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Title: quantitative comparision questions.
Description: Best ever Quant comparision NoteS.
Description: Best ever Quant comparision NoteS.
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501
Quantitative Comparison
Questions
501
Quantitative Comparison
Questions
NEW YORK
Copyright © 2003 LearningExpress, LLC
...
Published in the United States by LearningExpress, LLC, New York
...
—1st ed
...
cm
...
)
1
...
2
...
3
...
I
...
II
...
QA43
...
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
For more information or to place an order, contact LearningExpress at:
900 Broadway
Suite 604
New York, NY 10003
Or visit us at:
www
...
com
The LearningExpress Skill Builder in Focus Writing Team is comprised of experts in test preparation, as well as educators and teachers who
specialize in language arts and math
...
James, New York
Colleen Schultz
Middle School Math Teacher, Grade 8
Vestal Central School District
Math Tutor
Vestal, New York
Contents
Introduction
1
2
3
4
ix
Arithmetic
Questions
Answers and Explanations
1
11
Algebra
Questions
Answers and Explanations
25
42
Geometry
Questions
Answers and Explanations
59
85
Data Analysis
Questions
Answers and Explanations
113
129
Introduction
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
...
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
...
Gaining familiarity with this specialized question type is a proven technique for increasing test scores
...
Math topics included in this
volume are arithmetic, algebra, geometry, and data analysis
...
Of these questions, 12 are quantitative comparisons
...
A Note about the SAT Assessment
On the SAT exam, there are two 30-minute math sections, for a total of 60
questions
...
(There is also a 15-minute section of 5-choice math questions
...
This book focuses on each of
xi
501 Quantitative Comparison Questions
these areas of math so that you can get the targeted practice necessary to ace
the quantitative comparison sections in the SAT exam
...
Since you don’t
know which section is experimental, it’s best to be prepared to answer many
of these question types
...
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
...
Overall, the percentage 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
...
These are math concepts usually studied in high school, and this book
specifically targets these areas
...
Questions, Questions, Questions
You have just read about the math topics covered in this book
...
For
example, in the arithmetic section, you may find a question or two on exponents, then three questions on square roots, followed by three questions
involving decimals and fractions
...
This way, you are preparing for any type of arithmetic question that
you could encounter on the real exam
...
Each chapter begins with an information and instruction overview describing the
mathematical concepts covered in that particular section
...
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
501 Quantitative Comparison Questions
because calculators are allowed in the testing center
...
This goes for everyone—even GRE test-takers
...
And if you are unfamiliar with prime numbers, use a list of them so you won’t waste time trying
to factor numbers that can’t be factored
...
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
...
Don’t
worry! If you are getting most of the questions correct, you are probably in
good shape for your test
...
The questions in this book can help you prepare for your test in many
ways
...
Quantitative comparisons usually involve less reading,
take less time to answer, and require less computation than regular 5-choice
questions
...
If the quantity in column A is larger, you
will select the letter A on your answer grid
...
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
...
Second, quantitative comparison practice will get you thinking of values
in terms of equalities, inequalities, and estimation
...
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
...
xiii
501 Quantitative Comparison Questions
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
...
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 further study
...
The
answer keys give you not only the right answer, but also an explanation of
why the answer is correct
...
If you find yourself getting stuck solving a problem, you can look at the answer explanation and use it as a personal tutor
...
You have shown your commitment by purchasing this
book
...
You can even work in pencil and do the exercises again to reinforce what you have learned
...
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
...
Look alikes
...
For example, make sure all units are equal
...
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
...
Eliminate any information the
columns share
...
For example, you are given the two quantities 5(x + 1) and 3(x + 1), with the proviso that x is greater than 0
...
That leaves you to decide which is
greater, 5 or 3
...
Plug it in
...
If you can do this
quickly, many comparisons will become straightforward
...
Always remember to simplify
the equation or expression as much as possible before you plug in a value
...
Try not to get stuck doing complicated computations
...
There is often more than one way to solve a problem
...
No assumptions necessary
...
If the question asks you to make assumptions,
then choose D
...
Remember that x2 will have
two roots, one positive and one negative
...
Do not let the test fool you
...
A parenthetical aside
...
This is a simple technique that can make
xv
501 Quantitative Comparison Questions
✔
✔
a large difference in the similarity of the two quantities
...
Let’s play Operation™
...
This is especially useful
when working with fractions
...
Just
keep in mind that, like working in an equation, the operation performed
must be exactly the same in each column
...
Use your time wisely
...
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 a good idea to practice timing yourself as you solve the practice problems
...
xvi
501 Quantitative Comparison Questions
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
xvii
501 Quantitative Comparison Questions
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
501 Quantitative Comparison Questions
Reference Sheet
■
■
■
The sum of the interior angles of a triangle is 180°
...
There are 360 degrees of arc in a circle
...
Figures: Figures that accompany questions are intended to provide information useful in answering the questions
...
are in the order shown; angle measures
are positive; lines shown as straight are straight; and figures lie in a plane
...
Common Information: In a question, information concerning one or both
of the quantities to be compared is centered above the two columns
...
Directions: Each of the following questions consists of two quantities, one
in Column A and one in Column B
...
if the quantity in Column A is greater
b
...
if the two quantities are equal
d
...
ᎏᎏ of 25
5
5
ᎏᎏ
2
of 2
The answer is c
...
2
2
...
Regardless of the value of a, adding 7 will always
result in a higher number than adding 5
...
the number of even integers
the number of even integers
between 2 and 14
between 1 and 13
2
...
16
ෆ
͙0
...
Amy, Megan, and Sharon divided a batch of cookies among them-
selves
...
Amy ate ᎏ3ᎏ of the cookies she took and Sharon ate ᎏ4ᎏ of the
cookies she took
...
p − 8
p+8
5
...
a2
a3
w
7
...
the percent increase from
the percent increase from
54 cm to 58 cm
10 cm to 14 cm
...
ᎏᎏ
25
10
...
5 + ͙32
ෆ
͙23 + 4
ෆ
12
...
at this rate, the cost of
x candy bars
x dollars
13
...
8
...
18 − ᎏᎏ + ᎏᎏ
5
2
18 − ᎏ2ᎏ + ᎏ5ᎏ
1
16
...
7n
4n
18
...
(85 − 93)(22 − 8)
(42 − 95)(11 − 17)
x≥0
20
...
5 × (3 + 1) ÷ 2
5×3+1÷2
22
...
75 of the people are men,
60 are women, and the remainder are children
...
75%
...
The ratio of dogs to cats in a pet store is 5:3
...
the number of dogs
in the pet store
65
25
...
a
b
27
...
5 × 103
1
...
ᎏᎏ of ᎏᎏ
4
11
6
ᎏ1
1ᎏ
x<0
29
...
ab = 30
...
a+b
32
0
31
...
0001
ෆ
͙
...
123 × 10−3
12
...
x
and ᎏ6ᎏ are positive integers
...
5
...
014
x > 20
36
...
the average rate of speed
the average rate of speed
needed to drive 780 km
in 12 hrs
needed to drive 350 km
in 5 hrs
38
...
282 + 422
(28 + 42)2
40
...
n(n − 1)
n2
a>b
42
...
1
4
ᎏᎏ
7
+ ᎏ7ᎏ
13
5
44
...
ᎏxᎏ + 3y
7y + ᎏ1ᎏ
x
46
...
47
...
0 < y < x
...
x
ᎏᎏ
y
remainder when xy
is divided by 2
x>0
49
...
x
ᎏᎏ
5
4x
−1
51
...
x2 + 1
x+1
0
53
...
54
...
(84 + 12)(15 + 91)
(74 + 22)(20 + 86)
56
...
63
...
remainder when 4x
is divided by 2
0
59
...
͙17 + ͙5
ෆ
ෆ
͙22
ෆ
61
...
ᎏᎏ%
4
...
x and y are prime numbers and x + y = 12
...
(b − a)(b − a)
(a + b)(a + b)
65
...
number of years
number of years
from 1631 to 1809
from 1625 to 1812
67
...
76 × 14 + 14 × 26
(76 + 26)14
p<0
69
...
8
k
71
...
231 × 93
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
...
x−5
5−x
3
1
5
74
...
75
...
(26 − 31)(296 + 134)
(31 − 26)(296 + 134)
3
77
...
ԽxԽ + ԽyԽ
x+y
79
...
4͙2
ෆ
ෆ
͙35
81
...
Wendy runs w miles in 1 hour
...
r5
32
m>1
83
...
x > 0 and y > 0
...
3x
y
85
...
This year they sold 20 percent more
...
956 + 274 + 189
200 + 275 + 970
87
...
a−b
a+b
Lisa drove 117 miles between 8:15 A
...
and 10:30 A
...
without stopping
...
60
Lisa’s average speed
in miles per hour
90
...
He has completed ᎏ2ᎏ of his trip in 2
...
5
91
...
the sum of the integers
the sum of the integers
from −5 to 10
from −10 to 5
93
...
683 ×
...
1
8
501 Quantitative Comparison Questions
Column A
Column B
95
...
45
ᎏᎏ
r
45r
97
...
59
254
1
3
7
99
...
2 + 82 − 6 − 10
× ᎏ7ᎏ × ᎏ1ᎏ
9
4
(2 + 8)2 − 6 − 10
x>0
101
...
ᎏᎏ of x
3
35% of x
1
ᎏᎏx
3
103
...
b (a + 1)
ba
x>y>0
105
...
1 − ᎏᎏ
30
1
9
ᎏᎏ
10
107
...
the number of primes
the number of primes
between 1 and 6
between 40 and 50
11
+ ᎏ0ᎏ
1 0
x > 0 and y > 0
109
...
Kendra is driving at a steady rate of 56 miles per hour
...
{x, y} represents the remainder when x is divided by y
...
x = ᎏ2ᎏm and y = ᎏ5ᎏm and z = ᎏ90 y
5
3
1ᎏ
112
...
z
15
Ίᎏ2ᎏ
3
ᎏෆ
ᎏ
͙3
h<0
114
...
x
y
116
...
(−21)12
(−31)13
118
...
ᎏᎏ
5
x−7
120
...
16
...
4
...
Խx − 8Խ
Խ8 − xԽ
124
...
Each pizza used ᎏ2ᎏ a package of cheese
...
x
ᎏᎏ
y
y
ᎏᎏ
x
10
501 Quantitative Comparison Questions
Answer Explanations
The following explanations show one way in which each problem can be
solved
...
1
...
There are 6 even integers between 1 and 13
...
There are 5 even integers between 2 and 14
...
2
...
The square root of 0
...
4
...
0016 is
0
...
1
3
...
Amy took 30% of the cookies and ate ᎏᎏ of those, which is 10%
3
of the original number of cookies
...
Since both women ate 10% of the original
number of cookies, they ate the same amount
...
b
...
5
...
The relationship cannot be determined
...
If the value of x is positive, then 35x
is greater than 31x
...
6
...
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
...
7
...
The relationship cannot be determined
...
If w = 0, x = 1, y = 2, and z = 3, then wx = 0 and yz = 6,
so quantity B is greater
...
8
...
Both quantities increased by 4, but quantity A increased 4 from
10, or 40%, and quantity B increased 4 from 54, or 7
...
40%
is greater than 7
...
11
501 Quantitative Comparison Questions
1
9
...
Change ᎏ5 to a decimal by dividing 1 by 25 to get
...
Change
2ᎏ
...
016 by 4 to get
...
04 is greater
than
...
10
...
Compare ͙8 to ͙9 and find that ͙9 is greater and comes
ෆ
ෆ
ෆ
from quantity B
...
Since both parts of quantity
B are greater than the parts of quantity A, quantity B is greater
...
a
...
Compare ͙32 to ͙23 and find that ͙32 is greater
ෆ
ෆ
ෆ
and comes from quantity A
...
12
...
