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1

Explanation for Correct Answer E :

Choice (E) is correct. If and then so

Explanation for Incorrect Answer A :

Choice (A) is not correc t. If were equal to thenSolving for gives However, notTherefore, cannot be equal to

Explanation for Incorrect Answer B :

Choice (B) is not correct. If were equal to then

Solving for gives However, not

Therefore, cannot be equal to

Explanation for Incorrect Answer C :

Choice (C) is not correct. If were equal to thenSolving for gives However, notTherefore, cannot be equal to

Explanation for Incorrect Answer D :

Choice (D) is not correc t. If were equal to then

Solving for gives However, not

View Answers and ExplanationsBack to Score Report

Online - Practice Test #3

If and wha t is the v alue of

(A )

(B)

(C)

(D)

(E)

A NSWERS AND E XPLANATIONS

Test Sections

Section 1

Section 3

Section 4

Section 5

Section 6

Section 7

Section 8

Section 9

Sect ion 1 0
https://satonlinecourse.collegeboard.org/SR/fptscorereportaction.do?whereFrom=scoreNow&assessmentResultId=10979468https://satonlinecourse.collegeboard.org/SR/showExamAnswerExplanation.do?assessmentResultId=10979468&assessmentLocator=3CF8A603-66B4-9F5F-0099-8B08F172FC7B-1::AS&section=3CFB40E9-66B4-9F5F-9BE4-B7C62645AE24-1::S&fromPage=examscorereport&resultId=nullhttps://satonlinecourse.collegeboard.org/SR/showExamAnswerExplanation.do?assessmentResultId=10979468&assessmentLocator=3CF8A603-66B4-9F5F-0099-8B08F172FC7B-1::AS&section=3CFB40E9-66B4-9F5F-9BE4-B7C646DE3384-1::S&fromPage=examscorereport&resultId=nullhttps://satonlinecourse.collegeboard.org/SR/showExamAnswerExplanation.do?assessmentResultId=10979468&assessmentLocator=3CF8A603-66B4-9F5F-0099-8B08F172FC7B-1::AS&section=3CFB40D9-66B4-9F5F-9BE4-B7C6C098A518-1::S&fromPage=examscorereport&resultId=nullhttps://satonlinecourse.collegeboard.org/SR/showExamAnswerExplanation.do?assessmentResultId=10979468&assessmentLocator=3CF8A603-66B4-9F5F-0099-8B08F172FC7B-1::AS&section=3CFB40D9-66B4-9F5F-9BE4-B7C6008F7F8E-1::S&fromPage=examscorereport&resultId=nullhttps://satonlinecourse.collegeboard.org/SR/showExamAnswerExplanation.do?assessmentResultId=10979468&assessmentLocator=3CF8A603-66B4-9F5F-0099-8B08F172FC7B-1::AS&section=3CFB40D9-66B4-9F5F-9BE4-B7C651E87543-1::S&fromPage=examscorereport&resultId=nullhttps://satonlinecourse.collegeboard.org/SR/showExamAnswerExplanation.do?assessmentResultId=10979468&assessmentLocator=3CF8A603-66B4-9F5F-0099-8B08F172FC7B-1::AS&section=3CFB40D9-66B4-9F5F-9BE4-B7C6A4FC1CB5-1::S&fromPage=examscorereport&resultId=nullhttps://satonlinecourse.collegeboard.org/SR/showExamAnswerExplanation.do?assessmentResultId=10979468&assessmentLocator=3CF8A603-66B4-9F5F-0099-8B08F172FC7B-1::AS&section=3CFB40D9-66B4-9F5F-9BE4-B7C66FD69DDB-1::S&fromPage=examscorereport&resultId=nullhttps://satonlinecourse.collegeboard.org/SR/showExamAnswerExplanation.do?assessmentResultId=10979468&assessmentLocator=3CF8A603-66B4-9F5F-0099-8B08F172FC7B-1::AS&section=3CFB40D9-66B4-9F5F-9BE4-B7C62C0C2B44-1::S&fromPage=examscorereport&resultId=nullhttps://satonlinecourse.collegeboard.org/SR/showExamAnswerExplanation.do?assessmentResultId=10979468&assessmentLocator=3CF8A603-66B4-9F5F-0099-8B08F172FC7B-1::AS&section=3CFB40D9-66B4-9F5F-9BE4-B7C68A022024-1::S&fromPage=examscorereport&resultId=nullhttps://satonlinecourse.collegeboard.org/SR/fptscorereportaction.do?whereFrom=scoreNow&assessmentResultId=10979468https://satonlinecourse.collegeboard.org/SR/professionalreportshomeaction.dohttps://satonlinecourse.collegeboard.org/SR/examsaction.dohttps://satonlinecourse.collegeboard.org/SR/quizzesaction.dohttps://satonlinecourse.collegeboard.org/SR/lessonsaction.dohttps://satonlinecourse.collegeboard.org/SR/gettingstartedaction.dohttps://satonlinecourse.collegeboard.org/SR/studenthomeaction.dohttps://satonlinecourse.collegeboard.org/SR/logoff.dohttp://popup%28%27/SR/mybookmarkaction.do'%20,600%20,600)https://satonlinecourse.collegeboard.org/SR/viewMyOrganizer.dohttps://satonlinecourse.collegeboard.org/SR/studentProfileAction.dohttps://satonlinecourse.collegeboard.org/SR/helpAction.do


