know T the total number of students, we cannot solve for n – b.


Take-Aways

• The group formula, 1 2Total Group Group neither both= + + − , is a useful way to solve questions about 2 overlapping groups.


No rephrase needed. What is the distance from K to L?

(1) This statement doesn’t give any information about the location of K


(2) This statement doesn’t give any information about the location of L


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MERGE STATEMENTS

Together, the statements fix the distance between J and L (21) and between K and M (26), but they still do not fix the distance between K and L as illustrated in two possible configurations below



Take-Aways



is odd only if n is even.

Our rephrase “Is n even?”

(1) Consecutive integers alternate between even and odd. Since is even, is odd and n is even.


(2) Consecutive integers alternate between even and odd. Since is odd, n must be even




1n +

2n + 1n +

1n −

J K L M J K L M

21

26

21

26


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The annual rate of return is the annual percent growth or interest rate. The interest rate is 100interestcapital

⋅ . For

example, an investment that yields 20% in interest produces $20 for each $100 of capital. 20 100 20%

100⋅ = .

Rephrase “Which investment has a greater 100interestcapital

⋅ ?”

(1) J returns $115 per $1,000 of capital, so the interest rate of J is 115 100

1,000⋅ . The interest rate of K is

300 1002,500

⋅ . Since we have the data needed to find the interest rate of each investment, we have enough

information to determine which investment has a greater rate of return. It would be a waste of time to actually do the math.


(2) This tells us nothing about investment J




Let o and g be the number of orange and grapefruit crates respectively. Each orange crate costs $15. Each grapefruit crate costs $18. How many orange crates are there?

Rephrase “What is o?”

(1) This gives us a relationship between the number of orange crates and the number of grapefruit crates. Thus if we know one number we could figure out the other. However, without knowing either number there is no way to find the number of orange crates.


(2) The total cost of all crates is $38,700. There could be lots of orange crates and few grapefruit crates, but there could instead be lots of grapefruit crates and few orange crates. There are many ways to get to the $38,700 total and we have no way to know exactly how many orange crates were purchased.



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MERGE STATEMENTS

Each statement gives us one relationship between o the number of orange crates and g the number of grapefruit crates. In algebra, a relationship is an equation. Thus without doing any work, you can be pretty sure that you will have 2 equations and 2 variables and thus enough information to find the value of each variable. You can stop here if you understand. Continue to see the equations.

(1) tells us that “orange crates were 20 more than 2x grapefruit crates” 2 20o g= + . (2) tells us that the total cost is $38,700. Each orange crate costs $15, so the cost of oranges is 15o. Each grapefruit crate costs $18, so the cost of all grapefruits is 18g. Statement (2) can be translated to 15 18 38,700o g+ = . Because we have as many independent linear equations as variables, we have enough information to find o.


Take-Aways

• If you have as many independent linear equations as you have variables, you will be able to solve for all your variables. Equations are independent if it is impossible to manipulate them to make them look identical.


The question needs no rephrasing. $600 in savings. What is the total income?

(1) . As the figure demonstrates, the $600 saved are equivalent to one-sixth of the total income. Thus the total income is $600×6.


(2) Since Pat saved $600, he must have paid $1,200 in taxes, so his income must be at least $1,800. However because we don’t know how much money he actually had to spend (after taxes and savings), we cannot determine the total income.




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We are asked to find the rate at which the water is increasing in the tank. Water increases in the tank because it is pumped in faster than it is drained. If the inlet pipe rate is only slightly faster than the outlet, then the water will accumulate in the tank at a low rate. However, if the inlet pipe is much faster, the water will accumulate much faster. We can deduce that it is the difference between the rates that will determine the rate of increase of the water in the tank.

Our rephrase “How much faster is the inlet than the outlet pipe?”

(1) The amount of water in the tank at the start alone gives us no information on pipe rates.


(2) This statement directly gives us the rates of each pipe. Inlet pipe: 10 gallons per minute. Outlet pipe: 10 gallons per 2.5 minutes. Since we have both rates we can find the difference between rates and answer our rephrase.




Our rephrase “Is ?”

(1) Subract 9x from both sides to get . This answers our rephrase


(2) , so . We cannot compare x to 0, as it could be -2 or +2




0x <

0 x>

3 0x + > 3x > −


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To get an even sum, the two integers need to be either both even or both odd, because a mix of .

Our rephrase “Are i and j both even or both odd?”

(1) This gives us no information about j, so we don’t know enough about


(2) i and j are equal, so they must be both even or both odd. We can answer our rephrase




We know the exact cost of each minute: $0.42 per minute for the first three minutes and $0.18 per minute for the remaining minutes. Thus the length of the call will determine the total cost of the call. To find out how long the call lasted we need to be able to determine how much it cost.

Rephrase “What was the total cost of the call?”

(1) Since we know that the first three minutes cost $0.42 each, we can find out how much all three minutes cost. This statement tells us that the remaining minutes cost $0.36 less than the first three combined. With the cost of the first three minutes combined, we can find the cost of the remaining minutes (just subtract $0.36). Therefore this statement gives us a way to find the total cost and answer our rephrase.


(2) This statement directly answers our rephrase. There can only be a unique number of minutes that result in a cost of $2.88 adding or subtracting minutes would change the total cost.




We know the length of the bridge (0.5 miles). How long it takes Car X to cross depends on its speed.

even odd odd+ =

i j+


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Rephrase “What is the speed of X?”

(1) This gives us no information about speeds. However because X got unto the bridge 3 seconds after Y and exited the bridge only 2 seconds after Y, we know that X was on the bridge for 1 second less than Y. Still we don’t know whether the time they were actually on the bridge lasted minutes or hours.


(2) This gives us no information about Car X


MERGE STATEMENTS

Statement (2) gives us the speed of Y. Since we also know the length of the bridge (0.5 miles) we can determine

how long it took Y to cross the bridge using the rate formula: distancedistance rate time time

rate= ⋅ → = .

Statement (1) tells us that X spends 1 second less than Y on the bridge, so once we find Y’s time we can find X’s time by subtracting 1 second.




Let’s isolate what we’re looking for:


(1) We can plug in 10 for n in our rephrase to simplify it. Instead of “What is ?” we now need to answer “what is ?” Without m, we still do not have enough information.


(2) By subtracting 10 and n from both sides, we get . This statement answers our rephrase.




k m n= −

m n−

m n−10m −

10m n− = −


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The question needs no rephrasing

(1) Put the data in an equation and isolate x. . x is the double of an integer. The

double of an integer must be another integer, so x must be an integer. We can definitively answer the question


(2) Put the data in an equation and isolate x. . x is half of an unknown integer. If int

is even, then x will be an integer. However, if int is odd, then x will not be an integer. We cannot say with certainty whether x is an integer.


Take-Aways • When solving number properties, it’s often helpful to insert the known property as a variable in equations.

For instance, if you know that an unknown is an integer, you may refer to it as int. A prime number would be prime, an even number 2k or even, an odd number 2k+1 or odd, a multiple of 7 would be 7k… Doing so allows you to manipulate the unknown value while accounting for its property



The question needs no rephrase.

(1) The sum is 4 11 3 15P P P P+ + + + = + . Because this sum is even and 15 is odd, we can write: 3 3 3P odd even P even odd P odd+ = → = − → = . What values of P would make 3P odd? P cannot be even because any integer times an even equals even. P must be an odd number (1, 3, 5..). We have a definitive answer.


( )int 2 int2x x= → =

int2 int 2

x x= → =


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(2) This is very similar to (1) so it is probably sufficient. The sum is 3 11 3 8P P P P− + + + = + . This sum is odd and 8 is even, so we can write: 3 3 3P even odd P odd even P odd+ = → = − → = . This statement gives us the same information as (1) did. We can find the property of P.






The number of blocks that will fit inside the box depends on the size of the box.

Rephrase “What are the dimensions of the box?”

(1) We know that at least 25 blocks will fit in the box. However these 25 blocks only make up the bottom layer. We have no idea how many layers there are. If there are just two layers, then 50 blocks will fit in the box, but if there are more layers then more blocks will fit.


(2) This statement directly answers our rephrase. We know the exact dimensions of the blocks (from the prompt) and we’re now given the exact dimensions of box X. We could (but we shouldn’t waste the time to) find out how many blocks will fit in the box.




The question needs no rephrase. What is the 192nd term?

(1) First term is -40. Without knowing the relationship between terms, we cannot use the first term to find the 192nd term.


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(2) Because each term is 3 less than the preceding term, we can determine the difference between any two terms. Still we have no concrete values to work with, no starting point, so we cannot find the value of any term.


MERGE STATEMENTS

We know the 1st term (-40) and we know that each term is 3 less than the term before it. We can setup:

1st 2nd 3rd 4th 5th -40 40 3 43− − = − 43 3 46− − = − 46 3 49− − = − …

We could continue all the way to the 192nd term. We have all we need to answer the question




In a triangle, the greatest angle is opposite the longest side, so we need to find the longest side. We already know that QR is longer than PQ because . So to find the longest side, we need to know whether the last side, PR, is longer than QR. In other words, we need to know whether


(1) Since , we can say that . We can definitively answer our rephrase


(2) This tells us nothing about y so we cannot compare y to .



2x x+ >2y x> +

2y x> +

3y x= + 2y x> +

2x +


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Our rephrase “What is females in club

all club members?”

(1) This statement tells us that 40% of all females are in the drama club. It gives us the value of females in club 0.40

all females in school= . This is not what we’re asked for. We have no way to determine what

fraction of the club membership these females account for.


(2) This statement tells us that 25% of all males are in the drama club. It gives us the value of males in club 0.25

all males in school= . The statement gives us no information about the females at the school.


MERGE STATEMENTS

We know that 40% of females and 20% of males are in the club. However because we have no idea how large the groups of males and females are relative to one another, we have no way to compare 40% of females to 20% of males.

If there are a lot more females than males in the school, then the 40% of females who are part of the club will be a lot more than the 25% of males who are in the club. For example if there are 1,000 females and just 4 males in the school, then there will be 400 females and 1 male in the club (a large percentage of club members are female)

On the other hand, if there are a lot more males than females in the school then the 40% of females who are part of the club will be a fewer than the 25% of males who are in the club. For example if there are 10 females and just 1,000 males in the school, then there will be 4 females and 250 males in the club (a small percentage of club members are female).




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Rephrase The cost per ounce is cost

weight. “What is the cost-to-weight ratio for the family-size box?”

(1) This gives no information about cost.


(2) This gives us no information about the weight.


MERGE STATEMENTS

Although we know that the family-size is 10 ounces heavier than the regular-size, we have no way to find the weight of the family-size because we don’t know the regular-size weight. Thus we cannot find the cost-to-weight

ratio for the family-size. At best we can say that it equals 5.40

regular weight 10+




Profit rises with sales, though not proportionally. This means that we can be certain that every additional sale will increase profit. However, we cannot predict what the increase will be. Still, if we learn that profit reached the $4 million mark before sales reached 380,000, we would be sure that by 380,000 sales, the profit will have exceeded that $4 million mark.

Our rephrase “Did profit exceed $4 million on or before the sale of 380,000 units?”

(1) Because profit doesn’t rise proportionally with sales, we cannot predict what the profit will after 380,000 sales. This statement cannot answer our rephrase



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(2) This statement clearly answers our rephrase. We can deduce with certainty that when sales reach 380,000 units, profit will have gone even higher (because every additional sale increases profit)





(1) 2 2 21 1 n odd n odd n even− = → = + → = . For the square of n to be even, n itself must be an even number.

Another approach is to factor: ( )( )2 1 1 1n odd n n odd− = → − + = . This means that n-1 and

n+1 are odd, so n must be even.


(2) 3 4 3 3 3n even n even even n even even n even+ = → + = → = − → = . The only way to get an even product is to have at least one even factor. Since 3 is odd, n must be even.






We know how long Carmen works now (30hrs) and we know what the wage increase will be ($1.50). Because pay is a function of hours worked and hourly wage, the only thing we’re missing is the hourly wage. If we had Carmen’s hourly wage, we would be able to find her current pay and determine how many hours she would have to work with her new wage ($1.50 more) to keep the pay constant.

Rephrase “What is Carmen’s hourly wage?”


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(1) ( )( )pay hours wage= . We already knew that she works 30hrs. This statement tells us that her

current pay is $225. By plugging these into the pay equation, we can find her hourly wage and answer our rephrase.


(2) Since the wage increase of $1.50 is equal to a 20% increase, we know that $1.50 is the same as 20% of the wage: ( )1.50 0.20 wage= . With this information we can find the wage and answer our rephrase






In the given formula 900

1 2 tnc −=

+, n is the number of units, t is the month, and c is a constant (a value that never

changes). We’re asked to find the number of units for month 6. By replacing t with 6, we get 6900

1 2n

c −=+

. To

find n, all we need is the value of c.

Rephrase “What is c?”

(1) n is 180 when t is 1. Putting this in the equation gives us 1900180

1 2c −=+

. Notice that this is an equation

with only one variable. Because it’s a linear equation (the variable has no exponent), we can solve for the exact value of c. Of course it would be a waste of time to actually find c. It’s enough to determine that we have enough information to answer our rephrase.



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(2) This statement is the mirror image of (1). n is 300 when t is 2. Putting that in the equation gives us

2900300

1 2c −=+

. Once again, because c is the only variable in this linear equation, we can solve for it

and answer our rephrase.






(1) This tells us that the starting amount (dashed line) is at the halfway mark (drawing is obviously not to scale). As the figure shows, the starting amount is 3/7 of capacity plus 200 gallons. Since this is equal to

half of capacity, we can write: 3 12007 2

C C+ = where C is the capacity. This equation can be solved

for the exact value of the capacity.


(2) This statement tells us that to fill the tank back up to capacity would take 1,600 gallons. Since as the drawing shows the oil level is currently at 3/7 of capacity, filling the tank requires filling the remaining

4/7 of capacity. Thus we can write: 4 1,6007

C = and solve for the exact capacity.



The dashed line represents the starting oil level. We don’t know how much oil was in the tank at the start. After 200 gallons are removed, what’s left is 3/7 of capacity. To find out what the capacity is, it would be sufficient to know exactly where the dashed line is (how many gallons) or to know how many gallons correspond to 3/7 of capacity.


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Rephrase The average salary of all employees is sum of all salaries

all employees. What is this ratio?

(1) The company wide average will depend not only on employees of Division R but also on employees from other divisions. We need information about the other employees as well.


(2) The company-wide average salary depends on the salaries of all employees. Without any information about the salaries of division R, we cannot determine the company-wide average.


MERGE STATEMENTS

Employees of R earn $30,000 while the other employees earn $35,000. Thus the company-wide average will be between 30 and 35. Exactly what value the overall average has will depend on the sizes of the two groups of employees. We know that division R has 1,000 employees. If the other employees also number 1,000 then the average will be exactly in the middle of 30-35. However, if there are ten million employees outside of division R then the salaries of division R will be statistically insignificant to the whole employee body and the overall average will be close to the $35,000 that the other employees earn. To find the exact average salary, we need to be able to compare the sizes of the two groups.

Algebraically, if we call x the number of employees outside of R, the company-wide average is

( ) ( )1,000 30,000 35,000sum of all salaries R's salaries other salariesall employees all employees 1,000

xx

++= =+

. We cannot find this

value without x.




The surface of the band is determined by how tall it is (the value of x) and by how large the cylinder is (determined by the radius of its base)


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Our rephrase “What is the radius and what is the value of x?”

(1) We cannot determine the radius from x alone.


(2) The height of the tube neither gives us x, nor gives us the radius.


MERGE STATEMENTS

Together, the statements still give us no way to find how large the radius is.



#30, pg. 275 Difficulty Level: 500-600 Topics: Translations & Manipulations; Exponents & Roots


(1) At first glance, it may seem that n has to be 2, but if you notice that multiplying through would give a quadratic, you might suspect that there might be more than just one solution. Expand the left side to get

. By factoring, we get , so . We do not

have a unique value of n.


(2) Only , so and . We have a unique solution.



Take-Aways

• To solve quadratic equations (equations that contain a variable square) always set them to zero and factor. Other methods may cause you to miss one possible solution.

2 26 6 0n n n n+ = → + − = ( )( )3 2 0n n+ − = 2 or 3n = −

42 16= 2 4n = 2n =


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The question needs no rephrase. What is t in 0.43 7t ?

(1) Decimals that equal 0.44 when rounded to the nearest hundredth are 0.435 to 0.445. Since the hundredths digit of d is 3, we can be sure that 0.435 0.43 7 0.439t≤ ≤ so t can be 5, 6, 7, 8 or 9. We can’t find a unique value of t


(2) Decimals that equal 0.436 when rounded to the nearest thousandth are 0.4355 to 0.4365, so our decimal is in that range. Since our decimal ends with a 7, it can only be 0.4357. It cannot be 0.4367 because that falls outside the range. Because we know the exact value of our decimal, we have enough data to find t.




The amount paid in tax is 7% of the purchase amount for any item that cost at least $100. This means that to find out how much was paid in taxes, we need to know the prices of the items that cost at least $100.

Rephrase “What is the total price of all items that cost at least $100?”

(1) Coat = $125. Since $365 was spent on the 7 items, the remaining 6 items cost $240 total. We can find the tax paid on the coat (since the coat cost at least $100). However, because we don’t know whether any other item cost at least $100 there is no way to know whether more tax was paid on another item. We can’t find the total tax.


(2) Since the 6 items average $40, their sum must be 6 40 $240⋅ = . Since $365 was spent on the 7 items, with this statement we can find the price of the coat. However, because we don’t know whether any other item cost at least $100 there is no way to know whether more tax was paid on another item. We can’t find the total tax.


MERGE STATEMENTS


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The statements tell us exactly the same thing (coat = $165, other 6 = $240 total) so putting the statements together will not allow us to find any new information.



Take-Aways

• In data sufficiency, when the data from one statement can be logically deduced from the other statement, C cannot be the right answer. When one statement offers the same data (no more, no less) as the other statement, the correct answer must be D or E.


The question needs no rephrasing. What is cost?

(1) Profit is 20% of selling price. This gives us no concrete dollar amount with which to work. We could have Selling Price = $10 and Profit = $2, but we could also have Selling price = $100 and Profit = $20. The cost of the appliance would be different in each case.


(2) The difference between the selling price and the purchase price is $50. This is the profit. Knowing the profit alone doesn’t tell us what the purchase price (cost) was. The appliance could have cost $10, $100 or $1,000. All we know is that it was resold at a $50 profit.


MERGE STATEMENTS

Statement (1) tells us that profit is 20% of the selling price. Since the cost and profit add up to the selling price, this means that the cost is 80% of the selling price (see picture below). So cost is 4 times profit. Statement (2) tells us that profit is $50. Since the cost is 4 times the profit, the cost must be 50 4 $200⋅ =


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k is the number of cylindrical cans that will fit in the carton. Since we know the exact size of the carton as well as the height of the cans, the only unknown that will affect how many cans can fit is how wide the cans are. How wide is determined by the radius of the cans.

Our rephrase “What is r, the radius of the cans?”

(1) This directly answers our rephrase.


(2) We know the length of the carton (48). The distance from edge to edge of a can is its diameter, d. “6 of the cans fit along the length” means that . We can find the diameter, so we know the radius. The statement answers our rephrase






Variables x and t stand in the way of finding z in the 1st and 3rd equations respectively. If we had one of these values, we would just have to plug it in the appropriate equation to find z.

Our rephrase “What is x?” OR “What is t?”

(1) This directly answers our rephrase



6 48d =


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Definition: ( ) nf n a= To apply, replace n with the values that take its place. (a is a constant so its value doesn’t

change)

Application: What is ( )1f ? Just replace n with 1 in the definition: ( ) 11f a a= =

Rephrase “What is a”?

(1) ( )2 100f = . By putting 2 for n in the definition, we can determine that ( ) 22f a= . Thus this

statement is really telling us that 2100 10 or 10a a= → = − . We have two possible values of a. Since we don’t know which is the actual value, we cannot answer the question.


(2) ( )3 1,000f = − . By putting 3 for n in the definition, we can determine that ( ) 33f a= . Thus this

statement is telling us that 31,000 10a a− = → = − . Because there is only one value of a that can satisfy the equation, we have answered our rephrase.





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(1) Markup is 25%, or a quarter of costs. Putting this in the diagram shows that the markup is clearly one fifth, or 20% of the sell price.


(2) Selling price is $250. This tells us nothing about the markup (except that the markup is less than $250). The markup could be $25 (10% of selling price) but could also be $50 (20% of sell price)



Take-Aways • One great technique for solving ratios is to express all parts of the ratio in terms of the same variable. For

instance, if a problem tells you that the ratio of boys to girls is 4:5, you should use 4x and 5x as the number of boys and girls. By writing all values in terms of the same variable, you greatly simplify the problem (in general, the fewer variables the easier a problem will be to solve).


The ratio of population to representatives is 1

1

pr

for district 1 and 2

2

pr

for district 2. The question is, which of

these is greater? This question can be written as “is ratio 1 > ratio 2?” or as “is ratio 2 > ratio 1?” Either rephrase

Selling price is the sum of cost & markup. We’re looking for markup as a percentage of price.


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is acceptable because if you have enough information to answer 1, you’ll be able to answer the other. “Is

1 2

1 2

p pr r> ?” Cross multiply to get rid of fractions:

Rephrase “Is 1 2 2 1p r p r> ?”

(1) 1 2p p> This tells us nothing about the r’s, so we cannot compare the two sides in the rephrase.


(2) 2 1r r> This tells us nothing about the p’s, so we cannot compare the two sides in the rephrase.


MERGE STATEMENTS

Since 1 2p p> and 2 1r r> the left side of our rephrase “Is 1 2 2 1p r p r> ?” has the two larger values. Because all

values are positive, we can be sure that the product of two larger positive numbers ( 1 2p r ) will be greater than the

product of two smaller positive numbers ( 2 1p r ). The rephrase can be answered with a definitive YES.


Notes: In rephrasing, the reason we can go from 1 2

1 2

p pr r> to 1 2 2 1p r p r> is that we know all values are positive

(they represent number of people). If 1r or 2r could be negative, we would have to account for the possibility

that cross multiplying might flip the direction of the inequality sign.

When combining statements, we only have enough information because all values are positive. If values could be negative, 1 2p p> and 2 1r r> would not necessarily lead to 1 2 2 1p r p r> . For example, 1 2− > − and

3 4− > − but ( )( )1 3− − is not bigger than ( )( )2 4− −



There are 80 people, some are grads and the rest are non-grads (no one can belong to both groups so the sum of the groups must equal the total of 80. How many grads are there?


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(1) The ratio of non-graduates to graduates is 3 to 1. We can use 3x and x to represent each group.


(2) There are 40 more non-grads than grads. We can use x + 40 and x to represent each group.



Take-Aways • One great technique for solving ratios is to express all parts of the ratio in terms of the same variable. For

instance, if a problem tells you that the ratio of boys to girls is 4:5, you should use 4x and 5x as the number of boys and girls. By writing all values in terms of the same variable, you greatly simplify the problem (in general, the fewer variables the easier a problem will be to solve).


Our rephrase “What is x?” OR “What is the distance between R and T?”

(1) This tells us nothing of the distance between R and T.


(2) . We can find a unique value for x.



1.5 69 1.5TU RT x= → =

The diagram clearly shows that the total, 80, is equal to 4x. So we can write 80 4x= and solve for x, the number of graduates

Since the x grads and x + 40 non-grads total 80 people, we can write 40 80x x+ + = and solve for x, the number of graduates


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The question needs no rephrasing

(1) x does not have a unique value. It could be 12 or 30.


(2) x does not have a unique value. It could be 12 or 30.


MERGE STATEMENTS

Together, the statements are no more useful than they were alone. Specifically, a number whose sum of digits equals 3 MUST BE divisible by 3, so statement (2) is already implied by statement (1). x can be any two-digit number whose digits add up to 3.



Take-Aways • In data sufficiency, when the data from one statement can be logically deduced from the other statement,

C cannot be the right answer. When one statement offers the same data (no more, no less) as the other statement, the correct answer must be D or E.


The question needs no rephrasing. We’re looking for the value of r


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(1) “The ratio of t – r to r is 0.15” means 0.15t rr− = . In addition we know that 0.3t r− = . Since we

have 2 linear independent equations and we have 2 variables, we can solve for the value of each variable. Thus we can find out what r is (just replace t – r with 0.3 in the ratio)


(2) The area of the cross section is the shaded area. This is the difference between the big circle and the small circle. Knowing this area cannot help us find the circles because the same 1.29π difference can exist between two pretty circles or between two pretty small circles




The tenth digit is the digit immediately after the decimal point. The question is already simplified

(1) . The tenth digit could be 3, 2, 1 or anything else, if the number is negative


(2) . The tenth digit could be 2, 3…


MERGE STATEMENTS

Combining the statements gives . The tenth digit is could still be 2 or 3.



0.33n <

0.25n >

0.25 0.33n< <

In these circles, the difference between the areas of the larger and the smaller circle is the same. However the radius of the small circle is not the same. Knowing the difference in areas is not enough to tell us the exact sizes of our circles.


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Take-Aways




According to the prompt, xrateXrrateZ

= and yrateYrrateZ

= .

Rephrase “Is rateZ the greatest?”

(1) x yr r< . This notation is not helpful because we’re interested in specific rates, not in xr or yr . We

should rewrite the statement as rateX rateYrateZ rateZ

< . Because rates are positive, we can get rid of the

denominators by multiplying both sides by rateZ. We would end up with rateX rateY< . Thus we know that Y’s rate is greater than X’s, but we have no idea where Z fits, so we can’t answer the question.


(2) 1yr < . Because we’re interested in specific rates, not in yr , we should rewrite the statement as

1rateYrateZ

< . Because rates are positive, we can cross-multiply by rateZ and get rateY rateZ< . Thus

we know that Z’s rate is greater than Y’s, but we cannot determine whether Z is the greatest because we don’t know where X fits.


MERGE STATEMENTS

We learned in (1) that Y is faster than X. Statement (2) told us that Z is faster than Y. Merging the statements, we can be certain that rateX rateY rateZ< < , so we know that Z has the greatest rate




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Take-Aways • On the GMAT, fractions in inequalities are tricky, especially when the sign of the denominator is

unknown. Why tricky? If you cross-multiply by a negative the inequality sign flips but if you cross-multiply by a positive, the inequality sign stays the same. So if you multiply by an unknown value, you can’t tell what happens to the sign and you should account for both cases (we don’t have to worry about that in this case because rates are inherently positive, so there is no possibility that the sign will flip).

#45, pg. 276 Difficulty Level: 600-700 Topics: Functions & Sequences In sequences, the subscript (small number below the main line) is an index that identifies each term. For example,

1a is the 1st term and 3a is the 3rd term. In this notation, notation na is the nth term, 1na − is 1 term before the

nth, and 2na + is 2 terms after the nth.

We’re asked “for how many values of n is 100na < ?” In other words, “how many terms are less than 100?”

According to the function given, the value of the nth term na rn= . Thus the terms of this sequence are:

1st term 2nd term 3rd term …

1a r= 2 2a r= 3 3a r= …

Thus to find every term in the sequence and determine how many terms are less than 100, we only need the value of r

Rephrase “What is r?”

(1) 50 500a = . According to the sequence formula, the 50th term is 50 50a r= . Thus we can write

500 50 10r r= → = . We can directly answer our rephrase. With the value of r we can find every term in the sequence and answer the question


(2) 100 105 2,050a a+ = . According to the sequence formula, the 100th term is 100 100a r= and the 105th

term is 105 105a r= , so their sum must be 100 105 100 105a a r r+ = + . Thus we can equate the two

expressions: 2,050 100 105r r= + . This is enough information to solve for r and answer the rephrase.




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Take-Aways

• In Data Sufficiency, if your rephrase is a good one, data sufficient to answer your rephrase is necessarily sufficient to answer the original question. Consequently, while taking your test, there is no need to check whether you have sufficient info to answer the original question if you’ve already checked whether you can answer your rephrase.


0. 5r t= , what is t?

(1) 1 0.333

r r< → < . This means that r could be 0.25, 0.15 or 0.05. Therefore t could be 0, 1 or 2


(2) 1 0.10

10r r< → < . This means that r can only be 0.05 and t must be 0.



Take-Aways




There is a height of 9ft between floors. The number of steps needed is determined only by the height of each step. The figure below shows that if the height of each step is fixed, the same number of steps will link the two floors regardless of how wide each step is.


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Our rephrase “What is the average height of the steps?”



(2) This gives no information on the height of the steps.




Rephrase “What is juneXjuneY

?“

(1) This statement compares juneX to mayY. We are given no information about juneY.


(2) This statement gives us no information about juneX.


MERGE STATEMENTS

Statement (2) tells us how mayY and juneY are related by percent change. Statement (1) tells us how mayY and juneX are related by percent change. By merging the two statements, we’ll be able to link juneX and juneY by percent change as well. If this makes sense to you, you can stop. The statements work together to link juneX and juneY.

The width of the steps doesn’t matter. The number of steps needed is determined by their height.


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If we wanted to solve: (1) tells us that juneX is 50% greater than mayY, so we can plug-in 100mayY = and

150juneY = . (2) tells us that juneY is 25% more than mayY; since we plugged in 100 for mayY, we can

determine that 125juneY = . By merging the statements, we get 150125

juneXjuneY

=

Algebraically, (1) tells us that juneX is 50% greater than mayY, so 1.5juneXmayY

= . (2) tells us that juneY is 25%

more than mayY, so 1.25juneYmayY

= . We can very quickly find what we’re asked, juneXjuneY

, by dividing the first

ratio by the second: 1.5 1.5 1.25 1.25

juneXjuneX mayY juneXmayY

juneY mayY juneY juneYmayY

= → ⋅ → =


Take-Aways



We know that x is between a and b while y is between c and d. The question itself needs no rephrase: is x y< ?

(1) a c< . Without knowing whether x is less than c, we cannot determine whether x is less than y


(2) We already knew from the prompt that x b< and c y< . This statement tells us that b c< , so we can

conclude that x b c y< < < , so x y<




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This is an overlapping groups question. The group formula might come in handy: . In our case, the question deals with directors only, so we can

ignore neither. Our equation is and we want both. Isolate both to get


(1) This gives us Total and simplifies our rephrase from “What is ?” to “What is?” However we cannot tell the value of , so we cannot answer our rephrase.


(2) This gives us K and R and simplifies our rephrase from “What is ?” to “What is?” However, we cannot tell the value of Total, so we cannot answer our rephrase.


MERGE STATEMENTS

Together, the statements give us all parts of our rephrase. By plugging in the given values, our rephrase becomes “What is ?” We can certainly answer that!







y is positive, so we can simplify the question “Is ?” by multiplying by y

1 2Total Group Group neither both= + + −Total K R both= + −

both K R Total= + −

K R Total+ −

K R Total+ − 17K R+ −K R+

K R Total+ −12 8 Total+ −

12 8 17+ −

1xy>


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(1) Knowing that their product is larger than 1 doesn’t help us compare x to y. We cannot answer our rephrase


(2) Add y to each side to get . This directly answers our rephrase



Take-Aways • On the GMAT, fractions in inequalities are tricky, especially when the sign of the denominator is unknown.

Why tricky? If you cross-multiply by a negative the inequality sign flips but if you cross-multiply by a positive, the inequality sign stays the same. So if you multiply by an unknown value, you can’t tell what happens to the sign and you should account for both cases (in this problem we know that y is positive, so we can rephrase the question without worrying about whether the sign should flip).


Rephrase “What is yx

?”

(1) The difference between y and x cannot help us know the ratio of y/x. Selling price (y) could be $30 while the cost (x) is $10. In this case, profit would be $20, or 200% of cost. On the other hand, selling price (y) could be $40 while the cost (x) is $20. In this case, profit would still be $20, but in this case it’s exactly 100% of cost. The percentage that profit is of cost cannot be determined.





x y>

x y>

Sell price is the sum of cost and profit, so profit must be y – x. The percentage that profit is of cost can be expressed

as profit 1cost

y x y x yx x x x

=− = − → − . The only thing

standing in our way is the ratio y/x


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It would make no sense to leave the expression as is. Always simplify expressions:

( ) ( ) ( ) ( ) 4n x n y n z n k n x y z k− + − + − + − = − − − −

Rephrase “What is 4n x y z k− − − − ?”

(1) Average is sum divided by number of values, so this statement tells us that 4

x y z k n+ + + = . Cross-

multiplying gives us 4x y z k n+ + + = . Thus if we replace 4n with its equivalent, the rephrased

question will become: “What is x y z k x y z k+ + + − − − − ?” We don’t need any more data to determine the answer. Variables will cancel each other out; the expression equals 0.


(2) By itself, this statement gives us no idea how large or small the consecutive integers could be. Furthermore, this statement tells us nothing about n, so we cannot compute 4n x y z k− − − −




Since the first mile cost f and each of the other nine miles cost m, the cost of the trip was 9f m+

Rephrase “What is 9f m+ ?”

(1) The 2-mile ride must have cost f m+ . Thus we know that $0.90f m+ = (Eq.1). There is no way to

manipulate this equation to get the value of 9f m+ . Logically, all we know is that f and m must add up


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to 0.90. If f were large (0.80 for example) and m were small (0.10), then 9f m+ would be smaller than

if f were small (0.10) and m were large (0.80). We can’t find 9f m+


(2) This statement can be handled exactly as (1) above. This tells us that the first mile and 3 additional miles cost $1.20. So 3 $1.20f m+ = (Eq.2). There is no way to manipulate this equation to get the value of

9f m+ .


MERGE STATEMENTS

Together, the statements give us (Eq.1) and (Eq.2). Because we have 2 linear independent equations and 2 variables, we’ll be able to solve for the exact value of each variable and find 9f m+



Take-Aways • If you have as many independent linear equations as you have variables, you will be able to solve for all

your variables. Equations are independent if it is impossible to manipulate them to make them look identical.

#55, pg. 277 Difficulty Level: 600-700 Topics: FDPs & Ratios; Weighted Averages

(1) The net income depends on both the gross and the deductions. This statement tells us nothing about the deductions, so we have no way to tell how the net income was affected. The net and deductions could have each increased equally (a 4% increase in each would result in the 4% increase in the total), or one might have decreased while the other increased.


(2) This tells us nothing about the net income, so we can’t answer the question.

Statement (2) is NOT SUFFFICIENT

net gross deduct gross net deduct= − → = + .

The diagram on the left expresses this relationship. The question itself needs no rephrase. What is the percent change in net income?


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MERGE STATEMENTS

Because all the data given is in percentages and we’re asked to find another percentage, we are not restricted to any concrete dollar amounts. One easy way to determine whether the statements allow us to find a unique percentage change in net income is to plug in two sets of values that agree with the information we know and see what the change in net income would be. Whatever we plug in, the gross income must increase by 4% and the deductions must increase by 15%

Case 1: before after Case 2 Gross (4% increase) 200 208 300 312 Deductions (15% increase) 100 115 200 230 Net (g – d) 100 93 (a 7% drop) 100 82 (an 18% drop)

The table shows that if we change the starting amounts for gross income and deductions, the percentage change in net income varies. This is because since gross is made up of net and deductions, it’s the changes in these two parts that determine how the gross changes as a whole. We know that deductions increase by 15%. If the gross income was made up entirely of deductions, then the gross income would have increased by 15% as well. However, gross income only increases by 4% because the net income either increases only a little (less than 4%) or decreases. Basically, the change in the gross is a weighted average of the changes in its parts (net and deductions) and without knowing how big each part is, we can’t figure out exactly how much they contributed to the gross change.

Notes: Whenever you plug in values, choose values that will make your work easy. In this case, notice that I plugged in gross incomes and deductions that would always make the net income $100 at the start. This way it is very easy to quickly determine the percentage change in net income. No math is needed to determine that if net income goes from 100 to 82, it has decreased by 18%



numbers, you can plug in a value that will be easy to deal with and use it to solve the problem..


Whenever you’re dealing with triangle angles, remember that the sum of angles in any triangle is 180. . To find z, we need x+ y

Our rephrase “What is x+ y?”


( )180 180x y z z x y+ + = → = − +


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(2) We cannot solve for z, or for x+ y from this statement. We could solve for x alone, but that’s not helpful.






If we call x and y the number of $5 and $20 bills respectively, 5x and 20y will be the total dollar amount for each type of bill. Since Max has $125 total, we can write 5 20 125 4 25x y x y+ = → + = (Eq.1). We’re interested in finding x, the number of $5 bills.

Rephrase “What is x?”

(1) 5x < . Since x and y are the number of bills, they must be integers. Therefore, if x < 5, it could be 1, 2, 3 or 4. Now consider (Eq.1)

If x = 4 4 4 25 4 21y y+ = → = y is not an integer If x = 3 3 4 25 4 22y y+ = → = y is not an integer If x = 2 2 4 25 4 23y y+ = → = y is not an integer If x = 1 1 4 25 4 24y y+ = → = y = 6

Since y must be an integer, this statement actually limits us to a unique solution. There is one $5 bill and there are six $20 bills


(2) Max’s total amount is $125. If he has more than five $20 bills, he can only have six of them (worth $120) because any more $20 bills would increase the total beyond $125. This statement guarantees that there be six $20 bills. As a result, there must be one $5 to make the total $125




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We know that M and P have the same ratio of teachers to students. What we need is the ratio of M’s students to P’s students.

(1) M and P could have 20,000 and 10,000 students respectively (2-to-1 ratio), but they could instead have 30,000 and 20,000 students (3-to-2 ratio). The difference between student numbers doesn’t help us find student to student ratios. Algebraically, the best we can do is call 10,000s + and s the number of students in M and P (M has 10,000 more students)


(2) This gives us the ratio of teachers to students in both M & P (remember they have the same teacher to student ratio). M could have 20 students and 1 teacher while P has the same. This would make the student-to-student ratio 1-to-1. On the other hand, M could have 20 students and 1 teacher while P has 200 students and 10 teachers. This 2nd case would result in a different student-to-student ratio. This statement gives us no way to compare one group of students to the other.


MERGE STATEMENTS

We can quickly determine whether we have sufficiency by checking whether different plug-ins always give us the same answer. Whatever we plug-in here must agree with both statements: (1) M has 10,000 more students than P and (2) student-to-teacher ratio is 20-to-1 in both districts

M students M teachers P students P teachers Student-to-Student Case 1 20,000 1,000 10,000 500 20,000 to 10,000 2 to 1 Case 2 40,000 2,000 30,000 600 40,000 to 30,000 4 to 3

The table conclusively proves that the ratio of student to student is not fixed, even when we take both statements in account. We cannot definitively answer the question (we have no unique solution)



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#59, pg. 278 Difficulty Level: 500-600 Topics: Sets & Groups; Weighted Averages

84 people are split into 2 sections of the course. The question needs no rephrasing. How many females?

(1) This tells us nothing about section 2, so we can’t determine what fraction of the 84 people over the two sections is female.


(2) This tells us nothing about section 1, so we only have a partial picture and can’t determine how many of the 84 people are female.


MERGE STATEMENTS

Section 1 is 2/3 (66%) female. Section 2 is 50% female. When the sections are combined, the percentage who are female will depend on the relative sizes of the sections. For instance, if section



The only thing standing in our way is the fact that we don’t know the value of s

Our rephrase “What is n?” OR “What is s?”




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(2) Cross multiply to get . This is a 2nd linear equation. By replacing n with 4s in the original equation (found in the prompt), we could solve for s and then for n




your variables. Equations are independent if it is impossible to manipulate them to make them look identical


Because each guest was served one time, the number of guests who received double-scoops is equal to the number of double-scoops that were served

Our rephrase “How many double-scoops were served?”

(1) Without a concrete number of scoops (or number of guests), we cannot find out how many double scoops were served.


(2) Without knowing the ratio of single to double scoops, we can’t find out how many of the 120 scoops were served as double-scoops.


MERGE STATEMENTS

By merging the statements we have not only the total number of scoops, but also the ratio of single to double scoops. We could solve to find out exactly how many double scoops were served.


Note: Because many students have asked me how to actually solve for the number of double scoops, I’ve included two solutions below. To make it easier to understand, I’ve renamed double-scoop to “double serving”:

4n s=


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• Since 60% were double servings and 40% single servings, the ratio of double to single is 6:4. However, each double serving has 2 scoops, so the ratio of scoops in double servings to scoops in single servings is 12:4 (simplify to 3x:1x). Since a total of 120 scoops were served, scoops were served as double servings. Hence, 45 double servings (or double scoops) were served.

• Consider that where scoops is the total number of scoops, s is the number of single servings, and d is the number of double servings. Since 60% of servings were double servings, 40% were

single servings and we can say that . Isolate s to find that . Statement (2) gives us our total

scoops so the equation becomes . Solving for d, the number of double servings (or

double-scoops), gives



The room costs $120 for the 1st person, and x for each additional person. What is x?

(1) If 4 people must each pay $45, the total cost for the 4 must be 45 4 $180⋅ = . We know that the first person cost $120, so the other 3 must cost a total of 180 120 $60− = . The cost of each additional person will be a third of that, or $20.


(2) This statement basically tells us the difference in costs between 2 people and 4 people. With this, it will definitely be possible to find the additional cost of each person. No work is needed here. If you wanted to find x however, you could setup an algebraic equation. The 1st person costs $120 and each additional costs x, so the total cost of two people is 120 x+ and the total cost for four people is

120 3x+ . The cost per person for 2 people is 120

2x+

and the cost per person of 4 people is 120 3

4x+

. This statement tells us that the cost per person of 2 people is $25 more than the cost per person for 4

people, so 120 120 3 25

2 4x x+ += + . This can be solved for x



4 120 3 90x x= → =

2scoops s d= +

64

ds=

23ds =

2120 23d d= +

45d =


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We don’t know any price or percentage discount yet so we have nothing to start with. The question needs no rephrase. Is discount price of M less than discount price of L?

(1) This statement gives no way to determine what dollar amounts we’re working with. A greater percentage discount doesn’t necessarily mean the final discount price is lower. A 70% discount price on a house is still more expensive than a 1% discount price on a chocolate bar. We need concrete dollar amounts before we can compare prices.


(2) L has a $5 discount while M has a $6 discount. Without knowing more info about the regular prices, we cannot determine which store has the lower discount price. If the regular prices were equal, then M’s discount price would be smaller (because of the greater discount of $6). On the other hand, if M’s regular price was $1,000 more expensive, even the $6 discount would not bring the discount price below L’s discount price.


MERGE STATEMENTS

Statement (1) tells us what the discounts are as a percentage of the regular prices. (2) tells us what the dollar amounts equivalent to these percentages are. By merging the statements, we learn that L’s $5 discount is 10% of regular price (5 0.10L= where L is the regular price). We also learn that M’s $6 discount is 15% of regular price ( 6 0.15M= where M is the regular price). With this data we can find the regular prices, and subtract the discounts to find the discount prices and answer the question.




The question needs no rephrase. Is 0.5d ≥ ?

(1) What rounded to the nearest tenth would equal 0.50? . We cannot tell how d compares to 0.5


0.45 0.54d≤ ≤


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(2) What rounded to the nearest integer would equal 1? . All possible values of d are greater than or equal to 0.5, so




On the number line, the only thing that will determine how many integers there are between r and s is how far apart r is from s. For instance, the number of integers between 10 and 15 (4 integers) is equal to the number of integers between 1021 and 1026 because the two pairs of numbers are equally far apart. So to rephrase our question, we need to ask how far apart r is from s. Algebraically, the distance between two numbers is the positive difference between them (the absolute value of the difference).

Our rephrase “How far is r from s?” OR “What is ?”

(1) This statement tells us that r and s are 10 units apart. It answers our first rephrase.


(2) The interval between and is equal to the interval between r and s. The size of the interval doesn’t change because each end of the interval was increased by 1. Therefore, there must be 9 integers between r and s





• In data sufficiency, when the data from one statement can be logically deduced from the other statement,

C cannot be the right answer. When one statement offers the same data (no more, no less) as the other statement, the correct answer must be either D or E.


0.5 1.4d≤ ≤0.5d ≥

r s−

1r + 1s +


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n can only be a factor of t if t can be evenly divided by n.

Rephrase “Is integertn= ?”

(1) 23nn −= . This statement tells us nothing about t, so don’t waste your time here.


(2) 3nt = . Let’s try some values.

n N Is n a factor of t? 1 13 3= Yes, 1 is a factor of 3 2 23 9= No, 2 is not a factor of 9


MERGE STATEMENTS

By replacing n with 23n− and by replacing t with 3n , we can rewrite our rephrase to “Is 23 integer

3

n

n− = ?”

According to rules of exponents, a

a bb

x xx

−= so ( )2 2

23 3

33

n n nn

− −− = = . Thus we have enough data to determine

that 9tn= . Since the result of the division is an integer, n must be a factor of t




Total loans scholars n b= + + − . Since the question isn’t concerned with specific numbers of people (only percentages), we can ignore the given total and just use 100% as the total, 30% with loans and 40% with scholarships. Putting this in the equation gives us 100 30 40 n b= + + − . The only thing that stands in our way of knowing neither (n) is the value of b.

Rephrase “What is b?” or “What is n?”


Depending on the values we use, n may or may not be a factor of t. We cannot definitively answer the question


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(1) The people with scholars but no loans are those people who belong to the scholars group only: all 40 scholars minus those who belong to both groups. This statement tells us that 40 25b− = . This information is sufficient to find b and answer our rephrase.


(2) This tells us that half of those who had received loans also received scholarships. Since 30% received loans, half of them, or 15% of the total, also received scholarships. These are the people who belong to both groups. This statement tells us that 15b = , thus it answers the rephrase.



Take-Aways



#68, pg. 278 Difficulty Level: 700+ Topics: Rates & Work

Two concepts of rates are very important to memorize. First, the rate is the inverse of the time it takes to complete 1 job. For example, if a machine can build a car in 3 hours, then its rate is 1/3 car per hour. Secondly, if machines work together, their combined rate is the sum of the individual rates.

In this case, since the machines take 24 minutes when working together, their combined rate will be the inverse,

or 124

. We can write 124

rateK rateM rateP+ + = (Eq.1). We’re interested in finding how long it takes K,

so we want timeK . Because the time it takes to do 1 job and the rate are inverses, 1timeK

rateK= . In other

words, if we had a way to find rateK, we would be able to find timeK

Rephrase “What is rateK?” or “What is timeK”?

(1) Since 36 minutes is the combined time of M and P, we know that their combined rates would be the

inverse of 36. 136

rateM rateP+ = (Eq.2). We can combine this with (Eq.1) above:


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( ) ( ) ( ) ( ) 1 1Eq.1 Eq.2 24 36

rateK rateM rateP rateM rateP rateK−− → + + + → = −

We are able to find the rate of K and answer our rephrase (the time of K is the inverse of the rate) Logically, this statement must be sufficient because the prompt tells us how long all 3 machines together take to do the job, so it tells us how quickly they work when all three are operating. This statement tells us how quickly M and P alone can do the job. The loss rate of work must be the rate of the missing machine, K. Since we can find the rate of K, we can figure out how fast it will do the job on its own (rate is the inverse of time)


(2) This statement tells us how quickly K and P work together, but we have no way to pry them apart to find out how much K is doing by itself. Algebraically, this statement tells us that the combined times of K and P is 48 minutes, so the combined

rates must be the inverse: 148

rateK rateP+ = (Eq.3). By combining this with (Eq.1), we could find

the rate of M, but there is never any way to isolate the rate or time of K.


Notes: Don’t ever make the mistake of thinking that you can simply combine times timeA timeB combinedTime+ ≠


Take-Aways

• The rate is the inverse of the time it takes to complete 1 job: 1rate

time= and

1timerate

=

• If machines work together, their combined rate is the sum of the individual rates.


The question needs no rephrasing. All we know is the order of the variables. Is r closest to 0?


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(1) 0 must be halfway between s and –s. Because q and –s are the same, we know that 0 is halfway between

q and s. . This means that 0 is exactly as far away from q as it is from s. Since r is between q and s, wherever we place r it will be closest to 0.


(2) t q− < and q t< , so –t must be less than t. This statement simply tells us that t is positive and t− is negative. 0 is exactly halfway between t and t− , but without knowing where t− is located (how far to the left of q), we have no idea where 0 is. Therefore we cannot determine which variable is closest to zero. The variables could be anything as long as t is positive.