Each candy bar costs more than $1 (divide $4 by 3)
...
13
...
The relationship cannot be determined
...
If 6 is substituted for x, quantity A is 64
[(6 + 2)2 = 64] and quantity B is 16 [(6 − 2)2 = 16]
...
The relationship cannot be determined
...
b
...
In quantity B, 368 is
being multiplied by a larger number, making quantity B greater
than quantity A
...
b
...
Subtract 18 from both sides, leaving −ᎏ4ᎏ + ᎏ1ᎏ = −ᎏ1ᎏ + ᎏ4ᎏ
...
A positive
number is always greater than a negative number, so quantity B
is greater
...
c
...
Therefore, the two quantities are
20
equal
...
b
...
Try a couple of negative numbers to see
the pattern
...
Substitute in −
...
5) = −3
...
5) = −2
...
12
501 Quantitative Comparison Questions
18
...
Any number squared is positive
...
A positive number is always greater than a
negative number
...
19
...
Both sets of parentheses in quantity B are negative
...
Quantity A has one
negative set of parentheses and one positive
...
Since quantity B is
always positive and quantity A is always negative, quantity A is
greater
...
d
...
If x < 1, for example
...
If x = 1, then both quantities are equal
(both 1)
...
21
...
Use the order of operations to simplify
...
5 =
15
...
c
...
10 = 10%
...
23
...
Change
...
75% = 0
...
24
...
Use the equation 5x + 3x = 96 and solve for x
...
Since x = 12, 5x = 60
...
13
501 Quantitative Comparison Questions
25
...
Notice that 45% of 104 would be more than 45% of 100
...
26
...
Simplify the exponents on the left-hand side of the equation by
multiplying
...
5b − 5 = 5 + a − 5
5b − 5 = a
The variables must be integers greater than 1 (so the smallest
possible value is 2)
...
As the value of b gets larger, so does the value of a
...
27
...
Multiplying by 103 moves the decimal in 16
...
Multiplying by 104 moves
the decimal in 1
...
28
...
“Of ” means multiply
...
Quantity A is equal to quantity B
...
a
...
x3 is also negative because three
negatives multiply to a negative answer
...
The value of y doesn’t matter because it is positive, making it
greater than quantity B
...
b
...
Of these factor pairs, (1,30) has the largest sum;
1 + 30 = 31
...
31
...
x is a positive number less than 1
...
If x = 0
...
5
ᎏ
ᎏ
= 1 ÷ 0
...
Quantity B is always
0
...
5)
greater than quantity A
...
c
...
1 and ͙
...
1; the two quantities are equal
...
a
...
000123
...
0000123
...
000123 > 0
...
x
x
34
...
In order for ᎏᎏ and ᎏᎏ to be integers, x must be evenly divisible by
5
6
5 and 6
...
An even number
does not have a remainder when divided by 2
...
35
...
Add two zeros to the end of quantity A to compare it to
quantity B
...
100 > 5
...
36
...
If x was 20, then quantity B would be 4
...
37
...
Use this formula: rate × time = distance
...
If 12r = 780, then r = 65 (quantity B)
...
b
...
To find the number of seconds in a day,
multiply 60 seconds by 60 minutes by 24 hours to get 86,400
seconds
...
b
...
(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
...
c
...
Quantity
A and quantity B are the same, just in a different order
...
41
...
Distribute the n in quantity A to get n2 − n
...
Quantity A becomes −n and quantity B
becomes 0
...
Any negative
number is less than 0; therefore, quantity B is greater
...
d
...
If a
is 200 and b is 100, 30% of 200 is 60 and 50% of 100 is 50,
making quantity A greater
...
1
43
...
Both fractions in quantity B are greater than ᎏᎏ
...
5
1
44
...
“Of” means multiply; ᎏᎏ × 12 = 10 and ᎏᎏ × 60 = 10
...
d
...
Subtract ᎏxᎏ from both
quantities
...
If y is negative, 3y is
greater
...
46
...
(n2)4 = n8 and (n4)2 = n8
...
3
47
...
The first integer that n can be is 2
...
5, quantity B is
2
greater
...
48
...
The product of two odd numbers is odd
...
Therefore, quantity
A is always 1
...
49
...
Square both quantities to compare without the square roots:
(5͙3x 2 = 25(3x) = 75x; (3͙5x 2 = 9(5x) = 45x
...
50
...
Try a number less than 1 such as 0
...
5) = 2 and
0
...
1
...
Try a number greater
2
ᎏ
than 1 such as 20
...
Whenever a positive
number is multiplied by 4, it will be greater than if that positive
number was divided by 5
...
b
...
Quantity A is now x − 5 and
quantity B is now 0
...
A is now x and B
is now 5
...
Therefore, it is less
than 5
...
d
...
If x = 0
...
5)2 + 1 =
...
25 and quantity B is 0
...
5
...
5, quantity B is greater
...
Quantity A is greater when
x = 6
...
a
...
Since x < y and x is positive, ᎏyᎏ < 1
...
b
...
Therefore,
the ratio of failed to passed is 30 to 70, which simplifies to ᎏ3ᎏ
...
5
7
55
...
Add the numbers inside the parentheses
...
56
...
“Of” means multiply; 75% of 30 =
...
5 and 30% of
75 =
...
5
...
a
...
3 > ᎏ1ᎏ
...
c
...
The
remainder when an even number is divided by 2 is 0
...
b
...
ෆ
ෆ
60
...
͙17 > 4 and ͙5 > 2, therefore, ͙17 + ͙5 > 6
...
Quantity A is greater
...
a
...
5
7
3ᎏ 35
3
3
62
...
ᎏᎏ% = 0
...
0075; ᎏᎏ as a decimal is 0
...
To change 0
...
...
75
...
b
...
Therefore, x and y are 5 and 7; xy = (5)(7) = 35
...
64
...
For any positive numbers a and b, a + b > b − a
...
65
...
Combine the terms of quantity A; ͙10 + ͙10 = 2͙10 To get
ෆ
ෆ
ෆ
...
40 < 49
...
a
...
Therefore,
it is a longer period of time
...
67
...
Square both quantities to get rid of the square roots; (͙63 2 =
ෆ)
63 and (3͙7)2 = 9(7) = 63
...
ෆ
68
...
Use the distributive property to multiply quantity B out
...
Quantity B is equivalent to
quantity A
...
a
...
When the number is negative,
multiplying by 9 yields a number that is “more negative” or less
than that negative multiplied by 5
...
d
...
Solve the inequality for
57
8k
6
ᎏ
k by dividing both sides by 8; ᎏ8ᎏ < ᎏ8ᎏ < ᎏ87
...
125 < k < 8
...
It
cannot be determined whether k is less than, greater than, or
equal to 8
...
a
...
Therefore, the
product for quantity A is greater than the product for
quantity B
...
c
...
First, combine
like terms on the left-hand side of the equation; 12x = 48
...
x = 4
...
73
...
x is between 0 and 5
...
Since x is less than 5, 5 − x must be positive
...
1
5
3
74
...
Subtract ᎏᎏ and ᎏᎏ from both quantities
...
Quantity A is greater than quantity B
...
b
...
Let Joel’s rate be 1
...
In 80 minutes, Sue’s distance is 1(80) = d or
80 = d
...
5(60) = d or 90 = d
...
b
...
Both
parentheses in quantity B are positive, so quantity B is positive
...
40
2 3
2
77
...
40% is equivalent to ᎏᎏ = ᎏᎏ; ᎏᎏ of 80 is greater than ᎏᎏ of 80
...
a
...
Therefore, the sum of x and y is
negative (quantity B)
...
Any positive
number is greater than any negative number
...
d
...
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
...
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
...
80
...
Square both quantities to get rid of the square roots; (4͙2)2 =
ෆ
(16)(2) = 32 and (͙35 2 = 35
...
Quantity B is greater
...
d
...
There is not enough
information to determine an answer
...
c
...
25 = 32
...
83
...
͙m12 = m6 and (m2)5 = m10; since m is greater than 1, quantity B
ෆ
is greater than quantity A
...
b
...
4y
...
4y for x in quantity B; 3(
...
2y
...
2y
...
85
...
Find 20% of 120 by multiplying 120 by
...
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
...
b
...
Since each of
the numbers in quantity A is less than the corresponding
number in quantity B, quantity B is greater
...
a
...
Take the
second largest pair (2 and 99) and add them together (101)
...
There are 50
pairs whose sum is 101
...
88
...
Subtract a from both quantities
...
Since b is negative, −b is positive and therefore
greater than b
...
a
...
Lisa’s distance is 117
miles and her time is 2
...
25r
...
2
1 7
by 2
...
The rate is 52 miles per hour
...
ᎏ
5
2
...
b
...
There is
an odd number of numbers for quantity A and an even number
of numbers for quantity B
...
When multiplying an
even number of negatives, you get a positive answer
...
2
91
...
Find the number of miles that John has driven by finding ᎏᎏ of
5
350; ᎏ2ᎏ × 350 = 140
...
5 hours
...
The distance is 140 and the time is 2
...
140 = 2
...
Solve for r
1 0
2
...
5; ᎏ4ᎏ = ᎏ
...
5
2ᎏ
5
r = 56 mph
...
b
...
So the only comparison that must be done is outside of those
numbers
...
The additional numbers
that quantity B has are all positive, which brings the sum up
...
93
...
Create a factor tree to help visualize the factors of 122 and 503
...
The only
distinct prime factors of 503 are 5 and 2
...
94
...
683 ×
...
15 and 683 ×
...
3
20
501 Quantitative Comparison Questions
95
...
|−83| = 83 and |83| = 83
96
...
The relationship cannot be determined
...
1, 45(−0
...
5 and ᎏ5 = −450
...
If r
−0ᎏ
...
In this example, quantity B is
−5
greater
...
c
...
a
...
25 can be written as 52
...
Compare the exponents of quantities A and B
...
Quantity A
is greater
...
c
...
The numbers in the numerators in quantity
A and quantity B are the same (1, 3, 7), just in a different order
...
When multiplied out, both quantities
2
are ᎏ1
...
b
...
Quantity A is 2 + 82 − 6 − 10 = 2
+ 64 − 6 − 10 = 50
...
101
...
Square both quantities to get rid of the square roots; (7͙x)2 =
ෆ
49x and (͙3x 2 = 3x
...
Quantity A is
ෆ)
greater than quantity B
...
d
...
When x is positive,
quantity B is greater
...
1
1
1
103
...
ᎏᎏx + ᎏᎏx + ᎏᎏx = 1x; therefore, x = 9
...
3
3
3
104
...
Since the bases are the same (b) and b > 1, the exponents can be
compared
...
105
...
Simplify both fractions by canceling the x’s
...
y is positive, so −y is negative
...
21
501 Quantitative Comparison Questions
106
...
Quantity A is less than 1 and quantity B is greater than 1
...
For quantity B, create a
1 1
9
1
common denominator of 100; ᎏ000 + ᎏ010 = ᎏ0ᎏ
...
107
...
Use rules of exponents to simplify quantities
...
The quantities are equal
...
c
...
The primes
between 1 and 6 are 2, 3, and 5
...
109
...
Square both quantities to get rid of some of the square roots
...
110
...
Use the formula distance = rate × time
...
Solve the equation by dividing by
4
5
56; ᎏ2 = ᎏ6t ; t =
...
75 hours is equivalent to 45 minutes
...
a
...
When it is divided by 10, there will
be a remainder of 5
...
Any power of
10 ends in 0
...
2
112
...
Compare x and z in terms of m; x = ᎏᎏm; z can be rewritten in
5
5
ᎏᎏm
3
terms of m by substituting
for y; z = ᎏ9ᎏ(ᎏ5ᎏm) = ᎏ3ᎏm
...
15
113
...
Get rid of the square roots by squaring both fractions; (Ίᎏᎏ)2 =
2
15
ᎏᎏ
4
3
ᎏ
= 3
...
