1

APTITUDE 1

[Aptitude] Alligation: Advanced applications in Interest rates, Profit-loss, Average Wages (Wine-Water Concept)

1. Case: Loan taken from two banks2. Case: Profit on Single commodity3. Case: No. of Employees based on Avg Salary4. Case: Profit on Two commodities5. Summary6. Mock Questions

Consider these four questions

1. Dr.Haathi has taken some loan from State bank of India at 8% interest rate per annum.He has also borrowed some money from ICICI at 10% interest.The total loan amount is Rs.10,000. He pays total Rs.950 as interest rate at the end of first year.How much money did he borrow from the State bank of India?

2. Jethalal purchased some mobile phone memory cards for Rs.1000. He sold memory cards worth rupees worth Rs.400 at 10% profit.At what percent profit should he sell the remaining memory cards, to gain overall profit of 20%?

3. Roshan Singh Sodhi is the owner of a big hotel. The average salary of all employees is Rs.8500. The average salary of 7 cooks is Rs.10,000.If the average salary of remaining employees, is Rs. 7800. Find the total number of people employed by Roshan Singh Sodhi.

4. Bhide master bought a set of laptop and printer for Rs.20000.He sold the laptop for 20% profit and printer for 10% loss.Overall, he made 2% profit. Find the cost price of laptop.

Superficially, ^these questions seem different from one another. But if you pay close attention, these are all problems of weighted average.

We can quickly solve them using the allegation (wine and water mixture) method.

Case #1: Loan taken from two banks

Dr.Haathi has taken some loan from SBI at 8% interest and ICICI at 10% interest.The total loan amount is Rs.10,000. He pays total Rs.950 as interest rate at the end of first year. How much money did he borrow from the SBI

The question is mentioning two interest rates: 8% and 10%the effective interest rate, that Dr Haathi has to pay, will be in between these two numbers 8 and 10.The question also says that he paid Rs.950 as interest rate.So in terms of percentage he has paid(950/10000) x 100 =9.5% interest rate (overall).

Arrange these numbers in to ascending (increasing) order

Interest rate% 8 9.5 10

BankSBI

Overall

ICICI

Now apply the “visual move” of Wine and water mixture

2

APTITUDE 1

10-9.5=SBI …eq 1Similarly9.5-8=ICICI….eq 2 (donot make mistake of 8-9.5 because we donot want negative difference)

Divide eq. 1 with eq 2

If you simplify this further, you get

SBI / ICICI =1/3

Apply Componendo at bottom

But we know that SBI + ICICI=Rs.10,000 that Dr.Haathi has borrowed!

Plug in the valueSBI /10000 = ¼Therefore, SBI =10,000/4=Rs.2500It means Dr.Haathi has borrowed Rs.2500 from SBI Bank.

Verifying the answer and concept

Amount borrowed Interest rate

SBI 2500 8% of 2500=200ICICI 10k-2500=7500 10% of 7500=750Total 8500 200+750=Rs.950 paid in interest = matches the question Description!

0

Case: Profit on Single commodity

Jethalal purchased some mobile phone memory cards for Rs.1000. He sold memory cards worth rupees worth Rs.400 at 10% profit. At what percent profit should he sell the remaining memory cards, to gain overall profit of 20%?

He already made 10% profit on part of his card.If he wants to have overall (average) profit of 20% then he must sell the remaining cards at a profit higher than 20%The question says he sold cards worth Rs.400 (cost price). Meaning the cost price of remaining cards is 1000-400=Rs.600Draw a table.

Profit10% 20% M%

Overall

Cards worth (cost price)

400 1000-400=600

3

APTITUDE 1

Apply the visual move of wine-water mixtureM-20=400….eq120-10=600….eq2 (donot make mistake of 10-20 because we don’t want negative)Now Divide equation 1 with equation 2

Meaning, if Jethalal wants to make overall profit of 20%, he must sell the remaining memory cards @26 (2/3)% interest rate.

Verifying the answer

Cards already sold @10% interest 10% x 400=Rs.40

Remaining cards sell @26 (2/3) 26 (2/3)% x 600=Rs.160

Total profitRs 40 +160 =200. And 200 is 20% of Rs.1000 so it matches our question condition.

Case: No. of Employees based on Avg Salary

Roshan Singh Sodhi is the owner of a big hotel. The average salary of all employees is Rs.8500. The average salary of 7 cooks is Rs.10,000. If the average salary of remaining employees, is Rs. 7800. Find the total number of people employed by

Sodhi.

The question is talking about three sets of salaries7800 > 8500 (average) >10000Suppose total number of employees =m.And we already know there are 7 cooks.So remaining employees = total minus 7 = m-7.

Draw the table.

Salary (Rs.) 7800 8500 10000

Overall

Employees (total =m) m-7 7

cooks

Apply the visual move10,000 minus 8500 = (m-7)…eq18500 minus 7800=7…eq2Now Divide equation 1 with equation 2

When you simplify this further, you get m=22Therefore, Total number of employees is 22.

Verification

4

APTITUDE 1

Number Avg. salary

Total cost to Sodhi

Cooks 7 10000 7 x 10k=70kOthers

15 7800 1,17,000

Total 15+7=22 – 70k+1,17,000=1,87,00

0

Avg salary of all employees=(Total salary paid)/(Total number of employees)=187000/22=Rs.8500 = matches the question condition.

Case: Profit on Two commodities

Bhide master bought a set of laptop and printer for Rs.20000. He sold the laptop for 20% profit and printer for 10% loss. Overall, he made 2% profit. Find the cost price of laptop.

Approach

We’ll take 10% loss =-10%Now the question is talking about three rates : -10% , +2% (avg) and +20%

Profit/loss -10% 2% 20%

Overall

Cost price Printer

Laptop

That’s it. Just apply the visual move BUT with care!20 minus 2=printer…eq 1

Be Careful while approaching second equation2 minus (-10)=laptop2-(-10)=laptopHence 2+10=laptop …eq2

Rest is easy. Just divide eq.1 with eq.2

When you simplify it further, you get the ratioprinter/laptop=3/2we’re interested in the Cost price of laptop. So we’ll apply componendo @top

But question already said hat total investment (cost price) of printer +laptop =Rs.20000Plug in the value20000/laptop=5/2Therefore cost price of laptop=(2/5) x 20000=Rs.8000

5

APTITUDE 1

Verification

Cost price Profit/lossLaptop

8,000 20% profit on 8k=1600

Printer

20,000-8000=12,000 10% loss on 12k=Minus 1200

Total 20,001600-1200=+400 Rs. as profit

So overall he received Rs.400 profit on the investment of Rs.20,000=(400/20000) x 100=2% profit= matches the condition given in question statement.

Summary

Weighted average method works when we’ve 3 rates (including one average) and 2 quantities. First arrange the rates in ascending order. Then draw the table with corresponding rates and quantities. Apply “visual” move and get the answer.

Mock Questions

Case: Journalist Popatlal took some loan from Matkaa-King Mohanlal @36% interest rate and some loan from Taarak Mehta @10% interest rate. Total money borrowed Rs.6000. At the end of first year, Popat paid total interest of Rs.912. Answer following questions

1. Total amount borrowed2. Money borrowed from Mehta3. Money borrowed from Mohan4. Effective interest rate on total borrowing?

Case: Jethalal bought a chair+table set @Rs.6000. Later he sold Chair @10% loss and table @10 profit. Overall he gained Rs.100. Answer following questions

5. Cost price of Chair6. Cost price of Table7. Selling price of Chair8. Selling Price of Table

Case: In Tappu’s class, the average weight of boys and girls is 77.5kg, 70kg respectively. Overall average weight of all students is 74 kg.

9. Find Ratio of Boys: girls10. If total number of students is 30, how many boys study in Tappu’s class?11. At present, total number of students is 60. If school administration wants to have equal number of boys

and girls in this class, how many new girls should be admitted?

6

APTITUDE 1

Case: In a state, total number of Desi-liquor dens and Land mafias is 100. They pay total Rs.1650 crores bribe to home minister every year. Average bribe paid by each Desi-liquor den and land mafia is 15 crores and 20 crores respectively.

12. Find total number of land mafias in this state13. Final total bribe paid by these land mafias14. Owing to media-pressure, the home minister orders police to close down some of the desi-liquor dens.

Now his total bribe income declined by 20%. What is the new ratio of liquor den: land mafias?

Case: A parking lot has only cars and bikes. Average number of tires per vehicle is 2.9. There are 55 bikes in this parking plot.

15. Find total number of cars16. If 5 cars are added and 5 bikes are removed, what’ll be the new average number of tires per vehicle?

Answers

Popat in debt

1. Total borrowed Rs. 60002. 4800 from Mehta3. 1200 from Mohanlal.4. 15.2% interest rate (overall)

Jetha

5. Cost price Chair =25006. CP Table= 3500.7. Selling price of Table (+10% profit)=1.1 x 3500=38508. Selling price of Chair (-10% loss)=0.9 x2500=2250

School

9. Boys/Girls=8/710. 16 boys and 14 girls.11. 4 new girls need to be admitted.

Home minister

12. Avg. annual bribe is 1650/100=16.5 Now the rates are 15>>16.5>>20. Ratio of liquor-dens:mafia=7:3. Since total are 100, number of Mafia=3/10 x 100=30.

13. 30 mafia x avg 20 crores=600 crores paid by them.14. 20% of 1650=330. And 330/15=22 liquor dens were closed down. So new ratio is (70-

22):30=48:30=8/5

Parking lot15. Number of cars=45.16. New avg. no. of tires per vehicle=3

7

APTITUDE 1

[Aptitude] Averages: Entry / Removal of 1 element => Increase / Decrease in Average Shortcut method explained

1. Situation: Entry of Element-Increase in Avg

1. Case: Avg age of Students + Teacher

1. Approach #1: convert into money problem2. Approach #2: Formula method

2. Case: Avg salary workers+manager3. Case: Avg weight students+teachers4. Case: original average not given

2. Situation: Entry of 1 element – Decrease in Avg

1. Case: younger Kid joins: average goes down2. Case: Average Runs decline (original not given)

3. Situation: Removal of 1 element -increase in avg4. Situation: Removal of 1 element -decrease in avg5. Mock questions

Situation: Entry of Element-Increase in Avg

Let’s start with really simple case. In Gokuldham society, the average age of four kids is 14 years. If Mr. Champak is included in this group, the average age increases by 11 years. How old is Mr.Champak?

Convert this into money problem

Four kids got Rs.14 each. Champak makes a new gang with rule, “everyone must have Rs.25 (bcoz 14+11=25)” But those kids got only Rs.14. so how much cash should Champak bring, to form this new gang?

Let’s think about it

8

APTITUDE 1

Those kids got only Rs.14 each and want to join the new group. But each of them would need Rs.11 extra. So Champak must give them Rs. 11 x 4 kid = Rs.44 And Champak also wants to be a part of this gang, so he should also bring Rs.25 for himself. Therefore, Champak must bring Rs. 44 (to give to those kids) + 25 (for himself) =44+25=69 Hence final answer: age of Champak is 69 years.

Observe how numbers are moving while the average changes.

Case: Avg age of Students + Teacher

Average age of 36 students is 14 years. If teacher’s age is included, then average increases by one. Find the age of teacher. (RBI)

We can solve it by two approaches

1. Convert into Money problem (easy, quick)2. Formula method. (lengthy, time consuming)

Approach #1: convert into money problem

Whenever you face such question on average, try to convert it into a money problem, then it becomes very easy to visualize.

Converting into money problem

36 kids had Rs.14 each 1 teacher comes. New average is 15 (original 14+ increases by 1) How much cash does teacher have?

Ok let’s think about it.

There is a gang of 36 kids. And it has rule: everyone must have Rs.14. Teacher wants to form a new gang. And it has rule: everyone must have Rs.15. But those kids have only Rs.14 each. So teacher must give Rs.1 to each of those 36 kids. Meaning teacher must bring Rs.1 x 36 =Rs.36 And teacher should keep Rs.15 aside for himself. (because new gang rule is “everyone must have Rs.15) So how much cash did teacher bring in? =Rs.36 (to give to kids) PLUS Rs.15 (for himself) =36+15=Rs.51 that’s our final answer. Age of Teacher is 51 years.