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Therefore, cannot be equal to

2

Explanation for Correct Answer B :

Choice (B) is correct. Since Cliff earns dollars for each hour thathe works as a television salesperson, his salary for hours of work is

dollars. Since Cliff earns dollars for each television he sells,his commission for selling televisions is dollars. Therefore,the total amount, in dollars, Cliff earns if he sells televisions while working hours is


Choice (A) is not correc t. If Cliff sold televisions while workinghours, he would earn dollars. But if Cliff sellstelevisions while working hours, he would earn dollars.


Choice (C) is not correct. Multiplying the average of Cliffscommission rate and hourly pay by the sum of the number of hourshe worked and the number of televisions he sold does not give thetotal amount Cliff earned.


Choice (D) is not correct. Multiplying the sum of Cliffs commission

rate and hourly pay by the sum of the number of hours he workedand the number of televisions he sold does not give the total amountCliff earned.

Explanation for Incorrect Answer E :

Choice (E) is not correct. Multiplying the sum of Cliffs commissionrate and hourly pay by the product of the number of hours he worked and the number of televisions he sold does not give the total

amount Cliff earned.

Cliff earns per hour for each hour tha t he works as a telev ision salesperson

plus an additional for each telev ision tha t he sells. Which of the following

represents the am ount Cliff earns, in dollars, if he sells telev isions while working hours?

(A )

(B)

(C)

(D)

(E)



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3

Explanation for Correct Answer A :

Choice (A) is correct. If a nonzero number is multiplied by a negativenumber, its sign will change. Since the numbers on the original listare all nonzero and each would be multiplied by to form the new list, there would be as many negative numbers on the new list as thereare positive numbers on the original list. Counting positive numberson the original list thus shows that two of the numbers on the new list would be negative.


Choice (B) is not correct. There would be only two negative numbers

on the new list: and


Choice (C) is not correct. There are four negative numbers on theoriginal list, but there would be only two negative numbers on thenew list.


Choice (D) is not correct. There would be only two negative numberson the new list: and


Choice (E) is not correct. The new list would have six numbers, butonly two of these would be negative numbers.

4

If each nu mber in the list abov e were to be mu ltiplied by a new list would be

form ed consisting of the six products. How m any of the six n um bers on t he new list would be n egat iv e num ber s?

(A) Two

(B) Three

(C) Fou r

(D) Fiv e

(E) Six


The recta ngu lar ca rd pictur ed abov e has a black dot on one side, and t he other side


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Explanation for Correct Answer C :

Choice (C) is correct. In all three figures, the cards have sides parallelto the sides of the original rectangle, so to determine if any of thefigures can be obtained by rotating the card, we only need to

consider successive rotations of in a single direction. When therectangular card is rotated in the clockwise direction, the resultis the rectangle in figure I. When the card is again rotated in theclockwise direction, the result is the rectangle in figure II. When thecard completes a third rotation in the clockwise direction, theresult is a vertical rectangle with the black dot in the upper-leftcorner. The card in figure III has the black dot in the upper-rightcorner, so this rotation does not produce any of the three figuresshown. When the rectangular card completes one more rotationin the clockwise direction, the result is the orientation of the originalcard. Therefore, only figures I and II could be obtained by rotating

the card.


