501 quantitative comparison_questions

501
Quantitative Comparison
Questions

501Quantitative Comparison
Questions
N E W Y O R K

Copyright © 2003 LearningExpress, LLC.
All rights reserved under International and Pan-American Copyright Conventions.
Published in the United States by LearningExpress, LLC, New York.
Library of Congress Cataloging-in-Publication Data:
501 quantitative comparison questions.—1st ed.
p. cm.—(Skill builder in focus)
ISBN 1-57685-434-5 (pbk.)
1. Mathematics—Examinations, questions, etc. 2. Achievement
tests—Study guides. 3. Graduate Management Admissions Test—Study
guides. I. Title: Five hundred and one quantitative comparison
questions. II. Series.
QA43 .A116 2003
510'.76—dc21 2002011852
Printed in the United States of America
9 8 7 6 5 4 3 2 1
First Edition
ISBN 1-57685-434-5
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The LearningExpress Skill Builder in Focus Writing Team is com-
prised of experts in test preparation, as well as educators and teachers who
specialize in language arts and math.
LearningExpress Skill Builder in Focus Writing Team
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(continued on next page)

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Introduction ix
1 Arithmetic
Questions 1
Answers and Explanations 11
2 Algebra
Questions 25
3 Geometry
Questions 59
4 Data Analysis
Questions 113
Contents

ix
Welcome to 501 Quantitative Comparison Questions! This
book is designed to help you prepare for a specialized math section of a
select few of the most important assessment exams. By completing the exer-
cises in this book, you will also increase your math knowledge and reﬁne
your logic and analytical skills.
Key academic aptitude tests produced by the Educational Testing Service
(ETS) for the College Board—the Preliminary Scholastic Aptitude
Test/National Merit Scholarship Qualifying Test (PSAT/NMSQT) exam,
the Scholastic Aptitude Test (SAT) assessment, and the Graduate Records
Examination (GRE) test—include a quantitative comparison section
within the math portion of the exam, so for college-bound students and
many graduate students, mastering this question type is essential for getting
into their school of choice.
What makes this book different than other math practice books? It’s sim-
ple—this math practice book contains only questions that ask you which
column contains the item of greater value, if the values are the same, or if
the value cannot be determined by the information given. Gaining famil-
iarity with this specialized question type is a proven technique for increas-
ing test scores.
Introduction

In order to solve a quantitative comparison problem, you must ﬁrst com-
pare the quantities in the two columns and then decide whether:
■ one quantity is greater than the other
■ the two quantities are equal
■ the relationship cannot be determined from the information
given
On the PSAT exam, SAT assessment, or GRE test, your answer always
should be:
■ a if the value in Column A is greater
■ b if the value in Column B is greater
■ c if the two values are equal, or
■ d if the relationship cannot be determined from the
information given in the question
Math Topics
The book is divided into four content sections. Math topics included in this
volume are arithmetic, algebra, geometry, and data analysis. These cate-
gories are further divided to include the following concepts:
Arithmetic
■ Absolute Value
■ Decimals
■ Exponents and Square Roots
■ Fractions
■ Integers
■ Ordering and the Real Number Line
■ Percent
■ Ratio
Algebra
■ Applications
■ Coordinate Geometry
■ Inequalities
■ Operations with Algebraic Expressions
■ Rules of Exponents
x
501 Quantitative Comparison Questions

xi
■ Solving Linear Equations
■ Solving Quadratic Equations in One Variable
■ Translating Words into Algebraic Expressions
Geometry
■ Circles
■ Lines and Angles
■ Polygons
■ Quadrilaterals
■ Three-Dimensional Figures
■ Triangles
Data Analysis
■ Counting
■ Sequences
■ Data Representation and Interpretation
■ Frequency Distributions
■ Measures of Central Tendency
■ Measures of Dispersion
■ Probability
A Note about the PSAT/NMSQT Exam
On the PSAT/NMSQT exam, there are two 25-minute math sections, for
a total of 40 questions. Of these questions, 12 are quantitative comparisons.
This math section requires a basic knowledge of arithmetic, algebra, and
geometry, so practicing with this book is ideal, as you will improve your
PSAT/NMSQT exam score and start getting familiar with the SAT exam
that you will encounter in the next school year.
A Note about the SAT Assessment
On the SAT exam, there are two 30-minute math sections, for a total of 60
questions. Of these questions, 30 are quantitative comparison questions.
(There is also a 15-minute section of 5-choice math questions.) Math concepts
tested include arithmetic, algebra, geometry, and other topics, such as log-
ical reasoning, probability, and counting. This book focuses on each of

these areas of math so that you can get the targeted practice necessary to ace
the quantitative comparison sections in the SAT exam.
In addition, each SAT exam contains a 30-minute experimental verbal or
math section, used for equating, that does not count toward your score, so
there may be 15 more quantitative comparisons on your exam (though you
will only be scored for 30 of the potential 45 questions). Since you don’t
know which section is experimental, it’s best to be prepared to answer many
of these question types. You are allowed to use a calculator to answer math
questions on the exam, if you wish.
A Note about the GRE Test
You will be given 45 minutes to complete the 28 quantitative comparison
questions on the computer-based GRE test. However, there are 60 quan-
titative comparison questions on the paper-based exam. Overall, the per-
centage of questions on the GRE test that are quantitative comparison are
the same for both versions of the test: approximately 30% of both tests are
quantitative comparison questions. The content areas included in the quan-
titative sections of the test are arithmetic, algebra, geometry, and data analy-
sis. These are math concepts usually studied in high school, and this book
speciﬁcally targets these areas. Calculators are not permitted.
Questions, Questions, Questions
You have just read about the math topics covered in this book. To mimic a
real test environment, math concepts are mixed within each section. For
example, in the arithmetic section, you may ﬁnd a question or two on expo-
nents, then three questions on square roots, followed by three questions
involving decimals and fractions. Then the next question is about square
roots. This way, you are preparing for any type of arithmetic question that
you could encounter on the real exam. This helps you to be prepared for
whatever questions the College Board/ETS throws your way. Each chap-
ter begins with an information and instruction overview describing the
mathematical concepts covered in that particular section. Then you are pre-
sented with a variety of problems awaiting your solutions.
If you are practicing for the PSAT assessment or SAT exam, you should
have a calculator on hand as you work through some of the chapters,
xii

xiii
because calculators are allowed in the testing center. Even if you don’t use
it to arrive at solutions, it’s a good idea to use it to check your calculations.
This goes for everyone—even GRE test-takers. If you have difficulty fac-
toring numbers, a multiplication chart may help you. And if you are unfa-
miliar with prime numbers, use a list of them so you won’t waste time trying
to factor numbers that can’t be factored. Don’t forget to keep lots of scrap
paper on hand for working out complex problems.
Practice Makes Perfect
Because this book is designed for many levels of test takers, you may find
that some of the more advanced questions are beyond your ability. If you
are using this book to study for the PSAT assessment or SAT exam, there
is a chance that you may get a few of the toughest questions wrong. Don’t
worry! If you are getting most of the questions correct, you are probably in
good shape for your test. However, if you are studying for the GRE test, the
full range of questions presented is appropriate for your level.
The questions in this book can help you prepare for your test in many
ways. First, completing these practice exercises will make you familiar with
the question format. Quantitative comparisons usually involve less reading,
take less time to answer, and require less computation than regular 5-choice
questions. Remember, there are only four choices from which to select the
correct answer: A, B, C, and D. If the quantity in column A is larger, you
will select the letter A on your answer grid. If column B contains the greater
quantity, then you will select B. Select the answer C if the two quantities are
equal, and the letter D if you cannot determine which is larger from the
choices and information given. When you take the test, be careful not to
mark an “E” on the answer sheet because it is never the correct choice.
Second, quantitative comparison practice will get you thinking of values
in terms of equalities, inequalities, and estimation. Likewise, you will get
accustomed to knowing how to logically assess whether or not you have
enough information in the question to assign value. However, it is not
always necessary to find the exact value of the two quantities, and often, it
is important not to waste time doing so. Remember, you have a limited
amount of time to arrive at correct answers, so it is important to use esti-
mating, rounding, and the elimination of irrelevant information to deter-
mine the relationship between the information in Column A and the
information in Column B.

Third, in the test-taking environment, it can be difﬁcult to switch gears
from regular math questions to quantitative comparisons; completing the
exercises in this book will make these mental gymnastics more comfortable
as you grow familiar with the question format. Also, your performance on
these questions will help you assess your overall math skill level. Because the
quantitative comparison questions assess a wide variety of math topics, these
exercises will help you pinpoint the areas of math for which you need fur-
ther study.
Finally, each question is fully explained at the end of each chapter. The
answer keys give you not only the right answer, but also an explanation of
why the answer is correct. In other words, for every problem, there is a
complete explanation of the solution. If you ﬁnd yourself getting stuck solv-
ing a problem, you can look at the answer explanation and use it as a per-
sonal tutor.
You have already taken an important step toward improving your math
skills and your score. You have shown your commitment by purchasing this
book. Now all you need to do is to complete each exercise, study the answer
explanations, and watch your math skills increase. You can even work in pen-
cil and do the exercises again to reinforce what you have learned. Good luck!
xiv

xv
Tip Sheet
✔ Not exactly. It is not always necessary to ﬁnd the exact value of the two
quantities, and often, it is important not to waste time doing so. Use esti-
mating, rounding, and the elimination of excess information to deter-
mine the relationship and avoid wasting time.
✔ Look alikes. Attempt to make the two columns look as similar as possi-
ble. For example, make sure all units are equal. This is also true if one of
the answer choices is a fraction or a decimal. If this is the case, then make
the other answer into an improper fraction or a decimal, in order to make
the choices look the most similar.
✔ It’s not necessarily nice to share. Eliminate any information the
columns share. This will leave you with an easier comparison. For exam-
ple, you are given the two quantities 5(x + 1) and 3(x + 1), with the pro-
viso that x is greater than 0. In this case, you would select the ﬁrst
quantity because, since you know that x is a positive quantity, you can
eliminate the (x + 1) from both. That leaves you to decide which is
greater, 5 or 3. This has become a very easy problem by eliminating the
information that the two quantities shared.
✔ Plug it in. Assign values for unknowns or variables. If you can do this
quickly, many comparisons will become straightforward. Plug in num-
bers for variables whenever they are given. Always remember to simplify
the equation or expression as much as possible before you plug in a value.
✔ Sticky situations. Try not to get stuck doing complicated computations.
If you feel yourself doing a lot of computations, stop and try another
method. There is often more than one way to solve a problem. Try to
pick the easier one.
✔ No assumptions necessary. Make no assumptions about the informa-
tion listed in the columns. If the question asks you to make assumptions,
then choose D. For example, if one of the answer choices is x2
, you can-
not assume that the answer is a positive root. Remember that x2
will have
two roots, one positive and one negative. Many times the test-maker will
try to trick you into assuming that the answer to such a problem is
known. Do not let the test fool you. Be aware of the possibility of mul-
tiple answers.
✔ A parenthetical aside. If one or both of the expressions being compared
have parentheses, be sure to remove the parentheses by completing the
calculations before proceeding. This is a simple technique that can make

a large difference in the similarity of the two quantities. For example, if
you move the parentheses from the two quantities (x − 2)(x − 2) and
x2
− 4x + 4 by squaring (x − 2), you can clearly see that they are equal.
✔ Let’s play Operation™. Sometimes the best way to solve the question
is to perform an operation on both columns. This is especially useful
when working with fractions. Often, on ﬁnding a LCD and multiplying
that number in both columns helps to make the comparison easier. Just
keep in mind that, like working in an equation, the operation performed
must be exactly the same in each column.
✔ Timing is everything. Use your time wisely. Try to solve each problem,
or be close to a solution, after one minute. It is not necessary to complete
every problem on the test unless you want to be the next math genius
honored by the Educational Testing Service. It is best just to focus on the
problems you know how to do. It is a good idea to practice timing your-
self as you solve the practice problems. See how close you can get to a
solution in one minute per problem.
xvi

xvii
Multiplication Table
× 2 3 4 5 6 7 8 9 10 11 12
2 4 6 8 10 12 14 16 18 20 22 24
3 6 9 12 15 18 21 24 27 30 33 36
4 8 12 16 20 24 28 32 36 40 44 48
5 10 15 20 25 30 35 40 45 50 55 60
6 12 18 24 30 36 42 48 54 60 66 72
7 14 21 28 35 42 49 56 63 70 77 84
8 16 24 32 40 48 56 64 72 80 88 96
9 18 27 36 45 54 63 72 81 90 99 108
10 20 30 40 50 60 70 80 90 100 110 120
11 22 33 44 55 66 77 88 99 110 121 132
12 24 36 48 60 72 84 96 108 120 132 144

Commonly Used Prime Numbers
2 3 5 7 11 13 17 19 23 29
31 37 41 43 47 53 59 61 67 71
73 79 83 89 97 101 103 107 109 113
127 131 137 139 149 151 157 163 167 173
179 181 191 193 197 199 211 223 227 229
233 239 241 251 257 263 269 271 277 281
283 293 307 311 313 317 331 337 347 349
353 359 367 373 379 383 389 397 401 409
419 421 431 433 439 443 449 457 461 463
467 479 487 491 499 503 509 521 523 541
547 557 563 569 571 577 587 593 599 601
607 613 617 619 631 641 643 647 653 659
661 673 677 683 691 701 709 719 727 733
739 743 751 757 761 769 773 787 797 809
811 821 823 827 829 839 853 857 859 863
877 881 883 887 907 911 919 929 937 941
947 953 967 971 977 983 991 997 1009 1013
xviii

xix
Reference Sheet
■ The sum of the interior angles of a triangle is 180°.
■ The measure of a straight angle is 180°.
■ There are 360 degrees of arc in a circle.
r
A = πr2
C = 2πr
r
V = πr2h
h
l
w
h
l
w
A = lw
h
b
A = 1
2
bh
45°
45°
S
S
S
√¯¯¯2
√¯¯¯¯¯3x
60°
30°
x
2x
Special Right Triangles

1
In this chapter, the following math concepts will be the subject of the 125
arithmetic-based quantitative comparison questions:
■ Absolute Value
■ Decimals
■ Exponents and Square Roots
■ Fractions
■ Integers
■ Ordering and the Real Number Line
■ Percent
■ Ratio
Some important guidelines:
Numbers: All numbers used are real numbers.
Figures: Figures that accompany questions are intended to provide infor-
mation useful in answering the questions. Unless otherwise indicated, posi-
tions of points, angles, regions, etc. are in the order shown; angle measures
are positive; lines shown as straight are straight; and ﬁgures lie in a plane.
1Arithmetic

Unless a note states that a ﬁgure is drawn to scale, you should NOT solve
these problems by estimating or by measurement, but by using your knowl-
edge of mathematics.
Common Information: In a question, information concerning one or both
of the quantities to be compared is centered above the two columns. A sym-
bol that appears in both columns represents the same thing in Column A
as it does in Column B.
Directions: Each of the following questions consists of two quantities, one
in Column A and one in Column B. Compare the two quantities and
choose:
a. if the quantity in Column A is greater
b. if the quantity in Column B is greater
c. if the two quantities are equal
d. if the relationship cannot be determined from the information given
Examples:
Column A Column B
1. ᎏ
1
5
ᎏ of 25 ᎏ
5
2
ᎏ of 2
The answer is c. The quantities are the same: ᎏ
1
5
ᎏ × 25 = 5, and
ᎏ
5
2
ᎏ × 2 = 5.
2. a + 5 a + 7
The answer is b. Regardless of the value of a, adding 7 will always
result in a higher number than adding 5.
2

3
Questions
Column A Column B
1. the number of even integers the number of even integers
between 1 and 13 between 2 and 14
2. ͙0.16ෆ ͙0.0016ෆ
3. Amy, Megan, and Sharon divided a batch of cookies among them-
selves. Amy took 30% of the cookies and Sharon took 40% of the
cookies. Amy ate ᎏ
1
3ᎏ of the cookies she took and Sharon ate ᎏ
1
4ᎏ of the
cookies she took.
number of cookies Amy ate number of cookies Sharon ate
4. p − 8 p + 8
5. 31x 35x
6. a < 0
a2
a3
7. w < x < y < z
wx yz
8. the percent increase from the percent increase from
10 cm to 14 cm 54 cm to 58 cm
9. ᎏ
2
1
5
ᎏ ᎏ
.0
4
16
ᎏ
10. ͙8ෆ + ͙13ෆ ͙19ෆ + ͙9ෆ
11. 5 + ͙32ෆ ͙23ෆ + 4
12. Marvin sells candy bars at a rate of 3 bars for $4.
at this rate, the cost of x dollars
x candy bars
13. (x + 2)2
(x − 2)2
14. 8.7 × 368 9 × 368
15. 18 − ᎏ
4
5ᎏ + ᎏ
1
2ᎏ 18 − ᎏ
1
2ᎏ + ᎏ
4
5ᎏ
16. ᎏ
2
1
0
ᎏ of 800 5% of 800

Column A Column B
17. n < 0
7n 4n
18. p2
−p2
19. (85 − 93)(22 − 8) (42 − 95)(11 − 17)
20. x ≥ 0
x2
x3
21. 5 × (3 + 1) ÷ 2 5 × 3 + 1 ÷ 2
22. There are 150 people in a movie theater. 75 of the people are men,
60 are women, and the remainder are children.
percent of people in the 10%
theater that are children
23. .75% .0075
24. The ratio of dogs to cats in a pet store is 5:3. There are 96 dogs
and cats in the store.
the number of dogs 65
in the pet store
25. 45% of 104 43
26. a, b, and c are integers greater than 1 and (c b
)5
= c5 + a
a b
27. 16.5 × 103
1.65 × 104
28. ᎏ
3
4ᎏ of ᎏ1
8
1ᎏ ᎏ1
6
1ᎏ
29. x < 0 < y
−x3
y 0
30. a and b are integers. ab = 30.
a + b 32
31. 0 < x < 1
ᎏ
1
xᎏ ᎏ
x
1
2ᎏ
32. ͙
4
.0001ෆ ͙.01ෆ
4

5
Column A Column B
33. .123 × 10−3
12.3 × 10−6
34. ᎏ
5
x
ᎏ and ᎏ
6
x
ᎏ are positive integers.
the remainder when x 1
is divided by 2
35. 5.1 5.014
36. x > 20
4 24 − x
37. the average rate of speed the average rate of speed
needed to drive 350 km needed to drive 780 km
in 5 hrs in 12 hrs
38. hours in a year seconds in a day
39. 282
+ 422
(28 + 42)2
40. 7 × 9 × 12 × 3 12 × 3 × 9 × 7
41. n > 0
n(n − 1) n2
42. a > b
30% of a 50% of b
43. 1 ᎏ
4
7ᎏ + ᎏ
1
7
3
ᎏ
44. ᎏ
5
6ᎏ of 12 ᎏ
1
6ᎏ of 60
45. ᎏ
1
xᎏ + 3y 7y + ᎏ
1
xᎏ
46. (n2
)4
(n4
)2
47. n > 1 and n is an integer.
ᎏ
3
nᎏ n
48. 0 < y < x. x and y are odd integers.
remainder when xy ᎏ
x
yᎏ
is divided by 2
49. x > 0
5͙3xෆ 3͙5xෆ

Column A Column B
50. x > 0
4x ᎏ5
x
ᎏ
51. − 1 < x < 1
3x −5 2x
52. x2
+ 1 x + 1
53. 0 < x < y
ᎏ
x
xᎏ ᎏ
x
yᎏ
54. 70% of the students enrolled in a
chemistry class passed the ﬁnal exam.
the ratio of those ᎏ
3
5ᎏ
who failed to those
who passed
55. (84 + 12)(15 + 91) (74 + 22)(20 + 86)
56. 75% of 30 30% of 75
57. .63
.6
ᎏ
1
3
ᎏ
58. x is an integer
remainder when 4x 0
is divided by 2
59. 2 ͙5ෆ
60. ͙17ෆ + ͙5ෆ ͙22ෆ
61. ᎏ
a
b
ᎏ = ᎏ
4
5
ᎏ and ᎏ
b
c
ᎏ = ᎏ
9
7
ᎏ
ᎏ
a
cᎏ 1
62. ᎏ
3
4ᎏ% .75
63. x and y are prime numbers and x + y = 12.
xy 38
64. 0 < a < b
(a + b)(a + b) (b − a)(b − a)
65. ͙10ෆ + ͙10ෆ 7
6

7
Column A Column B
66. number of years number of years
from 1625 to 1812 from 1631 to 1809
67. ͙63ෆ 3͙7ෆ
68. 76 × 14 + 14 × 26 (76 + 26)14
69. p < 0
5p 9p
70. 57 < 8k < 67
8 k
71. 243 × 96 231 × 93
72. The Spartans played a total of 48 games and had
a win to loss ratio of 7 to 5, and no game ended in a tie.
number of wins 28
73. 0 < x < 5
x − 5 5 − x
74. ᎏ
3
4ᎏ + ᎏ
1
9ᎏ + ᎏ
5
8ᎏ ᎏ
1
9ᎏ + ᎏ
5
8ᎏ + ᎏ
1
2ᎏ
75. Joel swims laps 50% faster than Sue.
number of laps Sue swims number of laps Joel swims
in 80 minutes in 60 minutes
76. (26 − 31)(296 + 134) (31 − 26)(296 + 134)
77. ᎏ
3
5ᎏ of 80 40% of 80
78. x < 0 and y < 0
ԽxԽ + ԽyԽ x + y
79. 7(3x + 1) 12(3x + 1)
80. 4͙2ෆ ͙35ෆ
81. Dan runs d miles in 30 minutes. Wendy runs w miles in 1 hour.
number of miles Dan runs number of miles Wendy runs
in 1 hour in 1 hour

Column A Column B
82. r6
= 64
32 r5
83. m > 1
͙m12ෆ (m2
)5
84. x > 0 and y > 0. x is 40 percent of y.
y 3x
85. An appliance store sold 120 refrigerators last year.
This year they sold 20 percent more.
number of refrigerators 145
sold this year
86. 956 + 274 + 189 200 + 275 + 970
87. sum of integers 1,000
from 1 to 100
88. − 1 < a < b < 0
a + b a − b
89. Lisa drove 117 miles between 8:15 A.M.
and 10:30 A.M. without stopping.
60 Lisa’s average speed
in miles per hour
90. the product of the integers the product of the integers
from −21 to −73 from −45 to −72
91. John plans to drive a total of 350 miles.
He has completed ᎏ
2
5ᎏ of his trip in 2.5 hours.
60 John’s average speed
in miles per hour
92. the sum of the integers the sum of the integers
from −10 to 5 from −5 to 10
93. the number of distinct the number of distinct
prime factors of 122
prime factors of 503
94. 683 × .05 683 × .1
8

9
Column A Column B
95. Խ−83Խ Խ83Խ
96. r < 0
45r ᎏ
4
r
5
ᎏ
97. −11 + (−4) −11 − 4
98. 59
254
99. ᎏ
1
2ᎏ × ᎏ
3
4ᎏ × ᎏ
7
9ᎏ ᎏ
3
2ᎏ × ᎏ
7
9ᎏ × ᎏ
1
4ᎏ
100. 2 + 82
− 6 − 10 (2 + 8)2
− 6 − 10
101. x > 0
7͙xෆ ͙3xෆ
102. ᎏ
1
3ᎏ of x 35% of x
103. ᎏ
1
3ᎏx + ᎏ
1
3ᎏx + ᎏ
1
3ᎏx = 9
x 9
104. a > 1 and b > 1
b(a + 1)
ba
105. x > y > 0
ᎏ
−
x
xy
ᎏ ᎏ
x
x
y
ᎏ
106. 1 − ᎏ
3
1
0
ᎏ ᎏ
1
9
0
ᎏ + ᎏ
1
1
0
1
0
ᎏ
107. (t6
)2
t8
t4
108. the number of primes the number of primes
between 40 and 50 between 1 and 6
109. x > 0 and y > 0
͙xෆ + ͙yෆ ͙x + yෆ
110. Kendra is driving at a steady rate of 56 miles per hour.
number of minutes it will 45 minutes
take Kendra to drive 42 miles
111. {x,y} represents the remainder when x is divided by y.
{56
,10} {104
,5}

Column A Column B
112. x, y, z, and m are positive integers.
x = ᎏ
2
5ᎏm and y = ᎏ
5
3ᎏm and z = ᎏ1
9
0ᎏy
x z
113. Ίᎏ
1
2
5
ᎏ๶ ᎏ͙
3
3ෆ
ᎏ
114. h < 0
5h h4
115. −(x − y) = x − y
x y
116. the product of the integers the product of the integers
from −5 to 6 −8 to 1
117. (−21)12
(−31)13
118. (m3
)6
͙m36ෆ
119. ᎏ
5x −
5
35
ᎏ x − 7
120. the number of primes the number of primes
that are divisible by 7 that are divisible by 11
121. .16 .0989
122. 4.25 × 105
42,500,000 ÷ 102
123. Խx − 8Խ Խ8 − xԽ
124. The sophomore class used 60 packages of cheese to make pizzas.
Each pizza used ᎏ
2
3ᎏ a package of cheese.
number of pizzas they made 100
125. x = ᎏ
1
4ᎏ and y = ᎏ
1
5ᎏ
ᎏ
x
yᎏ ᎏ
x
y
ᎏ
10

11
Answer Explanations
The following explanations show one way in which each problem can be
solved. You may have another method for solving these problems.
1. a. There are 6 even integers between 1 and 13. They are 2, 4, 6, 8,
10, and 12. There are 5 even integers between 2 and 14. They
are 4, 6, 8, 10, and 12.
2. a. The square root of 0.16 is 0.4. The square root of 0.0016 is
0.04.
3. c. Amy took 30% of the cookies and ate ᎏ
1
3ᎏ of those, which is 10%
of the original number of cookies. Sharon took 40% of the
cookies and ate ᎏ
1
4ᎏ of those, which is also 10% of the original
number of cookies. Since both women ate 10% of the original
number of cookies, they ate the same amount.
4. b. 8 more than any number (p + 8) is more than 8 less than that
number (p − 8).
5. d. The relationship cannot be determined. If the value of x is 0,
then both quantities are 0. If the value of x is positive, then 35x
is greater than 31x. If the value of x is negative, then 31x is
greater than 35x.
6. a. Since a is a negative number (a < 0), a2
is a positive number
because a negative times a negative is a positive; a3
is a negative
number because three negatives multiply to a negative answer.
A positive is always greater than a negative, so quantity A is
greater than quantity B.
7. d. The relationship cannot be determined. If w = −10, x = −9,
y = 1, and z = 2, then wx = 90 and yz = 2, so quantity A is
greater. If w = 0, x = 1, y = 2, and z = 3, then wx = 0 and yz = 6,
so quantity B is greater. Either quantity can be greater,
depending on the choice of variables.
8. a. Both quantities increased by 4, but quantity A increased 4 from
10, or 40%, and quantity B increased 4 from 54, or 7.4%. 40%
is greater than 7.4%.