9

APTITUDE 1

Approach #2: Formula method

Formula for average: average = sum of observation divided by number of observation

Situation 1: without teacher

36=(sum of kids)/14Therefore, sum of kids(s)=36 x 14S=504

Situation 2: with teacher

37 =((sum of kids)+age of teacher))/15Therefore, S + teacher=37 x 15Now plugin the value of s=504 from previous equation504+teacher =555Teacher =555-504=51 that’s our final answer.

As you can see, the problem with this “formula” method = you’ve to do lengthy multiplications (37 x 15 and 36 x 14). Thus

1. it breaks your momentum, decreases your speed in the exam.2. you might make some silly mistake in multiplications or during “plugging” the values in equations.

So better just try to solve such questions using “money problem” method.

Case: Avg salary workers+manager

The average salary of 20 workers is Rs.1500. If manager’s salary is added, this average increases by Rs.100. Find the salary of Manager.(RRB)

It is already in “money problem” format.

20 guys have Rs.1500 each Now Manager wants to increase this average by Rs.100. (new avg =1500+100=1600) So manager must give Rs. 100 to each of these 20 guys= manager must bring total 100 x 20 =Rs.2000

with him. But this new group has a rule “everybody must have Rs.1600”. Hence, If manager wants to be part of this

“new” group, he must keep Rs.1600 aside for himself. So, how much cash should manager bring? =Rs.2000 (to give to workers) + Rs.1600 for himself =2000+1600 =3600 this is the final answer.

Case: Avg weight students+teachers

There are 24 kids in a class with average weight of 35 kilograms. If Teacher’s weight included, the average increases by 400 grams. Find weight of teacher.

Convert this into money problem

10

APTITUDE 1

1. 24 kids have 35 rupees each2. Teacher wants to form new gang: “everyone must have 35.4 rupees each” (because 400 gms =400/1000

kg =0.4 kg)

Ok let’s think about it

Teacher must give 0.4 rupees to each of those 24 kids=24 x 0.4 =9.6 And he must bring Rs.35.4 for himself. so how much cash should teacher bring in? =9.6 (for kids) + 35.4 (for himself) =9.6+35.4 =45. That’s our final answer: weight of teacher is 45 kilos.

Case: original average not given

In his 11th inning, a batsman scored 216 runs and his average increased by 12 runs. What’s his new average? (Bank)

Money problem

1. 10 guys had Rs.”m’ each2. New guy (11th) has Rs.216 and he forms a new gang.3. In this new gang of 11 guys, each person has Rs.(m+12) each. Find value of m+12.

Let’s think about it:

New guy wanted to form a new gang where everyone had Rs.m+12. But those 10 guys from old gang, had only Rs.m That means new guy should give each of those 10 people Rs.12 So new guy gave away Rs.12 x 10 =Rs.120 So new guy is left with Rs. 216 (he originally had) MINUS Rs.120 (that he gave away) 216-120=Rs.96 that’s our answer.

so far we saw “Average: Entry of 1 Element=>leading to Increase in Average”.

Now let’s check “Average: Entry of 1 Element => leading to decrease in average”

Situation: Entry of 1 element – Decrease in Avg

Case: younger Kid joins: average goes down

The average age of six kids is 12 years. A new kid joins the gang and average decreased by one year. What is the age of this new kid?

Money problem

Six kids got 12 rupees each. New kid wants to form a new gang with rule “eachone must have only Rs.11”

11

APTITUDE 1

Ok let’s think about it

Since average is declining, we are sure that new kid has less money than Rs.11. But he wants everyone to have (12-1)=11 rupees. He tells those 6 kids, “each of you give me 1 rupees.” = 6 x 1 = Rs.6 borrowed. So how much money did that new kid bring? Well, Now he is part of new gang so obviously he has Rs.11 . but he has borrowed Rs.6 from others. Meaning, he originally had 11-6 =Rs.5 Final answer: age of new kid is 5 years.

Let’s try a case where one element is removed.

Case: Average Runs decline (original not given)

A cricketer made 90 runs in his 12th inning, and his average declined by 5 runs. Find new average.

Money problem

11 guys had average Rs.m each 12th guy has Rs.90 and he makes new gang, with rule “everyone must have Rs.(m-5)”

Let’s think about it

Since average is declining, we can be sure that new guy has less money than previous average (“m”). But he wants everyone to have (m-5) rupees (he should also have m minus 5 rupees.) So he tells those 11 people, give 5 rupees each to me. = 11 x 5 = Rs.55 How much cash does the new guy have now?= 90 (he originally had) + Rs.55 (that he

borrowed)=Rs.90+55=145 that’s our new average. Final answer: new average runs after 12 innings is 145.

Situation: Removal of 1 element -increase in avg

A batsman played 30 innings with an average of 78 runs.

If we ignore the score of his last innings, then average increases by 2 runs. How many runs did he score in last inning?

Money problem

30 guys with Rs.78 each 1 guy (Jethalal) leaves this gang after settling his debts. Average of remaining 29 guys is (78+2)=80 runs.

Ok let’s proceed

Since average is increases after member leaves the gang= this member (Jetha) had less money than others.

That means other people had lended him some money (to raise him to the average of 78.) When Jethalal leaves the gang, remaining guys get 2 rupees each.

12

APTITUDE 1

That means Jetha gave them Rs. 2 to each of those 29 remaining guys.=2 x 29= Rs.58 And once upon a time, Jethalal was part of this gang, so he had Rs.78. It means Jetha’s original money = Rs.78 minus Rs.58 (that he borrowed)=78-58=Rs.20 Final answer: batsman’s score in last inning was 20 runs.

Situation: Removal of 1 element -decrease in avg

Four members of a family earn average Rs.7350. One of them, leaves and average goes down to Rs.6500. How much did that person earn, who has now left the family? (SSC)

Money problem in our format:

1. 4 guys with Rs.7350 each2. One of them (known as Jethalal) steals money from others and leaves.3. Now 3 guys left and they’ve Rs.6500 each

Ok let’s think about it

Jetha wants to bring down average from 7350 to 6500 =decline of 850. So he must steal Rs.850 from each of the remaining three guys =Rs.850 x 3=2550. And since Jetha was once part of this “gang”, he too used to earn Rs.7350 himself. Therefore total cash with Jethalal =7350 (he originally had) + 2550 (that he stole) =Rs.9900.

Final answer: the income of person who left family, is Rs.9900

Mock questions

1. Average age of 30 students is 11 .When teacher’s age is added, average increases by 1 year. Find the age of teacher.

2. Average of 24 wrestlers is 90. A new wrestler joins the gym and average increases by 2 kilos. Find weight of new wrestler.

3. Average annual income of 25 families in a society is 1.9 lakhs. A new family comes to live in this society and the new average annual income of the whole society is increased by 10,000 rupees. Find the monthly income of this new family.

4. Average salary of 21 workers+1 manager is Rs.2000. When manager’s income is not considered, the average declines by 25%. How much does Manager earn?

5. A batsman makes average 50 runs in 36 matches. After 37th match, his average increased by 1 run. How many runs did he score in last match?

6. A batsman made 102 runs in 21st inning and his average increased by 2 runs. Find total runs scored by him in first 20 matches.

7. A batsman scored 199 runs in his last match and his average increased from 44 to 49. Find total number of matches played by him.

8. A thief steals average 2 cars per night for 13 nights continuously. But after 14th night’s “adventure”, his average increases to 3 cars. How many cars did he steal in that last night?

9. Average salary of 21 workers is 2000. A new worker joins the company and overall average declines by 2.5%. What is the salary of new worker?

10. A batsman made average 80 runs in 29 matches. After 30th match, his average declined by 2 runs. How many runs did he score in that last match?

13

APTITUDE 1

11. In a dance troupe, average weight of 20 junior dancers is 60 kilos. If the weight of lead dancer is also considered, the average declines by 500 gms. How much does the lead dancer weigh?

12. In a society, average annual income of 30 families is 9 lakhs. One new family starts to live here and average declines to 8.8 lakhs. Find the annual income of this new family.

Answers

1. 422. 1403. 4.5 lakhs annually / 12 = 37500 monthly income.4. Decline of Rs.500. So manager had Rs.500 x 21 + 2000= Rs.125005. 87 runs.6. 20 x 60 = 1200 runs.7. 30 matches. he kept 49 for himself and gave 5 rupees each to “m” number of people. 199=49+5m. so

m=30 matches.8. 16 cars. (Rs.3 for himself + 1 x 13 donated=16 cars stolen last night)9. 2.5% decline= 50 rupees on 2000. so 21 workers gave him 21 x 50 =1050. and he is part of new gang= he

has 1950 minus 1050 borrowed =His salary is 900.10. 20 runs.11. 49.5 kilos.12. 2.8 lakh.

14

APTITUDE 1

[Aptitude] Averages: More than two elements added / removed : shortcut technique explained

In the previous article, we saw how to approach the concepts for “Averages-> Entry/removal of One element”. (click me) Now extending that concept for Entry/removal of multiple elements.

1. Situation: Removing 2 Elements

1. Case: highest score in (40/38) innings? (SSC CGL)2. Case: Highest marks in (10 / 8) papers (CAT-1997)3. Case: average price of (12/10) books (SSC CGL)

2. Situation: More than Two elements

1. Approach #1: Money problem2. Approach #2: allegation method3. Case: Average run rate (10/40) overs (MAT)4. Case: Avg age of boys (18/30)

1. Approach: money problem2. Approach: alligation

3. Mock questions

Situation: Removing 2 Elements

Case: highest score in (40/38) innings? (SSC CGL)

A Batsman played 40 innings with average score of 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, then score average score of his remaining innings is 48 runs. What is the highest score of this batsman? (SSC CGL 2011)Convert this into a Money problem

1. Old gang has 40 guys and each of them got Rs.502. Two guys leave the gang. One of them has Rs.172 more than the other guy.3. Now Average of the old gang declined by 50-48=Rs.2

Ok now let’s think about it.

Initially everyone has Rs.50 each. So two guys who left the game, had total Rs.50 x 2 =Rs.100.

15

APTITUDE 1

We are left with 40-2=38 people. Two guys left the group. If they wanted to decrease the avg. of those 38 people by Rs.2. So they must

have got away with 38 x 2 =Rs.76 So the two guys had total cash of Rs.100 (theirs) + Rs.76 (they stole from others) =100+76=176. But the question says difference between highest and lowest is 172. And now we know that total is 176. Meaning lowest was 2 and highest was 174. (because 2 + 174 =176)** Therefore, final answer=174 was the highest score.

**in case you’re confused:

Suppose the two observations were “a” and “b” where (a>b).i) total of those two observations is 176 (we found this via “money problem” approach): a + b =176ii) difference between them is 172 (already given in question): a-b =172

Solve these two equations and you get: 2a=348=> a=174

A similar question was asked in CAT-1997

Case: Highest marks in (10 / 8) papers (CAT-1997)

The average marks of a student in 10 papers are 80. If highest and lowest scores are not considered, then the average score is 81. If the highest score is 92, then what is the lowest score?Convert this into a Money problem.

1. 10 kids have 80 rupees each.2. 2 kids run away, so 8 kids remain. (10=8-2)3. Average of remaining 8 kids is 81. (Meaning average increased by 1 (81-80=1))

Ok, now let’s think about it.

10 kids had 80 rupees each. Meaning those two kids also had 80 each So two kids originally had 80 x 2 =160 rupees. They wanted to raise the average of remaining eight kids by 1. So they (two kids) gave 1 rupee to each of those eight remaining kids. Meaning two kids gave away Rs.1 x 8kids = 8 rupees. So now two kids are left with 160 minus 8 =152 rupees. But the question says highest of those kids was 92. So the other kid had 152 minus 92 =Rs.60 that’s our final answer.

Case: average price of (12/10) books (SSC CGL)

Average price of 10 books is Rs.12. Average price of 8 books is Rs.11.75.Of the remaining two books, if one book has price 60% higher than the other book, find the price of both books.Convert this into a Money problem. (although it is already a money problem but ‘men’ are not included).

1. 10 guys had 12 rupees each.2. 2 guys left. Average goes down by 25 paisa. (12 minus 11.75=0.25)

16

APTITUDE 1

Ok now let’s think about it.