If n were 3, then Mary would have persuaded 3 friends, and each of them would have persuaded 3 of their friends so the “bottom level” friends would number 3 3 9⋅ = . If n were 5, there would be 5 top level friends, each of whom would persuade 5 of their friends so the “bottom level” would number 5 5 25⋅ = . Essentially, there are n

initial donors and 2n “bottom level” donors. The total number of donors must be 2n n+ and since each person

donated $500, the total amount donated was ( ) ( )( )2500 500 1n n n n=+ + . The question needs no rephrase:

what is n?

(1) Since each person donated the same amount, the amount donated by the first group is proportional to the

number of people in the group. In other words, if the first n people donated 1

16 of the total amount, then

these people must number 1

16 of all donors. Algebraically,

( ) ( ) 1 16 1 1 161 16

n n n n nn n

→= ⋅ = + → + =+

. We can find n.


(2) ( )( ) ( )500 1 120,000 1 240n n n n+ = → + = . Because n must be a positive number (# of

people), there can only be one solution. There must be a unique pair of consecutive positive integers (n


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and n+1) whose product is 240; using a greater or smaller pair would result in a different product. Therefore, this statement limits n to a unique value. It would be a waste of time to actually find it.


Notes: If n could be a negative number, then statement 2 would actually have two solutions because there must be a pair of negative numbers that also fits the statement (the product of the pair will equal the product of the pair of positive numbers). We can ignore the negative solution in this case because we’re dealing with people. To avoid missing solutions, solve quadratic equations (highest exponent is 2) by setting them to zero and factoring:

( ) ( )( )21 240 240 0 16 15 0 15 or 16n n n n n n n+ = → + − = → + − = → = − .



Let s be the short route, and l be the long route

Our rephrase “What is s?”

(1) . We cannot solve for s


(2) . This says nothing of the short route.


MERGE STATEMENTS

If we plug 23 for l in statement (1), we get . We can solve for s, the short route.




The question needs no rephrase. Is x > y?

42s l+ =

2 46 23l l= → =

23 42s + =


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(1) If you must add 2 to y to get x, then x must be greater than y


(2) Cross multiply to get . This data allows is to simplify the question from “Is ?” to “Is

?” Simplifying this question further leads us to “Is ?” We can’t answer this question. This statement is good enough to simplify the question but not sufficient to answer it.




Algebraically, an odd number is one that can be expressed as an even number plus 1 (as 2 1k + where k is an integer)

Rephrase “Is integer m odd?” or “Is 2 1m k= + ?”

(1) Do not fall into the trap of assuming that “NOT an even integer” is the same as “odd integer” because

“NOT even” could also mean “not an integer”. could be 3, in which case m is even, but could also

be 2.5 in which case m is odd.


(2) I found statement (2) easier. Since , we can say that . By adding 3 to an even, we end up with an odd. Thus m must be odd.



Take-Aways

• When solving number properties, it’s often helpful to insert the known property as a variable in equations. For instance, if you know that an unknown is an integer, you may refer to it as int. A prime number would be prime, an even number 2k or even, an odd number 2k+1 or odd, a multiple of 7 would be 7k… Doing so allows you to manipulate the unknown value while accounting for its property.


2 2x y= − x y>2 2y y− > 2y >

2m

2m

3m even− = 3m even= +


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The area of a triangle is . To find it, we need the numerator.


(1) Because the angle at D is 90 degrees, AC is the base and BD is the height. This statement directly gives us .


(2) A degree measure alone is not enough to figure out how large the triangle is. Degree measures only determine the shape of triangles (how acute the corners are) but many triangles of different sizes can have the same shape.




The standard format for the equation of a line is y mx b= + where m is the slope and b is the y-intercept. When working with lines in coordinate geometry, use the standard format. In this case, we should manipulate the

equation we’re given to reflect this format: 0 a cax by c by ax c y xb b

+ + = → = − − → = − − . The

slope (what takes the place of m) is ab

− . Since we’re also told that the slope is 2/3, we can write

2 3 23

a a bb

− = → − = . This means that to find the value of b, it would be sufficient to know a

Rephrase “What is a?”

(1) 4a = This directly answers our rephrase


(2) 6c = − This tells us nothing about a or about b



2b h⋅

b h⋅

b h⋅


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Take-Aways

• The standard format for the equation of a line is y mx b= + where m is the slope and b is the y-intercept. • In Data Sufficiency, if your rephrase is a good one, data sufficient to answer your rephrase is necessarily



For a product of integers to be even, at least one factor must be even.

Rephrase “Is one or more variable even?”

(1) Merge p’s to get 2t m p+ = . Since 2p is even (it has an even factor), t m+ is also even. Because

odd odd even+ = and even even even+ = , this statement only tells us that t and m are similar (they’re both odd or they’re both even). Since we don’t know which and since we have no information about p, we cannot determine whether one or more variables are even. they could all be odd but they could also all be even.


(2) odd odd even− = and even even even− = , so this statement only tells us that t and m are similar (they’re both odd or they’re both even). Since we don’t know which is the case, we cannot definitively answer the question: we could have all odds or all evens.


MERGE STATEMENTS

Each statement tells us that t and m are similar. Combining the two statements cannot be more helpful because they don’t add to each other. They say the same thing.




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Take-Aways

• In data sufficiency, when the data from one statement can be logically deduced from the other statement, C cannot be the right answer. When one statement offers the same data (no more, no less) as the other statement, the correct answer must be D or E.


The pay is $300 + 5% of the sales over $1,000. Because the pay depends on sales, to determine how much he was paid we need to know the amount of sales he made.

Rephrase “What were sales?”

(1) There is a base pay of $300 on the first $1,000 of sales. If sales were exactly $1,000, the pay would thus be 30% of sales. Sales above $1,000 result in 5% commission. Notice that as he makes more and more in sales, the total pay approaches 5% of total sales because most of the pay will come from the extra sales (above $1,000). If the salesman made an extra 10 million in sales, his total pay would be 300 base pay plus 5% of 10 million (500,000). His total pay would be $500,300, or about 5% of his total sales. There must be an exact unique amount in sales that would result in a pay equal to 10% of sales. There is no need to find that amount of sales. Just realize that this statement allows us to find the sales and thus answer the rephrase.


(2) This directly answers the rephrase.



Take-Aways



We know how much was invested (60,000) and we know that overall the interest earned for the year was $4080. This means that we could (but shouldn’t spend the time) find out the overall percentage rate. Let’s call it 8%


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(because 4,000 is a bit less than 6,000 which would’ve been 10%). Since one portion was invested at x% and the other at y% , the final 8% interest is a weighted average of x and y. The 8% must be determined by two things: (a) the values of x and y and (b) how the 60,000 were broken up. If the 60,000 were broken in 2 equal pieces, then 8% would be the exact average of x and y. On the other hand, if twice as much money was invested at x% than at y%, then the final interest rate (8%) will be closer to x than to y. The question asks for the value of x.

Rephrase “what is y AND how were the $60,000 broken up?” or “What is x?”

(1) 3 3 4 4y xx

y= → = . This gives us a ratio of x to y which means if we can find y it will be easy to

find x. However, without information about how the money was broken up, we cannot determine which interest rates would result in a final rate of 8%


(2) This tells us how the money was broken up the ratio of the amount invested in x to that invested in y is 3-to-2. So 3 out of 5 dollars were invested at x%. However, this only gives us a partial picture. To find x we also need y.


MERGE STATEMENTS

Statement (2) tells us that out of every $5, $3 are invested at x% and $2 at y%. The result is an 8% return. Statement (1) tells us that the ratio of x to y is 3 to 4. This means that we could say that 3x i= and 4y i= . Thanks to this ratio, we only need to find 1 variable, i, to know everything. Since we have only 1 variable and there is definitely an equation in here somewhere, we can be certain that we will be able to solve without doing the work. On the exam, stop here. To see the math, continue

We know that if you invest $3 at x (which is 3i) percent and invest $2 at y (which is 4i) percent, the resulting

interest is 8% of your starting $5. Algebraically, this means that ( ) ( ) ( )3 4 8$3 $2 $5100 100 100

i i+ = . We could

(but shouldn’t) solve for i, then find x x (which is 3i)



#79, pg. 279 Difficulty Level: 600-700 Topics: FDPs & Ratios; Inequalities & Absolute Values


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The first option (A) is to pay 1p and add a tax of 1t percent. “Percent” means “divided by 100”, so the exact

amount of the tax is ( )11100

t p . The total cost of the first option is ( )11 1100

tp p+ . The cost of the 2nd option

(B) is ( )22 2100

tp p+ . We’re asked to determine whether the first option costs more than the 2nd.

Rephrase “Is ( ) ( )1 21 1 2 2100 100

t tp p p p>+ + ?”

(1) This statement tells us which option has a higher tax rate. However information to help compare the cost, we cannot determine which option costs more. A sandwich may have a higher tax rate than a car; doesn’t mean that the total cost of the sandwich is greater.


(2) 1 1 2 2p t p t> . This can help us compare only the tax portion of the rephrase: ( )11100

t p must be greater

than ( )22100

t p . However, we have no way to determine how 1p and 2p compare. We need more data

Logically, this statement compares the base cost multiplied by the tax rate. This is really a comparison on the tax dollars. For example if an item costs $100 and is taxed at 5%, the amount of tax will be the product: ( )100 0.05 . Knowing that the dollar of taxes is greater for one item than for the other doesn’t

help us compare total costs. This is because the item with the fewer tax dollars could still have the greater total cost if that item has a high base cost and a low tax percentage. For instance item 1 could cost $100 and have a 50% tax rate. Its tax would be $50. Item 2 could cost $200 and have a 1% tax rate. Its tax would be $2. Even though item 1 has a greater tax bill, it can have a smaller overall cost ($150 vs $202)


MERGE STATEMENTS

Together, the statements tell us that item 1 has a higher tax rate (statement 1) and that item 1 has more tax dollars (statement 2). The easiest way to merge these statements is to plug in values and test whether we are limited to a unique answer to the question “does item 1 has a greater total cost?”

It would be easy to make item 1 have the greater total cost. If it has a greater base cost, given that it has a greater tax rate, it will have a greater total cost. But what if item 2 has the greater base cost? Well the example we used in statement (2) can be reused here: if item 1 has a base cost of $100 and a tax rate of 50% while item 2 has a base cost of $200 and a tax rate of 1%, item 1will have the greater tax amount ($50 vs $2) even though item 2 will have the greater total cost ($202 vs $150). To be able to make the comparison between total cost, we must have information about the base costs



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Take-Aways

• One way to check whether a statement or two statements together are sufficient is to plug-in a couple of values that fit the data given in those statements. If the values yield more than one unique answer to the question, then the statement or statements together are NOT sufficient. On the other hand, if all values that you plug-in yield a unique answer to the question, then the statement or statements together may (not must) be sufficient.


Because the variables are positive, we can multiply both sides by r and s without worrying about the inequality

sign flipping: 2 2? ?r s r ss r< → < For a number to have a greater square, it needs to be farther away from

0. For example 2 23 2> because 3 is farther from 0 than 2. Similarly, ( ) ( )2 25 4− > − because -5 is farther

from 0. So the question is, is s farther from 0 than r? Since both are positive we can simply ask:

Rephrase “Is s r> ?”

(1) 1 3

3 4 4r rs s= → = . This statement tells us that r-to-s is a ratio of 3-to-4. Since they are both

positive, s must be greater than r. We can definitively answer our rephrase.


(2) 4s r= + . To get s, we must add 4 to r. Therefore, s must be greater than r.


Notes: Statement 1 is only sufficient because we know that our variables are positive. If they could be negative, it would be possible for r to be bigger than s because r and s could have been -3 and -4.



What is n in the list “k, n, 12, 6, 17”?

(1) k n< This gives us no specific values to work with.


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(2) The median is the middle of an ordered list, or the average of the two middle values if there is no single middle value. In this case, there are 5 values, so the 3rd greatest value is the median. Since 2 values are unknown we cannot order the values and determine which of k or n equals 10.


MERGE STATEMENTS

The known values are 6, 10, 12, and 17. We know that 10 is the median, so 10 is the middle of the list. This means that one variable equals 10 and the other variable must be smaller (to make 10 the 3rd greatest). Since statement (1) tells us that k n< , it’s guaranteed that 10n =




A multiple of 6 is the result of the product of 6 and an integer. Algebraically, “multiple of 6” can be represented as 6 int⋅ . Because this is a number properties question, it makes sense to use prime factorizations . “x is a multiple of 6” means 6int 2 3 intx x= → = ⋅ ⋅ where “int” is an unknown integer. Similarly, “y is a multiple of 14” means 2 7 inty = ⋅ ⋅ . By putting the product of x and y therefore must be of the format

22 3 7 intxy = ⋅ ⋅ ⋅ (Eq.1). In other words, we know for sure that xy has 22 , 3 and 7 among its factors. We don’t know what else it may have

The question asks whether “xy is a multiple of 105”. This can be rephrased as is 105 intxy = ⋅ ? By breaking

105 into its prime factors we get “is 3 5 7 intxy = ⋅ ⋅ ⋅ ?” In other words, we’re asked to determine whether xy has 3, 5 and 7 among its factors. (Eq.1) above told us that xy has 3 and 7 among its factors. Therefore to answer the question, we need to know whether xy also has 5 among its factors

Rephrase “Does xy have 5 among its factors?”

(1) This tells us that x has 23 among its factors. Because we have no idea what other factors it may have, we cannot determine whether 5 is among the factors of xy.



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(2) This tells us that y has 25 among its factors. This guarantees that y has 5 among its factors, so xy will also have 5. This directly answers our rephrase.



Take-Aways


connections.

• When solving number properties, it’s often helpful to insert the known property as a variable in equations. For instance, if you know that an unknown is an integer, you may refer to it as int. A prime number would be prime, an even number 2k or even, an odd number 2k+1 or odd, a multiple of 7 would be 7k… Doing so allows you to manipulate the unknown value while accounting for its property




(1) There are too many variables and no way to cancel out a and d.


(2) Tells us nothing of b+ c.


MERGE STATEMENTS

Statement (1) can be factored to isolate b+ c as follows:

Regroup

Factor out b & c

Factor out from each bracket

( ) ( ) 6ab bd ac cd+ + + =

( ) ( ) 6b a d c a d + + + =

( )a d+ ( )( ) 6a d b c+ + =


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To find b+ c, we need a value for a+ d. That’s where statement (2) comes in. By replacing a+ d with 4, our

equation becomes




. To find the average, we need the numerator


(1)


(2)




Paula & Sandy sold 100 tickets means . It’s important to note that there were other people selling tickets. The only thing standing in the way of p is s.

Our rephrase “What is p?” OR “What is s?”

(1) This statement gives us a 2nd independent linear equation: Replacing s with in the equation

would allow us to solve for p


( ) ( ) 64 6 4

b c b c+ = → + =

2j kavg +

=

j k+

2 4 11 6 22 162

j k j k j k+ + += → + + = → + =

14 10 14 30 163

j k j k j k+ += → + + = → + =

100p s+ =

23

s p=23

p

100p s+ =


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(2) We have no way to know how many tickets Sandy sold, since we do not know how many tickets were sold in total (when you take other sellers into account). We’re only told the number that Sandy and Paula sold together.




Each person wrote down a number between 1 and 30 (inclusive).

Rephrase “Was any number written down twice?”

(1) There are only 30 numbers to pick from, so if there are more than 40 people, at least 10 people picked a number that someone else had already written down. It must be the case that at least one integer was written down by multiple persons.


(2) I found statement (2) easier. There were fewer than 70 people. This really doesn’t help. There could have been just two people, and they each wrote down the same integer, or there could have been 20 people but they wrote down 20 different integers. We cannot say whether the same integer was written down by more than one person



#87, pg. 280 Difficulty Level: 500-600 Topics: Inequalities & Absolute Values; Rates & Work

The distance equation is d rt= where d, r and t are distance, rate and time. This question asks us to compare

the time it takes. Manipulating the equation for time gives us dtt

= . Traveling 1d feet at 1r speed will take

1

1

dr

. Traveling 2d feet at 2r speed will take 2

2

dr

. We’re asked whether the 1st time is greater than the 2nd . “Is

2

2

11 2 2 1

1

d d d r d rr r

→> > ?”


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(1) 1 2 30d d= + . This gives us no information about the rates, so we can’t compare how long it will take to

cover the two distances.


(2) 1 2 30r r= + . This gives us no information about the distances. So knowing that one rate is greater

doesn’t help us compare how long it will take to cover the distances.


MERGE STATEMENTS

By replacing 1r and 1d with their equivalent inside our rephrase, we can rewrite the rephrase from:

“Is 1 2 2 1d r d r> ?” “Is ( ) ( )2 2 2 230 30d r d r+ > + ?”

Distribute Is 2 2 2 2 2 230 30d r r d r d+ > + ? Is 2 230 30r d> ?

Because we have no way to compare 2r and 2d , we cannot answer the rephrase.

Notes: Getting rid of fractions in the rephrase (by cross-multiplying) makes the question easier. In general, it’s a good idea to get rid of fractions in Data Sufficiency because doing so makes data easier to evaluate.



(1) taxes & insurance together were 33% of mortgage. Express all values in terms of x


mortgage + taxes + insurance = 12,000

Rephrase How much are taxes?

Since the total of 12,000 = 133x, we can find x. However, we don’t know exactly how many x represent taxes so we can’t find out the exact amount in taxes.


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(2) taxes were 20% of mortgage & insurance combined. Express all values in terms of y.



Take-Aways

• One great technique for solving ratios is to express all parts of the ratio in terms of the same variable. For instance, if a problem tells you that the ratio of boys to girls is 4:5, you should use 4x and 5x as the number of boys and girls. By writing all values in terms of the same variable, you greatly simplify the problem (in general, the fewer variables the easier a problem will be to solve). Percentages and ratios are the same thing, just written in different ways. For instance, x is 20% of y is the same as x-to-y is 20-to-100.


Let’s call x and y the total number of Club X and Club Y members

Rephrase “Is x y> ?”

(1) This tells us that those who belong to both clubs are just 20% of all club X members. If we call b the number who belong to both clubs, we could write 0.20b x= . This statement doesn’t tell us anything about the people who belong to club Y only, so we cannot compare x to y.


(2) This is very similar to statement 1. Only, it tells us that of all who belong to Y, 30% belong to both groups: 0.30b y= . Without any information about how many people belong to X only, we cannot compare x to y


MERGE STATEMENTS

Since the total of 12,000 = 120y, we can find y. Since taxes = 20y, once we find y, we can find the exact amount paid in taxes.


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0.20b x= and 0.30b y= . Essentially we’re told that the same number, b, is 20% of group x but 30% of group y. Since the same number makes up a bigger portion of y, y must be a smaller group than x. y x< . We’ve answered the question.




We should simplify the equation given, 6 4 12k m t+ = , by getting rid of fractions. Multiply both sides by 12 to

get 2 3k m t+ = (Eq.1). The question is “does t have a factor in common with 12 other than 1?” The factors of 12 are 1, 2, 3, 4 and 6.

Rephrase “Is t a multiple of 2, 3, 4 or 6?”

(1) Algebraically, n is a multiple of p is equivalent to intn p= ⋅ where “int” is any integer. Since k is a

multiple of 3, (Eq.1) can be rewritten from 2 3k m t+ = to ( )2 3 int 3m t⋅ + = where “int” is an

unknown integer. Note that we can factor 3 out of the left side to get ( )3 2int m t+ = . Essentially, this

statement tells us that t is 3 times an integer, so t must be a multiple of 3 and therefore must have a factor in common with 12.


(2) Algebraically, 3 intm = ⋅ . This allows us to rewrite (Eq.1) from 2 3k m t+ = to ( )2 3 3 intk t+ ⋅ = .

It’s not possible to factor the left side to get ( )intt f= where f is a known number. Without more

information, we cannot determine any factor of t. Alternatively, we could replace k and m with a couple of values and try to get this statement to answer the question both ways. If we’re able to accomplish this, we would have proven that this statement is not sufficient. If we always get the same answer to the question “do t & 12 have a common factor?”, then we’ll be forced to assume that we have sufficiency. When plugging-in in data sufficiency, be sure that you only plug in values that agree with the statement(s) you’re checking. In this case, m must be a multiple of 3. • Try to get a YES: If m = 3 and k = 3, then 2 3 15k m t t+ = → = . In this case, t and 12 have

a common factor (3).


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• Try to get a NO: If m = 3 and k = 1, then 2 3 10k m t t+ = → = . in this case, t and 12 will have no common factor.

We’re able to get conflicting answers with values that respect the statement’s data.


Notes: Getting rid of fractions in the makes the question easier. In general, it’s a good idea to get rid of fractions in Data Sufficiency because doing so makes data easier to evaluate.


Take-Aways

• One great technique for solving ratios is to express all parts of the ratio in terms of the same variable. For instance, if a problem tells you that the ratio of boys to girls is 4:5, you should use 4x and 5x as the number of boys and girls. By writing all values in terms of the same variable, you greatly simplify the problem (in general, the fewer variables the easier a problem will be to solve). Percentages and ratios are the same thing, just written in different ways. For instance, x is 20% of y is the same as x-to-y is 20-to-100.

#91, pg. 280 Difficulty Level: 500-600 Topics: Geometry; Inequalities & Absolute Values

The question needs no rephrase. Is segment CD longer than segment BC?

(1) The total length doesn’t help us compare individual segments.


(2) This tells us nothing of BC. It could take up most of the line, or only a tiny part, while AB remains equal to CD. We cannot compare BC to CD


MERGE STATEMENTS

Total length is 20, and the the first and last segment equals each other. However, we still cannot say whether CD > BC. Consider the following configurations:


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Let the total number of employees be t. Let’s organize the initial info into equations:

“50% are college graduates” “College graduates equal ”

“60% are over 40yrs…30% of those have master’s” means that employees over 40yrs old who also have a

master’s equal 30% of 60% of total “Over 40yrs with master’s equal ”. This is the quantity we are

asked to find. The only thing standing in our way is t, the total number of employees.

Our rephrase “What is t — the total number of employees?”

(1) Since we already know that college graduates equal , this statement gives us . We can solve

for t and answer our rephrase


(2) This statement doesn’t give us any concrete number of employees to work with, so it isn’t possible to find the total number of employees



12

t

30 60100 100

t⋅ ⋅

12

t 1 1002

t =

A B C D

4 12 4

A B C D

4 12 12

CD < BC

CD > BC


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A sequence in which there is a constant difference between adjacent values is an arithmetic sequence. Some examples include:

• A set of consecutive integers: 12, 13, 14… Difference is 1 • A set of consecutive even integers: 2, 4, 6… Difference is 2 • A set of consecutive multiples of seven: 7, 14, 21… Difference is 7.

The average of an arithmetic sequence is the average of its highest and lowest values; the average is also the median (middle value). p, q, r, s and t are consecutive even integers, so they make up an arithmetic sequence. The mean of the sequence is the middle integer, or r

Rephrase “What is r?”

(1) 24q s+ = . s is 2 even integers after q, so the difference between s and q is 4. We can write 4s q= + .

This equation along with 24q s+ = gives us 2 independent linear equations with 2 variables. We’ll be able to solve for the values of s and q. Once we know any of the integers we’ll be able to find the others because they are consecutive evens. Thus we can find r


(2) 112

q r+ = . Since q and r are consecutive evens, we can write 2r q= + . As in statement 1, we have

two linear independent equations and two variables. We have enough data to find q and r so we can definitively answer our rephrase



Take-Aways • A set of numbers in which the difference between each value and the next higher value is constant is

called an arithmetic sequence. • The average of an arithmetic sequence is the average of its highest and lowest values. The average is

always equal to the median of this sequence.


• If you have as many independent linear equations as you have variables, you will be able to solve for all your variables. Equations are independent if it is impossible to manipulate them to make them look identical


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The standard format for the equation of a line is y mx b= + where m is the slope and b is the y-intercept. This we’re given an equation already in the standard format and asked for the slope we can rephrase.

Rephrase “What is m?”

(1) Two lines are parallel if they have the same slope. The slope of the line is the coefficient of x (the number in front of x) when the line is in the standard format (with y isolated). We’re told that our line, y mx b= + , is parallel to ( )1 1y m x b= − + + . The “1” added at the end is there to confuse you.

1b+ together make up the number at the end (the y intercept). Think of this equation as

( ) ( )1 1y m x b= − + + . The coefficient of x is (1 – m), so this is the slope. Because our original line

and the line from this statement are parallel, they must have the same slope. Thus we can write: “original slope” = “new slope” ( )1m m= − . This gives us enough data to find m, the original slope.


(2) This statement tells where line k intercepts a known line.



#95, pg. 281 Difficulty Level: 600-700 Topics: Translations & Manipulations; Number Properties

The question needs no rephrase. Is 1rst = ?

(1) Tells us nothing about t, so we cannot deduce anything about the value of rst


Basically we’re simply told that line k goes through the point (2,7). Either of the dotted lines, and an infinite number of other lines that go through the given point could be line k. The slope of line k (remember slope measures direction) cannot be determined.


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(2) Tells us nothing about r, so we cannot deduce anything about the value of rst


‘

MERGE STATEMENTS

By merging the statements, we can tell that . To compare this data to what is asked, divide both

sides by s to get .

With this data, we can rephrase the question from “Is ?” to “Is ?” Without the value of s, we still do

not have sufficient information to answer.



#96, pg. 281 Difficulty Level: 400-500 Topics: Geometry; Statistics

Since we’re dealing with a circular pie, the whole pie is 360⁰. The expenses of division R are determined by how big a piece it represents. The size of the piece in turn depends on the degree measure of x

Our rephrase “What is x?”

(1) This directly answers our rephrase; don’t bother solving.


(2) Although we can build a ratio, 21

S TR+ = , we can’t find the value of x unless all angles are involved in

the ratio, or unless we are told what the angles included add up to.




( )( ) 1 1rs st = ⋅

1rsts

=

1rst = 1 1s=


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We know that 0x < ; Is 3x < − ?

(1) 2 9x > . What type of numbers have squares larger than 9? Positive numbers bigger than 3 (such as 2 24 ,5 ,... ) and negative numbers smaller than -3 (such as ( ) ( )2 24 , 5 ,...− − ). So this statement is telling

us that 3x > or 3x < − . Since we know from the prompt that x is negative, x cannot be bigger than 3. At the end of the day, 3x < − . This statement directly answers the question.


(2) 3 9x < − . What types of numbers have cubes smaller than -9? Well x must be a negative number. To answer the question (Is 3x < − ?), we need to be more specific. Because -9 is a small number, we can

just test for possible values of x: ( )31 1− = − , ( )32 8− = − , ( )33 27− = − , ( )34 64− = − . Notice that

as the numbers get smaller, the cube also gets smaller. This means that x could be less than -3. -4 is an example. However do we have a unique answer? Does x HAVE to be less than -3? Because

( )32 8− = − and ( )33 27− = − , there must be values between -2 and -3 whose squares are less than -9.

I’m sure ( )32.99− is really close to -27, so x could be less than -2.99 but x could also be -4. We cannot

definitively determine whether x is less than -3.




When an integer is divided by 7, the possible values of the remainder are {0, 1, 2, 3, 4, 5 or 6}. In this case 7 numbers were each divided by 7. What is the sum of the remainders?

(1) Range = 6. Range only gives the difference between the highest and smallest values in a set. Because range gives no information about all the other values in the set, it’s impossible to find the sum from the range.


(2) To know how this is relevant, you need to understand a property of division and remainders. Basically, as the number at the top of the division increases, so does the remainder. For example, 10 5÷ has remainder 0; 11 5÷ has remainder 1; 12 5÷ has remainder 2; 13 5÷ has remainder 3… The remainder is increased by the same amount as the number at the top of the fraction. However, the biggest remainder of division by 5 is 4. The remainder resets to 0 each time you reach another multiple of the divisor. 14 5÷ has remainder 4; 15 5÷ has remainder 0! How does this help us? Well if in our problem, the 7 numbers


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that were divided by 7 are consecutive, they will have 7 different remainders: {0, 1, 2, 3, 4, 5 and 6}. Because we know all remainders we can find the sum of the remainders


Notes: Range actually gives very little information – nothing beyond the difference between the greatest and smallest values. From range we cannot determine sum, average, median, mode, deviation…

In statement (2) we cannot tell what the remainder of the smallest number divided by 7 is. The first remainder doesn’t have to be 0; It could be 5. However, even so the remainders will be {5, 6, 0, 1, 2, 3, and 4} because remainders reset to 0 when they reach 6.



Remember that each row and each column (not diagonals) must contain a 1, a 2 and a 3. What is r?

(1) 6v z+ = . For two numbers in the table to add up to 6, each number must equal 3. So: r s t u 3 w x y 3


(2) 6s t u x+ + + = . None of the 4 variables in the addition can be a 3. This is because if one of them were 3, the other three variables would have to add up to 3, so they would all have to be 1’s. However since s and t are in the same row, they cannot both equal 1. Likewise since u and x are in the same column, they cannot both equal 1. Thus s, t, u and x must all be a number other than 3. r not 3 not 3 not 3 v w not 3 y z




Each row must contain a 3. s cannot equal 3 because in its column there is already a 3. Likewise t cannot equal 3. That means that in the first row, only r can equal 3.

Since the first row must contain a 3 and s and t cannot equal 3, it must be true that r equals 3.


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Definition: x is “the greatest integer less than or equal to x”. This simply means that the brackets will round x

down to the nearest integer (or leave x alone if x is already an integer).

Application: We’re asked “Is 0x = ?”

Rephrase “Does x round down to 0?”

(1) 5 1 3 2x x+ = + This linear equation can be solved for x. Once we know x we can answer our rephrase (we’ll know what it rounds down to).


(2) 0 1x< < . x is a decimal such as 0.1, 0.3… These decimals all round down to 0. This statement directly gives us the answer to our rephrase.



#101, pg. 281 Difficulty Level: 500-600 Topics: Weighted Averages

The cost of the overall mixture (K) depends on how much of each material is put into the mix.

By itself, A costs $3 per kilograms. x kilograms of the material are in the mixture.

By itself, B costs $5 per kilograms. y kilograms of the material are in the mixture.

If the mixture is 99% material A, then the cost of the mixture will be really close to $3 per kilogram. On the other hand, if the mixture is 99% B the cost of the mixture will be really close to $5 per kilogram. If the same amount of A and B are put into the mixture, then the cost of the mixture will be exactly in the middle of the two costs; it would be $4 per kilogram. Because x and y are the amounts of the $3 and $5 material respectively, x y> only if the cost of the mixture is closer to $3 per kilo than to $5 per kilo

Rephrase “Is the cost of mixture K less than $4 per kilogram?”

(1) 4y > . Because the entire mixture is 10 kilos, we know that 10x y+ = . Knowing that 4y > doesn’t tell us whether x is greater than y. They could be equal (x,y) = (5,5) or they could be (1,9). We can’t tell whether x is greater.


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(2) Cost of the 10 kilos is less than $40. This means that the mixture costs less than $4 per kilogram. This statement directly answers our rephrase (since the cost is closer to $3 than to $5, there must be more of material A in the mix; x > y).



Take-Aways



To find out how much time it would take for X to get 2 miles ahead, we need to know how many extra miles X covers each minute. This “extra mileage” depends on how much faster X travels.

Our rephrase “X is how much faster than Y?”

(1) This directly answers our rephrase by telling us that X is 10miles per hour faster.


(2) We already know that X is 1 mile ahead now. This statement tells us that 3 minutes ago, X was 0.5miles ahead. We can deduce that every 3 minutes, X gains an extra half mile. We can calculate how many more minutes it will take X to get 2 miles ahead.



#103, pg. 281 Difficulty Level: 600-700 Topics: FDPs & Ratios; Rates & Work

The work/rate formula is . We already know the work that needs to be

done: 17,280 frames. What we need is the rate

workwork rate time timerate

= ⋅ → =


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Our rephrase “What is the rate?”

(1) This explicitly tells us the rate and thus answers our rephrase.


(2) We’re given the ratio of with a total time of 7x. Since this total time is also 14 minutes,

we could write and find the run time (6x)



Take-Aways

• One great technique for solving ratios is to express all parts of the ratio in terms of the same variable. For instance, if a problem tells you that the ratio of boys to girls is 4:5, you should use 4x and 5x as the number of boys and girls. By writing all values in terms of the same variable, you greatly simplify the problem (in general, the fewer variables the easier a problem will be to solve).


We need to know the speed at which the train is traveling, not averaged over a part of the trip, but in a specific instant. The speed of the train at the instant it reaches the halfway mark. It’s very unlikely that any statement will help us determine this.

(1) With the distance and time it took to travel the trip, we can find the average speed during the trip, but there is no way to find the speed at any specific point of the trip because we don’t know whether speed was constant or whether there were spikes and dips in speed during the trip.


(2) This is no more helpful than the previous statement because it gives us the average speed of the train, but no way to find the exact speed at a particular point during the trip. We just don’t know how fast the train was going the instant it reached the halfway mark.


MERGE STATEMENTS

We can deduce statement 2 (the average speed) from the information in statement 1, so merging the statements will be no more helpful than using statement 1 alone.

run time 6rewind

xx

=

7 14 minutesx =


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Together the statements are NOT SUFFICIENT


Take-Aways

• In data sufficiency, when the data from one statement can be logically deduced from the other statement, C cannot be the right answer. When one statement offers the same data (no more, no less) as the other statement, the correct answer must be D or E


The three houses average 120. Median is the middle value (or the average of the 2 middle values when there is an even number of values).

Rephrase “what’s the middle house?”

(1) Tom = 110. Since the average is 120, there must be at least one house above 120 to balance out Tom’s low price. Therefore, all we know is that Tom cannot have the most expensive house. However it’s impossible to tell which is the middle house.


(2) Jane’s house equals the overall average of 120. There are just two possibilities: • If one of the other houses is greater than 120, the third house will have to be less than 120 to keep the

overall average at 120. • On the other hand if one of the other houses equals 120, the third house will also have to equal 120 to

keep the overall average at 120.

Either way, the median (middle price) is always 120. Thus we know the median.




x and y are integers. For the product xy to be even, one or both integers must be even.

Rephrase “Is at least one of the variables even?”


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(1) 1x y= + . This tells us that x and y are consecutive integers. If x is even (2, 4, 6…) then y will be odd (1, 3, 5). On the other hand, if x is odd (3, 5, 7) then y will be even (2, 4, 6…). Either way, the answer to the question is a definitive YES. At least one of the variables is even.


(2) An even number is a multiple of 2. This statement can be rewritten as 2x ky= where k is an integer (so

2k is an even number). Simplify: ( )( )2 2x k x k yy= → = . We have no information about y, but

we can see that x is the product of y and an even number. The product of any integer and an even integer will be even, so we can be certain that x is even. This answers our rephrase. At least one of the variables is even



Take-Aways

• When solving number properties, it’s often helpful to insert the known property as a variable in equations. For instance, if you know that an unknown is an integer, you may refer to it as int. A prime number would be prime, an even number 2k or even, an odd number 2k+1 or odd, a multiple of 7 would be 7k… Doing so allows you to manipulate the unknown value while accounting for its property


Let P(w), P(b) and P(r) be the probabilities of selecting a white, blue or red chip respectively. All probabilities add up to 1. . The probability we want (white or blue) is

Our rephrase “What is ?” or “What is ?”

(1) . With this we can simplify our rephrase from “What is ?” to “What is

?” We don’t have sufficient data to answer the question


( ) ( )( ) 1P w P b P r+ + = ( ) ( ) ( )1P w P b P r+ = −

( ) ( )P w P b+ ( )1 P r−

( ) 15

P b = ( ) ( )P w P b+

( ) 15

P w +


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(2) . With this we can answer our 2nd rephrase.



Take-Aways • Assuming A and B are independent events (the likelihood that one happens is not affected by whether the

other happened), Probability of A or B is . Probability of A and B is


Note that the first tick mark is 0. To find y, we need to know the scale of the figure. For instance, if each interval (distance between two adjacent tick marks) has length 1, then y would be 7. On the other hand, if each interval has length 3, then y would be 21.

Our rephrase “How long is each interval?”

(1) x corresponds to 3 intervals, so by dividing the value of x by 3, we can find the length of each interval and answer our rephrase.


(2) y – x corresponds to 4 intervals. By dividing the value of y – x by 4, we can find the length of each interval and answer our rephrase.



#109, pg. 282 Difficulty Level: 700+ Topics: Geometry Draw a triangle that matches the description given. Note that the side lengths of each smaller triangle are half of the side lengths of the triangles that contain it.

( ) 13

P r =

( ) ( )P A P B+ ( ) ( )P A P B⋅


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Rephrase “What is 12

bh ?”. Actually since we know the value of 1/2, all we need to know is bh.

(1) Area of ABX is 32. Draw!


(2) An “altitude” is a height. Area depends not only on height, but on height and base. This statement doesn’t give us any information about any base, so we have no way to determine how large any of the triangles is.



Note that we can express all bases and heights in terms of the same variable. The base of the entire ABC triangle is 4b and its height is 4h. The triangle whose area we want (RCS) has a base of b and a height of h. The area of a triangle is half the base times the height

Notice that ABX has a base of 2b and a height equal to the entire height, of 4h. Since the area of ABX is 32 we

can write: ( )( )1 2 4 322

b h = . From

this equation we can find bh and answer our rephrase.


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Take-Aways



We’re told that the product of the three digits is 96. Because this problem is about factors it makes sense to

express our numbers in their prime factorization. Thus the product is 596 32 3 2 3= ⋅ = ⋅ . What is the units digit?

(1) Since m is odd, the units digit cannot be even (it cannot contain 2 as a factor). Since all the prime factors

of the digits combined are 52 3⋅ and the units digit must be odd, the units digit must be 3. Note that the units digit can’t be 1 because it would then be impossible to make the other digits large enough for all three digits to have the given product of 96. The units digit of m is 3.


(2) Since the product of all three digits is 96 and the first digit is 8, the last two digits must have a product of 96 8 12÷ = . The last two digits could be (4,3), (3,4), (6,2) or (2,6). There is no way to know the value of the units digit.



Take-Aways


connections.


Each person received x pens, y pencils and z pads. How many people are there?

(1) The ratio x to y to z is 2:3:4. This gives us information about the package that each person receives, but gives us no information about the number of people. In general, ratio information by itself is never sufficient to determine exact number of items because ratios don’t tell us how large or small the numbers are. There could be 2 pens, 3 pencils and 4 pads but there could also be 20 pens 30 pencils and 40 pads


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per person. Even if we knew how much each received, we wouldn’t know how many people there are in total.


(2) 18 pens, 27 pencils and 36 pads were distributed in total. Without knowing how much each person received we can’t determine how many people there are. For instance, if each person received 6 pens, 9 pencils and 12 pads, there would be 3 people total. On the other hand if each person received 2 pens, 3 pencils and 4 pads, there would be 9 people total. Because there are at least 2 ways to equally distribute the 18 pens, 27 pencils and 36 pads, we have no idea how they were distributed and how many people there are.


MERGE STATEMENTS

Statement (2) told us that there are 18 pens, 27 pencils and 36 pads total. This is a ratio of 2:3:4. Because each person received the same number of items, each person must have received items in this ratio. Therefore, the information in statement 1 can be deduced from statement 2. For that reason, adding statement 1 to 2 is no more useful than using statement 2 by itself.



Take-Aways

• In data sufficiency, when the data from one statement can be logically deduced from the other statement, C cannot be the right answer. When one statement offers the same data (no more, no less) as the other statement, the correct answer must be D or E


Rephrase “what fraction of the lot did X fill in 4hrs?”

X filled some of the lot in 4hrs and Y filled the rest in 3hrs. To know how long it would have taken X to do the whole lot, we need to know what fraction of the lot it filled in the 4hrs. If X had filled 1/2 the lot in the 4hrs that it worked, then it would have taken X 8hrs to do the whole lot.


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(1) This gives us X’s rate (30 bottles/min). With this we can determine exactly how many bottles X produced in 4hrs. However because we don’t know how many bottles it would take to fill the lot, there is no way to determine how long X will take at its rate to fill the lot.


(2) This gives a ratio of 2-to-1 between what X did and what Y did. We can express both amounts in terms of the same variable.

If you wanted to actually find the time asked (a waste of time on the test), note that the number of bottles produced is directly proportional to the time spent working, so we can setup a proportion:time to produce 2k time to produce 3k 4hrs 6hrs

2 3 2 3t t

k k k k= → = → =



Take-Aways

• One great technique for solving ratios is to express all parts of the ratio in terms of the same variable. For instance, if a problem tells you that the ratio of boys to girls is 4:5, you should use 4x and 5x as the number of boys and girls. By writing all values in terms of the same variable, you greatly simplify the problem (in general, the fewer variables the easier a problem will be to solve)


Let t be the total number of passengers.

• “Two thirds of passengers were employees and the rest were their guests” means “Employees and

Guests ”

• “Three fourths of employees were managers” means “Managers ”.

• “What was the number of employees who were NOT managers?” This is all employees minus the

managers:

23

t=

13

t=

3 2 14 3 2

t t = =

2 13 2

t t −

So X produced 2k bottles in 4hrs. To fill the lot, X will need to produce 3k bottles. Thus we can answer our rephrase. In 4hrs, X filled 2/3 of the lot. STOP HERE on the test.


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The only thing that stands in the way of finding employees who were NOT managers is t

Our rephrase “What is t, the total number of passengers?”



(2) Since Guests , this statement tells us that 230 . We can solve for t and answer our rephrase.






“The length of the edging of K is half the length of the edging of G” means the circumference of K is half the

circumference of G. Let and be the radii of K and G. We can write . To find the area of

K, we need its radius. Isolate in the equation to get . The only thing that stands between us and the

radius of K is the radius of G.

Our rephrase “What is , the radius of G?”

(1) . We can solve for and answer our rephrase.


(2) . We can solve for and answer our rephrase.



13

t=13

t=

Kr Gr12 22K Gr rπ π=

Kr12K Gr r=

Gr

2 25Grπ π= Gr

2 10Grπ π= Gr


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• Area of a circle is and circumference is where r is the radius and d the diameter


min(10,w) is the smaller of the two values.

Rephrase “What is the smaller value between 10 & w?”

(1) ( )max 20,w z= . ( )max 20, z is whichever is greater. This value cannot be less than 20. Since this

max is w, it follows that w cannot be less than 20. As a consequence, ( )min 10, w must equal 10

because 10 20 w< ≤ . We can definitively answer the question. ( )min 10, 10w =


(2) ( )max 10,w w= . Since w is the bigger (the max) of the two values, then 10 must be the smaller value.

( )min 10, 10w =




If a question deals with the average and the number of items, realize that the key to solving is to focus on the total because it’s the total that links the other two. In this case there are 6 days and we’re asked whether the average is

greater than 90. If average were exactly 90 then total 90 540

6T= → = . For the average to be greater

than 90, the total would have to be greater than 540.

Rephrase “Is the total greater than 540?”

(1) The top 4 days had 100 people on average. So the total for these days is 400. What about the smallest 2 days? Well we know from the prompt that the only restriction is that the smallest value is 80, so the

2rπ 2 r dπ π=


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smallest that these two days can add up to is 80 2 160⋅ = . Thus the smallest possible total for the 6 days is 400 160 560+ = . Since the smallest total possible is greater than 540, we can answer the rephrase with a definitive YES. Total is greater than 540.


(2) The 3 smallest days had 85 on average, so the total for these days is 85 3 255⋅ = . What about the greatest 3 days? Well the smallest average that they can have is 85 since they can’t have fewer people than the smallest 3. In other words, the smallest possible total for the greatest 3 days is 255. The overall total is at a minimum 255 255 510+ = . Since there is no upper limit on how many people the highest day can have, the overall total could be above 540 as well. So this statement doesn’t allow us to know for sure whether the total over the 6 days is greater than 540.


Notes: in statement 2, as soon as I realized that it would be possible to get a total under 540, I knew that the statement would be insufficient. This is because statements can never contradict one another, so since statement 1 proved that the total is at least 560, statement 2 must also allow this total. Thus the fact that statement 2 also allowed us to have a total under 540 meant that it would be insufficient.



(1) 3AB = and 2BC = . AB gives us the radius of the small circle, so we can definitely find the area of the small circle. To find the area of the big circle, we need its radius, AC. Because we have both AB and BC, and because these segments are joined to make AC, we can determine that the radius of the big circle is 3 2AC = + . Thus we have all we need to find each area. With these we can do big area – small area to get the shaded area.


(2) 1CD = and 4DE = . By joining these segments, we can find the radius of the big circle: 1 4CE = + , so we can find the big area. To find the shaded area, we still need the smaller area. Can we find the smaller circle’s radius? It may not seem so at first but there is a way to find its diameter (AD) and the radius is half of the diameter:

We’re looking for the shaded area. This is the area of the big circle, minus the area of the small circle.

Rephrase “What is the difference between areas?”


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The small circle’s diameter is 6, so the small radius is 3. This statement gives us botht the big radius (5) and the small radius (3), so we can find the area of each circle and the shaded area (difference between areas)




handle. Always make a drawing of your own for geometry questions, even if one is already provided • Area of a circle is and circumference is where r is the radius and d the diameter


To determine how much the employee was paid we need know three things, so the rephrase is a little lengthy:

• How many overtime hours did she work? AND

• How many of those hours were on Sunday (paid at the higher rate)? AND

• What is her hourly rate?

(1) This tells us nothing about the first two parts of our rephrase. We know neither number of overtime hours, nor the number of Sunday hours.


(2) This tells us nothing about her hourly rate.


MERGE STATEMENTS

2rπ 2 r dπ π=

Image not drawn to scale. AC = CE = 5, so the big diameter (AE) is 10. Note that the small circle’s diameter equals the big diameter minus the last segment (DE). Therefore the small diameter is 10 – 4 = 6.


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Together, the statements answer the 1st and 3rd parts of our rephrase. However, we do not know how many hours were worked on Sunday. The employee could have worked 6hrs on Sunday and 8hours the rest of the week (a total of 54 hours). On the other hand, she would have received a higher pay if she had worked 8hours on Sunday and split the other 46hrs among the other days of the week.




The revenue from full-priced tickets is the number of tickets sold times the full price. Similarly, the revenue from reduced priced tickets is the number of reduced-priced tickets sold times the reduced price. So to find total revenue, we need the number of tickets sold of each type, and we need to know the two prices.

Our rephrase “How many tickets were sold at each price AND what are the two prices?”

(1) This doesn’t give us the prices, so we cannot calculate total revenue.


(2) This doesn’t give us the reduced price, nor does it tell us how many tickets were sold at each price. We cannot calculate the total revenue.


MERGE STATEMENTS

Together, the statements give us almost all we need. We can find how many tickets were sold at each price from statement (1) and we can find the full price from statement (2). Unfortunately we have no way of finding the reduced price.





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by 5%, you would multiply it by 0.95, or 51

100 −

. Algebraically, to increase a value by x%, multiply it by

1100

x +

. To decrease a value by y%, multiply it by 1100

y −

. This is worth memorizing. In our case,

because the question deals only with percentages without limiting us to a concrete value, we can plug in a starting amount. Let the starting value be $1. Below are the values for each year:

1997 1998 (increase previous year by x%) 1999 (decrease previous year by y%)

1 1100

x +

1 1100 100

x y + −

Let’s simplify 1999 by the FOIL operation: 1 1 1100 100 100 100 10,000

x y y x xy + − → − + −

. The question

asks whether 1999’s value is greater than 1997’s: “Is 1 1100 100 10,000

y x xy− + − > ?” Subtract 1 from each side

and multiply everything by 10,000 to get rid of fractions. You will get: “Is 100 100 0 100 100y x xy x y xy− + − > → − > ?’

Rephrase “Is 100 100x y xy− > ?”

(1) 0x y x y> → − > This statement tells us that the left side of our rephrase, 100x – 100y, is positive. However it gives us no information about xy, so we can’t answer our rephrase.


(2) Multiply by 100 to get rid of fractions: 100 100100xy x y xy x y< − → < − . This directly answers

our rephrase.