9. a. Change ᎏ2
1
5ᎏ to a decimal by dividing 1 by 25 to get .04. Change
ᎏ
.0
4
16
ᎏ to a decimal by dividing .016 by 4 to get .004. .04 is greater
than .004.
10. b. Compare ͙8ෆ to ͙9ෆ and find that ͙9ෆ is greater and comes
from quantity B. Compare ͙13ෆ to ͙19ෆ and find that ͙19ෆ is
greater and comes from quantity B. Since both parts of quantity
B are greater than the parts of quantity A, quantity B is greater.
11. a. Compare 5 to 4 and find that 5 is greater and comes from
quantity A. Compare ͙32ෆ to ͙23ෆ and find that ͙32ෆ is greater
and comes from quantity A. Since both parts of quantity A are
greater than the parts of quantity B, quantity A is greater.
12. a. Each candy bar costs more than $1 (divide $4 by 3). Therefore,
the cost of x candy bars is more than x dollars.
13. d. The relationship cannot be determined. If 0 is substituted for x,
both quantities are 4. If 6 is substituted for x, quantity A is 64
[(6 + 2)2
= 64] and quantity B is 16 [(6 − 2)2
= 16]. In the first
example, the quantities are equal and in the second example,
quantity A is greater. The relationship cannot be determined.
14. b. Both quantities contain the number 368. In quantity B, 368 is
being multiplied by a larger number, making quantity B greater
than quantity A.
15. b. Consider there to be an equal sign between the two columns.
Subtract 18 from both sides, leaving −ᎏ
4
5ᎏ + ᎏ
1
2ᎏ = −ᎏ
1
2ᎏ + ᎏ
4
5ᎏ. Since ᎏ
4
5ᎏ is
larger than ᎏ
1
2ᎏ, the left side of the equation (quantity A) is
negative and the right side (quantity B) is positive. A positive
number is always greater than a negative number, so quantity B
is greater.
16. c. ᎏ
2
1
0
ᎏ and 5% are the same thing. Therefore, the two quantities are
equal.
17. b. n is a negative number. Try a couple of negative numbers to see
the pattern. Substitute in −2; (7)(−2) = −14 and (4)(−2) = −8,
quantity B is greater. Substitute in −.5; (7)(−.5) = −3.5 and
(4)(−.5) = −2. Again, quantity B is greater.
12

13
18. a. Any number squared is positive. Therefore, p2
is positive and
(−p)2
is negative. A positive number is always greater than a
negative number. Quantity A is greater.
19. b. Both sets of parentheses in quantity B are negative. Two neg-
atives multiplied yield a positive answer. Quantity A has one
negative set of parentheses and one positive. A negative multi-
plied by a positive yields a negative answer. Since quantity B is
always positive and quantity A is always negative, quantity A is
greater.
20. d. The relationship cannot be determined. If x < 1, for example .5,
then quantity A is larger. If x = 1, then both quantities are equal
(both 1). If x > 1, for example 3, then quantity B is larger.
21. b. Use the order of operations to simplify.
5 × (3 + 1) ÷ 2 =
5 × 4 ÷ 2 =
20 ÷ 2 =
10 = quantity A
5 × 3 + 1 ÷ 2 =
15 + .5 =
15.5 = quantity B
22. c. There are 15 children out of 150 people; ᎏ
1
1
5
5
0
ᎏ = 0.10 = 10%. The
percentage of people in the theater that are children is 10%.
23. c. Change .75% to a decimal by moving the decimal point two
places to the left. .75% = 0.0075.
24. b. Use the equation 5x + 3x = 96 and solve for x.
5x + 3x = 96
8x = 96
ᎏ
8
8
x
ᎏ = ᎏ
9
8
6
ᎏ
x = 12
5x of the animals are dogs. Since x = 12, 5x = 60. There are 60
dogs, which is less than quantity B.

25. a. Notice that 45% of 104 would be more than 45% of 100. Since
45% of 100 is 45, quantity A is greater than 45 which is greater
than quantity B.
26. a. Simplify the exponents on the left-hand side of the equation by
multiplying. Then, since the bases are the same (c), the
exponents can be set equal to each other:
(cb
)5
= c5 + a
c5b
= c5 + a
5b = 5 + a
Isolate a by subtracting 5 on both sides of the equation.
5b − 5 = 5 + a − 5
5b − 5 = a
The variables must be integers greater than 1 (so the smallest
possible value is 2). When 2 is substituted in for b, a is equal to
5. As the value of b gets larger, so does the value of a. Quantity
A is always greater than quantity B.
27. c. Multiplying by 103
moves the decimal in 16.5 three places to
the right to get 16,500 (quantity A). Multiplying by 104
moves
the decimal in 1.65 four places to the right to get 16,500
(quantity B).
28. c. “Of” means multiply. Multiply the fractions in quantity A;
ᎏ
3
4ᎏ × ᎏ
1
8
1
ᎏ = ᎏ
2
4
4
4ᎏ = ᎏ
1
6
1
ᎏ. Quantity A is equal to quantity B.
29. a. The value of x is negative. x3
is also negative because three
negatives multiply to a negative answer. The negative in front
of −x3
y negates the negative of x3
, making the quantity positive.
The value of y doesn’t matter because it is positive, making it
30. b. The possible integer factor pairs of 30 are (1,30), (2,15), (3,10),
and (5,6). Of these factor pairs, (1,30) has the largest sum;
1 + 30 = 31. 31 is not larger than 32, so quantity B is greater.
31. b. x is a positive number less than 1. The easiest way to see this
solution is to try different numbers for x. If x = 0.5, then ᎏ
1
xᎏ = ᎏ0
1
.5ᎏ
= 1 ÷ 0.5 = 2 and ᎏ
x
1
2ᎏ = ᎏ
(0.
1
5)2ᎏ = ᎏ
0.
1
25
ᎏ = 4. Quantity B is always
greater than quantity A.
14

15
32. c. ͙
4
.0001ෆ = .1 and ͙.01ෆ = .1; the two quantities are equal.
33. a. In quantity A, multiplying by 10−3
moves the decimal 3 places to
the left to yield 0.000123. In quantity B, multiplying by 10−6
moves the decimal to the left 6 places to yield 0.0000123.
0.000123 > 0.0000123.
34. b. In order for ᎏ
5
x
ᎏ and ᎏ
6
x
ᎏ to be integers, x must be evenly divisible by
5 and 6. If a number is divisible by 6, it is even. An even number
does not have a remainder when divided by 2. Quantity A is 0
which is less than quantity B.
35. a. Add two zeros to the end of quantity A to compare it to
quantity B. 5.100 > 5.014.
36. a. If x was 20, then quantity B would be 4. Since x is greater than
20, quantity B is less than 4 and less than quantity A.
37. a. Use this formula: rate × time = distance. If 5r = 350, then r = 70
(quantity A). If 12r = 780, then r = 65 (quantity B).
38. b. To ﬁnd the number of hours in a year multiply 24 hours by 365
days to get 8,760 hours. To ﬁnd the number of seconds in a day,
multiply 60 seconds by 60 minutes by 24 hours to get 86,400
seconds.
39. b. Multiply (28 + 42)2
out using FOIL.
(28 + 42)2
= (28 + 42)(28 + 42)
= (28)2
+ (28)(42) + (42)(28) + (42)2
Since (28 + 42)2
has the two middle terms in addition to (28)2
and (42)2
, it is greater than quantity A.
40. c. Multiplication is commutative (order doesn’t matter). Quantity
A and quantity B are the same, just in a different order.
Therefore, they are equal.
41. b. Distribute the n in quantity A to get n2
− n. Next, subtract n2
from both quantities. Quantity A becomes −n and quantity B
becomes 0. Since n is positive, −n is negative. Any negative
number is less than 0; therefore, quantity B is greater.

42. d. The relationship cannot be determined; a is greater than b. If a
is 200 and b is 100, 30% of 200 is 60 and 50% of 100 is 50,
making quantity A greater. If a is 120 and b is 118, 30% of 120
is 36 and 50% of 118 is 59, making quantity B greater.
43. b. Both fractions in quantity B are greater than ᎏ
1
2ᎏ. When they are
added together, they will make more than 1 whole.
44. c. “Of” means multiply; ᎏ
5
6ᎏ × 12 = 10 and ᎏ
1
6ᎏ × 60 = 10.
45. d. The relationship cannot be determined. Subtract ᎏ
1
xᎏ from both
quantities. Now compare 3y and 7y. If y is negative, 3y is
greater. If y is positive, 7y is greater.
46. c. (n2
)4
= n8
and (n4
)2
= n8
. The quantities are equal.
47. b. The ﬁrst integer that n can be is 2. Since ᎏ
3
2ᎏ = 1.5, quantity B is
greater. Quickly plugging in other integers shows that this is
always true.
48. b. The product of two odd numbers is odd. The remainder when
an odd number is divided by 2 is always 1. Therefore, quantity
A is always 1. Since x is larger than y, ᎏ
x
yᎏ is always greater than 1
because y always goes into x more than one time.
49. a. Square both quantities to compare without the square roots:
(5͙3xෆ)2
= 25(3x) = 75x; (3͙5xෆ)2
= 9(5x) = 45x. Since x is
positive, 75x is always greater than 45x.
50. a. Try a number less than 1 such as 0.5 for x; 4(0.5) = 2 and
ᎏ
0
5
.5
ᎏ = .1. In this case, quantity A is greater. Try a number greater
than 1 such as 20. 4(20) = 80 and ᎏ
2
5
0
ᎏ = 4. Whenever a positive
number is multiplied by 4, it will be greater than if that positive
number was divided by 5.
51. b. Subtract 2x from both quantities. Quantity A is now x − 5 and
quantity B is now 0. Add 5 to both quantities. A is now x and B
is now 5. x is deﬁned as between −1 and 1. Therefore, it is less
than 5.
16

17
52. d. The relationship cannot be determined. If x = 0.5, quantity A is
(0.5)2
+ 1 = .25 + 1 = 1.25 and quantity B is 0.5 + 1 = 1.5. When
x = 0.5, quantity B is greater. If x = 6, quantity A is 62
+ 1 = 36 +
1 = 37 and quantity B is 6 + 1 = 7. Quantity A is greater when
x = 6.
53. a. ᎏ
x
xᎏ always equals 1. Since x < y and x is positive, ᎏ
x
yᎏ < 1.
54. b. If 70% of students passed, 30% of students failed. Therefore,
the ratio of failed to passed is 30 to 70, which simpliﬁes to ᎏ
3
7ᎏ.
ᎏ
3
5ᎏ > ᎏ
3
7ᎏ.
55. c. Add the numbers inside the parentheses. (84 + 12)(15 + 91) =
(96)(106) and (74 + 22)(20 + 86) = (96)(106).
56. c. “Of” means multiply; 75% of 30 = .75 × 30 = 22.5 and 30% of
75 = .30 × 75 = 22.5.
57. a. Since the bases are equal, just compare the exponents to
compare the expressions. 3 > ᎏ
1
3ᎏ.
58. c. 4x is an even number because any multiple of 4 is even. The
remainder when an even number is divided by 2 is 0.
59. b. ͙4ෆ = 2, so ͙5ෆ > 2.
60. a. ͙17ෆ > 4 and ͙5ෆ > 2, therefore, ͙17ෆ + ͙5ෆ > 6. Since ͙25ෆ = 5,
͙22ෆ < 5. Quantity A is greater.
61. a. ᎏ
a
bᎏ × ᎏ
b
cᎏ = ᎏ
a
cᎏ and ᎏ
4
5ᎏ × ᎏ
9
7ᎏ = ᎏ
3
3
6
5ᎏ; ᎏ
3
3
6
5ᎏ > 1.
62. b. ᎏ
3
4ᎏ% = 0.75% = .0075; ᎏ
3
4ᎏ as a decimal is 0.75. To change 0.75%
to a decimal, move the decimal point two places to the left.
.0075 < .75.
63. b. The only pair of numbers that adds to 12 and is prime is 7 and
5. Therefore, x and y are 5 and 7; xy = (5)(7) = 35. 35 < 38.
64. a. For any positive numbers a and b, a + b > b − a. Therefore,
(a + b)(a + b) > (b − a)(b − a).
65. b. Combine the terms of quantity A; ͙10ෆ + ͙10ෆ = 2͙10ෆ. To get
rid of the square root, square both quantities (A and B);
(2͙10ෆ)2
= 4(10) = 40 and 72
= 49. 40 < 49.

66. a. Notice that quantity A starts earlier and ends later. Therefore,
it is a longer period of time. You do not need to do the
subtraction.
67. c. Square both quantities to get rid of the square roots; (͙63ෆ)2
=
63 and (3͙7ෆ)2
= 9(7) = 63. The two quantities are the same.
68. c. Use the distributive property to multiply quantity B out.
(76 + 26)14 = 76 × 14 + 26 × 14. Quantity B is equivalent to
quantity A.
69. a. Any positive number multiplied by 9 is greater than that same
number multiplied by 5. When the number is negative,
multiplying by 9 yields a number that is “more negative” or less
than that negative multiplied by 5.
70. d. The relationship cannot be determined. Solve the inequality for
k by dividing both sides by 8; ᎏ
5
8
7
ᎏ < ᎏ
8
8
k
ᎏ < ᎏ
6
8
7
ᎏ. 7.125 < k < 8.375. It
cannot be determined whether k is less than, greater than, or
equal to 8.
71. a. Both numbers in quantity A are greater than the corresponding
number in quantity B; 243 > 231 and 96 > 93. Therefore, the
product for quantity A is greater than the product for
quantity B.
72. c. Use the equation 7x + 5x = 48 and solve for x. First, combine
like terms on the left-hand side of the equation; 12x = 48.
Divide both sides by 12 to solve for x; ᎏ
1
1
2
2
x
ᎏ = ᎏ
4
1
8
2ᎏ. x = 4. The wins
are 7x or 28.
73. b. x is between 0 and 5. Since x is less than 5, x − 5 must be
negative. Since x is less than 5, 5 − x must be positive. Any
positive number is greater than any negative number.
74. a. Subtract ᎏ
1
9ᎏ and ᎏ
5
8ᎏ from both quantities. Quantity A is then just ᎏ
3
4ᎏ
and quantity B is just ᎏ
1
2ᎏ. Quantity A is greater than quantity B.
75. b. Use the formula distance = rate × time. Let Joel’s rate be 1.5
and Sue’s rate be 1. In 80 minutes, Sue’s distance is 1(80) = d or
80 = d. In 60 minutes, Joel’s distance is 1.5(60) = d or 90 = d.
18

19
76. b. (26 − 31) is negative and therefore, when multiplied by a
positive (296 + 134) it yields a negative answer. Both
parentheses in quantity B are positive, so quantity B is positive.
Any positive number is greater than any negative number.
77. a. 40% is equivalent to ᎏ1
4
0
0
0ᎏ = ᎏ
2
5ᎏ; ᎏ
3
5ᎏ of 80 is greater than ᎏ
2
5ᎏ of 80.
78. a. Both x and y are negative. Therefore, the sum of x and y is
negative (quantity B). The absolute value of x and y are both
positive, so their sum is positive (quantity A). Any positive
number is greater than any negative number.
79. d. The relationship cannot be determined. If x = −2, then quantity
A is 7(3(−2) + 1) = 7(−6 + 1) = 7(−5) = −35 and quantity B is
12(3(−2) + 1) = 12(−6 + 1) = 12(−5) = −60. In this case, quantity
A is greater. If x = 2, then quantity A is 7(3(2) + 1) = 7(6 + 1) =
7(7) = 49 and quantity B is 12(3(2) + 1) = 12(6 + 1) = 12(7) = 84.
In this case, quantity B is greater.
80. b. Square both quantities to get rid of the square roots; (4͙2ෆ)2
=
(16)(2) = 32 and (͙35ෆ)2
= 35. 32 < 35. Quantity B is greater.
81. d. The relationship cannot be determined. There is not enough
information to determine an answer.
82. c. 26
= 64; therefore, r = 2. 25
= 32. Both quantities are equal to 32.
83. b. ͙m12ෆ = m6
and (m2
)5
= m10
; since m is greater than 1, quantity B
is greater than quantity A.
84. b. “x is 40 percent of y” translates into the equation x = .4y.
Substitute .4y for x in quantity B; 3(.4y) = 1.2y. y < 1.2y.
Quantity B is greater.
85. b. Find 20% of 120 by multiplying 120 by .20 to get 24 more
refrigerators sold. Add 24 to the number of refrigerators sold
last year to get the number sold this year; 120 + 24 = 144; 144
refrigerators were sold this year.
86. b. Compare each number in quantity A to a different number in
quantity B; 956 < 970, 274 < 275, and 189 < 200. Since each of
the numbers in quantity A is less than the corresponding
number in quantity B, quantity B is greater.

87. a. Pair the largest number with the smallest number (1 and 100)
from the list of integers and add them together (101). Take the
second largest pair (2 and 99) and add them together (101). The
sum of the third largest pair (3 and 98) is also 101. There are 50
pairs whose sum is 101. To find the sum of the integers,
multiply 101 by 50 to get a sum of 5,050, which is larger than
quantity B.
88. b. Subtract a from both quantities. Quantity A is then b and
quantity B is −b. Since b is negative, −b is positive and therefore
greater than b.
89. a. Use the formula distance = rate × time. Lisa’s distance is 117
miles and her time is 2.25 hours; 117 = 2.25r. Divide both sides
by 2.25; ᎏ
2
1
.
1
2
7
5
ᎏ = ᎏ
2
2
.
.
2
2
5
5
r
ᎏ, so 52 = r. The rate is 52 miles per hour.
90. b. All the numbers involved in the problem are negative. There is
an odd number of numbers for quantity A and an even number
of numbers for quantity B. When multiplying an odd number
of negatives, you get a negative answer. When multiplying an
even number of negatives, you get a positive answer. Any
91. a. Find the number of miles that John has driven by finding ᎏ
2
5ᎏ of
350; ᎏ
2
5ᎏ × 350 = 140. 140 miles have been driven in 2.5 hours. To
find the average speed, use the formula distance = rate × time.
The distance is 140 and the time is 2.5. 140 = 2.5r. Solve for r
by dividing both sides of the equation by 2.5; ᎏ
1
2
4
.5
0
ᎏ = ᎏ
2
2
.
.
5
5
r
ᎏ, so
r = 56 mph.
92. b. Quantity A and quantity B both contain the integers from −5 to
5. So the only comparison that must be done is outside of those
numbers. The additional numbers that quantity A has are all
negative, which brings the sum down. The additional numbers
that quantity B has are all positive, which brings the sum up.
Quantity B is therefore greater.
93. c. Create a factor tree to help visualize the factors of 122
and 503
.
The only distinct prime factors of 123
are 3 and 2. The only
distinct prime factors of 503
are 5 and 2. Since they both have 2
distinct prime factors, the quantities are equal.
94. b. 683 × .05 = 34.15 and 683 × .1 = 68.3
20

21
95. c. |−83| = 83 and |83| = 83
96. d. The relationship cannot be determined. If r = − 0.1, 45(−0.1) =
−4.5 and ᎏ−
4
0
5
.1ᎏ = −450. In this example, quantity A is greater. If r
= −5, 45(−5) = −225 and ᎏ−
45
5ᎏ = −9. In this example, quantity B is
greater.
97. c. −11 + (−4) = −11 − 4 = −15
98. a. In order to compare the two quantities, the bases must be the
same. 25 can be written as 52
. Therefore, 254
= (52
)4
= 58
.
Compare the exponents of quantities A and B. 9 > 8. Quantity A
is greater.
99. c. To multiply fractions, simply multiply across the numerators
and denominators. The numbers in the numerators in quantity
A and quantity B are the same (1, 3, 7), just in a different order.
The quantities in the denominators are also the same (2, 4, 9),
just in a different order. When multiplied out, both quantities
are ᎏ
2
7
1
2ᎏ.
100. b. Follow the order of operations. Quantity A is 2 + 82
− 6 − 10 = 2
+ 64 − 6 − 10 = 50. Quantity B is (2 + 8)2
− 6 − 10 = 102
− 6 − 10
= 100 − 6 − 10 = 84.
101. a. Square both quantities to get rid of the square roots; (7͙xෆ)2
=
49x and (͙3xෆ)2
= 3x. Since x is positive, 49x > 3x. Quantity A is
102. d. The relationship cannot be determined. When x is positive,
quantity B is greater. When x is negative, quantity A is greater.
103. c. ᎏ
1
3ᎏx + ᎏ
1
3ᎏx + ᎏ
1
3ᎏx = 1x; therefore, x = 9. The quantities are equal.
104. a. Since the bases are the same (b) and b > 1, the exponents can be
compared. a + 1 > a since a is positive.
105. b. Simplify both fractions by canceling the x’s. The fractions then
become just −y and y. y is positive, so −y is negative. Any

106. b. Quantity A is less than 1 and quantity B is greater than 1. This
is easily seen by the fact that ᎏ3
1
0ᎏ is being taken from 1, which
brings the value below 1 for quantity A. For quantity B, create a
common denominator of 100; ᎏ1
9
0
0
0ᎏ + ᎏ1
1
0
1
0ᎏ = ᎏ
1
1
0
0
1
0ᎏ. This fraction is
greater than 1.
107. c. Use rules of exponents to simplify quantities. (t6
)2
= t12
and
t8
t4
= t12
. The quantities are equal.
108. c. The primes between 40 and 50 are 41, 43, and 47. The primes
between 1 and 6 are 2, 3, and 5. Each quantity equals 3.
109. a. Square both quantities to get rid of some of the square roots.
(͙xෆ + ͙yෆ)2
= (͙xෆ + ͙yෆ)(͙xෆ + ͙yෆ) = x + 2͙xyෆ + y
(͙x + yෆ)2
= x + y
Since quantity A has the extra term 2͙xyෆ, which is positive, it is
greater.
110. c. Use the formula distance = rate × time. Kendra’s rate is 56 and
her distance is 42; 42 = 56t. Solve the equation by dividing by
56; ᎏ
4
5
2
6ᎏ = ᎏ
5
5
6
6
t
ᎏ; t = .75 hr; .75 hours is equivalent to 45 minutes.
111. a. Any power of 5 ends in 5. When it is divided by 10, there will
be a remainder of 5. Therefore, quantity A is 5. Any power of
10 ends in 0. When it is divided by 5, there will be a remainder
of 0.
112. b. Compare x and z in terms of m; x = ᎏ
2
5ᎏm; z can be rewritten in
terms of m by substituting ᎏ
5
3ᎏm for y; z = ᎏ
1
9
0
ᎏ(ᎏ
5
3ᎏm) = ᎏ
3
2ᎏm. ᎏ
2
5ᎏ < ᎏ
3
2ᎏ,
therefore, quantity B is greater.
113. a. Get rid of the square roots by squaring both fractions; (Ίᎏ
1
2
5
ᎏ๶)2
=
ᎏ
1
4
5
ᎏ = 3.75 and (ᎏ
͙
3
3ෆ
ᎏ)2
= ᎏ
9
3ᎏ = 3.
114. b. h is negative, so 5 times a negative is a negative. Quantity A is
negative. A negative multiplied by itself 4 times is a positive.
Quantity B is positive. Any positive number is greater than any
negative number.
22

23
115. c. Simplify the equation by distributing the negative.
−(x − y) = x − y
−x + y = x − y
Next, add x to both sides. Then add y to both sides.
−x + x + y = x + x − y
y + y = 2x − y + y
2y = 2x
Divide both sides by 2.
ᎏ
2
2
y
ᎏ = ᎏ
2
2
x
ᎏ
y = x
116. c. Zero is included in both lists. Zero multiplied by anything
yields a zero. Both products are zero.
117. a. Quantity A is positive because a negative number to an even
power yields a positive answer. Quantity B is negative because a
negative number to an odd power yields a negative answer. Any
118. c. Use the rules of exponents to simplify the quantities; (m3
)6
=
m18
; ͙m36ෆ = m18
. Both quantities are equivalent to m18
.
119. c. Divide both terms in the numerator of quantity A by 5. This
yields x − 7, which is equivalent to quantity B.
120. c. The only prime divisible by 7 is 7. It is the same for 11—the
only prime divisible by 11 is 11. If any other number was
divisible by 7 or 11, it would not be prime.
121. a. Add zeros on to the end of quantity A to make it easier to
compare to quantity B; .16 = .1600; .1600 > .0989.
122. c. When multiplying by 105
, move the decimal 5 places to the
right to get 425,000. When dividing by 102
, move the decimal 2
places to the left to get 425,000. The values in A and B are the
same.