Originally two guys had total 2 x 12 = 24 rupees. They wanted to bring down the average of remaining eight guys by 0.25 paisa. So two guys stole 25 paisa each from the remaining eight people. Thus two guys stole: 8 x 0.25 =Rs.2 Now two guys have total rupees = Rs.24 (they originally had) PLUS Rs. 2 (they stole) =24+2=26 rupees.

But the question says, price of one book is 60% higher than the other one.Meaning cheap book is Rs”m” then expensive book is “1.6 times m”. **

**side concept

Expensive book is 60% higher than cheap book.Therefore, price of expensive book= 100% (of cheap book) + 60% of cheap book=160% of cheap book=1.6 of cheap book. (because %=1 upon 100 so 160%=160/100=1.6)=1.6mAnyways back to the main process: We found that total is Rs.26Therefore Rs. 26 = Cheap book + expensive book=1.0m + 1.6m=2.6mIf 26=2.6m, then m=26/2.6=Rs.10And if m=10 then expensive book=1.6 x 10 = Rs.16.So final answer isPrice of those two books are Rs.10 and Rs.16I hope the concept of “Averages: Entry/Removal of Two elements” is clear by now.Now let’s check “more than two elements.”

Situation: More than Two elements

Raynolds produces average 4000 pens per month for the first three months. How many pens should this company produce on an average for remaining months to get yearly production average of 4375?There are two ways to solve this problem

1. Convert in Money problem2. Wine and Water mixture (alligation method)

Approach #1: Money problem

Convert this into a money problem

12 guys have avg. Rs.4375 each (because 1 years= 12 months) 9 guys leave the gang. (because we are given avg of 3 months=12-3=9) 3 guys remain and their average is Rs.4000 each. (meaning average declined by 4375 minus 4000=375) Find the average money held by those 9 guys.

Ok now let’s think about it.

17

APTITUDE 1

12 guys have avg Rs.4375 each. Meaning originally those 9 guys had total Rs. 4375 x 9 They want to decrease the average of remaining 3 guys by Rs.375. So they steal Rs. 375 x 3 rupees from those three guys. Now total money with 9 guys = 4375 x 9 (they originally had) + 375 x 3 (they stole)

But the question is asking us to find the average money held by these 9 guys=total money with 9 guys divided by 9=(4375×9)+(375×3)/9=Rs.4500

Even more shortcut!

Recall what we do with “Average: errors” concept.New average = original average plus or minus average change.In this case original average of 9 guys =Rs.4375And they stole Rs.375 x 3 (=increase)So what’s the average increase= 375 x 3 / 9 =Rs.+125 (this is plus because they gained money)And new average = original average (Rs.4375) + average change (+125)=4375+125=Rs.4500

Approach #2: allegation method

If you don’t know the allegation (visual move) technique, then go through my old article (click me)Company wants average yearly production of 4375 (this is our “middle” value)For 3 months , they had average of 4000That means for the remaining 9 months, average production must be higher than 4375. (let’s assume this is “m”)Make the table.

Avg. production

4000

4375 M

Months 3 9

That’s it. Apply the visual move.m-4375=3….eq14375-4000=9…eq2

Divide eq1 by 2 and solve itYou get m=4500 that’s our final answer.

Case: Average run rate (10/40) overs (MAT)

In a cricket match, team Pakistan makes 281 runs. Now team India starts batting. In the first ten overs, we have (bogus) run rate of 3.2 only. If we want to win this match, what must be

the run rate for remaining 40 overs?

18

APTITUDE 1

I suggest you try the money problem method here on your own. Interestingly, we can solve this problem via alligiation (wine water mixture) as well! How? Since Pakistan made 281 runs, we must make 282 runs to win the game. And we want to do this in 50 overs (40+10) So what should be our “overall” runrate= 282/50=5.64 For the first ten overs we had run rate of only 3.2. so for remaining 40 overs, we must make run rate

higher than 5.64.

Run rate

3.2

5.64 M

Weight 1040

That’s it, just apply the visual move!M-5.64=10….eq15.64-3.2=40…eq2Divide equation 1 by equation 2

When you solve this equation, you get m=25/4=6.25Final answer: India must get run rate of 6.25 in the remaining 40 overs, to win the match.

Case: Avg age of boys (18/30)

A group of 30 boys have average age of 13. Later, 18 boys, with an average of 15 years left the group. What is the average age of remaining 12 boys?

Approach: money problem

1. 30 boys had Rs.13 each2. 18 of them left, with Rs.15 each.

Let’s think about it.

Originally those 18 boys had Rs.13 (when they were part of the gang) Meaning remaining 12 boys also had 12 x 13 = 156 rupees. Now 18 boys form a new gang with Rs.15 each. So, each of these 18 boys need Rs.2 more (15-13=2). So

total they need Rs.2 x 18 = Rs.36 They steal this money from those remaining 12 boys. So how much money is left with those 12 boys? = 156 they originally had MINUS Rs.36 stolen= Rs.120 And since we are asked “average” of those 12 boys So we’ll divide 120 by 12=Rs.10 is our final answer.

Approach: alligation

19

APTITUDE 1

I’m directly constructing the table

Avg age M 13

15

No of boys

12

18

Apply the visual move, you get two equations.

When you solve them, you get m=10.Final answer: average age of remaining 12 boys is 10 years.

Mock questions

1. A class has 35 people with avg weight of 62 kilos. When we donot consider the weight of teacher and class Secretary, the average goes down to 61.5 kilos. If the weight of class Secretary is 60 kilos, how much does the teacher weight?

2. A batsman made average 85 runs in 50 matches. When we donot consider his last two matches, average falls down to 82 runs. If he made 150 runs in his second last match, how many runs did he score in his last match?

3. Gokuldham Society has 30 families with average annual income of 8 lakh rupees. If we ignore the families of Jethalal and Bhide, the average declines by 1.25%. If Jetalal earns 740% higher than Bhide, what is the annual income of Bhide?

4. In a bookstore, there are 100 books with average price of Rs.300. Two of them are removed and average declines by 10 rupees. of the two books that were removed, if the expensive book cost Rs.1330, find the price of cheaper book.

5. In a mobile store, there are 200 handsets with an average MRP of Rs.8000. When we ignore a Nokia Lumia and a Samsung Galaxy note2, the average of remaining phones goes down to Rs.7850. If Nokia Lumia costs Rs.9999, find the price of that Samsung phone.

Answers

1. 80.5kg2. 164 runs3. 2 lakh rupees. 1.25% decline in average = 0.0125×8 lakhs=10,000 rupees. Meaning Jetha+Bhide had total

money of [10k x 28 + (2x8lakh)]=18.8lakhs. And Jetha+Bhide(B)=8.4B+1B=18.8lakhs. Hence Bhide(B)=2 lakhs.

4. Rs.2505. Rs.35701

20

APTITUDE 1

[Aptitude] Concepts of Marked Price and Successive Discounts (Profit-Loss) without (stupid) formulasProfit, Loss, Discount, Marked Price = all of them are mere extension of Percentage calculation concept. And they can be solved effortlessly, without mugging up any formulas.

1. What is Discount?2. What is successive discount?3. Case: Successive vs One Time Discount?4. What is marked price?5. Case: Finding Marked Price from Profit %6. Case: Finding marked price from Selling price7. Case: Finding % Profit / loss in MP-Discount8. MOCK TEST9. Answers with explanations and shortcuts

What is Discount?

In the garment stores, you’ve come across this sign many times: “HURRY! 20% OFF!!”That 20% is called discount.Suppose a Music CD has price tag of Rs.100 and the shopkeeper is offering 20% discount.That means 20% less on the price tag.The price you’ve to pay=Rs.100 minus 20% of 100=100-(20×100/100)=100-20=Rs. 80 you’ve to pay, to buy that music CD.But in exam, we don’t have no time to do such lengthy calculations.So just think in your head:20 percent discount means you’ve to pay 100 percent minus 20 percent=80 percent of the price tag.% means one upon hundred (1/100)So 80% means (80/100=0.8)So you’ve to pay 0.8 times the original price.=0.8 x 100=Rs. 80

What is successive discount?

It means discount on the discount. (just like Compound Interest rate= interest on interest)The original price of a music CD is Rs.100. A shopkeeper offers 10% discount on this music CD and then again offers 20% discount on the new price. How much will you have to pay, finally?

Phase I: 10% discount

21

APTITUDE 1

So new price = 0.9 x original price. (by the way, how did we get 0.9 ? well 10% discount means you’ve to pay 90%. And 90% means 0.9)

Phase II: 20% discount

=0.8 x new price=0.8 x 0.9 x original price.It means, after successive discount of 10% and 20%, the final price you’ve to pay=0.72 x original price.=0.72 x 100=Rs.72If it was successive discount of 20% and then 10%? Still answer will remain same. ORDER or sequence doesn’t matter.Let’s test our skill with an actual question from SSC/FCI Exam taken in Nov 2012.Q1. A dealer buys a table listed at Rs.1,500 and gets successive discounts of 20% and 10%. He spends Rs.20 on transportation and sells it at a profit of 20%. The selling price of the table is

A. Rs.1,320B. Rs.1,350C. Rs.1,360D. Rs.1,380

Think in your head:20% discount =0.8 x original price (this is our new price)10% successive discount= 0.9 x New price=0.9 x 0.8 x original priceFinally dealer had to pay: 0.72 x original price.Then he invested Rs.20 on transport, so his total investment (cost price) is(0.72 x original price) + Rs.20And ultimately he sold It @20% profit.20% profit means 100%+20%=120%=1.2 times the investment (cost price).=1.2 times [(0.72 x original price) + Rs.20]In the examhall,only write following line and do the math, don’t waste time writing everything or multiplying numbers @every stage.=1.2 times [(0.72 x 1500) + Rs.20]=1.2 [1080+20]=1.2 [1100]=Rs. 1320(I hope you know quick multiplication with 11, basically 12×11= in 12 you do:1|(1+2)|2=132)

Case: Successive vs One Time Discount?

Q. Jethalal, has a mobile phone with price tag Rs.12,399/-. He is offering two schemes to you

a. Buy it at one time discount of 30%b. Buy it at two successive discounts of 15% and 15%

Which of the above scheme is more beneficial to you (customer)?

Long cut (tiresome and boring calculation)

22

APTITUDE 1

Scheme A Sheme B

Discount offered=30% of 12,399=0.3 x 12,399=3719.7

Phase I15% of 12399=1859.85 Rs. (discount I)So new price : 12399 MINUS 1859.85=10539.15Phase II: (second discount)=15% of 10539.15=0.15 x 10539.15=1580.8725 (discount II)Total discount=1859.85 +1580.8725=3440.7225

We are getting more discount in Scheme A compared to Scheme B.So, from customer’s point of view, Scheme A is more beneficial.^Very lengthy and tiresome calculation isn’t it?Well, if you look at it carefully, the price tag Rs.12,399/- is irrelevant here.

Shortcut (almost none calculation)

Scheme A Scheme B

Discount offered=30%Final price you’ve to pay=0.7 times original price.

Successive discount 15% (two times)Final price you’ve to pay=0.85 x 0.85 x original price=0.7225 x original price

In Scheme A, you’ve to pay less price compared to Scheme B. Because 0.7 < 0.7225So, from customer’s point of view, Scheme A is more beneficial.Sidenote: you can do 85×85 with Vedic shortcut (square of numbers ending with five)Square of 85=8|5=(8x(8+1)|25=8*9|25=7225I hope discount and successive discount is clear. Moving to the next topic.

What is marked price?

Before that, answer this question:

Why do shopkeepers give discount?

Well it’s a marketing propaganda to seduce customers.I’m running a jeans store. I bought a pair of jean for Rs.100 (cost price).Now I put a new tag on it Rs.500 (marked price) and then put a huge poster outside my shop: “50% OFF on any item!”From buyer’s point of view it looks very attractive “wow 50% discount!!”But in reality, I’m making huge profit. Because 50% discount on Rs.500= You’ve to pay Rs.250, while I had

23

APTITUDE 1

invested only Rs.100! so my profit=(250-100)=Rs.150Marked price=this new price tag Rs.500. Discount is given on marked price. If we want to visualize it, here is the diagram:Mind it: Profit or loss is always calculated on actual investment (Cost price), unless they specifically instruct you do something else.Time to test the skill with actual questions

Case: Finding Marked Price from Profit %

Q. Jethalal bought a Nokia mobile phone @ Rs. 800. He sold it to Master Bhide after allowing a discount of 10%. In this deal, Jethalal made a profit of 12.5%. What was the marked price?