Notes: Getting rid of fractions in the rephrase (by cross-multiplying) makes the question easier. In general, it’s a good idea to get rid of fractions in Data Sufficiency because doing so makes data easier to evaluate.


There is an x% increase, then a y% decrease. Suppose you wanted to

increase a value by 10%. You would multiply it by 1.10 or 101

100 +

.

Similarly, if you wanted to decrease a value


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Take-Aways


• To increase a value by x percent, multiply it by 1100

x +

. To decrease a value by y percent, multiply it

by 1100

y −

#121, pg. 283 Difficulty Level: 700+ Topics: Coordinate Geometry; Inequalities & Absolute Values

Does the point (r,s) lie in the region 2 3 6x y+ ≤ ? For the point to be in the region, the coordinates of the point must fit the equation.

Rephrase “Is 2 3 6r s+ ≤ ?”

(1) 3 2 6r s+ = . The easiest way to approach the statement is to plug in values, trying our best to force the statement to give us conflicting answers to the question. If we find out that the statement provides conflicting answers, we would have proven that it’s not sufficient. On the other hand, if it’s not possible to get conflicting answers, we will be forced to conclude that the statement is sufficient. Remember that whatever you plug in must agree with the statement:

• To get the statement to answer the rephrase with a YES, we need to make 2 3r s+ small, which means we need to pick a small value for s. If 0s = and 2r = (this statement is respected), the question “Is 2 3 6r s+ ≤ ?” will be answered with a YES

• To get the statement to answer the rephrase with a NO we need to make 2 3r s+ large, which means we need to pick a large value for s. If 3s = and 0r = (this statement is respected), the question “Is 2 3 6r s+ ≤ ?” will be answered with a NO

Because we’re able to get conflicting answers, we don’t have enough data to answer the question


(2) 3r ≤ and 2s ≤ . My strategy here would be the same as in the last statement. Plug in values that agree with the statement in an attempt to get conflicting answers to the original question. When employing this strategy, always check whether values you’ve used before can be used again with the current statement. for instance, we can re-use 0s = and 2r = from the 1st plugin in the previous statement.

• If 0s = and 2r = , the question will be answered with a YES (no need to go through the work again. We’ve already used these values so we know how they answer the Q – this is why it’s good to recycle as much as possible. Saves time.).


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• Next, we should try to get the statement to answer with a NO. To accomplish this, we need to make 2 3r s+ large, so let’s pick the largest values available. If 2s = and 3r = , the question “Is 2 3 6r s+ ≤ ?” will be answered with a NO

Because we’re able to get conflicting answers, we don’t have enough data to answer the question


MERGE STATEMENTS

Once again, try to get conflicting answers. This time however, the values we use must agree with both statements because we’re merging statements. As I mentioned in statement 2, you should first check whether something you’ve used before can be used again. In particular 0s = and 2r = worked for both statements so it can be used here.

• If 0s = and 2r = , the question will be answered with a YES (no need to go through the work again; these are recycled values)

• Is it possible to get a NO? To accomplish this, we need to make 2 3r s+ large so it’s important to maximize the value of s. According to statement 2, 2s ≤ so let’s use 2s = . According to statement 1,

3 2 6r s+ = , so if we use 2s = we must use ( ) 23 2 2 6 3

r r+ = → = . If 2s = and 23

r = , the

question “Is 2 3 6r s+ ≤ ?” will be answered with a NO





The size of the box is the product of its three dimensions: Length, width, and height.

Rephrase “What is l w h⋅ ⋅ ?”

(1) Boxes are made up of 6 faces grouped into pairs: front & back, left & right, top & bottom. Each pair has two identical faces. This statement gives us the size of two adjacent faces so in fact it tells us about two of the three pairs.


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(2) This statement gives us information about one pair of faces, but nothing about the other two pairs, so we have no idea how large the box is. We can’t find its volume.


MERGE STATEMENTS

Together, the statements allow us to find the exact size of each face.


Notes: The official guide does a great job of showing how to actually solve for the volume. On the exam though I wouldn’t take the time to do this.





To avoid working with decimals, convert dollars to cents. Some stamps cost 15 cents and others cost 29 cents. How many 15 cents stamps were purchased?

This information gives us no idea how large the last pair of sides is so we have an incomplete picture and cannot determine the size of the box. We can’t find its volume.

It’s a waste of time to actually find the volume. Because we know the shape (rectangle) and size of each face, we know how big the box is and can answer the question


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(1) Let x and y be the number of 15 and 20 cent stamps respectively. Joanna must have spent 15x and 29y

on the stamps. Since she spent a total of $4.40, we can write 15 29 440x y+ = . Because this is an equation with two variables, it seems we won’t be able to solve. However, because the variables must be integers (they represent number of stamps) the possible values are limited. The question is, are they limited to a unique solution? Well in algebra, if you’re looking to solve an equation in the format above, you will have sufficient information to solve if: • The variables are integers • The total is less than the Least Common Multiple of the coefficients + the smaller coefficient.

In our case, the least common multiple of 15 and 29 is 15 29 435⋅ = and the smaller coefficient is 15. Since the total, 440, is less than 435 15+ , and since our variables are integers, we can be certain without doing any work that there is only 1 set of values that will work. This statement limits us to a unique value for x, the number of 15 cent stamps.


(2) x y= The same number of each stamp was purchased. We have no idea what that number is; we can’t find the number of 15 cent stamps.

Statement (2) is NOT SUFFICIEN



(1) Since 40 people did not respond Favorable to either, the remaining 60 people responded “Favorable” to one or both candidates. According to the table, there were a total of 70 favorable responses. How is this possible? Well that’s simply because there must be 10 people who were counted twice. To understand why it has to be so, consider three voters: A, B and C. If A and B are favorable to candidate 1, and if B and C are favorable to candidate 2, there will be 4 favorable responses even though there are only three people. That’s because 1 person (voter B) is counted twice. Thus when we’re told that 60 people responded favorable to one or both candidates and according to the table there are 70 favorable responses, we can conclude that 10 people are counted twice because they responded “favorable” for both candidates. The answer to the question is 10.


How many of the 100 voters responded “favorable” to both candidates?


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(2) Knowing that 10 people responded “unfavorable” to both candidates would allow us to find out the total number of people who responded unfavorable. But even once we find that, there is no way to figure out information about the “Favorable” group because we have the “Not Sure” group to consider as well. Without being able to determine how many people responded “favorable” to either candidate, we cannot find how many responded “favorable” to both candidates.




Of the three possible operations which one(s) can satisfy ? Only

multiplication would work

Our rephrase “Is ‘ ’equal to multiplication?”

(1) 1 1k k° ≠ ° Since is always equal to , we know from this statement that ‘ ’is not multiplication. We can answer our rephrase.


(2) The statement directly answers our rephrase. ‘ ’is not multiplication.



#126, pg. 284 Difficulty Level: 700+ Topics: Sets & Groups

The overlapping group formula is . In our case, let P and S be the

number of cars with power windows and stereos respectively. . We’re asked to find neither.

Our rephrase “What is neither?”

(1) The number of cars with stereos but no power windows is all cars with stereos minus cars with both. This results in cars with stereos only: . We can plug this data into our initial equation to get

( ), , and + − × ( ) ( ) ( )k l m k l k m° + = ° + °

°

1k × 1 k× °

°


60 P S neither both= + + −

20S both− =


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. Without a value for P, we cannot solve

for neither


(2) Statement 2 gives us so we can plug this data in our initial equation to get. Without a value for we cannot find neither.


MERGE STATEMENTS

Together, the statements aren’t any more helpful than they were alone because we still cannot find P to plug into our initial equation. Without P we cannot solve for neither.

Another approach is to observe that statement (1) tells us that , and statement (2) tells us that, so we can deduce that . If we plug all the data we have into the original equation, it goes

from . Without P, we cannot solve for neither.



Take-Aways • The group formula, 1 2Total Group Group neither both= + + − , is a useful way to solve questions about



The group formula is . In our case, we know there are 300 students

and that everyone studies one or both languages, so we can ignore neither. Our equation iswhere F and S are the French and Spanish students.

“100 do not study French” refers to Spanish-only. This is found by taking all Spanish minus both. “”. We are asked to solve for both. The only thing that stands in our way of finding both is S

Our rephrase “What is S, the number of Spanish students?”

(1) “60 do not study Spanish” means that the rest, or 240 study Spanish: . This statement directly answers our rephrase.


( )60 60 20P neither S both P neither= + + − → = + +

30both =60 30P S neither= + + − P S+

20S both− =30both = 50S =

60 60 50 20P S neither both P neither= + + − → = + + −


300 F S both= + −

100S both− =

240S =


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(2) . This statement directly answers our rephrase (gives us data identical to statement (1)).





• In data sufficiency, when the data from one statement can be logically deduced from the other statement,

C cannot be the right answer. When one statement offers the same data (no more, no less) as the other statement, the correct answer must be D or E


For n students to be evenly assigned to m classrooms, n must be divisible by m. For example if there are 3 classrooms, there would need to be 3, 6, 9… students. Therefore the question is “Is n divisible by m (does the division have an integer result)?”

Rephrase “Is intnm= ?”

(1) For 3n students to be evenly assigned to m classrooms, 3n must be divisible by m. We can write 3 intnm

= . So 3n is divisible by m, but is n itself a multiple of m? The prompt told us that

3 13m n< < < . For 3nm

to divide evenly, it could be that nm

divides evenly. For example, if n=30 and

m=6, then n would be divisible by m. On the other hand, if 3 and m have a common factor, it’s possible

for nm

to not be an integer. For example, if n=20 and m=6, 3nm

would be an integer, but n would not be

divisible by m. Because we get conflicting answers to the question, this statement is not sufficient.


240S =


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(2) 13n students can be evenly assigned to m rooms, so 13n is divisible by m: 13 intnm

= . The prompt told us

that 3 13m n< < < , so m is a number between 3 and 13. For 13nm

to divide evenly, it must be true that

nm

divide evenly because 13 and m have no common factors.






The median is the middle value. What is the median number of employees?

(1) 25% of projects have 4 or more employees. This gives us information about the top 25% of values. The median is the value that is above the bottom 50% but below the top 50%. This statement gives us no information about what that value might be, except that it must be less than 4.


(2) 35% of projects have 2 or fewer employees. This gives us information about the bottom 35% of values, but no information about the top 65%. The median is at the 50% mark, so we need more information to find out the median number of employees. What this statement does is guarantee that the median be greater than 2.


MERGE STATEMENTS

Statement (1) tells us that median is less than 4. Statement (2) tells us that median is greater than 2. Together, the statements guarantee that the median = 3.




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No rephrase needed. Was the appointment on Wednesday?

(1) Let’s call t, the time of the appointment. We are told that . Since the question is

about days of the week, it would be smart to convert 60hrs into days. .

Therefore, .

The appointment can be either Wednesday after 12pm or Thursday before 12pm.


(2) This statement gives no information related to any specific day of the week.


MERGE STATEMENTS

Statement (1) told us that the appointment is either Wednesday after 12pm or Thursday before 12pm. Since the appointment is between 1pm and 9pm (from statement (2)), we can conclude that it must be on Wednesday.




Let h and t be the number of heads and tails respectively. There are “4 more heads than tails” means “ ”. We’re asked to find h. The only thing that stands in our way is t.

Our rephrase “What is h?” OR “What is t?”

60hrs Mondayt − =1 day60hrs 2.5 days24hrs⋅ =

2.5 days Monday Monday 2.5 dayst t− = → = +

4h t= +

60hrs or 2.5 days

Monday Tuesday Wednesday Thursday

Moving Monday forward by 2.5 days gives us a t between noon Wednesday and noon Thursday


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(1) . This statement gives us a 2nd independent linear equation and is therefore sufficient to solve for h and t. Don’t bother solving (though you could by replacing h with t+4 thanks to the equation in the prompt)


(2) . This statement gives us a 2nd independent linear equation and is therefore sufficient to solve for h and t. Don’t bother solving (though you could be replacing h with t+4)






(1) 95w = From this we cannot find w z+ and answer the rephrase, or find x y+


(2) 125z = From this we cannot find w z+ and answer the rephrase, or find x y+


MERGE STATEMENTS

Together, the statements tell us that 95 125w z+ = + , so they answer our rephrase.


24h t+ =

3 52h t+ =

Because each variable makes a straight line with an interior angle, we can express each interior angle in terms of that variable. The interior angles must add up to 360⁰ because they make up a 4-sided polygon.

180 180 180 180 360x y w z− + − + − + − = . This simplifies to

1,080x y w z+ + + = . The question asks for the value of x y+ . The only thing standing in our way is w z+

Rephrase “What is w z+ ?”


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Are all 15 numbers equal?

(1) The sum gives us a way to find the average (60 415

= ), but no way to know whether all numbers equal

that average or whether there is variation in values.


(2) The sum of any 3 numbers is 12. This means that the average of any group of 3 numbers is 4. Think of the list of 15 numbers as five groups of 3 numbers. Every group has an average of 4. This means that the average of the entire list as must be 4. Because each group must average 4, the smallest term in each group must be 4 or less. If any term from any group is less than 4, then the sum of that term and the smallest terms from two other groups will be

( ) ( ) ( )less than 4 4 or less 4 or less+ + . This sum would be “less than 12” and would therefore

contradict this statement. Thus it’s not possible for any term to be less than 4 without contradicting the statement. Furthermore, because the average of the list as a whole must be 4, if no term can be less than 4, then no term can be greater than 4 either. Every term equals 4.


Notes: If you were to guess, you can safely eliminate answer C because statement 1 (average of the whole list = 4) can be deduced from statement 2 (average of any 3 numbers is 4). As a result combining the statements would be no more helpful than looking at statement 2 alone. You should be able to get it down to B or E.




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No rephrase needed. What is standard deviation?

(1) Standard deviation is a measure of the average distance between each value and the mean. This statement gives us the mean, but no way to calculate whether the values are close to the mean or spread far away from the mean. We cannot determine the deviation of values.


(2) If all values are the same, there is no deviation. Standard deviation = 0



Take-Aways • For the purposes of the GMAT, you do not need to memorize the formula for standard deviation.

However, you need to have a good understanding of what standard deviation means. Broadly, standard deviation is a measure of how tightly/loosely values are grouped around the average. The more dispersed the values, the higher the standard deviation. If all values are the same, standard deviation is 0


There are two ways to find the area of a trapezoid.

• Typically, one would use the area formula. The area of a trapezoid is . We already have

one base (ST) as well as the height. The only thing standing in our way of solving this way is the 2nd base (RU).

• A 2nd way to find the area is to divide the trapezoid into two triangles and one rectangle, and add up the areas of all three parts. I’ve added point X for illustrative purposes.

Let’s resume; in order to solve using the area formula, we need RU. To solve by summing the three smaller areas, we need XU.

Our rephrase “What is RU?” OR “What is XU?”

1 2

2base base h+

⋅

R

S T

U W X

15

60 60

45 We already have the data to find the areas of the left-hand triangle, and the middle rectangle. What we lack is the area of TUX, or more specifically its base, XU


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(1) 80RU = . This statement directly answers our rephrase


(2) TU gives us the hypotenuse of triangle TUX as illustrated above. Since we already have the height (TX), we can use the Pythagorean Theorem to solve for XU:

Once we have XU, we can answer our 2nd rephrase.





• There are two ways to find the area of a trapezoid. One way is to use the area formula, .

The 2nd way is to divide the trapezoid into two triangles and the rectangle in the middle, and add up the areas of these three parts.


No rephrase needed. Average is 75. How many numbers equal 75?

(1) Since there is no number under 75, there must not be any number over 75, otherwise (if one or more numbers were over 75) the average would be bigger than 75. All 6 numbers must equal 75


(2) Since there is no number over 75, there must not be any number under 75, otherwise (if one or more numbers were under 75) the average would be smaller than 75. All 6 numbers must equal 75




( )22 2 2 2 2 60 20 10TX XU TU TU+ = → + =

1 2

2

base baseh

+⋅


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Let d and b be the price of a donut and bagel respectively.

Rephrase “What is 5d + 3b?”

(1) 10 6 12.90d b+ = (Eq.1). This gives us the price of twice what we want. To find 5 3d b+ , just divide (Eq.1) by 2.


(2) Bagels cost $0.15 more than donuts. The difference between prices gives us no idea how expensive these items are. If bagels and donuts cost $0.50 and $0.65 each, the price of 5 donuts and 3 bagels would be a lot less than if bagels and donuts cost $5.00 and $5.15 each.




To find the total revenue, we need 3 pieces of information

• The full price • The reduced price (the value of x) • How many tickets were sold at each price

(1) This tells us neither how many tickets of each price were sold, nor what the full price is.


(2) This tells us neither how many tickets of each price were sold, nor what the reduced price is. Statement (2) is NOT SUFFICIENT

MERGE STATEMENTS

Together, the statements tell us the full price and the reduced price, but fail to tell us how many tickets of each price were sold.


Notes: The formula for finding the total revenue is where n is the number of

tickets sold at full price and full is the full price.

( )400100

xR n full n full = ⋅ + −


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The question needs no rephrase. Is rs

a terminating decimal?

(1) 90 100r< < . This statement tells us nothing about the denominator s. If 955

rs= , it will be a

terminating decimal. On the other hand if 953

rs= it will never terminate (decimal results of division by

three are either .333… or .666…) We cannot determine whether is a terminating decimal.


(2) The decimal portion of is either depending on the remainder of the division. All these

options are terminating decimals, so must be a terminating decimal.




(1) This statement tells us nothing about y, or even about triangle ADC.


rs

4r 1 2 30, , , or

4 4 4rs

Notice that y x> . However, since the two triangles must have 180⁰, the difference between y and x must be made up by the two bottom corner angles. Specifically, y x− must be equal to

BAD BCD∠ +∠

The question needs no rephrase. What is x y+ ?


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(2) This statement is not helpful in finding any degree measures.


MERGE STATEMENTS

Since and the big triangle is isosceles, must each equal 55 degrees. This restriction however doesn’t help us find the value of y. Consider the two scenarios below:

Both of these scenarios agree with the statements, but they yield different values for .



#141, pg. 285 Difficulty Level: 500-600 Topics: Weighted Averages

X and Y merge to form committee Z. Does X have more members than Y?

(1) This gives us the average age of X and of Y. We have no way to determine whether these are the average of a hundred people or a thousand people. Without knowing information about the total of all ages, we cannot determine or compare the number of people in the committees.


(2) This gives us the average of everyone combined. This data gives us no way to split the whole group into X and Y. There could be 1,000 people evenly divided into X and Y, or there could be 800 in X and 200 in Y. We only know what the overall average is.


MERGE STATEMENTS

70x = and BAC BCA∠ ∠

x y+

70

120

30 30

25 25

70

90

45 45

10 10


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Statement (1) tells us that the people of X are 25.7 years old while the people of Y are 29.3 years old. If everyone or almost everyone was from X, then the average of the entire population would be equal to or almost equal to 25.7. On the other hand, if almost everyone was from Y, then the average of the entire population would be very close to 29.3. Finally, if the same number of people belonged to X and to Y, the overall average would be exactly in the middle. This is the concept of weighted averages: the overall average is closer to the average of the largest group. Because statement (2) gives us the overall average (26.6), and because this average is closer to X=25.7 than to Y=29.3, it must be true that X is the larger group. X has more members.




The question needs no rephrase. What was earning from commissions?

(1) Commission = 5% of sales. To answer the question we need to know the total sales.


(2) This gives us no information about the commission


MERGE STATEMENTS

Statement (2) tells us that she sold $10,000 more per month in the 2nd half of the year. Each half-year has 6 months, so statement (2) tells us that sales in the 2nd half were $60,000 more than sales in the first half. If she sold $1,000,000 and the first six months and $1,060,000 in the last six months she would have earned a lot of money in commission. On the other hand if she sold $100,000 in the first six months and $160,000 in the last six months, she would have earned a lot less in commissions. To find her commission we need to know more about the actual amount of sales.




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Let’s represent all values algebraically. Let x and y be the original values of stocks X and Y. After the 10% price

changes, the increased price of X is 1.1x and the decreased price of Y is 0.9y. Percentage is , so “The

reduced price of Y is what percentage of the original price of X?” can be rephrased as “What is ?” The

only thing that stands in our way is .


(1) Algebraically, this statement states that . We can divide both sides by x to find and answer

our rephrase.


(2) The 10% increase in x is equivalent to an increase of 0.1x. Similarly, the 10% decrease in y is a decrease

of 0.1y. Algebraically, the statement states that . This equation can be manipulated to

find and answer our rephrase.






100partwhole

⋅

0.9 100yx

⋅

yx

yx

1.1x y= yx

( )100.1 0.111

x y= ⋅

yx

Let the sides of the quadrilaterals touching the triangle be a, b, and c.

The area of the triangle, . The side we are asked

for– a – is the hypotenuse of the right triangle and can be found using the Pythagorean Theorem


4 82

b c b c⋅= → ⋅ =

2 2 2a b c= +

2 2b c+


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(1) The area of a square is . This statement states that . We can plug this

value of b into , find c, then solve for


(2) The area of a square is . This statement states that 2 64 8 9 3

c c= → = . We can plug this

value of c into , find b, then solve for


Notes: We can ignore negative roots because b and c are lengths and must therefore be positive.





Let s and b be Sara and Bill’s ages. We’re given one linear equation: The only thing standing in our way of solving for s is b.

Our rephrase “What is s?” OR “What is b?”

(1) This statement gives a 2nd independent linear equation and is therefore sufficient. 4 years ago, Sarah was and Bill was . This statement can be written as . Combine it with 2s b=

to find Sara’s age.


(2) This statement gives a 2nd independent linear equation and is therefore sufficient. In 8 years, Saray will be and Bill will be . This statement can be written as Combine it with

2s b= to find Sara’s age.



side side⋅ 2 9 3b b= → =8bc = 2 2b c+

side side⋅

8bc = 2 2b c+

2s b=

4s − 4b − ( )4 3 4s b− = −

8s + 8b + ( )8 1.5 8s b+ = +


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#146, pg. 286 Difficulty Level: 600-700 Topics: Translations & Manipulations; Inequalities & Absolute Values

Average words per paragraph is . Algebraically, the question can be written as “Is

?” Since we know that the first 23 paragraphs contain

2,600 words, we can further simplify the question to the rephrase below:

Rephrase “Is the total number of words in the 2-paragraph preface smaller than 400?”

(1) This means the total number of words in the 2-paragraph preface is more than 200. It may be 250 or 500 words. We cannot compare this total to 400, so we do not have sufficient data to answer our rephrase.


(2) This means that the total number of words in 2-paragraph preface is smaller than 300. Therefore, this total must be smaller than 400. We have sufficient data to answer our rephrase.




Once all values are ordered, the median is the middle value, or the average of the two middle values if there are an even number of values. For example, the median of {6, 3, 9} is 6 and the median of {1, 2, 3, 4} is 2.5

Rephrase “What is median?”

(1) Upper shelf has 25 books. Since we have no idea how many books there are in total, we cannot determine what the median is.


(2) Lower shelf has 24 books. There could be 25 books total (if top shelf has 1 book) or there could be 1,235 books total. We just don’t know, so we can’t even begin to determine what the median (middle value) is.

total # of words# of paragraphs

total # of words 120 Is total # of words 3,00025

< → <


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MERGE STATEMENTS

Together, the statements tell us that there are 25 24 49+ = books. The middle value (the median) is the 25th value because there are 24 values to either side of it: 24 1 24 49+ + =

To find the median we need the 25th value. The top shelf has 25 books, so we need the value of the last book in the top shelf. The prompt gives it to us: “On the upper shelf the book with the greatest number of pages has 400 pages”. Median is 400




(1) We have no idea with side is 120. If the shortest side is x, then 120x = . On the other hand, it’s also possible that 60 120x+ = or 3 120x = . We can’t find the exact value of x.


(2) The shortest sides have length x. The other sides are 3x and 60x+ . This statement tells us that one side is twice as long as x. This side can’t be 3x (because 3x would be three times as long as x), so this statement must be referring to 60x+ . “ 60x+ is twice as long as x” 60 2x x+ = . This gives us enough data to solve x, and find the length of each side and the total length of the path.



The total distance of the path is the sum of all sides. The only thing preventing us from finding this sum is the value of x

Rephrase “What is x?”


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#149, pg. 286 Difficulty Level: 700+ Topics: Coordinate Geometry

is significant because it lets us know that in the figure below

Area is . “Is area greater than 48?” means “Is ?”


(1) Let’s put this data in our drawing

This statement establishes that , thus answering our rephrase


(2) This statement gives us the base. Because it tells us nothing of the height, we cannot find out how the area compares to 48.



OP PQ< OR RQ<

base height2 2

OQ PR⋅ ⋅= 48

2OQ PR⋅

>

96OQ PR⋅ >

96OQ PR⋅ >

O

P (6, 8)

Q R

The y-coordinate of P is the height. SinceOR RQ< , 6RQ > and the base 12OQ > .

With this data we can say that OQ PR⋅ is

( )( ) ( )more than 12 8 more than 96= 8

6 More than 6

O

P

Q R

OP and PQ have the same elevation (their elevation is the height of the triangle). Since OP is shorter, it must cover a shorter distance along the x-axis. This distance, OR, is shorter than the RQ.


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handle. Always make a drawing of your own for geometry questions, even if one is already provided


Simplify the equation

The only thing that stands in our way of solving for S is the value of


(1) Divide by n to get


(2) Plugging this data in our rephrase allows us to simplify the question from “What is ?” to “What is

?” We do not have sufficient data to answer the question.




n is a positive integer and 5.1 10nk = × . This means that k is a number such as 5.1 10× , 5.1 100× … so k could be 51, 510, 5100, 51000… The only thing standing in the way of finding k is n

Rephrase “What is n?” or “What is k?”

2 22 3 6

3 2 5 5 53 3 3

x xn nSn n

x x x

= = = ⋅ = ⋅+

xn

xn

2xn=

xn 1

2

x


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(1) 6,000 500,000k< < . When you consider the type of number that k must be (51, 510, 5100…) the only number that fits this statement is 51,000. 5,100 and 510,000 are too small and too large respectively.

51,000k =


(2) 2 92.601 10k = × . This gives us the value of 2k . Typically, knowing the square wouldn’t be sufficient

because there would be two solutions (for example, 2 4 2 or 2x x= → = − ) In this question

however, the prompt tells us that 5.1 10nk = × so we know that k is a positive number. We don’t have

to worry about the negative root. 92.601 10k = + ×




Let c and r be the number of tapes that Carmen and Rafael have. If Carmen had 12 more tapes, she would have 12c+ . We’re told that if she had that many tapes, she would have twice as many as Rafael, so 12 2 2 12c r c r+ = → = − (Eq.1). “Does Carmen have fewer tapes than Rafael” can be written as “Is

c r< ?” If we replace c with its equivalent from (Eq.1), the rephrase would become “Is 2 12r r− < ?” Simplify.

Rephrase “Is 12r < ?” Does Rafael have fewer than 12 tapes?

(1) 5r > . Knowing this doesn’t tell us whether Rafael has fewer than 12 tapes. He may have 6 tapes or 30 tapes.


(2) 12c < . What does this tell us about r? Well thanks to (Eq.1), we can replace c with 2 12r − and rewrite this statement as 12 2 12 12c r< → − < . Simplify to 2 24 12r r< → < . This statement directly answers our rephrase


Notes: To combine equations (=) and inequalities (<, >), isolate a variable in your equation and replace it in the inequality with its equivalent.



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The question could be rephrased but rephrasing may not be worth the time. Is 2xx x < ?

(1) 0x < . Since x is positive and x is negative, the left side, ( )( )x x negative positive= , so the left

side is negative. By contrast, the right side, 2x , is always positive. Even a negative exponent won’t

change the sign. For instance, 22

122

− = . Thus, this statement tells us that 2xx x < because the left

side is negative and the right side positive.


(2) 10x = − We could plug in -10 for x in the question to figure out whether .



#154, pg. 287 Difficulty Level: 700+ Topics: Exponents & Roots; Inequalities & Absolute Values

To understand this question, remember that and also that . Algebraically, we’re

asked “Is ?” By distributing 2, we get “Is ?”


(1) By inputting statement (1)’s data into our rephrase, we can further simplify the question to “Is ?”

Add to both sides “Is ?”

Divide both sides by “Is ?”

Simplify “Is ?”

We don’t need any more information to know that the answer to the question is YES.

2xx x <

m n m nx x x +⋅ =m

m nn

x xx

−=

( )2 3 2n nb a− ≥ − 2 3 2 2n nb a− ≥ ⋅ − ⋅

12 3 2n nb a +− ≥ ⋅ −

1 1 13 2 2 3 2n n n n+ + +− ≥ ⋅ −

12n+ 13 2 3n n+ ≥ ⋅

3n13 2 3

3 3

n n

n n

+ ⋅≥

3 2≥


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(2) This statement gives us no information about a or b




“For every dollar spent in 1970, $3.56 had to be spent in 1989” means that if p had been the price in 1970, then 3.56p was the price in 1989. We’re asked for the price of the mixer in 1970.

Our rephrase “What is p?”

(1) The difference between the two prices, . We can solve for p and answer our rephrase.


(2) This tells us that 1989’s price, . We can solve for p and answer our rephrase.







(1) Divide both sides by 5 to get 15 3,000 5 600

5 5

kk

+

> → > . Still, we don’t know whether is

between 600 and 1,000 or is greater than 1,000

3.56 $102.40p p− =

3.56 $142.40p =

5 1000k <

5k


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(2) Multiply both sides by 5 to get

Distribute the 5 on the left side

Isolate 2500 (think of the right side as )

Simplify

Statement (2) gives us the value of , so it is sufficient to answer the question.


#157, pg. 287 Difficulty Level: 700+ Topics: Translations & Manipulations; Geometry

(1) Area of a triangle is 12

bh . In a right triangle, the base and height are the perpendicular sides. In our case

we can write 1 25 502

ab ab= → = (Eq.2). Solving quickly requires noticing that (Eq.1) and

(Eq.2) give us two parts of the same quadratic: ( )2 2 22a b a ab b+ = + + . Since 2 2 210a b+ = (Eq.1)

and since 50ab = (Eq.2), we can re-write the quadratic as ( ) ( )2 10 2 50a b+ = + . With this we can

find the value of ( )2a b+ as well as a b+ (because a and b are distances, we can ignore any negative

solution).


( ) ( ) ( )15 5 5 5 5 500k k− = −

( )5 5 5 2500k k= −

( )2500 5 5 5k k= − 5 4x x x− =

( ) 25002500 4 5 54

k k= → =

5k

Since we’re dealing with a right triangle of hypotenuse 10, we can

write 2 2 210a b+ = (Eq.1)

The perimeter is 10a b+ + . The only thing standing in our way of finding the perimeter is the value of a b+

Rephrase “What is a b+ ?”


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(2) A right triangle that is also isosceles (two equal sides) must be a 45⁰-45⁰-90⁰ triangle. The sides of these

triangles are of the ratio : : 2s s s . Thanks to this ratio we only need to know one side to find the

lengths of the other two. The prompt gives us a side (the hypotenuse): 2 10s = . This is sufficient to find all sides and solve for the perimeter.






( )22 22x xy y x y− + = − and ( )( )2 2x y x y x y− = + −

• Memorize the ratios of the lengths of the sides of 30-60-90 degree triangles ( ): 3 : 2x x x as well as 45-

45-90 degree triangles also called isosceles right triangle ( ): : 2x x x


Because each member contributes the same amount toward the $60, we can say that where n is the number of members and c is the amount of each contribution. To find the number of members, we need to know c.

Our rephrase “What is c, the contribution of each member?”

(1) This statement directly answers our rephrase. We could find n by solving


(2) This statement gives us a second relationship (equation) between c and n. Because c and n must be positive values (money and people), it is very likely that the statement will be sufficient. If we had to consider negative values, we would need to definitely do the work to find out whether we end up with a positive and negative solution.

Anyway, the statement states that if we decrease the number of contributing members by 5, we would need to increase each contribution by $2 to purchase this $60 gift. Algebraically, . We want n, so

let’s replace c with its equivalent. The original data told us that

$60n c⋅ =

4 60n ⋅ =

( )( )5 2 60n c− + =

6060 n c cn

⋅ = → =


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Replace c with

Expand (FOIL)

Subtract 60, then multiply by n

Divide by 2

Don’t bother to factor. Notice that the last term is negative, so we will have one positive and one negative solution. We can ignore the negative solution since n represents people and must be positive. Solving would give us a unique value for n, the number of members.

Statement (2) is SUFFICIENT. Notes: Statement (2)’s quadratic factors into



If 0x < , is 0y > ?

(1) means that x and y have opposite signs. Since x is negative, y must be positive.


(2) Add x to get . The statement tells us that y is bigger than an unknown negative value. This means that y could be positive, but it could also be a bigger negative (a value between x and zero).




60n

( ) 605 2 60nn

− + =

30060 2 10 60nn

+ − − =

22 300 10 0n n− − =

2 5 150 0n n− − =

( )( )15 10 0 15 or 10n n n− + = → = −

0xy<

y x>


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Rephrase “What is any one side of the triangle?”

(1) Perimeter of triangle is 20 10 2+ . The triangle is a 45⁰-45⁰-90⁰ with side lengths of ratio : : 2s s s .

This statement allows us to write 2 20 10 2s s s+ + = + and solve for s, the length of a triangle leg (this is also the radius)


(2) The length of an arc is directly proportional to its degree measure. XYZ is a 90⁰ arc and the circle as a

whole is 360⁰ so we can setup a proportion to find C, the circumference: . We have enough

data to find the circumference.



handle. Always make a drawing of your own for geometry questions, even if one is already provided. • Memorize the ratios of the lengths of the sides of 30-60-90 degree triangles ( ): 3 : 2x x x as well as 45-

45-90 degree triangles also called isosceles right triangle ( ): : 2x x x

#161, pg. 287 Difficulty Level: 700+ Topics: Statistics

In the middle of each month the balance changes by either +120 or -50 depending on whether Carl made a deposit or withdrawal that month. We know that he made a string of deposits first, then a string of withdrawals the rest of the year. The question is “what was the range of balances?”

Range is the difference between the smallest and greatest values. If Mark made deposits from January through November and only made 1 withdrawal, then the range of balances will be large because each month would increase the range by $120 (except December). On the other hand, if Mark made only 1 deposit, then withdrew money from February to December, then the range of balances will be much lower because each month would

3605 90Cπ

°=

°

Because OZ and OX are both radii of the circle, they must be equal to each other. This means that the triangle must be an isosceles right triangle. This is a 45⁰-45⁰-90⁰ triangle. The sides of these triangles are of the ratio

: : 2s s s , thus if we find one side, we will be able to find any of the others.

To find the circumference of the circle ( 2 rπ ) we need the radius. Thanks to the ratio discussed above, with any side of the triangle we can find the radius.


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only add $50 to the range (deposits decrease the balance by $50). Therefore, the range depends on how many months deposits were made.

Rephrase “How many months were deposits made?” or “When was the last deposit?’

(1) We know that May’s closing balance was $2,600. There are only two possibilities for April:


(2) We know that May’s closing balance was $2,600. There are only two possibilities for June:


MERGE STATEMENTS

The 1st scenario is accurate if Mark deposited $120 in May. In this case April’s closing balance must have been $2,480. The 2nd scenario is accurate if Mark withdrew $50 in May. In this case April’s closing balance must have been $2,650

Because this statement tells us that April’s closing balance was less than $2,625 we know that the 2nd scenario cannot be true. In short, this statement tells us that Mark deposited money in May. This doesn’t tell us how many months the deposits were made because we don’t know whether he kept depositing after May.

The 1st scenario is accurate if Mark deposited $120 in June. In this case June’s closing balance must have been $2,720. The 2nd scenario is accurate if Mark withdrew $50 in June. In this case June’s closing balance must have been $2,550

Because this statement tells us that June’s closing balance was less than $2,675 we know that the 1st scenario cannot be true. In short, this statement tells us that Mark withdrew money in June. This doesn’t tell us how many months the deposits were made because he could have stopped depositing in February, March, April or May.


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Statement (1) tells us that Mark deposited money in May. Statement (2) tells us that he withdrew money in June. Together, the statements guarantee that the last deposit was in May, so he deposited for 5 months. We can definitively answer our rephrase. Don’t waste time to actually find all balances and compute the range!






Rephrase the question by squaring both sides: “Is 2n k n+ > ?” “Is 4n k n+ > ?”. Simplify

Rephrase “Is 3k n> ?”

(1) 3k n> . This directly answers our rephrase.


(2) Isolate k: 3 2n k n k n+ > → > . Because n is positive, 3 2n n> . Although we know that 2k n>, we have no way to determine whether 3k n> . We cannot compare k to 3n, so we cannot answer our rephrase.


Notes: The transformation 2 2 a b a b> → > can only be done if we know that we’re dealing with positives. If negative values are possible, this transformation is not necessarily true. For instance, 2 5− > − but

( ) ( )2 22 5− < − .




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p & e are directly proportional, and e & i are directly proportional, so we can write and where

are constant. By multiplying these equations, we can deduce that , so p and i are also

directly proportional.

When . To find p when i is 70, we need


(1) Input the data in the relationship between e and i to get . The statement gives us

no way to find or to find


(2) Input the given data in the relationship between p and i to get . The

statement gives us a value for and thus answers our rephrase

Alternatively, since p and i are directly proportional, you could set up a proportion to solve for

p when i is 70.





• Two values are directly proportional if . As one value increases (or decreases), the other

must follow suit to maintain the constant. Two values are inversely proportional if . As one value decreases, the other must increase to maintain the constant.

#164, pg. 288 Difficulty Level: 700+ Topics: Coordinate Geometry

1p ke= 2

e ki=

1 2and k k 1 2p k ki= ⋅

1 270, 70pi k k= = ⋅ 1 2k k⋅

1 2k k⋅

2 20.5 60

e k ki= → =

1k 1 2k k⋅

1 2 1 22

50p k k k ki= ⋅ → ⋅ =

1 2k k⋅2

50 70p

=

value 1constant

value 2=

( )( )value 1 value 2 constant=


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To find the distance between any point and the origin, build a right triangle

By asking whether (r, s) and (u, v) are equidistant from the origin, the question is asking…


(1) This statement doesn’t give us any data about u or v.


(2) Inputting this data into our rephrase allows us to rewrite the question from “Is ?” to “Is

?”

Expand “Is ?”

Simplify “Is ?”

Subtract “Is ?”

Divide by 2 & simplify “Is ?”

The question has been greatly simplified, but we need more data before we can answer it.


MERGE STATEMENTS

Statement (2) simplified the question to “Is ?”. Statement (1) provides the info needed to answer this simplified question.



2 2 2 2r s u v+ = +

2 2 2 2r s u v+ = +

( ) ( )2 22 2 1 1r s r s+ = − + −

( ) ( )2 2 2 21 2 1 2r s r r s s+ = − + + − +

2 2 2 22 2 2r s r s r s+ = − − + +

2 2r s+ 0 2 2 2r s= − −

0 1 ? Is 1r s r s= − − → + =

1r s+ =

(a, b)

b

a

a

According to the Pythagorean Theorem, the distance, d, between the origin and any point (a, b) is denoted by 2 2 2d a b= +


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Is 9 9x x b−+ = ?

(1) Let’s isolate b to see what we can learn from this statement:

Square both sides

Expand

Put the x’s outside the parentheses


Simplify


(2) 0x > . This statement tells us nothing about b


Take-Aways • Before combining, or evaluating data, try to write all your equations in a similar format


To more easily compare these values, let’s format them to look similar:

Original question “Is ?”

( ) ( )223 3 2x x b−+ = +

( ) ( ) ( )2 23 2 3 3 3 2x x x x b− −+ ⋅ + = +

( ) ( ) ( )2 0 23 2 3 3 2x x

b−

+ + − =

9 9 2 2x x b−+ + − =

9 9x x b−+ =

1 0.0110

n <


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Change 0.01 “Is ?”

A proper fraction gets smaller as it is raised to a higher exponent. So the left side of the inequality would be smaller if n were the higher exponent.


(1) 2n > . This statement directly answers our rephrase


(2) Manipulate the data to get . We are dealing with values between 0 and 1, so the

higher the exponent the lower the result. In the inequality above, the left side is smaller, so n – 1 is the higher exponent. . This data answers our rephrase





• Before combining, or evaluating data, try to write all your equations in a similar format


n is a positive integer. Let n be XYZ where X, Y, and Z are the hundreds, tens, and units digit.

Rephrase “What is Y?”

(1) To multiply an integer by 10, just add a zero on the right (move the decimal point one position to the right). 10n is XYZ0. The hundreds digit of the number is Y. The statement tells us that the hundreds digit is 6, so we can write 6Y = . We can definitively answer our rephrase.


(2) Knowing the units digit of n+1 doesn’t tell us the tens digit. The only thing we can be sure of is that the tens digit of n is either 6 or 7. n could be 69 or it could be 70


21 1 1 1? Is 10 100 10 10

n n < → <

2n >

1 11 110 10

n− <

1 1 2n n− > → >


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What is 2t t x

t x+ −−

? We can break this into two fractions: 2 2 1t t x t

t x t x t x−+ → +

− − −. The only thing

standing in the way of knowing the value of this expression is 2t

t x−.

Rephrase “What is 2t

t x−?”

(1) 2 3t

t x=

−. This statement directly answers our rephrase.


(2) 5t x− = . This gives us the denominator of the rephrase and allows us to further simplify the question

from “What is 2t

t x−?” to “What is

25t

?” Without the value of t, we still cannot answer the question.


Guessing Notes: You should be suspicious of answer C here, especially if you suspect that this may be a hard question. This is because C is so clearly sufficient (statements give you 2 linear equations and you have 2 variables)


#169, pg. 288 Difficulty Level: 500-600 Topics: Exponents & Roots; Number Properties


(1) Take the square root: . The root of an integer could be an integer (ie. )

or a non-integer (ie. )


(2) . n is the square of an integer, so it must also be an integer.

intn =

2 int intn n= → = 25

7

2int intn n= → =


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Factor: . This is the product of 3 consecutive integers,

Our rephrase “Is divisible by ?”

(1) n is a multiple of 2 plus 1, so n must be odd. This means that is (even)(odd)(even).

This product must be divisible by 2 twice, so 3n n− is divisible by


(2) Factor: is divisible by , so we know that either n or is even. If ( )1n +

were even, would also be even, and would be divisible by . 4 5 6× × is

an example

On the other hand, if n were even, would be odd, and may not be divisible by

4. 1 2 3× × is an example.





• Most of the time, factoring equations makes them easier to simplify and allows you to isolate variables.

Therefore, in data sufficiency, consider factoring any initial equation when possible.

( ) ( )( )3 2 1 1 1n n n n n n n− = − = − +

1, , and 1n n n− +

( )( )( )1 1n n n− + 2 2⋅

( )( )( )1 1n n n− +

2 2⋅

( )2 1n n n n+ = + 2 3⋅ 1n +

( )1n − ( )( )( )1 1n n n− + 2 2⋅

( )1n − ( )( )( )1 1n n n− +


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What is the tens digit of positive integer x?

(1) This means that x is a multiple of 100 plus 30. Possible values of x include … The tens digit of x must be 3.


(2) This means that x is a multiple of 110 plus 30. Possible values of x include . There is more than one possible value for the tens digit of

x.


Notes: “x divided by 100 has a remainder of 30” can be algebraically expressed as . The result

is an integer, plus a fraction. Solving for x would yield . This is why I pointed our earlier that “x is a multiple of 100 plus 30.”



would be odd only if one was even and the other was odd.

Our rephrase “Between x and y, is one even and the other odd?”

(1) This statement tells us nothing about y


(2) This statement tells us nothing about x


MERGE STATEMENTS

By merging the statements we can replace x with and replace y with

( ) ( )100 1 30 130, 100 2 30 230, 330+ = + =

( ) ( )110 1 30 140 and 110 2 30 250+ = + =

30int100 100

x= +

100int 30x = +

x y−

2z ( )21z −


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Rephrase x – y

Simplify . 2z – 1 is a multiple of 2 minus 1, so it must be odd. Therefore, is odd.




Since is inscribed in a semi-circle, . Let’s draw the figure, and label all lengths using the Pythagorean Theorem

Let’s simplify the Pythagorean equation above to see what it says about a and b.

Expand both sides

Subtract

We’re asked to find the diameter. The length of the diameter is


(1) By inputting this data into , we get . Since we know the values of a and b, we can find .


(2) By inputting this data into , we get . Since we know the values of a and b,

we can find .


( ) ( )22 2 2 2 21 2 1 2 1x y z z z z z z z z− = − − = − − + = − + −

2 1x y z− = − x y−

PQR 90Q∠ = °

( ) ( )2 2 2 2 2 22 2 2a b a ab b+ + + = + +

2 2 and a b 2 22 2 2 4ab ab+ = → =

a b+

a b+

4ab = 4 4 1b b= → =a b+

4ab = ( )1 4 4a a= → =

a b+

a b

2 2 22b + 2 22a +

P

Q

R

Since PQR is a right triangle, 2 2 2PQ QR PR+ = . Substitute the

actual lengths to get

( ) ( ) ( )2 2

2 2 2 22 2a b a b+ + + = +


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Notes: If you couldn’t solve a similar question, consider the fact that together, the statements are so blatantly sufficient to find . Because we’re dealing with a difficult question, it is unlikely that solving would be that straightforward. It is much more likely that one or both statements are sufficient alone. In a similar case, you could probably safely cross off C and E.



handle. Always make a drawing of your own for geometry questions, even if one is already provided • It is generally a good idea to use as few variables as you have to, and avoid needlessly introducing new

variables. • In Data Sufficiency, if your rephrase is a good one, data sufficient to answer your rephrase is necessarily



Let C be the total capacity of the bucket.

Our rephrase “What is C?”

(1) This statement tells us nothing about the total capacity of the bucket


(2) The bucket starts half full, so the current quantity is 2C

. If we add 3 liters of water, the quantity will

increase by a third. So 3 liters equals a third of the current quantity. 133 2

C = ⋅

. We can solve for C



a b+

Patrick Siewe - GMATFix.com OG Companion 12th Advanced Drills P

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Advanced Speed Drills

• 10 Drills test all GMAT math topics • Master The Official Guide before attempting these drills • Designed for high math scorers (top 20%) • Follow the timing instructions and watch your pacing • Invest time to study the solutions carefully

Patrick Siewe - GMATFix.com OG Companion 12th Advanced Drills

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Advanced Speed Drill I (15Qs in 45 mins)

1. Marco would like to buy 10 widgets from either store A or B. Individual widgets cost $20 in each store. At store A, there is a 15% discount on every widget purchased after the 4th. At store B, there is a 25% discount on every widget purchased after the 6th. Which store offers the cheapest deal on 10 widgets, and by how much?

(A) Store A is $1 cheaper (B) Store B is $1 cheaper (C) The two are equally cheap (D) Store A is $2 cheaper (E) Store B is $2 cheaper

2. At a manufacturing plant, 60% of workers are men and the rest are women. During a round of layoffs, 18 workers leave the plant and 10 of them are men. The ratio of women to men among the remaining workers is 3 to 5. How many more men than women were there before the layoffs?

(A) 10 (B) 9 (C) 8 (D) 7 (E) 6

3. Do most employees of Company A earn less than the average salary of $60,000?

(1) The mode of all salaries is less than the average salary

(2) John whose salary is $62,000 earns $500 less than the median salary

4. How many intersection points are possible between a circle and a rectangle?

(A) 0 through 4 (B) 0 through 6, except 5 (C) 0 through 6 (D) 0 through 8 (E) 0 through 8, except 7

5. Cube A has ½ the volume of cube B. By what percentage must the length of the side of A increase so that the two cubes have the same volume?

(A) 3 2 percent

(B) ( )3 2 1

percent100

−

(C) ( )31 2

percent100

−

(D) ( )3100 2 1 percent−

(E) ( )3100 1 2 percent−

6. What is the remainder when p is divided by 42?

(1) 36 is a factor of 3p (2) 10p is a multiple of 56

7. The sum of all multiples of 3 between 20 and 79 is how much greater than the sum of all multiples of 5 between 9 and 99?

(A) 42 (B) 45 (C) 48 (D) 51 (E) 54

8. An investment fund grows at a constant percentage every year. After 10 years, the fund has doubled in size. If the initial amount invested was $1500, approximately how much money was in the fund after the first 5 years?