123. c. The absolute value of opposites is the same number. For
example, |5| = 5 and |−5| = 5; x − 8 is the opposite of 8 − x. This
can be shown by multiplying either expression by −1. The result
is the remaining expression; −1(x − 8) = −x + 8 = 8 − x.
Therefore, the two values are equal.
124. b. Divide the number of packages of cheese by the amount used
per pizza to ﬁnd the number of pizzas made; 60 ÷ ᎏ
2
3ᎏ = ᎏ
6
1
0
ᎏ × ᎏ
3
2ᎏ =
90. They made 90 pizzas. Quantity B is greater.
125. a. Set up the complex fractions and then divide. Quantity A is
ᎏ
x
yᎏ = = ᎏ
1
4ᎏ ÷ ᎏ
1
5ᎏ = ᎏ
1
4ᎏ × ᎏ
5
1ᎏ = ᎏ
5
4ᎏ, so Quantity A is ᎏ
5
4ᎏ; Quantity B is
ᎏ
x
y
ᎏ = = ᎏ
1
5ᎏ ÷ ᎏ
1
4ᎏ = ᎏ
1
5ᎏ × ᎏ
4
1ᎏ = ᎏ
4
5ᎏ, so Quantity B is ᎏ
4
5ᎏ.
Since quantity A is greater than 1 and quantity B is less than 1,
quantity A is greater.
ᎏ1
5
ᎏ
ᎏ
ᎏ1
4
ᎏ
ᎏ1
4
ᎏ
ᎏ
ᎏ1
5
ᎏ
24

25
algebra-based quantitative comparison questions:
■ Applications
■ Coordinate Geometry
■ Inequalities
■ Operations with Algebraic Expressions
■ Rules of Exponents
■ Solving Linear Equations
■ Solving Quadratic Equations in One Variable
■ Translating Words into Algebraic Expressions
Some important guidelines:
2Algebra

choose:
d. if the relationship cannot be determined from the information given.
Examples
Column A Column B
1. x + 5 = 9
4 x
The correct answer is c. Subtract 5 on both sides of the equation. x
+ 5 − 5 = 9 − 5. Simplifying this gives a solution of x = 4. The
columns are equal.
2. a < 0
a2
a3
The correct answer is a. Since a is less than zero, it represents a
negative number. A negative number raised to an even numbered
power will be a positive result, but a negative number raised to an
odd numbered power will be a negative result. Column A is larger.
26

27
Questions
Column A Column B
126. 117
+ 23
25
127. a is a positive integer.
ᎏ
a
1
7ᎏ ᎏ
a
1
10ᎏ
128. (a4
)2
a8
129. 33
9
130. a is an integer.
a2
a3
a6
131. (ᎏ
3
4ᎏ)2
− (ᎏ
1
4ᎏ)2
(ᎏ
1
2ᎏ)2
132. 4x
= 64
x 4
133. 212
= 8x
3 x
134. a < 0
a2
a9
135. y = 5x
x is a positive integer.
5x + 1
5y
136. x > 0
(2x + 4)(x + 1) 2x2
+ 5x + 4
137. x2
− 4x − 21 = 0
sum of the roots product of the roots
138. The prom committee orders an arch for the entrance to the
dance ﬂoor. The arch follows the equation y = 2x − .1x2
where y is the height of the arch, in feet.
maximum arch height 10 feet

Column A Column B
139. x < 0
x(x + 7) x2
+ 7
140. x > 0
(x + 3)2
x2
+ 9
141. y = x2
+ 6x + 9
minimum y value 9
of the function
142. y = 4x2
+ 4x − 8
smaller root −2
143. y = −x2
+ 6x
larger root 6
144. 9x2
= 6x − 1
x 1
145. (x + 5)(x − 5) x2
+ 10x − 25
146. xy < 0
(x + y)2
x2
+ y2
147. 2(x + 3) + 6 = 4x
6 x
148. Julie is 5 years older than Ravi. Three years ago,
Julie was twice as old as Ravi.
Ravi’s age now 5
149. The sum of 3 consecutive integers is 37 more
than the largest integer.
the middle integer 19
150. The sale price for a snowboard is $63.00.
This price reﬂects a 30% discount.
the original price $90.00
of the snowboard
28

29
Column A Column B
151. The ratio of rabbits to squirrels is 2:3.
There are a total of 225 rabbits and squirrels.
the number of squirrels 90
152. To mail an envelope first class costs $0.34 for the first ounce,
plus $0.17 for each additional ounce. The cost to mail an
envelope was $1.53.
8 ounces weight of the envelope
153. Carlos has 14 coins in his pocket, consisting of quarters
and nickels only. The monetary value of these coins is $2.30.
the number of quarters the number of nickels
154. The drama club collected $907.50 from the sale of tickets.
Adult tickets cost $5.00 and student tickets cost $2.50.
They sold 63 more student tickets than adult.
the number of student tickets 163
155. There are 146 athletes and 8 coaches taking a trip to
a competition. They travel in buses that seat 48 people.
the number of buses needed 4
156. Monique had 68 points correct out of a total of
85 points on her math test.
Monique’s percentage grade 68%
157. A bus leaves the station traveling at a constant speed of 45 miles
per hour. A second bus leaves the same station heading
in the same direction one hour later traveling at a constant
speed of 50 miles per hour.
8 hours The number of hours after first
bus left that the buses will pass
each other
158. ᎏ
8 +
x
2x
ᎏ = 50
x ᎏ
1
6ᎏ
159. 5x + 3 = 18
x 3

Column A Column B
160. x + 4 = −14
−10 x
161. 4(x − 2) = 8
4 x
162. p + q = 16
3p + 2q = 44
p q
163. 2a + 2b = 20
4a + 2b = 14
b 13
164. ᎏ
1
6ᎏx + 6 = 12
15 x
165. c + 4d = 11
6c − 2d = 40
c d
166. 9c + 12 = d
ᎏ
d −
9
12
ᎏ c
167. ᎏ
a
bᎏ < 0
ab 0
168. a = 2; b = 6; c = 4
c + b ÷ a 7
169. 3x − (4 + x) 2x − 4
170. b < 0
(−2b)2
−2b2
171. a < 0
2a2
a3
172. abc > 0
b + c 0
30

31
Column A Column B
173. x ≠ 0
(͙x4ෆ)(͙16ෆ) 8x2
174. a ≥ 0
͙(a + 4)ෆ2
ෆ a + 4
175. a ≠ 0
a + b2
(a + b)2
− 2ab
176. x > 5
ﬁve less than x 5 − x
177. twice the sum of 14 and b, b + 14
divided by 2
178. the product of a and b |ab|
179. one third of a variable x, ᎏ
1
x
2
ᎏ
divided by four
180. x > 1
the square root of x, x3
raised to the fourth power
181. Let d = number of dimes.
Let p = number of pennies.
0.10d + 0.01p 10d + p
182. x > y > 4
four divided by the ᎏ
x
4
y
ᎏ
product of x and y
183. the cost of a $50.00 sweater, $30.00
on sale for 20% off
184. the cost of a $28.00 28 + .6(28)
basketball, including
6% sales tax

Column A Column B
185. x is a positive odd integer.
the product of two x(x + 2)
consecutive positive odd
integers
186. 3b + 3 > 18
b 5
187. 2(5 − x) < 70
−30 x
188. 0 < a < b
ᎏ
1
aᎏ ᎏ
1
bᎏ
189. b + 7 < −5
b −2
190. x < 5 ; y < z
x + y z + 5
191. ᎏ
1
2ᎏx − ᎏ
1
3ᎏx ≥ 4
x 20
192. a > b > 0
ᎏ
1
aᎏ ᎏ
1
bᎏ
32

33
Column A Column B
Use the following ﬁgure to answer questions 193–194.
193. the x-coordinate of point A the x-coordinate of point B
194. 1 the slope of line AB
-5 5 10
-5
5
10
A
B

Column A Column B
195. 5 the length of segment BC
196. the area of triangle ABC 12
197. A circle with center at the origin passes through the point (0,8).
4 the radius of the circle
198. the slope of the line that 2
passes through the points
(1,2) and (2,4)
-5 5
-5
5
AB
C
(0,0)
(0,4)
(-3,0)
34

35
Column A Column B
199. a − b 0
200. c −9
5
−5
5
10
10
−10
−5−10
(4,3)
(a,b)
(−8,c)

Column A Column B
Use the following ﬁgure to answer question 201.
201. p − q q − p
202. ab < 0
b > 0
a b
203. ᎏ
1
y
ᎏ y
204. ᎏ
3
4
x
x
2
2
y
y
ᎏ 0.75
205. x and y are positive integers.
10y 10x
206. (ᎏ
1
2
ᎏ)2
(ᎏ
1
3
ᎏ)0
207. x > 1
2x2
3x3
208. 8x2 + x − 2
= 1
x2
+ x −2 0
209. a < 0
b > 0
(ab)3
0
210. a > 1
a6 × a2
(a6
)2
5
5
(q,3)
D
(5,p)
C
B
(5,1)
A
(2,1)
36

37
Column A Column B
211. 0 < b < 1
b2
b3
212. a ≠ 0
−a2
(−a)2
213. (ᎏ
2
3
ᎏ)2
(0.667)3
214. 2 < c < d < 3
ᎏ
1
c
ᎏ ᎏ
1
d
ᎏ
215. y = x + 8
x + 4 y − 4
216. 2(x − 5) − 4 = 10
x 12
217. 6x + 3 = −4x − 7
−x x2
218. 11x + 3 = 42
11x + 3 11x − 3
219. ᎏ
4
3
x
ᎏ + 4 = 2x
6x x2
220. x − y = 6
y + x = 4
3x xπ
221. x + y = 5
x − y = 1
x y
222. c + d = 3
c − d = 3
c 3
223. y = 5x − 1
the value of x when y = 0 the value of y when x = 0

Column A Column B
224. 8x − 2y = 30
x = ᎏ
3
2
ᎏy
x y
225. x + y = 16
x − y = 8
x2
− y2
(x + y)(x + y)
226. The degree of the quadratic The degree of a cubic term
term of a polynomial of a polynomial
227. 3x2
− 27 = 0
the positive value of x 3
228. x2
− 3x − 10 = 0
the negative value of x −3
229. −y( y − 4) = 10
y2
+ 10 4y
230. y > 2x − 1
2x y + 1
231. 4 < 2x − 2 < 8
x 6
232. 5 < y + 1
y 4
233. 6(x − 1) > 30
6 x
234. −3x − 1 > 14
−5 x
235. Point (x, y) is located in Quadrant IV.
x y
236. the slope of the line y = 2x − 3 the slope of the line y = ᎏ
1
2
ᎏx + 3
38

39
Column A Column B
237. Point (x, y) is located in Quadrant II.
the opposite of x the reciprocal of y
238. The equation of line l is y = ᎏ
2
3
ᎏx − 1.
Line m is perpendicular to the line y = − ᎏ
3
2
ᎏx + 1.
the slope of line l the slope of line m
239. Points (4, c) and (0, d) are on line n.
The slope of line n is ᎏ
3
4
ᎏ.
c − d 3
240. y = −2x − 3
the y-intercept of the equation the x-intercept of the equation
241. the slope of the equation the slope of the equation
2y − 4x = 6 −3y + 3x = 9
242. the distance between the 5
points (0, 0) and (−3, 4)
243. the difference between 26 the sum of 8 and 2
and the product of 4 and 3
244. six less than nine the square root of four
245. The square of a number is four.
the square root of the number 4
246. the quotient of ten and two the quotient of sixty-ﬁve and
thirteen
247. One-half of y is x. One-half of z is y. x + y + z = 35.
z 15
248. Two cars leave the same city traveling in opposite directions.
Car A is traveling at 60 miles per hour and car B is
traveling at 55 miles per hour.
the number of hours it takes 4
for the cars to be 460 miles
apart

Column A Column B
249. The sum of two consecutive integers is 83.
23 less than three times 16 more than two times
the smaller integer the greater integer
250. A vending machine has exactly $1.15 in quarters and dimes.
the number of quarters the number of dimes
40

41
Answer Explanations
126. b. The number 1 to any power is 1; 23
is 2 × 2 × 2 which is 8;
1 + 8 = 9. The quantity 25, in column B, is greater.
127. a. Since the variable a is a positive integer, both choices are
positive, and a10
> a7
. These are fractions with the same
numerator. When two fractions are being compared with the
same numerators, the smaller the denominator, the larger the
number. Column A is greater.
128. c. By the laws of exponents, (a4
)2
= a4 × 2
= a8
. This is true for any
real number a. Therefore, the quantities in Column A and
Column B are equal.
129. a. 33
is equal to 3 × 3 × 3, which is 27. 27 is greater than 9, so
column A is greater.
130. d. The answer cannot be determined. This problem involves a law
of exponents that is true for any real number: a2
× a3
= a2+3
= a5
.
For most integers, a6
> a5
. Note that this is true even for
negative integers, since 6 is an even number, and 5 is an odd
number. There are two exceptions, however, that would make
these choices equal. They are when a = 0 or a = 1.
131. a. When a fraction is squared, you square both the numerator, and
the denominator, so (ᎏ
3
4
ᎏ)2
= ᎏ
3
4
2
2ᎏ = ᎏ
1
9
6
ᎏ and (ᎏ
1
4
ᎏ)2
= ᎏ
1
1
6
ᎏ and ᎏ
1
9
6
ᎏ − ᎏ
1
1
6
ᎏ = ᎏ
1
8
6
ᎏ
= ᎏ
1
2
ᎏ in lowest terms. Likewise, (ᎏ
1
2
ᎏ)2
= ᎏ
1
4
ᎏ. One half is greater than
one fourth.
132. b. Four to the x power means that 4 is multiplied by itself “x”
times. By trial and error, 4 × 4 × 4 = 64, so x is equal to 3.
Choice B is greater.
133. b. For exponential equations, you must ﬁrst rewrite the equation
to have the same bases when possible. Since 2 × 2 × 2 = 8,
23
= 8. Two is the common base. Rewrite the equation as
212
= (23
)x
. By the laws of exponents, (23
)x
= 23x
. The equation is
now 212
= 23x
. Since the bases are now the same, this becomes a

simple equation to solve, by setting the exponents equal to each
other: 12 = 3x. Divide both sides of this equation by 3, and it
becomes x = 4.
134. a. When a is less than zero, a is negative. A negative number to
any even power is a positive number, while a negative number
to any odd power is a negative number. This is a case where
even though the exponent of 2 is smaller, the quantity will be
greater.
135. c. Since y = 5x
, multiply each side of this equation by 5, to get
5y = (51
)(5x
). (51
)(5x
) = 5x + 1
, which is the value of column A.
136. a. Using the distributive property, (2x + 4)(x + 1) = 2x2
+ 2x +
4x + 4. Combine like terms, and this is equal to 2x2
+ 6x + 4.
Subtracting 2x2
and 4 from both columns leaves 6x in column A
and 5x in column B. Since x > 0, column A is greater.
137. a. To ﬁnd the roots of the equation, factor the left hand side into
two binomials; x2
− 4x − 21 = (x − 7)(x + 3), so the equation
becomes (x − 7)(x + 3) = 0. Either x − 7 = 0 or x + 3 = 0 to make
the equation true. So x = 7 or x = −3. The sum of the roots is
7 + −3 = 4. The product of the roots is (−7)(3) = −21.
138. c. The arch is in the shape of a parabola, and the maximum arch
height (the y value) is the height at the vertex. When a quadratic
is in the form ax2
+ bx + c (a, b, c are real numbers), the
x-coordinate of the vertex is given by the formula ᎏ2
−
a
b
ᎏ = ᎏ−
−
0
2
.2ᎏ = 10.
When x is 10, y = 2(10) − 0.1(10)2
. So y = 20 − 0.1(100) =
20 − 10 = 10. So the maximum arch height is 10 feet.
139. b. By the distributive property, x(x + 7) = x2
+ 7x for column A.
Since x < 0, x is negative, and therefore 7x is negative, x2
+ 7
will be greater in this case.
140. a. The quantity (x + 3)2
= (x + 3)(x + 3). By the distributive
property, this equals x2
+ 3x + 3x + 9 = x2
+ 6x + 9. You can
subtract x2
and 9 from each column and you are left with 6x in
column A and zero in column B. Since x > 0, column A is
greater.
42

43
141. b. The given equation is a quadratic in the form ax2
+ bx + c, where
a, b, and c are real numbers. The minimum y value of the
function is the y value at the vertex. The x-coordinate of the
vertex is given by the formula ᎏ
2
−
a
b
ᎏ = ᎏ
−
2
6
ᎏ = −3. When x = −3,
y = (−3)2
+ 6(−3) + 9 = 9 − 18 + 9 = 0. Column B is greater.
142. c. To find the roots of this equation, first factor out the common
factor of 4; 4x2
+ 4x − 8 = 4(x2
+ x − 2). Next, factor this into two
binomials; 4(x2
+ x − 2) = 4(x + 2)(x − 1). The roots are the
values of x that make the y value equal to zero. The equation
will equal zero when x = −2 or x = 1. The smaller root is −2.
143. c. To find the roots of this equation, factor out the common factor
of −x. The equation becomes y = −x(x − 6). The roots are the
values of x that make the y value equal to zero. The equation
will equal zero when x = 0 or x = 6. The larger root is 6.
144. b. To find the values of x, move all terms to the left side so that it
is a quadratic equation set equal to zero. Subtracting 6x and −1
from both sides makes the equation 9x2
− 6x + 1 = 0. Factor this
quadratic into two binomials; 9x2
− 6x + 1 = (3x − 1)(3x − 1).
Now the equation is (3x − 1)(3x − 1) = 0. This will be true when
3x − 1 = 0; add one to both sides to give 3x = 1; divide both
sides by 3 and x = ᎏ
1
3
ᎏ.
145. d. The answer cannot be determined. The binomials in column A
are the difference of two squares, so (x + 5)(x − 5) =
x2
− 5x + 5x − 25 = x2
− 25. Since there is no indication as to
whether x is positive or negative, the term 10x in column B
could be either positive or negative and the answer cannot be
determined.
146. b. Using the distributive property and combining like terms,
column A is (x + y)2
= x2
+ xy + xy + y2
= x2
+ 2xy + y2
. Since
xy < 0, 2xy is negative, and thus column B is greater.
147. c. To solve this equation, first use the distributive property, and
combine like terms, on the left hand side. The left side becomes
2(x + 3) + 6 = 2x + 6 + 6 = 2x + 12. Now, 2x + 12 = 4x. Subtract
2x from each side, so 12 = 2x. Divide both sides by 2, and x = 6.

148. a. Set up an equation and let x represent Ravi’s age now. Julie’s age
now is represented by x + 5. Three years ago, Ravi’s age was
x − 3 and Julie’s age was x + 5 − 3 = x + 2. Three years ago, Julie
was twice as old as Ravi. So x + 2 = 2(x − 3). Use the distributive
property on the right hand side to get x + 2 = 2x − 6. Add six to
both sides, yielding x + 8 = 2x. Subtract x from both sides, and
x = 8. Ravi’s age now is 8.
149. c. Set up an equation where x represents the ﬁrst consecutive
integer. Therefore the second consecutive integer is x + 1, and
the third is x + 2. The sum of these integers is represented by
x + x + 1 + x + 2. Combining like terms, the sum is represented
by 3x + 3. This sum is 37 more than the largest integer, so
3x + 3 = x + 2 + 37. Combining like terms on the right hand side
yields the equation 3x + 3 = x + 39. Subtracting x from both
sides yields 2x + 3 = 39. Subtract 3 from both sides gives
2x = 36. Divide both sides by 2, and x = 18. Now, x represents
the smallest integer, so the middle integer is x + 1 = 19.
150. c. $63.00, which is the sale price, is 70% of the original price.
Therefore, let x represent the original price of the snowboard.
So 0.70x = 63.00. Divide both sides of this equation by 0.70,
and x = $90.00.
151. a. Since the ratio of rabbits to squirrels is 2:3, there are 2 rabbits
for every 3 squirrels. Let 2x represent the number of rabbits,
and then 3x represents the number of squirrels. So 2x + 3x =
225. Combine like terms: 5x = 225. Divide both sides by 5, and
then x = 45. The number of squirrels is 3x = 3(45) = 135.
152. c. Let x represent the total weight in ounces. So x − 1 will
represent the additional ounces over the ﬁrst ounce. The cost
is then represented by 0.34 + 0.17(x − 1) = 1.53. Use the
distributive property on the left hand side to get 0.34 + 0.17x −
0.17 = 1.53. Combine like terms: 0.17 + 0.17x = 1.53. Subtract
0.17 from both sides: 0.17x = 1.36. Divide both sides by 0.17,
and x = 8. The total weight is 8 ounces.
153. a. Let q represent the number of quarters. Since there are a total
of 14 quarters and nickels, 14 − q will represent the number of
nickels. Set up an equation to represent the monetary amount:
.25q + .05(14 − q) = 2.30. Use the distributive property, and
44

45
distribute .05: .25q + .70 − .05q = 2.30. Combine like terms to
get .20q + .70 = 2.30. Subtract .70 from both sides, to get
.20q = 1.60. Divide both sides by .20, so q = 8. Since there are
14 coins, the number of nickels is 14 − q = 14 − 8 = 6.
154. c. Since there were 63 more student tickets sold than adult tickets,
let a represent the number of adult tickets sold. So a + 63 will
represent the number of student tickets sold. The total ticket
sales can be represented by 5a + 2.5(a + 63) = 907.50. Use the
distributive property to distribute 2.5 to the terms in
parentheses: 5a + 2.5a + 157.50 = 907.50. Combine like terms,
to get 7.5a + 157.50 = 907.50. Subtract 157.50 from both sides
of the equation to get 7.5a = 750. Divide both sides by 7.5, and
a = 100. The number of adult tickets is therefore 100; the
number of student tickets is 100 + 63 = 163.
155. c. The buses need to seat 146 athletes, plus 8 coaches. This is 154
people total. Three buses will hold 144 people. Four buses are
needed.
156. a. Monique’s percentage grade is the ratio of the number correct
to the total number of points on the test. This is ᎏ
6
8
8
5
ᎏ = 0.8 and
0.8 is equal to 80%.
157. b. Let x represent the number of hours after the ﬁrst bus left in
which they will pass. Since the second bus left one hour later,
the number of hours after the second bus leaves is x − 1. The
buses will pass when their distances are the same. Since the ﬁrst
bus is traveling 45 miles per hour, 45x represents the distance
the bus has gone in x hours, because distance = rate × time. For
the second bus, since it is traveling at 50 miles per hour, its
distance is represented by 50(x − 1). Set up the equation to
represent that the two distances are equal: 45x = 50(x − 1).
Use the distributive property on the right hand side to get
45x = 50x − 50. Subtract 45x from both sides, and add 50 to
both sides, and 5x = 50. Divide both sides by 5, and x = 10
hours.
158. c. For this fractional equation, the best way to simplify is to
multiply both sides by x. This will leave 8 + 2x = 50x. Subtract
2x from both sides to get 8 = 48x. Divide both sides by 48, so
x = ᎏ
4
8
8
ᎏ = ᎏ
1
6
ᎏ in simplest form.

159. c. For the given equation, subtract three from both sides to get
5x = 15. Divide both sides by 5 to get x = 3.
160. a. For the given equation, subtract four from both sides to get
x = −18.
161. c. For the given equation, first apply the distributive property, and
distribute four to each term on the left. The equation now is
4x − 8 = 8. Now add eight to both sides of the equation to get
4x = 16. Divide both sides by 4 and x = 4.
162. a. One way to solve this system of equations is to multiply each
term in the first equation by 2, to get 2p + 2q = 32. Line up the
equations, and subtract the bottom terms from the top terms:
2p + 2q = −32)
−(3p + 2q = −44)
−p = −12)
Divide both sides now by −1, and p = 12. Going back to the
original first equation, if p = 12, then 12 + q = 16. Therefore, by
subtracting twelve from both sides, q = 4.
163. c. For this given system of equations, both have the term 2b, so
subtract the bottom terms from the top terms:
2a + 2b = 20)
−(4a + 2b = 14)
−2a = 6)
Divide both sides by −2, and a = −3. Going back to the original
first equation, if a = −3, then 2(−3) + 2b = 20, or −6 + 2b = 20.
Add six to both sides of this equation to get 2b = 26. Divide
both sides by two, and b = 13.
164. b. For the given equation, first subtract six from both sides to get
ᎏ
1
6
ᎏx = 6. Next, multiply both sides by the reciprocal of ᎏ
1
6
ᎏ, which is
six. The equation now is x = 36.
165. a. For the given system of equations, multiply each of the terms in
the second equation by two, so that both equations have a
common term of 4d:
c + 4d = 11
12c − 4d = 80
Add the two equations to get 13c + 4d = 91.
46

47
Divide both sides of this resulting equation by thirteen to get
c = 7. Using the original first equation and substituting in seven
for c, 7 + 4d = 11. Subtract seven from both sides, and divide
both sides by four, and d = 1.
166. c. Since the choice in column B is c, solve the given equation for c,
by subtracting twelve from both sides, and then dividing both
sides by nine. The quantities are equal.
167. b. It is given that a divided by b is less than zero, so by the rules of
positive/negative arithmetic, either a is negative or b is negative,
but not both. In this case, the term ab will be negative. Zero is
larger.
168. c. For the given values of a, b, and c, column A is 4 + 6 ÷ 2. By
order of operations, first compute six divided by two, which is
three. Now, four plus three is equal to seven.
169. c. Use the distributive property on the expression in column A,
and distribute negative one to both terms in parentheses. Now
column A is 3x − 4 − x. Combining like terms leaves 2x − 4 in
column A, which is now identical to column B.
170. a. It is given that b is less than zero, which means that b is a
negative number. For the expression in column A, do the
quantities in parentheses first. Negative two times any negative
b results in a positive number. Any positive number squared is
also positive. In column B, order of operations says to do the
exponent first. A negative number squared is positive. Now,
however, this number is multiplied by −2, which always results
in a negative number. Any positive number is always greater
than a negative number.
171. a. It is given that a is less than zero, which means that a is a
negative number. By the rules of multiplying with negative
numbers, the quantity in column A is two times a negative
number squared, which is positive. The quantity in column B
is a negative times a negative (a positive result) which is then
multiplied by another negative number. This results in a
negative number.