A. Rs. 1,000B. Rs. 1,100C. Rs. 1,200D. Rs. 1,300

Profit is always calculated on actual investment (cost price).Cost Price: Rs.800The question says, profit =12.5%. What is the profit in rupee terms?

Long cut Shortcut=12.5% x 800=(125/10)x (1/100) x 800=Rs. 100

12.5% =1/8 (percentage to fraction conversion table).So, 12.5% x800= (1/8) x 800= Rs. 100

Anyways in either method, profit is Rs.100…Fact (i)Jethalal gave 10% discount on marked price. So customer Bhide Master had to pay only90% of the Marked Price.=0.9 times the marked price=this is the money Jethalal received from customer.

Jethalal’s actual profit

=the money he received from customer MINUS his original investment.=0.9 times Marked price MINUS Rs.800But recall fact (i), Jethalal’s profit is Rs.100Therefore100=0.9 times Marked price – Rs.800100+800=0.9 times marked price900=0.9 x MPMP=900 x 10/9MP=Rs. 1000.Final Answer: marked price is Rs.1000

In the exam hall, directly frame the equation: 0.125 x800 =(0.9 xMP)-800Let’s try another one.

24

APTITUDE 1

Case: Finding marked price from Selling price

Q. Master Bhide, bought two pens for Rs.37.40 at a discount of 15%. What must be the marked price of each of pen?

A. Rs.11B. Rs.44C. Rs.33D. Rs.22

Discount of 15%, means Master Bhide (customer) paid (100-15)=85% of the marked price.=0.85 times the marked priceBut question also says he paid Rs.37.40, Therefore37.40=0.85 x MP.MP=3740/85=(374x 10)/ (17×5)=(374 x 2)/17=(22×2)=44. (Sidenote, if you don’t know the multiplication table of 17, then use approximation method explained in earlier article. CLICK ME)Most important part: Rs.44 is the marked price of two pens. So marked price of each pen =44/2=Rs.22Answer is (D)In the actual exam, rough paper calculation: =37.4/(0.85 x 2)

Case: Finding % Profit / loss in MP-Discount

Jethalal marks the price tags of all his mobile phones, 10% above the cost price. And then He offers 10% discount on their marked prices. What is his profit or loss?

A. 1% profitB. 1% lossC. 5% profitD. No profit No loss.

Assume his total investment is Rs.100 (=cost price). Well we can assume total investment as “CP” or “x” or “y” or anything but assuming Rs.100 saves the time of calculating percentages later on.Marked price=10% above cost priceSo price tag (marked price) is100%+10% of Cost price=110% of cost price=1.1 x cost price=1.1 x 100=Rs. 110Now he offers 10% discount on marked price, so customer has to pay only0.9 times the marked price=0.9 x 110=Rs.99Verdict: His investment is Rs.100 and he receives only Rs.99 from customers so he is making lossHow much loss?

25

APTITUDE 1

Long cut

Loss % formula =(100-99)/100 x 100=1%

Shortcut

We assumed Rs.100 initially for the very reason that to prevent long cut formula!Since he receives one rupee less (100-99), his loss is 1%.

In the actual exam, rough paper:

Invested Received

1000.9 x 1.1 x 100=99

So, loss is 1%.I hope the concept of marked price is clear by now. So it is time for a

MOCK TEST

Time: 20 MinutesCorrect =1mWrong= minus 0.33m1. When Jethalal offers discount on cost price, he makes

A. ProfitB. LossC. No Profit no LossD. Profit or loss, depending on situation.

2. Jethalal’s offering to sell an LCD TV priced @Rs.1,79,299/- under two offers

A. One time discount of 40%B. Successive Discount of 36% and 4%

Which offer is more beneficial from customer’s point of view?3. Jethalal offered a scheme: “two successive discounts 10% and then 10%”. Which of the following scheme will give same effect?

A. One time discount of 15%B. One time discount of 19%C. One time discount of 20%D. One time discount of 25%

4. Which of the following, is a better deal from customer’s point of view?

A. Three successive discounts of 10,20,30%B. Two successive discounts of 20,20%C. One time discount of 50%

5. Jethalal offers two deals

26

APTITUDE 1

1. Successive discounts: first 10% and then 20%2. Successive discounts: first 20% and then 10%.

Which of the following are true?

A. Customer will save more money in Deal 1B. Customer will save more money in Deal 1C. Customer will save same amount of money in both deals.

6. Jethalal fixes the sale price of his mobiles @15% above the cost price. He sells all his mobiles @12% less than the fixed price. What is his profit percentage?

A. 2.5%B. 1.2%C. 1.5%D. 2%

7. Jethalal sells his mobiles on 10% discount on marked price. He bought a mobile @Rs.900 and wants to get 10% profit. What should be the marked price?

A. 1275B. 1250C. 1175D. 1100

8. Jethalal wants to give 10% discount on all his mobile phones. But at the same time, he also wants to make 20% profit. At what percent above the cost price, should he mark the price of his phones?

A. 25%B. 30%C. 33.33%D. 37.5%

9. Consider these statementsI. Marked Price is always higher than Cost priceII. Marked price is never lower than Selling price.Which of above statements are correct?

a. Only 1b. Only 2c. Both 1 and 2d. None

10. How can one become Master of aptitude?

A. By searching for a competitive exam, that doesn’t have aptitude questions.B. By thinking like a loser, “My maths is not good and it can never be improved. I’ll just hope to clear some

exam with luck by chance.”C. By watching India-England Cricket test series because those cricketers will give money to start some

business, should you fail in the CSAT/IBPS/SSC/CMAT/CAT or some other stupid exam.

27

APTITUDE 1

D. By uploading motivational/funny photos on Facebook and tagging random friends in it.E. By practicing maximum questions at home.

Answers and Explanations

Q.Ans. Explanation

1 BDiscount on cost price=he’ll receive less money than his original investment=always loss.

2 A 0.6<0.64×0.96

3 B Successive: 0.9×0.9=81One time: 100-81=19

4 C

A. 0.9*0.8*0.7=0.504B. 0.8*0.8=0.64C. 0.5

Customer has to pay minimum in situation “C”.

5 C Order or sequence doesn’t matter in successive discounts.

6 B

Don’t be confused with vocabulary. Fixed price, marked price concept is same.1.15×0.88×100=101.2Profit % is 101.2 MINUS 100=1.2%

7 DProfit of 10% on 900=0.9 x Marked price MINUS 9000.1*900=0.9*M-900M=1100

8 C

Assume cost price=Rs.100Profit of 20% on 100=0.9 x Marked price MINUS 1000.2*100=0.9*m-100M=133.33So marked price is 133.33-100=33.33% above the cost price.

9 C Both statements are correct.10 ?? Lolz

28

APTITUDE 1

[Aptitude] Data Interpretation (DI): Bar and Pie chart practice Questions with Explanation

.Time: 30 mins. Correct Ans: 1 m, Wrong answer=-0.33 m.

1. DI Set#1: Pie Chart2. DI Set#2: Pie-Chart CO2 Emission3. DI Set #3: Bar graph on Profit-loss4. Solution Set#1: Pie Chart with Shortcuts5. Solution: Set #2: CO2 Emission Pie Chart6. Solution:Set #3 Bar Chart

DI Set#1: Pie Chart

The following pie chart cost of constructing one house. The total cost was Rs. 6 lakhs. (source FCI/SSC Nov. 2012 Paper)

1. The amount spent on cement is

A. Rs.2,00,000B. Rs.1,60,000C. Rs.1,20,000D. Rs.1,00,000

2. The amount spent on labour exceeds the amount spent on steel by

A. 5% of the total costB. 10% of the total costC. 12% of the total costD. 15% of the total cost

3. The amount spent on cement, steel and supervision is what percent of the total cost of construction?

4. The amount spent on labour exceeds the amount spent on supervision by

A. Rs.2,00,000B. Rs.16,000C. Rs.1,20,000D. Rs.60,000

5. The amount spent on Timber is

A. Rs.60,000B. Rs.90,000C. Rs.1,20,000

Rs.36,000

29

APTITUDE 1

A. 40%B. 45%C. 50%

55%

DI Set#2: Pie-Chart CO2 Emission

The total annual CO2 emissions from various sectors are 5 mmt. In the Pie Chart given below, the percentage contribution to CO2 emissions from various sectors is indicated.

1. Which of the following sectors together emit 2.5 mmt of CO2 every year?

a. Thermal and Transportb. Domestic and Commercialc. Transport and Commericiald. Commercial and Thermal

2. Which of the following sectors have emission difference of 1 mmt between them.

a. Domestic and Commercialb. Transport and Commercialc. Thermal and Domesticd. Thermal and Transport.

3. Emission of Domestic sector is how much % of Transport and Commercial sector combined?

a. 20%b. 15%c. 30%d. 35%

4. In the next year, if emission from Commercial sector decreases by 20% of its current emission, while other sectors continue to emit same amount of CO2 as earlier then What will be the new % contribution of Thermal power sector (approximatly)?

a. 31.9%b. 39.8%c. 42.7%d. 36.5%

5. Which of the following is incorrectly matched?

a) Domestic=54 degrees.b) Thermal=125 degrees.c) commercial =72 degrees.d) transport=108 degrees.

DI Set #3: Bar graph on Profit-loss

Observe this graph and answer questions (source FCI/SSC Nov. 2012 Paper)

1. The amount of maximum profit as seen from the 3. The ratio of the maximum profit earned to the minimum

30

APTITUDE 1

diagram is

A. 2 thousand rupeesB. 1.5 thousand rupeesC. 1 thousand rupeesD. 4.5 thousand rupees

2. The amount of total loss incurred during 2005-2010 is

A. 3 thousand rupeesB. 2.5 thousand rupeesC. 1.5 thousand rupees

2 thousand rupees

loss suffered is

A. 4 : 3B. 3 : 4C. 9 : 2D. 2 : 9

4. If the loss is x% of the profit for the years under study, then x is

A. 15B. 25C. 20

19

Solutions for the DI questions

Solution DI Set#1: Pie Chart with Explainations/Shortcuts

You might be tempted to transform degrees into percentages and find absolute Rupee value of each item (timber, labour etc). But in the exam, don’t waste time by calculating everything. Only calculate the stuff that is asked.

1. The amount spent on cement is

a. Rs.2,00,000b. Rs.1,60,000c. Rs.1,20,000d. Rs.1,00,000

Total is Rs. 6,00,000 (=360 degrees of the circle)Cement is 72 Degrees.Apply the ratio principle

Degrees Value72 (Cement)

M?

360(total) 6 lakhs

Divide left column on one side and right column on the other side.72/360= M/6 lakhsM=(72*6)/360=1.2 lakhs.2. The amount spent on labour exceeds the amount spent on steel by

A. 5% of the total costB. 10% of the total costC. 12% of the total cost

31

APTITUDE 1

D. 15% of the total cost

There is no need to find absolute values. Just observe the degrees.

Item Degrees

Labour 90Steel (Base %) 54

Difference 36

So the difference between Labour and Steel is 36 degrees.And total cost is (6lakh=) 360 degrees.So the percentage= 36/360 x 100 =10% of the total cost.Answer (B)3. The amount spent on cement, steel and supervision is what percent of the total cost of construction?

A. 40%B. 45%C. 50%D. 55%

Again no need to find absolute values. Just observe the degrees.

degrees

Cement 72Steel 54Supervision 54

Total 180

He is asking, “what percent of the total cost of construction?”So total cost (360 degrees) is the “base”%=(180/360) x 100=50%4. The amount spent on labour exceeds the amount spent on supervision by

A. Rs.2,00,000B. Rs.16,000C. Rs.1,20,000D. Rs.60,000

Again no need to find absolute values for both items. Just observe the degrees.Difference between labour and supervision=90-54=36 degreesShortcut: 36 degrees =10% of a circle. (because total is 360 degrees)

32

APTITUDE 1

So the amount spend = 10% of total (6,00,000)= one decimal leftwards=60,000.If shortcut is not clicking your mind, no problem. Go by ratio principle

Degrees Value36 (difference between labor and supervision) M?