3
͙ෆ
114
...
h is negative, so 5 times a negative is a negative
...
A negative multiplied by itself 4 times is a positive
...
Any positive number is greater than any
negative number
...
c
...
−(x − y) = x − y
−x + y = x − y
Next, add x to both sides
...
−x + x + y = x + x − y
y + y = 2x − y + y
2y = 2x
Divide both sides by 2
...
c
...
Zero multiplied by anything
yields a zero
...
117
...
Quantity A is positive because a negative number to an even
power yields a positive answer
...
Any
positive number is greater than any negative number
...
c
...
Both quantities are equivalent to m18
...
c
...
This
yields x − 7, which is equivalent to quantity B
...
c
...
It is the same for 11—the
only prime divisible by 11 is 11
...
121
...
Add zeros on to the end of quantity A to make it easier to
compare to quantity B;
...
1600;
...
0989
...
c
...
When dividing by 102, move the decimal 2
places to the left to get 425,000
...
23
501 Quantitative Comparison Questions
123
...
The absolute value of opposites is the same number
...
This
can be shown by multiplying either expression by −1
...
Therefore, the two values are equal
...
b
...
They made 90 pizzas
...
125
...
Set up the complex fractions and then divide
...
5
4
5
1
5
5
Since quantity A is greater than 1 and quantity B is less than 1,
quantity A is greater
...
Figures: Figures that accompany questions are intended to provide information useful in answering the questions
...
are in the order shown; angle measures
are positive; lines shown as straight are straight; and figures lie in a plane
...
Common Information: In a question, information concerning one or both
of the quantities to be compared is centered above the two columns
...
Directions: Each of the following questions consists of two quantities, one
in Column A and one in Column B
...
if the quantity in Column A is greater
b
...
if the two quantities are equal
d
...
Examples
Column A
Column B
x+5=9
1
...
Subtract 5 on both sides of the equation
...
Simplifying this gives a solution of x = 4
...
a<0
2
...
Since a is less than zero, it represents a
negative number
...
Column A is larger
...
117 + 23
25
a is a positive integer
...
1
ᎏᎏ
a7
1
ᎏ10
ᎏ
a
128
...
33
9
a is an integer
...
a2a3
3
a6
1
(ᎏ1ᎏ)2
2
131
...
4
x
212 = 8x
133
...
2
a9
a
y = 5x
x is a positive integer
...
5x + 1
5y
x>0
136
...
x2 − 4x − 21 = 0
sum of the roots
138
...
The arch follows the equation y = 2x −
...
maximum arch height
10 feet
27
501 Quantitative Comparison Questions
Column A
Column B
x<0
139
...
(x + 3)2
x2 + 9
y = x2 + 6x + 9
141
...
−2
smaller root
y = −x2 + 6x
143
...
1
x
145
...
(x + y)
2
x2 + y2
2(x + 3) + 6 = 4x
147
...
x
Julie is 5 years older than Ravi
...
Ravi’s age now
149
...
the middle integer
150
...
00
...
the original price
of the snowboard
$90
...
Column B
The ratio of rabbits to squirrels is 2:3
...
the number of squirrels
152
...
34 for the first ounce,
plus $0
...
The cost to mail an
envelope was $1
...
8 ounces
153
...
The monetary value of these coins is $2
...
the number of quarters
the number of nickels
The drama club collected $907
...
Adult tickets cost $5
...
50
...
154
...
There are 146 athletes and 8 coaches taking a trip to
a competition
...
the number of buses needed
156
...
Monique’s percentage grade
157
...
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
...
= 50
1
ᎏᎏ
6
x
5x + 3 = 18
159
...
−10
x
4(x − 2) = 8
161
...
p
q
2a + 2b = 20
4a + 2b = 14
163
...
+ 6 = 12
15
x
c + 4d = 11
6c − 2d = 40
165
...
d − 12
ᎏᎏ
9
c
a
ᎏᎏ
b
167
...
c+b÷a
7
169
...
−2b2
(−2b)2
a<0
171
...
b+c
0
30
501 Quantitative Comparison Questions
Column A
Column B
x≠0
173
...
ෆ2
͙(a + 4) ෆ
a+4
a≠0
175
...
5−x
five less than x
177
...
the product of a and b
|ab|
179
...
x3
the square root of x,
raised to the fourth power
181
...
Let p = number of pennies
...
10d + 0
...
10d + p
x>y>4
xy
ᎏᎏ
4
four divided by the
product of x and y
183
...
00 sweater,
$30
...
the cost of a $28
...
6(28)
basketball, including
6% sales tax
31
501 Quantitative Comparison Questions
Column A
Column B
x is a positive odd integer
...
the product of two
consecutive positive odd
integers
x(x + 2)
3b + 3 > 18
186
...
−30
x
0
188
...
−2
b
x<5;y
190
...
− ᎏ1ᎏx ≥ 4
3
20
x
a>b>0
192
...
10
A
5
B
-5
5
-5
193
...
1
the slope of line AB
33
10
501 Quantitative Comparison Questions
Column A
Column B
Use the following figure to answer questions 195–196
...
5
the length of segment BC
196
...
A circle with center at the origin passes through the point (0,8)
...
the slope of the line that
2
passes through the points
(1,2) and (2,4)
34
501 Quantitative Comparison Questions
Column A
Column B
Use the following figure to answer questions 199–200
...
a − b
0
200
...
5
(q,3)
D
(5,p)
C
A
(2,1)
B
(5,1)
5
201
...
a
b
1
203
...
ᎏᎏ
4x2y
0
...
205
...
(ᎏᎏ)2
2
x>1
207
...
x−2
=1
x + x −2
2
0
a<0
b>0
209
...
a6 × a2
(a6)2
36
501 Quantitative Comparison Questions
Column A
Column B
0
211
...
−a2
(−a)2
2
213
...
667)3
2
214
...
y−4
x+4
2(x − 5) − 4 = 10
216
...
−x
x2
11x + 3 = 42
218
...
+ 4 = 2x
x2
6x
x−y=6
y+x=4
220
...
x
y
c+d=3
c−d=3
222
...
y = 5x − 1
the value of x when y = 0
the value of y when x = 0
37
501 Quantitative Comparison Questions
Column A
Column B
8x − 2y = 30
3
x = ᎏ2ᎏy
224
...
x2 − y2
(x + y)(x + y)
226
...
the positive value of x
3
x2 − 3x − 10 = 0
228
...
y2 + 10
4y
y > 2x − 1
230
...
6
x
5
232
...
6
x
−3x − 1 > 14
234
...
Point (x, y) is located in Quadrant IV
...
the slope of the line y = 2x − 3
38
1
the slope of the line y = ᎏ2ᎏx + 3
501 Quantitative Comparison Questions
Column A
Column B
237
...
the opposite of x
238
...
3
Line m is perpendicular to the line y = − ᎏ2ᎏx + 1
...
the slope of line m
Points (4, c) and (0, d) are on line n
...
c−d
3
y = −2x − 3
240
...
the slope of the equation
the x-intercept of the equation
the slope of the equation
−3y + 3x = 9
2y − 4x = 6
242
...
the difference between 26
the sum of 8 and 2
and the product of 4 and 3
244
...
245
...
the quotient of ten and two
4
the quotient of sixty-five and
thirteen
247
...
One-half of z is y
...
15
z
248
...
Car A is traveling at 60 miles per hour and car B is
traveling at 55 miles per hour
...
Column B
The sum of two consecutive integers is 83
...
A vending machine has exactly $1
...
the number of quarters
the number of dimes
40
501 Quantitative Comparison Questions
Answer Explanations
The following explanations show one way in which each problem can be
solved
...
126
...
The number 1 to any power is 1; 23 is 2 × 2 × 2 which is 8;
1 + 8 = 9
...
127
...
Since the variable a is a positive integer, both choices are
positive, and a10 > a7
...
When two fractions are being compared with the
same numerators, the smaller the denominator, the larger the
number
...
128
...
By the laws of exponents, (a4)2 = a4 × 2 = a8
...
Therefore, the quantities in Column A and
Column B are equal
...
a
...
27 is greater than 9, so
column A is greater
...
d
...
This problem involves a law
of exponents that is true for any real number: a2 × a3 = a2+3 = a5
...
Note that this is true even for
negative integers, since 6 is an even number, and 5 is an odd
number
...
They are when a = 0 or a = 1
...
a
...
Likewise, (ᎏ2ᎏ)2 = ᎏ4ᎏ
...
132
...
Four to the x power means that 4 is multiplied by itself “x”
times
...
Choice B is greater
...
b
...
Since 2 × 2 × 2 = 8,
23 = 8
...
Rewrite the equation as
212 = (23)x
...
The equation is
now 212 = 23x
...
Divide both sides of this equation by 3, and it
becomes x = 4
...
a
...
A negative number to
any even power is a positive number, while a negative number
to any odd power is a negative number
...
135
...
Since y = 5x, multiply each side of this equation by 5, to get
5y = (51)(5x)
...
136
...
Using the distributive property, (2x + 4)(x + 1) = 2x2 + 2x +
4x + 4
...
Subtracting 2x2 and 4 from both columns leaves 6x in column A
and 5x in column B
...
137
...
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
...
So x = 7 or x = −3
...
The product of the roots is (−7)(3) = −21
...
c
...
When a quadratic
is in the form ax2 + bx + c (a, b, c are real numbers), the
−b
−2
x-coordinate of the vertex is given by the formula ᎏᎏ = ᎏᎏ = 10
...
2
When x is 10, y = 2(10) − 0
...
So y = 20 − 0
...
So the maximum arch height is 10 feet
...
b
...
Since x < 0, x is negative, and therefore 7x is negative, x2 + 7
will be greater in this case
...
a
...
By the distributive
property, this equals x2 + 3x + 3x + 9 = x2 + 6x + 9
...
Since x > 0, column A is
greater
...
b
...
The minimum y value of the
function is the y value at the vertex
...
When x = −3,
2a
y = (−3)2 + 6(−3) + 9 = 9 − 18 + 9 = 0
...
142
...
To find the roots of this equation, first factor out the common
factor of 4; 4x2 + 4x − 8 = 4(x2 + x − 2)
...
The roots are the
values of x that make the y value equal to zero
...
The smaller root is −2
...
c
...
The equation becomes y = −x(x − 6)
...
The equation
will equal zero when x = 0 or x = 6
...
144
...
To find the values of x, move all terms to the left side so that it
is a quadratic equation set equal to zero
...
Factor this
quadratic into two binomials; 9x2 − 6x + 1 = (3x − 1)(3x − 1)
...
This will be true when
3x − 1 = 0; add one to both sides to give 3x = 1; divide both
1
sides by 3 and x = ᎏ3ᎏ
...
d
...
The binomials in column A
are the difference of two squares, so (x + 5)(x − 5) =
x2 − 5x + 5x − 25 = x2 − 25
...
146
...
Using the distributive property and combining like terms,
column A is (x + y)2 = x2 + xy + xy + y2 = x2 + 2xy + y2
...
147
...
To solve this equation, first use the distributive property, and
combine like terms, on the left hand side
...
Now, 2x + 12 = 4x
...
Divide both sides by 2, and x = 6
...
a
...
Julie’s age
now is represented by x + 5
...
Three years ago, Julie
was twice as old as Ravi
...
Use the distributive
property on the right hand side to get x + 2 = 2x − 6
...
Subtract x from both sides, and
x = 8
...
149
...
Set up an equation where x represents the first consecutive
integer
...
The sum of these integers is represented by
x + x + 1 + x + 2
...
This sum is 37 more than the largest integer, so
3x + 3 = x + 2 + 37
...
Subtracting x from both
sides yields 2x + 3 = 39
...
Divide both sides by 2, and x = 18
...
150
...
$63
...