Choice (A) is not correct. While it is true that figure I could beobtained by rotating the card, figure II could also be obtained.


Choice (B) is not correct. While it is true that figure II could beobtained by rotating the card, figure I could also be obtained.


Choice (D) is not correct. While it is true that figure I could beobtained by rotating the card, figure III could not be obtained.


Choice (E) is not correct. While it is true that figures I and II could beobtained by rotating the card, figure III could not be obtained.

s an . c o e gures e ow cou e o a ne y ro a ng e car

(A ) I on ly

(B) II on ly

(C) I and II only (D) I and III only

(E) I, II, and III



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5


Choice (C) is correct. From their positions on the number line, the value of appears to be approximately and the value of

appears to be approximately Then is approximatelyso is approximately Therefore, of the

choices given, is closest to


Choice (A) is not correct. From their positions on the number line,the value of appears to be approximately and the value ofappears to be approximately Then is approximately

so, of the choices given, is closest to

However, the question asked for the best approximation of

not of


Choice (B) is not correct. From their positions on the number line,the value of appears to be approximately and the value of

appears to be approximately Then, of the choices given, isclosest to However, the question asked for the bestapproximation of not of

Explanation for Incorrect Answer D : Choice (D) is not correc t. If the value of was approximatelyand the value of was approximately then the value of

would be approximately However, from their positions

on the number line, the value of appears to be approximatelyand the value of appears to be approximately and

Thus, of the choices given, is a better

approximation of than is

Ex lanation for Incorrect Answer E :

On the n um ber line above, w hich of the following is the best approximat ion of

(A )

(B)

(C)(D)

(E)



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Choice (E) is not correct. If the value of was approximatelyand the value of was approximately then the value of

would be approximately However, from their

positions on the number line, the value of appears to beapproximately and the value of appears to beapproximately and Thus, of the choices

given, is a better approx imation of than is

6

Explanation for Correct Answer B : Choice (B) is correct. When a transversal intersects a pair of parallellines, the alternate interior angles are supplementary (have measures

that add up to ). Thus or

Therefore, if it follows that


Choice (A) is not correc t. If and the sum of the

measures of the three labeled angles would beHowever, the sum of the measures of these

angles must be because when a transversal intersects a pair of parallel lines, the alternate interior angles are supplementary.Therefore, the value of cannot be


Choice (C) is not correct. If and the sum of themeasures of the three labeled angles would be

However, the sum of the measures of these

In the figure above, line is parallel to line If wha t is the v alue of

(A )

(B)

(C)

(D)

(E)



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parallel lines, the alternate interior angles are supplementary.Therefore, the value of cannot be


Choice (D) is not correct. If and the sum of the


angles must be because when a transversal intersects a pair of parallel lines, the alternate interior angles are supplementary.Therefore, the value of cannot be


Choice (E) is not correct. If and the sum of the


angles must be because when a transversal intersects a pair of parallel lines, the alternate interior angles are supplementary.

Therefore, the value of cannot be

7

Explanation for Correct Answer D :

Choice (D) is correc t. Let be the total number of compact discsJaime has. Then of Jaimes discs cost and discs cost

The total cost, in dollars, of the discs is thusTherefore, the

average cost, in dollars, per disc is or

Alternatively, one can see that the question is asking for the weightedaverage of two different kinds of Jaimes discs, those that cost

each, and the others that cost each, with respective weightsand Therefore, the average cost per disc, in dollars, of Jaimes

compact discs is or

If percent of Jai m es compact discs cost each and if percent of her

compact discs cost each, wha t is the av erage (arith metic mean ) cost per disc?

(A )

(B)

(C)(D)

(E)



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Choice (A) is not correct. If Jaime had an equal number of discsand discs, the average cost per disc would simply be the average

of and which is But the number of discs isnot the same as the number of discs.