(A) $1950 (B) $2100 (C) $2250 (D) $2400 (E) $2550

9. If p is a positive integer and p r≠ − , is 2

2 ?n np r

>+

(1) 0n < (2) r r≠


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10. If a, b, c, d, and e are distinct positive integers, which of the five integers is the median?

(1) 2

b dc +=

(2) 2c a e< <

11. A and B are each two-digit integers with the same digits in reverse order. If 36A B− = , what is the difference between the two digits?

(A) 7 (B) 6 (C) 5 (D) 4 (E) 3

12. PQ is the diameter of the circle above, and x and y are the lengths of the two segments of PQ. What is the product of x and y?

(A) 18 (B) 24 (C) 36 (D) 45 (E) It cannot be determined

13. If 2z z< and y is a negative integer, is x greater than the average of y and z?

(1) x y z− > (2) 0xz >

14. Bonny left Philadelphia at 10am Monday and headed for Los Angeles by car. At 3pm on the same day, Clyde left and followed the same route. If they each traveled at a constant speed, when did Clyde catch up with Bonny?

(1) Clyde’s car traveled 25% faster than Bonny’s car

(2) When he caught Bonny, Clyde had spent 20% less time traveling than Bonny had

15. In the figure above, line m has the equation 4y x= − . Lines l and k are parallel. The x-

intercept of line l is -8 and the x-intercept of line k is -2. What is the distance between points a and b?

(A) 2 2 (B) 2 3 (C) 4 (D) 3 2 (E) 5

l

k

m

x

y

a

b

P

Q

x

y

6

R


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Advanced Speed Drill II (15Qs in 45 mins)

1. In a Junior class of 100 students, all students take either History, Chemistry, or Math. No student takes Math and Chemistry only, but five students take all three courses. The number of students who take History and Chemistry only is one half of the number who take History and Math only. 65 students take only one of the three courses, and 15 of those take History only. If enrollment in any one course is limited to a maximum of 55 students, what is the least number of students enrolled in Chemistry?

(A) 20 (B) 30 (C) 35 (D) 40 (E) 55

2. In Company A, the average salary of all employees is 20% more than the average salary of employees without an advanced degree. If employees with an advanced degree earn on average twice as much as employees without an advanced degree earn, what percentage of workers in the company have an advanced degree?

(A) 20% (B) 35% (C) 50% (D) 65% (E) 80%

3. If a b c+ > , is b c c b− > − ?

(1) 0a c− >

(2) 1bc<

4. In a decreasing sequence of 12 consecutive odd integers, the sum of the first six integers is 84. What is the sum of the last 6 integers in the sequence?

(A) 5 (B) 12 (C) 51 (D) 123 (E) 156

5. A group of 2 boys and 4 girls stand side by side in line to take a photograph. If the boys must not stand next to each other, how many arrangements are possible?

(A) 120 (B) 240 (C) 360 (D) 480 (E) 600

6. Sanjay has 10¢ coins and 25¢ coins only. How many 10¢ coins does he have? (100¢ = $1)

(1) Sanjay has 27 coins (2) Sanjay has a total of $4.35 in coins

7. Company A and Company B were founded in 1872. The quarterly revenue, R, for each company is a function of t, the number of quarters the company has been in operation. The revenue of Company A is 5 100AR t= − and

the revenue of Company B is 53003BR t= − .

When was Company B’s revenue greater than Company A’s revenue?

(A) B’s revenue has never been greater than A’s

(B) Before 1887 and after 1902 (C) From 1932 to 1992 (D) Before 1932 and after 1992 (E) From 1887 to 1902

8. In a 200-member association consisting of men and women, exactly 20% of men and exactly 25% of women are homeowners. What is the least possible number of members who are homeowners?

(A) 49 (B) 47 (C) 45 (D) 43 (E) 41


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9.An investment bank of 2,400 employees has two offices: one office in London and a second office in Zurich. How many employees of the Zurich office earn less than $80,000 per year?

(1) There are twice as many London employees earning under $80,000 per year as Zurich employees earning at least $80,000 per year

(2) Four out of every ten employees earn at least $80,000 per year, and three-fourths of those work at the London office

10. If a, b, and c are integers, is c even?

(1) ab

is an odd integer

(2) b

ac+ is an even integer

11. The sum of the ages of Patrick and Michelle is x years. Fourteen years ago, Michelle was four times as old as Patrick. In how many years, in terms of x, will Michelle be twice as old as Patrick?

(A) 42

5x +

(B) 2 126

5x −

(C) 4 42

5x −

(D) 4 126

5x −

(E) 8 126

5x −

12. What is the value of angle u in the figure above?

(1) z is 30 more than w (2) The ratio of v to z is 3 to 1

13. Which of the following corresponds to the shaded segment of the number line above?

(A) 65 92 2

x + <

(B) 65 92 2

x + >

(C) 65 92 2

x − <

(D) 65 92 2

x − >

(E) 28 37x< <

14. 235 45 80n k nA = ⋅ ⋅ . What is n?

(1) 1k =

(2) 2 3

3k n

k+ −

=

15. If a, b, c, d, and e are consecutive integers, which of the following must be true?

I. The mean of a, and b is smaller than the mean of d and e

II. 5

a b c d en

+ + + + is an integer if n is an

integer III. The product of the mean and the median

of the five integers is a positive integer

(A) None (B) I only (C) I and II (D) I and III (E) I, II, and III

x

28 37

x

y

z

u

v w


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Advanced Speed Drill III (15Qs in 50 mins)

1. What is the largest possible volume of a rectangular box placed inside cylinder C?

(1) C has a height of 1 foot (2) C has a volume of π cubic feet

2. Which of the following is equivalent to 7 7

1 5 2 5 88 4

− + −

?

(A) 319332

(B) 319364

(C) 32 14

5

(D) 32 14

5−

(E) 52 7

3. In Fake City, the population density, x, is inversely proportional to the square of the satisfaction indicator, y, and y is directly proportional to the square root of the graduation rate, z. If the graduation rate in Fake City doubled, how would the population density have to change for the satisfaction indicator to remain unchanged?

(A) 75% increase (B) 50% increase (C) No change (D) 50% decrease (E) 75% decrease

4. Parallel lines k and l have x-intercepts at 146 and 1

212

respectively. If line l has a slope of 43

− , what is the

shortest distance from line k to line l?

(A) 9 (B) 1

38

(C) 7247

(D) 146

(E) 5

5. On the same day, Mario and Luigi each open an investment account and each make an initial deposit. No other deposit or withdrawal was made on either account. If each account grows at p percent per year, the size of Mario’s account will be what percent of the size of Luigi’s account after y years?

(1) Mario’s initial deposit is $10,000 and 5y =

(2) Luigi’s initial deposit is 40% greater than Mario’s initial deposit.

6. If

252

2 2x =

− and

25 2 256

y+

= , what is the

ratio of x to y?

(A) 3 to 5 (B) 2 2 to 6 (C) 1 to 2 (D) 3 2− to 4 (E) 5 2 to 5

7. 2 0x bx c+ + = has two integer solutions for the values of x. b is an integer constant and c is a prime number constant. Is 2x > ?

(1) b is odd (2) c is even

8. At a 225-workers manufacturing plant, experienced workers can build a widget in 40 minutes while novice workers spend 90 minutes to build a widget. The average rate of all

workers is 56

widgets per hour. If 25% of the

novice workers resign, how many experienced workers must be added to the workforce so that production levels remain the same?

(A) 9 (B) 16 (C) 20 (D) 30 (E) 45


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9. a is an odd integer and a b≠ − . Is 2a a b

b a> −

+?

(1) 0ab = (2) 3a a>

10. What is the sum of all prime numbers between 110 and 130?

(A) 236 (B) 240 (C) 347 (D) 351 (E) 359

11. 3 5 643 3 3 3 3+ + + + is

(A) between 1 and 2 (B) between 2 and 3 (C) between 3 and 4 (D) between 4 and 5 (E) greater than 5

12. When integer n is divided by integer p, the remainder is r. What is the sum, in terms of p and r, of all possible distinct remainders when n is divided by 3p?

(A) 3r + 3p (B) r + 2p (C) 3r

(D) 3p – r (E) 3r + p

13. A store sells 3 items either separately or bundled. When the items are bundled, the most expensive item costs 20% above its regular price while the 2nd most expensive and the least expensive are each discounted by 15%. Is it cheaper to buy the 3 items in a bundle than to buy them separately at their regular prices?

(1) The 2nd most expensive item costs $10 (2) The least expensive item costs $3

14. If 2 0yx

> , is 0xy > ?

(1) x y x y+ > +

(2) 0x x− >

15. What is the smaller angle formed by the minutes and hours hands of a 12-hour clock when the time reads 2:40?

(A) 180° (B) 160° (C) 155° (D) 140° (E) 135°


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Advanced Speed Drill IV (15Qs in 50 mins)

1. Quarter circle O is centered at the origin and is tangent to line l. If quarter-circle O has an area of 25

4π

, what is the area of the shaded region?

(A) 25 3 25

4π−

(B) 50 10 3

3π −

(C) 50 25 3

6π −

(D) 50 3 25

12π−

(E) 25 3

12π −

2. x, q, and r are positive integers and

remainder 17x

q r= . What is the remainder when x

is divided by 34?

(1) q is odd (2) 5r =

3. A company has 3 salary levels. Level II’s employees earn 50% more than Level I’s, and Level III’s earn 30% more than Level II’s. If 46% of all employees are on Level II, what is the average salary of all employees?

(1) If 10% of Level I employees were promoted to Level III, there would be twice as many Level III employees as Level I employees, and those promoted would receive a $38,000 increase in their salary.

(2) 20% of all employees are Level I employees with a $40,000 salary

4. What is the perimeter of triangle ABC?

(1) Side AC has length 3, and side AB has length 4 (2) Triangle ABC can be inscribed in a circle such

that side BC is the diameter

5. If x and y are distinct non-zero integers, is x y x y+ = − ?

(1) 2xy < (2) 0x y− >

6. 25% of US residents have no car, and 25

of these are

citizens. There are 1333 % more citizens with one car

than foreigners with multiple cars. For every 5 US residents with no car, 6 citizens have multiple cars. The ratio of citizens with multiple cars to citizens with one car is 3 to 2. What percentage of US residents are foreigners with one car?

(A) 10% (B) 15% (C) 20% (D) 30% (E) 40%

7. In the figure above, circles A and B are congruent, and circles C and D are congruent. If the sum of the areas of the four circles is 80π , what is the area of trapezoid ABCD?

(1) The sum of the radii of circles A and C is 8 (2) The product of the radii of circles B and D is

12

A B

C

D


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8. Of the following, what is the closest

approximation to ( )8.87 10.15 5.80

3.91+

is

(A) 3 (B) 6 (C) 9 (D) 12 (E) 15

9. It takes Abe 4510 hours to paint a house

while Brian can paint 7 houses in 54hours. Working together, Abe and Brian start painting at the same time. Once ½ the house is painted, Brian is replaced by Cory without delay. What must Cory’s rate be so that it takes Abe and Cory 1

333 % less time to paint the second half of the house than it took Abe and Brian to paint the first half of the house?

(A) 4 houses per day (B) 1

35 houses per cay

(C) 795 houses per day

(D) 8 houses per day (E) 2

98 houses per day

10. Is x more than 65% of the sum of y and z?

(1) x is more than 20% greater than y (2) The ratio of z to y is 45 to 52

11. In 1978, 20% of all goods sold in the USA were foreign. In 1987, 25% of all goods sold in the USA were foreign. If the number of foreign goods sold in the USA increased by 80% from 1978 to 1987, by what percentage did the total number of goods sold in the USA increase over the same period?

(A) 32% (B) 44% (C) 56% (D) 65% (E) 70%

12. A room contains only 5yr olds and 20yr olds, with a combined average age of 15yrs. If eleven 5yr olds and ten 20yr olds were to enter the room, 37.5% of the people in the room would be 5yrs old. How many people in the room are 20yrs old?

(A) 25 (B) 38 (C) 50 (D) 60 (E) 68

13. In the figure above, the sum of z and w is equal to the sum of z and y. What is the sum of x and y?

(A) 80° (B) 90° (C) 100° (D) 110° (E) 120°

14. A machine printed a repeating sequence of dots colored red, white, blue, green, orange, and yellow in that order. If the first dot was blue and the last dot was red, which of the following could be the total number of dots printed?

(A) 112 (B) 123 (C) 126 (D) 132 (E) 149

15. 0xy ≠ and y x> . If 2

x yz

+= , which of the

following must be FALSE? (A) x y z+ < (B) xy z< (C) z x y< − < (D) y x z− < < (E) z x<

x

y z

100°

w


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Advanced Speed Drill V (15Qs in 50 mins)

1. If x is an integer, what is the sum of all distinct positive factors of x ?

(1) x has exactly three distinct positive factors (2) 2 1 3x k− = , where k is an odd integer

a c 2 4 1 1 2 b 7 9 6 c 2 0 a

2. Each cell in the table above is the positive distance on the number line between the column header and the row header of that cell. For instance, a and b are 7 units apart. What is the value of a b c+ + ?

(A) -2 (B) 4 (C) 6 (D) 12 (E) 18

3. ( )* 3x x= when x is an integer and

( )2

*2xx = when x is a non-integer. Which of

the following is equivalent to ( )

( )( )4* 3 2 3

* * 27?

(A) *(7) (B) *(6) (C) *(5) (D) *(4) (E) *(3)

4. The ratio of managers to workers at a company must be between 5:72 and 3:22. If there are 8 managers at this company, the possible number of workers ranges from

(A) 59 to 115 (B) 58 to 115 (C) 59 to 116 (D) 58 to 116 (E) 60 to 115

5. In the figure above, the diagonals of both squares are parallel to the x-axis and y-axis, and both squares are centered on the point ( )15,0− . What is the area of the shaded region?

(A) 22 (B) 36 2 25− (C) 144 25 2− (D) 94 (E) 110

6. N is a finite set of distinct positive integers. How many even integers does N contain?

(1) 6.25% of the product of all numbers in N is an odd integer

(2) The sum of all even integers in N is 10

7. In the figure above, x is how many degrees greater than b?

(1) 60d = (2) 100a c+ =

a b c

d

x

30

(-15,-5)

( )6 2 15, 0−


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8. On Monday, Hillary and Barrack leave from the same location and travel in different directions. On that day, Hillary travels half as fast as Barrack, but she spends 20% more time on the road. On Tuesday, Barrack remains stationary while Hillary travels 12 miles in a direction perpendicular to her previous day’s direction. If Hillary ends up exactly where Barrack is located, how many combined miles did Hillary and Barrack travel on Monday?

(A) 9 (B) 15 (C) 24 (D) 36 (E) 48

9. In a class of boys and girls, the average age is 8.5 years for boys and 9 years for girls. What percentage of people in the class are girls?

(1) If 8 ten year old girls and 17 eight year old boys joined the class, the average age of all children in the class would be 8.72 years.

(2) If 4 nine year old girls and 1 eight year old boy joined the class, the average age of all children in the class would be 8.8 years.

10. A snail traveled at 75 meters per hour. Was the distance traveled shorter than 25 meters?

(1) It took the snail fewer than 21 minutes to travel the distance.

(2) If the snail’s rate had been 25% slower, it would have taken fewer than 24 minutes to travel the same distance.

11. What is the shortest distance between the line 5 2y x= − and the origin?

(A) 1 (B) 2 (C) 3

(D) 5

(E) 52

12. Is 4% of x a multiple of 60?

(1) 30% of x is a multiple of 5

(2) 760

x is an integer

13. x is inversely proportional to the square root of y, and y is directly proportional to the square of z. If the value of z drops by 30%, the new value of

2x will be how many times the original value of 2x ?

(A) 100

9

(B) 103

(C) 1710

(D) 10049

(E) 107

14. The sequence a1, a2, a3…an is such that

1 22n n na a a− −= − for all 3n ≥ . a7 is 20%

greater than a6. If 4 3a = , what is the value of

100 200a a+ ?

(A) 298 (B) 299 (C) 300 (D) 301 (E) 302

15. If ( ) 2 16f x x x= − − and ( ) 110

g xx

=+

,

which of the following CANNOT be the value of ( )( )g f x ?

(A) 1 (B) 0 (C) -2 (D) -3 (E) -10


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Advanced Speed Drill VI (15Qs in 50 mins)

1. The Least Common Multiple of x and y is 300, and the Greatest Common Factor of x and y is 15. What is the

greatest possible value of xy

?

(A) 4 (B) 10 (C) 20 (D) 60 (E) 100

2. In the xy-plane above, the slope of line l is 54

and lines l

and k intersect when 4x = . If the y-intercept of line k is ( )0, 3− , which of the following corresponds to the shaded region?

(A) 1 5

3 102 4

x y x− − < ≤ −

(B) 4 5

3 85 4

x y x− − ≥ > +

(C) 4 5

3 85 4

x y x− − < ≤ +

(D) 1 5

3 102 4

x y x− − ≤ < −

(E) 5 1

10 or 34 2

y x y x≥ − > − −

3. In the xy-plane, at which point does the graph of ( )( )y x m x n= + + intersect the y-axis?

(1) The graph intersects the x-axis at ( )5,0−

(2) 215

m n+ =

4. Is 0ab > ?

(1) a b a b+ = − (2) a b>

5. In which quadrant of the xy-plane is the point ( ),p q located?

(1) 3

32

p q>

(2) 8 0p q− <

6. A company has 150 workers. Working together, all the workers can complete a job in 3hrs. If the job starts at 9:37am and must be completed by 2:52pm, what is the maximum number of workers that can be absent?

(A) 63 (B) 64 (C) 65 (D) 66 (E) 67

7. In a room of 25 people, the median age is 50 years and the average age is 42 years. Which of the following statements must be true?

I. At least one person is older than 50 years old II. The number of people over 50 years old is equal

to the number of people under 50 years old III. At least one person is younger than 34 years old

(A) None of the above (B) I only (C) III only (D) I and II (E) I, II and III

(8,0) l

k


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8. How many four letter strings are possible if two of the letters must be the same, and the other two must each be unique within the string?

(A) 15,600 (B) 16,250 (C) 89,700 (D) 93,600 (E) 358,800

9. The Richter magnitude scale quantifies the amount of energy released by an earthquake. An earthquake of magnitude 1m + releases 900% more energy than an earthquake of magnitude m. The energy released by an earthquake of magnitude 4.7 is approximately how many times the energy released by an earthquake of magnitude 3.2?

(A) 150 (B) 100 (C) 55 (D) 32 (E) 15

10. At a bookstore, the number of books sold, n, is related to the price of each book sold, p, according to the formula 140 10n p= − . George and Dick are two sales agents at this store. George’s pay is a $29 base pay plus 5% of sales revenue in excess of $100. Dick’s pay is 10% of sales revenue. If George and Dick receive equal pay, which of the following could be the price of each book sold?

(A) $12 (B) $11 (C) $10 (D) $9 (E) $8

11. At a party, the number of people who own a car but not a home is double the number of people who own a home but not a car. The number of people who own neither a home nor a car is 2 more than half the number of people who own both a home and a car. If 32 people do not own a home, how many people are at the party?

(1) 40 people own both a home and a car (2) 50 people own a car

12. In a group of men and women, each person is German or Spanish, but not both. Is the ratio of German men to German women smaller than the ratio of German men and Spanish women to German women and Spanish men?

(1) The ratio of Spanish women to Spanish men is smaller than the ratio of German men and Spanish women to German women and Spanish men

(2) The ratio of Spanish women to Spanish men is greater than the ratio of women to men

13. The dollar amount of interest, I, earned by an investment account in the first n years is

1 1100

nrI P= + −

where P is the initial

deposit and r is the annual interest rate. Raj opened an investment account in 1990. If the only transaction was an initial deposit of $15,000, is the annual interest rate less than 5%?

(1) The total interest accrued by the end of 1997 was $5,000

(2) Rounded to the nearest integer, the total interest accrued increased by $742 from January 1994 to January 1995

14. A bowl contains 34 marbles colored red, black or green. The number of red marbles is fewer than a third of the number of black marbles. There are more than twice as many green marbles as red marbles. What is the least possible sum of black marbles and green marbles?

(A) 25 (B) 26 (C) 27 (D) 28 (E) 29

15. Last year, the 5 managers of a company received a total of $208,000 in salaries. If the highest salary was 20% greater than the lowest, what is the maximum possible value of the highest salary?

(A) $42,000 (B) $44,000 (C) $46,000 (D) $48,000 (E) $50,000


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Advanced Speed Drill VII (15Qs in 50 mins)

1. The test scores of 1050 students have a normal distribution. If the average test score is 73 and the standard deviation is 10.5, approximately how many students have a test score less than or equal to 62.5?

(A) 147 (B) 168 (C) 205 (D) 288 (E) 357

2. In the figure, the line is tangent to both circles. If the area of circle Q is 200 and the area of PQR is 40% of the area of OPQR, what is the area of circle O?

(A) 800

9

(B) 200 (C) 300 (D) 400 (E) 450

3. The greatest of five numbers is 12. What is the median of the five?

(1) The standard deviation of the five numbers is not positive

(2) The sum of the five numbers is 60

4. If 1 0z− < < , which of the following expressions has the second greatest value?

(A) z (B) 3z− (C) 2z−

(D) 2z (E) 2z

5. The sequence a1, a2, a3,… ,an is such that

1 3

3n n

na aa − −+

= for all 4n ≥ . 6 14a = and

3 27a = . If the median of the second and third terms is 27, what is the value of a7?

(A) 2310

(B) 14 (C) 15 (D) 18 (E) 27

6. Each day in September, John spent the same fraction of his daily pay to buy food. Every other day, he also bought gasoline with some of his pay. The amount spent on gasoline in September is 25% less than the amount spent on

food that month. If 1320

of John’s September

pay was not spent on food or gasoline, what fraction of his pay was spent on food?

(A) 1

10

(B) 15

(C) 14

(D) 720

(E) 12

7. A toy factory has 48 experienced workers and 72 inexperienced workers. Each experienced worker builds toys 25% faster than each inexperienced worker does. If all workers work the same number of hours, approximately what percentage of all toys do inexperienced workers build?

(A) 35% (B) 40% (C) 45% (D) 50% (E) 55%

O

P

Q

R


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8. Five people wrote down their names on five pieces of paper, and put the pieces in a bag. If each person then randomly took out one piece of paper, what is the probability that exactly two people took the pieces with their names on them?

(A) 1

120

(B) 160

(C) 130

(D) 1

12

(E) 16

9. At a store, Mark bought three items. What fraction of his total expense was used to buy the first item?

(1) The first item cost $20 more than the second item did.

(2) Mark spent twice as much on the second item as he did on the third

10. If 0x y+ < , is 0y < ?

(1) ( ) ( )3 22 0 2 1y x y x− < < − <

(2) 1

1 4 02

y x− < − + <

11. In the figure above, the circle is centered at O, the origin. What is the value of c?

(1) 3 3b = (2) a b=

12. N is a set of consecutive integers. Is the sum of all members of N is equal to the sum of the absolute values of each member of N?

(1) The quotient of the product of all members of N and the sum of all members of N is negative

(2) At least one member of N is different from its absolute value

13. It took Sarah t hours to travel from home to school at a rate of r miles per hour. Her return trip took an hour longer because her rate was 3 miles per hour slower. In terms of t, what was the total distance traveled?

(A) 23 32

t t+

(B) 23 3t t+ (C) 26 6t t+ (D) 26 12t t+ (E) 212 12t t+

14. Points ( ),P a b and ( ),Q b c are in the xy-plane. Is the distance between P and Q greater than the distance between P and the origin?

(1) 0a c+ = (2) P and Q are distinct points

15. In a room, some students take Math and some take Chemistry. The number of students who take both subjects equals the number of students who take neither subject. If a student is to be chosen at random, is the probability that he take only one of these two subjects greater than the probability that he take Chemistry?

(1) The probability that the student selected take neither or take both subjects is smaller than the probability that he take Math.

(2) The number of students who take Math is equal to the number of students who take Chemistry.

Q (-3, b)

R (a, c)

O


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Advanced Speed Drill VIII (15Qs in 50 mins)

1. x and y are positive odd integers. What is the remainder when the product xy Is divided by 18?

(1) When x is divided by 9, the remainder is 3 (2) y – 1 is a multiple of 6.

2. On the number line, the distance between point P and point Q is 8 and the distance between point Q and point S is 4. If S is to the right of P and point R is closer to Q than to P, which of the following must be true?

I. Point P is the smallest of the four points II. Point S is not the greatest of the four points III. Point Q is the greatest or second greatest amount the

four points

(A) None (B) I only (C) I and II (D) I and III (E) II and III

3. At a factory, each employee’s daily pay, S, is defined by

754ny

S = + where n is the number of widgets built by the

employee that day and y is the number of years of employment. Abe, Bob and Cindy have 3, 6, and 8 years of employment respectively. Each day, Abe works twice as long as Bob and twice as long as Cindy, but Bob builds widgets 50% faster than Abe and 40% slower than Cindy. If Abe, Bob and Cindy’s combined daily pay is $675, how many widgets do Abe and Bob build together each day?

(A) 150 (B) 160 (C) 170 (D) 180 (E) 190

4. Does 0n = ?

(1) 22m mn= (2) m is half of n

5. If the diagonal of cube A is 56

of the diagonal of

cube B, the volume of B is approximately what percent greater than the volume of cube A?

(A) 20% (B) 44% (C) 61% (D) 73% (E) 80%

6. 2

24

0.07 10 0.5 100.00014 10

x

y

+

−

×= ×

× What is y – x?

(A) 7 (B) 3 (C) 0 (D) -3 (E) -7

7. In the figure shown, ABC is an isosceles triangle. The area of BDE is what fraction of the area of ABC ?

(1) BE BD=

(2) 23

DCDE

=

8. Patrick must store boxes of dimensions 3 by 3 by 5 inches into a rectangular container measuring 9 by 18 by 21 inches. What is the maximum number of boxes that can fit in this container?

(A) 42 (B) 63 (C) 72 (D) 90 (E) 126

A

B

C D E


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9. The age of a fossil is defined by 2 kyc s−= ⋅where c is the current concentration of compound X in the fossil, s is the starting concentration of compound X, k is a constant, and y is 410− times the age of the fossil in years. All newborn sharks have a 64% concentration of compound X and a 60,000 year old shark fossil has a compound X concentration of 32%. Approximately how old is a shark fossil with a compound X concentration of 2

15 %2 ?

(A) 153,000 years (B) 294,000 years (C) 305,000 years (D) 326,000 years (E) 358,000 years

10. One in five of the 660 students at City High School is a boy. If x boys were to enroll and no other changes occur, what value of x would increase the percent of boys in the school to 45%?

(A) 132 (B) 165 (C) 226 (D) 297 (E) 300

11. A bookseller sells books at p dollars each. How much revenue, in dollars, did the bookseller expect to collect last week?

(1) 40% more books than expected were sold last week

(2) Last week, revenue from book sales was $200 above expectations

12. A, B, and C are the digits of integer ABC. Is 399ABC > ?

(1) 2B A− < (2) The hundreds digit of 110% of ABC is

different from the hundreds digit of ABC

13. In the figure shown, each of the smaller squares is inscribed in a larger square such that each diagonal is parallel to the side of another square. The shaded region is what fraction of the area of the largest square?

(A) 1

12

(B) 1

16

(C) 120

(D) 128

(E) 132

14. For numbers x such that 1x ≠ − , if f(x) is defined

by ( ) 11

xf x

x−

=+

, then ( )( ) ( )( ) 11f x f x−− =⋅

(A) 2

2

11

xx−+

(B) ( )( )

2

2

1

1

x

x

+

−

(C) ( )( )

2

2

1

1

x

x

−−

+

(D) 1 (E) – 1

15. ( )2

7 48 7 48+ + − =

(A) 18 (B) 16 (C) 7 4 3+

(D) 7 4 3− (E) 1


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Advanced Speed Drill IX (15Qs in 45 mins)

1. 5 5 2 2 2 2 2 2 2 22 2 3 3 3 5 5 5 5 5+ + + + + + + + + = (A) 6 3 32 3 5+ + (B) 3 32 5⋅ (C) 2 22 19⋅ (D) 10 4 42 3 5+ + (E) 5 2 22 3 5⋅ ⋅

2. A recipe for cornbread batter requires 32

cups

of cornmeal, 52

cups of milk, 2 cups of flour,

and 23

cups of sugar. If 103

cups of cornbread

batter can feed 5 people, how many gallons of milk are required to make enough cornbread to feed 52 people? (1 gallon = 16 cups)

(A) 45

(B) 1316

(C) 1310

(D) 13 (E) 26

3. In the figure, the circle with center 0 bisects

side AB and points A and C lie on the circle. If 4BC = , what is the maximum possible area of

ABC ? (A) 4 (B) 4 3 (C) 8 (D) 8 3 (E) 16

4. A movie producer wants to hire 4 men and 4 women from the actors who auditioned. If 7 women auditioned and 525 combinations are possible, how many men auditioned?

(A) 5 (B) 6 (C) 7 (D) 8 (E) 9

5. PQR is inscribed in a circle with diameter PR. What is the measure of R∠ ?

(1) PQR is isosceles (2) 90Q∠ = °

6. What is the average salary of all employees of company X?

(1) The bottom 60% of employees earn an average of $40,000

(2) The top 60% of employees earn an average of $80,000

7. 6! 6! – 5! 5! =

(A) 36! – 25!

(B) ( )25 5!⋅

(C) ( )235 5!⋅

(D) ( )26! - 5! (E) 6! – 5!

8. If x y≠ and 1y ≠ − , is 2 2

x y xx y y

−<

− +?

(1) 0x y> > (2) 1y < −

A

B

C O


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9. Working together, Rihanna and Jen can build 2366 % more toys in 8 hours than Patrick and

Jen can build in 6 hours of working together. How many toys can Jen build alone in 1 hour?

(1) In 1 hour, Rihanna can build two more toys than Patrick can

(2) Rihanna can build three more toys in 4 hours than Patrick can build in 5 hours.

10. A string of length 12 6 2+ is cut in two pieces. One piece is used to form an isosceles right triangle with shortest side of length s, and the other piece is used to form a circle. Which of the following represents the radius of the circle in terms of s?

(A) ( )14 5 2

2

s

π

− ⋅

(B) ( )( )6 3 2 2sπ

π

+ +

(C) ( )( )2 2 6

2

s

π

+ −

(D) ( )2 2 s+ ⋅

(E) 2 6 2sπ +

11. 236

6xy

x−=−

. If x is an integer, which

value of x would yield the least possible value of y?

(A) –6 (B) –5 (C) 0 (D) 5 (E) 6

12. Three solutions are 30%, 45%, and 90% alcohol by volume. If a ounces of the 30% solution, b ounces of the 45% solution, and c ounces of the 90% solution are mixed to create a solution of 50% alcohol, what is b in terms of a and c?

(A) 3a + 9c (B) 4a + 2c (C) 3c – 9a (D) 8c + a (E) 8c – 4a

13. If k is a positive integer, what is the remainder when 2 4k − is divided by 12?

(1) k is even (2) k is not divisible by 3

14. Does a cylinder with surface area s and radius r have a height of x?

(1) ( )( )( )22 122 0s rr x r s

xππ

− −+ − =

(2) ( )22 2 02

srx s r r xr

π ππ

− − − − =

15. 20 machines working together at the same constant rate and without a break can complete a certain job in 2 days. If 15 new machines must

complete 25

of the job in 16 hours, the rate of

each new machine must be how much faster than the rate of each original machine?

(A) 45% faster (B) 50% faster (C) 55% faster (D) 60% faster (E) 65% faster


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Advanced Speed Drill X (15Qs in 50 mins)

1. John leaves a certain location at 1pm traveling at 45 miles per hour. At 3:40pm, Sally leaves the same location and takes the same exact route traveling at 60 miles per hour. At what time will the distance between John and Sally be half of the distance that John has traveled?

(A) 4:25pm (B) 5:16pm (C) 7:56pm (D) 8:22pm (E) 11:40pm

2. In the figure shown, the circle with center A and area 27π is tangent to an equilateral triangle at points B and D. What is the area of ABCD?

(A) 27 3

2

(B) 27 15

4π +

(C) 9 6π +

(D) 9 27

2π +

(E) 27 3

3. Last Tuesday, Allie, Brianna and Chevone were paid according to their hourly wages and the length of time they worked. Who was paid the least?

(1) Allie’s hourly wage is $3 more than Chevone’s, and Allie worked two hours longer than Brianna worked

(2) Chevone’s hourly pay is $1 less than Brianna’s, and Chevone spent more time working than Allie and Brianna spent combined

4. m and n are positive integers, and the quotient of m and 5 is the integer q. What is the remainder when m n+ is divided by 15?

(1) n – 3 is a multiple of 15 (2) When q is divided by 3, the remainder is 2

5. On Monday, the probability of rain in New York is 20%, in Paris is 50%, and in Moscow is 80%. If these probabilities are independent, what is the probability that it rains on Monday at exactly two of these three locations?

(A) 150

(B) 825

(C) 25

(D) 2150

(E) 12

6. If x is greater than 0, is 2 2 ?2xy x y y x− + + > −

(1) 2y y n− > (2) 0x n> >

7. 0p q− > . Is q p pr q r−

> −+

?

(1) r < 0 (2) q < 0

8. N is a set of four consecutive even integers whose sum is positive. O is a set of four consecutive odd integers. Is the median of N greater than the average of O?

(1) The product of all elements of O is negative (2) The product of the average of N and the

median of O is negative

A

B

C

D


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9. Michele, Sarah and John have marbles in a ratio of 5 to 12 to 6. If Sarah gave 5 marbles to Michele and 2 to John, the ratio of Michele to Sarah to John’s marbles would be 15 to 17 to 14. How many marbles do Michele and John have together?

(A) 10 (B) 12 (C) 22 (D) 29 (E) 33

10. If 0x y+ > , is 0x > ?

(1) x x y> +

(2) 2 2x y>

11. What is the sum of all multiples of 3 between 1,000 and 2,000?

(A) 496,500 (B) 498,000 (C) 499,500 (D) 501,000 (E) 502,500

12. At 3:30am train A leaves its station traveling at 180 miles per hour. At 4:45am train B leaves another station 1,665 miles away heading in the opposite direction on the same track. If the two trains cross each other’s paths at 10:05am, what is train B’s speed?

(A) 90 miles per hour (B) 95 miles per hour (C) 100 miles per hour (D) 105 miles per hour (E) 110 miles per hour

13. On the xy-plane, is the distance between the points (a, b) and (b, -c) greater than the distance between the points (-c, a) and (b, 5)?

(1) b > 5

(2) 112

a <

14. Country A’s GDP increased by x percent from 2006 to 2007, and increased by y percent from 2007 to 2008. What is the ratio, in terms of x and y, of country A’s 2008 GDP to its 2006 GDP?

(A) 1100 10,000

x xy+ +

(B) 100 10,000x y xy+

+

(C) 100

y xy+

(D) 1100 10,000x y xy+

+ +

(E) 1100

x y xy+ ++

15. If the perimeter of rhombus ABCD is 24, what is its area?

(A) 24 (B) 16 2 (C) 18 3 (D) 36 (E) 36 3

60°

A

B

C

D

Patrick Siewe - GMATFix.com OG Companion 12th Drills - Solutions

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Advanced Drills Solutions

Drill I 1. E

2. A

3. B

4. D

5. D

6. C

7. B

8. B

9. E

10. E

11. D

12. C

13. D

14. D

15. A

Drill II

1. C

2. A

3. A

4. B

5. D

6. C

7. E

8. E

9. C

10. E

11. B

12. A

13. C

14. B

15 A

Drill III

1. B

2. A

3. D

4. E

5. B

6. A

7. D

8. C

9. C

10. B

11. E

12. A

13. C

14. D

15. B

Drill IV

1. D

2. C

3. D

4. C

5. E

6. A

7. D

8. B

9. C

10. E

11. B

12. C

13. A

14. E

15. E

Drill V

1. C

2. B

3. E

4. A

5. D

6. C

7. A

8. C

9. B

10. B

11. D

12. E

13. D

14. A

15. B

Drill VI

1. C

2. A

3. C

4. A

5. C

6. B

7. C

8. D

9. D

10. E

11. D

12. A

13. D

14. E

15. D

Drill VII

1. B

2. E

3. D

4. D

5. A

6. B

7. E

8. E

9. E

10. D

11. B

12. D

13. C

14. C

15. A

Drill VIII

1. C

2. B

3. D

4. E

5. D

6. A

7. E

8. C

9. B

10. E

11. C

12. C

13. B

14. E

15. B

Drill IX

1. A

2. B

3. C

4. B

5. A

6. E

7. C

8. C

9. B

10. C

11. A

12. E

13. C

14. C

15. D

Drill X

1. B

2. E

3. E

4. C

5. D

6. A

7. E

8. B

9. C

10. B

11. C

12. A

13. E

14. D

15. C


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Advanced Speed Drill I (15Qs in 45 mins)

Drill I: #1 Difficulty Level: 500-600 Topics: FDPs & Ratios

Marco would like to buy 10 widgets from either store A or B. Individual widgets cost $20 in each store. At store A, there is a 15% discount on every widget purchased after the 4th. At store B, there is a 25% discount on every widget purchased after the 6th. Which store offers the cheapest deal on 10 widgets, and by how much?

(A) Store A is $1 cheaper

(B) Store B is $1 cheaper

(C) The two are equally cheap

(D) Store A is $2 cheaper

(E) Store B is $2 cheaper

Solution

Find out how much Marco will spend at each store. At Store A, the first 4 widgets are regularly priced. The remaining 6 widgets are at a 15% discount, so Marco would only pay 85% of their regular price. At Store B, the first 6 widgets are regularly priced. The remaining 4 widgets are at a 25% discount, so Marco would only pay 75% of their regular price.

Regular widgets Discounted widgets Total Cost

Store A 4 widgets for $80 ( ) 8585% $20 6 20 6 $102100

⋅ ⋅ = ⋅ ⋅ = $182

Store B 6 widgets for $120 ( ) 7575% $20 4 20 4 $60

100⋅ ⋅ = ⋅ ⋅ = $180

Store B is cheaper, by $2


Drill I: #2 Difficulty Level: 600-700 Topics: FDPs & Ratios

At a manufacturing plant, 60% of workers are men and the rest are women. During a round of layoffs, 18 workers leave the plant and 10 of them are men. The ratio of women to men among the remaining workers is 3 to 5. How many more men than women were there before the layoffs?

(A) 10 (B) 9 (C) 8 (D) 7 (E) 6

Solution

Before the layoffs, 40% of workers were women and 60% of workers were men. This is a ratio of 2:3. So we can represent their numbers as 2women x= and 3men x= . The difference between the number of men and women is 3 2x x x− = . After 8 women and 10 men are taken away, there are 2 8x − women and 3 10x − men left and the

resulting ratio is 3:5. This can be expressed in the following equation: 2 8 3 103 10 5

x xx−

= → =−


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Drill I: #3 Difficulty Level: 600-700 Topics: Statistics

Do most employees of Company A earn less than the average salary of $60,000?

(1) The mode of all salaries is less than the average salary (2) John whose salary is $62,000 earns $500 less than the median salary

Solution

The question is already simplified

(1) The mode is the value that appears the most in a set. From this statement we know that this value is less than the average. There could be two people with a salary equal to the mode, or there could be 100 people with a salary equal to the mode. Statement (1) gives us no indication of whether most people earn less than $60,000


(2) This statement tells us that the median is $62,500. So ½ the employees earn above this figure and above the average of $60,000. John whose salary is $62,000 also earns above the average of $60,000. Thus we can say that at least ½ plus one earn above the average. We have enough information to definitively determine that the answer to the original question “Do most…earn less than the average” is NO.



Drill I: #4 Difficulty Level: 600-700 Topics: Geometry

How many intersection points are possible between a circle and a rectangle?

(A) 0 through 4 (B) 0 through 6, except 5

(C) 0 through 6 (D) 0 through 8 (E) 0 through 8, except 7

Intuitive Solution

Take a look at the answer choices. To do this quickly, figure out the number of intersections that would help you quickly eliminate answers. Ignore 0 through 4 since all the answer choices have these options. Note that only one answer choice, D, has 7 intersections. If we could draw a 7-intersection diagram, we would know right away that D is correct.


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Take-Aways Drawing figures reveals connections between data and often makes the problem easier to understand and handle.

Always make a drawing of your own for geometry questions, even if one is already provided.

Drill I: #5 Difficulty Level: 700+ Topics: FDPs & Ratios; Geometry

Cube A has ½ the volume of cube B. By what percentage must the length of the side of A increase so that the two cubes have the same volume?

(A) 3 2 % (B) ( )3 2 1

%100

− (C)

( )31 2%

100

− (D) ( )3100 2 1 %− (E) ( )3100 1 2 %−

30 Seconds Hack

Cube B is double cube A in terms of volume, so the side of A will have to increase by a substantial percentage for the cubes to be of equal volume. Based on this simple fact, you can eliminate most wrong answers:

(A) 3 2 percent . This value is smaller than the root of 2, which is about 1.4. This answer choice is probably wrong because a side increase of less than 1.4% isn’t enough to double the volume

(B) ( )3 2 1

percent100

−. This is a tiny increase. The top of the fraction is smaller than 1, so the whole

fraction is smaller than 0.01. 0.01% increase in the side is probably not enough to double the volume

(C) ( )31 2

percent100

−. This is a negative value (numerator is negative), so it cannot be the correct

percentage increase. (D) ( )3100 2 1 percent−

(E) ( )3100 1 2 percent− . This is a negative value, so it cannot be the correct percentage increase.

Guess D

This figure has 7 points of intersection between circle and rectangle. Since answer choice D is the only one that allows for this scenario, it must be

t


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Solution The volumes will be equal if the sides are equal. To solve, we need to figure out what the side of B is in terms of A (or vice versa) so that we will know how much to increase Aside by to make it equal Bside .

For cubes, 3Volume side= , so since cube B has double the volume of cube A, we can write that 3 32B Aside side=

. By taking the cube root of both sides, we end up with 3 2B Aside side= ⋅

We now know that for the cubes to be the same volume, Aside must increase to 3 2 Aside⋅ but what percentage

increase is that? The formula used to find percentage change is positive difference

100starting value

⋅ . In this case, the

percentage increase is 3

1002 A A

Asideside side

⋅⋅ −

. Factor out and simplify by Aside and we end up with

( )3 1002 1 ⋅−


Take-Aways

To calculate percentage change, be it increase or decrease, do positive difference

100starting value

⋅ .

Drill I: #6 Difficulty Level: 600-700 Topics: Number Properties

What is the remainder when p is divided by 42?

(1) 36 is a factor of 3p (2) 10p is a multiple of 56

Solution


(1) 3p is a multiple of 36, so we can write that 3 36 intp = ⋅ where int is an unknown integer. Divide both sides

by 3 to get 212 int 2 3 intp p= → = ⋅ ⋅ .

So we know p is a multiple of 12, but we cannot tell what its remainder would be when divided by 42. If p equaled 48, the remainder would be 6. However, if p equaled 60, the remainder would be 18.


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(2) Algebraically, 10 56int 5 28intp p= → = . The prime factorization of 28 reveals that 25 2 int7p = ⋅ ⋅ .

Since the two sides of the equation must be equal, the left side must somehow contain 22 7⋅ . Since this is not contained in 5, it must be in p.

This statement tells us that 22 7 intp = ⋅ ⋅ (we must include “int” because we don’t know what other factors p may contain). Knowing that p is a multiple of 28 doesn’t help us answer our question however. If p equaled 56, the remainder when dividing by 42 would be 14. However, if p equaled 84, the remainder would be 0.


MERGE STATEMENTS

From statement (1), we know that 22 3 intp = ⋅ ⋅

From statement (2), we know that 22 7 intp = ⋅ ⋅

The minimum data that must make both statements true is 22 3 7 intp = ⋅ ⋅ ⋅ meaning that p is a multiple of 84. Since p must therefore be a multiple of 42, we know that the remainder when we divide p by 42 is 0


Take-Aways When solving a Number Properties question, express all values in their prime factorizations (eg.


Drill I: #7 Difficulty Level: 600-700 Topics: Sets & Groups

The sum of all multiples of 3 between 20 and 79 is how much greater than the sum of all multiples of 5 between 9 and 99?

(A) 42 (B) 45 (C) 48 (D) 51 (E) 54

Solution

According to the average formula, ( )( )sumavg sum avg # of items

# of items= → = . We need the average and

number of items to find the sum of each list. To find the average of each list, average the highest and lowest values.


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The multiples of 3 in question go from 21 to 78, and their average is 21 78

49.52+

= . The list goes up to the 26th

multiple of three ( 3 26 78⋅ = ), but excludes the first 6 multiples, so it is a list of 20 numbers. The sum of all multiples of 3 between 20 and 79 is 49.5 20 990⋅ =

The multiples of 5 in question go from 10 to 95, and their average is 10 95

52.52+

= . The list goes up to the 19th

multiple of five ( 5 19 95⋅ = ), but excludes the first multiple (5 itself), so it is a list of 18 numbers. The sum of all multiples of 5 between 9 and 99 is 52.5 18 945⋅ =

The difference between the two lists is 990 945 45− =


Take-Aways A set of numbers in which the difference between each value and the next higher value is constant is called an

arithmetic sequence.

The average of an arithmetic sequence is the average of its highest and lowest values. This average is always equal to the median of the sequence.

The sum of an arithmetic sequence is its average multiplied by the number of numbers in the sequence

Drill I: #8 Difficulty Level: 700+ Topics: FDPs & Ratios

An investment fund grows at a constant percentage every year. After 10 years, the fund has doubled in size. If the initial amount invested was $1500, approximately how much money was in the fund after the first 5 years?

(A) $1950 (B) $2100 (C) $2250 (D) $2400 (E) $2550

30 Seconds Hack

A constant percentage increase over a period means that there is a greater growth in dollar terms during the 2nd half of the period than during the first. For instance, if we started with $100 increasing at 10% over two years, we would have a $10 increase in the first year, but an $11 increase in the 2nd year. This is because we are taking a percentage of a bigger and bigger amount. Let’s use this to our advantage.

In this problem, the fund doubled, or increased by $1500, after 10 years. We can conclude that most of the increase (more than $750) happened after the halfway mark. So at the 5yr mark, the amount of money in the fund will be less than $1500 $750 $2250+ = . We can safely eliminate answers C, D and E

Guess A or B

Solution


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Let p be the percentage increase. Each year, the amount in the fund will be multiplied by 1100

p+

to find the

new amount. Since the fund doubled in size after 10 years, we can say that: 10 10

1500 1 3000 1 2100 100

p p+ = → + =

. Don’t waste your time trying to solve for p. The GMAT will

never require that much work.

The value of the fund after 5 years is 5

$1500 1100

p ⋅ +

. Note that 5 10

1 1 2100 100

p p + = + =

, so we

simply need to do $1500 2⋅

2 1.4≈ so $1500 2 $1500 1.4 $2100⋅ = ⋅ =


Drill I: #9 Difficulty Level: 700+ Topics: Number Properties

If p is a positive integer and p r≠ − , is 2

2 ?n np r

>+

(1) 0n < (2) r r≠

Solution

It would be difficult to simplify the fraction without more information about signs, because multiplying both sides by a negative would flip the inequality

(1) Since n is negative, we know that 2n is positive (is not zero), so we can divide both sides by it to get the

rephrase “is 1

1?p r

>+

”

If the denominator is bigger than 1, then the left side will be smaller, but if the denominator is between 1 and 0 then the left side will be larger. We cannot answer the question.


(2) Since r is different from its absolute value, it must be negative. Knowing the sign of r doesn’t help answer the original question however because p r+ could still be anything.


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If p r+ is negative, then the answer to the original question would be NO. On the other hand, if p r+ is between 0 and 1, then the answer to the original question would be YES.


MERGE STATEMENTS

Merging the two statements doesn’t help us determine the sign or even a range of values for p r+ , so the logic we used to show that statement (2) was not sufficient can be used to demonstrate that together the statements are still not sufficient



Drill I: #10 Difficulty Level: 600-700 Topics: Statistics

If a, b, c, d, and e are distinct positive integers, which of the five integers is the median?