Therefore, let x represent the original price of the snowboard
...
70x = 63
...
Divide both sides of this equation by 0
...
00
...
a
...
Let 2x represent the number of rabbits,
and then 3x represents the number of squirrels
...
Combine like terms: 5x = 225
...
The number of squirrels is 3x = 3(45) = 135
...
c
...
So x − 1 will
represent the additional ounces over the first ounce
...
34 + 0
...
53
...
34 + 0
...
17 = 1
...
Combine like terms: 0
...
17x = 1
...
Subtract
0
...
17x = 1
...
Divide both sides by 0
...
The total weight is 8 ounces
...
a
...
Since there are a total
of 14 quarters and nickels, 14 − q will represent the number of
nickels
...
25q +
...
30
...
05:
...
70 −
...
30
...
20q +
...
30
...
70 from both sides, to get
...
60
...
20, so q = 8
...
154
...
Since there were 63 more student tickets sold than adult tickets,
let a represent the number of adult tickets sold
...
The total ticket
sales can be represented by 5a + 2
...
50
...
5 to the terms in
parentheses: 5a + 2
...
50 = 907
...
Combine like terms,
to get 7
...
50 = 907
...
Subtract 157
...
5a = 750
...
5, and
a = 100
...
155
...
The buses need to seat 146 athletes, plus 8 coaches
...
Three buses will hold 144 people
...
156
...
Monique’s percentage grade is the ratio of the number correct
68
to the total number of points on the test
...
8 and
85
0
...
157
...
Let x represent the number of hours after the first bus left in
which they will pass
...
The
buses will pass when their distances are the same
...
For
the second bus, since it is traveling at 50 miles per hour, its
distance is represented by 50(x − 1)
...
Use the distributive property on the right hand side to get
45x = 50x − 50
...
Divide both sides by 5, and x = 10
hours
...
c
...
This will leave 8 + 2x = 50x
...
Divide both sides by 48, so
8
1
x = ᎏ8 = ᎏ6ᎏ in simplest form
...
c
...
Divide both sides by 5 to get x = 3
...
a
...
161
...
For the given equation, first apply the distributive property, and
distribute four to each term on the left
...
Now add eight to both sides of the equation to get
4x = 16
...
162
...
One way to solve this system of equations is to multiply each
term in the first equation by 2, to get 2p + 2q = 32
...
Going back to the
original first equation, if p = 12, then 12 + q = 16
...
163
...
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
...
Add six to both sides of this equation to get 2b = 26
...
164
...
For the given equation, first subtract six from both sides to get
1
ᎏᎏx
6
1
= 6
...
The equation now is x = 36
...
a
...
46
501 Quantitative Comparison Questions
Divide both sides of this resulting equation by thirteen to get
c = 7
...
Subtract seven from both sides, and divide
both sides by four, and d = 1
...
c
...
The quantities are equal
...
b
...
In this case, the term ab will be negative
...
168
...
For the given values of a, b, and c, column A is 4 + 6 ÷ 2
...
Now, four plus three is equal to seven
...
c
...
Now
column A is 3x − 4 − x
...
170
...
It is given that b is less than zero, which means that b is a
negative number
...
Negative two times any negative
b results in a positive number
...
In column B, order of operations says to do the
exponent first
...
Now,
however, this number is multiplied by −2, which always results
in a negative number
...
171
...
It is given that a is less than zero, which means that a is a
negative number
...
The quantity in column B
is a negative times a negative (a positive result) which is then
multiplied by another negative number
...
47
501 Quantitative Comparison Questions
172
...
The answer cannot be determined
...
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,
however, both b and c were negative, then the quantity in
column A would be less than zero, the value of column B
...
b
...
The square root of x to
the fourth power is x squared
...
Also, the square root
of 16 is four
...
Therefore, the quantity in
column B, 8x2, is greater than 4x2
...
c
...
175
...
The answer cannot be determined
...
Column B is now a2 + 2ab + b2 − 2ab
...
Since a is not equal to
zero, and any number squared is positive, the quantity in
column B is greater if a < 0
...
If,
however, a = 1, column A is equal to column B
...
176
...
The words in column A translate in algebra to x − 5
...
The quantity in column B will be a
negative number
...
c
...
The twos can
cancel, leaving b + 14
...
d
...
For column A, the product
of a and b is ab
...
Since there is no indication as to whether a or b is
positive or negative, the answer cannot be determined
...
c
...
12
48
501 Quantitative Comparison Questions
180
...
The words in column A translate into algebra as (͙x)4 = x2
ෆ
since (͙x)4 = (͙x)2 (͙x)2 = x × x = x2
...
181
...
Since the variables d and p refer to the number of dimes and
pennies respectively, they must be positive whole numbers (it’s
1
impossible to have −3 dimes, or ᎏ4ᎏ of a penny)
...
Column B will always
be greater
...
b
...
Since
xy
x > y > 4, both x and y are greater than four
...
183
...
The cost of a $50
...
00
...
00)(
...
00
...
b
...
00 basketball, including 6 percent sales tax, is
28 +
...
The decimal in column B representing sales tax is
actually equal to 60 percent, not 6 percent
...
c
...
The next (consecutive) odd
integer is thus x + 2
...
186
...
To solve the given inequality, first subtract three from both
sides, to get 3b > 15
...
Since b is greater than five, column A is greater
...
b
...
The inequality
becomes 10 − 2x < 70
...
Now, divide both sides by negative two, noting that
when you divide by a negative number with an inequality, you
switch the inequality symbol
...
1
1
188
...
Since it is given that 0 < a < b, then ᎏaᎏ > ᎏbᎏ since, for fractions
with the same numerator, the smaller the denominator, the
larger the value of the fraction
...
b
...
Since this is true, −2 is greater
than any allowed value for b
...
b
...
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
...
a
...
Now, multiply both
sides by six to isolate the variable
...
192
...
It is given that a > b > 0, so both a and b are positive, and a is
larger than b
...
193
...
The x-coordinate of an ordered pair determines how far to the
left or right a point is plotted on the coordinate plane
...
Therefore, the x-coordinate of point B is greater
...
a
...
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
...
195
...
Notice that the points A, B, and C form a right triangle in the
figure
...
Use these
lengths and the Pythagorean theorem to find the length of
segment BC
...
Segment BA
is 3 units long, by counting
...
Substituting in for a and b gives 32 + 42 = c2
...
Take the square root of both sides and c = 5
...
b
...
The base of this triangle is 3 units (by counting), and the
1
height is 4 units
...
1
This is ᎏ2ᎏ(3)(4), which is six units
...
b
...
198
...
The slope of a line that passes through two given points is
change in y
y −y
4−2
2
2
1
determined by the formula ᎏᎏ = ᎏᎏ = ᎏᎏ = ᎏ1ᎏ = 2
...
b
...
This indicates that a = 0 and that b is positive
...
Column B is
greater
...
a
...
Determine the slope of this line, using the points (0,0) and (4,3)
...
The
change in y
y2 − y1
3−0
3
slope of the line can be found by ᎏᎏ = ᎏᎏ = ᎏᎏ = ᎏ4ᎏ
...
Using
−8
cross multiplication, 4c = −24
...
201
...
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
...
Similarly,
the y-coordinate of point C equals the y-coordinate of point D,
and p = 3
...
Column B is 2 − 3 = −1
...
b
...
Since b > 0, b is a positive
number
...
Therefore b, the positive value, is larger
...
d
...
If y < −1, column A is
1
greater
...
If 0 < y < 1, column
−4
1
1
A is greater
...
If y = 1,
1
ᎏᎏ
3
then both columns simplify to 1
...
51
501 Quantitative Comparison Questions
204
...
After canceling out the factors of x2y in column A, you are left
3
with ᎏ4ᎏ, which is equal to 0
...
205
...
The relationship cannot be determined
...
If y is greater than x, column A is
greater
...
If x and y are
equal, then the columns are equal
...
b
...
One
2
4
fourth is less than 1
...
207
...
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
...
Column B is greater
...
c
...
Column A and column B are
equal
...
b
...
A negative value raised to an odd numbered
power, like three, also results in a negative answer
...
210
...
Since a is greater than one, use the rules of exponents to
determine the larger value
...
a6 × a2 = a8
...
(a6)2 = a12
...
Column B is
greater
...
a
...
212
...
−a2 means a2 times −1, which will result in a negative number
...
Therefore
column B is larger
...
a
...
6, which is very close in value to 0
...
Since these values
3
are between zero and one, raising them to the third power will
be smaller than raising them to the second power
...
a
...
3 and d = 2
...
3 = 0
...
5 = 0
...
0
...
4
...
c
...
Now
column B says x + 8 − 4 which simplies to x + 4, making column
A equal to column B
...
c
...
Combine like terms to get 2x − 14 = 10
...
2x = 24
...
x = 12
...
217
...
In order to solve for x, add 4x to both sides of the equation
...
This simplifies to 10x + 3 = −7
...
Divide both sides by 10 and the
result is x = −1
...
Both columns equal 1
...
a
...
Subtracting three again on both
sides results in 11x − 3 = 36
...
4x
219
...
Multiply both sides of the equation by 3; 3(ᎏᎏ + 4) = 3(2x)
...
Subtract 4x from both sides of the
equation; 4x − 4x + 12 = 6x − 4x, which is 12 = 2x
...
The columns are equal because 6 × 6 =
36 and 62 = 36
...
b
...
Combine like
terms vertically to simplify to 2x = 10
...
Therefore 3 × 5 = 15 and π
(approximately equal to 3
...
221
...
Solving the equations for x and y by adding them together
vertically to simplify to 2x = 6
...
Substituting x = 3 into the first equation
results in 3 + y = 5, so y = 2
...
222
...
Solve for c by adding the equations together vertically to
simplify to 2c = 6
...
The columns are equal
...
a
...
Adding 1 to
both sides results in 0 + 1 = 5x − 1 + 1, which is equal to 1 = 5x
...
Substituting
x = 0 gives y = 5(0) − 1 which becomes y = 0 − 1 = − 1
...
5
224
...
Substitute the second equation into the first for x to get
3
3
8(ᎏ2ᎏ)y − 2y = 30; 8(ᎏ2ᎏ)y simplifies to 12y so the equation becomes
12y − 2y = 30
...
y = 3
...
5
...
225
...
x2 − y2 is the difference between two squares and factors to
(x − y)(x + y)
...
Column B is (x + y)(x + y) which is 16 ×
16 = 256
...
226
...
The degree of a quadratic term is 2 and the degree of a cubic
term is 3, so column B is greater
...
c
...
This is
the difference between two perfect squares which factors to
(x − 3)(x + 3) = 0
...
Since 3 is the positive solution for
x, the columns are equal
...
a
...
Setting each factor equal to zero is x − 5 = 0 or x + 2 = 0, which
results in a solution of 5 or −2
...
229
...
Multiplying using the distributive property on the left side gives
−y2 + 4y = 10
...
Since this equation states that 4y is equal to y2 + 10,
the columns have the same value
...
b
...
This
simplifies to y + 1 > 2x
...
54
501 Quantitative Comparison Questions
231
...
Take the compound inequality and add 2 to each section;
4 + 2 < 2x − 2 + 2 < 8 + 2
...
Dividing all sections by 2 gives a result of 3 < x < 5
...
232
...
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
...
233
...
Use the distributive property on the left side of the inequality
to get 6x − 6 > 30
...
6x − 6 + 6 > 30 + 6
...
Divide both
sides by 6 to get a result of x > 6
...
234
...
Add one to both sides of the inequality; −3x − 1 + 1 > 14 = 1
...
Divide both sides of the inequality
by −3 to get a solution of x > −5
...
Since x is less than −5, the
answer is column A
...
a
...
Therefore, the x-values are
greater than the y-values
...
236
...
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
1
1
the slope of the line in column B is ᎏ2ᎏ
...