Choice (B) is not correct. If of Jaimes discs cost and the rest

cost the average cost per disc would be But of

Jaimes discs cost , not


Choice (C) is not correct. If of Jaimes discs cost and the rest

cost the average cost per disc would be But of



Choice (E) is not correct. If of Jaimes discs cost and the

rest cost the average cost per disc would be But of


8


Choice (B) is correct. By the laws of exponents, the equation

can be rewritten as and can be

If and are positive integers, and which of the

following could be the v alue of

(A )

(B)

(C)

(D)

(E)



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rewr tten as owever, can e wr tten as an

can be written as Hence and Since the

base on each side of both exponential equations is it follows thatand The integers and are the only positive

integers whose sum is and whose product is Therefore,could be either or Thus, of the integers listed, only could bethe value of


Choice (A) is not correc t. If were equal to then, since

the value of would have to be It is

also given that so However,

Therefore, the value of cannot be


Choice (C) is not correct. The values of and must be positiveintegers and must satisfy the two equations and If

were equal to to satisfy the equation would have

to be Since is not a positive integer, the value of cannot beIt follows that the value of cannot be


Choice (D) is not correc t. The values of and must be positiveintegers and must satisfy the two equations and If




Choice (E) is not correct. The values of and must be positiveintegers and must satisfy the two equations and If



9If what is the v alue

(A )

(B)

(C)

(D)

(E)


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Choice (E) is correct. We have so

so


Choice (A) is not correc t. If were equal to then would

be equal to It would follow that was equal to

However, not Therefore, cannot be

equal to


Choice (B) is not correct. If were equal to then would

be equal to It would follow that was equal to

However, not Therefore, cannot be equal to


Choice (C) is not correct. If were equal to then

would be equal to It would follow that was equal

to However, not Therefore, cannot be

equal to


Choice (D) is not correc t. If were equal to then

would be equal to It would follow that was equal

to However, not Therefore, cannot be

equal to

10


In Ms. Murray s art class, one-third of the students hav e taken a t least one previous

art class of those one-third hav e taken two or more rev ious art classes. Which of


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Explanation for Correct Answer E : Choice (E) is correct. Since one-third of the students in Ms. Murraysclass have taken at least one previous art class, it follows that two-thirds of the students have taken no previous art classes. Thus thecorrect circle graph must have one sector that makes up two-thirdsof the graph, so that it represents the students who have taken noprevious art classes. Only the graphs in options (D) and (E) have sucha sector.

Of the students who have taken at least one art class, one-third havetaken two or more art classes, and so two-thirds have taken exactly

one class. Thus the remaining third of the correct circle graph must be divided into two sectors, one of which is twice as large as theother. T he graph in (D) has the remaining third divided into sectorsof equal size; hence it is not the correct graph. The graph in (E) hasthe remaining third divided into two sectors, one of which is twice aslarge as the other.

Therefore, the circle graph in (E) could represent the students in Ms.

Murrays class divided into different groups: those who have takentwo or more previous art classes, one previous art class, or noprevious art classes.

the following circle gr aphs could represent Ms. Murr ay s class divided into

different g roups: those who hav e taken two or more prev ious art classes, oneprev ious art class, or no prev ious art class?

(A )

(B)

(C)

(D)

(E)



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Choice (A) is not correct. Since one-third of the students in Ms.Murrays class have taken at least one previous art class, it followsthat two-thirds of the students have taken no previous art classes.Thus the correct c ircle graph must have one sector that makes uptwo-thirds of the graph, so that it represents the students who havetaken no previous art classes. The circle graph in (A) has no such

sector.


Choice (B) is not correct. Since one-third of the students in Ms.Murrays class have taken at least one previous art class, it followsthat two-thirds of the students have taken no previous art classes.Thus the correct c ircle graph must have one sector that makes uptwo-thirds of the graph, so that it represents the students who havetaken no previous art classes. The circle graph in (B) has no suchsector.


Choice (C) is not correct. Since one-third of the students in Ms.Murrays class have taken at least one previous art class, it followsthat two-thirds of the students have taken no previous art classes.Thus the correct c ircle graph must have one sector that makes uptwo-thirds of the graph, so that it represents the students who havetaken no previous art classes. The circle graph in (C) has no suchsector.


Choice (D) is not correct. Since one-third of the students in Ms.Murrays class have taken at least one previous art class, it followsthat two-thirds of the students have taken no previous art classes.Thus the correct c ircle graph must have one sector that makes uptwo-thirds of the graph, so that it represents the students who havetaken no previous art classes. Only the graphs in options (D) and (E)have such a sector.