172. d. The answer cannot be determined. It is given that the quantity
a multiplied by b multiplied by c is greater than zero. This can
be true if all of a, b, and c are positive numbers, or if any two,
but not all, of the three variables are negative. If all three
variables were positive, then column A would be greater. If,
however, both b and c were negative, then the quantity in
column A would be less than zero, the value of column B.
173. b. Simplify the expression in column A. The square root of x to
the fourth power is x squared. Any number squared, whether
positive or negative, is a positive number. Also, the square root
of 16 is four. Column A is 4x2
. Therefore, the quantity in
column B, 8x2
, is greater than 4x2
.
174. c. The quantity is column A becomes a + 4 when simplified, since
squaring and square root are inverse operations.
175. d. The answer cannot be determined. Simplify the expression in
column B, by first applying the distributive property to (a + b)2
,
to get a2
+ 2ab + b2
. Column B is now a2
+ 2ab + b2
− 2ab.
Combining like terms results in a2
+ b2
. Since a is not equal to
zero, and any number squared is positive, the quantity in
column B is greater if a < 0. If a > 1, column B is also greater. If,
however, a = 1, column A is equal to column B. If 0 < a < 1,
176. a. The words in column A translate in algebra to x − 5. Since it is
given that x is greater than five, the quantity in column A will
be a positive number. The quantity in column B will be a
negative number.
177. c. The words in column A translate into algebra as ᎏ
2(b +
2
14)
ᎏ since it
is twice the sum of b and 14, then divided by 2. The twos can
cancel, leaving b + 14.
178. d. The answer cannot be determined. For column A, the product
of a and b is ab. The expression in column B is the absolute
value of ab, which is always positive, regardless of the sign of the
quantity ab. Since there is no indication as to whether a or b is
positive or negative, the answer cannot be determined.
179. c. The words in column A translate into algebra as
ᎏ
1
3
ᎏx ÷ 4 = ᎏ
3
x
ᎏ ÷ ᎏ
4
1
ᎏ = ᎏ
3
x
ᎏ × ᎏ
1
4
ᎏ = ᎏ
1
x
2
ᎏ.
48

49
180. b. The words in column A translate into algebra as (͙xෆ)4
= x2
since (͙xෆ)4
= (͙xෆ)2
(͙xෆ)2
= x × x = x2
. Since x > 1, the quantity
in Column B is greater.
181. b. Since the variables d and p refer to the number of dimes and
pennies respectively, they must be positive whole numbers (it’s
impossible to have −3 dimes, or ᎏ
1
4
ᎏ of a penny). Regardless of
what numbers the variables represent, ᎏ
1
1
0
ᎏ of any positive
number plus ᎏ
1
1
00
ᎏ of another positive number will always be less
than the sum of the two whole numbers. Column B will always
be greater.
182. b. The words in column A translate into 4 ÷ xy = ᎏ
x
4
y
ᎏ. Since
x > y > 4, both x and y are greater than four. The quantity in
column B is a number larger than one, and the quantity in
column A is a number between zero and one.
183. a. The cost of a $50.00 sweater, on sale for 20% off is 80% of the
$50.00. This is (50.00)(.80) = $40.00.
184. b. The cost of a $28.00 basketball, including 6 percent sales tax, is
28 + .06(28). The decimal in column B representing sales tax is
actually equal to 60 percent, not 6 percent.
185. c. Let x represent an odd integer. The next (consecutive) odd
integer is thus x + 2. The product of these two integers is
x(x + 2).
186. a. To solve the given inequality, first subtract three from both
sides, to get 3b > 15. Now, divide both sides by three, to get
b > 5. Since b is greater than five, column A is greater.
187. b. To solve the given inequality, first use the distributive property
and distribute two to the terms in parentheses. The inequality
becomes 10 − 2x < 70. Subtract ten from both sides to get
−2x < 60. Now, divide both sides by negative two, noting that
when you divide by a negative number with an inequality, you
switch the inequality symbol. The inequality thus becomes
x > −30.
188. a. Since it is given that 0 < a < b, then ᎏ
1
a
ᎏ > ᎏ
1
b
ᎏ since, for fractions
with the same numerator, the smaller the denominator, the
larger the value of the fraction.

189. b. Solve the given inequality by subtracting seven from both sides
of the inequality to get b < −12. Since this is true, −2 is greater
than any allowed value for b.
190. b. The expression in column B can be written as 5 + z, because of
the commutative property of addition. Since it is given that
x < 5 and also that y < z, then the quantity in column A, x + y is
less than the quantity in column B, z + 5, because of the one to
one correspondence of the terms and the fact that you are
adding.
191. a. To solve the given inequality, first combine the like terms on
the left side of the inequality, to get ᎏ
1
6
ᎏx ≥ 4. Now, multiply both
sides by six to isolate the variable. The inequality is now x ≥ 24,
which is greater than the quantity in column B.
192. b. It is given that a > b > 0, so both a and b are positive, and a is
larger than b. For fractions with the same numerator, the
smaller the denominator, the larger the value of the fraction.
193. b. The x-coordinate of an ordered pair determines how far to the
left or right a point is plotted on the coordinate plane. Just by
looking at the drawing, point B is to the right of point A.
Therefore, the x-coordinate of point B is greater.
194. a. The slope of a line is the “steepness” of the graphed line. Lines
that go “downhill” when read from left to right have a negative
slope; lines that go “uphill” when read from left to right have a
positive slope. The line in the figure is going downhill, and
therefore has a negative slope.
195. c. Notice that the points A, B, and C form a right triangle in the
figure. To determine the length of segment BC, you count the
length of segment BA and the length of segment AC. Use these
lengths and the Pythagorean theorem to find the length of
segment BC. The Pythagorean theorem is a2
+ b2
= c2
, where a
and b are the lengths of the legs of a right triangle. Segment BA
is 3 units long, by counting. Segment AC is 4 units long.
Substituting in for a and b gives 32
+ 42
= c2
. This is 9 + 16 = c2
,
or 25 = c2
. Take the square root of both sides and c = 5.
50

51
196. b. Notice that the points A, B, and C form a right triangle in the
figure. The base of this triangle is 3 units (by counting), and the
height is 4 units. The formula for the area of a triangle is ᎏ
1
2
ᎏbh.
This is ᎏ
1
2
ᎏ(3)(4), which is six units.
197. b. If the circle has its center at the origin and passes through the
point (0,8), then the radius of the circle is 8.
198. c. The slope of a line that passes through two given points is
determined by the formula ᎏ
c
c
h
h
a
a
n
n
g
g
e
e
i
i
n
n
x
y
ᎏ = ᎏ
x
y2
2
−
−
y
x
1
1
ᎏ = ᎏ
4
2
−
−
2
1
ᎏ = ᎏ
2
1
ᎏ = 2.
199. b. The point given by the coordinates (a,b) is on the y-axis, and
above the x-axis. This indicates that a = 0 and that b is positive.
Therefore the quantity in column A is 0 − b, which will be a
negative number since b is greater than zero. Column B is
greater.
200. a. The points (−8,c) and (0,0) and (4,3) all lie on the same line.
Determine the slope of this line, using the points (0,0) and (4,3).
After the slope is determined, use it to find the value of c. The
slope of the line can be found by ᎏc
c
h
h
a
a
n
n
g
g
e
e
i
i
n
n
x
y
ᎏ = ᎏx
y2
2
−
−
y
x
1
1
ᎏ = ᎏ
3
4
−
−
0
0
ᎏ = ᎏ
3
4
ᎏ. To
find the value of c, use the given slope, which is ᎏ
3
4
ᎏ = ᎏ−
c
8ᎏ. Using
cross multiplication, 4c = −24. Divide both sides by 4, and
c = −6.
201. a. Since the given figure is a rectangle, and a rectangle has right
angles, then as the figure is drawn, the x-coordinate of point D
equals the x-coordinate of point A. Therefore, q = 2. Similarly,
the y-coordinate of point C equals the y-coordinate of point D,
and p = 3. Column A will be 3 − 2 = 1. Column B is 2 − 3 = −1.
202. b. If ab < 0, then a × b is negative. Since b > 0, b is a positive
number. Thus, a must be a negative number to make the
product negative. Therefore b, the positive value, is larger.
203. d. The relationship cannot be determined. If y < −1, column A is
greater. For example, ᎏ
−
1
4
ᎏ is greater than −4. If 0 < y < 1, column
A is greater. For example, = 3 which is larger than ᎏ
1
3
ᎏ. If y = 1,
then both columns simplify to 1. If y > 1, then column B is
greater; ᎏ
1
5
ᎏ < 5.
1
ᎏ
ᎏ
1
3
ᎏ

204. c. After canceling out the factors of x2
y in column A, you are left
with ᎏ
3
4
ᎏ, which is equal to 0.75.
205. d. The relationship cannot be determined. The relationship
between x and y is not stated. If y is greater than x, column A is
greater. If x is greater than y, column B is greater. If x and y are
equal, then the columns are equal.
206. b. (ᎏ
1
2
ᎏ)2
is equal to ᎏ
1
4
ᎏ, and anything to the zero power is 1. One
fourth is less than 1. Column B is greater.
207. b. Since x > 1, any value of x raised to the third power and
multiplied by three will be larger than the same value squared
and multiplied by two. For example, if x = 3, then 2 × 32
= 2 × 9
= 18 and 3 (33
= 3 × 27 = 81. Column B is greater.
208. c. Any base number to the zero power is equal to one, so x2
+ x − 2
in column A must equal zero. Column A and column B are
equal.
209. b. Since a is negative and b is positive, multiplying a × b results in a
negative value. A negative value raised to an odd numbered
power, like three, also results in a negative answer. Therefore,
Column B is larger.
210. b. Since a is greater than one, use the rules of exponents to
determine the larger value. In column A, when multiplying like
bases add the exponents. a6
× a2
= a8
. In column B, when raising
a power to another power, multiply the exponents. (a6
)2
= a12
. A
value greater than one raised to the twelfth power will be
greater than the same value raised to the eighth. Column B is
greater.
211. a. Since b is between 0 and 1, the value of b could be a fraction
like ᎏ
1
4
ᎏ; (ᎏ
1
4
ᎏ)2
= ᎏ
1
4
2
2ᎏ = ᎏ
1
1
6
ᎏ and (ᎏ
1
4
ᎏ)3
= ᎏ
1
4
3
3ᎏ = ᎏ
6
1
4
ᎏ; ᎏ
1
1
6
ᎏ is larger than ᎏ
6
1
4
ᎏ, so
212. b. −a2
means a2
times −1, which will result in a negative number.
(−a)2
equals (−a)(−a) which will give a positive result. Therefore
column B is larger.
213. a. ᎏ
2
3
ᎏ = 0.6
–
, which is very close in value to 0.667. Since these values
are between zero and one, raising them to the third power will
be smaller than raising them to the second power.
52

53
214. a. Since c and d are between 2 and 3, and c is less than d, substitute
values like c = 2.3 and d = 2.5. ᎏ
2
1
.3
ᎏ = 0.4348 and ᎏ
2
1
.5
ᎏ = 0.4. 0.4348
is larger than 0.4.
215. c. Since y = x + 8, substitute x + 8 in for y in column B. Now
column B says x + 8 − 4 which simplies to x + 4, making column
A equal to column B.
216. c. Multiply using the distributive property to get 2x − 10 − 4 = 10.
Combine like terms to get 2x − 14 = 10. Add 14 to both sides;
2x − 14 + 14 = 10 + 14. 2x = 24. Divide both sides by 2. x = 12.
Therefore, the columns are equal.
217. c. In order to solve for x, add 4x to both sides of the equation.
6x + 4x + 3 = −4x + 4x − 7. This simplifies to 10x + 3 = −7.
Subtract 3 from both sides of the equation 10x + 3 − 3 = −7 − 3
which simplifies to 10x = −10. Divide both sides by 10 and the
result is x = −1. Therefore, −x = −(−1) = 1 and x2
= (−1)2
= 1.
Both columns equal 1.
218. a. Since 11x + 3 = 42, subtracting 3 from both sides of the
equation results in 11x = 39. Subtracting three again on both
sides results in 11x − 3 = 36. Thus, column A is larger.
219. c. Multiply both sides of the equation by 3; 3(ᎏ
4
3
x
ᎏ + 4) = 3(2x). This
simplifies to 4x + 12 = 6x. Subtract 4x from both sides of the
equation; 4x − 4x + 12 = 6x − 4x, which is 12 = 2x. Divide both
sides by 2 to get x = 6. The columns are equal because 6 × 6 =
36 and 62
= 36.
220. b. Rearrange the second equation to be x + y = 4. Combine like
terms vertically to simplify to 2x = 10. Divide both sides of the
equation by 2 to get x = 5. Therefore 3 × 5 = 15 and π
(approximately equal to 3.14) × 5 would be greater than 15.
221. a. Solving the equations for x and y by adding them together
vertically to simplify to 2x = 6. Dividing both sides of the
equation gives x = 3. Substituting x = 3 into the first equation
results in 3 + y = 5, so y = 2. Since 3 > 2, the answer is column A.
222. c. Solve for c by adding the equations together vertically to
simplify to 2c = 6. Dividing each side of the equation by 2 gives
c = 3. The columns are equal.

223. a. Substituting y = 0 gives the equation 0 = 5x − 1. Adding 1 to
both sides results in 0 + 1 = 5x − 1 + 1, which is equal to 1 = 5x.
Divide both sides by 5 to get an x-value of ᎏ
1
5
ᎏ. Substituting
x = 0 gives y = 5(0) − 1 which becomes y = 0 − 1 = − 1. Since
ᎏ
1
5
ᎏ > −1, column A is greater.
224. a. Substitute the second equation into the first for x to get
8(ᎏ
3
2
ᎏ)y − 2y = 30; 8(ᎏ
3
2
ᎏ)y simplifies to 12y so the equation becomes
12y − 2y = 30. Subtract to get 10y = 30 and then divide by 10
on both sides; ᎏ
1
1
0
0
y
ᎏ = ᎏ
3
1
0
0
ᎏ. y = 3. Using the second equation,
x = ᎏ
3
2
ᎏ × 3 = 4.5. Thus, x is larger than y.
225. b. x2
− y2
is the difference between two squares and factors to
(x − y)(x + y). Since x + y = 16 and x − y = 8, column A then
becomes 16 × 8 = 128. Column B is (x + y)(x + y) which is 16 ×
16 = 256. Column B is greater.
226. b. The degree of a quadratic term is 2 and the degree of a cubic
term is 3, so column B is greater.
227. c. Divide each term by a factor of three to get x2
− 9 = 0. This is
the difference between two perfect squares which factors to
(x − 3)(x + 3) = 0. Setting each factor equal to zero and solving
results in x-values of 3 or − 3. Since 3 is the positive solution for
x, the columns are equal.
228. a. Factoring the left side of the equation gives (x − 5)(x + 2) = 0.
Setting each factor equal to zero is x − 5 = 0 or x + 2 = 0, which
results in a solution of 5 or −2. Since the negative result is −2,
and −2 is larger than −3, column A is larger.
229. c. Multiplying using the distributive property on the left side gives
−y2
+ 4y = 10. Adding y2
to both sides results in the equation
4y = y2
+ 10. Since this equation states that 4y is equal to y2
+ 10,
the columns have the same value.
230. b. Add 1 to both sides of the inequality; y + 1 > 2x − 1 + 1. This
simplifies to y + 1 > 2x. Since the inequality states that y + 1 is
greater than 2x, then column B is greater.
54

55
231. b. Take the compound inequality and add 2 to each section;
4 + 2 < 2x − 2 + 2 < 8 + 2. Simplified this becomes 6 < 2x < 10.
Dividing all sections by 2 gives a result of 3 < x < 5. Therefore,
the value of x is between 3 and 5, so column B is larger.
232. a. If you subtract one from both sides of the first inequality, it
yields a result of 4 < y, which means y is greater than 4. Thus,
column A is larger.
233. b. Use the distributive property on the left side of the inequality
to get 6x − 6 > 30. Add 6 to both sides of the inequality.
6x − 6 + 6 > 30 + 6. This simplifies to 6x > 36. Divide both
sides by 6 to get a result of x > 6. Column B is larger.
234. a. Add one to both sides of the inequality; −3x − 1 + 1 > 14 = 1.
This results in −3x > 15. Divide both sides of the inequality
by −3 to get a solution of x > −5. Remember that dividing both
sides of an inequality by a negative number changes the
direction of the inequality symbol. Since x is less than −5, the
answer is column A.
235. a. Since point (x,y) is located in Quadrant IV, the x-values are
positive and y-values are negative. Therefore, the x-values are
greater than the y-values. Column A is greater.
236. a. Using slope-intercept ( y = mx + b) form where m is the slope of
the linear equation, the slope of the line in column A is 2 and
the slope of the line in column B is ᎏ
1
2
ᎏ. Two is greater than ᎏ
1
2
ᎏ, so
column A is greater than column B.
237. d. This relationship cannot be determined. Since point (x,y) is
located in Quadrant II, x-values are negative and y-values are
positive. The opposite of any x-values, which are negative in
Quadrant II, would be positive. The reciprocals of any y-values
would also be positive and could be greater than, less than, or
equal to any of the x-values.
238. c. Using slope-intercept ( y = mx + b) form where m is the slope of
the linear equation, the slope of line l is ᎏ
2
3
ᎏ. Since line m is
perpendicular to a line that has a slope of −ᎏ
3
2
ᎏ, the slope of line m
is the negative reciprocal of this, or ᎏ
2
3
ᎏ. The slope of line l is
equal to the slope of line m.

239. c. To find the slope of a line, calculate the change in the y-values
over the change in the x-values. Therefore, ᎏ
4
c −
−
d
0
ᎏ = ᎏ
3
4
ᎏ which
simplifies to ᎏ
c −
4
d
ᎏ = ᎏ
3
4
ᎏ. Since the denominators are equal, set the
numerators equal to each other. Then, c − d = 3. The columns
are equal.
240. b. From the equation y = mx + b, b is the y-intercept. The
y-intercept for this equation is −3. The x-intercept is found by
substituting zero for y and solving for x. Starting with
0 = −2x −3, then adding 3 to both sides results in 3 = −2x.
Dividing both sides by −2 gives ᎏ
−
3
2
ᎏ or −1.5 = x. Since −3 < −1.5,
column B is greater.
241. a. Converting to slope-intercept form, the first equation becomes
2y = 4x + 6 by adding 4x to both sides. Dividing both sides by 2
gives ᎏ
2
2
y
ᎏ = ᎏ
4
2
x
ᎏ + ᎏ
6
2
ᎏ which simplifies to y = 2x + 3. The slope is 2.
The second equation is −3y = −3x + 9 after subtracting 3x from
both sides. Dividing both sides by −3 to get y by itself results in
ᎏ
−
−
3
3
y
ᎏ = ᎏ
−
−
3
3
x
ᎏ + ᎏ−
9
3ᎏ which simplifies to y = x −3. The slope of the
second equation is 1. Therefore column A is greater.
242. c. Using the distance formula, ͙(x1 − xෆ2)2
+ (yෆ1 − y2 )ෆ2
ෆ, the distance
between (−3,4) and (0,0) is ͙(−3 − 0ෆ)2
+ (4ෆ− 0)2
ෆ. This simplifies
to ͙(−3)2
+ෆ (4)2
ෆ = ͙(9 + 16ෆ)ෆ = ͙25ෆ = 5. The distance between
the point (0,0) and (−3,4) is 5 units, so the columns are equal.
243. a. “The difference between 26 and the product of 4 and 3”
translates to the expression 26 − (4 × 3), which simplifies to
26 − 12 = 14. “The sum of 8 and 2” is equal to 8 + 2 = 10. 14 is
larger than 10.
244. a. “Six less than nine” translates to 9 − 6 = 3. The square root of
four is 2. Column A is larger.
245. b. If the square of a number is four, then the number is 2. The
square root of 2 is approximately equal to 1.41, which is less
than 4.
246. c. The key word quotient tells you to divide; 10 ÷ 2 = 5 and
65 ÷ 13 = 5. Thus, column A is equal to column B.
56

57
247. a. If one-half of y is x, let y = 2x. If one-half of y is z, let z = 2y. By
substitution, z = 4x. Using the given equation x + y + z = 35,
substituting gives an equation of x + 2x + 4x = 35. Combine like
terms to get 7x = 35, which results in x = 5. Since z = 4x then
z = 4(5) = 20.
248. c. Use distance = rate × time. The distance of car A can be
expressed as 60t and the distance of car B can be expressed as
55t. Add the two distances and set the result equal to 460 miles.
60t + 55t = 460. Combine like terms. 115t = 460. Divide both
sides by 115; ᎏ
1
1
1
1
5
5
t
ᎏ = ᎏ
4
1
6
1
0
5
ᎏ = 4; t = 4 hours. The columns are equal.
249. c. Two consecutive integers are integers that are one number apart
like 4 and 5 or −22 and −23. Two consecutive integers whose
sum is 83 are 41 and 42. Twenty-three less than three times the
smaller is 100 and 16 more than two times the greater is 100.
The columns are equal.
250. b. The only possibilities have an odd number of quarters because
the total amount ends in a 5, which is impossible to get with
dimes. There can be 1 quarter and 9 dimes, or 3 quarters and 4
dimes. Five quarters is too much money. Either way, there are
more dimes than quarters, so the answer is column B.

59
geometry-based quantitative comparison questions:
■ Circles
■ Lines and Angles
■ Polygons
■ Quadrilaterals
■ Three-Dimensional Figures
■ Triangles
Some important information:
3Geometry

choose:
Examples:
Column A Column B
1. the sum of the measures 90°
of two acute angles
The answer is d. The only thing you can infer about the acute
angles in column A is that they are, by deﬁnition, each less than 90
degrees. However, depending on their measures, column A could
be smaller (two 30-degree angles = 60 degrees) or larger (an 80-
degree and a 45-degree angle = 125 degrees) than column B. The
answer cannot be determined from the information you are given
here.
2. the circumference of a circle the area of a circle with
with radius 2 radius 1.5
The answer is a. This problem is solved by substituting the values
into the formulas for the circumference and area of a circle—2πr
and πr2
, respectively. In column A, 2(π)(2) = 4π. In column B,
π(1.5)2
= 2.25π. 4π is greater than 2.25π, so a is correct.
60

61
Questions
Column A Column B
251. the area of ΔQ 8 sq m
252. the slope of the line in ᎏ
3
2
ᎏ
the graph
-3 -2 -1 1 2 3
-1
-2
-3
3
y
x
2
1
(2,3)
(0,0)
4
5
3Q

Column A Column B
253. the area of the circle in the area of a circle with
the ﬁgure diameter 3y
254. x ≥ 3
the volume of the box the volume of the cylinder
on the left on the right
h = x
r = x
4
6
x
box with length = 6,
width = 4, height = x
cylinder with radius = x, height = x
2x
62

63
Column A Column B
255. x z
256. 9 x
257. 6b the sum of the interior angles of
the polygon above
80° b
10
6
x
x z
y
Lines AB and CD are parallel.
A
C
B
D

Column A Column B
258. the area of the largest circle the area of the largest circle that
that can be cut out of a can be cut out of a rectangular
square piece of paper with piece of paper with length of 3.2″
sides of 3.4″ and a width of 5″
259. AC 4
260. ΔQRS is an isosceles triangle with angles Q = 45° and R = 45°
and line segments QS = 8 and QR = x. Polygon DEFGH has
sides DE = 3 and GH = y and polygon LMNOP has sides
LM = 1 and OP = 2.
x y
261. the measure of ∠AOD the measure of any right angle
262. the measure of ∠AOB the measure of an acute angle
180°
135°
100°
70°
20°
0°
F
E
D C
B
AO
60°
30°
2x
B
A
C2 3√¯¯¯
x
64

65
Column A Column B
263. the measure of a reﬂex angle the measure of ∠FOA
264. the measure of an angle the measure of an angle
supplementary to ∠BOC supplementary to ∠FOE
265. the sum of the measures of the sum of the measures of the
∠AOF, ∠AOD, and ∠BOD interior angles in a square
266. the measure of ∠1 the measure of ∠3
269. the sum of the measures the sum of the measures of
of angles 5 and 8 angles 2 and 3
272. The measure of ∠1 is 100°.
75° the measure of ∠8
the measure of ∠6 77°
1
4 3
2
5 6
8 7
m
n
t

Column A Column B
the measure of ∠4 the measure of ∠8
110° the measure of ∠11
277. The measure of ∠1 is x.
the measure of ∠8 2x
278. The sum of the measures of ∠13 and ∠10 is 160°.
a b
1 2
5 6
9 10
13 14 15
11 12
16
87
3 4
d
c
66

67
Column A Column B
279. the number of sides the number of sides
of this polygon of a triangle
280. the sum of the interior 360
angles of this polygon
281. the sum of the interior the sum of the exterior
angles of this polygon angles of this polygon
282. the number of sides the number of sides in
in this polygon a heptagon
283. the sum of the interior the sum of the interior
angles of this polygon angles of a hexagon
284. the sum of the interior the sum of the interior angles
angles of an 8-sided polygon of this ﬁgure
285. ᎏ
1
2
ᎏ of the sum of the interior the sum of the interior angles
angles of this ﬁgure of a triangle

Column A Column B
286. the area of this polygon the area of a convex polygon
if all sides have a length of 8 whose interior angles measure
900.
287. A convex polygon has 5 sides. This polygon also has
three right angles and two congruent angles.
the measure of one of 130°
the congruent angles
The ﬁgure is a parallelogram.
288. the measure of ∠A the measure of ∠B
289. the length of line segment 6
BC
290. the length of line segment the length of line segment
AE CE
A
C
B
D
E
A
50°
CD
6
9 B
68

69
Column A Column B
291. the length of line segment PR three times the length of line
segment SO
The ﬁgure is a rectangle.
The ﬁgure is a rhombus.
292. 25° the measure of ∠JKN
293. the measure of ∠KNL 80°
294. the length of line the length of line
segment NK segment JK
N
J
M
K
L
25°
O
13
25°
P Q
RS