237
...
This relationship cannot be determined
...
The opposite of any x-values, which are negative in
Quadrant II, would be positive
...
238
...
Using slope-intercept ( y = mx + b) form where m is the slope of
2
the linear equation, the slope of line l is ᎏ3ᎏ
...
The slope of line l is
equal to the slope of line m
...
c
...
Therefore, ᎏᎏ = ᎏ4ᎏ which
4−0
c−d
3
simplifies to ᎏ4ᎏ = ᎏ4ᎏ
...
Then, c − d = 3
...
240
...
From the equation y = mx + b, b is the y-intercept
...
The x-intercept is found by
substituting zero for y and solving for x
...
3
Dividing both sides by −2 gives ᎏᎏ or −1
...
Since −3 < −1
...
241
...
Converting to slope-intercept form, the first equation becomes
2y = 4x + 6 by adding 4x to both sides
...
The slope is 2
...
Dividing both sides by −3 to get y by itself results in
9
−3y
−3x
ᎏᎏ = ᎏᎏ + ᎏᎏ which simplifies to y = x −3
...
Therefore column A is greater
...
c
...
between (−3,4) and (0,0) is ͙(−3 − 0ෆ− 0)2 This simplifies
ෆ (4)2
ෆ)
ෆ
to ͙(−3)2 +ෆ = ͙(9 + 16ෆ = ͙25 = 5
...
243
...
“The difference between 26 and the product of 4 and 3”
translates to the expression 26 − (4 × 3), which simplifies to
26 − 12 = 14
...
14 is
larger than 10
...
a
...
The square root of
four is 2
...
245
...
If the square of a number is four, then the number is 2
...
41, which is less
than 4
...
c
...
Thus, column A is equal to column B
...
a
...
If one-half of y is z, let z = 2y
...
Using the given equation x + y + z = 35,
substituting gives an equation of x + 2x + 4x = 35
...
Since z = 4x then
z = 4(5) = 20
...
c
...
The distance of car A can be
expressed as 60t and the distance of car B can be expressed as
55t
...
60t + 55t = 460
...
115t = 460
...
The columns are equal
...
c
...
Two consecutive integers whose
sum is 83 are 41 and 42
...
The columns are equal
...
b
...
There can be 1 quarter and 9 dimes, or 3 quarters and 4
dimes
...
Either way, there are
more dimes than quarters, so the answer is column B
...
Figures: Figures that accompany questions are intended to provide information useful in answering the questions
...
are in the order shown; angle measures
are positive; lines shown as straight are straight; and figures lie in a plane
...
Common Information: In a question, information concerning one or both
of the quantities to be compared is centered above the two columns
...
Directions: Each of the following questions consists of two quantities, one
in Column A and one in Column B
...
if the quantity in Column A is greater
b
...
if the two quantities are equal
d
...
the sum of the measures
90°
of two acute angles
The answer is d
...
However, depending on their measures, column A could
be smaller (two 30-degree angles = 60 degrees) or larger (an 80degree and a 45-degree angle = 125 degrees) than column B
...
2
...
5
with radius 2
The answer is a
...
In column A, 2(π)(2) = 4π
...
5)2 = 2
...
4π is greater than 2
...
60
501 Quantitative Comparison Questions
Questions
Column A
Column B
5
4
Q
3
251
...
the slope of the line in
the graph
61
501 Quantitative Comparison Questions
Column A
Column B
2x
253
...
the volume of the box
on the left
the volume of the cylinder
on the right
62
501 Quantitative Comparison Questions
Column A
Column B
A
x
C
B
z
D
y
Lines AB and CD are parallel
...
x
z
10
6
x
256
...
6b
the sum of the interior angles of
the polygon above
63
501 Quantitative Comparison Questions
Column A
Column B
258
...
2″
and a width of 5″
that can be cut out of a
square piece of paper with
sides of 3
...
AC
260
...
Polygon DEFGH has
sides DE = 3 and GH = y and polygon LMNOP has sides
LM = 1 and OP = 2
...
D 100°
C
70°
E
135°
B
20°
180°
0°
F
O
A
261
...
the measure of ∠AOB
the measure of an acute angle
64
501 Quantitative Comparison Questions
Column A
Column B
263
...
the measure of an angle
the measure of an angle
supplementary to ∠FOE
supplementary to ∠BOC
265
...
2
1
4
5
8
n
3
m
6
7
t
266
...
the measure of ∠1
the measure of ∠5
268
...
the sum of the measures
the sum of the measures of
angles 2 and 3
of angles 5 and 8
270
...
the measure of ∠3
the measure of ∠6
The measure of ∠1 is 100°
...
the measure of ∠8
75°
273
...
the measure of ∠6
77°
65
501 Quantitative Comparison Questions
Column A
Column B
Use the following figure to answer questions 274–278
...
275
...
the measure of ∠16
277
...
the measure of ∠8
278
...
274
...
c
2x
The sum of the measures of ∠13 and ∠10 is 160°
...
66
501 Quantitative Comparison Questions
Column A
Column B
279
...
the sum of the interior
360
angles of this polygon
Use the following figure to answer questions 281–283
...
the sum of the interior
the sum of the exterior
angles of this polygon
angles of this polygon
282
...
the sum of the interior
the sum of the interior
angles of a hexagon
angles of this polygon
Use the following figure to answer questions 284–286
...
the sum of the interior
the sum of the interior angles
of this figure
angles of an 8-sided polygon
1
285
...
the area of this polygon
the area of a convex polygon
whose interior angles measure
900
...
A convex polygon has 5 sides
...
the measure of one of
the congruent angles
130°
Use the following figure to answer questions 288–289
...
288
...
the length of line segment
6
BC
A
B
E
C
D
290
...
the length of line segment PR
three times the length of line
segment SO
P
Q
25°
O
13
S
R
The figure is a rectangle
...
J
K
25°
N
M
L
The figure is a rhombus
...
25°
the measure of ∠JKN
293
...
the length of line
the length of line
segment JK
segment NK
69
501 Quantitative Comparison Questions
Column A
Column B
Use the following figure to answer questions 295–297
...
the measure of ∠AEB
the measure of ∠BCD
296
...
the ratio of
3
ᎏᎏ
4
the length of line segment AB
ᎏᎏᎏᎏ
the length of line segment DB
to the ratio of
the length of line segment AE
ᎏᎏᎏᎏ
the length of line segment DC
298
...
At noon, the
flagpole casts a shadow on the ground that is 6 feet long
...
The woman is also
perpendicular to the ground
...
the height of the flagpole
9 feet
70
501 Quantitative Comparison Questions
Column A
299
...
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 telescope
pointing at building A is 250 feet away from the base of the
building
...
the height of building A
The height of building B
Use the following figure to answer questions 300–302
...
the length of line segment TS
7
Recreate the diagram substituting 3 for the length of line
segment RT and 4 for the line segment TS
...
301
...
Recreate the diagram substituting 3 for the length of line segment
RT and 5 for the length of line segment RS
...
4 in
7 in
The figure is a rectangle
...
21 in
the perimeter of the figure
304
...
5 ft
305
...
25 sq ft
the area of this square if each side
is 6 in longer than indicated in
the diagram
307
...
the area of a square whose
the area of a rectangle with
length 2x and width x
sides have a length of 3x
72
501 Quantitative Comparison Questions
Column A
Column B
309
...
5 cm
3 cm
8 cm
310
...
the area of the parallelogram
42 sq cm
in the figure
Use the following figure to answer questions 312–313
...
42 sq in
the area of the triangle in the
figure
313
...
Column B
Mr
...
The pool will
be a rectangle with a length of 12 feet and a width of 8 feet
...
120 sq ft
315
...
She has
48 square inches of matting board to work with, and a stack
of five rectangular photos each measuring 3″ × 5″
...
the number of photos she
can mount
316
...
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
...
the perimeter of the Stop sign
317
...
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
...
3
ᎏᎏ
4
the area of the largest
circular cake that can be cut
out of the square cake
the total area of the
square cake
74
501 Quantitative Comparison Questions
Column A
Column B
318
...
2 in
the surface area of this cube
24 sq in
320
...
3 cm
6 cm
6 cm
100 cubic cm
the volume of the prism in the
diagram
75
501 Quantitative Comparison Questions
Column A
Column B
322
...
the volume of a cube with
twice the volume of a cube with
an edge of 2 meters
an edge of 4 meters
Use the following figure to answer questions 324–325
...
the volume of this pyramid
1,200 cubic in
325
...
h = 5 cm
51 cm2
80 cm3
326
...
the volume of the pyramid
the volume of a regular pyramid
with area of the base = 60 cm2
and height = 5 cm
in the diagram if the height
is lengthened to 6 cm
328
...
the circumference of a circle
21 cm
with diameter d = 7 cm
330
...
4 m
circumference C = 25
...
10 ft
331
...
the area of the circle in the diagram
15
...
Column B
A circle has a radius of r = 4 in
...
the area of the circle
A circle has a diameter of d = 8 cm
...
1m
12 m
82 cm2
the total surface area of the
cylinder in the diagram
336
...
9 cm
4 cm
1,350 cm3
the volume of the cylinder
in the diagram
78
501 Quantitative Comparison Questions
Column A
Column B
338
...
A cylinder has a radius r = 5 cm and height h = 10 cm
...
the volume of this cylinder
A cone has a radius r = 5 cm and height h = 10 cm
...
341
...
16 sq cm
Use the following figure to answer questions 342–343
...
the surface area of this sphere
615 sq m
343
...
the volume of a sphere with
the volume of a sphere with
a radius of 3 ft
a radius of 2 ft
345
...
y
x
1
−ᎏ3ᎏ
the slope of the line in
the diagram
347
...
y
x
2
ᎏᎏ
3
the slope of the line in
the diagram
349
...
350
...
0
352
...
He knows the hill rises along a line measured
to have a slope of ᎏ2ᎏ
...
35 ft above sea level
the vertical elevation of the peak
of the hill
81
501 Quantitative Comparison Questions
Column A
Column B
Use the following figure to answer questions 353–354
...
the y-intercept of the line in
0
the diagram
354
...
y
x
355
...
the y-intercept of the line in
the y-intercept of a line defined
by the equation y = x − 1
the diagram
82
501 Quantitative Comparison Questions
Column A
Column B
357
...
the y-intercept of a line
the y-intercept of a line
defined by the equation
y = 9x − 4
defined by the equation
y = x + ᎏ2ᎏ
5
359
...
the y-intercept of a line
the y-intercept of a line
defined by the equation
6=x−y
defined by the equation
3
y = ᎏ2ᎏx − 6
361
...
the y-intercept of a line
the y-intercept of a line
defined by the equation
xy = x2 + 4x
defined by the equation
2y = 10
Use the following figure to answer questions 363–366
...
363
...
The measure of ∠X is 45°
...
The circumference of circle O is 10 cm
...
the length of arc ABC
365
...
The measure of ∠X is 45°
...
The length of arc ABC is 3 cm
...
the measure of ∠x
35°
367
...
the sum of the interior
the sum of the interior angles
of an isosceles right triangle
angles of a 30-60-90 triangle
369
...
the hypotenuse of a right
the hypotenuse of a right
triangle with shorter sides
of length 6 and 8
triangle with shorter sides
of length 3 and 4
Use the following figure to answer questions 371–372
...
the measure of ∠3
the measure of ∠1
372
...
A line is represented by the equation y = 3x + 2
...
Column B
ΔABC is a right triangle with shorter sides of 3 and 6
...
the ratio of the lengths of
the two triangles’ hypotenuses
1
ᎏᎏ
2
375
...
The slices are all cut starting at the center of the pie out to
the edge of the crust
...
5″
85
501 Quantitative Comparison Questions
Answer Explanations
The following explanations show one way in which each problem can be
solved
...
1
251
...