Of the students who have taken at least one art class, one-third havetaken two or more art classes, and so two-thirds have taken exactly one class. Thus the remaining third of the correct circle graph must be divided into two sectors, one of which is twice as large as theother. T he graph in (D) has the remaining third divided into sectorsof equal size. Therefore, it is not the correct graph.

11 Three consecutiv e integers are such t hat t wo times the largest is thr ee times thesmallest. What is the smallest of these thr ee integers?

(A )

(B)


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Choice (D) is correc t. Let be the smallest of the consecutiveintegers. Then the other two integers are and It is giventhat two times the largest integer is three times the smallest integer:

Therefore, and so the value of thesmallest of the three integers, is


Choice (A) is not correct. If the smallest of the three integers werethen the largest would be But

Explanation for Incorrect Answer B : Choice (B) is not correct. If the smallest of the three integers were

then the largest would be But


Choice (C) is not correct. If the smallest of the three integers werethen the largest would be But


Choice (E) is not correct. The largest of the three integers is butthe question asks for the smallest, which is

12


Choice (B) is correct. A circ le with radius inches has acircumference of inches. If a circle has a circumference ofinches, then and so It follows that any circ le in

(C)

(D)

(E)


Point lies in plane How ma ny circles are there in plane that hav e center and a circu mference of inches?

(A) None

(B) One

(C) Two(D) Fou r

(E) More than four



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have a radius of inches. No two distinct circ les in the same planecan have the same center and radius, so there is exactly one circle inplane that has center and a circumference of inches.


Choice (A) is not correc t. There is exactly one circle in plane thathas center and a circumference of inches.


Choice (C) is not correct. There cannot be two different circles inplane that have center and a circumference of inches.There is exactly one such circle.


Choice (D) is not correct. There cannot be four different circles inplane that have center and a circumference of inches.There is exactly one such circle.


Choice (E) is not correc t. T here cannot be more than four differentcircles in plane that have center and a circumference ofinches. There is exactly one such circle.

13

The table abov e giv es the temperatu re of five different mixtu res at seconds afterthe mix tur e was form ed. The temperat ure of which m ixtu re falls the great estnum ber of degrees during the period from to


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Explanation for Correct Answer A :

Choice (A) is correct. According to the table, the temperature of mixture is degrees seconds after the mixture

was formed and is degrees seconds after themixture was formed. So, the temperature of mixture falls

degrees during the period from to

According to the table, the temperature of mixture is

degrees seconds after the mixture was formed

and is degrees seconds after the mixture was

formed. So, the temperature of mixture falls degreeduring the period from to


degree seconds after the mixture was formed

and is degrees seconds after the mixture was

formed. So, the temperature of mixture falls degree

during the period from to


degree seconds after the mixture was formed and is

degrees seconds after the mixture was formed.

So, the temperature of mixture rises degree during theperiod from to

According to the table, the temperature of mixture isdegrees seconds after the mixture was formed and

is degrees seconds after the mixture was formed.

So, the temperature of mixture rises degrees duringthe period from to

So, of the five mixtures, the temperatures of mixtures andfall during the period from to and, for these threemixtures, the temperature of mixture falls the greatest number of degrees during this period.

(A )

(B)

(C)

(D)

(E)



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Choice (C) is not correct. For example, if then oris less than or But not


Choice (D) is not correct. The difference

simplifies to not


Choice (E) is not correct. For example, if then oris less than or But not

16


Choice (D) is correct. Since the measure of arc is themeasure of the central angle is All radii of a circ le are of

equal length, so sides and of are of equal length. Itfollows that the two angles opposite these sides of are of equal measure, and so The sum of themeasures of the angles of is so

and


Choice (A) is not correct. The sum of the measures of the angles ofis so if were equal to then would

In the figure above, the circle has center If the measure of arc is

wha t i s th e v al ue of

(A )

(B)(C)

(D)

(E)



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, ,the value of cannot be


Choice (B) is not correct. The sum of the measures of the angles ofis so if were equal to then would

be equal to However, not Therefore,the value of cannot be


Choice (C) is not correct. The sum of the measures of the angles of

is so if were equal to then would be equal to However, not Therefore,the value of cannot be