(1) 2

b dc +=

(2) 2c a e< <

Solution


(1) c is the average of b and d, so we know that either b c d< < or d c b< < . Without information about a and e, we cannot tell which integer is the median


(2) Without information about b and d, we cannot determine which integer is the median


MERGE STATEMENTS

Since 2a c> , and 2c b d= + , we know that a is bigger than b, c, and d. By merging the statements, we can limit possibilities to: b c d a e< < < < or d c b a e< < < < . In the first case, the median is d, whereas in the second case the median is b




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Drill I: #11 Difficulty Level: 600-700 Topics: Number Properties

A and B are each two-digit integers with the same digits in reverse order. If 36A B− = , what is the difference

between the two digits?

(A) 7 (B) 6 (C) 5 (D) 4 (E) 3


Since 36A B− = , we know that the bigger smaller 36− = . Knowing that the units digit of this difference is 6

limits us to a small set of possibilities for the units digits of our two integers. Let’s try them until we find the two integers whose difference is 36

The units digits could be 7 and 1. In this case, the integers would be 71 and 17, but the difference between the two is bigger than 36.

The units digits could be 8 and 2. In this case, the integers would be 82 and 28, but the difference between the two is bigger than 36

The units digits could be 9 and 3. In this case, the integers would be 93 and 39, but the difference between the two is bigger than 36

The units digits could be 1 and 5. In this case, the integers would be 51 and 15. These may be the right integers because 51 15 36− = . The difference between the two digits is 5 1 4− =


Solution B

Any two-digit integer can be written as 10 t u⋅ + where t is the tens and u is the units digit. For instance, 47 is 10 4 7⋅ +

Knowing that the positive difference between our integers is 36, we can write ( ) ( )10 10 36t u u t+ − + = . Note

that because the integers use the same digits in reverse order, t goes from being the tens to being the units digit, and u does the reverse. Simplifying this equation, we get 9 9 36 4t u t u− = → − = .

Thus the difference between our digits is 4.



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Drill I: #12 Difficulty Level: 700+ Topics: Geometry

PQ is the diameter of the circle above, and x and y are the lengths of the two segments of PQ. What is the product of x and y?

(A) 18 (B) 24 (C) 36 (D) 45 (E) It cannot be determined

Solution

Any triangle inscribed in a circle must be a right triangle. In this case, angle R is 90 degrees. With this information, we can use the Pythagorean Theorem to Find all segments in terms of x and y

2 2 26y PR+ = (Eq. 1) 2 2 26x QR+ = (Eq. 2) ( )22 2PR QR x y+ = + (Eq. 3)

Using the first 2 equations, we can rewrite (Eq. 3) as:

Rewrite Eq 3 ( ) ( ) ( )22 2 2 26 6y x x y+ + + = +

Remove parentheses 2 2 2 272 2y x x xy y+ + = + +

Subtract like terms 72 2xy=

Find the product of x and y 36xy =


Drill I: #13 Difficulty Level: 700+ Topics: Inequalities & Absolute Values; Statistics

If 2z z< and y is a negative integer, is x greater than the average of y and z?

P

Q

x

y

6

R


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(1) x y z− > (2) 0xz >

Solution

Our rephrase “Is ?2

y zx

+> ”

Since 2z z< , we can determine that 0 1z< < . Since y is a negative integer, z y> .

z is bigger than y, so if we found out that x was bigger than z (thus bigger than both), we would know that it is bigger than the average of y and z.

(1) Rewrite this statement as x z y> + . Since x is bigger than z plus another positive value, it must be true

that x z> . z is positive and y is negative, so x is larger than both y and z, and must be larger than their average


(2) This statement tells us that x and z have the same sign. 2z z< , so z is a positive fraction. Therefore, x is positive.

0 1z< < . Since y is a negative integer, 1y ≤ − and y z+ is negative. We have enough information to

determine that 2

y zx

+> because the left side is positive and the right side is negative.



Drill I: #14 Difficulty Level: 700+ Topics: Rates & Work

Bonny left Philadelphia at 10am Monday and headed for Los Angeles by car. At 3pm on the same day, Clyde left and followed the same route. If they each traveled at a constant speed, when did Clyde catch up with Bonny?

(1) Clyde’s car traveled 25% faster than Bonny’s car (2) When he caught Bonny, Clyde had spent 20% less time traveling than Bonny had

Solution

t Number of hours Bonny has spent on the road

5t − Clyde leaves 5 hours later, so his travel time is Bonny’s minus 5.


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br Bonny’s rate

cr Clyde’s rate

When Clyde catches Bonny, the two have traveled the same distance. Since distance rate time= ⋅ , to set the distances equal to each other, we can write ( )5cbr t r t⋅ = ⋅ − (Eq. 1).

To find out when Clyde caught Bonny, we need to find t, the number of hours that Bonny has been on the road. Let’s isolate t in (Eq. 1) above

Eq. 1 ( )5cbr t r t⋅ = −

Expand 5c cbr t r t r= −

Bring t’s to one side 5 c c br r t r t= −

Factor out and isolate t 5 c

c b

r tr r

=−

With this new expression for t, we can write our rephrase:

Our rephrase “What is 5 c

c b

rr r

−

?”

(1) Algebraically, this tells us that 1.25c br r= . Putting this into our rephrase takes us from “What is

5 c

c b

rr r

−

?” to “What is 1.25

51.25

b

b b

rr r

⋅ −

?” We don’t need any more data to answer the question

since this fraction simplifies to ( )

1.25 1.255 5 251.25 1 0.25

b

b

rr

⋅ = ⋅ = −

25t = , so Clyde caught up to Bonny 25hrs after Bonny departed.


(2) Time and rate are inversely related. As you decrease time, the rate must increase to maintain the distance traveled. If Clyde’s time is 20% less than Bonny’s, then Clyde’s time is 0.8t. Because they traveled the same distance, we can write 0.8cbr t r t= . Divide both sides by t to get 0.8 cbr r=


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Putting this data into our rephrase takes us from “What is 5 c

c b

rr r

−

?” to “What is 50.8c

c c

rr r

⋅

− ?”

We don’t need any more data to answer the question since this fraction simplifies to

( )15 5 25

1 0.8 0.2c

c

rr

⋅ = ⋅ = −

25t = , so Clyde caught up to Bonny 25hrs after Bonny departed.



Drill I: #15 Difficulty Level: 700+ Topics: Coordinate Geometry

In the figure above, line m has the equation 4y x= − . Lines l and k are parallel. The x-intercept of line l is -8 and the x-intercept of line k is -2. What is the distance between points a and b?

(A) 2 2 (B) 2 3 (C) 4 (D) 3 2 (E) 5

Solution

Draw, and put in the given data. From the equation of m, 4y x= − , we know that the y-intercept of that line is 4,

so point a is at ( )0, 4

l

k

m

x

y

a

b


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Note that since lines k and l are parallel, they have the same slope. The slope of l, riserun

can be found by

observing that between the (-8, 0) and (0, 4), rise 4run 8

= so the slope of lines l and k is ½. Let’s use this to find the

equation of line k

Standard equation of a line y mx b= +

Plug in a slope of k (slope is ½) 12

y x b= +

Plug in the given point (-2, 0) ( )10 2 12

b b= − + → =

Equation of line k 1

12

y x= +

To find point b, the intersect of k and m, set the two equations of the line equal to each other: 14 1 22

x x x− = + → = . Now plug this value of x into either equation to find y: 4 4 2 2y x y= − → = − = .

We now have the coordinates of b: (2, 2)

The final step is to find the distance between a and b, between (0, 4) and (2, 2). To do this, draw up a right triangle and use the Pythagorean Theorem:

k

l m

x

y

a (0,4)

b

-8 -2

To find the distance from a to b, we need the coordinates of b, the intersection of lines m and k.

Since we are given the equation of line m, all we need is the equation of line k to determine where the two intersect.


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Advanced Speed Drill II (15Qs in 45 mins)

Drill II: #1 Difficulty Level: 700+ Topics: Sets & Groups

In a Junior class of 100 students, all students take either History, Chemistry, or Math. No student takes Math and Chemistry only, but five students take all three courses. The number of students who take History and Chemistry only is one half of the number who take History and Math only. 65 students take only one of the three courses, and 15 of those take History only. If enrollment in any one course is limited to a maximum of 55 students, what is the least number of students enrolled in Chemistry?

(A) 20 (B) 30 (C) 35 (D) 40 (E) 55

Solution

This overlapping sets problem has 3 sets. Let’s drawn them in a Venn Diagram and fill in the info given.

(0,4)

(2,2)

2

2

x

y The distance between the two points is the hypotenuse of our right triangle.

You should recognize that since we have a right triangle with two equal legs, it is a 45-45-90 degree triangle and the hypotenuse will be

2 2

You can find the hypotenuse by using the

Pythagorean theorem: 2 2 22 2 d+ = . 2 2d =


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When selecting your answer, be careful not to pick 20, the least number of students enrolled in Chemistry only. We’re asked for the least enrolled in Chemistry, and this will include Chem & Hist, Chem & Math as well as all three courses. The answer is 35


Drill II: #2 Difficulty Level: 600-700 Topics: Translations & Manipulations; FDPs & Ratios; Statistics

Hist. Chem.

Math

5 20

10

0

• 65 students take only one course means that 35 take two or more courses. So 2 5 0 35x x+ + + = . Solve to get 10, 2 20x x= =

• Of the 65 who take only one course, 15 take H. only. This means that the remaining 50 take C only or M only.

• _ _ 50C only M only+ = . Since enrollment in a course is limited to a 55 maximum, M_only is 30 at most so C_only is at least 20

• At a minimum, C has 35 students ( )20 10 5+ +

15

_ 30M only ≤

_ 20C only ≥

Hist. Chem.

Math

5 2x

x

0

• No student takes M. & C. only

• 5 students take all three courses

• H & C is half of H & M, so we can use x and 2x


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In Company A, the average salary of all employees is 20% more than the average salary of employees without an advanced degree. If employees with an advanced degree earn on average twice as much as employees without an advanced degree earn, what percentage of workers in the company have an advanced degree?

(A) 20% (B) 35% (C) 50% (D) 65% (E) 80%

30 Seconds Hack

Since advanced degree holders earn the double of non-degree holders, we would expect the average salary to be in the middle, 50% above the non-degree holders’ salary. However, the fact that the average salary of all employees is only 20% above the non-degree holders’ salary indicates that the advanced degree employees have only a relatively small impact on overall salary. This can only be because these advanced degrees are the minority of employees. Eliminate C, D, and E

Guess A or B

Solution

Let’s define our variables

n - percentage of advanced degree employees

100 n− - percentage of employees without an advanced degree (the two percentages add up to 100%)

X - average salary of employees without an advanced degree

2X - average salary of advanced degree employees is the double of the salary above

To find the overall average, we must add up all the salaries and divide them by 100. “The average salary of all employees is 20% more than the average salary of employees without an advanced degree”

( )( ) ( )( )2 100advanced degrees salaries non-advanced salaries

Avg. Salary 1.2All employees 100

X n XX

n + −+= = =

Expand and multiply by 100 2 100 120nX X nX X+ − =

Divide by X 2 100 120n n+ − =

Solve for n 20n =


Drill II: #3 Difficulty Level: 700+ Topics: Inequalities & Absolute Values

If a b c+ > , is b c c b− > − ?


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(1) 0a c− >

(2) 1bc<

Solution

Note that b c c b− = − since the two expressions are just negative of each other. The question can be written as

“is c b c b− > − ?” The only way for the absolute value of a number to be larger than the original number is if

that number is negative, so we can further simplify the question:

Our rephrase “Is 0c b− < ” or even better “Is c b< ?”

(1) This statement tells us that a− has the same sign as c . Since they’re both positive, a itself is negative.

Now consider the first inequality given: a b c+ > . This inequality tells us that b plus a negative value (a) is bigger than c. So b must be greater than c. We have enough data to answer our rephrase. “Is c b<?” the answer is YES.


(1) If c is positive, this statement can be rephrased as b c< . On the other hand, if c is negative, the inequality sign will flip and the statement can be rephrased as b c> . We don’t have enough information to compare the two and answer our rephrase.





Drill II: #4 Difficulty Level: 600-700 Topics: Translations & Manipulations

In a decreasing sequence of 12 consecutive odd integers, the sum of the first six integers is 84. What is the sum of the last 6 integers in the sequence?

(A) 5 (B) 12 (C) 51 (D) 123 (E) 156

30 Seconds Hack

Note that it is a decreasing sequence, so the sum of the last 6 integers will be smaller than the sum of the first 6. Eliminate D and E


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Guess A, B or C


The 12 consecutive odd integers have been split into two groups. Imagine writing each group in decreasing order, and stacking the two groups as in the figure below

Since each integer in the top row is 12 more than its corresponding integer in the bottom row, the sum of the top row is 12 6 72⋅ = more than the sum of the bottom row. Consequently, since the first six integers add up to 84, the last six integers will add up to 84 72 12− = .


Solution B

In an arithmetic sequence (a sequence in which the difference between two adjacent values is always the same) such as consecutive odd integers, the median is equal to the average of the list. In our case, the first six integers

add up to 84, so the median is 84

146= . So the first three integers are above 14, and the next three are below.

From this information, we can determine our entire list: 19, 17, 15, 13, 11, 9 and 7, 5, 3, 1, -1, -3. Find the sum of the last 6 integers: 7 5 3 1 1 3 12+ + + − − =


Drill II: #5 Difficulty Level: 700+ Topics: Combinatorics

A group of 2 boys and 4 girls stand side by side in line to take a photograph. If the boys must not stand next to each other, how many arrangements are possible?

(A) 120 (B) 240 (C) 360 (D) 480 (E) 600

30 Seconds Hack

Approximate. If there were no restrictions, the six children could stand in 6! 720= possible arrangements. We’re told however that the boys must not stand next to each other. With 6 positions possible, there are more ways for



int1 is the largest, and the difference between adjacent integers is always 2. (we’re dealing with ODDs only), so the difference between each linked pair is 12

int6

int12


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them to stand apart than to stand together. Depending on where the first boy stands, there will be one or two spots next to him, but there will be three or four spots away from him. Since there are more ways for the boys to stand apart, most of the 720 arrangements are still possible. Eliminate A, B and C

Guess D or E

Solution

One way to solve the problem is to break it into two smaller problems. First, find out how many ways the boys can stand apart if the first boy (we’ll call him A) stands at one end of the line. Then find out how many ways the boys can stand apart if A doesn’t stand at the end. Finally, add the two numbers to find how many total arrangements are possible when the boys stand apart.

The total number of valid arrangements is 192 288 480+ =


Drill II: #6 Difficulty Level: 500-600 Topics: Translations & Manipulations

Sanjay has 10¢ coins and 25¢ coins only. How many 10¢ coins does he have? (100¢ = $1)

(1) Sanjay has 27 coins (2) Sanjay has a total of $4.35 in coins

Solution


A

A

• For each of the 2 end positions boy A takes, B must pick from the 4 positions that are not next to A. Once the boys are placed, the four girls are not restricted and can take any of 4! arrangements. So if A stands at the end, there will be 2 4 4! 192⋅ ⋅ = valid arrangements

• For each of 4 middle positions boy A takes, B must pick from the 3 positions that are not next to A. Once the boys are placed, the four girls are not restricted and can take any of 4! Arrangements. So if A doesn’t stand at the end, there will be 4 3 4! 288⋅ ⋅ = valid options


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(1) The total number of coins doesn’t help us determine how many 10¢ coins there are.


(2) $4.35 can be reached using different combinations of 25¢ and 10¢ coins since any group of five 10¢ coins can be replaced with two 25¢ coins. We cannot determine how many 10¢ there are.


MERGE STATEMENTS

Let d be the number of 10¢ coins. Since there are 27 coins, the number of 25¢ coins will be 27 d− . By merging the statements, we can write up an equation for the total amount of 435¢: ( )10 25 27 435d d+ − = . This is a

linear equation that can be solved for d, the number of 10¢ coins



Drill II: #7 Difficulty Level: 600-700 Topics: Functions & Sequences; Inequalities & Absolute Values

Company A and Company B were founded in 1872. The quarterly revenue, R, for each company is a function of t, the number of quarters the company has been in operation. The revenue of Company A is 5 100AR t= − and

the revenue of Company B is 53003BR t= − . When was Company B’s revenue greater than Company A’s

revenue?

(A) B’s revenue has never been greater than A’s

(B) Before 1887 and after 1902

(C) From 1932 to 1992

(D) Before 1932 and after 1992

(E) From 1887 to 1902

Solution

To find when B’s revenue is greater than A’s, we need to find for which values of t is B AR R> .

Setup the inequality 5

300 5 1003

t t− > −

Remove the absolute value ( ) ( )( )5 5300 5 100 or... 300 5 100

3 3t t t t− > − − > ⋅ − −

Note: 100t − simplifies to ( )100 or 100t t− − − . Absolute value reverses the sign if what was inside had a

negative value. Since we don’t know the sign of 100t − , we must consider both possibilities.

Multiply both sides by 3 900 5 15 1500 or... 900 5 15 1500t t t t− > − − > − +


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Isolate t 2400 20 or... 10 600t t> >

Solve 120 or... 60t t< >

60 120t< < . Since t represents “the number of quarters the company has been in operation”, we can deduce that

B’s revenue is greater than A’s from 60

154= years to

12030

4= years after creation. Since the companies were

founded in 1872, this interval refers to 1887 to 1902


Drill II: #8 Difficulty Level: 600-700 Topics: Sets & Groups

In a 200-member association consisting of men and women, exactly 20% of men and exactly 25% of women are homeowners. What is the least possible number of members who are homeowners?

(A) 49 (B) 47 (C) 45 (D) 43 (E) 41

30 Seconds Hack

If the association was men-only, there would be 20% of 200 members or 40 homeowners. On the other hand, if it were women-only, there would be 25% of 200 members or 50 homeowners. We actually have a mix of men and women, but because we want the least number of homeowners possible, we will maximize the number of male members. This will allow us to have a number of homeowners closer to 40 and farther from 50. Eliminate A, B and C

Guess D or E

Solution

Because 20% and 25% correspond to 15

and 14

respectively, the number of men must be a multiple of 5, and

women must be a multiple of 4. Since a smaller percentage of men are homeowners, to minimize the number of homeowners we need to have as many men as possible.

• There cannot be 200 men because we are told that there are some women in the association. • There cannot be 195 men because that would leave 5 women, which isn’t a multiple of 4 • There cannot be 190 men because that would leave 10 women, which isn’t a multiple of 4 • There cannot be 195 men because that would leave 15 women, which isn’t a multiple of 4 • 180 is the highest possible number of men, and that leaves 20 women in the association.

20% of men and 25% of women are homeowners. The number of homeowners is 1 1

180 20 415 4⋅ + ⋅ =


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Note: It is no coincidence that there are 20 women. A quick way to get this number is to find the least common multiple of 4 and 5. Because the number of men and of women must be multiples of 5 and 4 respectively, 20 is the smallest number that allows us to replace a group of men with a group of women.


Drill II: #9 Difficulty Level: 600-700 Topics: Sets & Groups

An investment bank of 2,400 employees has two offices: one office in London and a second office in Zurich. How many employees of the Zurich office earn less than $80,000 per year?

(1) There are twice as many London employees earning under $80,000 per year as Zurich employees earning at least $80,000 per year

(2) Four out of every ten employees earn at least $80,000 per year, and three-fourths of those work at the London office

Solution


We need to know one of two things: either the other two values in the Zurich column, or the other two values in the 80k- row. Either of these would allow me to solve for x

(1)

Without concrete values, I don’t have enough information to solve for x


(2)

“Four out of every ten” is 40%. Three-fourths of 40% is 30%. We can deduce that 10% of employees are 80k+ earners in Zurich. We can also determine that since 40% earn 80k+, then 60% earn 80k-. However, there is still no way to solve for x


MERGE STATEMENTS

London Zurich Total 80k+ 80k- x Total 100%

London Zurich Total 80k+ n 80k- 2n x Total 100%

London Zurich Total 80k+ 30% 10% 40% 80k- x 60% Total 100%

London Zurich Total


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By combining the statements, we find that n is 10% and 2n is 20%. This allows us to use the 80k- row to setup 20% 60%x+ = and find that x is 40% of the total 2,400 employees. Don’t bother finding it. We have enough data.



Drill II: #10 Difficulty Level: 600-700 Topics: Number Properties

If a, b, and c are integers, is c even?

(1) ab

is an odd integer

(2) b

ac+ is an even integer

Solution


(1) Multiply by b to get odda b= ⋅ . If b is odd, then a will be odd. If b is even, then a will be even. However, this statement tells us nothing about c


(2) Multiply by c to get evenb ac+ = . This means that b and ac are similar: either they are both odd, or they are both even. If b is odd, then a and c will also be odd. However if b is even, c could be even. We cannot determine exactly the nature of c


MERGE STATEMENTS

Even when the statements are taken together, if a and b are even then c is still not restricted. For example, 18, 6, 3a b c= = = satisfy both statements. On the other hand 12, 4, 2a b c= = = also satisfy both statements.



Drill II: #11 Difficulty Level: 700+ Topics: Translations & Manipulations

80k+ 30% 10% 40% 80k- 20% x 60% Total 100%


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The sum of the ages of Patrick and Michelle is x years. Fourteen years ago, Michelle was four times as old as Patrick. In how many years, in terms of x, will Michelle be twice as old as Patrick?

(A) 42

5x +

(B) 2 126

5x −

(C) 4 42

5x −

(D) 4 126

5x −

(E) 8 126

5x −

Solution

Today, the sum of the ages is x. To find the sum of their ages 14 years ago, we need to take today’s sum and subtract 14 twice, because each person was 14 years younger. So 14 years ago, the sum of the ages was 28x − .

Since Michelle was four times Patrick’s age, Michelle was 80% of the sum, ( )428

5x − and Patrick was 20% of

the sum, ( )128

5x − . Today, 14 years later, Michelle is ( )4

28 145

x − + and Patrick is ( )128 14

5x − +

Let n be the number of years in which Michelle will be twice as old as Patrick. This is what we’re asked to find.

( ) ( ) ( )4

1Michelle 2 Patrick 28 14 2 28 14

5 5n n x n x n+ = + → − + + = − + +

Distribute the 2 ( ) ( )428 14

52 28 28 25

x n x n− + + = − + +

Multiply both sides by 5 ( ) ( )4 28 70 5 2 28 140 10x n x n− + + = − + +

Expand 4 112 70 5 2 56 140 10x n x n− + + = − + +

Simplify 2 126

5x

n−

=


Drill II: #12 Difficulty Level: 700+ Topics: Geometry

What is the measure of angle u in the figure above?

(1) z is 30 more than w

x

y

z

u

v w


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(2) The ratio of v to z is 3 to 1

Solution


(1) 180u v w+ + = , and 180v z+ = , so z u w= + (Eq. 1).

This statement tells us that 30z w= + . By combining this information with (Eq. 1), we can write 30u w w+ = + 30u =


(2) This statement tells us that 3v z= . Since 180v z+ = , we can write 4 180z = and find values for z and for v. Knowing v, we can find u w+ , but we cannot isolate a unique value for u.



Drill II: #13 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values

Which of the following corresponds to the shaded segment of the number line above?

(A) 65 92 2

x + < (B) 65 92 2

x + < (C) 65 92 2

x − < (D) 65 92 2

x − > (E) 28 37x< <

Solution

To quickly derive the absolute value equation that corresponds to a number line graph, follow the following steps:

1. Find the center of the shaded graph. x c− will be the absolute value component of the equation, where c is

the center of the shaded graph. For example, here the center of our graph is +32.5, so ( )32.5x − + will be the absolute value component. Since the answer choices use fractions, we should write the expression as

652

x −

2. Figure out the distance from the center to the limits of the shaded graph. Let’s call this distance d. If the shaded points are between the center and the limits, the equation will be x c d− < or x c d− ≤ . Otherwise,

x

28 37


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if the shaded points are away from the center, the equation will be x c d− > or x c d− ≥ . In our case, the

distance from the center to the limits is 4.5 or 92

, and the shaded points are between the center and the limits.

As a result, the correct equation in our case is 65 92 2

x − <


Drill II: #14 Difficulty Level: 600-700 Topics: Exponents & Roots

235 45 80n k nA = ⋅ ⋅ . What is n? (1) 1k =

(2) 2 3

3k n

k+ −

=

Solution

The question itself is already simplified.

(1) The statement tells us nothing about n


(2) Square both sides

2 33k n

k+ −

=

Multiply by 3 3 2 3k k n= + −

Subtract 3k 0 2 n= −

We don’t need any more info to find the value of n



Drill II: #15 Difficulty Level: 600-700 Topics: Number Properties; Statistics

If a, b, c, d, and e are consecutive integers, which of the following must be true?

I. The mean of a, and b is smaller than the mean of d and e


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II. 5

a b c d en

+ + + + is an integer if n is an integer

III. The product of the mean and the median of the five integers is a positive integer

(A) None (B) I only (C) I and II (D) I and III (E) I, II, and III

Solution

I. This would be false if a and b were larger than d and e. Although we’re dealing with consecutive integers, we cannot assume that they have been ordered for us

II. While it is true that the sum of five consecutive integers must be a multiple of 5, 5

a b c d en

+ + + + would

not be an integer if the bottom were larger than the top (ie. if n were large enough) III. In a list of consecutive integers, the mean and the median equal the middle integer, so they are equal. If

the middle integer equals 0, then the product of the mean and the median would be 0, so III does not have to be true


Advanced Speed Drill III (15Qs in 50 mins)

Drill III: #1 Difficulty Level: 700+ Topics: Geometry

What is the largest possible volume of a rectangular box placed inside cylinder C?

(1) C has a height of 1 foot (2) C has a volume of π cubic feet

Solution

The box and the cylinder will have the same height. To maximize the volume of the box, we’ll need to use a square base.

s

s 2s

• The base of the box is a square, so its diagonal is 2s where s is the side of the square. Use the

Pythagorean theorem 2 2 2s s d+ = or memorize the ratio of the sides of a 45-45-90 degree triangle:

: : 2s s s • 2s is also the diameter of the circular base, so

22

sr = .


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Regardless of the exact measurements, the base of the box will always be the same percentage of the base of the

cylinder. 2 2 2

2 2 2base of box 2

base of cylinder 2242

s s sr ssπ π

ππ

= = = =

. In other words, the base of the box is always

2base of cylinder

π⋅ . Let BV be the volume of the box, and CV be the volume of the cylinder. Since their heights

will be the same, and volume is base height⋅ , we can conclude that 2

B CV Vπ

= . We’ve just demonstrated that to

find the volume of the box, all we need is the volume of the cylinder.

Our rephrase “What is the volume of the cylinder?”

(1) This isn’t sufficient to find the volume of the cylinder


(2) This statement directly answers our rephrase. We could find the volume of the box using 2

B CV Vπ

=





Drill III: #2 Difficulty Level: 600-700 Topics: Translations & Manipulations; Exponents & Roots

Which of the following is equivalent to 7 71 5 2 5 8

8 4− + −

?

(A) 319332

(B) 319364

(C) 32 14

5 (D)

32 145

− (E) 52 7

Solution

There are several ways to get to the right answer.


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Original equation 7 71 5 2 5 8

2 4− + −

FOIL ( )7 7 7 71 5 2 5 2 5 8 5 8

4 8 4 8− ⋅ + ⋅ − + ⋅ + ⋅ −

Multiply each term ( )14 7 141 25 4

4 32 45 5

− + − ⋅ + + −

Combine like terms 7

1 10032

− − +

Simplify 3200 7 3193 3193

1 132 32 32

− + −− = − =


Drill III: #3 Difficulty Level: 700+ Topics: FDPs & Ratios

In Fake City, the population density, x, is inversely proportional to the square of the satisfaction indicator, y, and y is directly proportional to the square root of the graduation rate, z. If the graduation rate in Fake City doubled, how would the population density have to change for the satisfaction indicator to remain constant?

(A) 75% increase (B) 50% increase (C) No change (D) 50% decrease (E) 75% decrease Solution

• x is inversely proportional to 2y means 2 constantx y⋅ =

• y is directly proportional to z means constantyz=

When z is doubled, the bottom of the second proportion will become 2 2z z= ⋅ . Because the bottom has

been multiplied by 2 the top must be multiplied by the same thing to maintain the constant. So when z is

doubled, y is multiplied by 2 and thus becomes 2y . The question is, what should we do to x to bring y back to its original value?

When y becomes 2y , the left side of the first proportion becomes ( )2 22 2x y xy⋅ = . To return everything

back to normal and maintain the constant, x will be multiplied by 12

to cancel out the effect of 2. So if the

graduation rate z doubles, the population density x will be halved (reduced by 50%).



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Take-Aways

• Two values are directly proportional ifvalue 1

constantvalue 2

= . As one value increases (or decreases), the other

must follow suit to maintain the constant. Two values are inversely proportional if ( )( )value 1 value 2 constant= . As one value decreases, the other must increase to maintain the constant.

Drill III: #4 Difficulty Level: 700+ Topics: Coordinate Geometry

Parallel lines k and l have x-intercepts at 146 and 1

212 respectively. If line l has a slope of 43

− , what is the

shortest distance from line k to line l?

(A) 9 (B) 138 (C) 7

247 (D) 146 (E) 5

30 Seconds Hack

The shortest distance between two parallel lines is the distance between the points of intersection of those lines with the perpendicular line that crosses them. Since we are given x-intercepts 1

46 and 1212 , we know that the

points of intersection between the x-axis and the two lines are a distance of 1 1 12 4 412 6 6− = apart. The shortest

distance between the lines will have to be shorter than this 146 .

Guess E

Solution

The Hack above is the best way to tackle this question. However, we can solve. First, find the equations of lines

k and l. For each line, we are given a point (x-intercept) and the same slope of 43

− (they’re parallel). Start with

the standard equation of a line y mx b= + where m is the slope and b is the y-intercept, and input your data:

• Line k: put the point ( )146 ,0 into

43

y x b= − + and get 14

4 250 6 3 3

b b= − ⋅ + → = . So the equation for

line k is 4 253 3

y x= − +

• Line l: put the point ( )1212 ,0 into

43

y x b= − + and get 12

4 500 12 3 3

b b= − ⋅ + → = . So the equation

for line l is 4 503 3

y x= − +


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The shortest distance between two parallel lines is the distance between the points of intersection of those lines with the perpendicular line that crosses them. Perpendicular lines must have negative reciprocal slopes. Since

lines l and k have a slope of 43

− , their perpendicular will have a slope of 34

.

Find the coordinates of a: 3 4 25

44 3 3

x x x= − + → = . To find the y-coordinate, plug the value of x back into

the equation of one of the two lines that intersect at a. I used 34

y x= because it is simpler. When we plug in 4

for x, it becomes 3

4 34

y = ⋅ = , so the coordinates of a are ( )4,3

Find the coordinates of b: 3 4 50

84 3 3

x x x= − + → = . Plug this x into 34

y x= to find 3

8 64

y = ⋅ = . The

coordinates of b are ( )8,6 . Finally, we need to find the distance between a and b. To do this, draw up a right

triangle and use the Pythagorean Theorem:


b (8,6)

(8,3)

3

4

The distance between the two points is the hypotenuse of our right triangle. You should recognize the Pythagorean triple 3-4-5 and know right away that the distance is 5.

You can also find the hypotenuse by using the Pythagorean theorem: 2 2 23 4 d+ = . 5d =

a (4,3)

a

b

• The distance between points a and b is the shortest distance between lines k and l. To find the intersection between two lines, set their equations equal to each other.

• To find the coordinates of a set 3 4 254 3 3

x x= − +

• To find the coordinates of b, set 3 4 504 3 3

x x= − +

34

y x=

Line l

Line k


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• Parallel lines have the same slope but the slopes of two perpendicular lines are negative reciprocal. If the

first line has slope m, its perpendicular will have slope 1

m−


On the same day, Mario and Luigi each open an investment account and each make an initial deposit. No other deposit or withdrawal was made on either account. If each account grows at p percent per year, the size of Mario’s account will be what percent of the size of Luigi’s account after y years?

(1) Mario’s initial deposit is $10,000 and 5y = (2) Luigi’s initial deposit is 40% greater than Mario’s initial deposit.

Solution

Let M be Mario’s initial deposit. For each year that passes, the value of the account will be multiplied by

1100

p+

. For example, if p is 15%, each year, the value of the account will be multiplied by 1.15. After y

years, the value of Mario’s account will be 1100

ypM ⋅ +

and the value in Luigi’s account will be

1100

ypL ⋅ +

where L is Luigi’s initial deposit. The size of Mario’s account as a percentage of Luigi’s is

Mario's account 100Luigi's account

⋅ . Using the formulas above, this percentage is 100 1001

100

1100

y

yML

pM

pL⋅ = ⋅

⋅ + ⋅ +

. The

only thing standing in our way of finding this percentage is the ratio of M to L

Our rephrase “What is ML

?”

(1) This statement tells us nothing about Luigi’s deposit, so we can’t figure out how much bigger or smaller it is than Mario’s deposit.



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(2) 1

1.4 1.4

ML M

L= → = . The statement gives us enough data to find the answer to our rephrase.





Drill III: #6 Difficulty Level: 600-700 Topics: Exponents & Roots

If

252

2 2x =

− and

25 2 256

y+

= , what is the ratio of x to y?

(A) 3 to 5 (B) 2 2 to 6 (C) 1 to 2 (D) 3 2− to 4 (E) 5 2 to 5

Solution

First, simplify x

5252 2

2 2 2 2x = =

− −

Multiply top and bottom by 2 5

2 2 2x =

−

Multiply top and bottom by 2 2 2+ ( )

( )( )5 2 2 2 10 2 10 5 2 5

8 4 22 2 2 2 2 2x

⋅ + + += = =

−− +

Next, setup the ratio xy

( )5 2 5

3 5 2 55 2 5 62225 2 25 25 2 25 25 2 25

6

xy=

+++

= ⋅ =+ + +

Factor out and simplify ( )( )

3 5 2 1 3525 2 1

xy

⋅ ⋅ += =

⋅ +



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Drill III: #7 Difficulty Level: 700+ Topics: Translations & Manipulations; Number Properties

2 0x bx c+ + = has two integer solutions for the values of x. b is an integer constant and c is a prime number constant. Is 2x > ?

(1) b is odd (2) c is even

Solution

The values of x can be found by factoring the quadratic into ( )( ) 0x m x n+ + = where m n c⋅ = and m n b+ = .

For example, 2 2 8 0x x− − = can be factored into ( )( )2 4 0x x+ − = where 2 4 8⋅ − = − and ( )2 4 2+ − = − .

We’re told that c is prime, so primem n⋅ = . Because we’re dealing with integers only – we’re told that x has

integer solutions – this means that 1m = ± , and primen = ± (or vice-versa).

(1) m n b+ = , so ( ) ( )1 prime odd± ±+ = . Since odd even odd+ = , this statement tells us that n is even.

The only even prime is 2, so either 1, 2m n= = or 1, 2m n= − = − . Note that the solutions of

( )( ) 0x m x n+ + = are –m and –n. Based on the possible values of m and n, we can determine that

1 or 2x = , so we have enough data to answer the question “Is 2x > ?”. It is NOT greater than 2.


(2) We already know that c is prime, so this statement tells us that 2c = . m n c⋅ = and m and n are integers – we’re told that x has integer solutions – so 1m = ± and 2n = ± . Note that the solutions of ( )( ) 0x m x n+ + = are –m and –n. Based on the possible values of m and n, we can determine that

1 or 2x = , so we have enough data to answer the question “Is 2x > ?”. It is NOT greater than 2.



Drill III: #8 Difficulty Level: 700+ Topics: Rates & Work


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At a 225-workers manufacturing plant, experienced workers can build a widget in 40 minutes while novice

workers spend 90 minutes to build a widget. The average rate of all workers is 56

widgets per hour. If 25% of the

novice workers resign, how many experienced workers must be added to the workforce so that production levels remain the same?

(A) 9 (B) 16 (C) 20 (D) 30 (E) 45

Solution

To find out how many experienced workers must be added, we need to know how many novice workers left, so we need to find out the total number of novice workers. The novice workers’ rate of 1 widget in 90 minutes is

equivalent to 46

widgets per hour. The experienced workers’ rate of 1 widget in 40 minutes is equivalent to 96

widgets per hour. The overall average rate is 56

widgets per hour

As the graph above demonstrates, the weighted average rate is four times as far from the experienced rate as it is from the novice rate. In practice, this means that if we split the worker population in five parts, they would be

four parts novice workers and one part experienced workers. 225

4 1805

⋅ = novice workers.

You can also find the number of novice workers, N, by using the weighted average formula. In this formula, I

simplified the rates. Novice workers’ rate is 4 26 3= widgets per hour and experienced workers’ rate is

9 36 2=

widgets per hour.

( )2252 3

sum of novice rates sum of experienced rates 53 2weighted averageall workers 225 6

N N+ −+

= = =

Simplify to get 180 novice workersN = .

25% of novice workers resign, so the plant has lost 180

454

= novice workers. We need to replace them with

experienced workers who will produce the same number of widgets so we will equate the hourly production of the

46

96

Overall average

Novice rate Experienced rate 56


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45 who left with the hourly production of x experienced workers. The value of x will tell us how many

experienced workers we need: 2 3 2 2

45 w/hr w/hr 45 203 2 3 3

x x⋅ = ⋅ → = ⋅ ⋅ =


Drill III: #9 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values; Number Properties

a is an odd integer and a b≠ − . Is 2a a b

b a> −

+?

(1) 0ab = (2) 3a a>

Solution

To simplify absolute value, remove the symbol and divide your problem into two possibilities. x x= if x is

positive, but x x= − if x is negative. In our case, after cross-multiplying we get “Is ( )2a a b b a> − + ?”

When we remove the absolute value, we’re left with:

• If 0b a+ > , then our rephrase “Is ( )( )2a a b a b> − + ?”

• If 0b a+ < , then our rephrase “Is ( ) ( )2a a b a b> − ⋅ − − ?”

In short, we cannot solve until we know the sign of b a+

(1) We already know that a is an odd integer, so this statement tells us that b equals 0. However, because a can be either positive or negative, we still do not know the sign of b a+ . 1, 0a b= = and 1, 0a b= − = would result in different answers to the original question.


(2) This statement tells us that 1 0a− < < or 1a > . Since we already know that a is an integer, it must be the case that 1a > . However, since we don’t know anything about b, we cannot determine the sign of b a+ .

3, 4a b= = − and 3, 2a b= = would result in different answers to the original question.


MERGE STATEMENTS


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By merging the statements, we can determine that b equals 0 and 1a > , so 0b a+ > . With this information, we

can further simplify the question: Is 2a

a bb a

> −+

?

Cross multiply “Is ( )2a a b a b> − + ?”

Remove absolute value “Is ( )( )2a a b a b> − + ?”

FOIL “Is 2 2 2a a b> − ?”

Isolate b “Is 2 0b > ?”

Since we know that b equals 0, we have enough data to answer the simplified question.


Drill III: #10 Difficulty Level: 500-600 Topics: Number Properties

What is the sum of all prime numbers between 110 and 130?

(A) 236 (B) 240 (C) 347 (D) 351 (E) 359

Solution

To determine whether a number N is a prime, check whether smaller prime numbers are factors of N, from 2 (the

smallest prime) up to N . To do this quickly, it is vital to know the rules of divisibility by 2, 3, and 5. A number is divisible by 3 if the sum of its digits is a multiple of 3. For instance, 126 is divisible by 3 since 1 2 6 9+ + = which is a multiple of 3. Let’s check our numbers. I’ve ignored evens since they cannot be prime, as well as all numbers that end in a 5 since they’re multiples of 5.

There are only 2 primes between 110 and 130, and their sum is 113 127 240+ =


Drill III: #11 Difficulty Level: 700+ Topics: Number Properties

Number Divisible by Number Divisible by 111 3 113 Prime! 117 3 119 7 121 11 123 3 127 Prime! 129 3


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3 5 643 3 3 3 3+ + + + is

(A) between 1 and 2 (B) between 2 and 3 (C) between 3 and 4 (D) between 4 and 5 (E) greater than 5

Solution

Each of these roots is greater than 1, because a number between 0 and 1 only gets smaller when it is raised to a positive power. As a result, such a number would never reach 3 if it were raised to the 2nd, 3rd, 4th or even 5th power. It would only result in a smaller fraction.

Because each of the roots is greater than 1, the sum of the five roots must be greater than 5


Drill III: #12 Difficulty Level: 700+ Topics: Number Properties

When integer n is divided by integer p, the remainder is r. What is the sum, in terms of p and r, of all possible distinct remainders when n is divided by 3p?

(A) 3 3r p+ (B) 2r p+ (C) 3r (D) 3p r− (E) 3r p+

Solution A – Plug-In

The easiest way to do this question is to plug in some values that correspond to the information given.

Our plug-in 5p = 3r =

With these plug-ins, the question becomes: “When integer n is divided by 5, the remainder is 3. What is the sum of all possible distinct remainders when n is divided by 15?” We’ll try different values of n and find all possible remainders:

All possible values of the remainder when n is divided by 15 are 3, 8, and 13. Their sum is 24. Finally, find which answer choice gives us 24 when we replace p with 5 and r with 3. Only answer A fits the bill.

(A) ( ) ( )3 3 3 3 3 5 24r p+ = + =


Solution B

n Remainder of

5n np= Remainder of

3 15n np=

18 3 3 23 3 8 28 3 13 33 3 3 38 3 8 43 3 13 48 3 3 53 3 8 58 3 13


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“When integer n is divided by integer p, the remainder is r”, so on a number line n has to be positioned r units past a multiple of p. Every 3rd multiple of p is a multiple of 3p. On a number line, there are only 3 possible remainders when n is divided by 3p



A store sells 3 items either separately or bundled. When the items are bundled, the most expensive item costs 20% above its regular price while the 2nd most expensive and the least expensive are each discounted by 15%. Is it cheaper to buy the 3 items in a bundle than to buy them separately at their regular prices?

(1) The 2nd most expensive item costs $10 (2) The least expensive item costs $3

Solution

Let a, b, and c be the regular prices of the most expensive, 2nd most expensive, and least expensive items respectively. When bundled, the items cost 1.20 , 0.85 , and 0.85a b c . To find out whether it is cheaper to buy the items together, we need to determine if 1.20 0.85 0.85a b c a b c+ + < + + . Subtract 1a from both sides to get 0.20 0.85 0.85a b c b c+ + < + . Subtract 0.85b and 0.85c from both sides to get 0.20 0.15 0.15a b c< + . Multiply by 100 to clear decimals, then divide by 5 to get our rephrase.

Our rephrase: “Is 4 3 3a b c< + ?”

(1) 10b = . Putting this data into our rephrase allows us to rewrite the rephrase as “Is 4 30 3a c< + ?” We know that 4a is more than 40 (since 10a > ), but we still cannot find a definitive answer.


n 3px

r

px px 3px

n 3px

r p+

px px 3px

3px

2r p+

px px 3px n

px is a multiple of p, but not a multiple of 3p. Only every 3rd multiple of p is a multiple of 3p. On the graphs, 3px is a multiple of p that is also a multiple of 3p

All possible remainders when n is divided by 3p are r, r p+ , and

2r p+ . The sum of these remainders is

( ) ( )2 3 3r r p r p p r+ + + + = +


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(2) 3c = . Like statement (1), this will allow us to simplify the rephrase, but cannot give us a definitive answer since we have no way to compare a to b


MERGE STATEMENTS

Statement (1) simplified our rephrase to “Is 4 30 3a c< + ?” If we put 3c = into this rephrase, it becomes “Is 4 39a < ?” Since a must be greater than 10 ( 10b = ), 4a must be greater than 40. We have enough information to answer the simplified rephrase.





Drill III: #14 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values; Number Properties

If 2 0yx

> , is 0xy > ?

(1) x y x y+ > +

(2) 0x x− >

Solution

A fraction is positive if its top and bottom have the same sign. Since 2x must be positive, 20

yx

> tells us that y

is positive. The question “Is 0xy > ?” can thus be rephrased as:

Our rephrase “Is 0x > ?”

(1) This is a fancy way of saying that x and y do not have the same sign. If x and y had the same sign then in , they would add to each other, and even if they were each negative, the absolute value would make

the sum positive and would equal . On the other hand, because x and y have opposite

signs, in , they partially cancel each other out, and x y x y+ > + . You can try this with different

plug-in values. whereas

x y+

x y+ x y+

x y+

2 3 2 3− + − = − − 2 3 2 3− + > − +


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(2) Statement (2) easier. It tells us that –x and have the same sign. In other words, -x is positive, and x

must be negative. We can answer our rephrase



Drill III: #15 Difficulty Level: 500-600 Topics: Geometry

What is the smaller angle formed by the minutes and hours hands of a 12-hour clock when the time reads 2:40?

(A) 180° (B) 160° (C) 155° (D) 140° (E) 135°

Solution

A 12-hour clock is divided into 12 equal segments. The entire circle is 360°, so each

segment is 360

3012°= ° . At 2:40, the hours hand is a bit past the 2, and the minutes hand

is on the 8. The gap from 2 to 8 contains 6 segments, or 30 6 180° ⋅ = ° . From this figure, we need to subtract the portion between 2 and 3 that the hour hand has already passed.

Since we are 40minutes into the hour, 40mins 260mins 3

= of the 30°, or a total of 20° were

already passed. The angle between the hours and minutes hand of the clock is 180 20 160° − ° = °


Advanced Speed Drill IV (15Qs in 50 mins)

Drill IV: #1 Difficulty Level: 700+ Topics: Geometry

Quarter circle O is centered at the origin and is tangent to line l. If quarter-circle

O has an area of 25

4π

, what is the area of the shaded region?

x


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(A) 25 3 25

4π−

(B) 50 10 3

3π −

(C) 50 25 3

6π −

(D) 50 3 25

12π−

(E) 25 3

12π −

30 Seconds Hack

Approximate. Given the odd shape of the shaded area, it is difficult to estimate its area. However, if we don’t know how to solve, this is the next best thing. In my estimate, the shaded area is maybe 10% of the quarter circle.

Since the area of the quarter circle is 25

4π

, my best estimate for the shaded area is 25 1 25 3 5 3

24 10 4 10 4 2π

≈⋅ ⋅

⋅ ≈ ≈⋅ ⋅

.

Next, go through the answer choices and pick whichever answer is closest to our estimate. I used 3 1.5≈ and 3π ≈ to move faster through the answer choices.

(A) ( )25 1.5 25 325 3 25

4 4π ⋅ − ⋅−= Stop! This will be a negative number.

(B) ( ) ( )50 3 10 1.550 10 3 150 15

453 3 3

π ⋅ − ⋅− −≈ = = This is too large

(C) ( ) ( )150 37.550 25 3

196 6

π −−≈ ≈ . This is too large

(D) ( ) ( )75 7550 3 25

12 12π −−≈ This is small. Almost 0

(E) ( )25 3 1.525 3

712 12π ⋅ −−

≈ ≈

Answers D and E come closest to our estimate of 2.

Guess D or E

Solution

We’re given the quarter-circle’s area as 25

4π

, so the area of the entire circle is 25π and its radius is 5. The given

angle of 120° tells us that the supplementary angle is 60°. Since the line is tangent to the quarter circle, the radius and the line form a right angle.

Note that the area of the shaded region equals the area of the triangle minus

the area of the 30° sector. The sector is 30 1

360 12= of the circle, so its area

is 2512π

. The area of the triangle is a bit harder to find. Because the

Not drawn to scale


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triangle is a 30-60-90 degree triangle, the lengths of the sides will be in proportion : 3 : 2s s s . In our case

3 3

rr s s= → = .Therefore the sides proportion in terms of r is actually

2: :

3 3r r

r . By replacing r with

its value of 5, we can find that the lengths of the sides are 53

(the base), 5 (the height), and 10

3 (the

hypotenuse). The area of the triangle, 12

bh is 1 5 25 25 3

52 63 2 3⋅ ⋅ = = .