The area of a triangle = ᎏᎏbh
...
A = ᎏ2ᎏ(3(4)) = ᎏ2ᎏ(12) = 6
...
change in y
252
...
The slope of a line is defined as ᎏᎏ
...
This creates a rise (change in the y-value) of 3 and a run
3
(change in the x-value) of 2, which gives a slope of ᎏ2ᎏ
...
253
...
The area of a circle is defined as A = πr2
...
The area of a circle with diameter 3y would be
a = πr2 = (3y)2π = 9y2π
...
If, for instance, x = 5 and
y = 2, then quantity A would be greater
...
So there is not enough
information to evaluate the equations and the answer is d
...
b
...
So the volume
of the box at left = 4(6)x or 24x
...
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
...
78
...
255
...
When two parallel lines are cut by a line segment, the resulting
corresponding angles created are equal, so x = y
...
However, no
information is given about the relationship between x and z
other than the fact that they add up to 180
...
Therefore there is not enough information to further
evaluate the problem and the answer is d
...
a
...
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
...
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
...
b
...
Therefore, 6b = 6(80°) = 480°
...
The polygon at the
right can be divided into 3 triangles, so 3(180°) = 540°
...
258
...
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
...
This is because the radius extends
equally from the center of the circle in all directions
...
2″, which is smaller than the sides of the square in
column A (3
...
Thus a circle with a larger radius—and,
therefore, greater area—could be cut from the square in
column A
...
87
501 Quantitative Comparison Questions
259
...
Sketching ΔABC will be helpful here
...
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, x͙3 and 2x
...
) 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
...
Setting
ෆ
2͙3 = x͙3, and dividing both sides of the equation by ͙3
ෆ
ෆ
ෆ
yields x = 2
...
Thus the values of
column A and column B are the same, and the answer is c
...
d
...
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
...
When taking the test, be sure to
read through the problems before spending time sketching
shapes and solving equations
...
Note: If you were told that the polygons were, in fact, similar,
the problem could be solved
...
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 =
3
1
LM/OP
...
Since x > 8, x must be greater than y and therefore the answer
would be a
...
a
...
The
measure of any right angle is always 90°
...
262
...
The measure of ∠AOB as indicated in the diagram is 20°
...
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
...
263
...
A reflex angle is defined as an angle whose measure is between
180° and 360°
...
This is smaller
than a reflex angle, so the answer is a
...
b
...
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°
...
The supplement to ∠FOE is larger, so the answer is b
...
c
...
∠AOF as indicated in the diagram measures 180° and ∠AOD
measures 100°
...
This is the same as
the sum in column B, so the answer is c
...
c
...
Vertical angles are defined
as angles, formed by 2 intersecting lines, which are directly
across or opposite from each other
...
The answer is c
...
c
...
Corresponding
angles are defined as the angles on the same side of the
transversal and either both above or below the parallel lines
...
268
...
Angles 7 and 3 are corresponding and therefore congruent
...
269
...
Angles 5 and 8 are supplementary because they combine to
form a straight line
...
Supplementary angles always add up to 180°, so the measures of both
sets of angles are the same and the answer is c
...
c
...
However, angles 2 and 6 are corresponding, so their measures
are equal
...
This information can be used to determine that angles 2
and 8 are congruent because m∠2 = m∠6 = m∠8
...
Alternate exterior
angles are always congruent, so the answer is c
...
d Angles 3 and 6 are same side interior angles
...
Same side interior angles are always supplementary,
so their measures add up to 180°
...
Therefore, there is not enough
information to solve the problem and the answer is d
...
b
...
This means that they
are both outside the parallel lines and on the same side of the
transversal
...
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°
...
273
...
Angles 3 and 6 are same side interior angles, which means that
they are supplementary
...
The answer is b
...
b
...
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
...
b
...
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
...
So the measure
of ∠3 is 180° − 65° = 115°
...
So the measure of ∠11 is also
115°, which is greater than column A
...
276
...
The measure of ∠9 is information you actually don’t need to
solve this problem
...
Therefore, the answer is c because their
measures are the same
...
b
...
So the measure of ∠8 is x; x indicates
the measure of an angle, so it cannot be negative
...
278
...
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
...
1
So the measure of each angle is ᎏ2ᎏ of their sum, or 80°
...
Now find the measure of ∠4
...
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 answer is b
...
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
...
279
...
This polygon has four sides, making it a quadrilateral
...
A quadrilateral has more sides than a
triangle, so the answer is a
...
c
...
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
...
a
...
The exterior sum of any
convex polygon is always 360, so the answer is a
...
c
...
This polygon also has
seven sides, so the values in the two choices are equal
...
283
...
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
...
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
...
c
...
The answer is c
...
a
...
In either case, the answer is a
...
d
...
Though the math is a bit involved, you do have enough
information to determine its area
...
Therefore this problem cannot be fully solved and the answer
is d
...
a
...
Right angles
measure 90° each, so subtracting the sum of the three right
angles from 540° leaves 270° for the two remaining congruent
angles
...
The answer, therefore, is a
...
a
...
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°
...
Since
130° > 50°, ∠A > ∠B, so the answer is a
...
c
...
The length of line
segment AD is 6, so the length of BC must be 6, as well
...
290
...
This question modifies the diagram to show line segments AC
and BD, which are the diagonals of the parallelogram
...
Line segments AE and CE are therefore the two
halves of line segment AC, and have equal lengths
...
291
...
The figure shows a rectangle with diagonals PR and SQ that
intersect at point O
...
Three
–––
–––
times the length of SO, however, is longer than PR, so the
answer is b
...
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
...
c
...
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 answer, then, is c
...
a
...
90° > 80°, which is the value in
column B, so the answer is a
...
b
...
This means that
ΔJNK is a right triangle
...
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°)
...
295
...
∠AEB is indicated to have a measure of 85°
...
The values of the two choices
are therefore equal and the answer is c
...
d
...
However, no
information is given to indicate the specific length of any side of
either triangle
...
There is not enough information to solve this
problem, and the answer is d
...
a
...
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
...
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
...
a
...
5 ft
3 ft
6 ft
shadow
shadow
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
...
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:
x
ᎏᎏ
length of shadow
x
ᎏᎏ
6
=
height of woman
ᎏᎏ
length of shadow
5
= ᎏ3ᎏ
3x = 30
x = 10
The flagpole is 10 feet tall, which is greater than the value in
column B
...
299
...
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
...
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
96
501 Quantitative Comparison Questions
their respective telescopes
...
Let b represent the
height of building B in feet:
height of building A
ᎏᎏᎏ
distance to telescope A
1000
ᎏᎏ
250
=
b
ᎏᎏᎏ
distance to telescope B
b
= ᎏᎏ
300
250b = 3,000,000
b = 1,200
The height of building B is 1,200 feet
...
Note: It is not necessary to solve for b because it is evident that
b > 1,000 from looking at the proportion
...
a
...
Note that in the
equation a2 + b2 = c 2, 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
...
301
...
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
...
Therefore the answer is a
...
d
...
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
...
303
...
The figure is a rectangle
...
The answer is b
...
c
...
The answer is c
...
b
...
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
...
98
501 Quantitative Comparison Questions
306
...
The formula for calculating the area of a square is A = s2, where
s is the length of any one side
...
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
...
The
answer is b
...
c
...
The formula p = 4s can be
used to calculate the perimeter of this square as 4(5) = 20 in
...
308
...
The formula A = s2 can be used to calculate the area of this
square as A = (3x)2 or 9x2
...
309
...
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
...
a
...
The base is longer, so the answer is a
...
The height of a
parallelogram is sometimes referred to as its altitude
...
b
...
In this case, A = 8(3) = 24 sq cm, which is less than the
value in column B
...
1
312
...
The area of a triangle is calculated using the formula A = ᎏᎏbh,
2
where b is the base of the triangle and h is the height
...
The area of
the triangle is 21 sq in, which is less than the value in column A
...
313
...
The figure indicates that the base of this triangle is 7 cm
...
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:
1
A = ᎏ2ᎏbh
1
35 = ᎏ2ᎏb(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
...
314
...
To find the area of Mr
...
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
...
315
...
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″
...
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
...
100
501 Quantitative Comparison Questions
316
...
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 equilateral triangle has three sides of equal
length, so this special Yield sign has a perimeter of P = ns =
3(26″) = 78″
...
80″ > 78″,
so the Stop sign has a greater perimeter and the answer is a
...
a
...
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
...
5″
...
5″ and an area
calculated using the formula A = πr2, where r is the radius and
π = 3
...
5)2
A = 30
...
25(3
...
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,
3
so the area is 121 sq in
...
75 sq in
...
99 sq in > 90
...
3
ᎏᎏ
4
101
501 Quantitative Comparison Questions
318
...
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
...
319
...
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
...
320
...
To calculate the surface area of a rectangular prism, the length,
width, and height of the prism must be known
...
321
...
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 is 108 cm3
and so the answer is b
...
b
...
The
volume of the prism in the diagram, then, is V = Bh = 30(5) =
150 m3
...
Column B therefore yields a greater value and the answer is b
...
a
...
324
...
The volume of a pyramid is calculated using the formula
1
V = ᎏ3ᎏlwh
...
The volume of the pyramid is 400 cubic in,
so the answer is b
...
b
...
The value of column B is V = ᎏ3ᎏlwh = ᎏ3ᎏ(9 × 9 × 12)
1
= ᎏ3ᎏ(972) = 324 cubic in
...
326
...
The area of the base of the pyramid is given in the diagram, so
1
calculate the volume with the formula V = ᎏ3ᎏBh:
1
V = ᎏ3ᎏBh
1
V = ᎏ3ᎏ(51 × 5)
1
V = ᎏ3ᎏ(255)
V = 85 cm3
The value of column A is greater, so the answer is a
...
a
...
The value of column B is calculated as
3
1
1
1
V = ᎏ3ᎏBh = ᎏ3ᎏ(60 × 5) = ᎏ3ᎏ(300) = 100 cm3
...
328
...
The circumference of a circle is calculated using the formula
C = 2πr or C = πd
...
So the answer is b
...
a
...
14:
C = πd
C = (3
...
98 cm
21
...
330
...
Use the formula C = πd to find the diameter in column B:
C = πd
25
...
12 m = 3
...
4 m = 2(3
...
331
...
The diagram indicates that the circle has a diameter of 10 ft
...
332
...
The area of the circle is calculated using the formula A = πr2,
where r is the radius of the circle
...
Using the formula yields A = πr2 = 3
...
14(25) =
78
...
The area of the circle is 78
...
333
...
Any figure must be smaller than any other figure it can fit inside
of
...
334
...
Before doing any calculations, take a good look at both choices
to note all of the important information provided
...
Column B refers to
a circle cut out of a square of fabric with a total area of 16 cm2
...
The largest circle that could be cut out
of that square, then, would have a diameter of 4 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
...
335
...
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
...
14) + 24(3
...
28 + 75
...
64 cm2
The total surface area of the cylinder is 81
...
336
...
The surface area of this cylinder is calculated as:
SA = 2πr2 + 2πrh
SA = 2π72 + 2π7(8)
SA = 98π + 112π
SA = 98(3
...
14)
SA = 210(3
...
4 cm2
The surface area of the cube in column B is calculated as
SA = 6e2 = 6(92) = 6(81) = 486 cm2; 659
...
105
501 Quantitative Comparison Questions
337
...
The volume of a cylinder is calculated using the formula
V = πr2h, where h is the height of the cylinder
...
36 cm3
This is less than the value of column B, so the answer is b
...
a
...
Doubling the radius of a
cylinder will change the volume more significantly because of
the squaring involved, so the answer is a
...
b
...
340
...
The volume of a cone is calculated using the formula
1
V = ᎏ3ᎏ(πr2h), where h is the height of the cone
...
67 cm3
300 cm3 > 261
...
341
...