Column A Column B
295. the measure of ∠AEB the measure of ∠BCD
296. the area of ΔABE the area of ΔCBD
297. the ratio of ᎏ
3
4
ᎏ
to the ratio of
298. A flagpole stands perpendicular to the ground. At noon, the
flagpole casts a shadow on the ground that is 6 feet long. At the
same time, a 5 foot tall woman stands next to the flagpole and
casts a shadow that is 3 feet long. The woman is also
perpendicular to the ground. Assume that the triangles
created in both situations are similar.
the height of the flagpole 9 feet
the length of line segment AE
ᎏᎏᎏᎏ
the length of line segment DC
the length of line segment AB
ᎏᎏᎏᎏ
the length of line segment DB
A
C
D
E
B
1 75°
20°2
85°
70

71
Column A Column B
299. A surveyor is looking at two buildings from a distance, both of
which stand perpendicular to the ground. He sets two telescopes
up at 45° angles to the tops of both buildings, and both telescopes
an elevation of 100 feet above sea level. The surveyor knows that
the bases of both buildings are 100 feet above sea level and that the
top of building A is 1,100 feet above sea level. The telescope
pointing at building A is 250 feet away from the base of the
building. The telescope pointing at building B is 300 feet away
from the base of that building.
the height of building A The height of building B
300. the length of line segment TS 7
301. Recreate the diagram substituting 3 for the length of line
segment RT and 4 for the line segment TS. Assume that
the length of line segment RS is not given.
6 the length of line segment RS
302. Recreate the diagram substituting 3 for the length of line segment
RT and 5 for the length of line segment RS.
4 b
R
ST
10
6

Column A Column B
The figure is a rectangle.
303. 21 in the perimeter of the figure
304. 28 sq in the area of the figure
305. 10 ft the perimeter of the figure
306. 25 sq ft the area of this square if each side
is 6 in longer than indicated in
the diagram
307. the perimeter of a rectangle the perimeter of a square with
with length 4 in and width 6 in sides of length 5 in each
308. the area of a square whose the area of a rectangle with
sides have a length of 3x length 2x and width x
5 ft
7 in
4 in
72

73
Column A Column B
309.
13 in the perimeter of this polygon
310. the base of the parallelogram the height of the parallelogram
in the figure in the figure
311. the area of the parallelogram 42 sq cm
in the figure
312. 42 sq in the area of the triangle in the
figure
313. the base of the triangle the base of a triangle whose
in the figure area is 35 sq cm and height is 10
cm
7 ft
5 ft
h
3 cm
8 cm
5 cm
3 in

Column A Column B
314. Mr. Jenkins is installing a pool in his backyard. The pool will
be a rectangle with a length of 12 feet and a width of 8 feet.
There will also be a 36 square foot wooden deck built
adjacent to one side of the pool.
120 sq ft the total area of the pool and
deck
315. Sally is mounting photographs on matting board. She has
48 square inches of matting board to work with, and a stack
of five rectangular photos each measuring 3″ × 5″. Assume
that she is cutting the board to fit each photo exactly,
with no board wasted.
the number of photos she the number of photos she does
can mount not have enough board to mount
316. A standard Stop sign has 8 sides, each of which measures
10 inches in length. A special Yield sign built to be seen
from a great distance is an equilateral triangle with sides
of 26 inches each and three reflective circles attached
to the front, each with a radius of 1.5 inches.
the perimeter of the Stop sign the perimeter of the Yield sign
317. A baker has made a square sheet cake for a birthday party.
At the last minute, the woman hosting the party calls to say
that the cake needs to be circular in shape, and not a square.
The square cake has a side 11″ in length.
the area of the largest ᎏ
3
4
ᎏ the total area of the
circular cake that can be cut square cake
out of the square cake
74

75
Column A Column B
318.
the surface area of this 135 sq m
rectangular prism
319.
the surface area of this cube 24 sq in
320. the surface area of a cube the surface area of a rectangular
with an edge 4″ long prism with a length of 5″
321.
100 cubic cm the volume of the prism in the
diagram
6 cm
3 cm
6 cm
2 in
3 m
5 m
8 m

Column A Column B
322.
the volume of the prism in the volume of a prism with a
the diagram base of 40 m2
and height of 4 m
323. the volume of a cube with twice the volume of a cube with
an edge of 4 meters an edge of 2 meters
324. the volume of this pyramid 1,200 cubic in
325. the volume of the pyramid the volume of the pyramid in
in the diagram if the height the diagram if the length and
is changed to 9 in width are changed to 9 in each
h= 12 in
10 in
10 in
30 m2
5 m
76

77
Column A Column B
326. the volume of the pyramid 80 cm3
in the diagram
327. the volume of the pyramid the volume of a regular pyramid
in the diagram if the height with area of the base = 60 cm2
is lengthened to 6 cm and height = 5 cm
328. the circumference of a circle the circumference of a circle
with diameter d = 5 cm with diameter d = 7 cm
329. the circumference of a circle 21 cm
with diameter d = 7 cm
330. the diameter of a circle with the radius of a circle with
circumference C = 25.12 m circumference C = 31.4 m
331. the radius of the circle in the diagram 4 ft
332. the area of the circle in the diagram 15.7 ft2
10 ft
51 cm2
h = 5 cm

Column A Column B
333. A circle has a radius of r = 4 in.
the area of a square that the area of the circle
would ﬁt inside of the circle
334. A circle has a diameter of d = 8 cm.
the area of the circle the area of the largest circle that
could be cut out of a square of
fabric with an area of 16 sq cm
335.
the total surface area of the 82 cm2
cylinder in the diagram
336.
the total surface area of the total surface area of a cube
this cylinder with edges e = 9 cm
337.
the volume of the cylinder 1,350 cm3
in the diagram
9 cm
4 cm
7 in
8 in
1 m
12 m
78

79
Column A Column B
338. the volume of a cylinder if the volume of a cylinder if
the radius is doubled the height is doubled
339. A cylinder has a radius r = 5 cm and height h = 10 cm.
750 cm3
the volume of this cylinder
340. A cone has a radius r = 5 cm and height h = 10 cm.
the volume of the cone 300 cm3
341. A sphere has a radius of 6 cm.
the surface area of the sphere 452.16 sq cm
342. the surface area of this sphere 615 sq m
343. the volume of this sphere 2,000 cm3
344. the volume of a sphere with the volume of a sphere with
a radius of 2 ft a radius of 3 ft
345.
the slope of the line in the diagram ᎏ
1
3
ᎏ
x
y
7 m

Column A Column B
346.
the slope of the line in −ᎏ
1
3
ᎏ
the diagram
347.
the slope of the line in ᎏ
1
2
ᎏ
the diagram
x
y
x
y
80

81
Column A Column B
348.
the slope of the line in ᎏ
2
3
ᎏ
the diagram
349. 0 the slope of the line that passes
through the points (2,2) and (4,0)
350. A line passes through the points (0,0) and (−2,−3).
0 the slope of the line
351. 0 the slope of the line y = −1
352. Tommy is standing at the base of a hill at a vertical elevation of 0 ft
above sea level. He knows the hill rises along a line measured
to have a slope of ᎏ
2
5
ᎏ. He also knows he has to walk
100 horizontal feet to get to the peak.
35 ft above sea level the vertical elevation of the peak
of the hill
x
y

Column A Column B
353. the y-intercept of the line in 0
the diagram
354. the y-intercept of the line in the y-intercept of a line deﬁned
the diagram by the equation y = −ᎏ
1
2
ᎏx + 4
355. the y-intercept of the line in 0
the diagram
356. the y-intercept of the line in the y-intercept of a line deﬁned
the diagram by the equation y = x − 1
x
y
x
y
82

83
Column A Column B
357. the y-intercept of a line the y-intercept of a line defined
defined by the equation by the equation y = −ᎏ
1
4
ᎏx + 4
y = ᎏ
3
2
ᎏx − 6
358. the y-intercept of a line the y-intercept of a line
defined by the equation defined by the equation
y = x + ᎏ
2
5
ᎏ y = 9x − 4
2y = x + 8 y = 4x + 3
y = ᎏ
3
2
ᎏx − 6 6 = x − y
2y + ᎏ
1
2
ᎏx = 0 x = 9
2y = 10 xy = x2
+ 4x
O is the center of the circle.
363. The circumference of circle O is 20 cm. The measure of ∠X is 45°.
the length of arc ABC 2 cm
O
A
B
C
X

Column A Column B
364. The circumference of circle O is 10 cm. The measure of ∠X is 36°.
the length of arc ABC 1 cm
365. The length of arc ABC is 2 cm. The measure of ∠X is 45°.
20 cm the circumference of circle O
366. The circumference of circle O is 30 cm.
The length of arc ABC is 3 cm.
35° the measure of ∠x
367. the area of a rectangle the area of a square with
with sides of 2 cm and 3 cm sides of 2 cm
368. the sum of the interior the sum of the interior angles
angles of a 30-60-90 triangle of an isosceles right triangle
369. the volume of a cube with the volume of a cylinder with
side of length 3 cm radius 2 cm and height 3 cm
370. the hypotenuse of a right the hypotenuse of a right
triangle with shorter sides triangle with shorter sides
of length 3 and 4 of length 6 and 8
373. A line is represented by the equation y = 3x + 2.
the slope of the line the y-intercept of the line
l
t
m
1 2
34
5 6
78
84

85
Column A Column B
374. ΔABC is a right triangle with shorter sides of 3 and 6.
ΔBCD is a right triangle with shorter sides of 4 and 8.
the ratio of the lengths of ᎏ
1
2
ᎏ
the two triangles’ hypotenuses
375. A circular apple pie with a circumference of 18″ is cut into 9 equal
slices. The slices are all cut starting at the center of the pie out to
the edge of the crust.
the length of the arc formed 1.5″
by the crust of one slice at its
outer edge

Answer Explanations
251. b. The area of a triangle = ᎏ
1
2
ᎏbh. ΔQ is a right triangle, so we can
substitute 3 and 4 for b and h. A = ᎏ
1
2
ᎏ(3(4)) = ᎏ
1
2
ᎏ(12) = 6. The area
of ΔQ is 6 sq m, which is less than 8 sq m, so the answer is b.
252. c. The slope of a line is defined as ᎏ
c
c
h
h
a
a
n
n
g
g
e
e
i
i
n
n
x
y
ᎏ. The line in the graph
crosses the origin at (0,0) and also intersects the point (2,3).
This creates a rise (change in the y-value) of 3 and a run
(change in the x-value) of 2, which gives a slope of ᎏ
3
2
ᎏ. Thus, the
values in a and b are equal, so the answer is c.
253. d. The area of a circle is defined as A = πr2
. The radius of the
circle drawn is 2x, so we can calculate the area of the circle as
4x2
π. The area of a circle with diameter 3y would be
a = πr2
= (3y)2
π = 9y2
π. However, since the values of x and y are
not defined, it is impossible to evaluate whether quantity A
(4x2
π) or quantity B (9y2
π) is greater. If, for instance, x = 5 and
y = 2, then quantity A would be greater. But if x = 2 and y = 5,
quantity B would be greater. So there is not enough
information to evaluate the equations and the answer is d.
254. b. The formula for the volume of a box is A = lwh. So the volume
of the box at left = 4(6)x or 24x. The volume of a cylinder =
πr2
h, so the volume of the cylinder at right = πx2
x, or πx3
.
Substituting x = 3 into both equations, the volume of the box
becomes 24(3) = 72 and the volume of the cylinder becomes
π(3)3
= π(27) = 3.14(27) = 85.78. So long as x is greater than or
equal to 3, the volume of the cylinder is greater and the answer
is b.
255. d. When two parallel lines are cut by a line segment, the resulting
corresponding angles created are equal, so x = y. The sum of
two complementary angles is always 180, so x + z = 180, and
since x = y, it is also true that y + z = 180. However, no
information is given about the relationship between x and z
other than the fact that they add up to 180. Even though the
86

87
ﬁgure makes it look like x > z is true, this cannot be taken for
fact. Therefore there is not enough information to further
evaluate the problem and the answer is d.
256. a. The Pythagorean theorem states that in any right triangle,
a2
+ b2
= c2
where c is the hypotenuse of the triangle and a and b
are the lengths of the other sides. Since the side labeled 10 is
opposite the right angle, it is the hypotenuse of the triangle, so
the Pythagorean theorem can be used by substituting 6, x, and
10 for a, b, and c, respectively.
a2
+ b2
= c2
62
+ x2
= 102
36 + x2
= 100
x2
= 64
x = 8
x = 8 and therefore is less than 9, so the answer is a.
Note: If you recognize that 6, 8, 10 is a Pythagorean triple,
then you know that x must be equal to 8 and you can quickly
solve the problem.
257. b. When two line segments intersect, the resulting vertical angles
are always equal, so b = 80°. Therefore, 6b = 6(80°) = 480°. The
sum of the interior angles of a polygon can be found by drawing
all diagonals of the polygon from one vertex and multiplying
the number of triangles formed by 180°. The polygon at the
right can be divided into 3 triangles, so 3(180°) = 540°.
540° > 480°, so the answer is b.
258. a. The area of a circle is dependent on the length of its radius, so
the problem here is to determine which circle could have the
largest radius. To cut a circle out of a rectangular piece of paper,
you must draw a circle whose radius is no greater than any of
the sides of the rectangle. This is because the radius extends
equally from the center of the circle in all directions. So even
though the width of the rectangle in column B is 5″, its length is
only 3.2″, which is smaller than the sides of the square in
column A (3.4″). Thus a circle with a larger radius—and,
therefore, greater area—could be cut from the square in
column A. So the answer is a.

259. c. Sketching ΔABC will be helpful here. ∠B must be right because
the sum of angles A and C is 90°, and the sum of all three angles
in the triangle must add up to 180°. The fact that this triangle
has angles of 30°, 60° and 90° means that it is a 30-60-90 special
right triangle whose sides are x, x͙3ෆ and 2x. (You should have
the lengths of this and the other special right triangle—a
45-45-90 triangle—memorized.) Since AC is the hypotenuse of
this triangle, its value can be represented by 2x according to the
relative lengths of the sides of this kind of special right triangle.
BC measures 2͙3ෆ, and it is opposite the 60° angle (∠A), so it
can be represented as the x͙3ෆ side of the triangle. Setting
2͙3ෆ = x͙3ෆ, and dividing both sides of the equation by ͙3ෆ
yields x = 2. Plugging this value of x into AC = 2x gives line
segment AC a value of 2(2), which equals 4. Thus the values of
column A and column B are the same, and the answer is c.
260. d. Though the problem tempts you to sketch the shapes and use
your knowledge of isosceles triangles to determine x and your
knowledge of similar polygons to determine y, there is no need.
The problem does not state that polygons DEFGH and
LMNOP are similar, and since there is no information
indicating that their corresponding sides are in the same ratio
or that corresponding angles are equal, this cannot be
determined. There is not enough information to solve the
problem, so the answer is d. When taking the test, be sure to
read through the problems before spending time sketching
shapes and solving equations. Determine whether or not you
have enough information to solve the problem before delving
into it.
Note: If you were told that the polygons were, in fact, similar,
the problem could be solved. Line segment QR in ΔQRS is the
hypotenuse (by virtue of being opposite of ∠s, which must
equal 90°) and therefore has a length greater than 8. The fact
that corresponding sides in similar polygons have lengths of a
similar ratio could then be used to set up the ratio DE/GH =
LM/OP. Substituting would yield ᎏ
3
y
ᎏ = ᎏ
1
2
ᎏ, which solves to y = 6.
Since x > 8, x must be greater than y and therefore the answer
would be a.
88

89
261. a. The measure of ∠AOD as indicated in the diagram is 100°. The
measure of any right angle is always 90°. Therefore ∠AOD is
bigger and the answer is a.
262. d. The measure of ∠AOB as indicated in the diagram is 20°. An
acute angle is an angle with a measure less than 90°. Since an
acute angle could be less than 20°, equal to 20° or between 20°
and 90°, there is not enough information to say whether or not
an acute angle would be greater than ∠AOB. Therefore, the
answer is d.
263. a. A reflex angle is defined as an angle whose measure is between
180° and 360°. The measure of ∠FOA as indicated in the
diagram is 180° (which makes it a straight angle). This is smaller
than a reflex angle, so the answer is a.
264. b. Supplementary angles are angles whose measure adds up to
180°. The diagram does not directly indicate the measures of
angles BOC and FOE, but it does give enough information to
find these measures using subtraction:
∠BOC = ∠AOC − ∠AOB
∠BOC = 70° − 20°
∠BOC = 50°
∠FOE = ∠FOA − ∠EOA
∠FOE = 180° − 135°
∠FOE = 45°
∠BOC = 50°, so its supplement must equal 180° − 50°, or 130°.
∠FOE = 45°, so its supplement must equal 180° − 45°, or 135°.
The supplement to ∠FOE is larger, so the answer is b.
265. c. The interior angles in a square are all right angles, and since
there are four of them, the sum of their measures is always 360°.
∠AOF as indicated in the diagram measures 180° and ∠AOD
measures 100°. To find the measure of ∠BOD, use subtraction:
∠BOD = ∠AOD − ∠AOB
∠BOD = 100° − 20°
∠BOD = 80°

Therefore the measures of the three angles in column A are
180°, 100°, and 80°, which add up to 360°. This is the same as
the sum in column B, so the answer is c.
266. c. Vertical angles are congruent (equal). Vertical angles are defined
as angles, formed by 2 intersecting lines, which are directly
across or opposite from each other. Angles 1 and 3 are vertical
and therefore congruent. The answer is c.
267. c. When a transversal line intersects two parallel lines, the
resulting corresponding angles are congruent. Corresponding
angles are defined as the angles on the same side of the
transversal and either both above or below the parallel lines.
Angles 1 and 5 are corresponding and therefore congruent, so
the answer is c.
268. c. Angles 7 and 3 are corresponding and therefore congruent. The
answer is c.
269. c. Angles 5 and 8 are supplementary because they combine to
form a straight line. The same is true of angles 2 and 3. Supple-
mentary angles always add up to 180°, so the measures of both
sets of angles are the same and the answer is c.
270. c. Angles 2 and 8 are neither corresponding nor vertical.
However, angles 2 and 6 are corresponding, so their measures
are equal. Angles 6 and 8 are vertical, so their measures are also
equal. This information can be used to determine that angles 2
and 8 are congruent because m∠2 = m∠6 = m∠8. In fact, angles
2 and 8 are called alternate exterior angles. Alternate exterior
angles are always congruent, so the answer is c.
271. d Angles 3 and 6 are same side interior angles. This means that they
are both inside the parallel lines and on the same side of the
transversal. Same side interior angles are always supplementary,
so their measures add up to 180°. However, this relationship
says nothing about the specific values of each angle, and even
though the drawing makes it look like one angle might be larger
than the other, no information is given that could determine the
actual value of either angle. Therefore, there is not enough
information to solve the problem and the answer is d.
90

91
272. b. Angles 1 and 8 are same side exterior angles. This means that they
are both outside the parallel lines and on the same side of the
transversal. Same side exterior angles are always supplementary.
Supplementary angles add up to 180° and the measure of ∠1 is
given as 100°, so the measure of ∠8 must be 180° − 100° = 80°.
80° is greater than 75°, so the answer is b.
273. b. Angles 3 and 6 are same side interior angles, which means that
they are supplementary. Since the measure of ∠3 is 105°, the
measure of ∠6 must be 180° − 105° = 75°; 77° > 75°, so column
B is greater. The answer is b.
274. b. ∠3 and ∠4 are supplementary angles, so the sum of their
measures must add up to 180°. Therefore the measure of
∠4 = 180° − 100° = 80°; ∠3 and ∠8 are vertical angles and
therefore congruent, so the measure of ∠8 = 100°; 100° > 80°,
so the measure of ∠8 is larger than the measure of ∠4 and the
answer is b.
275. b. ∠2 and ∠11 have no direct relationship upon ﬁrst glance.
However, there are two sets of parallel lines in this diagram
(lines a and b, and lines c and d) so there are many related angles
to work with. ∠2 and ∠3 are same side exterior angles along
transversal c and therefore are supplementary. So the measure
of ∠3 is 180° − 65° = 115°. Angles 3 and 11 are corresponding
angles (along transversal b and both above the parallel lines c
and d) and so are congruent. So the measure of ∠11 is also
115°, which is greater than column A. The answer is b.
276. c. The measure of ∠9 is information you actually don’t need to
solve this problem. Some questions will provide extra
information like this in an attempt to throw you off, so don’t be
tricked if you’re sure of how to solve the problem! In this case,
angles 16 and 8 are corresponding angles (both along
transversal b and beneath parallel lines c and d, respectively) and
so must be congruent. Therefore, the answer is c because their
measures are the same.

277. b. ∠1 and ∠8 are alternate exterior angles (on opposite sides of
transversal c and on the outside of the parallel lines a and b), and
therefore are congruent. So the measure of ∠8 is x; x indicates
the measure of an angle, so it cannot be negative. Therefore,
2x > x and the answer is b.
278. b. This problem is actually not that difﬁcult to solve, but it does
require several steps to determine the measures of both angles
since there is no direct relationship between the two. First,
angles 13 and 10 are vertical angles and therefore congruent.
So the measure of each angle is ᎏ
1
2
ᎏ of their sum, or 80°. Next,
angles 10 and 11 are same side interior angles (along trans-
versal d and both on the inside of parallel lines a and b) and
therefore supplementary, so the measure of ∠11 must be
180° − 80° = 100°.
Now ﬁnd the measure of ∠4. ∠11 and ∠7 are same side interior
angles (along transversal b and both on the inside of parallel
lines c and d) and so are supplementary. Therefore the measure
of ∠7 = 180° − 100° = 80°; ∠7 and ∠4 are vertical angles, and so
the measure of ∠4 must also be 80.
The measure of ∠11 is greater than the measure of ∠4 because
100° > 80°. The answer is b.
With some practice, you will become familiar with the
relationships between the angles created by a transversal cutting
two parallel lines. This will make analyzing a system of multiple
parallel lines and transversals much easier, and you will be able
to quickly intuit the relationship between angles like these by
logically connecting different pairs of congruent and
supplementary angles. Once you know the basic rules it
becomes easier and easier to break down the components of a
complicated system of lines and angles to solve the problem.
279. a. This polygon has four sides, making it a quadrilateral. Triangles
are three-sided polygons. A quadrilateral has more sides than a
triangle, so the answer is a.
92

93
280. c. You can use the formula S = 180(n − 2) to find the sum of the
angles in a convex polygon, where n represents the number of
sides in the polygon. A four-sided polygon such as this one has
an angle sum of S = 180(4 − 2) = 180(2) = 360, which is equal to
the amount in column A, so the answer is c.
281. a. You can use the formula S = 180 (n − 2) to determine that the
angle sum of this polygon is 900. The exterior sum of any
convex polygon is always 360, so the answer is a.
282. c. A heptagon is a seven-sided polygon. This polygon also has
seven sides, so the values in the two choices are equal. The
answer is c.
283. a. Though the formula S = 180 (n − 2) can be used to determine
that the angle sum of this polygon is 900 and the angle sum of a
hexagon is 720, an understanding of the nature of convex
polygons provides an easier way to solve the problem. A
polygon’s angle sum increases as the number of sides of the
polygon increases. Since this polygon has more sides (7) than a
hexagon (6), the sum of its interior angle sum will be greater, so
the answer is a.
284. c. This figure is an 8-sided polygon, so the value of choices are
equal. The answer is c.
285. a. Though you can use the formula S = 180 (n − 2) to determine
that ᎏ
1
2
ᎏ of the angle sum of this polygon is 540 (the entire sum is
1,080) and the angle sum of a triangle is 180, you should more
quickly be able to determine that since this is an 8-sided
polygon, its angle sum will be more than double that of a
3-sided triangle. In either case, the answer is a.
286. d. This figure is an 8-sided polygon with all sides of length 8.
Though the math is a bit involved, you do have enough
information to determine its area. However, while the sum of
the interior angles of a convex polygon can be used to
determine how many sides the figure has, the area of the second
polygon cannot be calculated without more information.
Therefore this problem cannot be fully solved and the answer
is d.