360 (total)6 lakhs

Divide left column on one side and right column on the other side.36/360= M/6 lakhsM=(36*600000)/360M=60,000Answer is (D)5. The amount spent on Timber is

A. Rs.60,000B. Rs.90,000C. Rs.1,20,000D. Rs.36,000

Observe that degree/percentage of Timber is not given the chart.So, first task is to find degrees of timber.In a circle Total=360 degrees.So Timber’s degree=360 MINUS the degrees of remaining items=360-(54+54+54+72+90)=36 degrees.Now use the concept given in previous question,36 degrees= 10% of total cost=60,000 rupees.Answer (A)

Solution: Set #2: CO2 Emission Pie Chart

1. Which of the following sectors together emit 2.5 mmt of CO2 every year?

a. Thermal and Transportb. Domestic and Commercialc. Transport and Commerciald. Commercial and Thermal

Total emission is 5. And 2.5mmt is 50% (half) of 5 (total emission)

From the given circle, transport + commercial, represent half circle (50%) Similarly Thermal+Domestic is also 50%. But it is not given in options. So answer is (C): Transport + Commercial.

2. Which of the following sectors have emission difference of 1 mmt between them.

a. Domestic and Commercialb. Transport and Commercial

33

APTITUDE 1

c. Thermal and Domesticd. Thermal and Transport.

Total is 5mmt (100%) So 1mmt=1 x 100/5=20%. Find out which two sectors have % difference of 20%? Thermal minus Domestic=35-15=20%. So (C) is the answer.

3. Emission of Domestic sector is how much % of Transport and Commercial sector combined?

a. 20%b. 15%c. 30%d. 35%

Observe the chart

Required %

=(15/50) x 100

=30% answer is (C)

4. In the next year, if emission from Commercial sector decreases by 20% while other sectors continue to emit same amount of CO2 as earlier then What will be the new % contribution of Thermal power sector?

Right now Commercial sector emits 20% of 5mmt=1/5 x 5=1mm

Next year 20% decrease=20% of 1mm=0.2mm will be less emitted.

But other sectors remains the same.

So total emission next year=5 minus 0.2=4.8mm.

And Share of thermal power

=(35% of 5)/4.8

=0.35*5/4.8

=36.5% (if you had trouble in long-division, use the speed maths %approximation trick CLICK ME)

5. Which of the following is incorrectly matched?

a) Domestic=54 degrees.b) Thermal=125 degrees.c) commercial =72 degrees.d) transport=108 degrees.

34

APTITUDE 1

100%=360 degrees So 1%=3.6 degree.

Multiply every sector’s % with 3.6 and you’ll get corresponding degree.

Ans (B) because correct degree of Thermal should be 35 x 3.6=126 degrees.

Solution :Set #3 Bar Chart (With explaination)

Be careful about the ‘unit’, Thousand/lakhs.

1. The amount of maximum profit as seen from the diagram is

A. 2 thousand rupeesB. 1.5 thousand rupeesC. 1 thousand rupeesD. 4.5 thousand rupees

The bar of 2008 is longest among all. So 4.5 Ans (D)2. The amount of total loss incurred during 2005-2010 is

A. 3 thousand rupeesB. 2.5 thousand rupeesC. 1.5 thousand rupeesD. 2 thousand rupees

Loss occurs during 2006 and 2009.

Year Loss2006 1.5

2009 1.0

Total

2.5 answer (B)

3. The ratio of the maximum profit earned to the minimum loss suffered is

A. 4 : 3B. 3 : 4C. 9 : 2D. 2 : 9

Max profit (2008)

4.5

Min Loss (2009) 1

Ratio=4.5/1=45/10=9/2Answer (C)4. If the loss is x% of the profit for the years under study, then x is

35

APTITUDE 1

A. 15B. 25C. 20D. 19

We’ve to take the total of profits and losses made in all years so,

Profit

Loss

2005

2.5 0

06 0 1.507 3.5 008 4.5 009 0 110 2 0Total 12.5 2.5

Profit: Loss=2.5/12.5=1/5=20% (remember the fraction to percentage table: Click ME.)Answer is (C)

[Aptitude] LCM, HCF, GCD: Basic concept, calculation, applications explained

1. Introduction2. What is Prime number?3. What is LCM?

o LCM4 EXamo How to find LCM using Prime-Factorization?o LCM of two numbers (56, 96)o LCM of three numbers: (12,15,20)o LCM of prime numberso LCM of co-prime numbers

4. What is HCF or GCD?

o HCF finding: Prime Factorizationo HCF of two numbers (4, 6)o HCF of three numbers (12,24,36)o HCF of prime numbers (13,29)o HCF of co-prime numbers (12,25)o HCF vs LCM: #1 multiplicationo HCF vs LCM: #2 Magnitudeo HCF vs LCM: of fractions

5. for more practice on LCM, HCF

36

APTITUDE 1

Introduction

1. Concept of LCM, HCF important for number theory and remainder based problems (generally asked in SSC CGL, CAT.)

2. LCM is important for time and speed, time and work problems.3. LCM is also important for circular racetracks, bells, blinking lights, etc.4. HCF is important for largest size of tiles, largest size of tape to measure a land etc.

But before getting into LCM, HCF, let’s understand

What is Prime number?

Consider this number : 12. This number can be found in many multiplication tables for example 1 x 12=12. 2 x 6 =12 3 x 4=12 That means, 12 has many factors (1,2,3,4,6,12). Such number is called a composite number. On the other hand, consider this number: 29. You cannot find it in any table except 29 x 1 =29. Such

number is called a prime number. Let’s make a shortlist from exam point of view

Prime Non-prime (composite)

2,3,5,7,11,13,17,19,23,29 4,6,8,9,10,12,14,15….

Now hold this prime number thought in your mind for a while.

What is LCM?

First, let’s create multiplication tables of 4 and 6.

4 s ′table

multiple

6 s ′table

multiple

4 x 1 = 4 6 x 1 = 64 x 2 = 8 6 x 2 = 124 x 3 = 12 6 x 3 = 184 x 4 = 16 6 x 4 = 244 x 5 = 20 6 x 5 = 304 x 6 = 24 6 x 6 = 364 x 7 = 28 6 x 7 = 424 x 8 = 32 6 x 8 = 484 x 9 = 36 6 x 9 = 54

Do you see any common numbers in the multiples of 4 and 6? Yes I see 12, 24 and 36 are common in both tables. Let’s isolate them.

4 x 3 1 6 x 2 1

37

APTITUDE 1

= 2 = 24 x 6 =

24

6 x 4 =

24

4 x 9 =

36

6 x 6 =

36

Ok so 12, 24 and 36 are common multiples of 4 and 6. But what is the smallest of these multiples? Ans 12 is smallest.

In the exam, we’ve no time to make such ^big tables to find LCM. So how to quickly find LCM of two or three numbers? There are many tricks, the easiest one is prime-factorization. We’ll learn that in a bit, but before that:

LCM4 EXam

1. Suppose there is a circular race track. Tarak Mehta takes 4 minutes to finish it and Jethalal takes 6 minutes to finish it. Now both of them start running from the same point at the same time in the same direction. They’ll continue running on this track forever. So after how many minutes will they meet for the first time on the starting point? Ans. LCM of time = LCM (4,6)=12 minutes. They’ll meet again on the starting point after 12 minutes.

2. Two bells ring at an interval of 4 and 6 minutes respectively. After how many minutes will they ring together? Ans LCM (4,6)

3. Two traffic lights blink at an interval of 40 and 60 seconds respectively. After how many minutes will they link together? Ans LCM (40,60).

4. HCF is also important for remainder related questions. but I’ll cover that in a separate article.5. How to apply LCM in time-speed-distance/work, pipes-cistern etc questions, is already covered in old

articles. (Mrunal.org/aptitude)

How to find LCM using Prime-Factorization?

Suppose in the exam, we need to find LCM of 4 and 6.Make a table like this

Number

Factors

46

Now you need to find the prime factors of 4 and 6.

Number

Factors

4 2 x 26 2 x 3

38

APTITUDE 1

Express it in terms of “powers”. For example 2 x 2 =22

Number

factors

4 22

6 2 x 3

Now make the third row called “LCM”.

Number

factors

4 22

6 2 x 3LCM

Now write all prime numbers in this “LCM row”

Number

factors

4 22

6 2 x 3LCM 2, 3

Write maximum power of each prime number

Number

factors

4 22

6 2 x 3LCM 22, 3

As you can see, maximum power of 2 was 22 (in 4’s row).Now multiple the numbers given in LCM row

Number

factors

4 22

6 2 x 3

LCM 22 x 3 =12

That’s our answer. LCM (4,6)=12.If I plot this LCM situation on a Venn Diagram, it’ll look like this:

39

APTITUDE 1

Anyways, Let’s try a difficult one: 56 and 96.

LCM of two numbers (56, 96)

Numbers

Factors

5696

First recall, in which tables do they come? Well 56 comes in 8’s table and 96 comes in 12’s table.

Number

Factors

56 8 x 796 12 x 8

but we need factors in “prime number” format. 12 and 8 are not prime numbers. So let’s Simplify further. 56 = 8 x 7 = 23 x 7 (; because 8 = 4 x 2 = 2 x 2 x 2) 96 = 12 x 8 = (4×3)x(4×2)=( 22x3) (23)=25x3 (please note you have to do this things in your head, if you

start making every calculation on a piece of paper, you’ll run out of time in the exam).

Number

Factors

56 23 x 796 25x3

Now let’s make the LCM row. Write all prime numbers (2,3 and7) in ascending order.

Number

Factors

56 23 x 796 25x3LCM 2 3 7

Now write maximum powers of each prime number.

40

APTITUDE 1

Number

Factors

56 23 x 796 25x3LCM 25 3 7

Multiply these numbers

Number

Factors

56 23 x 796 25x3

LCM 25 x3x7=32×21=672

So LCM (56,96)=672let’s try finding LCM of three numbers.

LCM of three numbers: (12,15,20)

Approach is same. Make prime factors

Number

Prime factors

12 22 x 315 3 x 520 22 x 5

Make a new row, write all prime factors in ascending order.

Number

Prime factors

12 22 x 315 3 x 520 22 x 5LCM 2,3,5

In the last row, Write the maximum power of those prime numbers.

Number

Prime factors

12 22 x 315 3 x 520 22 x 5LCM 22, 3, 5

Now multiple the numbers in last row

41

APTITUDE 1

Number

Prime factors

12 22 x 315 3 x 520 22 x 5LCM 22x3x5=60

Therefore LCM (12,15,20)=60.You can also look at it in following way:

12 x 5 = 60 15 x 4 = 60 20 x 3 = 60.

So 60 is the least common multiple.

LCM of prime numbers

Find LCM of 7,11,13We already know these are prime numbers. So they’ll not have any common factors. We just have to multiply them together and we’ll get LCM. But for the sake of conceptual clarity

Numbers Factors

7 7 x 111 11 x 113 13 x 1

LCM 1x 7 x 11 x 13 =1001

So 1001 is the answer.

LCM of co-prime numbers

Co prime numbers are those numbers that donot have any common factors. For example, 14 and 15. Individually none of them is prime number because 14=2 x 7 and 15 = 3 x 5. But they (14 and 15) donot have any common factors. So they’re called co-prime numbers (when they’re

given together). Any two consecutive numbers are co-prime numbers. (e.g. 11,12 or 1548,1549). In case of co-prime numbers, just multiply them and you will get LCM. There is no need to find factors.

example

6 2 x 37 7LCM

2 x 3 x 7 = (6)x7 =42

42

APTITUDE 1

Advantages of this method?

1. Extremely fast when you’ve to find LCMs of two digit numbers for example 12,15,96.2. And usually in time speed work, pipe-cistern type questions have number in two digits (e.g. 12, 15, 96)…

so it is very easy to recall in which multiplication tables do they come.

Disadvantages?

3. Becomes tedious, as the number grows bigger, for example LCM (235, 512). There are other methods to solve those LCMs, but let’s not complicate this article any further. Let’s stick to this Prime-Factorization method for a while.

Ok so far we know what is LCM and how to find HCF/GCD?