The surface area of a sphere is calculated using the formula
SA = 4πr2
...
14)(62)
SA = 12
...
16 sq cm
The two values are equal, so the answer is c
...
a
...
In this case, the formula simplifies to SA = 4πr2
= 12
...
44 sq m
...
343
...
The volume of a sphere is calculated using the formula
4
V = ᎏ3ᎏ πr3
...
14)(73)
4
V = ᎏ3ᎏ(3
...
02)
V = 1,436
...
03 cm3, which is less than
2,000 cm3, so the answer is b
...
b
...
The sphere in column B has the
greater radius, so it will also have the greater volume
...
the change in the y-value of the line
345
...
The slope of a line is defined as ᎏᎏᎏᎏ
...
the change in the y-value of the line
346
...
The slope of a line is defined as ᎏᎏᎏᎏ
...
the change in the y-value of the line
347
...
The slope of a line is defined as ᎏᎏᎏᎏ
...
the change in the y-value of the line
348
...
The slope of a line is defined as ᎏᎏᎏᎏ
...
0−2
349
...
The line’s slope is ᎏᎏ = −1; 0 > −1, so the answer is a
...
b
...
−2
2 2
107
501 Quantitative Comparison Questions
351
...
The line y = −1 is a horizontal line
...
352
...
The problem gives enough information to calculate the
elevation of the peak of the hill
...
the change in the x-value of the line
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
...
40 ft > 35 ft, so the answer is b
...
a
...
354
...
As determined in the previous problem, the y-intercept of the
line in the diagram is 1
...
Thus, the y-intercept of this line is 4 and the
answer is b
...
a
...
Even though the exact value cannot be
determined, it is somewhere between 3 and 4 which is clearly
greater than zero
...
356
...
Even though the exact y-intercept of the line in the diagram
cannot be determined, it is clearly greater than zero and
therefore positive
...
is in slopeintercept 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
...
b
...
The
y-intercept of the line in column A is −6, and in column B it’s 4,
so the answer is b
...
a
...
The
2
y-intercept of the line in column A is ᎏ5ᎏ, and in column B it’s −4,
so the answer is a
...
a
...
Dividing both sides
1
of the equation by 2 yields the proper y = mx + b form as y = ᎏ2ᎏx
+ 4, so the y-intercept is 4
...
360
...
The equation in column B must be put into slope-intercept
form in order to determine its y-intercept
...
The y-intercept of the line in
column A is also −6 so the answer is c
...
d
...
Simplifying yields
1
y = −ᎏ4ᎏx, so the y-intercept is 0
...
The problem has no solution and so the answer
is d
...
a
...
The first
equation simplifies to y = 5, which represents a horizontal line
that has a y-intercept of 5
...
The y-intercept of
this line is 4, so the answer is a
...
a
...
In this
case, the formula is set up as follows:
x
length = ᎏᎏ(C)
360
45
length = ᎏ6ᎏ(20)
3 0
1
length = ᎏ8ᎏ(20)
length = 2
...
5 cm > 2 cm, so the answer is a
...
c
...
In this
case, the formula is set up as follows:
length = ᎏxᎏ(C)
360
3
length = ᎏ660 (10)
3ᎏ
length = ᎏ1ᎏ(10)
10
length = 1 cm
The values in both choices are equal, so the answer is c
...
a
...
In this case, the
formula is set up as follows:
x
length = ᎏᎏ(C)
360
45
2 = ᎏ6ᎏ(C)
3 0
1
2 = ᎏ8ᎏ(C)
C = 16 cm
20 cm > 16 cm, so the answer is a
...
b
...
In this case, the formula is set up as
follows:
x
length = ᎏᎏ(C)
360
x
3 = ᎏᎏ(30)
360
30x
3 = ᎏ6ᎏ
3 0
1,080 = 30x
36 = x
36° > 35°, so the answer is b
...
a
...
The area
of the square in column B is 22 = 4 cm2
...
c
...
The
values in both choices are equal so the answer is c
...
b
...
The
volume of the cylinder in column B is πr2(h) = 3
...
14(4)(3) = 3
...
68 cm3; 37
...
370
...
These triangles are 3, 4, 5 and 6, 8, 10 Pythagorean triples,
which means that the hypotenuses are 5 and 10, respectively
...
Either method will find that
the hypotenuse of the triangle in column B is longer
...
371
...
Angles 1 and 3 are vertical angles created by the intersection of
two lines, and so are equal
...
372
...
Angles 1 and 2 are exterior angles on the outside of two parallel
lines cut by a transversal, and so are supplementary
...
All that can be
determined is that the two angles add up to 180°, which is not
enough information to solve the problem
...
373
...
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
...
The slope is the greater quantity, so the answer is a
...
a
...
The
triangles are similar, as the lengths of shorter sides of triangles
are proportional
...
375
...
The pie is a circle, so it measures 360° around
...
The formula length = ᎏᎏ(C) can be used to
360
determine the length of the arc of one of the slices by
substituting the information given in the problem as follows:
x
length = ᎏᎏ(C)
360
40
length = ᎏ6ᎏ(18)
3 0
1
length = ᎏ9ᎏ(18)
length = 2″
The length of the arc formed by one slice of pie is 2″; 2″ > 1
...
112
4
Data Analysis
In this chapter, the following math concepts will be the subject of the 126
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:
Numbers: All numbers used are real numbers
...
Unless otherwise indicated, positions of points, angles, regions, etc
...
Unless a note states that a figure is drawn to scale, you should NOT solve
113
501 Quantitative Comparison Questions
these problems by estimating or by measurement, but by using your knowledge of mathematics
...
A symbol that appears in both columns represents the same thing in Column A
as it does in Column B
...
Compare the two quantities and
choose:
a
...
if the quantity in Column B is greater
c
...
if the relationship cannot be determined from the information given
Examples:
Column A
Column B
1
...
Remember to look carefully at the two
columns, even if they initially appear to be the same
...
In
column A, ͙32 + 42 is equal to ͙9 + 16 or ͙25 which is equal to
ෆ
ෆ,
ෆ,
5
...
7 is greater than 5, so the
ෆෆ
ෆ
correct answer is b
...
2
...
The only information you are given about
the two quantities is that they are integers, which tells you nothing
about their respective values
...
Since you cannot determine which
value is greater, the answer is d
...
the average (arithmetic
the average (arithmetic
mean) of 11, 2, and 8
mean) of 6, 5, 8, 7, and 9
377
...
the median of set A
the mode of set A
The mean of set B is 17
...
x
y
C = 3, 6, 11, 12, 10, 18, x
The mean of set C is 9
...
8
x
Use the following figure to answer questions 381–383
...
number of freshmen
22
382
...
number of seniors
120
115
501 Quantitative Comparison Questions
Column A
Column B
383
...
...
5,
...
25, 1
...
75, 2,
...
the 53rd term of the sequence
13
385
...
5
Use the following series to answer questions 386 and 387
...
+ 98 + 100
386
...
the sum of all the terms
2,550
in the series
Use the information below to answer questions 388–400
...
the mean of set D
the mean of set E
389
...
the mean of set D
391
...
the median of set E
x > 17 > y
the median of set D
393
...
the median of set D
the median of set E
116
501 Quantitative Comparison Questions
Column A
Column B
x=y
394
...
the median of set D
the median of set E
x
396
...
the range of set D
the range of set E
The means of the two sets are equal
...
x
y
399
...
x
400
...
12
x
Use the information below to answer questions 401–402
...
the mean of set F
the median of set G
402
...
A coin is tossed 3 times
...
the number of possible
the number of possible
outcomes containing
exactly 1 tail
outcomes containing
exactly 2 heads
404
...
the probability of tossing
3 tails
Use the following bar graph to answer questions 406–408
...
12
difference between the number
of students with brown hair and
those with black hair
407
...
total number of students
90
surveyed
Use the following experiment to answer questions 409–411
...
409
...
number of outcomes in
3
which an even number is
rolled on the cube, and a
head is tossed on the coin
118
501 Quantitative Comparison Questions
Column A
Column B
411
...
A coin is tossed 14 times
...
The second and thirteenth tosses are heads
...
maximum number of
8
heads that can occur in a row
413
...
−20 + −18 + −16 +
...
the sum of all terms in
50
the series
415
...
three terms
416
...
1, 1, 2, 3, 5, 8, 13, 21, 34,
...
the 11th term in the sequence
90
418
...
William bought 4 pairs of pants for $80
...
He spent an average of $22
...
119
501 Quantitative Comparison Questions
Column A
Column B
419
...
$30
average cost of each of the first
four pairs of pants
the cost of the fifth pair of pants
Use the graph below to answer question 421–423
...
percent of budget spent
percent of budget spent
on Clothing and Food
on Housing
422
...
percent of budget spent on
percent of budget spent on
Housing, Clothing and Savings
Food, Auto, and Other
Use the following situation to answer questions 424–426
...
Papa’s has 8 total toppings from which to choose
...
total number of possible
8
1 topping pizzas Papa’s makes
425
...
number of possible pizzas
containing pepperoni
8
120
501 Quantitative Comparison Questions
Column A
Column B
426
...
427
...
Set H contains five positive integers such that the mean, median,
mode, and range are all equal
...
428
...
10
the largest possible number in
set H
Use the following situation to answer questions 430 and 431
...
The range of the numbers in set J is 0,
and the sum of the numbers in set J is 40
...
highest element in set J
lowest element in set J
431
...
K = {8, x, y, 10}
The mean of set K is 12, there is no mode, and x > y
...
x + y
24
433
...
Age of College Freshmen
13 17 18
53
121
501 Quantitative Comparison Questions
Column A
Column B
434
...
quartile 3
median
Susan has 3 pairs of pants and 4 shirts
...
436
...
A card is drawn from a standard deck of 52 cards
...
probability of drawing a queen
probability of drawing a club
438
...
probability of drawing a
probability of drawing a heart
or a face card
black card
440
...
A bag contains 6 blue marbles and 4 red marbles
...
441
...
probability of drawing a
red followed by a blue marble
probability of drawing a
blue followed by a red marble
Use the following experiment description to answer questions
443 and 444
...
Two marbles
are selected at random, with the first selected marble being
replaced in the bag before the second marble is drawn
...
probability of drawing
probability of drawing
2 red marbles
2 blue marbles
444
...
A bag contains 3 blue marbles and 2 red marbles
...
445
...
probability of selecting a
probability of selecting a blue
marble, then a red marble
if the first selected marble is
not replaced
blue marble, then a red
marble if the first selected
marble is replaced
447
...
blue
green
yellow
red
The spinner above is spun twice, then a 6-sided number cube (die)
is rolled
...
number of outcomes
14
449
...
1
the outcome: blue, red, 4
123
501 Quantitative Comparison Questions
Column A
Column B
450
...
probability of obtaining
the outcome: not green, red,
factor of 2
probability of obtaining
the outcome: not blue, not
blue, multiple of 3
Use the frequency distribution table below to answer questions 452–455
...
mean of distribution
2
...
median of distribution
2
454
...
range of distribution
5
Use the following set to answer questions 456–458
...
456
...
range of set K
10
458
...
2, 4, 8, 16, 32,
...
the tenth term of the sequence
210
460
...
64, 32, 16, 8, 4,
...
the seventh term of the
0
sequence
2−20
462
...
A survey of high school seniors at Blake High School revealed that 35%
play an instrument, 43% participate in sports, and 29% do neither
...
percent of students who
78%
play an instrument and/or
participate in sports
464
...
M = {4, 6, 3, 7, x, x}
N = {1, 5, 8, y}
x > 0, y > 0
x>y
465
...
the median of set M
the median of set N
Use the following box plot to answer questions 467–469
...
number of data elements
number of data elements
between third quartile
and max
between first quartile
and median
468
...
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
...