287. a. The formula S = 180(n − 2) can be used to determine that the
sum of the interior angles of this polygon is 540. Right angles
measure 90° each, so subtracting the sum of the three right
angles from 540° leaves 270° for the two remaining congruent
angles. Congruent angles are angles with equal measures, so
dividing ᎏ
27
2
0°
ᎏ yields 135° for each angle, which is larger than the
value in column B (130°). The answer, therefore, is a.
288. a. This diagram shows a parallelogram. Opposite angles in a
parallelogram are always congruent, so the measure of ∠B is
equal to the measure of ∠D, which is indicated as 50°.
Consecutive angles in a parallelogram are supplementary, so the
measure of ∠A = 180° − m∠B, or 180° − 50° = 130°. Since
130° > 50°, ∠A > ∠B, so the answer is a.
289. c. Opposite sides in a parallelogram are congruent, so the lengths
of line segments AD and BC will be equal. The length of line
segment AD is 6, so the length of BC must be 6, as well. The
value in column B is also 6, so the answer is c.
290. c. This question modiﬁes the diagram to show line segments AC
and BD, which are the diagonals of the parallelogram. Point E
marks the intersection of these line segments, where they bisect
each other. Line segments AE and CE are therefore the two
halves of line segment AC, and have equal lengths. The answer
is c.
291. b. The ﬁgure shows a rectangle with diagonals PR and SQ that
intersect at point O. The diagonals of a rectangle are congruent
and bisect each other, so the length of PR
–––
is equal to twice the
length of SO
–––
(which is one half of the diagonal SQ
–––
). Three
times the length of SO
–––
, however, is longer than PR
–––
, so the
answer is b.
Note: Line segment SO represents a length and so cannot
reasonably have a negative value. Thus three times SO
–––
cannot
be a negative value, either, so three times the length of SO
–––
will
always be greater than the length of PR
–––
.
94

95
292. c. The ﬁgure shows a rhombus. The diagonals of a rhombus
bisect the angles of a rhombus, so angles JKN and NKL have
equal values because they comprise the two halves of the
bisected ∠JKL. The measure of ∠NKL is given as 25°, so the
values in both choices are equal. The answer, then, is c.
293. a. The diagonals of a rhombus are perpendicular, so ∠KNL
measures 90° because it is formed out of the intersection of the
diagonals of this rhombus. 90° > 80°, which is the value in
column B, so the answer is a.
294. b. The diagonals of a rhombus are perpendicular to each other, so
angles JNK and LNK are both right angles. This means that
ΔJNK is a right triangle. Line segment JK is the hypotenuse of
ΔJNK because it is opposite the right angle. Line segment NK
is also part of the same triangle, opposite one of the smaller
angles of the triangle (in this case, ∠KJN, which measures 65°).
The hypotenuse is always the longest of the three sides of a
triangle, so line segment JK is longer and the answer is b.
295. c. ∠AEB is indicated to have a measure of 85°. The sum of the
angles in a triangle is always 180°, so ∠BCD also can be
determined to have a measure of 85° (because the other two
angles in ΔBCD add up to 95°). The values of the two choices
are therefore equal and the answer is c.
296. d. There is enough information given to ascertain that the two
triangles are similar according to Angle-Angle. However, no
information is given to indicate the speciﬁc length of any side of
either triangle. Therefore, even though triangle ABE appears
larger in the diagram, the actual area of either triangle cannot
be calculated. There is not enough information to solve this
problem, and the answer is d.
297. a. ∠1 and ∠2 are vertical angles, and therefore congruent, so ∠1
measures 75°. This can be used to determine the measure of the
third angle of each triangle and consequently that triangles ABE
and CBD are similar by Angle-Angle. The lengths of two
corresponding sides of similar triangles are always
proportionate, so the ratio in column A works out to ᎏ
1
1
ᎏ; ᎏ
1
1
ᎏ > ᎏ
3
4
ᎏ,

so the answer is a. This problem is easy to solve if you know the
properties of similar triangles and are able to understand the
ratio described in column A.
298. a. Drawing a diagram may be helpful to solving this problem.
The relationships between the flagpole and its shadow and the
woman and her shadow can be depicted as a right triangles
because both the flagpole and the woman are perpendicular to
the ground. The problem states that the triangles are similar.
Similar triangles have sides of lengths proportionate to each
other, so a ratio can be set up between the ratio of the flagpole’s
height to the length of its shadow and the woman’s height to
the length of her shadow as such, where x represents the height
of the flagpole in feet:
=
ᎏ
6
x
ᎏ = ᎏ
5
3
ᎏ
3x = 30
x = 10
The flagpole is 10 feet tall, which is greater than the value in
column B. Therefore the answer is a.
299. b. The problem states that both buildings are built at the same
elevation and are perpendicular to the ground, and that both
telescopes are level with the bases of the buildings and pointed
at angles of 45° to the tops of the buildings. This means that
similar 45/45/90 right triangles are created because the third
angle in each triangle can only be 45°. A ratio can be set up to
determine the height of building B based on the height of
building A and the distance from the bases of the buildings to
height of woman
ᎏᎏ
length of shadow
x
ᎏᎏ
length of shadow
3 ft 6 ft
5 ft
shadow shadow
96

97
their respective telescopes. Note that the telescopes are 100 ft
above sea level and subtract that amount from the height of the
top of building A to get its actual height. Let b represent the
height of building B in feet:
=
ᎏ
1
2
0
5
0
0
0
ᎏ = ᎏ
30
b
0
ᎏ
250b = 3,000,000
b = 1,200
The height of building B is 1,200 feet. This is more than the
height of building A (1,000) so the answer is b.
Note: It is not necessary to solve for b because it is evident that
b > 1,000 from looking at the proportion.
300. a. ΔDFE is a right triangle and the lengths of two of its sides are
indicated, so the Pythagorean theorem can be used to
determine the length of the missing side. Note that in the
equation a2
+ b2
= c2
, c represents the length of the hypotenuse of
the triangle where a and b are the lengths of the other two sides:
a2
+ b2
= c2
62
+ b2
= 102
36 + b2
= 100
b2
= 64
b = 8
The length of line segment TS is 8; 8 > 7, so a is the answer.
Note: If you know your Pythagorean triples, you would have
recognized the 6, 8, 10 triple and have been able to solve the
problem quickly without the need for any calculations.
301. a. 3, 4, 5 is another Pythagorean triple, so if you recognize this
you can immediately tell that the length of the hypotenuse, RS
–––
,
is 5. Otherwise, use the formula as follows:
a2
+ b2
= c2
32
+ 42
= c2
9 + 16 = c2
25 = c2
5 = c
b
ᎏᎏᎏ
distance to telescope B
height of building A
ᎏᎏᎏ
distance to telescope A

The length of line segment RS is 5, which is less than the value
in column A. Therefore the answer is a.
302. d. Using the Pythagorean theorem will show that the length of
line segment TS is 4; 3, 4, 5 is also a Pythagorean triple that
you should be familiar with. Even though the variable b is
commonly used to denote one of the shorter sides of a triangle
in the Pythagorean theorem, this problem does not indicate any
specific value for b. As such, there is not enough information to
draw a comparison between 4 and b and the answer is d.
303. b. The figure is a rectangle. The perimeter of a rectangle can be
calculated using the formula p = 2(l + w), where l represents the
length and w is the width:
p = 2(l + w)
p = 2(4 + 7)
p = 2(11)
p = 22
The perimeter is 22 in, which is greater than the value in
column A. The answer is b.
304. c. The area of a rectangle can be found using the formula a = lw,
where l is the length and w is the width of the rectangle:
a = lw
a = 4(7)
a = 28
The area is 28 sq in, which is equal to the value in column A.
The answer is c.
305. b. The figure is a rhombus because it has four sides all of the same
length. The perimeter of a rhombus can be calculated using the
formula p = 4s, where s is the length of any one side:
p = 4s
p = 4(5)
p = 20
The perimeter of the rhombus is 20 ft, which is more than the
value in column A. The answer is b.
98

99
306. b. The formula for calculating the area of a square is A = s2
, where
s is the length of any one side. The area of the square as drawn
is 25 sq ft, as follows:
A = s2
A = 52
A = 25
If the square was to be redrawn with sides longer than indicated
in the diagram, the area of the square would increase. Since
column A represents the area of the square as indicated, the
value of column B must be larger, and there is no need to
calculate the area (though it does work out to 30.25 sq ft). The
answer is b.
307. c. The formula p = 2(l + w) can be used to calculate the perimeter
of this rectangle as 2(4 + 6) = 20 in. The formula p = 4s can be
used to calculate the perimeter of this square as 4(5) = 20 in.
The values are the same, so the answer is c.
308. a. The formula A = s2
can be used to calculate the area of this
square as A = (3x)2
or 9x2
. The formula A = lw can be used to
calculate the area of this rectangle as A = 2x(x) or 2x2
; 9x2
> 2x2
,
so the answer is a.
309. b. The diagram indicates that the sides of this ﬁve-sided polygon
are congruent, so the perimeter can be calculated using the
formula p = ns, where n is the number of sides and s is the length
of any one side:
p = ns
p = 5(3)
p = 15 in
15 in > 13 in, so the answer is b.
310. a. The base is indicated as being 8 cm, while the height is 3 cm.
The base is longer, so the answer is a. Remember that the
height of a parallelogram is always indicated by a line segment
drawn at a right angle to the base. The height of a
parallelogram is sometimes referred to as its altitude.

311. b. The area of a parallelogram can be calculated using the formula
A = bh, where b is the base of the parallelogram and h is the
height. In this case, A = 8(3) = 24 sq cm, which is less than the
value in column B. Therefore the answer is b.
312. a. The area of a triangle is calculated using the formula A = ᎏ
1
2
ᎏbh,
where b is the base of the triangle and h is the height. The base
of the triangle in the figure is 7 cm and the height is 6 cm, so
the formula works out to A = ᎏ
1
2
ᎏ(7)(6) = ᎏ
1
2
ᎏ(42) = 21. The area of
the triangle is 21 sq in, which is less than the value in column A.
The answer is a.
313. c. The figure indicates that the base of this triangle is 7 cm. To
find the base of the triangle in column B, use the formula
A = ᎏ
1
2
ᎏbh. The problem states that the area of this triangle is
35 sq cm and the height is 10 cm, so the base can be calculated
as follows:
A = ᎏ
1
2
ᎏbh
35 = ᎏ
1
2
ᎏb(10)
35 = 5b
7 = b
The base of the triangle in column B is 7 cm, which is the same
as the value of column A. Therefore the answer is c.
314. b. To find the area of Mr. Jenkins’ pool, use the formula a = lw to
find that the area will equal 12 ft × 8 ft, or 96 sq ft. Added to the
36 square foot area of the deck, the total area is 132 sq ft, which
is more than the value of column A. The answer is b.
315. a. Each photo will require 15 sq in of mounting board, as
determined by using the formula a = lw, where the length is 3″
and the width is 5″. Mounting three photos will use up 15 × 3,
or 45 sq in of matting board. Sally started with 48 sq in of
board, so using 45 sq in will leave her 3 sq in left over, which is
not enough to mount a fourth photo. Mounting 3 of the five
photos will leave her with 2 unmounted photos, so she was able
to mount more photos than not and the answer is a.
100

101
316. a. The perimeter of a polygon is found by adding up the lengths
of all its sides, or multiplying the length of one side by the
number of sides if all sides are of equal length. An 8-sided Stop
sign with sides of 10″ length has a perimeter of P = ns =
8(10″) = 80″. An equilateral triangle has three sides of equal
length, so this special Yield sign has a perimeter of P = ns =
3(26″) = 78″. Information regarding the reflectors attached the
front of the sign is irrelevant and should be ignored, as
perimeter is the measure of a shape’s exterior edges. 80″ > 78″,
so the Stop sign has a greater perimeter and the answer is a.
317. a. This problem is not difficult in terms of the calculations
required to solve it, but rather requires several steps to get to
the final answer. First, to find the largest circle that can be cut
from a square with sides of 11″, determine the largest possible
radius within the square. Since a radius extends from the center
of a circle to any edge, it can be no longer than ᎏ
1
2
ᎏ of the length
of any of the square, in this case 5.5″. So the largest possible
circle within this square would have a radius of 5.5″ and an area
calculated using the formula A = πr2
, where r is the radius and
π = 3.14:
A = πr2
A = π(5.5)2
A = 30.25π
A = 30.25(3.14)
A = 94.99 sq in
The area of the original square can be found using the formula
A = s2
, where s is the length of one side of the square; 112
= 121,
so the area is 121 sq in. The value in column B, however, is ᎏ
3
4
ᎏ of
the area of the square, and 121 × ᎏ
3
4
ᎏ = 90.75 sq in.
94.99 sq in > 90.75 sq in, so the area of the circle is greater than
ᎏ
3
4
ᎏ the area of the square, and the answer is a.

318. a. The surface area of a rectangular prism can be calculated using
the formula SA = 2(lw + wh + lh), where l, w, and h are the
length, width and height of the prism, respectively. In this case
the length is 3 m, the width is 5 m and the height is 8 m, so:
SA = 2(lw + wh + lh)
SA = 2(3(5) + 5(8) + 3(8))
SA = 2(15 + 40 + 24)
SA = 2(79)
SA = 158
The surface area of this prism is 158 sq m, which is larger than
the value in column B, so the answer is a.
319. c. The surface area of a cube can be calculated using the formula
SA = 6e2
, where e is the length of one edge of the cube. In this
case each edge measures 2 in long, so the formula simpliﬁes as
follows:
SA = 6e2
SA = 6(2)2
SA = 6(4)
SA = 24 sq in
The values in the two choices are equal, so the answer is c.
320. d. To calculate the surface area of a rectangular prism, the length,
width, and height of the prism must be known. There is not
enough information given here to solve the problem, and so the
answer is d.
321. b. The volume of a rectangular prism can be found using the
formula V = lwh, where l is the length, w is the width, and h is
the height. The volume of this prism, then, is calculated as
follows:
V = lwh
V = 6(6)(3)
V = 36(3)
V = 108 cubic cm
Don’t forget that the volume of any container is always
expressed in cubic units. The volume of this prism is 108 cm3
and so the answer is b.
102

103
322. b. The volume of any prism is calculated using the formula
V = Bh, where B is the area of the base and h is the height. The
volume of the prism in the diagram, then, is V = Bh = 30(5) =
150 m3
. The volume of the prism in column B is 40(4) = 160 m3
.
Column B therefore yields a greater value and the answer is b.
323. a. The formula used to calculate the volume of a cube is V = e3
,
where e is the length of one edge; 43
= 64, whereas 2(23
) = 2(8)
= 16, so the volume of the cube in column A is greater and the
answer is a.
324. b. The volume of a pyramid is calculated using the formula
V = ᎏ
1
3
ᎏlwh. The length and width of this pyramid are both 10 in
and the height is 12 in, so the formula simpliﬁes to ᎏ
1
3
ᎏ(10 × 10 ×
12) = ᎏ
1
3
ᎏ(1200) = 400. The volume of the pyramid is 400 cubic in,
so the answer is b.
325. b. The value of column A is V = ᎏ
1
3
ᎏlwh = ᎏ
1
3
ᎏ(10 × 10 × 9) = ᎏ
1
3
ᎏ (900) =
300 cubic in. The value of column B is V = ᎏ
1
3
ᎏlwh = ᎏ
1
3
ᎏ(9 × 9 × 12)
= ᎏ
1
3
ᎏ(972) = 324 cubic in. 324 > 200, so the answer is b.
326. a. The area of the base of the pyramid is given in the diagram, so
calculate the volume with the formula V = ᎏ
1
3
ᎏBh:
V = ᎏ
1
3
ᎏBh
V = ᎏ
1
3
ᎏ(51 × 5)
V = ᎏ
1
3
ᎏ(255)
V = 85 cm3
The value of column A is greater, so the answer is a.
327. a. The value of column A is calculated as V = ᎏ
1
3
ᎏBh = ᎏ
1
3
ᎏ(51 × 6) =
ᎏ
1
3
ᎏ(306) = 102 cm3
. The value of column B is calculated as
V = ᎏ
1
3
ᎏBh = ᎏ
1
3
ᎏ(60 × 5) = ᎏ
1
3
ᎏ(300) = 100 cm3
. The answer is a.
328. b. The circumference of a circle is calculated using the formula
C = 2πr or C = πd. Thus, the greater the diameter of a circle, the
greater its circumference. So the answer is b.

329. a. Use the formula C = πd to calculate the circumference of this
circle, where π = 3.14:
C = πd
C = (3.14)7
C = 21.98 cm
21.98 cm > 21 cm, so the answer is a.
330. a. Use the formula C = πd to find the diameter in column B:
C = πd
25.12 m = πd
25.12 m = 3.14d
8 m = d
Now use the formula C = 2πr to find the radius in column B:
C = 2πr
31.4 m = 2(3.14)r
5 m = r
8 m > 5 m, so the answer is a.
331. a. The diagram indicates that the circle has a diameter of 10 ft.
The radius is ᎏ
1
2
ᎏ of the length of the diameter, which in this case
equals 5 ft; 5 ft > 4 ft, so the answer is a.
332. a. The area of the circle is calculated using the formula A = πr2
,
where r is the radius of the circle. The radius is not given in the
diagram, but can be calculated as 5 ft by dividing the diameter
in half. Using the formula yields A = πr2
= 3.14(52
) = 3.14(25) =
78.5. The area of the circle is 78.5 ft2
, which is greater than the
value of column B, so the answer is a.
333. b. Any figure must be smaller than any other figure it can fit inside
of. Therefore a circle is larger than a square that would fit
inside of it, and the answer is b.
334. a. Before doing any calculations, take a good look at both choices
to note all of the important information provided. The largest
circle that can be cut from a square of fabric has a diameter
equal to the length of a side of that square. Column B refers to
a circle cut out of a square of fabric with a total area of 16 cm2
.
This can be used to infer that the edges of the square are 4 cm
104

105
each, because the area of a square = s2
, where s is the length of
one side of the square. The largest circle that could be cut out
of that square, then, would have a diameter of 4 cm. The circle
mentioned in column A has a diameter of 8 cm. Since the area
of a circle increases as its diameter increases (remember, d = 2r),
the correct choice here is the circle with the largest diameter.
So the answer is a.
335. b. The surface area of a cylinder is calculated using the formula
SA = 2πr2
+ 2πrh, where r is the radius of the base of the
cylinder and h is the height. In this case, the radius is indicated
as 1 m and the height is 12 m, so the formula simpliﬁes as
follows:
SA = 2πr2
+ 2πrh
SA = 2π12
+ 2π1(12)
SA = 2π + 24π
SA = 2(3.14) + 24(3.14)
SA = 6.28 + 75.36
SA = 81.64 cm2
The total surface area of the cylinder is 81.64 cm2
, which is less
than the value of column B, so the answer is b.
336. a. The surface area of this cylinder is calculated as:
SA = 2πr2
+ 2πrh
SA = 2π72
+ 2π7(8)
SA = 98π + 112π
SA = 98(3.14) + 112(3.14)
SA = 210(3.14)
SA = 659.4 cm2
The surface area of the cube in column B is calculated as
SA = 6e2
= 6(92
) = 6(81) = 486 cm2
; 659.4 cm2
> 486 cm2
, so the
answer is a.

337. b. The volume of a cylinder is calculated using the formula
V = πr2
h, where h is the height of the cylinder. In this case,
the radius is 9cm and the height is 4 cm, so the formula
simplifies as:
V = πr 2
h
V = π92
(4)
V = π81(4)
V = 1017.36 cm3
This is less than the value of column B, so the answer is b.
338. a. The formula for calculating the volume of a cylinder multiplies
the square of the radius by the height. Doubling the radius of a
cylinder will change the volume more significantly because of
the squaring involved, so the answer is a.
339. b. The volume of this cylinder is V = πr2
h = π(25)(10) = 250π
= 785 cm3
; 785 cm3
> 750 cm3
, so the answer is b.
340. b. The volume of a cone is calculated using the formula
V = ᎏ
1
3
ᎏ(πr2
h), where h is the height of the cone. In this case,
the radius is 5 cm and the height is 10 cm, so the formula
simplifies as:
V = ᎏ
1
3
ᎏπr2
h
V = ᎏ
1
3
ᎏπ52
(10)
V = ᎏ
1
3
ᎏ(250π)
V = 261.67 cm3
300 cm3
> 261.67 cm3
341. c. The surface area of a sphere is calculated using the formula
SA = 4πr2
. Substituting yields:
SA = 4πr2
SA = 4(3.14)(62
)
SA = 12.56(36)
SA = 452.16 sq cm
The two values are equal, so the answer is c.
106

107
342. a. The surface area of a sphere is calculated using the formula
SA = 4πr2
. In this case, the formula simplifies to SA = 4πr2
= 12.56(49) = 615.44 sq m. This is greater than the value in
column B, and so the answer is a.
343. b. The volume of a sphere is calculated using the formula
V = ᎏ
4
3
ᎏπr3
. The formula simplifies as follows:
V = ᎏ
4
3
ᎏπr3
V = ᎏ
4
3
ᎏ(3.14)(73
)
V = ᎏ
4
3
ᎏ(3.14)(343)
V = ᎏ
4
3
ᎏ(1077.02)
V = 1,436.03 cm3
The volume of this sphere is 1,436.03 cm3
, which is less than
2,000 cm3
344. b. As the radius of a sphere increases, so will its volume (and
surface area, for that matter). The sphere in column B has the
greater radius, so it will also have the greater volume. The
answer is b.
345. a. The slope of a line is defined as .
The line in the diagram intersects the points (0,1) and (2,2), so
its slope is ᎏ
1
2
ᎏ; ᎏ
1
2
ᎏ > ᎏ
1
3
ᎏ, so the answer is a.
346. b. The slope of a line is defined as .
The line in the diagram intersects the points (0,1) and (−1,3), so
its slope is −ᎏ
2
1
ᎏ = −2; −ᎏ
1
3
ᎏ > −2, so the answer is b.
347. b. The slope of a line is defined as .
The line in the diagram intersects the points (1,−1) and (0,3), so
its slope is −ᎏ
4
1
ᎏ = −4; −4 < ᎏ
1
2
ᎏ, so the answer is b.
348. a. The slope of a line is defined as .
The line in the diagram intersects the points (−1,−1) and (1,2),
so its slope is ᎏ
3
2
ᎏ; ᎏ
3
2
ᎏ > ᎏ
2
3
ᎏ, so the answer is a.
349. a. The line’s slope is ᎏ
0
4
−
−
2
2
ᎏ = −1; 0 > −1, so the answer is a.
350. b. The line’s slope is ᎏ
−
−2
3
−
−
0
0
ᎏ = ᎏ
3
2
ᎏ; ᎏ
3
2
ᎏ > 0, so the answer is b.
the change in the y-value of the line
ᎏᎏᎏᎏthe change in the x-value of the line

351. c. The line y = −1 is a horizontal line. Thus, its slope is zero and
the answer is c.
352. b. The problem gives enough information to calculate the
elevation of the peak of the hill. Slope is deﬁned as
. In this case, the y-value is the
change in vertical distance (elevation) of the hill, while the
x-value is the horizontal distance Tommy has to walk. While a
diagram may be helpful to you in solving the problem, the
simplest method is to set up a ratio comparing the slope and the
distances Tommy has to travel, where y represents the vertical
distance:
ᎏ
2
5
ᎏ = ᎏ10
y
0ᎏ
5y = 200
y = 40 ft
The vertical elevation of the hill is 40 feet above sea level.
40 ft > 35 ft, so the answer is b.
353. a. This line intersects the y-axis at the point (0,1), so its
y-intercept is 1 and the answer is a.
354. b. As determined in the previous problem, the y-intercept of the
line in the diagram is 1. The equation in column B is in slope-
intercept form, so the y-intercept is represented by the term
without a variable. Thus, the y-intercept of this line is 4 and the
answer is b.
355. a. This line intersects the y-axis somewhere between the points
(0,3) and (0,4). Even though the exact value cannot be
determined, it is somewhere between 3 and 4 which is clearly
greater than zero. Therefore, the answer is a.
356. a. Even though the exact y-intercept of the line in the diagram
cannot be determined, it is clearly greater than zero and
therefore positive. The equation in column B. is in slope-
intercept form, so the y-intercept is represented by the term
without a variable which in this case is −1; −1 is a negative
value, so the value of column A must be greater and the answer
is a.
108

109
357. b. Both equations are in slope-intercept form, so the y-intercepts
can be determined from the terms without variables. The
y-intercept of the line in column A is −6, and in column B it’s 4,
so the answer is b.
358. a. Both equations are in slope-intercept form, so the y-intercepts
can be determined from the terms without variables. The
y-intercept of the line in column A is ᎏ
2
5
ᎏ, and in column B it’s −4,
so the answer is a.
359. a. The equation in column A must be put into slope-intercept
form in order to determine its y-intercept. Dividing both sides
of the equation by 2 yields the proper y = mx + b form as y = ᎏ
1
2
ᎏx
+ 4, so the y-intercept is 4. The y-intercept of the line in
column B is 3, so the answer is a.
360. c. The equation in column B must be put into slope-intercept
form in order to determine its y-intercept. Subtracting 6 from
both sides and adding y to both sides yields the proper equation
y = x − 6, with a y-intercept of −6. The y-intercept of the line in
column A is also −6 so the answer is c.
361. d. The equation in column A must be put into slope-intercept
form in order to determine its y-intercept. Simplifying yields
y = −ᎏ
1
4
ᎏx, so the y-intercept is 0. The line in column B is a
vertical line running through the point (9,0), and so has no
y-intercept. The problem has no solution and so the answer
is d.
362. a. Both equations must be put into slope-intercept form. The first
equation simplifies to y = 5, which represents a horizontal line
that has a y-intercept of 5. The second equation simplifies by
dividing both sides by x, yielding y = x + 4. The y-intercept of
this line is 4, so the answer is a.

363. a. The length of arc ABC can be determined using the formula
length = ᎏ
36
x
0
ᎏ(C), where x is the measure of the angle whose rays
intersect the arc and C is the circumference of the circle. In this
case, the formula is set up as follows:
length = ᎏ
36
x
0
ᎏ(C)
length = ᎏ
3
4
6
5
0
ᎏ(20)
length = ᎏ
1
8
ᎏ(20)
length = 2.5 cm
2.5 cm > 2 cm, so the answer is a.
364. c. The length of arc ABC can be determined using the formula
length = ᎏ36
x
0ᎏ(C), where x is the measure of the angle whose rays
intersect the arc and C is the circumference of the circle. In this
case, the formula is set up as follows:
length = ᎏ
36
x
0
ᎏ(C)
length = ᎏ
3
3
6
6
0
ᎏ(10)
length = ᎏ
1
1
0
ᎏ(10)
length = 1 cm
The values in both choices are equal, so the answer is c.
365. a. The circumference of the circle can be determined using the
same formula as above but solving for C. In this case, the
formula is set up as follows:
length = ᎏ36
x
0ᎏ(C)
2 = ᎏ
3
4
6
5
0
ᎏ(C)
2 = ᎏ
1
8
ᎏ(C)
C = 16 cm
20 cm > 16 cm, so the answer is a.
110

111
366. b. The measure of ∠x can be determined using the same formula
as above but solving for x. In this case, the formula is set up as
follows:
length = ᎏ
36
x
0
ᎏ(C)
3 = ᎏ
36
x
0
ᎏ(30)
3 = ᎏ
3
3
6
0
0
x
ᎏ
1,080 = 30x
36 = x
36° > 35°, so the answer is b.
367. a. The area of the rectangle in column A is 2(3) = 6 cm2
. The area
of the square in column B is 22
= 4 cm2
.
368. c. The sum of the interior angles of any triangle is 180°. The
values in both choices are equal so the answer is c.
369. b. The volume of the cube in column A is s3
= 33
= 27 cm3
. The
volume of the cylinder in column B is πr2
(h) = 3.14(22
)(3) =
3.14(4)(3) = 3.14(12) = 37.68 cm3
; 37.68 > 27, so the answer is b.
370. b. These triangles are 3, 4, 5 and 6, 8, 10 Pythagorean triples,
which means that the hypotenuses are 5 and 10, respectively.
The Pythagorean theorem (a2
+ b2
= c2
) can also be used to solve
for each hypotenuse individually. Either method will ﬁnd that
the hypotenuse of the triangle in column B is longer. The
answer is b.
371. c. Angles 1 and 3 are vertical angles created by the intersection of
two lines, and so are equal. The answer is c.
372. d. Angles 1 and 2 are exterior angles on the outside of two parallel
lines cut by a transversal, and so are supplementary. Even
though ∠1 appears to be slightly larger in the diagram, there is
no indication as to either angle’s actual measure. All that can be
determined is that the two angles add up to 180°, which is not
enough information to solve the problem. The answer is d.
373. a. The equation is in slope-intercept form, so the slope is the
coefﬁcient of the x term and the y-intercept is the term without
a variable. In this case, the slope is 3 and the y-intercept is 2.
The slope is the greater quantity, so the answer is a.