The area of the shaded region is area of triangle minus area of sector, or 25 3 25 50 3 25

6 12 12π π−

− =




Drill IV: #2 Difficulty Level: 700+ Topics: Number Properties

x, q, and r are positive integers and remainder 17x

q r= . What is the remainder when x is divided by 34?

(1) q is odd (2) 5r =

Solution

remainder 17x

q r= means that x r− is a multiple of 17 (this multiple is 17q). For instance 55

3 remainder 417

=

means that on a number line, 55 4− is a multiple of 17 (this multiple is 17 3⋅ ). Because every other multiple of 17 is a multiple of 34, there is a 50% chance that x r− is also a multiple of 34. I’ve illustrated the two possible

outcomes of 34x

below


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To summarize, the remainder when x is divided by 34 is either r or 17 r+

We need to know 2 things: first, is the remainder we want equal to r or equal to 17 r+ ? Secondly, what is the value of r? Both of these pieces are necessary to answer the question.

(1) This statement tells us nothing about the value of r, so we cannot answer the question.


(2) I found statement (2) easier. It gives us r, but doesn’t tell us whether the remainder is r or for , so we cannot answer the question.


MERGE STATEMENTS

Since q is odd, 17q must be odd. Remember that 17q x r= − . In other words, 17q is the multiple of 17 just before x on our number line graph. Since this number is odd, it cannot be a multiple of 34. Of the two possibilities illustrated in Step I above, the 2nd must be correct., so the remainder must be 17 r+ . Statement (2) tells us that r is 5, so the remainder is 17 5+ . We have enough data.

A simpler, but more time consuming alternative is to try numbers for x that agree with both statements. We know

that remainder 517x

q= and that q has to be odd

x 17x

34x

Notes

22 1 remainder 5 0 remainder 22 39 2 remainder 5 -- q must be odd. We cannot use this example 56 3 remainder 5 1 remainder 22 73 4 remainder 5 -- q must be odd. We cannot use this example 90 5 remainder 5 2 remainder 22

As long as q is odd, and r is 5, the remainder of x divided by 34 is 22.



17 r+

34n = 17q

17n 34n

34n 17n =

17q

34n

x

x

r

17 r+

On these number lines, 34n is a multiple of 17 and of 34, while 17n is a multiple of 17 but not a multiple of 34.

• When x r− is a multiple of 34, the remainder of the division will be r

• When x r− is not a multiple of 34, the remainder of the division will be 17 r+

• In both cases, 17x r q− =


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Drill IV: #3 Difficulty Level: 700+ Topics: Statistics; Weighted Averages

A company has 3 salary levels. Level II’s employees earn 50% more than Level I’s, and Level III’s earn 30% more than Level II’s. If 46% of all employees are on Level II, what is the average salary of all employees?

(1) If 10% of Level I employees were promoted to Level III, there would be twice as many Level III employees as Level I employees, and those promoted would receive a $38,000 increase in their salary.

(2) 20% of all employees are Level I employees with a $40,000 salary

Solution

Let SI, SII and SIII be the salaries of Levels I, II, and III employees respectively. Let NI, NII and NIII be the percentage of employees of Levels I, II, and II respectively. The average salaries of all employees is a weighted

average determined by 100

I I II II III IIIN S N S N S⋅ + ⋅ + ⋅

“Level II earn 50% more than Level I” means 1.5II IS S= (Eq. 1)

“Level III earn 30% more than Level II” means ( )1.3 1.3 1.5III II IS S S= = (Eq. 2)

(1) This statement is more tricky. It states that if we took 10% of employees away from NI and added it to NIII, the new NIII would be double the new NI. Algebraically, . We already

knew that . Now we have 2 independent linear equations and two variables, so we can

solve for NI and NIII. On the GMAT, don’t waste your time solving. Just realize that you have enough to find the percentage of employees at each level.

The 2nd bit this statement gives us is that the difference in salaries between Levels I and III is $38,000. Algebraically, . We already knew that (Eq. 2), so again we have 2

independent linear equations; we can solve for the salaries. Once we have SI, we can use (Eq 1) to find SII. As statement (2) did, this statement gives us a way to solve for all unknowns


(2) By telling us that 20% of employees are Level I, this statement lets us know that 34% are level III (since the levels add up to 54% of employees). The statement also tells us that Level I salary is $40,000. Based on (Eq. 1) and (Eq. 2) above, we can find the salaries for Levels II and III. In effect, this statement allows us to solve for all unknowns.



( )2 0.9 0.1I III IN N N= +

54I IIIN N+ =

38,000III IS S− = ( )1.3 1.5III IS S=


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Drill IV: #4 Difficulty Level: 500-600 Topics: Geometry

What is the perimeter of triangle ABC?

(1) Side AC has length 3, and side AB has length 4 (2) Triangle ABC can be inscribed in a circle such that side BC is the diameter

Solution


(1) Don’t mistakenly assume that we are dealing with a 3:4:5 right triangle. That may be the case, but we don’t know. The third side could be shorter or longer than 5 depending on the angle measure at A.


(2) This tells us nothing about the lengths of the sides.


MERGE STATEMENTS



Drill IV: #5 Difficulty Level: 700+ Topics: Inequalities & Absolute Values

If x and y are distinct non-zero integers, is x y x y+ = − ?

(1) 2xy < (2) 0x y− >

Solution

A

C

B

3 4

5

A triangle inscribed in a circle with one side as the diameter must be a right triangle. We can now use the Pythagorean Theorem to find that the length of the hypotenuse (or realize that it is a 3:4:5 Pythagorean triple). Perimeter is 3 4 5+ +


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I thought rephrasing would take more time than is reasonable, so I went right to the statements.

Absolute values are basically inequalities. When solving these questions, you almost always need to know the signs of your variables, or how they compare to each other. This question is no exception.

(1) Since x and y are integers, xy must be an integer. However, xy cannot be 1 because the only ways to get 1 would be or but we know that x and y are distinct. Since they are non-zero, xy cannot be

0 either. In short, , so x and y have opposite signs. Let’s play with some values and see whether this data is enough to definitively answer our question.

Here it took a few more tries to discover that we still don’t have a definitive answer. When using this method with absolute value questions, be sure to alternate signs of each variable, and to give each variable the chance to have the greater absolute value. In this case, it is only when y has the greater absolute value that we discover a different answer


(2) Statement (2) tells us that x is greater than y. Let’s play with some values and see whether this data is enough to definitively answer our question.

When plugging in, make sure you respect the restriction of the statements you’re checking. Stop as soon as you get different answers. This means that you do not have enough information.


MERGE STATEMENTS

If we choose to plug in values to check the statements together, we must make sure that the values we plug in agree with both statements. Once again, we have no definitive answer.


Note: x y x y+ = − is true only if x and y have opposite signs, AND x y≥ (so that the right side can be

positive). The 2nd condition is not addressed by either statement, so we never have sufficient data to answer “Is x y x y+ = − ?”


Drill IV: #6 Difficulty Level: 700+ Topics: FDPs & Ratios

1 1⋅ ( ) ( )1 1− ⋅ −

0xy <

x y is x y x y+ = − ? -3 2 YES 3 -2 YES 1 -1 YES -2 3 NO

x y is x y x y+ = − ? 3 2 NO 3 -2 YES

x y is x y x y+ = − ? 3 -2 YES 2 -3 NO


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25% of US residents have no car, and 25

of these are citizens. There are 1333 % more citizens with one car than

foreigners with multiple cars. For every 5 US residents with no car, 6 citizens have multiple cars. The ratio of citizens with multiple cars to citizens with one car is 3 to 2. What percentage of US residents are foreigners with one car?

(A) 10% (B) 15% (C) 20% (D) 30% (E) 40%

Solution

Build a table and input the given data one piece at a time.

Start by putting down 100% for total at the bottom right. “25% of US residents have no car”, and two-fifths of these, 10%, are citizens. Next identify what is asked. X is the “percentage of US residents who are foreigners with one car”

“ 1333 % more citizens with one car than foreigners with multiple

cars” means that if foreigners with multiple cars is a, citizens

with one car will be 1 43 3

a a a+ =

“For every 5 US residents with no car, 6 citizens have multiple cars.” We know that 25% of the total are residents with no car,

so we can setup a proportion, 5 25%6 ?= , and solve to find that

30% of the total are citizens with multiple cars.

“citizens with multiple cars to citizens with one car is 3 to 2” We know that citizens with multiple cars are 30% of

total, so we can setup a proportion, 3 30%2 ?= , and find that

20% of the total are citizens with one car. Since this value

is 43

a , we can go further and find that a, the foreigners with multiple cars, is 15%

Multiple One No car Total Citizens 10 Foreigners X Total 25 100

Multiple One No car Total Citizens

43

a 10

Foreigners a X Total 25 100


30 43

a 10

Foreigners a X Total 25 100


30 4

203

a = 10

Foreigners 15a = X Total 25 100

Multiple One No car Total Citizens 30 20 10


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Since the bottom row must add up to 100, and we know that the “Multiple” column and the “no Car” column add up to 70, the column

containing X must add up to 30 and thus X must equal 10.


Drill IV: #7 Difficulty Level: 700+ Topics: Geometry

In the figure above, circles A and B are congruent, and circles C and D are congruent. If the sum of the areas of the four circles is 80π , what is the area of trapezoid ABCD?

(1) The sum of the radii of circles A and C is 8 (2) The product of the radii of circles B and D is 12

Solution

Let r be the radius of circles A and B and R be the radius of circles D and C. Area of a trapezoid is

1 2base base2

h+

⋅ . In this case, note that base1 is 2r and base2 is 2R. Although it’s a bit more tricky, you may also

see that the height of the trapezoid is the vertical distance from base1 to the x-axis (r) + the vertical distance from the x-axis to base2 (R). So the height of the trapezoid is r R+ . We can rewrite the area of the trapezoid in terms

of r and R as ( ) ( )( )2 2A

2r R

r R r R r R+

= ⋅ + = + +

Our rephrase “What is ( )2r R+ ?”

(1) 8r R+ = . This means that ( )2 64r R+ = . We can definitively answer our rephrase


(2) 12r R⋅ = . Since the sum of the areas of all circles is we can write

. We now know that 2 2 40r R+ = . We can determine that

( ) ( )2 2 22 40 2 12r R r r R R+ = + ⋅ + = + . We have enough data to answer our rephrase

80π2 2 2 22 2 80 40r R r Rπ π π+ = → + =

Foreigners 15 X Total 45 25 100

A B

C

D


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Drill IV: #8 Difficulty Level: 500-600 Topics: Exponents & Roots

Of the following, what is the closest approximation to ( )8.87 10.15 5.80

3.91+

is

(A) 3 (B) 6 (C) 9 (D) 12 (E) 15

Solution

Approximate!! ( )9 10 6 9 16 12

64 24

+ ⋅= = =



Drill IV: #9 Difficulty Level: 700+ Topics: Rates & Work

It takes Abe 4510 hours to paint a house while Brian can paint 7 houses in 54hours. Working together, Abe and

Brian start painting at the same time. Once ½ the house is painted, Brian is replaced by Cory without delay.

What must Cory’s rate be so that it takes Abe and Cory 1333 % less time to paint the second half of the house

than it took Abe and Brian to paint the first half of the house?

(A) 4 houses per day (B) houses per day

(C) houses per day

(D) 8 houses per day (E) houses per day

Solution

135 7

95 298


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Abe’s rate is 45

1 510 54

= houses per hour and Brian’s rate is 754

houses per hour. Their combined rate is 12 254 9

=

houses per hour. The question basically asks us how fast must Cory work so that Abe and Cory take a third less time to do the same work that Abe and Brian did.

Let R be Abe and Cory’s combined rate. If it took t hours for Abe and Brian to do ½ the house, it would take Abe

and Cory 1333 % less time, or

23

t . Since the amount of work done is equal, we can set the two work equations

( )work rate time= ⋅ equal to one another: Let’s input the data: 2 29 3

t R t⋅ = ⋅ . Simplify to get 13

R = houses per

hour.

We’ve now found R, the combined rate of Abe and Cory. To find Cory’s rate, just subtract Abe’s rate from R.

Cory’s rate is 1 5 133 54 54− = houses per hours. To match it to the correct answer, convert the rate to houses per

day: 79

13 5224 5

54 9⋅ = = houses per day


Drill IV: #10 Difficulty Level: 600-700 Topics: FDPs & Ratios

Is x more than 65% of the sum of y and z?

(1) x is more than 20% greater than y (2) The ratio of z to y is 45 to 52

Solution

65% is 65 13

100 20=

Our rephrase “Is 13 1320 20

x y z> + ?”

(1) 120 6

100 5

x y x y> → > This tells us nothing about z



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(2) 45 45

52 52

zz yy

→ == . This tells us nothing about x


MERGE STATEMENTS

(2) allows us to simplify our rephrase by replacing z with 4552

y :

Original rephrase “Is 13 1320 20

x y z> + ?”

Plug in for z “Is 13 13 4520 20 52

x y y> + ⋅ ?”

Simplify “Is 13 920 16

x y y> + ?”

Add fractions “Is 9780

x y> ?”

Statement (1) told us that 6 96

5 80

x y x y> → > . We still cannot tell whether x is larger than 9780

y .





Drill IV: #11 Difficulty Level: 700+ Topics: FDPs & Ratios

In 1978, 20% of all goods sold in the USA were foreign. In 1987, 25% of all goods sold in the USA were foreign. If the number of foreign goods sold in the USA increased by 80% from 1978 to 1987, by what percentage did the total number of goods sold in the USA increase over the same period?

(A) 32% (B) 44% (C) 56% (D) 65% (E) 70%

Solution


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Let x and y be the total number of goods in ’78 and ’87 respectively. Foreign goods are 20% of x in ’78 and 25% of y in ’87. Since foreign goods increased by 80%, 0.25y

is 80% more than 0.20x. Algebraically 180

0.25 0.20100

y x= ⋅ . Cross multiply to get

36

25 36 25

yy x x== → .

The total number of goods goes from x to 3625

x . %Change is difference

100original

⋅ . In this case, the percentage

increase in total goods is

36 253611 4425 2525 44%25 100

xx x

x x

−−= = = =


Drill IV: #12 Difficulty Level: 700+ Topics: Statistics; Weighted Averages

A room contains only 5yr olds and 20yr olds, with a combined average age of 15yrs. If eleven 5yr olds and ten 20yr olds were to enter the room, 37.5% of the people in the room would be 5yrs old. How many people in the room are 20yrs old?

(A) 25 (B) 38 (C) 50 (D) 60 (E) 68

Solution

Let’s define some variables:

%A – The percentage of people who are 20yrs old (adults)

( )100 %A− – The percentage of people who are 5yrs old (kids)

The weighted average, 15yrs, is ( ) ( )5 100 % 20 %

15100A A− +

= .

Cross multiply ( ) ( )5 100 % 20 % 15 100A A− + = ⋅

Remove parentheses ( ) ( )500 5 % 20 % 15 100A A− + = ⋅

Isolate %A ( )15 % 1,000A =

Find %A ( ) 13% 66A =

1978 1987 Foreign goods 0.20x 0.25y Total goods x y


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Since 1366 % of people are adults, the rest are kids, and the ratio of adults to kids is 2:1

Note: A quick alternative is to observe that since the overall average (15yrs) is twice as far from 5yrs as it is from 20yrs, there must be twice as many 20yr olds in the room. Had the average been 8yrs, it would have been four times as far from 20yrs as from 5yrs. In that case, there would have been four times as many 5yr olds in the room.

Now we have a classic ratio problem. Starting ratio of adults to kids is 2:1. After ten adults and eleven kids are

added, the ratio of adults to kids becomes 62.5% 125 537.5% 75 3

= = . Algebraically this relationship can be expressed as

2 10 5 25

11 3x

xx+

= → =+

. The number of 20yr olds (adults) in the room is 2x, or 50


Drill IV: #13 Difficulty Level: 500-600 Topics: Geometry

In the figure above, the sum of z and w is equal to the sum of z and y. What is the sum of x and y?

(A) 80° (B) 90° (C) 100° (D) 110° (E) 120°

Solution


x

y z

100

y

We’re told that z w z y+ = + , so y w= . Let’s make replace w with y in the figure.

In the highlighted triangle, 100 180x y+ + = , so 80x y+ =

x

y z

100°

w


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Drill IV: #14 Difficulty Level: 600-700 Topics: Functions & Sequences

A machine printed a repeating sequence of dots colored red, white, blue, green, orange, and yellow in that order. If the first dot was blue and the last dot was red, which of the following could be the total number of dots printed?

(A) 112 (B) 123 (C) 126 (D) 132 (E) 149

Solution

This is a pattern problem. To solve patterns, it is always best to draw out what is happening as this will increases your chances of spotting the pattern. Below, I have reproduced each cycle on its own line.

Although in the question stem the first color listed is red, the sequence of dots actually starts with a blue dot, so our solution must reflect this. This is why each cycle starts with Blue in the figure above. Note that because each cycle has 6 dots, the total number of dots at the end of a cycle must be a multiple of 6. The sequence ends with a red dot, which means the total number of dots must be a multiple of 6 (end of a cycle) plus 5 additional dots. We are looking for a multiple of 6 +5.

(A) ( )112 6 18 4= ⋅ +

(B) ( )123 6 20 3= ⋅ +

(C) ( )126 6 21 0= ⋅ +

Green Orange Yellow Red White Total

6

12

18

24

Start

Blue

6 5n +


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(D) ( )132 6 22 0= ⋅ +

(E) ( )149 6 24 5= ⋅ + This is the only answer that is a multiple of 6 plus 5




Drill IV: #15 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values

0xy ≠ and y x> . If 2

x yz

+= , which of the following must be FALSE?

(A) x y z+ < (B) xy z< (C) z x y< − < (D) y x z− < < (E) z x<

Solution

Since z is an absolute value, it must be at least 0

(A) If x and y are negative, x y+ will be negative and (A) will be true (B) If y is positive and x is negative, xy will be negative and (B) will be true (C) If y is positive and x is negative, (C) could be true. For example, 3, 5x y= − = (D) If x and y are positive, (D) will be true. For example, 2, 4x y= = (E) We know that x is smaller than y, so the average of x and y must be greater than x. Since z is equal to or

greater than the average of x and y, z must be greater than x. (E) must be FALSE


Advanced Speed Drill V (15Qs in 50 mins)

Drill V: #1 Difficulty Level: 700+ Topics: Number Properties

If x is an integer, what is the sum of all distinct positive factors of x ?

(1) x has exactly three distinct positive factors (2) 2 1 3x k− = , where k is an odd integer

Solution



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(1) Only integers such as 4, 9, and 25 have exactly three distinct positive factors. These are the squares of prime

numbers. The statement tells us that x is the square of a prime, and its factors are . However,

we still cannot determine the exact value of the factors of .


(2) I found statement (2) easier. By factoring the left side we can rewrite this statement as .

Since the product of two odds is odd and are odd, so x must be even. However, without

knowing the value of x, we cannot find the factors of x .


MERGE STATEMENTS

By merging the statements, we learn that x must be even, and its square root must be prime. Since the root of x

must also be even, 2x = . We don’t need any more information to answer the original question



Drill V: #2 Difficulty Level: 700+ Topics: Statistics

Each cell in the table above is the positive distance on the number line between the column header and the row header of that cell. For instance, a and b are 7 units apart. What is the value of a b c+ + ?

(A) -2 (B) 4 (C) 6 (D) 12 (E) 18

Solution

The positive distance between two values is the absolute value of their difference. Each line below takes all the previous lines into account.

Row 2, Column 2: 4 1a − = means that a is 3 or 5

Row 2, Column 3: 4 1c − = means that c is 3 or 5

Row 4, Column 2: 2c a− = means that if a is 3, c is 5 but if a is 5, c is 3

Row 4, Column 4: 2c a− = means that c must equal 5 and a must equal 3 (the reverse wouldn’t work)

, , and 1x x

x

( )( )1 1 oddx x− + =

( )1x − ( )1x +

a c 2 4 1 1 2 b 7 9 6 c 2 0 a


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Row 3, Column 2: 7b a− = . Since a equals 3, b is either 10 or -4

Row 3, Column 4: 2 6b − = means that b must equal -4 (since 8b = would not work with the previous line)

We’ve found that 3, 4, 5 so 4a b c a b c= = − = + + =


Drill V: #3 Difficulty Level: 600-700 Topics: Functions & Sequences

( )* 3x x= when x is an integer and ( )2

*2x

x = when x is a non-integer. Which of the following is equivalent to

( )( )( )

4* 3 2 3

* * 27?

(A) *(7) (B) *(6) (C) *(5) (D) *(4) (E) *(3)

Solution

Of the two functions defined, the 1st function applies to the bottom of the fraction, and the 2nd function applies to the top.

( )( )( )

( )( ) ( ) ( )

4

2 22 4

* 3 2 3

* * 27

3 2 3 9 2 39 2 3 1 9 3 1 9 32 2 3

* 9 2 * 9 1 3 9 3 3* 3 27

⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅

= = = ⋅ = ⋅ = =⋅⋅

Next, check the answer choices to find out which one equals 3. Answer E is the only one. ( )* 3 3 3 3= ⋅ =


Drill V: #4 Difficulty Level: 500-600 Topics: FDPs & Ratios

The ratio of managers to workers at a company must be between 5:72 and 3:22. If there are 8 managers at this company, the possible number of workers ranges from

(A) 59 to 115 (B) 58 to 115 (C) 59 to 116 (D) 58 to 116 (E) 60 to 115

Solution


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The higher ratio of manager to workers will have the least number of workers for 8 managers. To find the least

number of workers, setup a proportion: 3 822 x

= . Cross multiply and divide to find that x is 2358 . So there must

be at least 59 workers.

To find the greatest number of workers possible, setup a proportion with the smallest ratio: 5 8

72 x= . Cross

multiply and divide to find that x is 15115 . So there must be at most 115 workers. The possible number of

workers ranges from 59 to 115.


Drill V: #5 Difficulty Level: 700+ Topics: Geometry

In the figure above, the diagonals of both squares are parallel to the x-axis and y-axis, and both squares are centered on the point ( )15,0− . What is the area of the shaded region?

(A) 22 (B) 36 2 25− (C) 144 25 2− (D) 94 (E) 110

Solution

The shaded region is the area of the large square minus the area of the small square. To find the areas, we need to know the length of the sides or of the diagonals. A lesser known way to find the area of a square is diagonal_1 diagonal_2

2×

. Let’s find the diagonals of each square, starting with the smaller square

(-15,-5)

( )6 2 15, 0−


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Next, we need the diagonal of the larger square. Since it is centered on ( )15,0− and the x-coordinate of the right

corner is 15 6 2− + , the x-coordinate of the left corner must be 15 6 2− −

The area of the shaded region is the area of the large square minus the area of the small square: 144 50 94− =



N is a finite set of distinct positive integers. How many even integers does N contain?

(1) 6.25% of the product of all numbers in N is an odd integer (2) The sum of all even integers in N is 10

Solution


( )15 6 2 , 0− + ( )15 6 2 , 0− −

• The length of each diagonal is the distance between the left and right corners:

( ) ( )15 6 2 15 6 2 12 2− + − − − = .

• The area of the larger square is 12 2 12 2

1442

⋅=

12 2

(-15,-5)

(-15,5) The square is centered on ( )15,0− , so the point at

the top of the square is ( )15,5− , and the length of

each diagonal is 10. The area of the smaller square

is 10 10

502⋅

=

10


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(1) . This statement tells us that one sixteenth of the product of all integers in N is odd, but

we cannot determine how many even integers there are


(2) I found statement (2) easier. This does not tell us how many even integers there are. On one hand, 4 and 6 could be the only two evens in the set. However, 10 could be the only even.


MERGE STATEMENTS

The fact that one sixteenth of the product is odd means that if we were to write the product of all the integers in N

as a product of primes, there would be four 2’s ( 42 16= ). There cannot be another 2 anywhere because 4product

2

is odd.

Statement (2) tells us that these four 2’s must form even integers that add up to 10. The possible combinations of distinct positive even integers that add up to 10 are 2 & 8, 4 & 6, and 10. Of these three possibilities, only 2 & 8 use up all four 2’s (their product includes 42 ). The even integers in N are 2 and 8.



Drill V: #7 Difficulty Level: 600-700 Topics: Geometry

In the figure above, x is how many degrees greater than b?

(1) 60d = (2) 100a c+ =

Solution

6.25 25 1100 400 16

= =

a b c

d

x

30


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(1) Since x b d= + , x b d− = . By giving us d, this statement gives us the difference between x and b. Besides, it answers our rephrase.


(2) There is no way to derive a value for d, or any other way to find x b− from the data given in this statement



Drill V: #8 Difficulty Level: 700+ Topics: Translations & Manipulations; Geometry; Rates & Work

On Monday, Hillary and Barrack leave from the same location and travel in different directions. On that day, Hillary travels half as fast as Barrack, but she spends 20% more time on the road. On Tuesday, Barrack remains stationary while Hillary travels 12 miles in a direction perpendicular to her previous day’s direction. If Hillary ends up exactly where Barrack is located, how many combined miles did Hillary and Barrack travel on Monday?

(A) 9 (B) 15 (C) 24 (D) 36 (E) 48

30 Seconds Hack

On Tuesday, as Hillary gets underway, the two people are 12 miles apart. So the day before, they must have traveled at least 12 miles combined. Eliminate A.

Guess B, C, D or E

Solution

Suppose that Barrack’s speed and time of travel on Monday were r and t. This would mean that the distance he traveled that day was rt. Hillary traveled half as fast, but spent 20% more time on the road, so her speed and time

a b c

d

x

30

k

Consider the triangle I’ve highlighted (b-d-k). 180x k+ = , and 180b d k+ + = . Therefore, x b d= +

To answer the question “x is how many degrees greater than b?”, we simply need to know d.

Our rephrase “What is d?”


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of travel on Monday were 12

r and 65

t . This means that the distance Hillary traveled on Monday was

1 6 32 5 5

r t t⋅ = . We can thus determine that the ratio of Hillary to Barrack’s distance was

3355

t tt t= on Monday.

On Tuesday, Hillary turns 90° (perpendicular direction) and travels 12 miles to meet up with Barrack. Their paths thus draw a right triangle


Drill V: #9 Difficulty Level: 700+ Topics: Statistics; Weighted Averages

In a class of boys and girls, the average age is 8.5 years for boys and 9 years for girls. What percentage of people in the class are girls?

(1) If 8 ten year old girls and 17 eight year old boys joined the class, the average age of all children in the class would be 8.72 years.

(2) If 4 nine year old girls and 1 eight year old boy joined the class, the average age of all children in the class would be 8.8 years.

Solution


(1) To find the average age of all children, we divide the sum of all ages by the total number of children. Let

B be the number of boys, and G be the number of girls in the class, ( ) ( )8.5 9 10 8 8 17

8.728 17

B GB G+ + +

=+ + +

. In

the top of this fraction, 8.5B + 9G is the sum of the ages of the original children. 10(8) + 8(17) is the sum of the ages of the added children.

Cross multiply 8.5 9 216 8.72 8.72 218B G B G+ + = + +

Simplify 0.28 0.22 2G B= +

5t

3t

12

Start

Finish

• You ought to recognize right away that this is a 3:4:5 Pythagorean triplet with 3t = . Otherwise, we can set up the

Pythagorean theorem ( ) ( ) ( )2 2 23 12 5t t+ = and solve to find that 3t =

• The combined distance they traveled on Monday is 8t, or 24 miles


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There is no way to further reduce this to find the ratio of boys to girls, or the percentage of people who are girls.


(2) To find the average age of all children, we divide the sum of all ages by the total number of children. Let

B be the number of boys, and G be the number of girls in the class, ( ) ( )8.5 9 9 4 8 1

8.84 1

B GB G+ + +

=+ + +

. In the

top of this fraction, 8.5B + 9G is the sum of the ages of the original children. 9(4) + 8(1) is the sum of the ages of the added children.

Cross multiply 8.5 9 44 8.8 8.8 44B G B G+ + = + +

Simplify 0.2 0.3G B=

Put as a ratio 32

GB=

The ratio we found means that 3 out of every 5 children is a girl: 60% of children are girls.



Drill V: #10 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values; Rates & Work

A snail traveled at 75 meters per hour. Was the distance traveled shorter than 25 meters?

(1) It took the snail fewer than 21 minutes to travel the distance.

(2) If the snail’s rate had been 25% slower, it would have taken fewer than 24 minutes to travel the same distance.

Solution

Because the statements deal in minutes rather than hours, let’s convert the rate into meters per minute by dividing

it by 60. The snail’s rate is 75 560 4

= meters per minute. Distance is rate times time. So “Is 25d < ?” can be

rewritten as “Is 5

254

t⋅ < ?“ which simplifies into our rephrase:

Our rephrase “Is 20t < minutes?”

(1) 21t < . We still cannot tell whether t is greater than 20. It could be 20.5 or 18



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(2) Lowering the rate by 25% is equivalent to multiplying it by 34

. In distance time rate= ⋅ , time and rate

are inversely proportional because the distance does not change. If rate is multiplied by 34

, then time

must be multiplied by 43

so the two will cancel out. The statement says that this new time is fewer than

24 minutes. Algebraically, 4

24 183

t t< → < . We can definitely answer our rephrase. It must be true

that 20t <

An alternative approach is to consider that distance

timerate

= . We know from the question stem that the

distance traveled, rate times time, is 54

t . We are now told that if the rate had been 25% slower, or

3 54 4⋅

, the time of travel would have been shorter than 24minutes. Since distance

24slow rate

> , we can write

5424 24 18

33 544 4

t t

t> → > → >

. It must be true that 20t <





Drill V: #11 Difficulty Level: 700+ Topics: Coordinate Geometry

What is the shortest distance between the line 5 2y x= − and the origin?

(A) 1 (B) 2 (C) 3 (D) 5 (E) 52

30 Seconds Hack


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The x-intercept of the line (the value of x when 0y = ) is 2.5. So the shortest distance between the line and the origin must be less than 2.5. Eliminate E

Guess A, B, C, or D

Solution

Imagine a circle that starts at the origin and gradually expands. It gets bigger and bigger until it touches the line. At this point the radius of the circle is the shortest distance between the origin and the line. Since the line will be tangent to our imaginary circle, the line will be perpendicular to the radius. The radius perpendicular to

5 2y x= − must have a slope of 12

because perpendicular lines have negative reciprocal slopes.



Is 4% of x a multiple of 60?

(1) 30% of x is a multiple of 5

(2) 760

x is an integer

Solution

Is 4

60 int100

x = ⋅ ? (int is an unknown integer) Isolate x and use prime factorizations to rephrase.

2

1 d 2 2 22 1 5d d= + → =

O

( )2,1

5 2y x= − 12

y x= The radius lies on the line

12

y x= . The point at which the

radius and the line intersect is the point closest to the origin. To find this point, equate the two line equations:

15 2 2

2x x x− = → = . Plug 2 back in for x to find 1y = .

The shortest distance between the line and the origin is the distance between the point ( )2,1 and the origin.


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Our rephrase “Is 2 32 3 5 intx = ⋅ ⋅ ⋅ ?”

(1) 2305 int 3 2 5 int

100x x= ⋅ → = ⋅ ⋅ . Since 3 does not share any prime factor with 22 5⋅ , this statement

tells us that 22 5 intx = ⋅ ⋅ . However, we cannot tell what other factors x may have, so we cannot answer our rephrase.


(2) 27int 7 2 3 5 int

60x

x= → = ⋅ ⋅ ⋅ . Since 7 does not share any prime factor with 22 3 5⋅ ⋅ , this statement

tells us that 22 3 5 intx = ⋅ ⋅ ⋅ . However, we cannot tell what other factors x may have, so we cannot answer our rephrase.


MERGE STATEMENTS

When merging 22 5 intx = ⋅ ⋅ with 22 3 5 intx = ⋅ ⋅ ⋅ , be careful not to include duplicate factors. Only use the highest exponent available from each prime. By merging the two statements, we can determine that

2 22 3 5 intx = ⋅ ⋅ ⋅ . We still do not know whether other factors of x might include the additional 5 needed to answer the rephrase with a YES.



Drill V: #13 Difficulty Level: 700+ Topics: Exponents & Roots; FDPs & Ratios

x is inversely proportional to the square root of y, and y is directly proportional to the square of z. If the value of z drops by 30%, the new value of 2x will be how many times the original value of 2x ?

(A) 100

9 (B)

103

(C) 1710

(D) 10049

(E) 107

Solution

“x is inversely proportional to the square root of y” means x y c= where c is a constant.

“y is directly proportional to the square of z” means 22y

cz

= where c2 is a constant.


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• If z drops by 30%, the new z will be 7

10z and the new value of 2z will be

227 49

10 100z z=

.

• Since y is directly proportional to 2z , the new value of y will be 49

100y .

• The new value of y will be 49 7

100 10y y= . Since x is inversely proportional to y , the new value

of x will be 107

x

• Finally, the new value of 2x will be 2210 100

7 49x x=


Take-Aways

• Two values are directly proportional ifvalue 1

constantvalue 2

= . As one value increases (or decreases), the other

must follow suit to maintain the constant. Two values are inversely proportional if ( )( )value 1 value 2 constant= . As one value decreases, the other must increase to maintain the constant.

Drill V: #14 Difficulty Level: 700+ Topics: Functions & Sequences

The sequence a1, a2, a3…an is such that 1 22n n na a a− −= − for all 3n ≥ . a7 is 20% greater than a6. If 4 3a = ,

what is the value of 100 200a a+ ?

(A) 298 (B) 299 (C) 300 (D) 301 (E) 302

Solution

“a7 is 20% greater than a6” means that 7 665

a a= (Eq 1). By making this substitution into the sequence formula,

we can write 7 6 5 6 6 5 6 56 5

2 2 5 4

a a a a a a a a= − → = − → = (Eq 2).

Plug this newfound relationship as well as 4 3a = into the formula for a6.

6 5 4 5 5 552 2 3 44

a a a a a a= − → = − → = . Using a5 along with (Eq 2) and (Eq 1), we can find that a6

is 5, and a7 is 6. Notice a pattern?


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The formula for the sequence can be rephrased as 2

1 2n n

na aa −

−

+= . In other words, each term is equal to the

average of the term before it and the term after it. This is a property of an arithmetic sequence, a sequence in which terms are equally spaced. For instance, multiples of 5 and consecutive odd integers are arithmetic sequences. In our case, we have consecutive integers

a4 a5 a6 a7 … a100 … a200 3 4 5 6 … 99 … 199

100 200 298a a+ =


Drill V: #15 Difficulty Level: 600-700 Topics: Functions & Sequences

If ( ) 2 16f x x x= − − and ( ) 110

g xx

=+

, which of the following CANNOT be the value of ( )( )g f x ?

(A) 1 (B) 0 (C) -2 (D) -3 (E) -10

Solution

To find ( )( )g f x , work from the inside out. ( )( )( ) ( )2

1 110 16 10

g f xf x x x

= =+ − − +

. Because the only

way for a fraction to equal 0 is for the numerator to be 0, this fraction cannot equal 0.


Advanced Speed Drill VI (15Qs in 50 mins)

Drill VI: #1 Difficulty Level: 600-700 Topics: Number Properties

The Least Common Multiple of x and y is 300, and the Greatest Common Factor of x and y is 15. What is the

greatest possible value of xy

?

(A) 4 (B) 10 (C) 20 (D) 60 (E) 100


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Solution

Because this question has to do with factors, the first step is to break the numbers down into their prime factorizations.

• The GCF of x and y is 3 5⋅ . This means that in their prime factorizations, both x and y must have at a minimum 3 5⋅

• The LCM of x and y is 2 22 3 5⋅ ⋅ . This means that one or both of the two unknowns must have 22 , 3, and 25

Since we want to maximize the value of xy

, we will give x as many factors as possible. Since a number cannot be

greater than its own multiple, the biggest possible value of x is 2 22 3 5⋅ ⋅ . In contrast, since no number can be smaller than its own factor, the smallest possible value of y is 3 5⋅ . As a result, the biggest possible value of

2 22 3 520

3 5xy

⋅ ⋅= =

⋅




Drill VI: #2 Difficulty Level: 600-700 Topics: Coordinate Geometry; Inequalities & Absolute Values

In the xy-plane above, the slope of line l is 54

and lines l and k

intersect when 4x = . If the y-intercept of line k is ( )0, 3− ,

which of the following corresponds to the shaded region?

(A) 1 5

3 102 4

x y x− − < ≤ − (B) 4 5

3 85 4

x y x− − ≥ > + (C) 4 5

3 85 4

x y x− − < ≤ +

(D)

1 53 102 4

x y x− − ≤ < − (E)

5 110 or 3

4 2y x y x≥ − > − −

(8,0) l

k


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A look at the shaded region tells us that it is strictly greater than (above) a line that has a negative slope (going down from left to right). We can safely eliminate B, D, and E. D is incorrect because it states that y is greater than or equal to… However, a dashed line means that y is strictly greater than the line.

Next, note that both lines l and k have negative y-intercepts. In answer C, one of the lines has a positive intercept (+8), so it cannot be the correct answer. Eliminate C

Guess A

Solution B

To identify the shaded region, we need to find the equations of the two lines that border it. Since we have point

( )8,0 and slope 54

for line l, it may be easier to find out its equation first:


Input (8,0) and slope 54

5

0 84

b= +

Find b 10b = −

Equation of line l 5

104

y x= −

The shaded region is below line l, but it also includes the line because l is solid, not dashed. As a result, part of

the correct answer must be 5

104

y x≤ − (Eq. 1). Only answers A and E incorporate this equation, so B, C and D

must be incorrect.

To find the equation of line k, note that that line intersects line l when 4x = . By plugging this value of x into line

l, we can discover the y-coordinate of the point ( )54 10 5

4y y= − → = − . We now have two points on line

k. ( )4, 5− and the y-intercept we were given, ( )0, 3− . The slope of line k is ( ) ( )( ) ( )

2 1

2 1

3 5 10 4 2

y yx x

− − −−= = −

− −. We

can find line k the same way we found line l:


Input y-intercept of -3 and slope 12

− 1

32

y x= − −


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Since line k is dashed, it isn’t included in the shaded region. So 1

32

y x> − − (Eq. 2)

By combining (Eq 1) and (Eq 2), we get the final answer, 1 5

3 102 4

x y x− − < ≤ −


Take-Aways • The standard format for the equation of a line is y mx b= + where m is the slope and b is the y-intercept.

Drill VI: #3 Difficulty Level: 600-700 Topics: Coordinate Geometry

In the xy-plane, at which point does the graph of ( )( )y x m x n= + + intersect the y-axis?

(1) The graph intersects the x-axis at ( )5,0−

(2) 215

m n+ =

Solution

The graph will intercept the y-axis whenever x is 0. In the equation given, when x is 0, ( )( )0 0y m n mn= + + =

Our rephrase “What is mn?”

(1) Plug the given point into the equation to get . Since the equation is 0, either m or n

or both must be equal to 5. If they m and n are both equal to 5, then their product will be , but they could be different. We need more data to tell.


(2) I found statement (2) easier. Knowing the sum of m and n, there is no way to find their product.

Statement (2) is NOT SUFFFICIENT

MERGE STATEMENTS

By merging the statements, we know that one of the following two possibilities applies:

a. 5m = and 215

m n+ = , so 45

n = − . In this case, 4mn = −

b. 5n = and 215

m n+ = , so 45

m = − . In this case, 4mn = −

( )( )0 5 5m n= − + − +25


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Either way, we have a unique value for mn, so we can definitively answer our rephrase.




Drill VI: #4 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values

Is 0ab > ?

(1) a b a b+ = − (2) a b>

Solution


(1) Generally speaking, measures the distance between x and y on the number line. This statement

can be rewritten as . The distance from a to –b is the same as the distance from a to

b. This can only happen if a is exactly half-way between b and –b (in which case ), or if b and –b are really the same value (in which case ). Either way, , so we can answer the question.


(2) I found statement (2) easier. Knowing that gives me no information about the sign of ab. They could both be positive, or a could be positive and b negative.



Drill VI: #5 Difficulty Level: 600-700 Topics: Coordinate Geometry; Inequalities & Absolute Values

In which quadrant of the xy-plane is the point ( ),p q located?

x y−

( ) ( )a b a b− − = −

0a =0b b= − = 0ab =

a b>


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(1) 3

32

p q>

(2) 8 0p q− <

Solution

Quadrant placement is determined solely by the signs (+/-) of the coordinates.

Our rephrase “what are the signs of p and q?”

(1) Cross multiply and simplify to get 2p q> . This doesn’t allow us to determine the signs of p and q


(2) 8p q< . They could both be positive, or p could be negative and q positive. We cannot determine the signs of p and q


MERGE STATEMENTS

Combine the inequalities to get 2 8 2 8q p q q q< < → < . q has to be positive because when we divide both sides by q, the sign must not flip ( 2 8< ). p has to be positive because 2p q> means that p is bigger than a positive number.

By merging the statements we discover that both p and q are positive.




Drill VI: #6 Difficulty Level: 700+ Topics: Rates & Work A company has 150 workers. Working together, all the workers can complete a job in 3hrs. If the job starts at 9:37am and must be completed by 2:52pm, what is the maximum number of workers that can be absent?

(A) 63 (B) 64 (C) 65 (D) 66 (E) 67

Solution

Find the number of workers that must be present, and subtract it from the total to determine the most that can be

absent. The period from 9:37am to 2:52pm is 5hrs 15minutes, or 14

215

4= hours.


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Let W be the amount of work that each worker completes per hour. Each hour, 150 workers will complete 150Wwork. In three hours, 150 workers will complete 150 3W ⋅ . The question is, how many workers would it take to

complete the same amount of work in 214

hours? Let’s call this number of workers x. By setting the amounts of

work equal to each other, we can find x

In 214

hours, x workers can do what 150 do in 3 hrs 21

150 34

W x W⋅ = ⋅ ⋅

Isolate x, then solve 57

150 3 4 85

21 x x

⋅ ⋅= → =

If we had exactly 5785 workers, we would finish the job exactly at 2:52pm. Since we cannot have a fraction of a

worker, at least 86 workers must be present to complete the work. This means that a maximum of 150 86 64− =workers can be absent.


Drill VI: #7 Difficulty Level: 700+ Topics: Statistics

In a room of 25 people, the median age is 50 years and the average age is 42 years. Which of the following statements must be true?

I. At least one person is older than 50 years old II. The number of people over 50 years old is equal to the number of people under 50 years old III. At least one person is younger than 34 years old

(A) None of the above (B) I only (C) III only (D) I and II (E) I, II and III

Solution

Since there are 25 values and the median is 50, we know that the 13th highest value is equal to 50.

I. This doesn’t have to be true. The 13 highest values could all be equal to 50, and the median would still be 50.

II. This doesn’t have to be true. The 13 highest values could all be equal to 50, and the median would still be 50. In this case, there will be more people under 50yrs old than people over 50yrs old

III. There are a total of 25 people who are 42years old on average. So the sum of all ages is 42 25 1050⋅ = . Since the median is set at 50, we know that 13 people are at least 50years old, so the sum of their ages is at least 50 13 650⋅ = years. The remaining 12 people’s ages add up to a maximum of 1050 650 400− = years.


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These remaining 12 people must average 13

40033

12= years of age. Since this is the average of these 12

people, they can’t all be older than 1333 yrs old. At least one of them must be 1

333 yrs or younger. So statement III “At least one person is younger than 34 years old” must be true


Drill VI: #8 Difficulty Level: 700+ Topics: Combinatorics

How many four letter strings are possible if two of the letters must be the same, and the other two must each be unique within the string?

(A) 15,600 (B) 16,250 (C) 89,700 (D) 93,600 (E) 358,800

Solution

Each string has four letters. There are six possible ways to pick which two letters must be the same: 1st & 2nd, 1st & 3rd, 1st & 4th, 2nd & 3rd, 2nd & 4th, or 3rd & 4th letters. Each of these pairings has many possible permutations as outlined below

24 1st & 4th

26 25 1

No restriction

Any of the 25 letters not used by slot 1

Any of the 24 letters not used elsewhere

Must match slot 1

26 25 24 15,600⋅ ⋅ = possibilities

1 1st & 3rd

26 25 24

No restriction


Must match slot 1


26 25 24 15,600⋅ ⋅ = possibilities

25 1st & 2nd

26 1 24

No restriction Must match slot 1



26 25 24 15,600⋅ ⋅ = possibilities


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The total number of possibilities is 15,600 6 93,600⋅ =


Drill VI: #9 Difficulty Level: 600-700 Topics: Translations & Manipulations; Exponents & Roots

The Richter magnitude scale quantifies the amount of energy released by an earthquake. An earthquake of magnitude 1m + releases 900% more energy than an earthquake of magnitude m. The energy released by an earthquake of magnitude 4.7 is approximately how many times the energy released by an earthquake of magnitude 3.2?

(A) 150 (B) 100 (C) 55 (D) 32 (E) 15

30 Seconds Hack

Approximate. Each increase of +1 in magnitude is a tenfold increase in energy (900% increase is the same as multiplying by 10). If the energy of magnitude 3.2 is e, then the energy of a magnitude 4.2 will be 10e, and the magnitude of a 5.2 earthquake will be 100e. Since we are given a magnitude of 4.7, the energy will be between 10e and 100e, so we can safely eliminate answers A and B.

The energy increases exponentially (faster and faster), so the energy increase between magnitudes 4.2 and 4.7 will be less than the energy increase between magnitudes 4.7 and 5.2. An earthquake of magnitude 4.2 has 10e of

26 3rd & 4th

25 24 1




25 24 26 15,600⋅ ⋅ = possibilities

24 2nd & 4th

25 26 1


No restriction Any of the 24 letters not used elsewhere

Must match slot 2

25 26 24 15,600⋅ ⋅ = possibilities

1 2nd & 3rd

25 26 24




25 26 24 15,600⋅ ⋅ = possibilities


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energy, and an earthquake of magnitude 5.2 has 100e of energy, so magnitude 4.7 will be less than the average.

An earthquake of magnitude 4.7 will have less than 100 10

552+

= e. Eliminate C

Guess D or E

Solution

If an increase of +1 on the Richter scale corresponds to tenfold increase in energy (900% increase is the same as multiplying by 10), then an increase in +0.1 on the Richter scale must correspond to a multiplication of energy by 10 10 . This is because increasing by +1 is equivalent to increasing by +0.1 ten times just as multiplying by 10 is

the same as multiplying by 10 10 ten times. ( )1010 10 10= .

Increasing the magnitude from 3.2 to 4.7 is an increase of 1.5. This is equivalent to increasing by +0.1 fifteen

times, or multiplying by ( ) ( ) ( )515 10 5

10 10 10 1010 10 10 10 10 10 10 10 3.2 32= ⋅ = ⋅ = ≈ ⋅ ≈


Drill VI: #10 Difficulty Level: 700+ Topics: Translations & Manipulations

At a bookstore, the number of books sold, n, is related to the price of each book sold, p, according to the formula 140 10n p= − . George and Dick are two sales agents at this store. George’s pay is a $29 base pay plus 5% of

sales revenue in excess of $100. Dick’s pay is 10% of sales revenue. If George and Dick receive equal pay, which of the following could be the price of each book sold?

(A) $12 (B) $11 (C) $10 (D) $9 (E) $8

Solution

Let S be the sales revenue:

“George’s pay is a $29 base pay plus 5% of sales revenue in excess of $100” means ( )529 100

100S+ −

“Dick’s pay is 10% of sales revenue” means 10100

S

“George and Dick receive equal pay” ( )5 1029 100

100 100S S+ − =

Multiply both sides by 20 ( )580 100 2S S+ − =


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Solve for S 480S =

We’ve now discovered that the sales revenue is $480. Since revenue is price per book multiplied by the number of books sold, we can write 480np = . The question stem gives us another equation relating price to number of books sold: 140 10n p= − . By replacing n with its equivalent 480np = , can find the price:

Sales revenue equation 480np =

Plug in 140 10 p− for n ( )140 10 480p p− =

Distribute and set to 0 210 140 480 0p p− + − =

Divide by -10 2 14 48 0p p− + =

Factor and find p ( )( )6 8 0 6 or 8p p p− − = → =

The correct book price can be either $6 or $8. The question asked “which could be the price?” The answers have only one of the two possible prices: $8


Drill VI: #11 Difficulty Level: 700+ Topics: Sets & Groups

At a party, the number of people who own a car but not a home is double the number who own a home but not a car. The number of people who own neither a home nor a car is 2 more than half the number of people who own both a home and a car. If 32 people do not own a home, how many people are at the party?