What is HCF or GCD?

HCF= Highest common factors. GCD= Greatest common divisor. Names are different otherwise they’re one and same. Suppose you’ve to find the HCF of (4 and 6). I’ll write the tables of numbers that come before 4 and 6 (i.e. 1, 2 and 3.)

1 x 1 = 1 2 x 1

= 2 3 x 1 = 3

1 x 2 = 2 2 x 2

= 4 3 x 2 = 6

1 x 3 = 3 2 x 3

= 6 3 x 3 = 9

1 x 4 = 4 2 x 4

= 8 3 x 4 =

12

1 x 5 = 5 2 x 5

=10

3 x 5 =

15

1 x 6 = 6 2 x 6

=12

3 x 6 =

18

1 x 7 =

7 2 x 7 =

14

3 x 7 =

21

1 x 8 =

8 2 x 8 =

16

3 x 8 =

24

1 x 9 = 9 2 x 9

=18

3 x 9 =

27

Ok, in which number’s table (1, 2 or 3) do you see both 4 and 6 reappearing?There are two such tables 1’s table and 2’s table.

4 and 6 are common in 1’s table.

4 and 6 are common in 2’s table.

1 x 4=4 2 x 2=41 x 6=6 2 x 3=6.

43

APTITUDE 1

What does ^this mean?

If I divide 4 by 1, I get zero remainder. Similarly if I divide 6 by 1, I get zero remainder. In other words, 1 is the factor of both 4 and 6. In other words, 4 and 6 come in the table of 1.

Similarly, If I divide 4 by 2, I get zero remainder. Similarly if I divide 6 by 2, I get zero remainder. In other words, 2 is the factor of both 4 and 6. In other words, 4 and 6 come in the table of 2.

Thus, 4 and 6 have two common factors (1 and 2) but highest of these common factors is 2. Therefore HCF of (4,6)=2.

HCF 4 EXAM?

What is the highest number that’ll divide 4 and 6 evenly. Ans HCF (4,6) There is a 4 x 6m rectangular farm. Find the length of longest tape that can measure this field. Ans HCF

(4,6) There is a 4x 6cm floor. Find the length of largest square tile that can be evenly laid on it. Ans HCF (4,6) Two drums contain 400 and 600 liters of desi and foreign liquor respectively. What is the biggest measure

(cup) that can measure both of them exactly? Ans. HCF (400, 600). A teacher has 40 pens and 60 pencils. Find maximum number of students among whom she can

distribute these items evenly. HCF is also important for remainder related questions. but I’ll cover that in a separate article.

HCF finding: Prime Factorization

In the exam, we can’t make multiplication tables of every number preceding the given numbers! So here is the shortcut technique. We’ll use the same approach we’ve used in LCM method: prime factorization.

HCF of two numbers (4, 6)

First make prime factors of given numbers.

4 22

62 x 3

Now, make third row: HCF and write the prime numbers that are common in both numbers.

4 22

62 x 3

HCF

21

Therefore, HCF (4,6)=2If I’ve to plot the HCF of 4 and 6 on a Venn diagram, it’ll look like this:

44

APTITUDE 1

HCF of three numbers (12,24,36)

12

2 x 6

24

3 x 8

36

6 x 6

But I want them in prime format. So I’ll further simplify.

12 2 x 2 x 3=22 x 3

24 3 x 2 x 2 x 2=23 x 3

36

3 x 2 x 3 x 2=22 x 32

In the exam you’ve to do this in your ^head.

12 22 x 3

24 23 x 3

36 22 x 32

Now make a new row, write the prime numbers that are common in all of above.

12 22 x 324 23 x 3

45

APTITUDE 1

36 22 x 32

HCF

22x3

^in case you’re confused, let me rewrite and do it again

12 22 x 3

24 22 x 2 x 3

36 22 x 3 x 3

HCF 22x3

The numbers highlighted in bold are common. Therefore HCF = 22 x 3=12.

HCF of prime numbers (13,29)

Prime numbers donot have any common factors. So HCF of such numbers is always 1. But for the clarity let’s do it

13 13 x 1

29 29 x 1

HCF 1 (because 1 is common in both)

HCF of co-prime numbers (12,25)

Again same: 1, because co prime numbers donot have common factors.Similarly consecutive numbers (like 456,457) donot have common factors either.Therefore, in all such cases, HCF =1.

HCF vs LCM: #1 multiplication

If we’ve two numbers a and b. and their HCF and LCM are given thenHCF x LCM = a x b.But this relation only work for TWO numbers and not for more than two numbers.Let’s understand this with an example.You know that LCM (4,6)=12 and HCF (4,6)=2.

Left hand side (LCM x HCF) Right hand side (multiplication of given numbers)12 x 2 4 x 6=24 =24

So both sides match. Therefore, in case of two numbers (a and b)LCM X HCF = a x b.

46

APTITUDE 1

But this is not always true for three numbers. For example, Find LCM and HCF of 12,15,20. You’ll get HCF=1 and LCM=60.

Left hand side (LCM x HCF)

Right hand side (multiplication of given numbers)

60 x 1 12 x 15 x 20=60 =3600

In this case, both sides donot match.

HCF vs LCM: #2 Magnitude

For any given numbers, their LCM is always greater than or equal to the biggest number. For example

Numbers LCM12,15,20 60 so greater than biggest number (20)15,30 30. which is equal to the biggest number (30).

Similarly, for HCF, the HCF of given numbers is always less than or equal to the smallest number. For example

Numbers HCF

12,15,20 1 so it is smaller than smallest number 12

15,30 15. so it is equal to the smallest number 15.

Ok this is just the basic overview. In the next article, we’ll see the application of these concepts. In the mean time, try finding LCM and HCFs of following numbers

Question Answer (LCM, HCF)

91, 12 1092, 146, 69 138, 2369, 97 6693, 163, 33 693, 372, 58 2088, 25, 84 420, 191, 41 3731, 165, 57 3705, 174, 12 444, 244, 55 220, 118, 28, 175 1400, 1

LCM, HCF of fractions

Just observe the color pattern in following image:

47

APTITUDE 1

for more practice on LCM, HCF

Book Chapter no.

Quantitative Aptitude, R.S.Agarwal 2Fast track Arithmetic, Rajesh Verma 2

Quantam CAT, Sarvesh Kumar Ex.1.3, 1.4Arun Sharma (CAT) 1

[Aptitude] Long Division, Two-Digit Division, % Calculation without Tears (and without boring Vedic Maths) using % Approximation Method

One of the biggest problem with IBPS/Bank, SSC, CAT, CMAT type of exams= you’ll encounter long division and percentage (%) calculation every now and then, directly or indirectly (in Data interpretation questions).

And If you’re not good with speed maths, you’ll waste a lot of time in such stupid calculations.

For the the aptitude questions of Bank/MBA exams, Knowing the answer approach is not sufficient. You also need to get the answer quickly and accurately. Otherwise someone else will tick more answers and he’ll get the rank.

Consider these two questions

48

APTITUDE 1

QuestionIn a warehouse there are 230kg of wheat initially. But rats ate away 34 kg. How much % of wheat is left?

What is the answer of 196/23=?

Answer

a. 49.3%b. 60.1%c. 85.2%d. 85.7%

a. 4.93b. 6.01c. 8.52d. 8.57

Approach {(230-34)/230} x100=(196/230)x 100=answer

196/23= answer

Everyone, even with half hearted preparation, knows the approach. But The problem is actually in ‘doing’ that calculation or division (196/23). Here I’ll show a single method, to solve both type of calculations. As long as you know how to add two numbers, and how to multiply a number with 5, you can execute this method effortlessly.

First create a “Master Table” (don’t just read it, do this simultaneously using your own pen and paper)

100%

230

Now make a new row for 50%

100%

230

50%

So either divide 230 by 2=115 or multiply 230 with 5 and then shift one decimal point leftwards. (that is 1150 ==> to 115.0) In either case you get 50% of 230=115

100%230

50% (half of 230)

115

Now make a new row for 20% but keep it empty right now.

100%

230

20%

Create one more row for 10% and simply shift one decimal leftwards. i.e. 230–> 23.0

100% 23050% (half of 230) 11520%

49

APTITUDE 1

10% (one decimal point less)

23.0

Now double the 10% number (i.e.by adding 23 into 23 again) so you get 23+23=46. That’s our 20%. Fill up the table.

100% 23050% (half of 230) 11520% (double of 10%) 4610% (one decimal point less)

23.0

Our master table is ready, now Imagine there is a big water tank with total capacity of 196 lit.

We can fill it with buckets of size 10%, 20% and 50% only.We want to fill up the tank with minimum effort. So first take 50% (115), some space will be left.By this time you get the idea that

1. answer is more than 50% (if % value of 196/230 is asked)2. answer is more than 5 (if absolute value 196/23 is asked)

so eliminate answer options that donot meet these criteria.

Move on

Tank Filled

Buckets

196 115 50%Total 115 50%

There is still some space left in the tank so let’s throw a 20% bucket

Tank Filled

Buckets

196 115 50%046 20%

Total 161 70%

Or you can add 10% bucket two times, you’ll get same result.

It’s clear that our answer is bigger than 70%. So eliminate any options less than 70%

Hmm, so far we’ve filled 161, It can still accommodate another 10% bucket

Tank Fille Bucket

50

APTITUDE 1

d s196 161 70%

023 10%Total

184 80%

Now we are very close, only 196-184=12 lit. remains. But no bucket is that small!

Solution= move the decimal numbers, to create new small sized buckets.

Master Table Moving decimal numbers

100% 23050% (half of 230) 115 5% 11.5

20% (double of 10%)23×2= 46 2% 4.6

10% (one decimal point less) 23.0 1% 2.3

In the exam, you don’t have to actually write new columns of 5%, 2% and 1%, just visualize them in your head, by shifting the decimal to one point leftwards.

Recall that 12 lit is empty and Now we’ve a new 5% bucket that can almost fill it up.

Tank Filled Buckets

196 184 80%011.5 5%

Total

195.5 85%

By this time you get the idea that

1. answer is just a little higher than 85% (if % of 196/230 is asked)2. answer is just a little higher than 8.5 (if absolute value 196/23 is asked)

so eliminate any answer options that are not meeting this criteria.

Still if two or more options remain. For example

a. 8.52b. 8.57

^This situation usually happens in CAT Data Interpretation questions. Now what to do?

Well, Total capacity is 196 lit. and so far we filled up 195.5 so, 0.5 lit is still empty. But no bucket is small enough to carry water in this scale. Solution= create more buckets, by shifting decimal points in the “Master Table”.

51

APTITUDE 1

Master Table Moving decimal numbers

100% 230 Phase I Phase II50% (half of 230) 115 5% 11.5 0.5% 1.15

20% (double of 10%) 23×2= 46 2% 4.6 0.2% 0.46

10% (one decimal point less) 23.0 1% 2.3 0.1% 0.23

Recall that 0.5 lit is empty and from above table, it is clear that 0.2% bucket (0.46 is very close) so let’s use it.

Tank FilledBuckets

196 195.50 85%

000.46 0.2%

Total

195.96 85.2%

So the final answer is

196/23=8.52 196/230=85.2%

If you want even more accurate answer, create more buckets and proceed in the same manner.

Important sidenotes

1. Whenever you have to do long-division e.g. 256/29, always make the denominator (bottom number i.e. 29) very close to the top number (256) and take that as 100%. That is 290=100%. And then rephrase question: “256 is how much % of 290”, then proceed according to the method you just learned. You’ll get 88.27%. but our question was 256/29. Recall that you’ve added one zero more. (290)

So, 1%=1/100

Therefore, 88.27%=(88.27/100)

And from the ‘bottom’ we take back one zero that we had added earlier. So instead of 100, there remains only 10

88.27/10=8.827 is our answer for 256/29

2. If there is 7526/67 then? Again same method, 7526 is how much % of 6700? You’ll get 112.3%. this time we’ve added two zeros more (i.e.we used 6700 instead of 67).

So, 1%=1/100

52

APTITUDE 1

Therefore 112.3%=112.3/100

But take back those two zeros we had added earlier. So, instead of 100, there remains only 1

112.3/1=112.3 is our answer for 7526/67

This method looks awkward and tiresome initially, but once you’ve enough practice of doing mental addition then it’s way easier than the Vedic Maths’ concept of double or triple digit division (because in Vedic method, many a times you’ve to adjust and carry over the numbers= not very convenient).