374. a. It may be helpful to draw a diagram of the two triangles. The
triangles are similar, as the lengths of shorter sides of triangles
are proportional. The shorter sides are in a ratio of 3:4 so the
hypotenuses must also be in a ratio of 3:4; ᎏ
3
4
ᎏ > ᎏ
1
2
ᎏ, so the answer
is a.
375. a. The pie is a circle, so it measures 360° around. Since it is cut
into eight equal slices, each slice forms an angle of 40° at the
center of the pie. The formula length = ᎏ
36
x
0
ᎏ(C) can be used to
determine the length of the arc of one of the slices by
substituting the information given in the problem as follows:
length = ᎏ
36
x
0
ᎏ(C)
length = ᎏ3
4
6
0
0ᎏ(18)
length = ᎏ
1
9
ᎏ(18)
length = 2″
The length of the arc formed by one slice of pie is 2″; 2″ > 1.5″,
so the answer is a.
112

113
data analysis-based quantitative comparison questions:
■ Counting
■ Sequences
■ Data Representation and Interpretation
■ Frequency Distributions
■ Measures of Central Tendency
■ Measures of Dispersion
■ Probability
Some important information:
4Data Analysis

choose:
Examples:
Column A Column B
1. ͙32 + 42ෆ ͙(3 + 4)ෆ2ෆ
The correct answer is b. Remember to look carefully at the two
columns, even if they initially appear to be the same. In this case,
while the two quantities look similar, they are not equivalent. In
column A, ͙32 + 42ෆ is equal to ͙9 + 16ෆ, or ͙25ෆ, which is equal to
5. In column B, be sure the work out the calculations inside the
parentheses ﬁrst: ͙(3 + 4)ෆ2ෆ = ͙72ෆ = 7. 7 is greater than 5, so the
correct answer is b.
2. m and n are integers.
m3
n2
The correct answer is d. The only information you are given about
the two quantities is that they are integers, which tells you nothing
about their respective values. Depending on what values you
assign, column A could be larger than column B or vice versa, or
both values could be the same. Since you cannot determine which
value is greater, the answer is d.
114

115
Questions
Column A Column B
376. the average (arithmetic the average (arithmetic
mean) of 6, 5, 8, 7, and 9 mean) of 11, 2, and 8
377. the average (arithmetic mean) the average (arithmetic
of 16, 23, 30, 45, and 17 mean) of 23, 18, 17, 35, and 45
378. A = {15, 20, 20, 13}
the median of set A the mode of set A
379. The mean of set B is 17.
B = −5, −1, 12, 29, x, y
x y
380. C = 3, 6, 11, 12, 10, 18, x
The mean of set C is 9.
8 x
Brown High School Student Distribution
381. number of freshmen 22
382. There are 400 students enrolled at Brown High School.
number of seniors 120
29% 22%
24%
25%
freshmen
sophomores
juniors
seniors

Column A Column B
383. difference between the difference between the
quantity of freshmen quantity of juniors
and sophomores and seniors
Use the following sequence to answer questions 384 and 385.
.25, .5, .75, 1, 1.25, 1.5, 1.75, 2, . . .
384. the 53rd term of the sequence 13
385. the 78th term of the sequence 19.5
Use the following series to answer questions 386 and 387.
2 + 4 + 6 + 8 + . . . + 98 + 100
386. the sum of the ﬁrst 23 550
terms of the series
387. the sum of all the terms 2,550
in the series
Use the information below to answer questions 388–400.
D = {13, 22, 17, 24, x}
E = {13, 22, 17, 24, y}
x > 0, y > 0
388. the mean of set D the mean of set E
389. the median of set D the median of set E
390. x > y
the mean of set D the mean of set E
391. x > y > 24
the median of set D the median of set E
392. x > 17 > y
393. The ranges of the sets are equal.
116

117
Column A Column B
394. x = y
395. x = y
396. x < y
397. x < y
the range of set D the range of set E
398. The means of the two sets are equal.
x y
399. The mean of set D is greater than the mean of set E.
x y
400. The modes of the two sets are equal.
12 x
F = {28, 29, 30, 31, 32}
G = {10, 20, 30, 40, 50}
401. the mean of set F the median of set G
402. the standard deviation the standard deviation
of set F of set G
Use the following experiment to answer questions 402–405.
A coin is tossed 3 times.
403. the number of possible the number of possible
outcomes containing outcomes containing
exactly 2 heads exactly 1 tail
404. the total number of possible 5
outcomes

Column A Column B
405. the probability of tossing ᎏ
1
2
ᎏ
3 tails
Use the following bar graph to answer questions 406–408.
406. 12 difference between the number
of students with brown hair and
those with black hair
407. percent of students with percent of students with
red hair gray hair
408. total number of students 90
surveyed
A number cube (die) is rolled and a coin is tossed.
409. total number of possible 8
outcomes
410. number of outcomes in 3
which an even number is
rolled on the cube, and a
head is tossed on the coin
0
5
10
15
20
25
30
35
NumberofStudents
Hair Color of Students
Hair Color
blond brown black red gray
118

119
Column A Column B
411. number of outcomes in 5
which a factor of 2 is rolled
on the cube, and a head or
a tail is tossed on the coin
A coin is tossed 14 times.
Heads occurs 8 times and tails 6 times.
The second and thirteenth tosses are heads.
412. maximum number of 8
heads that can occur in a row
413. minimum number of heads 4
in the first 10 tosses
Use the following series to answer questions 414–416.
−20 + −18 + −16 + . . . + 18 + 20 + 22 + 24
414. the sum of all terms in 50
the series
415. the sum of the first the sum of the fourth, fifth,
three terms and sixth terms.
416. the sum of the 18th, 19th, the sum of the last two terms
and 20th terms
Use the following sequence to answer questions 417–418.
1, 1, 2, 3, 5, 8, 13, 21, 34, . . .
417. the 11th term in the sequence 90
418. the sum of the 41st and 42nd the 43rd term of the sequence
terms of the sequence
Use the following facts to answer questions 419 and 420.
William bought 4 pairs of pants for $80. The next day, he purchased
another pair of pants. He spent an average of $22.50 for the five
pairs of pants.

Column A Column B
419. amount of money William average cost of each of the first
spent on the fifth pair of pants four pairs of pants
420. $30 the cost of the fifth pair of pants
Use the graph below to answer question 421–423.
Monthly Budget
421. percent of budget spent percent of budget spent
on Housing on Clothing and Food
422. percent of budget not spent 86%
on Savings
423. percent of budget spent on percent of budget spent on
Food, Auto, and Other Housing, Clothing and Savings
Use the following situation to answer questions 424–426.
Papa’s Pizza offers any of up to 2 different toppings on their pies.
Papa’s has 8 total toppings from which to choose.
424. total number of possible 8
1 topping pizzas Papa’s makes
425. Pepperoni is an available topping.
number of possible pizzas 8
containing pepperoni
$695
$325
$967
$439
$261
$775
Food
Clothing
Housing
Savings
Auto expenses
Other costs
120

121
Column A Column B
426. total number of possible pizzas 12
Papa’s makes
427. The mean of ﬁve distinct positive integers is 10.
the largest possible value 46
of one of the integers
Use the following situation to answer questions 428 and 429.
Set H contains ﬁve positive integers such that the mean, median,
mode, and range are all equal. The sum of the data is 25.
428. the smallest possible number 6
in set H
429. 10 the largest possible number in
set H
Set J consists of 5 elements. The range of the numbers in set J is 0,
and the sum of the numbers in set J is 40.
430. highest element in set J lowest element in set J
431. mode of set J 16
K = {8, x, y, 10}
The mean of set K is 12, there is no mode, and x > y.
432. x + y 24
433. 2x 30
Use the following box plot to answer questions 434 and 435.
Age of College Freshmen
13 17 18 53

Column A Column B
434. quartile 1 quartile 3
435. quartile 3 median
436. Susan has 3 pairs of pants and 4 shirts.
Assume each pair of pants matches each shirt.
7 number of outfits Susan has
Use the following experiment description to answer questions 437–440.
A card is drawn from a standard deck of 52 cards.
437. probability of drawing a queen probability of drawing a club
438. probability of drawing a jack ᎏ
1
5
7
2
ᎏ
or a spade
439. probability of drawing a probability of drawing a heart
black card or a face card
440. probability of drawing a probability of drawing an eight
red card with an even or a nine
number on it
Use the following experiment description to answer questions
441 and 442.
A bag contains 6 blue marbles and 4 red marbles. Two marbles
are selected at random, one after the other (the first marble
is not replaced in the bag after it is drawn).
441. probability of drawing probability of drawing
2 blue marbles 2 red marbles
442. probability of drawing a probability of drawing a
red followed by a blue marble blue followed by a red marble
Use the following experiment description to answer questions
443 and 444.
are selected at random, with the first selected marble being
replaced in the bag before the second marble is drawn.
122

123
Column A Column B
443. probability of drawing probability of drawing
2 blue marbles 2 red marbles
444. probability of drawing a probability of drawing a
red followed by a blue marble blue followed by a red marble
Use the following experiment description to answer questions 445–447.
are selected at random.
445. probability of selecting probability of selecting
2 blue marbles if the first 2 blue marbles if the first
selected marble is replaced selected marble is not replaced
446. probability of selecting a probability of selecting a blue
blue marble, then a red marble, then a red marble
marble if the first selected if the first selected marble is
marble is replaced not replaced
447. probability of selecting 2 probability of selecting 2
red marbles if the first blue marbles if the first
selected marble is replaced selected marble is not replaced
Use the spinner below and the experiment description to answer
questions 448–451.
The spinner above is spun twice, then a 6-sided number cube (die)
is rolled.
448. number of outcomes 14
449. probability of obtaining .1
the outcome: blue, red, 4
green
blue
yellow
red

Column A Column B
450. probability of obtaining probability of obtaining
the outcome: blue or red, the outcome: green, green
red, even number or yellow, odd
451. probability of obtaining probability of obtaining
the outcome: not green, red, the outcome: not blue, not
factor of 2 blue, multiple of 3
Use the frequency distribution table below to answer questions 452–455.
Data Value Frequency
0 3
1 6
2 2
3 1
4 2
452. mean of distribution 2.5
453. median of distribution 2
454. mode of distribution 1
455. range of distribution 5
Use the following set to answer questions 456–458.
For x = 0, x = 1, x = 2, and x = 3,
let set K = {2x, x2
, x + 2}.
456. 4 mode of set K
457. range of set K 10
458. median of set K mean of set K
Use the sequence below to answer questions 459–460.
2, 4, 8, 16, 32, . . .
459. the tenth term of the sequence 210
460. the twentieth term of the twice the tenth term
sequence
124

125
Column A Column B
64, 32, 16, 8, 4, . . .
461. the seventh term of the 0
sequence
462. the thirtieth term of the 2−20
sequence
Use the situation below to answer questions 463–464.
A survey of high school seniors at Blake High School revealed that 35%
play an instrument, 43% participate in sports, and 29% do neither.
463. percent of students who 78%
play an instrument and/or
participate in sports
464. percent of students who 10%
play an instrument and
participate in sports
M = {4, 6, 3, 7, x, x}
N = {1, 5, 8, y}
x > 0, y > 0
465. x > y
the mean of set M the mean of set N
466. x > y > 10
the median of set M the median of set N
Use the following box plot to answer questions 467–469.
Birthweight of Babies at Center Hospital
4 6 7 10 18

Column A Column B
467. number of data elements number of data elements
between ﬁrst quartile between third quartile
and median and max
468. interquartile range 3
469. spread of data in third quarter spread of data in second quarter
Use the menu for Kelly’s Deli below to answer questions 470–472.
Bread Meat Cheese
Rye Ham American
Wheat Salami Swiss
White Turkey Provolone
Roast Beef Cheddar
Tuna
A sandwich from Kelly’s consists of one type of bread, one meat,
and one cheese.
470. number of possible 60
sandwiches Kelly’s makes
471. number of sandwiches that number of sandwiches that
can be made on rye bread can be made on wheat bread
472. number of sandwiches that number of sandwiches that
can be made with roast beef can be made with provolone
cheese
1, 4, 9, 16, 25, . . .
473. the seventh term in 36
the sequence
474. the twentieth term in 400
the sequence
475. the difference between the the difference between the
34th and 35th terms 10th and 13th terms
476. the 1000th term in the 1 × 109
sequence
126

127
Column A Column B
477. 50th percentile of data set W average (arithmetic mean) of data
set W
478. 50th percentile of data set W median of data set W
479. 25th percentile of data set W ﬁrst quartile of data set W
Use the sets below to answer questions 480 and 481.
P = {1, 2, 3, 4, 5}
Q = {2, 3, 6, 5}
480. standard deviation of set P standard deviation of set Q
481. mean of set P median of set Q
482. 100 meters 1 kilometer
483. 3.46 × 109
miles 34.6 × 108
miles
484. 1.5 × 10−6
millimeters .00000015 millimeters
0
1,000
2,000
3,000
4,000
5,000
1986 1988 1990 1992 1994
500
1,500
2,500
3,500
4,500
5,500
Profit
(indollars)
Profits of Company B
Year

Column A Column B
485. profit increase from profit increase from
1988 to 1990 1987 to 1989
486. profit decrease from profit decrease from
1990 to 1991 1988 to 1989
487. overall profit from $1,500
1987 to 1993
Use the following sequence to answer questions 488–489.
1, 2, 6, 24, . . .
488. the fifth term of the sequence 150
489. the sixth term of the sequence 720
Use the following series to answer questions 490–491.
2 + 12 + 22 + 32 + 42 + . . . + 102 + 112 + 122 + 132
490. the number of terms 13
in the sequence
491. the sum of the series 1,000
492. 4.23 × 10−5
.423 × 10−4
493. 234 milliliters 2.34 liters
494. 45 centigrams .45 grams
495. 12 square yards 36 square feet
496. 24 square inches 2 square feet
497. 1 cubic foot 123
cubic inches
128

129
Column A Column B
Use the frequency distribution below to answer questions 498–501.
Ages of Women Entered in Road Race
Classes Frequency
15–19 3
20–24 14
25–29 17
30–34 13
35–39 10
498. number of 17-year-olds number of 27-year-olds
entered in race entered in race
499. total number of entrants 55
500. median age of entrant 30
501. number of entrants under 30 28

Answer Explanations
376. c. The mean of the numbers in column A is 7: 6 + 5 + 8 + 7 + 9
= 35. 35 divided by 5 = 7. The mean of the numbers in
column B is also 7: 11 + 2 + 8 = 21. 21 divided by 3 = 7.
377. b. Of the 5 numbers in each set, 3 are the same (23, 17, and 45).
The remaining 2 numbers are greater in the set in column B
than in the set in column A, therefore, the mean of the set in
column B is greater than the mean of the set in column A.
378. b. The median of set A (the middle of the set when the numbers
are put in numerical order) is 17.5: 13, 15, 20, 20. Since there is
an even number of elements in the set, the average of the two
middle numbers, 15 and 20 is found; 15 + 20 = 35; ᎏ
3
2
5
ᎏ = 17.5.
The mode of set A (the most frequently occurring element in
the set) is 20 because it occurs twice. 20 is greater than 17.5.
379. d. The relationship cannot be determined. Since the mean of set B
is 17, and the set has 6 elements, the sum of the elements in the
set must be (17)(6) = 102. The total of the given elements is
−5 + −1 + 12 + 29 = 35. Therefore, x and y must total what’s left,
namely, 102 − 35 = 67. There is no way to know which of the
two might be greater, or if the two are equal.
380. a. Since the mean of set C is 9 and the set has 7 elements, the sum
of the elements in the set must be (9)(7) = 63. The total of the
given elements is 3 + 6 + 11 + 12 + 10 + 18 = 60. Therefore,
x must equal the remaining value, namely, 63 − 60 = 3. 8 is
greater than 3.
381. d. The relationship cannot be determined. Since the number of
students enrolled at Brown High School is not given for this
question, it is not known whether the number of freshmen
equals 22 (if there are 100 students, because 22% of 100 = 22),
is more than 22 (if there are > 100 students, because 22% of >
100 is > 22) or is less than 22 (if there are < 100 students,
because 22% of < 100 is < 22)
130

131
382. b. Since there are 400 students at Brown High School and 29%
are seniors, 29% of 400, or .29 × 400 = 116; 120 > 116, so the
answer is b.
383. b. The difference between the number of freshmen and
sophomores is 24% − 22% = 2% of 400; .02 × 400 = 8 students.
The difference between the number of juniors and seniors is
29% − 25% = 4%. 4% of 400 is 16. 16 > 8.
384. a. The pattern in this series (.25, .5, .75, 1, 1.25, 1.5 . . .) can also
be written as ᎏ
1
4
ᎏ, ᎏ
2
4
ᎏ, ᎏ
3
4
ᎏ, ᎏ
4
4
ᎏ, ᎏ
5
4
ᎏ, ᎏ
6
4
ᎏ . . . . Since each term given is
equivalent to the term number (its number in the sequence)
divided by 4, an nth term would be equal to ᎏ
n
4
ᎏ. The 53rd term is
equal to ᎏ
5
4
3
ᎏ; ᎏ
5
4
3
ᎏ = 13.25; 13.25 > 13, so the answer is a.
385. c. Use the same equation you established in the previous problem.
The 82nd term = ᎏ
8
4
2
ᎏ = 19.5. Column B is also 19.5, so the
correct answer is c.
386. a. Note that the ﬁrst term is 2, the second term is 4, the third
term is 6, etc. The term is equal to two times the term number.
Therefore, the 23rd term is 23 × 2 = 46 and the sum of the even
numbers from 2 to 46 is needed to answer the question. In
series questions, an easy shortcut is to rewrite the series below
the original series and add vertically to get consistent sums. In
this case,
2 + 4 + 6 + . . . + 44 + 46
46 + 44 + 42 + . . . + 4 + 2
48 + 48 + 48 + . . . + 48 + 48
Since 23 terms were added, there are 23 totals of 48. Each term
has been written twice, so 23 × 48 is double the total needed
and must be divided by two; 23 × (ᎏ
4
2
8
ᎏ) = 23 × 24 = 552;
552 > 550, so column A is greater.
387. c. Following the same procedure as in the previous question, the
last term is 100, so there are ᎏ
10
2
0
ᎏ, or 50 terms in the series. The
consistent sum would be 102 because the ﬁrst term plus the last
term = 2 + 100 = 102. There would be 50 sums of 102 which
would again be double the total needed; 50 × (ᎏ
10
2
2
ᎏ) = 50 × 51
= 2,550.

388. d. The answer cannot be determined. Since the value of neither x
nor y is known, the means of the sets cannot be found and
cannot, therefore, be compared.
389. d. The answer cannot be determined. Although the question
indicates that both x and y are greater than zero, and both sets
D and E have 5 elements, the elements are not ordered, so the
third element is not necessarily the median. For example, if
x < 13, the order of the set would be x, 13, 17, 22, 24 and the
median would be 17. If 17 < x < 22, the order of the set would
be 13, 17, x, 22, 24 and the median would be x.
390. a. Since x > y, the sum of the elements in set D > the sum of the
elements in set E. When the sums are divided by 5 (the number
of elements in each set), the mean of set D will be a larger
number.
391. c. Since x > y > 24, we know that in each of sets D and E, x and y
respectively are the largest elements. Therefore, set D in
numerical order is {13, 17, 22, 24, x} and set E in numerical
order is {13, 17, 22, 24, y}, so in each case, the median is the
middle number, 22.
392. a. Since x > 17, the median of set D will be x, 22, or 24, depending
on how large x is. The median will therefore be one of these
elements: x which is > 17, 22, or 24. Since y < 17, the median of
set E must be 17 because the third element will be 17; there are
2 elements less than 17, namely, y and 13.
393. d. The answer cannot be determined. Choose an arbitrary value
for the range, say 15. Therefore, x might be 9 since that would
enable the range to be 24 − 9 = 15. This would make the
median of set D be 17. y might have the same value, which
would make the medians equal, but y could also equal 28
because 28 − 13 = 15 as well. This would make the median
equal 22 as the third element of the set.
394. c. If x = y, then set D = set E, and it follows that the mean of set D
= the mean of set E.
395. c. If x = y, then set D = set E, and all statistics for the two sets
would be equal, including, but not limited to, mean, median,
mode, and range.
132

133
396. b. Since x < y, the sum of the elements in set D < the sum of the
elements in set E. When the sums are divided by 5 (the number
of elements in each set), the mean of set E will be a larger
number.
397. d. The answer cannot be determined. Let x = 5 and y = 32. Then
the range of set D = 24 − 5 = 19 and the range of set E = 32 − 13
= 19. The ranges are equal, which eliminates choices a and b.
Now let x = 12 and y = 15. Then the range of set D = 24 − 12
= 12 and the range of set E = 24 − 13 = 11. Since in both cases,
x < y, the answer cannot be determined.
398. c. The only way for the mean of the two sets to be equal is for the
sum of the two sets to also be equal. In order for this to happen,
x must equal y.
399. a. Since the mean of set D > the mean of set E, the sum of the
elements in set D must be > the sum of the elements in set E.
Therefore, since the remaining elements are all equal, the only
way to get a larger sum in set D is for x to be > y.
400. b. Because the mode of the two sets are equal, there must be a
mode in each set, meaning that one of the elements must
repeat. This implies that x and y are equal to one of the
elements already shown to be in sets D and E. Since all of these
elements are > 12, x must be > 12.
401. c. The mean of set F is 28 + 29 + 30 + 31 + 32 = 150; ᎏ
15
5
0
ᎏ = 30.
The median of set G, which is already in numerical order, is the
middle element, 30.
402. b. Standard deviation is a measure of the spread of the data from
the mean. Since the mean of each set is 30, the data in set F is
obviously clustered more closely to the mean than the data in
set G; therefore, the standard deviation of set G > the standard
deviation of set F.
403. c. The number of possible outcomes containing exactly 2 heads is
3: HHT, HTH, and THH. These are the same three outcomes
that contain exactly one tail.

404. a. Since there are two possible outcomes for each toss of the coin,
and the coin is being tossed three times, there are 23
= 8
possible outcomes. 8 > 5.
405. b. Only one of the 8 possible outcomes, TTT, contains three tails
so the probability of tossing three tails is ᎏ
1
8
ᎏ; ᎏ
1
8
ᎏ < ᎏ
1
2
ᎏ.
406. b. It appears that there are at most 17 students with black hair and
32 students with brown hair; 32 − 17 = 15. 15 > 12.
407. a. Since more students have red hair than have gray hair, the
percent of students with red hair must be higher than the
percent of students with gray hair. The actual percents do not
have to be determined.
408. b. Even with rough approximations done with the data, rounding
up, if there are about 30 blond students, 32 brown-haired
students, 18 black-haired students, 4 redheads and 3 with gray
hair, that total, 87, is < 90.
409. a. There are 6 possible outcomes on the number cube, {1, 2, 3, 4,
5, 6}, and 2 on the coin, {H, T}. According to the counting
principle for probability, there are therefore, 6 × 2 = 12 possible
outcomes. 12 > 8.
410. c. The outcomes in which an even number on the cube would be
followed by a head on the coin would be: (2, H), (4, H), (6, H).
There are three outcomes, so columns A and B are equivalent.
411. b. The factors of 2 are 1 and 2. The outcomes that meet the
condition that there is a factor of 2 on the number cube and a
head or a tail on the coin are (1, H), (1, T), (2, H), (2, T). There
are four outcomes, and 5 > 4, so the correct answer is b.
412. b. Since there are a total of 8 tosses that were heads, but the
second and thirteenth are known to be two of them, there can
only be a maximum of 7 heads in a row since there are more
than 8 tosses between the second and thirteenth tosses. For
example, the outcomes could be: H H H H H H H T T T T T
H T or T H T T T T T H H H H H H H.
134

135
413. c. Since there are 8 total heads and the objective is to have a
minimal quantity of them in the first ten tosses, the last four
tosses would have to all be heads. This would leave 8 − 4 = 4
tosses left for the first ten outcomes.
414. b. If the series is examined carefully, it can be noted that the first
41 terms will total zero since every nonzero number in the
series, up until 22, can be paired with its opposite. Therefore,
the sum of the series is simply the sum of the last two numbers,
22 and 24; 22 + 24 = 46. 50 > 46.
415. b. The sums do not need to be found. The first three terms are
smaller numbers than than the following three terms; therefore,
their sum will automatically be smaller than the sum of terms 4,
5, and 6.
416. a. The 18th, 19th, and 20th terms are 14, 16 and 18; 14 + 16 + 18
= 48. The last two terms are 22 and 24; 22 + 24 = 46. 48 > 46.
417. b. This is the Fibonacci sequence in which each term after the first
two is found by adding the previous two terms. There are nine
terms provided in the sequence. The tenth term is found be
adding the eighth and ninth terms: 21 + 34 = 55, so the tenth
term is 55. The eleventh term is found by adding the ninth and
tenth terms: 34 + 55 = 89. Since 90 > 89, column B is greater.
418. c. Since each term is found by adding the previous two terms, the
43rd term is the sum of the 41st and 42nd terms.
419. a. The answer can be found with minimal calculations: Since
William bought 4 pairs of pants for $80, each pair of pants cost
an average of $20. The fifth pair of pants managed to bring his
average cost per pair up to $22.50, so it must have cost more
than $20.
420. b. If the five pairs of pants averaged $22.50 per pair, William’s
total cost must be $22.50 × 5 = $112.50. This cost, minus
the $80 he paid for the first four pairs, leaves $112.50 − $80
= $32.50.
421. b. $967 was spent on Housing. $695 + $325 = $1,020 was spent on
Food and Clothing. Since more money was spent on Food and
Clothing, this accounts for a higher percent of the budget.