(1) 40 people own both a home and a car (2) 50 people own a car

Solution – Venn Diagrams

.

Car Home

b 2x x

• “Car only is double Home only”.

• Let b be both. “Neither is 2 more than half of both”

22

b+


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Note that we’ve rephrased every section in terms of x, so if we’re given anyone section, we can solve everything.

(1) This statement gives us both: 60 4 40x− = . This is sufficient to find x and unlock the entire diagram.


(2) In our diagram, car owners add up to 2 60 4x x+ − . This statement tells us that this sum equals 50. This is sufficient to find x and unlock the entire diagram.



Drill VI: #12 Difficulty Level: 700+ Topics: FDPs & Ratios; Inequalities & Absolute Values

In a group of men and women, each person is German or Spanish, but not both. Is the ratio of German men to German women smaller than the ratio of German men and Spanish women to German women and Spanish men?

(1) The ratio of Spanish women to Spanish men is smaller than the ratio of German men and Spanish women to German women and Spanish men

(2) The ratio of Spanish women to Spanish men is greater than the ratio of women to men

Solution

Let Sm, Sw, Gm, and Gw be the number of Spanish men, Spanish women, German men and German women

respectively. Algebraically, the question is: “Is Gm Gm SwGw Gw Sm

+<

+?”

Cross multiply “Is ( ) ( )Gm Gw Sm Gw Gm Sw+ < + ?”

Distribute “Is Gm Gw Gm Sm Gm Gw Gw Sw⋅ + ⋅ < ⋅ + ⋅ ?”

Subtract GmGw from both sides to get our rephrase:

Our rephrase “Is Gm Sm Gw Sw⋅ < ⋅ ?”

Car Home

60– 4x 2x x

• “32 do not own a home” means that everything outside of Home adds up to

32: 2 2 32 60 42b

x b x+ + = → = −

• Replacing b with 60 – 4x allows us to express both and neither in terms of x

32– 2x


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(1) Sw Gm Sw

Gw Sw Sm Sw Gm Sm Sm Sw Gw Sw Gm SmSm Gw Sm

+< → ⋅ + ⋅ < ⋅ + ⋅ → ⋅ < ⋅

+

Note that the simplified equation is exactly what we need in order to answer our rephrase. This statement tells us that the product of women is smaller than the product of men. The rephrase asks whether the product of men is smaller than the product of women.


(2) Sw Gw Sw

Gm Sw Sm Sw Gw Sm Sm Sw Gm Sw Gw SmSm Gm Sm

+> → ⋅ + ⋅ > ⋅ + ⋅ → ⋅ > ⋅

+

From this information, we cannot determine whether the product of men is greater than the product of women





Drill VI: #13 Difficulty Level: 700+ Topics: FDPs & Ratios; Exponents & Roots

The dollar amount of interest, I, earned by an investment account in the first n years is 1 1100

nrI P= + −

where P is the initial deposit and r is the annual interest rate. Raj opened an investment account in 1990. If the only transaction was an initial deposit of $15,000, is the annual interest rate less than 5%?

(1) The total interest accrued by the end of 1997 was $5,000

(2) Rounded to the nearest integer, the total interest accrued increased by $742 from January 1994 to January 1995

Solution

Our rephrase “Is 5r < ?”

(1) Since the account was opened in 1990, we know that 7n = , 15,000P = and 5,000I = . We have the values of 3 of the 4 unknown variables. We could (but we shouldn’t unless we want to be at it all day) solve for the last unknown, r.


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(2) 5% of $15,000 is $750. The concept of compound interest is that each year, the interest earned is higher than the previous year, because it is the same percentage of a bigger amount.

If r were 5%, then the interest earned the first year would be $750, and in subsequent years, the interest earned would be greater than $750 each year. Since the interest earned between Year 4 and Year 5 is only $742, the interest earned in Year 1 has to be smaller than 5% of $15,000. 5r <



Drill VI: #14 Difficulty Level: 600-700 Topics: Translations & Manipulations

A bowl contains 34 marbles colored red, black or green. The number of red marbles is fewer than a third of the number of black marbles. There are more than twice as many green marbles as red marbles. What is the least possible sum of black marbles and green marbles?

(A) 25 (B) 26 (C) 27 (D) 28 (E) 29

Solution

Let r, g, and b be the number of red, green and black marbles respectively

“A bowl contains 34 marbles” 34 34r g b b g r+ + = → + = − (Eq 1)

“Red is fewer than a third of black” 33b

r r b< → < (Eq 2)

“there are more than twice as many green as red” 2r g< (Eq 3)

We’re asked about b + g. By adding (Eq 2) and (Eq 3), we get 5b g r+ > . Since b+ g is 34 – r (from Eq.1), it

follows that 34 5r r− > . Simplify to get 235r < . Since we cannot have fractions of a marble, we should change

this to 5r ≤ .

There are 34 marbles in total. If 5r ≤ , then the remaining marbles must be at least 29. 29b g+ ≥


Take-Aways • The only way to safely combine inequalities is to make sure that the signs are facing the same direction, and

to add the inequalities as we did with (Eq 2) and (Eq 3). Never subtract. Only add.


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Drill VI: #15 Difficulty Level: 700+ Topics: FDPs & Ratios; Statistics

Last year, the 5 managers of a company received a total of $208,000 in salaries. If the highest salary was 20% greater than the lowest, what is the maximum possible value of the highest salary?

(A) $42,000 (B) $44,000 (C) $46,000 (D) $48,000 (E) $50,000

Solution

For every dollar earned above the average salary, there must be a dollar earned below the average. To maximize the highest salary, we should make it the only salary above average. In addition, we should make all the other salaries equal to each other. To understand why this is, consider a simpler example:

Suppose 3 people had a total of $30 (average is $10), and that the highest amount must be $6 more than the lowest. If one person had exactly the average of $10, then the lowest amount would be $7 and the highest $13. On the other hand, if we kept the two low salaries equal to each other, then they would each be $8 and the high would be $14. Keeping the low salaries equal maximizes the high salary.

In our case, to maximize the high salary, we need to make the low salaries equal to each other. Since the high is 20% more than the low, we can call the low salaries s and the high salary 1.2s. Because the 5 salaries must all

add up to $208,000, 208,000

4 1.2 208,000 40,0005.2

s s s+ = → = = . The low salaries are $40,000 and the

highest salary is 1.2 1.2 40,000 48,000s = ⋅ =

Note: To quickly do the division 208,000

5.2, consider changing 5.2 into

52 2610 5

=


Advanced Speed Drill VII (15Qs in 50 mins)

Drill VII: #1 Difficulty Level: 600-700 Topics: Statistics

The test scores of 1050 students have a normal distribution. If the average test score is 73 and the standard deviation is 10.5, approximately how many students have a test score less than or equal to 62.5?

(A) 147 (B) 168 (C) 205 (D) 288 (E) 357

Solution


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A normal, or bell-curve distribution means that scores are symmetrically spread around the mean. When dealing with normal distributions and standard deviation, remember the ratio 34:14:2. In a normal distribution, 34% of points are between the mean and +1 standard deviation, 14% of data points are between +1 and +2 standard deviations, and 2% of data points are above +2 standard deviation. These proportions also apply in the opposite direction as shown below.

Since the mean of our set is 73, and the standard deviation is 10.5, a score of 62.5 is equal to 1 73 10.5mean sd− = − . A total of 16% of scores are less than or

equal to 62.5. 16

1050 168100

⋅ =


Take-Aways • In a normal distribution, 34% of the data is between the average and +/-1 standard deviation, 14% of data is

between +/-1 and +/-2 standard deviation, and 2% of data is beyond +/-2 standard deviation. Remember the ratio 34:14:2.

Drill VII: #2 Difficulty Level: 700+ Topics: Geometry

In the figure, the line is tangent to both circles. If the area of circle Q is 200 and the area of PQR is 40% of the area of OPQR, what is the area of circle O?

O

P

Q

R


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(A) 800

9 (B) 200 (C) 300 (D) 400 (E) 450

Solution

A tangent line is perpendicular to the radius at the point where the line and the circle intersect.

Simplifying the equation yields 2

3

2

3QR

OPQR OP= → = . We’re asked for the area of O, which is 2OPπ ⋅ .

The area of Q is 2

2 2 2003

QR OPπ π ⋅ = ⋅ =

.

Remove parentheses 24 2009

OPπ ⋅ =

Solve for 2OPπ ⋅ 2 450OPπ ⋅ =


Drill VII: #3 Difficulty Level: 600-700 Topics: Statistics

The greatest of five numbers is 12. What is the median of the five?

(1) The standard deviation of the five numbers is not positive (2) The sum of the five numbers is 60

Solution


O

P

Q

R

Since PQR is 40% of OPQR, OPR is 60% of the trapezoid, and the ratio of areas of PQR to OPR is 4:6. PR is the base of both triangles, and the radii are the

heights, so algebraically

142

1 62

PR QR

PR OP

⋅=

⋅


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(1) Standard deviation cannot be negative. Since this standard deviation is not positive, it must be 0, which means that all the numbers are the same. The median must be 12


(2) The average must be 60

125= . Since the greatest number and the average are equal, all numbers must

equal the average. The median must be 12.



Drill VII: #4 Difficulty Level: 500-600 Topics: Number Properties

If 1 0z− < < , which of the following expressions has the second greatest value?

(A) z (B) 3z− (C) 2z− (D) 2z (E) 2z

Solution

Since z is negative, 2 2 2, , and z z z− are all positive, so the 2nd greatest value will be one of these. Plug in an easy value to test these expressions.

Our plug in 12

z = −

With this plugin, 2 1

4z = ,

22

14z

z− = = , and 2 1 1

4 2z = = . Of the three values, the second greatest is

2z


Drill VII: #5 Difficulty Level: 600-700 Topics: Functions & Sequences; Statistics

The sequence a1, a2, a3,… ,an is such that 1 3

3n n

na a

a − −+= for all 4n ≥ . 6 14a = and 3 27a = . If the median

of the second and third terms is 27, what is the value of a7?

(A) (B) 14 (C) 15 (D) 18 (E) 27 2310


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30 Seconds Hack

Because to find each term we must divide the sum of two previous terms by 3, we can expect that the terms of the sequence generally get smaller and smaller. Indeed, we are told that a3 is 27 and a6 is 14, so we can guess that a7 will be smaller than 14. Eliminate B through E

Guess A

Solution

According to the sequence formula, 6 4 47

143 3

a a aa + += = (Eq. 1). To get a7, we need to find a4.

Using a6 and a3, we can find a5: 5 3

6 5 6 3 5 3 3 14 27 153

a aa a a a a+= → = − = ⋅ − → =

Since the median of a2 and a3 is 27, and because a3 is 27, a2 must also be 27. Now we can find a4 using the

sequence formula for a5. 4 2

5 4 5 2 4 3 3 15 27 183

a aa a a a a+= → = − = ⋅ − → =

Let’s put the value of a4 into (Eq. 1) to find a7. 4 1

7 314 13 18 10

3 3aa + +

= = =


Drill VII: #6 Difficulty Level: 600-700 Topics: Translations & Manipulations

Each day in September, John spent the same fraction of his daily pay to buy food. Every other day, he also bought gasoline with some of his pay. The amount spent on gasoline in September is 25% less than the amount

spent on food that month. If 1320

of John’s September pay was not spent on food or gasoline, what fraction of his

pay was spent on food?

(A) 1

10 (B)

15

(C) 14

(D) 720

(E) 12

Solution

Let P and F be John’s daily pay and his daily food expense respectively. Let G be the amount John spends on gasoline every other day. In September (a 30 day period), John’s total pay was 30P, he spent a total of 30F on food, and 15G on gasoline.


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“The amount spent on gasoline in September is 25% less than the amount spent on food that month” means 75 3

15 30 100 2

G F G F= ⋅ → =

“1320

of John’s September pay was not spent on food or gasoline” means that food and gasoline for the month

account for 720

of total pay. Algebraically, ( )715 30 30

20G F P+ = . Replace G with

32

F , then simplify:

Replace G with 32

F ( )3 715 30 30

2 20F F P+ =

Remove parentheses 45 21

302 2

F F P+ =

Multiply the equation by 2 45 60 21F F P+ =

Solve for F 21 1

105 5

F P F P= → =

One fifth of pay is spent on food


Drill VII: #7 Difficulty Level: 700+ Topics: Rates & Work A toy factory has 48 experienced workers and 72 inexperienced workers. Each experienced worker builds toys 25% faster than each inexperienced worker does. If all workers work the same number of hours, approximately what percentage of all toys do inexperienced workers build?

(A) 35% (B) 40% (C) 45% (D) 50% (E) 55%

Solution

Let r be the rate of work of each inexperienced worker and t be the amount of time that each worker works. Since work rate time= ⋅ , the amount of work done by each inexperienced worker is rt, and the amount of work done by all 72 inexperienced workers is 72rt.

Since experienced workers have a 25% faster rate, the rate of each experienced worker is 1.25r, and the amount of work done by all 48 experienced workers is ( )48 1.25 60r t rt=

The percentage of all work done by inexperienced workers is72 6

54.5%72 60 11

rtrt rt

= =+



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Drill VII: #8 Difficulty Level: 700+ Topics: Combinatorics

Five people wrote down their names on five pieces of paper, and put the pieces in a bag. If each person then randomly took out one piece of paper, what is the probability that exactly two people took the pieces with their names on them?

(A) 1

120 (B)

160

(C) 130

(D) 1

12 (E)

16

Solution

The probability we’re looking for is # of ways 2 ppl pick their names

All possible outcomes. First, let’s find the number of possible

outcomes.

Since there are five people, the first person can take any of 5 five pieces of paper, the second can take any of 4 pieces, the third can take any of 3 pieces, the fourth person can take any of 2 pieces, and the last person must take the only remaining piece in the bag. So the total number of possible outcomes is 5 4 3 2 1 120⋅ ⋅ ⋅ ⋅ =

Now we need to find the number of ways for exactly 2 people to pick their names out of the bag. Let A-B-C-D-E represent the “perfect” outcome in which everyone picks his name. A-B-E-C-D is one outcome in which only the first two people pick out their names. How many such outcomes are there? If only the first two people pick their names, the only outcomes possible are A-B-E-C-D and A-B-D-E-C. On the other hand, if only the first and the last people pick their names, the only outcomes possible are A-C-D-B-E and A-D-B-C-E

In short, there are two outcomes possible for every combination of 2 out of 5 people. Therefore, the total number

of ways for exactly two people to pick their names is 5!

2 203!2!⋅ = ways. So the probability we seek,

# of ways 2 ppl pick their names 20 1All possible outcomes 120 6

= =


Drill VII: #9 Difficulty Level: 500-600 Topics: Translations & Manipulations

At a store, Mark bought three items. What fraction of his total expense was used to buy the first item?

(1) The first item cost $20 more than the second item did. (2) Mark spent twice as much on the second item as he did on the third


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Solution

Let f, s, and t be the costs of the first, second and third items respectively

Our rephrase “What is f

f s t+ +?“

(1) 20s f= − . This gives us no data on the 3rd item, so we cannot deduce what fraction of that total is taken up by f


(2) 2s t= . This gives us no data on the 1st item.


MERGE STATEMENTS

By merging the equations from the two statements, we can deduce that 1

20 2 102

f t t f− = → = − . So,

( ) 120 10

2

f ff s t f f f

=+ + + − + −

. This cannot be simplified into a known ratio



Drill VII: #10 Difficulty Level: 700+ Topics: Inequalities & Absolute Values; Number Properties

If 0x y+ < , is 0y < ?

(1) ( ) ( )3 22 0 2 1y x y x− < < − <

(2) 1

1 4 02

y x− < − + <

Solution

The question is already simplified. We’ll need a way to remove x from the equation so we can compare y to known numbers.

(1) Since ( )22y x− is smaller than 1, 2y x− is a fraction. However, since ( )32 0y x− < , 2y x− must be

negative. So the statement can be simplified to 1 2 0y x− < − < . To combine this inequality with the one given in the question stem so that we can eliminate x, focus on the 2nd part: 2 0y x− < .


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By adding 2 0y x− < to 0x y+ < , we get ( ) ( )2 0 3 0 0y x x y y y− + + < → < → <


(2) To combine this inequality with the one given in the question stem so that we can eliminate x, focus on

the 2nd part:1

4 02

y x− + <

By adding 1

4 02

y x− + < to 0x y+ < , we get ( )1 1 14 0 5 0

2 2 10y x x y y y− + + + < → + < → < −



Drill VII: #11 Difficulty Level: 700+ Topics: Coordinate Geometry

In the figure above, the circle is centered at O, the origin. What is the value of c?

(1) 3 3b = (2) a b=

Solution


(1) The value of b gives us the full coordinates of Q, so we can find the distance between Q and the origin. OQ is also the distance between O and R (they’re both radii). The distance between O and R is

( ) ( )2 2 2 20 0 r a c r a c= − + − → = + . Although we can find the radius, without the value of a, we

cannot find c.


(2)

Q (-3, b) R (a, c)

O


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Drill VII: #12 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values; Number Properties

N is a set of consecutive integers. Is the sum of all members of N is equal to the sum of the absolute values of each member of N?

(1) The quotient of the product of all members of N and the sum of all members of N is negative (2) At least one member of N is different from its absolute value

Solution

Let n1 through nn be the members of N. We are asked: is 1 2 3 1 2 3... ...n nn n n n n n n n+ + + = + + + ? This can

only be true if all the members of N are either positive or zero.

Our rephrase “Are all members of N greater than or equal to 0?”

(1) The product and the sum must have opposite signs. This statement also means that 0 is not a member of N because if it were, the product would equal 0, and so would the quotient mentioned in this statement. Therefore, either all members are positive or all members are negative (remember that N consists of consecutive integers)

If all members of N were positive, the product and the sum would have the same sign. We can conclude that all members of N must be negative. This data answers our rephrase.


(2) I found statement (2) easier. Only negative numbers are different from their absolute values, so this statement tells us that at least one member of N is negative. This answers our rephrase.


Q (-3, b)

R (b, c)

O

This statement allows us to replace a with b. The hypotenuses (radii) of both triangles are equal, so

2 2 2 23 b b c+ = + . We can see from the diagram that c is positive, therefore, 3c =


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Drill VII: #13 Difficulty Level: 600-700 Topics: Rates & Work

It took Sarah t hours to travel from home to school at a rate of r miles per hour. Her return trip took an hour longer because her rate was 3 miles per hour slower. In terms of t, what was the total distance traveled?

(A) (B) (C) (D) (E)

Solution

rate time distance⋅ =

Rate Time Distance Home to School r t rt School to Home r-3 t +1 (r-3)(t +1)

Because the distance is the same each way, the total distance traveled is 2rt. To express this in terms of t, we need to put r itself in terms of t. Since the two trips have equal distance, we can write

( )( )3 1 3 3rt r t rt rt r t= − + → = + − − . Isolate r to get 3 3r t= + .

Now we can put the total distance in terms of t. Total distance is ( ) 22 2 3 3 6 6rt t t t t= + = +


Drill VII: #14 Difficulty Level: 700+ Topics: Coordinate Geometry

Points ( ),P a b and ( ),Q b c are in the xy-plane. Is the distance between P and Q greater than the distance

between P and the origin?

(1) 0a c+ = (2) P and Q are distinct points

Solution

23 32

t t+ 23 3t t+ 26 6t t+ 26 12t t+ 212 12t t+


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Use the distance formula to simplify the question.

Distance formula “Is ( ) ( ) ( ) ( )2 2 2 20 0a b b c a b− + − > − + − ?“

Simplifying further “Is ( ) ( )2 2 2 2a b b c a b− + − > + ?”

FOIL the left side “Is 2 2 2 2 2 22 2a ab b b bc c a b− + + − + > + ?”

Combine similar terms “Is 2 2 2 2 0b c ab bc+ − − > ?”

Our rephrase “Is 2 2 2 2b c ab bc+ > + ?”

(1) Factoring the right side of our rephrase gives us “Is ( )2 2 2b c b a c+ > + ?” Since 0a c+ = we can

simplify the rephrase to “Is 2 2 0b c+ > ?” The answer depends on whether b and c both equal 0.


(2) On its own, this doesn’t help us answer our rephrase at all.


MERGE STATEMENTS

Statement (1) allowed us to simplify our rephrase to “Is 2 2 0b c+ > ?” From statement (1), we learn that a c= −, so our points are ( ),P c b− and ( ),Q b c . If b and c were both equal to 0, the two points would be the same.

Thanks to statement (2), we know that the points are distinct, so either 0b ≠ or 0c ≠ . As a result, we have

enough data to answer the rephrase. It must be true that 2 2 0b c+ >




Drill VII: #15 Difficulty Level: 600-700 Topics: Combinatorics; Inequalities & Absolute Values

In a room, some students take Math and some take Chemistry. The number of students who take both subjects equals the number of students who take neither subject. If a student is to be chosen at random, is the probability that he take only one of these two subjects greater than the probability that he take Chemistry?


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(1) The probability that the student selected take neither or take both subjects is smaller than the probability that he take Math.

(2) The number of students who take Math is equal to the number of students who take Chemistry.

Solution

This is an overlapping sets question. Total M C n b= + + − where M is the number of math students, C is the number of chemistry students, n is the number of students who take neither math nor chemistry, and b is the number of students who take both math and chemistry.

“The number of students who take both subjects equals the number of students who take neither subject” means that n = b. So Total M C b b Total M C= + + − → = +

A higher probability means a greater number of students. So the question can be rephrased as “Is math only + chem only chem> ?” Algebraically: “Is ( ) ( )M b C b C− + − > ?” Simplify to get our rephrase:

Our rephrase “Is 2M b> ?”

(1) n b M+ < . Since we know that n = b, this statement tells us that 2b M< and thus answers our rephrase


(2) M C= . We cannot use this data to compare M to 2b, so this statement does not help us answer our rephrase.





Advanced Speed Drill VIII (15Qs in 50 mins)

Drill VIII: #1 Difficulty Level: 700+ Topics: Number Properties

x and y are positive odd integers. What is the remainder when the product xy is divided by 18?

(1) When x is divided by 9, the remainder is 3 (2) y – 1 is a multiple of 6.


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Solution


(1) This statement tells us that x is a number such as 21 or 39 (x cannot be 12 since we know that x and y are odd). x is a multiple of 9 + 3. Algebraically, we can write 9 3x a= + where a is an integer. This doesn’t tell us anything about y. If x is 21 and y is 1, the remainder when xy is divided by 18 will be 3. On the other hand If x is 21 and y is 3, the remainder when xy is divided by 18 will be 9.


(2) Algebraically, 1 6 6 1y b y b− = → = + where b is an integer. This doesn’t tell us anything about x. If y is 7 and x is 3, the remainder when xy is divided by 18 will be 3. On the other hand, if y is 7 and x is 5, the remainder when xy is divided by 18 will be 17


MERGE STATEMENTS

Statement (1) told us that 9 3x a= + where a is an integer. Since x is odd, a must be even and 9a must be a multiple of 18. This means that 18 3x c= + where c is an integer.

Statement (2) told us that 6 1y b= + where b is an integer. The product ( )( )18 3 6 1xy c b= + +

FOIL 18 6 18 1 3 6 3xy bc c b= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ +

Note that all the terms are multiples of 18, except for 3. Since the sum of two multiples of 18 must also be a multiple of 18, we know that 18 3xy d= + where d is an integer. In other words, the remainder when xy is divided by 18 must be 3



Drill VIII: #2 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values; Number Properties

On the number line, the distance between point P and point Q is 8 and the distance between point Q and point S is 4. If S is to the right of P and point R is closer to Q than to P, which of the following must be true?

I. Point P is the smallest of the four points II. Point S is not the greatest of the four points III. Point Q is the greatest or second greatest amount the four points

(A) None (B) I only (C) I and II (D) I and III (E) II and III

Solution


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Build a number line and input the given data one piece at a time. Each step takes into account the previous steps, so that the last line below shows all possible configurations of points P, Q, R, and S.

I. Point P is the smallest points in both possible configurations. I must be true. II. In configuration 2, if R sat between Q and S, then S would be the greatest of the four points. II could be false. III. In configuration 2, if R were greater than Q, then Q would not be one of the two greatest points. III could be

false.


Drill VIII: #3 Difficulty Level: 700+ Topics: FDPs & Ratios; Rates & Work

At a factory, each employee’s daily pay, S, is defined by 754ny

S = + where n is the number of widgets built by

the employee that day and y is the number of years of employment. Abe, Bob and Cindy have 3, 6, and 8 years of employment respectively. Each day, Abe works twice as long as Bob and twice as long as Cindy, but Bob builds widgets 50% faster than Abe and 40% slower than Cindy. If Abe, Bob and Cindy’s combined daily pay is $675, how many widgets do Abe and Bob build together each day?

(A) 150 (B) 160 (C) 170 (D) 180 (E) 190

Solution

The number of widgets built is a function of rate and time: work rate time= ⋅ . Therefore, we need to find each worker’s rate and time.

1) Distance between P and Q is 8

P Q Q P or

2) Distance between Q and S is 4, and S is to the right of P P Q S

or P Q S

3) R is closer to Q than to P

P Q S

R

or P Q S

R

Configuration 1 Configuration 2


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“Each day, Abe works twice as long as Bob and twice as long as Cindy” means that if Bob’s time is t, Abe’s time will be 2t and Cindy’s time will be t.

If Cindy’s rate is 5r, Bob will be 3r (40% slower than Cindy’s) and Abe will be 2r (Bob is 50% faster than Abe). With this data, we can find n, the number of widgets that each build in a day:

What we are looking for, “how many widgets do Abe and Bob build each day”, is 4 3 7rt rt rt+ = .

Abe, Bob and Cindy have 3yrs, 6yrs and 8yrs of employment respectively, so

according to the pay formula, their total pay is: 4 3 3 6 5 8

75 75 75 $6754 4 4

rt rt rt⋅ ⋅ ⋅+ + + + + =

Subtract 75 three times from each side 12 18 40

675 2254 4 4rt rt rt+ + = −

Simplify by 2 6 9 20

4502 2 2rt rt rt+ + =

Multiply both sides by 2 35 900 7 180rt rt= → =

Stop. We’ve found 7rt, the number of widgets that Abe and Bob build each day.


Drill VIII: #4 Difficulty Level: 600-700 Topics: Number Properties

Does 0n = ?

(1) 22m mn= (2) m is half of n

Solution


(1) Solve this as you would any quadratic. Set the equation equal to 0, then factor. 22 0m mn− = , so ( )2 0m m n− = . Our solutions: 0 or 2m m n= = . We cannot tell whether n is 0.


(2) 22

nm m n= → = . If m is 0, then n will be 0. Otherwise, they will both be nonzero.


Abe Bob Cindy Rate 2r 3r 5r Time 2t t t Widgets 4rt 3rt 5rt


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MERGE STATEMENTS

Together, the statements tell us that 2m n= . m and n could both be zero, or they could be different from zero. For instance, 2m = and 4n = would also work. We can’t tell whether n equals 0.



Drill VIII: #5 Difficulty Level: 700+ Topics: FDPs & Ratios; Geometry

If the diagonal of cube A is 56

of the diagonal of cube B, the volume of B is approximately what percent greater

than the volume of cube A?

(A) 20% (B) 64% (C) 61% (D) 73% (E) 80%


In geometry, the ratio of 2-dimensional measures between figures is the square of the ratios of one-dimensional measures. For example, consider two squares with sides 2 and 3. The sides of the two squares have a ratio of 2:3,

but their areas will have a ratio of ( )22 : 3 4 : 9= . This is because length is one-dimensional and area is 2-

dimensional.

Similarly, the ratio of 3-dimensional measures between figures is the cube of the ratios of one-dimensional measures. For example, consider two cubes with sides 2 and 3. Again, the sides’ ratio is 2:3, but the volume ratio

is ( )32 : 3 8 : 27= . This is because length is one-dimensional and volume is 3-dimensional. This concept works

for triangles, circles, trapezoids, rectangles, boxes, etc…

In our case, the ratio of diagonal A to diagonal B is 56

, so the ratio of volume B to volume A (notice the order

was switched from A:B to B:A) will be ( )3

36 1.2 1.7285

= =

. Volume B is 1.728 of volume A, or 72.8%

greater than volume A.


Solution B

The diagonal of a box, d, is defined by 2 2 2d length width height= + + . Since we are dealing with two cubes,

we can simply write 23d s= where s is the side of the cube.


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In our case, let a be the side of cube A and b be the side of cube B. The volumes of our cubes will be 3a and 3b .

“The diagonal of cube A is 56

of the diagonal of cube B” means

( )3

32 23

5 5 63 3 3 3 1.2 1.7286 6 5

b ba b a ba a

→= = → = → = = . Volume B is 1.728 of volume

A, or 72.8% greater than volume A.


Take-Aways • To find the diagonal of a box, use the following formula, an extension of the Pythagorean Theorem:

2 2 2 2d l w h= + + .

Drill VIII: #6 Difficulty Level: 600-700 Topics: Exponents & Roots

22

4

0.07 10 0.5 100.00014 10

x

y

+

−

×= ×

× What is y – x?

(A) 7 (B) 3 (C) 0 (D) -3 (E) -7

Solution

On the left side, change 0.07 to 70 and adjust the power of 10 in the numerator. On the right side, change 0.00014

to 14 and adjust the power of 10 in the denominator: 2 3

24 5

70 10 0.5 1014 10

x

y

+ −

− −

⋅= ⋅

⋅.

Simplify the left side ( ) 22 3 4 55 10 0.5 10x y+ − − − −⋅ = ⋅

Simplify exponents and change the right side from 0.5 to 5 185 10 5 10x y− +⋅ = ⋅

All other things are equal, equate exponents 8 1 7 7x y x y y x− + = → − = − → − =


Drill VIII: #7 Difficulty Level: 700+ Topics: Geometry

A

B

C D E


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In the figure shown, ABC is an isosceles triangle. The area of BDE is what fraction of the area of ABC ?

(1) BE BD=

(2) 23

DCDE

=

The question is already simplified. BDE has the same height as ABC , so to find the fraction of area covered,

we need data about the ratio of the bases of the triangles – Remember that 12

A bh=

(1) This statement gives us no specific data on the ratios of the bases of BDE to ABC , so we cannot yet answer the question asked.


(2) Without knowing the length of AE, or the ratio of AE to the rest of side AC, we cannot figure out how

much of AC is taken up by ED. Consequently, we cannot find the ratio of the bases: EDAC


MERGE STATEMENTS

At first glance, it might seem that together the statements give us all we need.

A

B

C D E

The figure is drawn to make it seem like AB and BC are equal. If AB = BC, and BE BD= , then the angles along the base would be mirrored and AE = DC. If this

were the case, then 3 3

2 3 2 7DEAC

= =+ +

and

with a ratio of the bases, we could find the ratios of areas (because the heights are the same, base ratio equals area ratio) x° x° y° y° z° z°

2n 2n 3n

h


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Drill VIII: #8 Difficulty Level: 600-700 Topics: Geometry

Patrick must store boxes of dimensions 3 by 3 by 5 inches into a rectangular container measuring 9 by 18 by 21 inches. What is the maximum number of boxes that can fit in this container?

(A) 42 (B) 63 (C) 72 (D) 90 (E) 126

Solution

There are three possible ways to place the boxes inside the container. We can orient the container so that its height is 9, 18, or 21 inches. For each possible orientation, let’s find out how many boxes will fit in the containers, assuming that the length (3), width (3) and height (5) of the boxes are lined up with the x, y, and z directions of the container as shown below.

18

9

21

This configuration will fit 18

63= boxes in the x direction,

217

3=

boxes in the y direction, and 1 box in the z direction. The total number of boxes that will fit this configuration is 6 7 1 42⋅ ⋅ = boxes

x

y

z

A

B

C D E

Although we know that ABC is isosceles, we don’t know which two sides are equal to each other. If AB = AC, then even given the data from both statements, we could not determine the ratio

of the bases, EDAC

because we would have no

data about AE


h

2n 3n ?


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Drill VIII: #9 Difficulty Level: 700+ Topics: Exponents & Roots; FDPs & Ratios; Functions & Sequences

The age of a fossil is defined by 2 kyc s−= ⋅ where c is the current concentration of compound X in the fossil, s is

the starting concentration of compound X, k is a constant, and y is 410− times the age of the fossil in years. All newborn sharks have a 64% concentration of compound X and a 60,000 year old shark fossil has a compound X

concentration of 32%. Approximately how old is a shark fossil with a compound X concentration of 2152 % ?

(A) 153,000 years (B) 294,000 years (C) 305,000 years (D) 326,000 years (E) 358,000 years

Solution

“A 60,000 year old shark fossil has a compound X concentration of 32%” means that 632 2 64k−= ⋅ . In that

equation, y is 6 because we are told that y is 4 110

10,000ageage− = ⋅ . Since the age is 60,000, y will be 6. Let’s

solve for k

Divide by 64 and find k 61 1

2 6 1 2 6

k k k−= → − = − → =

Let’s use the newfound value of k to find the age of the fossil when compound X concentration is 2152 %

18

21

9



93

3=

boxes in the y direction, and 4 boxes in the z direction. The total number of boxes that will fit this configuration is 6 3 4 72⋅ ⋅ =boxes. This is the maximum number possible.

x

y

z

21

18

9



93

3=

boxes in the y direction, and 3 boxes in the z direction. The total number of boxes that will fit this configuration is 7 3 3 63⋅ ⋅ = boxes

x

y

z


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Setup the equation with 64 as the starting concentration 1

2 6152 2 64

y−= ⋅

Change the mixed number and get rid of the negative exponent 6

32 1 6415

2y= ⋅

Divide both sides by 64 6

61 1 30 2

302

y

y

= → =

Approximate. 52 32= so 6y

is a little less than 5, and y is a little less than 30. Since 10,000

agey = , the fossil’s

age must be a bit less than 300,000


Drill VIII: #10 Difficulty Level: 600-700 Topics: FDPs & Ratios

One in five of the 660 students at City High School is a boy. If x boys were to enroll and no other changes occur, what value of x would increase the percent of boys in the school to 45%?

(A) 132 (B) 165 (C) 226 (D) 297 (E) 300

Solution

There are currently 20% 660 132⋅ = boys. If x boys were to enroll, the percentage of boys would be 45%, and the

new total number of students would be 660 + x. Algebraically, 132 45660 100

xx

+=

+

Simplify, then cross multiply ( ) ( )132 9 132 20 660 9

660 20x

x xx

+= → + ⋅ = + ⋅

+

Distribute and solve for x 3300

2640 20 5940 9 30011

x x x+ = + → = =


Drill VIII: #11 Difficulty Level: 600-700 Topics: Translations & Manipulations


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A bookseller sells books at p dollars each. How much revenue, in dollars, did the bookseller expect to collect last week?

(1) 40% more books than expected were sold last week (2) Last week, revenue from book sales was $200 above expectations

Solution

Let S and Sx be the actual number of sales and the expected number of book sales respectively. The expected revenue can be expressed as xS p⋅

Our rephrase “What is xS p⋅ ?”

(1) 1.4 xS S= . This tells us nothing about book prices or any specific data in dollars


(2) 200xS p S p⋅ − ⋅ = . From this equation alone, it is not possible to solve for the value of xS p⋅ .


MERGE STATEMENTS

By plugging 1.4Sx for S in statement (2), we get 1.4 200 0.4 200x x xS p S p S p⋅ − ⋅ = → = . We could solve

for xS p⋅ by dividing both sides by 0.4





Drill VIII: #12 Difficulty Level: 700+ Topics: Inequalities & Absolute Values; Number Properties

A, B, and C are the digits of integer ABC. Is 399ABC > ?

(1) 2B A− < (2) The hundreds digit of 110% of ABC is different from the hundreds digit of ABC

Solution


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First thing to notice is that A, B, and C are digits, so ABC is not a product, but a three digit integer. On the GMAT, capital letters are sometimes used to identify digits. 399ABC > is only true if A is 4 or more

Our rephrase “Is 3A > ?”

(1) B and A could both be 2, or they could both be 7. We don’t know enough about A to answer the question.


(2) 110% of ABC is found by adding AB.C to ABC

Still we cannot tell whether A is greater than 3. The integer could be 546 or 299


MERGE STATEMENTS

Statement (1) told us that 2 B A> − while statement (2) told us that 8A B+ > . By adding these inequalities, we get 2 8 2 6 3A B B A A A+ + > − + → > → >





• The only way to safely combine inequalities is to make sure that the signs are facing the same direction, and to add the inequalities. Never subtract. Only add.

Drill VIII: #13 Difficulty Level: 700+ Topics: Geometry

A B C.0

A B.C +

Since the hundreds digit of this sum is not A, it means that a “1” is carried over from the sum of the tens. In other words, the sum of the tens, the sum of A and B, must be at least 9 ( A B+ could be 9 if a “1” was also carried over from the sum of the units digits. 546 + 54.6 is an example). This statement tells us that 8A B+ >


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In the figure shown, each of the smaller squares is inscribed in a larger square such that each diagonal is parallel to the side of another square. The shaded region is what fraction of the area of the largest square?

(A) 1

12 (B)

116

(C) 120

(D) 128

(E) 132

30 Seconds Hack

Approximate. From comparing the shaded area to the entire figure, you may be able to estimate that 20 shaded areas would probably be too many to fit inside the big square. Certainly, 28 shaded areas would not fit. If you can see that, eliminate C, D, and E

Guess A or B

Solution

Note that each of the triangles formed has a 90° angle, and two equal sides: each of the triangles is a 45-45-90°

triangle. The proportion of the side lengths of these triangles is : : 2s s s . Because the side of the innermost

square is the hypotenuse of the shaded region, let’s call this side 2s .

Using similar logic, we can now find the area of the largest square given that the side of the middle square is 2s

2s

s

The small square’s area is ( )2 22 2s s= . The

medium square’s side is 2s, and its area is ( )2 22 4s s=

. The shaded area has base s and height s, so the

shaded area is 2

2s


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Find the fraction of the largest square that is occupied by the shaded area:

2

2

shaded area 12total area 8 16

s

s= =


Drill VIII: #14 Difficulty Level: 700+ Topics: Functions & Sequences

For numbers x such that 1x ≠ − , if f(x) is defined by ( ) 11

xf x

x−

=+

, then ( )( ) ( )( ) 11f x f x−− ⋅ =

(A) (B) (C) (D) 1 (E) -1

Solution A

( )( ) ( )( )( )

11

111 1 1

1 11

1

f xf x f xxx f x

x x

−−−

⋅ = ⋅ = ⋅−

++

Combine fractions

11

1 1

xxx

x xx

−+

= ⋅+ −

Dividing by a fraction is multiplying by its reciprocal 1 1 1

11 1 1

x x x xx x x x− + −

= ⋅ ⋅ = = −+ − −


2

2

11

xx−

+

( )( )

2

2

1

1

x

x

+

−

( )( )

2

2

1

1

x

x

−−

+

We can deduce from the previous step that each square has double the area of the square inside it. So the large square will have an area of 28s .

Alternatively, find the area: the side of the large

square is 2

2s

, so its area is 2 2

24 168

22s s

s= =

.

2s

22s


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Plug in a value for x, then solve and select the answer choice that matches your solution

Our plug in 2x =

( )( ) ( )( ) ( )( )1

1 11

111 2 122

12 2 112

ff x f x f−

− −−− −

⋅ = ⋅ = ⋅++

Multiply top and bottom of left fraction by 2 11 2 1 1

3 11 2 3 3

−− −

= ⋅ = ⋅ = −+

Only answer E matches our solution

Notes: Had our solution been a number not shown in the answers, we would’ve had to plug our chosen value of x, 2, into the answers that are in terms of x and find the answer that, once simplified, turns into our solution.


Drill VIII: #15 Difficulty Level: 700+ Topics: Translations & Manipulations; Exponents & Roots

( )2

7 48 7 48+ + − =

(A) 18 (B) 16 (C) 7 4 3+ (D) 7 4 3− (E) 1

Solution

This is a quadratic of the form ( )2a b+ . Expand to 2 22a ab b+ + by foiling

FOIL ( ) ( )( ) ( )2 27 48 2 7 48 7 48 7 48+ + ⋅ + − + − =

Expand 227 48 2 7 48 7 48= + + − + −

Simplify 7 48 2 1 7 48= + + ⋅ + −

Simplify 14 2 16= + =


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Advanced Speed Drill IX (15Qs in 45 mins)

Drill IX: #1 Difficulty Level: 600-700 Topics: Exponents & Roots

5 5 2 2 2 2 2 2 2 22 2 3 3 3 5 5 5 5 5+ + + + + + + + + =

(A) (B) (C) (D) (E)

Solution

There are two 2’s, three 3’s and five 5’s in the sum: 5 5 2 2 2 2 2 2 2 22 2 3 3 3 5 5 5 5 5+ + + + + + + + + =

( ) ( ) ( )5 2 2 6 3 32 2 3 3 5 5 2 3 5+ + = + +


Drill IX: #2 Difficulty Level: 600-700 Topics: FDPs & Ratios

A recipe for cornbread batter requires 32

cups of cornmeal, 52

cups of milk, 2 cups of flour, and 23

cups of sugar.

If 103

cups of cornbread batter can feed 5 people, how many gallons of milk are required to make enough

cornbread to feed 52 people? (1 gallon = 16 cups)

(A) 45

(B) 1316

(C) 1310

(D) 13 (E) 26

Solution

Let c be the number of cups of batter needed to feel 52 people. If 103

cups of cornbread batter can feed 5 people,

how many cups of batter are required to feed 52?

6 3 32 3 5+ + 3 32 5⋅ 2 22 19⋅ 10 4 42 3 5+ + 5 2 22 3 5⋅ ⋅


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•

1010 52 1043 5

5 52 3 3c

c c⋅

= → = → =

We need to find out how many gallons of milk are needed to make 104

3 cups of batter. First, find the fraction of

batter that is milk:

5 5 55 6 32 2 2

3 5 2 9 15 12 4 40 2 40 822 2 3 6 6 6 6 6

= = = ⋅ =+ + + + + +

of the batter is milk.

The amount of milk, in cups, required to make 104

3 cups of batter is

3 10413

8 3⋅ = cups of milk. The question

however asks for the amount in gallons, so the final step is to convert 13 cups into gallons: 1 gal 13

13 cups gallons16 cups 16⋅ = of milk


Drill IX: #3 Difficulty Level: 700+ Topics: Geometry

In the figure, the circle with center 0 bisects side AB and points A and C lie on the circle. If 4BC = , what is the maximum possible area of ABC ?

(A) 4 (B) 4 3 (C) 8 (D) 8 3 (E) 16

Solution

Let’s call D, the point where circle O intersects AB.

A

B

C O


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Since the base of ADC is fixed ( 4AC = ), to maximize the area, we must maximize the height. The greatest

height possible is the radius (2). The maximum possible area of ADC is 1

4 2 42⋅ ⋅ = . The maximum possible

area of ABC is 2 4 8⋅ = .


Drill IX: #4 Difficulty Level: 700+ Topics: Combinatorics

A movie producer wants to hire 4 men and 4 women from the actors who auditioned. If 7 women auditioned and 525 combinations are possible, how many men auditioned?

(A) 5 (B) 6 (C) 7 (D) 8 (E) 9

Solution – Reverse Engineering

The hiring procedure can be divided into two independent steps. First, hiring 4 women from 7 candidates, then hiring 4 men from n candidates. The product of the number of combinations from each step should give us the total number of possible hiring combinations, 525.

Combinations of 4 women from 7 candidates 7!

354!3!

= combinations of women

The product of 35 and the combinations of men hired equals 525, so there are 525

1535

= combinations of men.

What we need to do is find out how many men, x, must be candidates so that there are in fact 15 combinations of

4 possible. The combinations formula with x for the number of male candidates is: ( )

!15

4! 4 !xx

=−

. One easy

A

B

C O

D

ADC is a right triangle because it is inscribed in the circle.

Since ADC and DBC have two pairs of identical sides ( AD DB= and DC belongs to both triangles) and congruent angles at D, these triangles are mirror images of each other, and

4AC BC= =

Because ADC DBC= , The maximum possible area of ABC is twice the maximum area of ADC . Let’s find the latter.

4

2 2


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way to find x is to try out different answer choices until we find out which one gives us 15 combinations of 4 men.

How many men auditioned? Combinations of 4 Notes

(A) 5 5!

54!1!

= Not enough men to have 15 combinations

(B) 6 6!

154!2!

= Yes! 6 men can form 15 combinations of 4

(C) 7 Don’t bother, we’ve found a match (D) 8 Same as above (E) 9 Same as above


Drill IX: #5 Difficulty Level: 500-600 Topics: Geometry

PQR is inscribed in a circle with diameter PR. What is the measure of R∠ ?

(1) PQR is isosceles (2) 90Q∠ = °

Solution

(1) PR must be the longest side since it is the hypotenuse of the triangle. Therefore, PQ QR= , and 45P R∠ = ∠ = °


(2) This statement doesn’t tell us anything we didn’t already know from the question stem. We cannot find R∠ .



P

Q

R

PQR has a right angle at Q. What is the measure of R∠ ?

Since , we know that , so given one of the two smaller angles, we could find the other.

90Q∠ = ° 90P R∠ +∠ = °


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Drill IX: #6 Difficulty Level: 600-700 Topics: Statistics; Weighted Averages

What is the average salary of all employees of company X?

(1) The bottom 60% of employees earn an average of $40,000 (2) The top 60% of employees earn an average of $80,000

Solution

The question is already simplified.

(1) Alone, this statement doesn’t tell us the whole story. The overall average will also depend on the top 40% of earners, but we have no data about them.


(2) Alone, this statement doesn’t tell us the whole story. The overall average will also depend on the bottom 40% of earners, but we have no data about them.


MERGE STATEMENTS

Even when we merge the two statements, we cannot determine the overall salary because we do not know how much overlap, in terms of dollars earned, there is between the bottom 60% and the top 60%. To demonstrate this, try out two scenarios supposing there are 5 employees, A, B, C, D and E such that A, B, and C are the bottom 60% earners while C, D, and E are the top 60% earners. Since the bottom 60% earn an average of $40,000, A, B, and C must earn a total of $120,000. Conversely, since the top 60% earn an average of $80,000, C, D, and E must earn a total of $240,000.

Scenario A • Let A = B = C = $40,000. • We’ve made C = $40,000. Let D = $80,000 and E = $120,000.

• The overall average would be 40,000 40,000 40,000 80,000 120,000

$64,0005

+ + + +=

Scenario B • Let A = $20,000, B = $20,000, and C = $80,000 • We’ve made C = $80,000. Let D = $80,000 and E = $80,000

• The overall average would be 20,000 20,000 80,000 80,000 80,000

$56,0005

+ + + +=

The average salary of all employees depends on the salary of the middle 20%




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Drill IX: #7 Difficulty Level: 600-700 Topics: Combinatorics

6! 6! – 5! 5! =

(A) 36! – 25! (B) (C) (D) (E) 6! – 5!

Solution

Note that 6! = 6 5!⋅

6! 6! – 5! 5! = ( )( ) ( )( ) ( ) ( )2 26 5! 6 5! 5! 5! 6 5! 5!⋅ ⋅ − = ⋅ −

Detach the 6 ( ) ( ) ( )2 2 236 5! 5! 35 5!= ⋅ − =


Drill IX: #8 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values

If x y≠ and 1y ≠ − , is 2 2

x y xx y y

−<

− +?

(1) 0x y> > (2) 1y < −

Solution

Although the question isn’t simple, we can’t do much with it. Be very careful when cross multiplying inequalities because signs sometimes flip. Without information about the sign of x – y and the sign of 2 + 2y, we cannot know how the inequality will be simplified.

(1) This tells us that x y− , the first denominator in the question, must be positive. However, without knowing the sign of the other denominator, we just don’t know which way the inequality sign will face once the question is simplified.


(2) 1 2 2 2 2 0y y y< − → < − → + < . This statement tells us that the 2nd denominator in the question is negative, but without information about x, we cannot solve.