This method can be used for three-digit, four digit divisions also. You can do any division as long as you can find out 10%, 20% and 50% of a number (and consequently

1%, 2%, 5%, by shifting decimal places.)

Applications of ^this method

Percentage calculation or Long Division has direct or indirect applications in following topics:

1. Profit Loss2. Data Interpretation, especially those based on Pie-charts.3. Compound interest, Simple Interest Rate, Population Growth: by the way, they can be solved without

mugging up formulas, click me to know how!4. Mixture-Alligiation, Wine-water, Metal alloys: can be solved without mugging up formulas, click me to

know how!5. Time-Speed-Distance, Time and Work, Pipes and Cisterns, Boats and Railways. All of them are based on

the STD formula: speed x time = distance. So there is almost always a situation where two variables are given and third is to be found= division. All of them can be solved by mugging only single STD formula. Techniques are scattered around on this page: Mrunal.org/aptitude

Test your skill

Question % Absolute Value Hint: 100% is

231/19 190181/29 290158/117 117

198/6767 (because 670 would be too far)

67/51 5151/29 29158/117 1178189/541 5410

53

APTITUDE 1

[Aptitude] Product Consistency: If Sugar price increases then consumption should be Decreased by What %, Time-Speed-Distance problems, shortcuts for SSC, IBPS, CSAT, CAT

1. Case: Sugar Consumption Budget don’t change2. Case: Salary comparison: How much more or less?3. Case: Increased consumption absolute value4. Case: Apples price decrease absolute value5. Case: Time-Speed-Distance6. Mock questions7. Food for thought

IF you want to master this method, then don’t just read the article, simultaneously do all calculations in your rough-note/paper.

Case: Sugar Consumption Budget don’t change

Q. if price of sugar is increased by 25%. But a family wants to keep its expenditure same as earlier. Then they should decrease their consumption by how much percentage?

Suppose initial price of Sugar was Rs.100 per kg And this family needs 2 kg per month. (=quantity consumed) So what is their budget (or Expenditure?)

Before price risePrice 100Quantity 2 kgBudget = Price x quantity

100 x 2=Rs.200 per month

Now the price of sugar is increased by 25%.So if previously it was Rs.100 per kg, now it is Rs.100+25=Rs.125 per kg.

BeforeAfter

Price 100 125Quantity 2 kgBudget = Price x quantity

100 x 2=Rs.200 per month

54

APTITUDE 1

If the family still wants to buy 2 kg sugar, they’ll have to pay Rs.125 x 2 = Rs.250. But they don’t want to increase their budget (Expenditure). Obviously they will have to cut down on their monthly consumption (Quantity of sugar.)To solve this problem, just take ratio of both prices100/125=(25×4)/(25×5)=4 / 5It means 100/125=4/5Now, reverse this ratio (4/5 =>5/4) and write it, in the table.Means, in 100’s column, I’ll write 5 and in 125’s column, I’ll write 4.

Before After

Price 100 125Ratio reversed? 5 4Demand 2 kgBudget = Price x Demand

100 x 2=Rs.200 per month

What is the percentage change in Ratio-Reversed?Well, before it was “5” and after it has decreased to “4”.% decrease from 5 to 4 is=(5-4)/5=1/5=20% decrease.That’s our answer.If price is increased by 25% then we must decrease our consumption by 20% to keep the budget same.Let’s check if our approach is correct.According to above technique, we’ve found that consumption /demand should decrease by 20%.So if earlier family consumed 2kgNow they should consume only 100% minus 20%=80% of original demand=80% of 2kg=0.8 x 2=1.6 kg

Update the table.

Before AfterPrice 100 125Quantity 5 4Budget = Price x quantity 2 kg 1.6

Price100 x 2=Rs.200 per month

125 x 1.6=Rs.200 per month

As you can see, if we decrease demand/consumption from 2 kg to 1.6 then budget remains same (Rs.200).

It proves that we’ve not made any mistake.

Another way is to use readymade formula

55

APTITUDE 1

Formula

Percentage decrease = 100m / (100+m), where m is the original percentage.ThereforePercentage decrease in sugar case=100 x 25 / (100+25)=100 x 25/125=20% decrease.

You can use whichever technique you like. But once you master product consistency, lot of profit-loss, time-speed-distance questions can be solved

in less than a minute. Anyways, lets try some more “easy” questions and then move to difficult ones.

Case: Salary comparison: How much more or less?

1. If Salary of Mr.Abdul is 25% more then Mr.Bhide, then Mr.Bhide’s salary is how much % less than Mr.Abdul’s?

Assume Bhide earn Rs.100 per month Then Abdul has to earn Rs.100+25=125Fill up the table.

Bhide

Abdul

Salary 100 125Ratio (reversed)

Now get the ratio100/125=4/5Reverse it (from 4/5 to 5/4) and plug it in the table

Bhide

Abdul

Salary 100 125Ratio (reversed)

5 4

That’s it. What is the decrease?(5-4)/5 x 100=1/5 x 100=20% lessFinal answer: Mr.Bhide’s salary is 20% less than Abdul’s.Ofcourse you could directly calculate: (125-100)/125=20% but that won’t help us quickly solve the ‘complex’ cases situation like following.

Case: Increased consumption absolute value

56

APTITUDE 1

Q. A family spends Rs.600 on sugar every month. The price of sugar is decreased by 20% and they’re able to buy 5kg sugar more. What was the original price of sugar (per kg)?

This is not at all complicated. Budget =Rs.600 (Constant) Assume that original price is Rs.100 (please note: this is not the answer, we are just ‘assuming). New price is 20% less = 100 minus 20 = Rs.80 per kg Assume that originally they used to buy “m” kilos of sugar every month.

Before

After

Price Rs.100

Rs.80

Ratio-reversedQuantity MBudget 600 600

So if price is decreased by 20% then consumption quantity should increase by what percentage?Just take ratio of price100/80=10/8=5/4Reverse It and fill up in the table

Before After

Price Rs.100

Rs.80

Ratio-reversed 4 5

Quantity mBudget

So what’s the increase in quantity ?Just look at the ratio reversed: 4 to 5. So increase is:(5-4)/4 x 100=1/4 x 100=25%

It means if sugar price decreases by 20% then we can buy 25% more sugar. So if earlier this family used to by “m” kilos of sugar, now they should be able to buy Total= (m + 25% of

m) kilos of sugar.

How much more sugar can they buy? 25% of mBut the question itself says that family is able to buy 5 kg sugar more.It means25% of m =5 kg(25/100) x m=5M=5 x (100/25)M=20kg

57

APTITUDE 1

Another way is (% to fraction)

25% when converted to fraction =1/4So 1/4th of m=5 kgSo m= 5 x 4=20 kg.In either way: it means originally family used to buy 20 kilos of sugar.And their budget was Rs.600. (given in the question)So what was the per kilo price of sugar?20kg=Rs.600So 1 kg=600/20=30Rs. per kiloFinal answer= original price of sugar was Rs.30 per kg.Let’s try a similar question, so our concept is crystal clear.

Case: Apples price decrease absolute value

This is also from SSC-CGL exam.Q. A man spends Rs.54 on apples every month. Price of Apple is decreased by 20% and this man can buy 10 apples more. What is the reduced price per dozen?If price is 20% decreased => consumption should increased by 25% (as seen in previous case)Suppose he used to buy “m” number of apples initially.Now he can by 25% more apples.But question itself says that he is buying 10 more apples.Therefore,25% of m =101/4 of m=10 (because 25%=25/100=1/4)1/4 of m =10M=40

So what was the original price?

Total 54 rupees spent on 40 apples=Rs.54/40And we’ve to find answer in “dozen” so multiply with 12Original price= Rs.(54/40) x 12Donot simplify yet.We’ve to find reduced price.Question says, price is reduced by 20%.So new price=100 Minus 20=80%of original price=0.8 x (54/40) x 12=Rs.12.96Final answer: the reduced price of apples is Rs.12.96 per dozen.

Another approach

Originally he bought 40 apples.Now he can buy 25% more=100+25=125% of original apples=1.25 x 40=50 apples.And his budget is Rs.54 means he buys 50 apples for 54 rupees,

58

APTITUDE 1

50 apples

= 54 rupees

12 apples

= how much?

How much = 54 x 12 / 50 =Rs.12.96 per dozen.Now let’s apply this concept in Time-Speed-Distance question

Case: Time-Speed-Distance

Again from old SSC-CGL examQ. Walking at 6/7th of his usual speed, a man is 25 minutes late for his destination.What is his usual time to cover this distance (in hours)?Speed x time = distanceThis is same as price x quantity = budget.Here distance remains the same, just like in “sugar-cases”, budget remains the same.We don’t know his speed, so let’s assume his usual speed = 1 kmphAnd his late speed = 6/7th of usual speed = 6/7 x 1= 6/7 kmph

Usual case

Late case

Speed 1 6/7Ratio reversed

Just take ratioUsual/late=1/(6/7)=7/6Reverse it (7/6 to 6/7 )and put it back in the table.

Usual case

Late case

Speed 1 6/7Ratio reversed 6 7

Time “T”

Ok so if speed is decreased to 6/7th then time is increased by what fraction (or percentage?)6 to 7=(7-6)/6=1/6 increase in time.Suppose his usual time was “T”, then it should be increased by 1/6th of original time “T”But we know that he is late by 25 minutes.It means 1/6th of T=25 minutes1/6 x T = 25T=25 x 6=150 minutesBut the question is asking time in hours.150 minutes=60 minutes + 60 minutes + 30 minutes=1hour + 1 hour + 30 minutes

59

APTITUDE 1

=2 hours and 30 minutesAnother way to convert minutes into hours60 minutes = 1 hr150 minutes = how many hours?Therefore: How many hours = 1 x 150 / 60When you divide 150 by “60” you 2 as quotient and 30 as remainder. So Correct answer is 2 hours and 30 minutes.Important: In Minutes to hour conversion via division method, You should not cut zeros, else you get wrong answer.For example, 150/60= 15/6 but when you divide 15 by 6, you get remainder 3 = answer comes as 2 hours and 3 minutes = wrong answer.

Mock questions

1. Journalist Popatlal’s income is 37.5% less than Dr. Iyyer’s. Then Dr.Iyyer’s income is how much % more than Popatlal’s income?

2. If price of Desi-liquor is increased by 20% then Mohan should cut down consumption by what %, to keep his budget unchanged?

3. Petrol price is doubled. But we don’t want to raise our expenditure. Then We should cut down our petrol consumption by what percentage?

4. If the speed of a motorboat is decreased by one fourth, then journey completion time should increase by what percentage?

5. Price of rice is increased from Rs. 6 to 7.5 per kg. If we don’t want to increase our Expenditure. We should decrease our rice-consumption by what percentage?

6. Mobile company increased the call charges by 50%. If I want to keep my budget unchanged, I should reduce talk-time by what percentage?

7. Dish TV has reduced the channel prices by 20% now I’m able to subscribe to 5 more channels in the same budget of Rs.400. How many channels did I subscribe earlier?

8. Writing at 3/4th of my usual speed, I finished the question paper 20 minutes late. Had I written the answers at my regular speed, I could have finished the whole paper in how many hours?

9. Earlier UPSC used to ask 579 questions for 2900 marks. Now they want to decrease number of questions by 20% but want to keep total marks same as earlier. So, they should increase the marks per question by what percentage?

10. If a politician’s bribe income is decreased by 10% then his anger increases by how much percentage? hahaha

Answers

1. Iyyer Popat 60% Perhaps this is the reason why Popat is unable to find a bride while Iyyer walked away with Babita-ji.

2. Desi liquor16.67% decrease (1/6)

3. Petrol Double50% decrease in consumption.(hint double = 100 to 200 =100% increase)

4. MotorboatOne fourth=25% decrease=> 33.33% increase in time.

5. Rice25% increase in price =>20% reduction in consumption.

6. Mobile talktime50% increase in call rates=>33.33% decrease in talktime to keep budget unchanged.

60

APTITUDE 1

7. Dish TV Earlier I had subscribed to 20 channels. 20% decrease in price= 25% increase in number of channels. so 1/4 of original channels=5, hence original number of channels=5 x 4 =20.