Choice (E) is not correct. The sum of the measures of the angles of

is so if were equal to then would be equal to However, not Therefore,the value of cannot be

17


Choice (C) is correct. We have for all real values of and so

for all real values of Therefore, if the function

has the property that for all real values of then could

be defined by


Choice (A) is not correc t. If for all real values of thencould not be defined by because


If for all real v alues of which of the following could be the function

(A )

(B)

(C)

(D)

(E)



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Choice (B) is not correct. If for all real values of thencould not be defined by because


Choice (D) is not correc t. If for all real values of then

could not be defined by because


Choice (E) is not correct. If for all real values of then

could not be defined by because

18


Choice (B) is correct. The number of possibilities can be found by finding the number of possibilities for each digit and then applyingthe Fundamental Theorem of Counting. The first digit must be sothere is possibility for the first digit. Since digits may be repeated(but do not have to be repeated), there are no restrictions on thesecond or third digits; thus, there are possibilities for the seconddigit and possibilities for the third digit. Since the last digit cannot

be there are possibilities for the last digit. Therefore, by theFundamental Theorem of Counting, there aredifferent positive four-digit integers satisfying the given conditions.


Choice (A) is not correct. This is the number of different positivefour-digit integers such that the first digit is no digit is anddigits may not be repeated. But this is not what the question asked.


How m any different positiv e four-digit integers can be form ed if the first digit m ust be th e la st digi t c an not be an d dig it s may be r epea ted?

(A )

(B)

(C)

(D)

(E)



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.four-digit integers with first digit However, there is the additionalcondition that the last digit cannot be


Choice (D) is not correct. This is the number of different positivefour-digit integers. However, there are two conditions: the first digitmust be and the last digit cannot be


Choice (E) is not correct. This is the number of different nonnegativeintegers with four or fewer digits. However, the number must haveexactly four digits and satisfy two additional conditions: the firstdigit must be and the last digit cannot be

19


Choice (D) is correc t. To find all possible values of that satisfy theconditions of the question, express in terms of using theequation given. Combining the terms of equation that contain onthe left side gives or Dividing both sides of the equation by (note that and so ) yields

This can be rewritten as

Since can be any number greater than it follows that can

be any number greater than Hence can be any number

greater than but less than Therefore, all possible values of

that satisfy under the condition are

given by


Choice (A) is not correc t. Since and it is clear

If wha t are all possible v alues of tha t satisfy the equation above?

(A )

(B)

(C)

(D)

(E)



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that cannot take negative values.


Choice (B) is not correct. Since and it is clear

that should be greater than


Choice (C) is not correct. Since and it is clear

that should be greater than but contains values ofless than


Choice (E) is not correct. Since and it is clear

that is less than

20


Choice (E) is correc t. T he five-sided figure can be divided into arectangle and a triangle as shown below.

Which of t he following r epr esents t he ar ea of th e fiv e-sided figu re shown a bov e?

(A )

(B)

(C)

(D)

(E)



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The area of the five-sided figure can be found by adding the area of the rectangle and the area of the triangle in the figure above. Therectangle has sides of length and and so its area is Thetriangle in the figure is a right triangle, and so its base isequal to its height. The height of the triangle is since it is equalto the length of the entire vertical segment ( ) minus the short side

of the rectangle ( ). Therefore, the area of the triangle is

Therefore, the area of the five-sided figure is


Choice (A) is not correct. As shown in the figure in the rationale forthe key, the five-sided figure can be divided into a rectangle and anisosceles right triangle. The rectangle has sides of length and

and so its area is not The legs of the isosceles right triangle

are each of length not and so its area is not

Therefore, the area of the f ive-sided figure is

not


Choice (B) is not correct. As shown in the figure in the rationale forthe key, the five-sided figure can be divided into a rectangle and anisosceles right triangle. The rectangle has sides of length andand so its area is The legs of the isosceles right triangle are each

of length not and so its area is not

Therefore, the area of the five-sided figure is not


Choice (C) is not correct. As shown in the figure in the rationale forthe key, the five-sided figure can be divided into a rectangle and anisosceles right triangle. The rectangle has sides of length andand so its area is The legs of the isosceles right triangle are each

of length and so its area is not


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