MERGE STATEMENTS

( )25 5!⋅ ( )235 5!⋅ ( )26! - 5!


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Since one of the denominators is negative, the sign of the inequality will flip when we cross multiply to get rid of

fractions. The original question is: “Is 2 2

x y xx y y

−<

− +?”

Cross multiply and flip the sign “Is ( ) ( ) ( )2 2x y y x x y+ > − −⋅ ?”

Expand “Is 2 22 2 2x xy x xy y+ > − + − ?”

Subtract 2xy “Is 2 22x x y> − − ?”

The left side is positive because x>0, and the right side is negative. We do not need any more data to definitively answer this rephrase.


Drill IX: #9 Difficulty Level: 700+ Topics: FDPs & Ratios; Rates & Work

Working together, Rihanna and Jen can build 2366 % more toys in 8 hours than Patrick and Jen can build in 6

hours of working together. How many toys can Jen build alone in 1 hour?

(1) In 1 hour, Rihanna can build two more toys than Patrick can (2) Rihanna can build three more toys in 4 hours than Patrick can build in 5 hours.

Solution

2366 % more is equivalent to

53

of the original amount. In 8hrs, Rihanna and Jen build 53

as many toys as Patrick

and Jen build in 6hrs. Using work rate time= ⋅ we can write ( ) ( )58 6

3R J P J+ = ⋅ + where R, J, and P are the

hourly rates of Rihanna, Jen, and Patrick respectively. This simplifies to 8 8 10 10R J P J+ = + and finally 4 5J R P= − (Eq 1)

Our rephrase “What is J?” OR “What is 4R – 5P?”

(1) R = P + 2. Putting this into (Eq 1) gives ( )4 5 4 2 5J R P J P P= − → = + − . We cannot get a

numerical value for J from this statement.


(2) 4 5 3R P= + . Subtract 5P from both sides to get 4 5 3R P− = . By putting this information into (Eq 1), we get 4 5 3J R P J= − → =



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A string of length 12 6 2+ is cut in two pieces. One piece is used to form an isosceles right triangle with shortest side of length s, and the other piece is used to form a circle. Which of the following represents the radius of the circle in terms of s?

(A) ( )14 5 2

2

s

π

− ⋅ (B)

( )( )6 3 2 2sπ

π

+ + (C)

( )( )2 2 6

2

s

π

+ − (D) ( )2 2 s+ ⋅ (E) 2 6 2sπ +

Solution

An isosceles right triangle is a 45-45-90 degrees triangle. The ratio of the lengths of its sides is : : 2s s s , where

s is the length of each of the shorter sides, and 2s is the hypotenuse

Formula for circumference ( )2 12 6 2 2 2r sπ = + − + ⋅

Divide both sides by 2π ( )12 6 2 2 2

2

sr

π

+ − +=

Factor ( )( ) ( )2 2 6 2 2

2

sr

π

+ ⋅ − +=

Factor out 2 2+ ( )( )2 2 6

2

sr

π

+ −=


s

s 2s

The perimeter of the triangle is ( )2 2 2 2s s s+ = + ⋅ . Since the

original length of the string was 12 6 2+ , the remainder will be used to form the circle. So the circumference of the circle, 2 rπ , can also be

expressed as ( )12 6 2 2 2 s+ − + ⋅ . Let’s find the radius


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Take-Aways • Memorize the ratios of the lengths of the sides of 30-60-90 degree triangles ( ): 3 : 2x x x as well as 45-45-

90 degree triangles also called isosceles right triangle ( ): : 2x x x .

Drill IX: #11 Difficulty Level: 600-700 Topics: Exponents & Roots

2366

xy

x−

=−

. If x is an integer, which value of x would yield the least possible value of y?

(A) -6 (B) -5 (C) 0 (D) 5 (E) 6

Solution

The top of the fraction is at least 0 since it is an even root. To make y negative, the bottom must be negative. If 6 0x− < , then 6x > . This however cannot be the case since it would make the value under the square root negative. Besides, there is no option greater than 6 in the answer choices.

Since we cannot make y negative, to minimize it we must make it equal 0. For a fraction to equal 0, the top must

equal 0. 236 0 6x x− = → = ± . x must be different from 6 because 6x = would make the bottom of the fraction equal 0. Therefore, y is 0 (its least possible value) when 6x = −


Drill IX: #12 Difficulty Level: 700+ Topics: Weighted Averages

Three solutions are 30%, 45%, and 90% alcohol by volume. If a ounces of the 30% solution, b ounces of the 45% solution, and c ounces of the 90% solution are mixed to create a solution of 50% alcohol, what is b in terms of a and c?

(A) 3a + 9c (B) 4a + 2c (C) 3c – 9a (D) 8c + a (E) 8c – 4a

Solution

The mixture is a total of a + b + c ounces. The volume of alcohol in the mixture is 30% of a ounces, 45% of b ounces and 90% of c ounces. This volume of alcohol is equivalent to 50% of the mixture. Algebraically,

( ).30 .45 .90 .50a b c a b c+ + = + +

Multiply by 100 30 45 90 50 50 50a b c a b c+ + = + +

Divide by 5 6 9 18 10 10 10a b c a b c+ + = + +


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Solve for b 8 4c a b− =


Drill IX: #13 Difficulty Level: 700+ Topics: Number Properties

If k is a positive integer, what is the remainder when 2 4k − is divided by 12?

(1) k is even (2) k is not divisible by 3

Solution

( )( )2 4 2 2k k k− = − + .

Our rephrase “What is the remainder when ( )( )2 2k k− + is divided by 12?”

If we knew the remainder when 2k is divided by 12, we would have enough data to find the remainder when 2 4k − is divided by 12. Of course, if we had a way to find the value of k, it would also be sufficient information.

(1) If k = 4, then ( )( )2 2 12k k− + = and the remainder of division by 12 would be 0. On the other hand, if

k = 6, then ( )( )2 2 32k k− + = and the remainder of division by 12 would be 8. What is interesting to

note however, is that since k is even, ( )2k − and ( )2k + must each be even, so ( )( )2 2k k− + must be

divisible by 4.


(2) Either k is a multiple of 3 + 1, or a multiple of 3 + 2. In the first case, for instance if k were 4 or 7, ( )2k + would be a divisible by 3. In the second case, for instance if k were 5 or 8, ( )2k − would be

divisible by 3. Either way, ( )( )2 2k k+ − must be divisible by 3. But what is its remainder when divided

by 12?

If k = 4, then ( )( )2 2 12k k− + = and the remainder of division by 12 would be 0. On the other hand, if

k = 5, then ( )( )2 2 21k k− + = and the remainder of division by 12 would be 9.


MERGE STATEMENTS


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From statement (1), we learned that ( )( )2 2k k− + must be divisible by 4. From statement (2), we learned that

( )( )2 2k k+ − must be divisible by 3. By putting the statements together, we can conclude that ( )( )2 2k k+ − is

divisible by 12, so the remainder of division by 12 is 0.




Does a cylinder with surface area s and radius r have a height of x?

(1) ( )( )( )22 12

2 0s r

r x r sx

ππ

− −+ − =

(2) ( )22 2 02

srx s r r x

rπ π

π− − − − =

Solution

The surface area of a cylinder is the sum of the areas of the two circular bases, and the area of its rectangular vertical side.

The formula for the surface area of a cylinder is 22 2r rhπ π+ . In our particular cylinder, the surface area is s, and the radius is r, and we’re asked whether the height, h, equals x.

Our rephrase “Is 22 2s r rxπ π= + ?”

(1) A product equals zero only if one or more of the factors is equal to zero. This statement tells us that

( )( )( )2 0r x r sπ + − = or 22 12

0s r

xπ− −

=

. Let’s isolate s in each of these equations so we can

compare them to our rephrase:

h

r 2 rπ

h

The vertical side of a cylinder unravels to form a rectangle whose base is the circumference of the cylinder’s base


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• ( )( )( )2 0r x r sπ + − = 22 2 0rx r sπ π+ − = 22 2s r rxπ π= + (Eq 1) OR…

• 22 12

0s r

xπ− −

=

22 12 0s rπ− − = 22 12s rπ= + (Eq 2)

If (Eq 1) were true, then we would know that the answer to our rephrase is a definitive YES. On the other hand, if (Eq 1) were false and (Eq 2) were true, then we would not know what the answer to our rephrase is. Because we do not know which of these two equations is true, we cannot definitively answer our rephrase.


(2) This statement tells us that ( )22 2 0rx s rπ π− − = or 02

sr x

rπ− − =

. Let’s isolate s in each of these

equations so we can compare them to our rephrase:

• ( )22 2 0rx s rπ π− − = 22 2s rx rπ π= − (Eq 3) OR…

• 02

sr x

rπ− − =

2

sr x

rπ= + 22 2s r rxπ π= + (Eq 4)

If (Eq 4) were true, then we would know that the answer to our rephrase is a definitive YES. On the other hand, if (Eq 4) were false and (Eq 3) were true, then we would not know what the answer to our rephrase is. Because we do not know which of these two equations is true, we cannot definitively answer our rephrase.


MERGE STATEMENTS

The only equation that agrees with both statements is 22 2s r rxπ π= + , so this equation must be true. In other words, the answer to our rephrase is a definitive YES.




Drill IX: #15 Difficulty Level: 700+ Topics: Rates & Work


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20 machines working together at the same constant rate and without a break can complete a certain job in 2 days.

If 15 new machines must complete 25

of the job in 16 hours, the rate of each new machine must be how much

faster than the rate of each original machine?

(A) 45% faster (B) 50% faster (C) 55% faster (D) 60% faster (E) 65% faster

Solution

work rate time= ⋅ , and when multiple people or machines work together, their combined rate is the sum of the individual rates. Let r be the rate of each of the original 20 machines, and R be the rate of each of the new 15 machines.

• 20 machines working together for 2 days (48 hours) will do 20 48r ⋅ units of work • 15 new machines working together for 16 hours will do 15 16R ⋅ units of work

• The new machines did 25

of the job that the old machines did: 2

15 16 20 485

R r⋅ = ⋅ ⋅

Divide by 16 2

15 20 35

R r= ⋅ ⋅

Divide by 3 2

5 20 5 85

R r R r= ⋅ → =

Solve for R 8 3

5 5

R r R r r= → = +

The rate of each new machine, R, is 35

or 60% faster than the rate of each original machine, r.


Advanced Speed Drill X (15Qs in 50 mins)

Drill X: #1 Difficulty Level: 700+ Topics: Rates & Work

John leaves a certain location at 1pm traveling at 45 miles per hour. At 3:40pm, Sally leaves the same location and takes the same exact route traveling at 60 miles per hour. At what time will the distance between John and Sally be half of the distance that John has traveled?

(A) 4:25pm (B) 5:16pm (C) 7:56pm (D) 8:22pm (E) 11:40pm

Solution


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Since Sally left 2 hours and 40 minutes after John, her travel time is 23

82

3= hours shorter than John’s travel

time. Build a rate table, and remember that distance rate time= ⋅

“The distance between John and Sally be half of the distance that John has traveled” means the point in time at which John has gone twice as far as Sally.

Algebraically, 8

45 2 603

t t= ⋅ −

Divide by 15 8

3 2 43

t t= ⋅ −

Simplify 415

64 64 643 8 5 4

3 3 15t t t t= − → = → = =

This time span is equal to 4 hours and 4

60 1615

⋅ = minutes. Adding 4hrs 16mins to the start (1pm) results in

5:16pm. We start counting from 1pm because t is the length of time that John has traveled.


Drill X: #2 Difficulty Level: 700+ Topics: Geometry

In the figure shown, the circle with center A and area 27π is tangent to an equilateral triangle at points B and D. What is the area of ABCD?

(A) 27 3

2 (B)

27 154π +

(C) 9 6π + (D) 9 27

2π +

(E) 27 3

Solution

The area of the circle is 2 27 3 327 rrπ π → = == . Since the sides of the triangle are tangent to the circle, 90B D∠ = ∠ = ° , and since the triangle is equilateral, 60C∠ = ° . The angles of any quadrilateral must add up to

360, so 360 90 90 60 120A∠ = − − − = ° . I have redrawn ABCD below

John Sally Rate 45 60

Time t 83

t −

Distance 45t 8

603

t −

A

B

C

D


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Drill X: #3 Difficulty Level: 700+ Topics: Inequalities & Absolute Values; Rates & Work

Last Tuesday, Allie, Brianna and Chevone were paid according to their hourly wages and the length of time they worked. Who was paid the least?

(1) Allie’s hourly wage is $3 more than Chevone’s, and Allie worked two hours longer than Brianna worked (2) Chevone’s hourly pay is $1 less than Brianna’s, and Chevone spent more time working than Allie and

Brianna spent combined

Solution

Let RA, RB, and RC be the hourly pay rates of Allie, Brianna and Chevone respectively. Let TA, TB, and TC be the time in hours that Allie, Brianna and Chevone worked. Finally, let PA, PB, and PC be the total pay of each of the three people respectively. We can write

PA = RATA (Eq 1) PB = RBTB (Eq 2) PC = RCTC (Eq 3)

Our rephrase “Which of PA, PB, and PC, is the least?”

(1) RA – 3 = RC and TA – 2 = TB. Substitute these into (Eq 2) and (Eq 3) above.

PA = RATA (Eq 1) PB = RB (TA – 2) (Eq 4) PC = (RA – 3) TC (Eq 5)

We don’t have enough data to find out which of the three salaries is the least.


(2) 1C BR R= − and C B AT T T> + . Using the first of these two pieces of data, we can rewrite (Eq 3) above

PA = RATA (Eq 1) PB = RBTB (Eq 2) PC = (RB – 1) TC (Eq 6)

60 60

30 30

3 3

9

A

B

C

D

ABCD can be split into two identical 30-60-90° triangles. All such

triangles have a side lengths ratio of : 3 : 2s s s . In this case, since

the base of each is 3 3 (the radius), the height will be 3 3 3 9⋅ = .

The area of each triangle is 3 3 9 27 3

2 2⋅

= . The area of ABCD is

27 32 27 3

2⋅ =


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With this data alone, we cannot find out which of the three salaries is the least.


MERGE STATEMENTS

By merging the two statements, we learn a bit more. Statement (1) told us that RA – 3 = RC and statement (2)

added 1C BR R= − . Adding these two equations yields 3 1 2A C C B B AR R R R R R− + = + − → = − .

Plugging this into (Eq 4) allows us to express PB in terms of RA and TA.

PA = RATA (Eq 1) PB = (RA – 2) (TA – 2) (Eq 7) PC = (RA – 3) TC (Eq 5)

We can tell that PA > PB, but there is no way to know where PC stands. This is because although we know that

C B AT T T> + so that Chevone’s time is the longest, it is also true that Chevone’s rate, 3AR − is the least. If you

understand this, pick answer E and move on. Otherwise, consider the following proof that we don’t have enough data:

Scenario 1 • ( )( )4 7 $28AP = = ( )( )4 2 7 2 $10BP = − − = ( )( )4 3 15 $15CP = − =

Scenario 2 • ( )( )3.1 3 $9.3AP = = ( )( )3.1 2 3 2 $1.1BP = − − = ( )( )3.1 3 5 $0.5CP = − =

We cannot tell who gets paid the least. In Scenario 1, it is Brianna, but in Scenario 2, it is Chevone. When using plugin values to evaluate statements, be certain that the values your plug in agree with all the restrictions of the statement (or statements) that you are evaluating.



Drill X: #4 Difficulty Level: 700+ Topics: Number Properties

m and n are positive integers, and the quotient of m and 5 is the integer q. What is the remainder when m n+ is divided by 15?

(1) n – 3 is a multiple of 15 (2) When q is divided by 3, the remainder is 2

Solution

“The quotient of m and 5 is q” means 55m

q m q= → = . Essentially, m is a multiple of 5.


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(1) 3 15 15 3n a n a− = → = + where a is an integer. Therefore, 5 15 3m n q a+ = + + . Depending on the value of q, we would get different remainders when dividing m n+ by 15. For example if 5q were 15, then ( )15 1 3m n a+ = + + and the remainder of division by 15 would be 3. On the other hand, if 5q were

10, than 15 13m n a+ = + and the remainder of division by 15 would be 13. We cannot find the exact value of the remainder.


(2) q divided by 3 has a remainder of 2, means that 3 2q b= + (b is an integer). We already know that

( )5 5 3 2 15 6m q m b m b= → = + → = + . Unfortunately, we are given no data about n.


MERGE STATEMENTS

Statement (1) gave us 15 3n a= + . Statement (2) gave us 15 6m b= + . Therefore ( )15 9m n a b+ = + + . Since

m n+ is a multiple of 15 plus 9, the remainder of division by 15 is 9.



Drill X: #5 Difficulty Level: 700+ Topics: Combinatorics

On Monday, the probability of rain in New York is 20%, in Paris is 50%, and in Moscow is 80%. If these probabilities are independent, what is the probability that it rains on Monday at exactly two of these three locations?

(A) 150

(B) 825

(C) 25

(D) 2150

(E) 12

Solution

The probability that it doesn’t rain is 1 minus the probability that it rains. There are three possible outcomes for rain in exactly 2 cities.

• Rain in New York, rain in Paris, no rain in Moscow: 1 1 1 15 2 5 50⋅ ⋅ =

• Rain in New York, no rain in Paris, rain in Moscow: 1 1 4

5 2 5

450

⋅ ⋅ =

• No rain in New York, rain in Paris, rain in Moscow: 4 1 4 165 2 5 50⋅ ⋅ =


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Only one of these three outcomes can happen, so we are looking for outcome 1 or outcome 2 or outcome 3. In

our case, the probability that it rains in exactly 2 of the three cities is 1 4 16 2150 50 50 50

+ + =


Take-Aways • Assuming A and B are independent events (the likelihood that one happens is not affected by whether the

other happened), Probability of A or B is ( ) ( )P A P B+ . Probability of A and B is ( ) ( )P A P B⋅

Drill X: #6 Difficulty Level: 700+ Topics: Exponents & Roots; Inequalities & Absolute Values

If x is greater than 0, is 2 22 ?xy x y y x− + + > −

(1) 2y y n− > (2) 0x n> >

Solution

( )22 22xy x y x y− + + = − . A square under a square root is the technical definition of the absolute value. It

would be incorrect to assume that they simply cancel each other out because a negative value inside the square would be turned into a positive anyway.

The question can be rephrased as “Is x y y x− > − ?” “Is ( )x y x y− > − − ?” Think of ( )x y− as one

number. The answer to this rephrase would be YES if ( )x y− were positive (try 2x y− = ) but the answer

would be NO if it were negative or 0 (try 2x y− = − )

Our rephrase “Is 0x y− > ?”

(1) We don’t know anything about n, but we do know that 2 0n ≥ . Therefore, 0y y− > . So –y and y

have the same sign: 0 0y y− > → < . Since we already know that 0x > , it must be true that 0x y− >(a positive minus a negative is the same as a positive plus a positive) We can definitively answer our rephrase


(2) With no information about y, we cannot determine the answer to our rephrase.




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Drill X: #7 Difficulty Level: 600-700 Topics: Inequalities & Absolute Values

0p q− > . Is q p pr q r−

> −+

?

(1) r < 0 (2) q < 0

Solution

Unfortunately, without knowing the signs of our denominators we cannot cross multiply and simplify the question because we don’t know whether the inequality should flip.

(1) We can multiply both sides of the inequality by r. The sign will flip because r is negative.

“Is q p pr q r−

> −+

?” “Is qr pr

pr q−

< −+

?” We don’t have enough data to answer this question


(2) We have no data about r


MERGE STATEMENTS

Statement (1) allowed us to rephrase to “Is qr pr

pr q−

< −+

?”. By merging the two statements, we can determine

that 0r q+ < , so cross multiplying by r q+ will cause the sign to flip once more.

“Is qr pr pr pq− > − − ?” “Is qr pq> − ?” Divide by q (and flip the sign) “Is r p< − ?”. From statement (1) we know that 0r < , and from the question stem we know that p q> . However, we have no way to compare r to – p.




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Drill X: #8 Difficulty Level: 600-700 Topics: Number Properties; Sets & Groups

N is a set of four consecutive even integers whose sum is positive. O is a set of four consecutive odd integers. Is the median of N greater than the average of O?

(1) The product of all elements of O is negative (2) The product of the average of N and the median of O is negative

Solution

The mean and the median are the same for a list of values such that the difference between two adjacent values is always the same. Consecutive odd and consecutive integers are two such lists.

Since the sum of the elements of N is positive, there must be more positive than negative integers in N. There are only four elements in N, so there cannot be more than one negative element in N. The smallest possible member of N is -2, so the smallest possible median of N is the median of {-2, 0, 2, 4}. The median of N is at least 1.

Our rephrase “Is the median of N greater than the median of O?”

(1) A product of values is negative only if there are an odd number of negative values. Of the four consecutive odd integers in O, either one or three must be negative. So O is either{-1, 1, 3, 5} or {-5, -3, -1, 1}. The median of O is either 2 or -2. Since the median of N could be equal to 1, we cannot tell which set has the greater median


(2) Within each set, the median of the set equals the average of the set. This statement informs us that the median of N and the median of O have opposite signs. The median of N is positive (at least 1), so the median of O must be negative. We have enough data to compare the two medians.



Drill X: #9 Difficulty Level: 600-700 Topics: FDPs & Ratios

Michele, Sarah and John have marbles in a ratio of 5 to 12 to 6. If Sarah gave 5 marbles to Michele and 2 to John, the ratio of Michele to Sarah to John’s marbles would be 15 to 17 to 14. How many marbles do Michele and John have together?

(A) 10 (B) 12 (C) 22 (D) 29 (E) 33

Solution


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Let the number of marbles for Michele, Sarah and John be 5x, 12x and 6x. If Sarah gave 5 marbles to Michele and 2 marbles to John, the number of marbles for Michele, Sarah and John would be 5 5x + , 12 7x − and 6 2x + and the ratio would be 15 to 17 to 14. Ignore Sarah for a second and set a proportion between Michele and John: 5 5 15

6 2 14 70 70 90 30 20 40 2

x

xx x x x

+=

+→ + = + → = → = .

Michele and John have 5x and 6x marbles, a total of 11x or 22.


Drill X: #10 Difficulty Level: 700+ Topics: Inequalities & Absolute Values

If 0x y+ > , is 0x > ?

(1) x x y> +

(2) 2 2x y>

Solution

The question is already simplified . Since we know that 0x y+ > , if we learned that 0y < , it would be enough to conclude that 0x > .

(1) Since 0x y+ > , x y x y+ = + so we can rewrite this statement as x x y> + (Eq 1).

• If 0x > , (Eq 1) simplifies to 0x x y y> + → < • If 0x < , (Eq 1) simplifies to 0 2x x y x y− > + → > + . More interestingly, because the question

stem tells us that 0x y+ > , if 0x < , then 0y >

This statement essentially tells us that x and y have opposite signs, but we don’t know which one is positive.


(2) ( )( )2 2 0 0x y x y x y− > → − + > . Since the question stem tells us that ( ) 0x y+ > , it must also be

true that ( ) 0x y− > . Add these two inequalities to get 2 0 0x x> → >



Drill X: #11 Difficulty Level: 600-700 Topics: Translations & Manipulations


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What is the sum of all multiples of 3 between 1,000 and 2,000?

(A) 496,500 (B) 498,000 (C) 499,500 (D) 501,000 (E) 502,500

Solution

The multiples of 3 included within that range go from 1,002 to 1,998. A sequence of multiples of 3 is an

arithmetic sequence, and the average value is 1,998 1,002

1,5002−

= . # of items

sumavg = so we need to know how

many multiples of 3 there are from 1,002 to 1,998 in order to find the sum.

1,002 is the 1, 002

3343

th= multiple of 3 while 1,998 is the 1,998

6663

th= multiple of 3. Therefore, there are

666 334 1 333− + = terms in the sequence (add 1 because the sequence includes both ends). The sum of all multiples of 3 between 1,000 and 2,000 is 1,500 333 499,500⋅ =



an arithmetic sequence.

• The average of an arithmetic sequence is the average of its highest and lowest values. This average is always equal to the median of the sequence.


Drill X: #12 Difficulty Level: 700+ Topics: Rates & Work

At 3:30am train A leaves its station traveling at 180 miles per hour. At 4:45am train B leaves another station 1,665 miles away heading in the opposite direction on the same track. If the two trains cross each other’s paths at 10:05am, what is train B’s speed?

(A) 90mph (B) 95mph (C) 100mph (D) 105mph (E) 110mph

Solution

When the trains meet at 10:05am, train A’s running time is 6 hours 35 minutes which equals 35 79

6 hrs60 12

+ = , and

train B’s running time is 5 hours 20 minutes which equals 20 16

5 hrs60 3

+ = . Let r be the rate of train B in miles

per hour. Build a rate table ( distance rate time= ⋅ ).

Train A Train B


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total distance traveled is 1,665 miles, so 16

1,185 1,6653

r+ = . The

Simplify to get 3

48016

r = ⋅ and solve for r to get 90r = miles

per hour.


Drill X: #13 Difficulty Level: 700+ Topics: Coordinate Geometry; Inequalities & Absolute Values

On the xy-plane, is the distance between the points (a, b) and (b, -c) greater than the distance between the points (-c, a) and (b, 5)?

(1) b > 5

(2) 112

a <

Solution

“Is ( ) ( ) ( ) ( )2 2 2 25b a c b b c a− + − − > + + − ?” “Is ( ) ( ) ( ) ( )2 2 2 25b a b c b c a− + − − > + + − ?”

( ) ( )2 2b c b c− − = + , so we can subtract it from each side of the equation

Our rephrase “Is ( ) ( )2 25b a a− > − ?”

(1) Without information about the value of a, we cannot answer our rephrase. If 6b = and 0a = , the answer would be YES. However, if 6b = and 6a = , the answer would be NO.


(2) 5.5a < . Without information about the value of b, we cannot answer our rephrase.


MERGE STATEMENTS

Even merging the statements does not limit us to a unique answer. If 6b = and 0a = , the answer to our rephrase would be YES. However, if 5.4b a= = , the answer to our rephrase would be NO.



Rate 180 r

Time 7912

163

Distance 79

180 118512⋅ =

163

r


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Drill X: #14 Difficulty Level: 700+ Topics: FDPs & Ratios

Country A’s GDP increased by x percent from 2006 to 2007, and increased by y percent from 2007 to 2008. What is the ratio, in terms of x and y, of country A’s 2008 GDP to its 2006 GDP?

(A) 1100 10,000

x xy+ + (B)

100 10,000x y xy+

+ (C) 100

y xy+

(D) 1100 10,000x y xy+

+ + (E) 1100

x y xy+ ++

30 Seconds Hack

If x and y both equaled 0, there would be no increase and the ratio of 2008’s to 2006’s GDP should equal 1. Plug in 0 for x and y and see which answers give you 1. Eliminate the others (B and C)

Guess A, D or E

Solution

Let g be 2006’s GDP. To increase an amount by p percent, multiply it by 100

100p+

. To get 2008’s GDP,

increase g by x percent, then by y percent: 100 100

100 100x y

g+ +

FOIL and find 2008’s GDP 10,000 100 100

110,000 100 10,000

y x xy y x xyg g

+ + + +⋅ = + +

Put 2008’s GDP over the original, g, to get the ratio. The g’s cancel out and we are left with 1100 10,000y x xy+

+ +


Drill X: #15 Difficulty Level: 700+ Topics: Geometry


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If the perimeter of rhombus ABCD is 24, what is its area?

(A) 24 (B) 16 2 (C) 18 3 (D) 36 (E) 36 3

Solution

A rhombus is a slanted square so all its sides are equal. It is a type of parallelogram, meaning that opposite sides are parallel and opposite angles are equal. Since the perimeter of ABCD is 24, each side length is 6. In addition,

60B D∠ = ∠ = ° . The four angles must add up to 360°, and angles B and D add up to 120°, so 120A C∠ = ∠ = ° .

.

30°

A

B

C

D

60°

60° 60°

6

3

This rhombus can be divided into four identical 30-60-90° triangles each with a base of 3 and a hypotenuse of 6. The ratio of the side

lengths of a 30-60-90° triangle is : 3 : 2s s s . In our case, the

smallest side is 3, so the height of each of the four triangles is 3 3 .

The area of each 30-60-90° triangle is 1 9 3

3 3 32 2⋅ ⋅ = . The area of

the entire rhombus is 9 3

4 18 32

⋅ = .


60°

A

B

C

D

Patrick Siewe - GMATFix.com OG Companion 12th Take Aways List

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Take Aways List The following is a list of Take-Away lessons from the OG and OG Companion questions

1. Do not rush to do long calculations. Give yourself a few seconds to think about whether there is a faster way or whether you can approximate. Because the GMAT only grants you 2 minutes per math question on average, you will seldom be required to complete these operations.

2. Always glance at the answer choices before you start solving. In many cases, this will help you avoid unnecessary work. For instance, you might notice that the answers are expressed as fractions and thus discover that there is no need to engage in long division.

3. A set of numbers in which the difference between each value and the next higher value is constant is called an arithmetic sequence.

4. The average of an arithmetic sequence is the average of its highest and lowest values. The average is always equal to the median of this sequence.

5. The sum of an arithmetic sequence is its average multiplied by the number of numbers in the sequence

6. When solving a probability question, you can often use your intuition to eliminate answers that are way off

7. When solving probability, percentages, or ratios, consider ignoring the total given and using your own to make the math easy. 100 makes a great total in percentage questions.

8. Drawing figures reveals connections between data and often makes the problem easier to understand and handle. Always make a drawing of your own for geometry questions, even if one is already provided

9. In 3-d Geometry, you are expected to know the volume of a cylinder, 2V r hπ= , and the volume of a box, V l w h= × × . Just remember (area of base)(height).

10. The group formula, 1 2Total Group Group neither both= + + − , is a useful way to solve questions about 2 overlapping groups.

11. When drawing a Venn diagram, always start with the center

12. Whenever you are asked to find a ratio, fraction, or a percentage and you aren’t given any concrete numbers, you can plug in a value that will be easy to deal with and use it to solve the problem.

13. Most geometric figures are drawn to scale. When stuck, you can approximate lengths, angles, etc…

14. If you find yourself needing an equation that involves angles in a triangle, remember to use the “180° sum” equation

15. When a question gives you non-variable answer choices along with a word problem, you can often use reverse engineering. Start with answer C, especially if it will take more than a few seconds to check each answer.

16. In the answer choices, the GMAT will often include values that you are likely to come across as you work the problem. For instance, you may be asked to solve for y, and one of the answer choices will be the value for x. It is generally a good idea to take an extra 3 or 5 seconds to re-read the end of each problem (what you’re asked to solve for) before making your final selection.

17. When solving a Number Properties question, express all values in their prime factorizations (ex: 212 2 3x x= → = ⋅). In this format, it is easier to deal with very large numbers, see patterns, and make connections.


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18. 3rd side rule of triangles: In any triangle, the length of the third side must be greater than the difference, but smaller than the sum of the other two sides. 1 2 3 1 2side side side side side− < < +

19. Generally, you should simplify fractions before you perform multiplications. This will keep your numbers from growing

needlessly large. For instance, suppose you’re faced with 21 27 1

9 7⋅

⋅ . You could multiply across top and bottom, and

then divide56763

. Solving this way might take you 90 seconds. On the other hand, you can simplify by crossing off

common factors in 21 & 7, as well as in 27 & 9. In this case you’ll be left with 3 31⋅

to divide. Solving this way might

take you 15-20 seconds, and you would’ve saved over 60 seconds to use on a more difficult problem.

20. To solve complex rate questions, try to draw out what is happening to help you visualize the problem. The key to solving these questions is to spot the relationships between the three parts of all rate/work equations: Distance, Rate, and Time

21. It is generally a good idea to use as few variables as you have to, and avoid needlessly introducing new variables (such as 1 2 and d d if they are the same distance; just use d).

22. Don’t spend more than a few seconds trying to anticipate what data would be sufficient. If you have a gut feeling , make a mental note. If you don’t, move on to evaluating the statements.

23. One way to check whether a statement or two statements together are sufficient is to plug-in a couple of values that fit the data given in those statements. If the values yield more than one unique answer to the question, then the statement or statements together are NOT sufficient. On the other hand, if all values that you plug-in yield a unique answer to the question, then the statement or statements together may (not must) be sufficient.

24. For the purposes of the GMAT, you do not need to memorize the formula for standard deviation. However, you need to have a good understanding of what standard deviation means. Broadly, standard deviation is a measure of how tightly/loosely values are grouped around the average. The more dispersed the values, the higher the standard deviation. If all values are the same, standard deviation is 0

25. In Data Sufficiency, if your rephrase is a good one, data sufficient to answer your rephrase is necessarily sufficient to answer the original question. Consequently, while taking your test, there is no need to check whether you have sufficient info to answer the original question if you’ve already checked whether you can answer your rephrase.

26. Before combining, or evaluating data, try to write all your equations in a similar format

27. On the GMAT, fractions in inequalities are tricky, especially when the sign of the denominator is unknown. Why tricky? If you cross-multiply by a negative the inequality sign flips but if you cross-multiply by a positive, the inequality sign stays the same. So if you multiply by an unknown value, you can’t tell what happens to the sign and you should account for both cases.

28. Do not rush to “do the math”. Instead, build your equations and see whether you can cross off common factors. On the GMAT, things tend to simplify to easy numbers.

29. When variables appear in the answer choices, you can often plug in easy values for those variables within the question, solve and then find out which answer choice agrees with your solution. Avoid plugging-in 0, 1, or the same value for different variables (unless the variables are equal). By coincidence, two answers may agree with your solution. In that case, just plug in another set of easy numbers and try again

30. Memorize your multiplication tables up to 12 times 12 before taking the GMAT. You should also know your perfect squares up to 220 and how to recognize at a glance whether a number is divisible by 2, 3, 4, 5, 6, 8, and 9.


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31. When the question itself asks you to approximate, do so because the math will likely be unpleasant.

32. Consider using fractions instead of decimals in your calculations because the former are easier to simplify.

33. Reverse Engineering is a great technique when the algebra is unclear or too difficult. Sometimes however, it takes longer than one might like. If you can conventionally solve a problem quickly, you should. If not, consider techniques such as approximation, plug-in, and reverse engineering.

34. To solve pattern problems, it is always best to draw out what is happening as this will increases your chances of spotting the pattern.

35. When evaluating one answer choice at a time, skip choices that would take long to work with, because in many cases, you will know the right answer almost as soon as you see it.

36. You should be able to recognize a Pythagorean triple when you see it. 3:4:5, and 5:12:13 along with their multiples are often used on the GMAT.

37. Memorize the ratios of the lengths of the sides of 30-60-90 degree triangles ( ): 3 : 2x x x as well as 45-45-90 degree

triangles also called isosceles right triangle ( ): : 2x x x

38. Whenever you are asked to find the probability that something happens “at least one time”, it is easier to find the probability that it doesn’t happen at all, and subtract it from 1

39. To find the diagonal of a box, use the following formula, an extension of the Pythagorean Theorem: 2 2 2 2d l w h= + +

40. The standard format for the equation of a line is y mx b= + where m is the slope and b is the y-intercept.

41. In data sufficiency, when the data from one statement can be logically deduced from the other statement, C cannot be the right answer. When one statement offers the same data (no more, no less) as the other statement, the correct answer must be D or E

42. Most of the time, factoring equations makes them easier to simplify and allows you to isolate variables. Therefore, in data sufficiency, consider factoring any initial equation when possible.

43. Assuming A and B are independent events (the likelihood that one happens is not affected by whether the other happened), Probability of A or B is ( ) ( )P A P B+ . Probability of A and B is ( ) ( )P A P B⋅

44. Area of a circle is 2rπ and circumference is 2 r dπ π= where r is the radius and d the diameter

45. Two values are directly proportional ifvalue 1

constantvalue 2

= . As one value increases (or decreases), the other must follow

suit to maintain the constant. Two values are inversely proportional if ( )( )value 1 value 2 constant= . As one value decreases, the other must increase to maintain the constant.

46. If you have as many independent linear equations as you have variables, you will be able to solve for all your variables. Equations are independent if it is impossible to manipulate them to make them look identical.

47. There are two ways to find the area of a trapezoid. One way is to use the area formula, 1 2

2

base baseh

+⋅ . The 2nd way is

to divide the trapezoid into two triangles and the rectangle in the middle, and add up the areas of these three parts. 48. The total number of positive factors of ...a b ck l m⋅ ⋅ where k, l, and m are different prime factors is the product

( )( )( )1 1 1 ...a b c+ + +


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49. In data sufficiency, If you decide to plug in values to evaluate an inequality question, be sure to try positives and negatives (as long as the values agree with the statement under evaluation)

50. To calculate percentage change, be it increase or decrease, do positive difference

100starting value

⋅

51. Parallel lines have the same slope but the slopes of two perpendicular lines are negative reciprocal. If the first line has

slope m, its perpendicular will have slope 1

m−

52. The only way to safely combine inequalities is to make sure that the signs are facing the same direction, and to add the inequalities. Never subtract. Only add.

53. In a normal distribution, 34% of the data is between the average and +/-1 standard deviation, 14% of data is between +/-1 and +/-2 standard deviation, and 2% of data is beyond +/-2 standard deviation. Remember the ratio 34:14:2.

54. To find the answer that must be something, you can eliminate all the other answers by showing that they don’t have to be that thing.

55. I recommend memorizing the decimal values of 1 1 1, ,...,2 3 10


56. Know the following quadratic factorizations by heart: ( )22 22x xy y x y+ + = + ; ( )22 22x xy y x y− + = − and

( )( )2 2x y x y x y− = + −

57. To simplify a square root, break it into a product of two square roots, one of which has an integer solution. Example:

63 9 7 3 7= =

58. In general, if you want to get rid of a variable in a system of equations, isolate that variable in one equation and replace it with its equivalent in the other.

59. When solving number properties, it’s often helpful to insert the known property as a variable in equations. For instance, if you know that an unknown is an integer, you may refer to it as int. A prime number would be prime, an even number 2k or even, an odd number 2k+1 or odd, a multiple of 7 would be 7k… Doing so allows you to manipulate the unknown value while accounting for its property

60. One great technique for solving ratios is to express all parts of the ratio in terms of the same variable. For instance, if a problem tells you that the ratio of boys to girls is 4:5, you should use 4x and 5x as the number of boys and girls. By writing all values in terms of the same variable, you greatly simplify the problem (in general, the fewer variables the easier a problem will be to solve)

61. The rate is the inverse of the time it takes to complete 1 job: 1rate

time= and

1timerate

=

62. If machines work together, their combined rate is the sum of the individual rates

63. To increase a value by x percent, multiply it by 1100

x + . To decrease a value by y percent, multiply it by

1100

y −

Patrick Siewe - GMATFix.com OG Companion 12th Reference

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Reference Tables

• A Questions Directory organizes all questions by type and difficulty

• The OG Questions Distribution shows which question types are the most common on the GMAT

• Scores to Percentiles Conversion Scales


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Questions By Topic The following table organizes all questions from the Official Guide for GMAT Review, 12th Edition as well as all questions from the Advanced Speed Drills in this book according to topic and difficulty rating. Questions numbered T1 through T48 are from the OG’s Diagnostic test. Questions numbered P1 to P230 are from the OG’s Problem Solving section. Questions D1 through D174 are from the OG’s Data Sufficiency section. Finally, questions R1.1 through R10.15 are from the Advanced Speed Drills section of this book. Each number corresponds to the drill and the question from that drill. For instance, question R2.10 is equivalent to Drill II, number 10.

200-400 400-500 500-600 600-700 700+

Algebraic Translations & Manipulations

T001, T035, P001, P002, P006, P017, P038, P044, P077, P100, P168, D033, D035, D039, D071

T027, P026, P058, P061, P070, P075, P076, P094, P151, P155, P158, P184, P196, P222, D013, D015, D040, D053, D054, D060

P059, P072, P083, P084, P090, P091, P097, P127, P134, P140, P144, P153, P172, P175, P215, D008, D023, D030, D037, D052, D057, D062, D063, D070, D077, D085, D088, D119, D131, D137, D138, D174, R2.6, R7.9

T048, P008, P041, P065, P081, P089, P115, P128, P165, P186, P192, P213, D025, D061, D092, D095, D118, D130, D145, D146, D158, D165, D168, R2.2, R2.4, R3.2, R6.9, R6.14, R7.6, R8.11

P103, P163, P166, P178, P182, P195, P204, P206, P208, P217, P221, P227, D083, D123, D143, D157, R2.11, R3.7, R5.8, R6.10, R8.15

Combinatorics P035, D003 P012, P158, R7.15, R9.7 T004, P105, D107 T007, P067, P135, P174, P214

T011, P116, P121, P191, P229, R2.5, R6.8, R7.8, R9.4, R10.5

Exponents & Roots

T017, P015, P046, P073, P164

P011, P051, P054, P112, P114, P133

T044, P032, P104, P107, D030, D169, R4.8

T033, P137, P216, P226, P230, D151, D156, R2.14, R3.2, R3.6, R6.9, R8.6, R9.1, R9.11

T015, T016, P108, P110, P117, P208, D154, R5.13, R6.13, R8.9, R8.15, R10.6

FDPs & Ratios P001, P006, P010, P015, P017, P019, P020, P024, P037, P043, P045, P052, P055, P060, P063, P095, P118, P169, D007, D009, D022, D039, D046, D064

T021, P047, P056, P061, P075, P076, P078, P094, P111, P151, P156, P162, P187, P218, D031, D032, D043, D150

T008, T040, P013, P014, P016, P028, P074, P079, P084, P092, P096, P105, P109, P124, P139, P175, P179, P225, D038, D044, D063, D085, D111, D142, D174, R1.1, R5.4

T012, P008, P081, P098, P115, P128, P131, P150, P181, P186, P192, P201, P202, P220, P223, D027, D055, D058, D061, D079, D092, D103, D155, R1.2, R2.2, R4.10, R4.11, R8.10, R9.2, R10.9

T046, P103, P138, P163, P166, D078, D120, D143, D163, R1.5, R1.8, R3.3, R3.5, R3.13, R4.6, R5.13, R6.12, R6.13, R6.15, R8.3, R8.5, R8.9, R9.9, R10.14

Functions & Sequences

P143, D019, D100 P042, P171, P203, D037 P068, P136, P188, D026, D115

T003, P087, P146, P157, P190, P228, D045, D125, R2.7, R4.14, R5.3, R5.15, R7.5

R5.14, R8.9, R8.14

Coordinate Geometry

P009, P025 P039, P205 P194 T039, P088, P210, D094, R6.2, R6.3, R6.5

P229, D121, D149, D164, R1.15, R3.4, R5.11, R7.11, R7.14, R10.13


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200-400 400-500 500-600 600-700 700+

Geometry T036, P004, P018, P033, D074

T020, T028, P053, D005, D018, D029, D047, D056, D096, D148

T022, P016, P048, P062, P102, P113, P147, D020, D034, D042, D091, D114, D140, D144, R3.15, R4.4, R4.13, R9.5

T005, T019, T048, P145, P152, P177, D122, D132, D135, D160, R1.4, R5.7, R8.8

P189, P197, P209, P212, D109, D117, D157, D173, R1.5, R1.12, R2.12, R3.1, R4.1, R4.7, R5.5, R5.8, R7.2, R8.5, R8.7, R8.13, R9.3, R9.10, R9.14, R10.2, R10.15

Inequalities & Absolute Values

P049, P071, P119, D001, D011

P050, P114, P141, D049, D080, D159

T037, P104, P129, D051, D072, D087, D091, D097

T030, T033, T038, T041, P125, P130, P173, D069, D079, D146, D152, D153, D156, D162, D167, R2.7, R2.13, R3.9, R3.14, R4.15, R5.10, R6.2, R6.4, R6.5, R7.12, R7.15, R8.2, R10.7, R10.11

D121, D154, R1.13, R2.3, R4.5, R6.12, R7.10, R8.12, R9.8, R10.3, R10.6, R10.10, R10.13

Number Properties

T025, T026, P003, P007, P022, P040, P073, D006, D012, D017, D106

T018, T023, P023, P029, P036, P159, P185, D024, D065, D076, D167, D171

T042, P072, P082, P219, D016, D041, D093, D110, D169, R3.10, R7.4

P125, P142, P173, P176, P190, P198, P224, D066, D073, D082, D090, D095, D133, D139, D153, D167, D172, R1.6, R1.11, R2.10, R2.15, R3.9, R3.14, R6.1, R7.12, R8.2, R8.4, R10.8

T013, P106, P110, P148, P149, P161, P182, P211, P217, D098, D128, D170, R1.9, R3.7, R3.11, R3.12, R4.2, R5.1, R5.6, R5.12, R7.10, R8.1, R8.12, R9.13, R10.4

Rates & Work D104 D010 P021, P031, P034, P080, P086, P126, P154, P183, D014, D087, D112

T038, P101, D102, D103, R5.10, R7.13

T024, P206, D068, R1.14, R3.8, R4.9, R5.8, R6.6, R7.7, R8.3, R9.9, R9.15, R10.1, R10.3, R10.12

Sets & Groups D004 T029, P124, P193, P200, D002, D021, D050, D059

T006, T034, T014, P167, D067, D089, D113, D116, D124, D127, R1.7, R2.8, R2.9, R10.8

T047, P178, P221, D126, R2.1, R6.11

Statistics P005, P030, P093 T009, T043, P027, P069, P099, P184, P028, D084, D096, D108, D134, D147

T002, T032, P014, P057, P132, P180, P199, D081, D129, D136

T031, T045, P207, D099, D105, R1.3, R1.10, R2.2, R7.1, R7.3, R7.5, R9.6

T046, D161, R1.13, R4.3, R4.12, R5.2, R5.9, R6.7, R6.15

Weighted Averages

D059, D101, D141 D055, R9.6 D078, R4.3, R4.12, R5.9, R9.12


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Topic Frequency The following is the percentages of questions from The Official Guide that test each topic. Since some questions test multiple topics, the total is greater than 100%. This distribution is roughly mirrored in the official GMAT exam, so use this graph to determine which questions are most common and therefore which question types to prioritize in your prep program.

1%

6%

5%

4%

12%

7%

3%

9%

4%

19%

6%

3%

21%

Weighted Averages

Statistics

Sets & Groups

Rates & Work

Number Properties

Inequalities & Absolute Values

Geometry, Coordinate

Geometry

Functions & Sequences

FDPs & Ratios

Exponents & Roots

Combinatorics

Algebraic Translations & Manipulations


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Scores to Percentiles Conversion Scales

Verbal Quantitative Scaled Score

Percentile Scaled Score

Percentile

46-60 99 51-60 99 45 98 50 94 43 96 49 88 42 95 48 84 41 92 47 79 40 89 46 77 39 87 45 75 38 83 44 70 37 81 43 68 36 78 42 63 35 74 41 61 34 69 40 59 33 66 39 55 32 64 38 53 31 58 37 51 30 56 36 46 29 53 35 42 28 48 34 40 27 43 33 38 26 40 32 34 25 36 31 30 24 33 30 29 23 29 29 25 22 27 28 20 21 23 27 19 20 19 26 19 19 16 25 16 18 15 24 15 17 12 23 13 16 10 22 11 15 8 21 10 14 7 20 9 13 5 19 8 12 4 18 7 11 3 17 5 10 2 16 4 7-9 1 15 4 0-6 0 14 4

13 3 11-12 2 7-10 1 0-6 0

GMAT Score

Percentile GMAT Score

Percentile

760-800 99 500 36 750 98 490 34 740 97 480 31 730 96 470 29 720 95 460 25 710 92 450 23 700 90 440 21 690 88 430 19 680 86 420 17 670 85 410 15 660 83 400 13 650 80 390 12 640 76 380 10 630 75 370 9 620 72 360 8 610 69 350 7 600 66 340 6 590 63 330 5 580 60 320 4 570 58 310 3 560 54 300 3 550 51 290 3 540 48 280 2 530 44 270 2 520 42 230-250 1 510 39 200-220 0

AWA Score Percentile

6.0 87 5.5 80 5.0 55 4.5 42 4.0 21 3.5 13 3.0 6 2.5 4 2.0 3 1.5 3 1.0 3 0.5 3 0.0 0

the official guide companion 12th [2010.11.17]

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