8. Writing speed 3/4 th of writing speed= reverse ratio is 4:3; and increase in time is 1/3 (=33.33%). So if 1/3 rd of original time= 20 minutes late.Then original time = 20 x 3 =60 minutes = 1 hour.

9. Anger11.11% increase in anger lolz

10. UPSC Absolute values 579 and 2900 are irrelevant! No need to waste time in dividing them. 20% decrease=> increase marks per question by 25% percentage.

Food for thought

Try these two questions: (You can solve them via STD table method but try to solve them via this product consistency method. You’ll get the answer much more quickly!)

1. Jethalal goes to shop at the speed 30 km/h, and he reaches six minutes early. Next day he goes at the speed of 24 km/h, and he reaches five minutes late. Find the distance between his home and shop. (Ans=22kms)

2. Tappu walks from home to school @5kmph and reaches 15 minutes early. After the school is over, he is very tired. So he walks back from school to home at a slow speed of just 3kmph and reaches 9 minutes late. Find distance between his home and school. (Ans=3 kms)

[Aptitude Q] STD table : Application in train man bridge, time and work problems

Not really an article, just solving some readers’ queries to keep the STD concept refreshed.

1. Case: Train, Man and Bridge

1. Conversion : kmph vs ms2. Finding length of train

2. Case: Time n Work : A and B

Case: Train, Man and Bridge

A man is standing on a 180m long bridge.He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train and its speed.Approach:

When train crosses the bridge When train crosses the man

Total distance=

length of train + width of bridge

length of train + width of bridge

“m”+180 “m”+0

61

APTITUDE 1

Because in comparison to train, man / telephone wire / bike-rider etc are considered zero in width.Now make table

Train + bridge

Train + man

Speed “s” “s”Time 20 8Distance M+180 M

Now apply STD formula and you get two equations

Train + bridgeTrain + man

Speed “s” “s”Time 20 8Distance M+180 MSpeed X time = Distance

S x 20 = (M+180) S x 8 =M

Just solve those two equations20s=M+180….(i)8s=M….(ii)So in first equation, replace the value of M=8s20s=8s+18020s-8s=18012s=180S=180/12S=15 m/s. that’s our final answer. But suppose, in the answer options, the speed is given in in km/h, then you need to convert m/s into km/h.

Conversion : kmph vs ms

1km =1000 meters.1 hr = 60 minutes = 60 x 60 seconds = 3600 seconds.Therefore, 1 km / hr = 1000m / 3600 seconds1km / hr = 5 / 18 (m/s)In otherwords,1 meter per second = 18 / 5 kmph.So 15 (m/s) = 15 x (18/5) km/h = 54 kmph. That’s our final answer.

Finding length of train

Moving to the next part: find the length of this train.From the second equation, we know thatS x 8 =MAnd we’ve found that speed of train= 15m/s.

62

APTITUDE 1

hence 15 x 8 =mm=120meters.Answer: length of train is 120 meters.

Case: Time n Work : A and B

Question: A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes the remaining work in 42 days. In how much time will A and B together finish the work?Make table

A BSpeed a bTime 80 bDistance = speed x time

a x 10=80a

It means total distance to be covered, is “80a”.

A works for 10 days

A B A worksSpeed a b aTime 80 10Distance = speed x time

a x 10=80a

10 x a =10a

See the last column. A covered distance of 10a, meaningSo 80a-10a=70a distance is yet to be covered.Now B comes and finishes this in 42 days

A B A works B finishesSpeed a b a bTime 80 10 42Distance = speed x time

a x 10=80a

a x 10 =10a. (meaning 70a remains).

b x 42= 42b

That means 42b=70a (because remaining work =70a and B finished this.)b=(70/42)aTherefore, b=5a/3Now if A+B worked together then, we can add their speed. And together they’ve to finish 80a distance

Update the table

A B A works B finishes A+B together

Speed ab=5a/3 a b a+5a/3=8a/3

Time 80 10 42 T

63

APTITUDE 1

Distance = speed x time

a x 10=80a

a x 10 =10a. (so 70a work remains).

b x 42= 42b

80a

Apply STD in last columnSpeed x time = distance8a/3 x T =80aTherefore T= 30 days.Final answer: if A and B work together, it’ll take them 30 days to finish the job.

[Aptitude] Time n Distance: Early and late to office (shortcut using product consistency method) for SSC, IBPS, CSAT, CAT, CMAT

1. Case : Early Late

o Approach #1: Product consistencyo Approach #2 (STD Table)

2. Case #2: Tappu’s school

64

APTITUDE 1

o Approach #1: Product consistencyo Approach #2 (STD Table)

3. Case: Pinku’s college (total time given)4. Mock Questions5. Answer and explanation

Before proceeding further, make sure your concept regarding “product-consistency method” Is clear. If not, then go through my previous article click me

Case : Early Late

Jethalal goes to shop at the speed 30 km/h, and he reaches six minutes early. Next day he goes at the speed of 24 km/h, and he reaches five minutes late. Find the distance between his home and shop.This can be solved with any of the two approaches

1. Approach #1: Product consistency2. Approach #2: STD table.

Approach #1: Product consistency

Let me rephrase the question:Price of sugar is increased from 24 per kg to 30 per kg and now Jethalal is buying 11/60 kilograms sugar less (in the same budget). What was his original consumption?

Does it ring any bell with previous sums of product Consistency? Yep, that’s our approach.Prepare this table, plug in the “speed” values in ascending order

Slow speed

Fast speed

Speed km/h 24 30Ratio-reversed (Time)

What is the time difference between these two cases?suppose on regular speed, Jethalal used to reach office @10 AMon slow speed, he is 5 minutes late=10.05AMon fast speed he is 6 minutes early=09.54 minutesso the time difference between slow speed and fast speed = 11 minutes.in the exam, just add the two minutes given to you (6+5)=11 and since speed is given in km/h, we’ve to convert 11 minutes into hours =11/60 hours.

Slow speed

Fast speed

Speed km/h 24 30Ratio-reversed (Time)

65

APTITUDE 1

Now apply the product consistency method:Take ratio of 24/30=(6 x 4)/ (6 x 5)=4/5Reverse it.=5/4. Update the table

Slow speed

Fast speed

Speed km/h 24 30Ratio-reversed (Time)

5 4

So, when speed is increased, what is the decrease in time?5 to 4=(5-4)/5 x 100=20% (or just keep it in fraction form of 1/5)Meaning new time is 20% less than time.suppose during slow speed, he took “M” time.Then in fast speed he’ll take M minus 20% of M time.That means difference between two situations is 20% of mbut we’ve already inferred that time difference between two situations is 11/60 hourstherefore 20% of m=11/60or in other words1/5 x m=11/60M=11×5/60M=11/12 hours.This is the time he takes during slow speed, to reach his destinationNow just apply STD formula(slow) Speed x time = distance24 x 11/12 = distanceHence distance = 22 kms.This technique looks “odd” but It is very fast once you practice.

Thought process in the exam

You don’t even need to draw table. Just think in your head, speed is decreased from 24 to 30 so reverse ratio is 30/24=5/4And hence decrease from 5 to 4 is (5-4)/5=1/5.It means 1/5th of (slow) time =(6+5)/60Hence time = 11x 5/60Hence distance = just multiple time with slow speed=11 x 5 x (24)/60=22 km.Now let’s try solving It, using the

Approach #2 (STD Table)

Case 1

Case 2

Speed 24 30

66

APTITUDE 1

Time ? ?Distance

D D

We’ve ssumed that in both cases, he has to cover same distance “D” kms.Apply STD formula in column 1 (case 1)Speed x time = distanceTherefore time = distance / speed = D/24Similarly for case2, we get time=D/30Update table

Case 1

Case 2

Speed 24 30Time D/24 D/30Distance D D

From the question, we can infer that time difference between two cases is (6+5=11 minutes =11/60 hours)ThereforeD/24-D/30=11/60Simplify this equation and you get D=22 kms.Please note: in the fractions, D/24 is >greater than> D/30That’s why I did D/24-D/30=11/60Let’s try second question with both methods

Case #2: Tappu’s school

Tappu walks from home to school @5kmph and reaches 15 minutes early. After the school is over, he walks back from school to home @3kmph and reaches 9 minutes late. Find distance between his home and school.

Approach #1: Product consistency

The question is talking about two times: 15 minutes early and 9 minutes late.Therefore total time difference between two situation =15+9=24 minutes=24/60 hrs.

Slow speed

Fast speed

Speed km/h 3 5Ratio-reversed (Time)

5 3

What is the percentage decrease in time?(5-3)/5=2/5 (=40% decrease)That’s it. If time taken during slow speed =”m”Then 2/5th of m=24/60 hours (the time difference between two cases)Hence M=1 hour (=time taken during slow speed)Now speed x time = distance

67

APTITUDE 1

3 (slow speed) x1= distanceTherefore distance between Tappu’s school and home is 3 kms.

Approach #2 (STD Table)

Slow speed

Fast speed

Speed km/h

3 5

Time ?? ??Distance D D

Apply STD formula in each column you getSpeed x time =distanceTime = distance / speedTime = D/3 in first case and D/5 in second case update table

Slow speed

Fast speed

Speed km/h 3 5

Time D/3 D/5Distance D D

The time difference between two situations is (15+9)=24 minutes=24/60 hoursThereforeD/3 – D/5=24/60Solve this equation and you get D=3 kmsMeaning distance between Tappu’s school and home is 3 kmsNow let’s try a bit complicated case

Case: Pinku’s college (total time given)

Pinku goes to college @ speed of 3 kmph and returns back @2kmph. He spends total 5 hours in walking. What is the distance between his home and college?Slow speed , fast speed = 2 and 3 km respectively.

Slow speed

Fast speed

Speed km/h 2 3Ratio-reversed (Time)

3 2

What is the decrease % in time? (3-2)/3= 1/3 (=33.33%)It means if Pinku take “M” hours during slow speed.He’d take M minus 33.33% of M hours during fast speed.

68

APTITUDE 1

Therefore, total time (taken to goto college and come back)=m + m -33.33% of m=2m-m/3 (because 33.33%=1/3)=(6m-m)/3=5m/3And we know that total time is 5 hourstherefore 5m/3=5 hourshence m=3 hours. (time taken during slow speed)Apply STDSpeed x time = distance2 (slow speed) x 3 (time)=distanceHence distance=6 km

Thought process in the exam

Speed increased from 2 to 3, therefore reverse ratio is 3/2 and %decrease in time is 1/3.Pinku’s “Total” time is given 5 hours, thereforeM + m -(1/5)m=5 hours. Solve it and multiple with slow speed, you’ll get the distance.

Mock Questions

1. Gogi walks from home to school @2.5kmph and he is 6 minutes late. Next day he increases speed by 1 kmph and reaches 6 minutes early. Find distance between home and school?

2. Sonu walks @6kmph and late to college by 5 minutes. If she walks @5kmph, she is late by 30 minutes. Find total distance. (please note: since she’s late in both cases the time difference is 30-5=25 minutes. rest approach is same)

Answer and explanation

1. Gogi school

Question is talking about two speeds : 2.5 and (2.5+1.0)=3.5 kmphs

Slow speed

Fast speed

Speed km/h 2.5 3.5Ratio-reversed (Time)

7 5

What’s the decrease in time %From 7 to 5,=(7-5)/7=2/7Suppose during slow speed case, Gogi takes “m” hours to reach school.In fast case, he’ll do it in less time =m – 2/7 of m.but from question, we already know that time difference between two cases =6+6=12 minutes=12/60 hrsit means 2/7 of m=12/60 hourstherefore m=7/10 hours :This is the time taken during slow speed.Multiply it with slow speed and you’ll get the distance.Distance

69

APTITUDE 1

= 7/10 x 2.5= 7/4 kms.

2. Sonu college

Slow speed

Fast speed

Speed km/h 5 6Ratio-reversed (Time) 6 5

So % decrease in time=(6-5)/6=1/6Therefore 1/6 of slow time (m)= 25/60 hrs.M=25 x 6/60 hrsMultiply it with slow speed (5) and you get distanceDistance=speed x time=5 x 25 x 6/60=25/2=12.5 km distance between home and college.