422. a. The total monthly budget is $775 + $695 + $325 + $967 + $439
+ $261 = $3462. The percent spent on Savings is ᎏ
$
$
3
4
,4
3
6
9
2
ᎏ = about
12%. Therefore, the amount not spent on Savings is 100% −
12% = 88%. 88% > 86%.
423. c. The amount of money spent on Food, Auto Expenses, and
Other Costs is $695 + $261 + $775 = $1,731, which is exactly
half, or 50% of the total amount of the budget, $3,462.
424. c. Since there are 8 toppings available, a one-topping pie would
consist of any one of these 8 toppings. There are 8 possibilities.
425. c. One 1-topping pizza would contain pepperoni. In addition, any
of the remaining seven toppings could be paired with pepperoni
to make seven more possible pies. Therefore, there are a total
of eight possible pies with pepperoni as a topping.
426. a. Papa’s offers one plain pizza, 8 one-topping pizzas, and 28 two-
topping pizzas. Call the available toppings A, B, C, D, E, F, G,
and H. Then A can be paired with any of the remaining 7
toppings. B can be paired with any of the remaining 6 toppings
starting with C because it’s already been paired with A. C can be
paired with D, E, F, G, or H since it’s already been paired with
A and B. And so on. In other words, there are 7 + 6 + 5 + 4 + 3 +
2 + 1 = 28 possible 2-topping pies. The total number of pies
Papa’s offers is 1 + 8 + 28 = 37; 37 > 12.
427. b. If the mean of ﬁve distinct (meaning different) positive integers
is 10, the sum of these integers must be 5 × 10 = 50. For one of
these integers to be as large as it can be, the remaining integers
must be as small as they can be, namely, 1, 2, 3, and 4. These
integers total 1 + 2 + 3 + 4 = 10. So, the largest one of the
integers can be is 40.
428. b. Since the sum of the ﬁve elements of set H is 25, the mean is
ᎏ
2
5
5
ᎏ = 5. Since the mean, median, mode, and range are all equal,
they are all equal to 5. Therefore, 6 must be larger than the
smallest element in set H.
429. a. If 10 is the largest possible number in set H, then, since the
range of the numbers in the set is also 5, the smallest number
would have to be 5. Even if the set consisted of four 5’s and just
the one 10, the sum of the numbers would still be too high. To
136

137
get a sum of 25 with a mode and a median of 5, at least 2 of the
numbers in the set must be 5 and the remainder must add up to
15 while not violating the range requirement, which is also 5.
One possibility is {2, 5, 5, 6, 7}. Another is {3, 4, 5, 5, 8}.
430. c. If the range of the elements in set J is 0, then all of the elements
must be the same since the range is derived by subtracting the
smallest element from the largest element. Therefore the
highest element in set J is equal to the lowest element in set J.
431. b. Since the sum of the elements in set J is 40, and there are 5
elements, the mean of the set is 8. Also, since the range of the
numbers in the set is zero, each of the numbers in the set must
be 8, making 8 the mode of the set; 16 > 8.
432. a. Since the mean of set K is 12, the sum of the 4 elements in the
set must be 4 × 12 = 48. The two given elements, 8 and 10 total
18. Therefore, the difference between the total and the given
elements is 48 − 18 = 30. The two remaining elements, x and y
must total 30, which is greater than 24.
433. a. The total of x and y must be 30, and x > y, therefore, x must
be > 15 and y must be < 15. Therefore, 2x > 30.
434. b. Quartile 1 is 17 and Quartile 3 is 18. Therefore, Quartile 1 <
Quartile 3.
435. d. The answer cannot be determined. The median may be 17 or
18, which would make it less than, or equal to, Quartile 3.
436. b. Since each pair of pants matches each shirt, Susan has 4 × 3 = 12
outﬁts; 12 > 7.
437. b. There are 4 queens in a deck, so the probability of drawing a
queen is ᎏ
5
4
2
ᎏ. There are 13 clubs in a deck, so the probability of
drawing a club is ᎏ
1
5
3
2
ᎏ; ᎏ
1
5
3
2
ᎏ > ᎏ
5
4
2
ᎏ.
438. b. There are 4 jacks and 13 spades, but one of these is the jack of
spades which cannot be counted twice. Therefore, there are
4 + 13 − 1 = 16 cards that are jacks or spades (or both). The
probability of drawing one of these cards is ᎏ
1
5
6
2
ᎏ; ᎏ
1
5
6
2
ᎏ < ᎏ
1
5
7
2
ᎏ.

439. a. There are 26 black cards (13 spades and 13 clubs) so the
probability of drawing one of these cards is ᎏ
2
5
6
2
ᎏ. There are 13
hearts and 12 face cards, but three of these face cards are hearts:
the jack, queen and king of hearts. These cards cannot be
counted twice, therefore, there are 13 + 12 − 3 = 22 cards that
are hearts or face cards (or both). The probability of drawing
one of these cards is ᎏ
2
5
2
2
ᎏ; ᎏ
2
5
6
2
ᎏ > ᎏ
2
5
2
2
ᎏ.
440. a. There are 10 red cards with even numbers on them: the 2, 4, 6,
8, and 10 of hearts and of diamonds. The probability of drawing
one of these cards is ᎏ
1
5
0
2
ᎏ. There are 4 eights and 4 nines, so a
total of 8 cards that are eights or nines, and the probability of
drawing one of these cards is ᎏ5
8
2ᎏ; ᎏ
1
5
0
2
ᎏ > ᎏ5
8
2ᎏ.
441. a. The probability of drawing 2 blue marbles comes from
multiplying the probability of drawing the first blue marble by
the probability of drawing the second. The probability of
drawing the first blue marble is ᎏ1
6
0ᎏ because there are 6 blue
marbles out of the 10 total marbles in the bag. The probability
of drawing the second blue marble is ᎏ
5
9
ᎏ because having drawn
the first blue marble, there are 5 blue marbles left out of a total
of 9 marbles remaining in the bag (since the experiment is
conducted without replacing the first selected marble in the
bag); ᎏ1
6
0ᎏ × ᎏ
5
9
ᎏ = ᎏ
3
9
0
0
ᎏ.
The probability of drawing 2 red marbles comes from
multiplying the probability of drawing the first red marble by
the probability of drawing the second. The probability of
drawing the first red marble is ᎏ1
4
0ᎏ because there are 4 red
marbles out of the 10 total marbles in the bag. The probability
of drawing the second red marble is ᎏ
3
9
ᎏ because having drawn the
first red marble, there are 3 red marbles left out of a total of 9
marbles remaining in the bag; ᎏ
1
4
0
ᎏ × ᎏ
3
9
ᎏ = ᎏ
1
9
2
0
ᎏ; ᎏ
3
9
0
0
ᎏ > ᎏ
1
9
2
0
ᎏ.
442. c. The probability of drawing a red followed by a blue marble is
ᎏ
1
4
0
ᎏ × ᎏ
6
9
ᎏ = ᎏ
2
9
4
0
ᎏ. The probability of drawing a blue followed by a red
marble is ᎏ
1
6
0
ᎏ × ᎏ
4
9
ᎏ = ᎏ
2
9
4
0
ᎏ.
138

139
443. b. In this case, the original selected marble is replaced in the bag
before the second marble is drawn, keeping the denominators
of the multiplied fractions the same. The probability of drawing
2 blue marbles is ᎏ
1
6
4
ᎏ × ᎏ
1
6
4
ᎏ = ᎏ
1
3
9
6
6
ᎏ. The probability of drawing 2
red marbles is ᎏ
1
8
4
ᎏ × ᎏ
1
8
4
ᎏ = ᎏ
1
6
9
4
6
ᎏ; ᎏ
1
6
9
4
6
ᎏ > ᎏ
1
3
9
6
6
ᎏ.
444. c. The probability of drawing a red, followed by a blue marble is
ᎏ
1
8
4
ᎏ × ᎏ
1
6
4
ᎏ = ᎏ
1
4
9
8
6
ᎏ. The probability of drawing a blue, followed by a
red marble is ᎏ
1
6
4
ᎏ × ᎏ
1
8
4
ᎏ = ᎏ
1
4
9
8
6
ᎏ.
445. a. The probability of selecting 2 blue marbles if the first selected
marble is replaced is ᎏ
3
5
ᎏ × ᎏ
3
5
ᎏ = ᎏ
2
9
5
ᎏ (or .36). The probability of
selecting 2 blue marbles if the first selected marble is not
replaced is ᎏ
3
5
ᎏ × ᎏ
2
4
ᎏ = ᎏ2
6
0ᎏ (or .30); .36 > .30.
446. b. The probability of selecting a blue, then a red marble if the first
selected marble is replaced is ᎏ
3
5
ᎏ × ᎏ
2
5
ᎏ = ᎏ2
6
5ᎏ (or .24). The
probability of selecting a blue, then a red marble if the first
selected marble is not replaced is ᎏ
3
5
ᎏ × ᎏ
2
4
ᎏ = ᎏ2
6
0ᎏ (or .30); .30 > .24.
447. b. The probability of selecting 2 red marbles if the first selected
marble is replaced is ᎏ
2
5
ᎏ × ᎏ
2
5
ᎏ = ᎏ2
4
5ᎏ (or .16). The probability of
selecting 2 blue marbles if the first selected marble is not
replaced is ᎏ
3
5
ᎏ × ᎏ
2
4
ᎏ = ᎏ2
6
0ᎏ (or .30); .30 > .16.
448. a. Each spin of the spinner has 4 possible outcomes, and the
number cube has 6 possible outcomes. Therefore the
experiment has 4 × 4 × 6 = 96 possible outcomes; 96 > 14.
449. b. There is only one outcome (blue, red, 4) out of the possible 96
outcomes. The probability of obtaining this outcome is ᎏ
9
1
6
ᎏ;
.1 = ᎏ
1
1
0
ᎏ; ᎏ
1
1
0
ᎏ > ᎏ
9
1
6
ᎏ
450. c. There are 6 outcomes that meet the condition (blue or red, red,
even number): (blue, red, 2), (blue, red, 4), (blue, red, 6), (red,
red, 2), (red, red, 4), (red, red, 6). The probability of obtaining
one of these outcomes is ᎏ
9
6
6
ᎏ. Similarly, there are 6 outcomes
that meet the condition (green, green or yellow, odd): (green,
green, 1), (green, green, 3), (green, green, 5), (green, yellow, 1),
(green, yellow, 3), (green, yellow, 5). The probability of
obtaining one of these outcomes is also ᎏ
9
6
6
ᎏ.

451. b. There are 6 outcomes that meet the condition (not green, red,
factor of 2). This number can also be found by multiplying the
number of ways to achieve this outcome: There are 3 outcomes
that meet the condition “not green,” 1 outcome that meets the
condition “red,” and 2 outcomes (1 and 2) that meet the
condition “factor of 2.” 3 × 1 × 2 = 6. The probability of
obtaining one of these 6 outcomes is ᎏ
9
6
6
ᎏ. Likewise, for the
outcome (not blue, not blue, multiple of 3) there are 3
outcomes that meet the condition “not blue” and 2 (3 and 6)
that meet the condition “multiple of 3.” 3 × 3 × 2 = 18. The
probability of obtaining one of these outcomes is ᎏ
1
9
8
6
ᎏ; ᎏ
1
9
8
6
ᎏ > ᎏ
9
6
6
ᎏ.
452. b. To ﬁnd the mean of a set of data displayed in a frequency table,
the sum of the data must be found by multiplying each data
value by its frequency (the number of times it occurs) and
adding the products: 0 occurs 3 times (3 × 0 = 0), 1 occurs 6
times (6 × 1 = 6), 2 occurs 2 times (2 × 2 = 4), 3 occurs 1 time
(1 × 3 = 3), and 4 occurs 2 times (2 × 4 = 8). The sum is:
0 + 6 + 4 + 3 + 8 = 21. There are 3 + 6 + 2 + 1 + 2 = 14 pieces of
data displayed in the table (the sum of the frequencies).
ᎏ
2
1
1
4
ᎏ = 1.5; 2.5 > 1.5.
453. b. Since there are 14 data items (an even number) in the frequency
table, the median will be the average of the middle two, namely
the seventh and the eighth pieces. The table contains the data
in numerical order, so the ﬁrst three elements are zeros and the
next six are ones. Therefore, the seventh and eighth pieces of
data will both be 1 and the median will be 1.
454. c. The mode of a data set is the element that occurs most
frequently, which is easy to spot in a frequency table. In this
case, there are 6 occurrences of the data value 1, making it the
mode.
455. b. The range of the distribution is the difference between the
largest and the smallest piece of data. In this case, the smallest
data value is zero and the largest is 4; 4 − 0 = 4; 5 > 4.
140

141
456. c. To answer any of questions 456–458, set K must be determined
by substituting the corresponding values for x into each of the
expressions given for set K:
when x = 0, 2x = 2 × 0 = 0, x2
= 02
= 0, x + 2 = 0 + 2 = 2;
when x = 1, 2x = 2 × 1 = 2, x2
= 12
= 1, x + 2 = 1 + 2 = 3;
when x = 2, 2x = 2 × 2 = 4, x2
= 22
= 4, x + 2 = 2 + 2 = 4;
when x = 3, 2x = 2 × 3 = 6, x2
= 32
= 9, x + 2 = 3 + 2 = 5.
Set K, therefore, is 0, 0, 2, 2, 1, 3, 4, 4, 4, 6, 9, 5. The mode is
the value that occurs the most, 4.
457. b. The range of set K is the difference between the largest and
smallest values in set K. The largest value in set K is 9, the
smallest is 0. 9 − 0 = 9; 10 > 9.
458. a. The median of set K is the middle number when the set is put
in numerical order: 0, 0, 1, 2, 2, 3, 4, 4, 4, 5, 6, 9. The median is
the average (arithmetic mean) of the sixth and seventh data
values, namely 3 and 4; ᎏ
(3 +
2
4)
ᎏ = 3.5. The mean is the sum of the
data divided by 12: 0 + 0 + 1 + 2 + 2 + 3 + 4 + 4 + 4 + 5 + 6 + 9
= 40; ᎏ
4
1
0
2
ᎏ = 3.33; 3.5 > 3.33.
459. c. Each term of the sequence is a power of 2. The ﬁrst term is 21
,
the second term is 22
, the third term is 23
, etc. Therefore, the
10th term of the sequence would be 210
.
460. a. The 20th term of the sequence is 220
. The 10th term of the
sequence is 210
. Twice the tenth term is 2 × 210
, which equals 211
.
Since 220
> 211
, 220
> twice the tenth term.
461. a. Each term in this sequence is also a power of 2, beginning with
64 which equals 26
. Since the ﬁrst term = 26
, the second = 25
, the
third = 24
, the fourth = 23
and so on, each term is equal to 2
raised to the 7−n power, where n = the term number (in other
words, the power of 2, plus the term number = 7). Therefore,
the seventh term of the sequence is 27−7
= 20
= 1; 1 > 0.
462. b. The thirtieth term of the sequence is 27−30
= 2−23
; 2−20
> 2−23
.
463. b. If 29% of seniors neither play an instrument, nor participate in
sports, this leaves 71% who must participate in one or both of
the activities; 78% > 71%.

464. b. 71% of seniors must participate in one or both of the activities.
43% participate in sports and 35% play an instrument, but 43%
+ 35% = 78% which is too high because those seniors who
participate in both activities have been counted twice.
78% − 71% = 7%. Therefore, 7% of students participate in
both activities; 10% > 7%.
465. a. The sum of the elements given in set M is 4 + 6 + 3 + 7 = 20.
ᎏ
2
4
0
ᎏ = 5 meaning that the four given elements have an average of
5 before the value of 2x is added to the sum. The sum of the
elements in set N is 1 + 5 + 8 = 14; ᎏ
1
3
4
ᎏ = 4.667 meaning that the
three given elements have an average of 4.667 before the value
of y is added to the sum. Since x > y, 2 larger numbers will be
added to the sum of set M than will be added to the sum of set
N. This will keep the mean of set M > the mean of set N.
466. c. Since both x and y are > 10, the elements of set M can be
ordered as follows: M = {3, 4, 6, 7, x, x} which makes the median
the average of the third and fourth terms; 6 + 7 = 13; ᎏ
1
2
3
ᎏ = 6.5.
Likewise, set N in order = 1, 5, 8, y which makes the median the
average of the second and third terms; 5 + 8 = 13; ᎏ
1
2
3
ᎏ = 6.5.
467. c. Because quartiles divide the data into four quarters such that
there are the same number of pieces of data in each quarter, the
quantity of data between the ﬁrst quartile and the median, and
the quantity of data between the third quartile and the max
must be equal.
468. a. The interquartile range is the difference between the ﬁrst and
the third quartiles; in this case 6 and 10; 10 − 6 = 4; 4 > 3.
469. a. The data in the third quarter are spread between the values 7
and 10. The data in the second quarter are spread between the
values 6 and 7. Therefore the data in the third quarter have a
larger spread than the data in the second quarter.
470. c. Kelly’s offers 3 types of bread, 5 types of meat, and 4 types of
cheese. Therefore, Kelly’s can make any of 3 × 5 × 4 = 60 types
of sandwiches.
142

143
471. c. Fixing the type of bread means Kelly’s can use 1 type of bread,
any of the 5 meats and any of the 4 cheeses on the sandwich.
5 × 4 = 20 types of sandwiches that can be made on wheat or rye
bread.
472. b. If a sandwich must contain roast beef, it can be made on any of
3 types of bread and with any of 4 types of cheese; 3 × 4 = 12
types of roast beef sandwich. If a sandwich must contain
provolone cheese, it can be made on any of 3 types of bread
with any of 5 types of meat; 3 × 5 = 15 types of provolone
cheese sandwich; 15 > 12.
473. a. This sequence contains all the perfect squares of the positive
integers. The first term = 1 × 1. The second term = 2 × 2. The
third term = 3 × 3 and so on. The seventh term, therefore, is
7 × 7 = 49; 49 > 36.
474. c. The twentieth term of the sequence is 20 × 20 = 400.
475. c. The 34th term is 34 × 34 = 1,156. The 35th term is 35 × 35 =
1,225; 1,225 − 1,156 = 69. The 10th term is 10 × 10 = 100. The
13th term is 13 × 13 = 169; 169 − 100 = 69.
476. b. The 1,000th term is 1,000 × 1,000 = 1,000,000 = 1 × 106
.
1 × 109
> 1 × 106
since 109
> 106
.
477. d. The answer cannot be determined. The fiftieth percentile of a
data set is the same as the median of the set. The median of a
set of data may be greater than, less than, or equal to the mean
of the set. For example, in the data set 2, 3, 4, the median, or
fiftieth percentile, is 3. The mean is also 3. In the data set 1, 5,
8, 9, the median, or fiftieth percentile is the average of 5 and 8.
5 + 8 = 13; ᎏ
1
2
3
ᎏ = 6.5. The mean is the sum, 23, divided by 4;
ᎏ
2
4
3
ᎏ = 5.75, so the median is greater.
478. c. The median of a data set is always equal to the fiftieth
percentile.
479. c. The 25th percentile is, by definition, the same as the first
quartile.
480. b. The standard deviation of set P can be found be finding the
mean of set P, then finding the square of the distance each
element of the set is from the mean. Finding the sum of these

squares, dividing by the number of elements in the set, then
finding the square root of this quantity. For set P: 1 + 2 + 3 + 4
+ 5 = 15; ᎏ
1
5
5
ᎏ = 3, which is the mean; 1 − 3 = −2; −2 × −2 = 4;
2 − 3 = −1; −1 × −1 = 1; 3 − 3 = 0; 0 × 0 = 0; 4 − 3 = 1; 1 × 1 = 1;
5 − 3 = 2; 2 × 2 = 4. The sum of these square differences is
4 + 1 + 0 + 1 + 4 = 10; ᎏ
1
5
0
ᎏ = 2. The standard deviation is ͙2ෆ.
For set Q: 2 + 3 + 6 + 5 = 16; ᎏ
1
4
6
ᎏ = 4, which is the mean;
2 − 4 = −2; −2 × −2 = 4; 3 − 4 = −1; −1 × −1 = 1; 6 − 4 = 2;
2 × 2 = 4; 5 − 4 = 1; 1 × 1 = 1. The sum of these square
differences is 1 + 4 + 4 + 1 = 10; ᎏ
1
4
0
ᎏ = 2.5. The standard
deviation is ͙2.5ෆ; ͙2.5ෆ > ͙2ෆ.
481. b. The mean of set P is 3. The median of set Q is the average of
the second and third elements, 3 and 5; 3 + 5 = 8; ᎏ
8
4
ᎏ = 2; 4 > 3.
482. b. 1 kilometer is equal to 1,000 meters.
483. c. Using scientific notation, 3.46 × 109
means the decimal point in
3.46 would be moved 9 places to the right (since the exponent
on 10 is a positive 9). This would yield 3,460,000,000. Likewise,
in 34.6 × 108
, the decimal point in 34.6 would be moved 8
places to the right to yield 3,460,000,000 as well.
484. a. 1.5 × 10−6
means the decimal point would move 6 places to the
left (since the exponent on the 10 is a negative 6) to yield
.0000015; .0000015 > .00000015.
485. a. The profit in 1988 was approximately $4,100. In 1990, the
profit was approximately $4,800. The profit increase was,
therefore, about $700. The profit in 1987 was approximately
$3,250. In 1989, the profit was approximately $3,600. The
profit increase was, therefore, about $350. $700 > $350.
486. b. The profit in 1990 was approximately $4,800. In 1991, the
profit was approximately $4,600. The profit decrease was,
therefore, about $200. The profit in 1988 was approximately
$4,100. In 1989, the profit was approximately $3600. The profit
decrease was, therefore, about $500. $500 > $200.
487. a. The profit in 1987 was approximately $3,250. In 1993, the
profit was approximately $5,250. The profit increase was,
therefore, about $2,000. $2,000 > $1,500.
144

145
488. b. Each term in the sequence, beginning with the second, is found
by multiplying the previous term by increasing integers. So, the
first term, 1, times 2 = 2, the second term. The second term, 2,
times 3 = 6, the third term. The third term, 6, times 4 = 24, the
fourth term. The fifth term is therefore, the fourth term, 24,
times 5 = 120. 150 > 120.
489. c. The sixth term is the fifth term, 120, times 6 = 720.
490. a. The number of each term in the sequence can be determined by
examining its tens and hundreds places. The first term is 2, so
there is a zero in the 10s place. The second term is 12 and there
is a 1 in the 10s place. The seventh term is 62 . . . there is a 6 in
the 10s place. In other words, the 10s place is always one less
than the term number. Note that the eleventh term should
therefore be 102 and the twelfth term should be 112. The last
term, 132, must be the fourteenth term.
491. b. Rewriting the series in backward order beneath the original
series, then adding vertically, gives 14 consistent sums of 134.
This is double the total desired since every term has been
written twice; 134 × 14 = 1,876; ᎏ
1,8
2
76
ᎏ = 938; 1,000 > 938.
492. c. 4.23 × 10−5
means the decimal point must move 5 decimal
places to the left yielding .0000423; .423 × 10−4
means the
decimal point must move 4 decimal places to the left yielding
.0000423 as well.
493. b. A milliliter is ᎏ1,0
1
00ᎏ of a liter. Therefore, 234 milliliters, divided
by 1,000, = .234 liters; 2.34 > .234
494. c. A centigram is ᎏ
1
1
00
ᎏ of a gram. 45 centigrams, divided by 100, is
.45 grams.
495. a. Each square yard measures 3 feet long by 3 feet wide and so, is
9 square feet. 12 square yards would be equal to 12 × 9 = 108
square feet; 108 > 36.
496. b. One square foot measures 12 inches long by 12 inches wide and
is equivalent to 144 square inches; 2 square feet is equal to
2 × 144 = 288 square inches; 288 square inches > 24 square
inches.

497. c. 1 cubic foot measures 12 inches in height, width, and length
and is equivalent therefore to 123
cubic inches.
498. d. The answer cannot be determined. Since the data in this
frequency table is organized into classes of uniform width, the
original data values cannot be determined. In the class of 15–19
year olds, for example, there are 3 women who may be 15, 16,
and 19, or who may all be 17. There is no way to know. There
are more women in the 25–29 age range than in the 15–19 age
range, but there is no way to know whether any or all of them
are of any one specific age within the range.
499. a. The total number of entrants is the total of the frequency
column; 3 + 14 + 17 + 13 + 10 = 57; 57 > 55.
500. b. The median age of an entrant cannot be determined exactly;
however, since there are 57 entrants, the median age would be
the 29th age when the data were in numerical order, as they are
arranged, vertically, in the table. Adding frequencies vertically
until arriving at the class that would contain the 29th data value:
3 + 14 = 17; 17 + 17 = 34 which is > 29. This means that the
29th data value is in the third class, between 25 and 29; 30 is
larger than every number in this class so the answer is b.
501. a. To find the number of women under 30, add together the three
age groups that are listed first in the table (15–19, 20–24, and
25–29): 3 + 14 + 17 = 34; 34 is greater than 28, so a is correct.
146

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