470
...
number of sandwiches that
number of sandwiches that
can be made on wheat bread
can be made on rye bread
472
...
1, 4, 9, 16, 25,
...
the seventh term in
36
the sequence
474
...
the difference between the
the difference between the
10th and 13th terms
34th and 35th terms
1 × 109
476
...
50th percentile of data set W
average (arithmetic mean) of data
set W
478
...
25th percentile of data set W
first quartile of data set W
Use the sets below to answer questions 480 and 481
...
standard deviation of set P
standard deviation of set Q
481
...
100 meters
1 kilometer
483
...
46 × 109 miles
34
...
1
...
00000015 millimeters
Use the following figure to answer questions 485–487
...
profit increase from
profit increase from
1987 to 1989
1988 to 1990
486
...
overall profit from
$1,500
1987 to 1993
Use the following sequence to answer questions 488–489
...
488
...
the sixth term of the sequence
720
Use the following series to answer questions 490–491
...
+ 102 + 112 + 122 + 132
490
...
the sum of the series
1,000
492
...
23 × 10−5
...
234 milliliters
2
...
45 centigrams
...
12 square yards
36 square feet
496
...
1 cubic foot
123 cubic inches
128
501 Quantitative Comparison Questions
Column A
Column B
Use the frequency distribution below to answer questions 498–501
...
number of 17-year-olds
number of 27-year-olds
entered in race
entered in race
499
...
median age of entrant
30
501
...
You may have another method for solving these problems
...
c
...
35 divided by 5 = 7
...
21 divided by 3 = 7
...
b
...
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
...
b
...
5: 13, 15, 20, 20
...
5
...
20 is greater than 17
...
379
...
The relationship cannot be determined
...
The total of the given elements is
−5 + −1 + 12 + 29 = 35
...
There is no way to know which of the
two might be greater, or if the two are equal
...
a
...
The total of the
given elements is 3 + 6 + 11 + 12 + 10 + 18 = 60
...
8 is
greater than 3
...
d
...
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
501 Quantitative Comparison Questions
382
...
Since there are 400 students at Brown High School and 29%
are seniors, 29% of 400, or
...
383
...
The difference between the number of freshmen and
sophomores is 24% − 22% = 2% of 400;
...
The difference between the number of juniors and seniors is
29% − 25% = 4%
...
16 > 8
...
a
...
25,
...
75, 1, 1
...
5
...
Since each term given is
equivalent to the term number (its number in the sequence)
n
divided by 4, an nth term would be equal to ᎏ4ᎏ
...
25; 13
...
385
...
Use the same equation you established in the previous problem
...
5
...
5, so the
correct answer is c
...
a
...
The term is equal to two times the term number
...
In
series questions, an easy shortcut is to rewrite the series below
the original series and add vertically to get consistent sums
...
+ 44 + 46
46 + 44 + 42 +
...
+ 48 + 48
Since 23 terms were added, there are 23 totals of 48
...
387
...
Following the same procedure as in the previous question, the
100
last term is 100, so there are ᎏ2ᎏ, or 50 terms in the series
...
There would be 50 sums of 102 which
102
would again be double the total needed; 50 × (ᎏ2ᎏ) = 50 × 51
= 2,550
...
d
...
Since the value of neither x
nor y is known, the means of the sets cannot be found and
cannot, therefore, be compared
...
d
...
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
...
If 17 < x < 22, the order of the set would
be 13, 17, x, 22, 24 and the median would be x
...
a
...
When the sums are divided by 5 (the number
of elements in each set), the mean of set D will be a larger
number
...
c
...
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
...
a
...
The median will therefore be one of these
elements: x which is > 17, 22, or 24
...
393
...
The answer cannot be determined
...
Therefore, x might be 9 since that would
enable the range to be 24 − 9 = 15
...
y might have the same value, which
would make the medians equal, but y could also equal 28
because 28 − 13 = 15 as well
...
394
...
If x = y, then set D = set E, and it follows that the mean of set D
= the mean of set E
...
c
...
132
501 Quantitative Comparison Questions
396
...
Since x < y, the sum of the elements in set D < the sum of the
elements in set E
...
397
...
The answer cannot be determined
...
Then
the range of set D = 24 − 5 = 19 and the range of set E = 32 − 13
= 19
...
Now let x = 12 and y = 15
...
Since in both cases,
x < y, the answer cannot be determined
...
c
...
In order for this to happen,
x must equal y
...
a
...
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
...
b
...
This implies that x and y are equal to one of the
elements already shown to be in sets D and E
...
150
401
...
The mean of set F is 28 + 29 + 30 + 31 + 32 = 150; ᎏᎏ = 30
...
402
...
Standard deviation is a measure of the spread of the data from
the mean
...
403
...
The number of possible outcomes containing exactly 2 heads is
3: HHT, HTH, and THH
...
133
501 Quantitative Comparison Questions
404
...
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
...
405
...
Only one of the 8 possible outcomes, TTT, contains three tails
1 1
1
so the probability of tossing three tails is ᎏ8ᎏ; ᎏ8ᎏ < ᎏ2ᎏ
...
b
...
15 > 12
...
a
...
The actual percents do not
have to be determined
...
b
...
409
...
There are 6 possible outcomes on the number cube, {1, 2, 3, 4,
5, 6}, and 2 on the coin, {H, T}
...
12 > 8
...
c
...
There are three outcomes, so columns A and B are equivalent
...
b
...
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)
...
412
...
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
...
134
501 Quantitative Comparison Questions
413
...
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
...
414
...
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
...
50 > 46
...
b
...
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
...
a
...
The last two terms are 22 and 24; 22 + 24 = 46
...
417
...
This is the Fibonacci sequence in which each term after the first
two is found by adding the previous two terms
...
The tenth term is found be
adding the eighth and ninth terms: 21 + 34 = 55, so the tenth
term is 55
...
Since 90 > 89, column B is greater
...
c
...
419
...
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
...
50, so it must have cost more
than $20
...
b
...
50 per pair, William’s
total cost must be $22
...
50
...
50 − $80
= $32
...
421
...
$967 was spent on Housing
...
Since more money was spent on Food and
Clothing, this accounts for a higher percent of the budget
...
a
...
The percent spent on Savings is ᎏᎏ = about
$3,462
12%
...
88% > 86%
...
c
...
424
...
Since there are 8 toppings available, a one-topping pie would
consist of any one of these 8 toppings
...
425
...
One 1-topping pizza would contain pepperoni
...
Therefore, there are a total
of eight possible pies with pepperoni as a topping
...
a
...
Call the available toppings A, B, C, D, E, F, G,
and H
...
B can be paired with any of the remaining 6 toppings
starting with C because it’s already been paired with A
...
And so on
...
The total number of pies
Papa’s offers is 1 + 8 + 28 = 37; 37 > 12
...
b
...
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
...
So, the largest one of the
integers can be is 40
...
b
...
Since the mean, median, mode, and range are all equal,
they are all equal to 5
...
429
...
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
...
To
136
501 Quantitative Comparison Questions
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
...
Another is {3, 4, 5, 5, 8}
...
c
...
Therefore the
highest element in set J is equal to the lowest element in set J
...
b
...
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
...
a
...
The two given elements, 8 and 10 total
18
...
The two remaining elements, x and y
must total 30, which is greater than 24
...
a
...
Therefore, 2x > 30
...
b
...
Therefore, Quartile 1 <
Quartile 3
...
d
...
The median may be 17 or
18, which would make it less than, or equal to, Quartile 3
...
b
...
437
...
There are 4 queens in a deck, so the probability of drawing a
4
queen is ᎏᎏ
...
52 52
52
438
...
There are 4 jacks and 13 spades, but one of these is the jack of
spades which cannot be counted twice
...
The
16 16
17
probability of drawing one of these cards is ᎏᎏ; ᎏᎏ < ᎏᎏ
...
a
...
There are 13
5ᎏ
hearts and 12 face cards, but three of these face cards are hearts:
the jack, queen and king of hearts
...
The probability of drawing
22 26
22
one of these cards is ᎏ2 ; ᎏᎏ > ᎏᎏ
...
a
...
The probability of drawing
10
one of these cards is ᎏ2
...
52 52
52
441
...
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
5
of drawing the second blue marble is ᎏ9ᎏ 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
6
5
30
bag); ᎏᎏ × ᎏ9ᎏ = ᎏᎏ
...
The probability of
4
drawing the first red marble is ᎏᎏ because there are 4 red
10
marbles out of the 10 total marbles in the bag
...
1ᎏ
90 90
9
442
...
The probability of drawing a red followed by a blue marble is
4
ᎏᎏ
10
6
24
ᎏ
× ᎏ9ᎏ = ᎏ0
...
1ᎏ
90
138
501 Quantitative Comparison Questions
443
...
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
14
14
1 6
8
8
64
64
36
red marbles is ᎏᎏ × ᎏᎏ = ᎏ9ᎏ; ᎏ9ᎏ > ᎏ9ᎏ
...
c
...
The probability of drawing a blue, followed by a
14
1 6
6
8
48
red marble is ᎏᎏ × ᎏᎏ = ᎏ9ᎏ
...
a
...
36)
...
30);
...
30
...
b
...
24)
...
30);
...
24
...
b
...
16)
...
30);
...
16
...
a
...
Therefore the
experiment has 4 × 4 × 6 = 96 possible outcomes; 96 > 14
...
b
...
The probability of obtaining this outcome is ᎏᎏ;
96
1 1
1
...
c
...
The probability of obtaining
6
one of these outcomes is ᎏᎏ
...
The probability of
6
obtaining one of these outcomes is also ᎏᎏ
...
b
...
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
...
The probability of
6
obtaining one of these 6 outcomes is ᎏ6
...
” 3 × 3 × 2 = 18
...
96 9ᎏ
96
452
...
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)
...
There are 3 + 6 + 2 + 1 + 2 = 14 pieces of
data displayed in the table (the sum of the frequencies)
...
5; 2
...
5
...
b
...
The table contains the data
in numerical order, so the first three elements are zeros and the
next six are ones
...
454
...
The mode of a data set is the element that occurs most
frequently, which is easy to spot in a frequency table
...
455
...
The range of the distribution is the difference between the
largest and the smallest piece of data
...
140
501 Quantitative Comparison Questions
456
...
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
...
The mode is
the value that occurs the most, 4
...
b
...
The largest value in set K is 9, the
smallest is 0
...
458
...
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
...
5
...
33; 3
...
33
...
c
...
The first term is 21,
the second term is 22, the third term is 23, etc
...
460
...
The 20th term of the sequence is 220
...
Twice the tenth term is 2 × 210, which equals 211
...
461
...
Each term in this sequence is also a power of 2, beginning with
64 which equals 26
...
Therefore,
the seventh term of the sequence is 27−7 = 20 = 1; 1 > 0
...
b
...
463
...
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%
...
b
...
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
...
Therefore, 7% of students participate in
both activities; 10% > 7%
...
a
...
20
ᎏᎏ
4
= 5 meaning that the four given elements have an average of
5 before the value of 2x is added to the sum
...
667 meaning that the
three given elements have an average of 4
...
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
...
466
...
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
13
the average of the third and fourth terms; 6 + 7 = 13; ᎏ2ᎏ = 6
...
Likewise, set N in order = 1, 5, 8, y which makes the median the
13
average of the second and third terms; 5 + 8 = 13; ᎏ2ᎏ = 6
...
467
...
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
...
a
...
469
...
The data in the third quarter are spread between the values 7
and 10
...
Therefore the data in the third quarter have a
larger spread than the data in the second quarter
...
c
...
Therefore, Kelly’s can make any of 3 × 5 × 4 = 60 types
of sandwiches
...
c
...
5 × 4 = 20 types of sandwiches that can be made on wheat or rye
bread
...
b
...
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
...
a
...
The first term = 1 × 1
...
The
third term = 3 × 3 and so on
...
474
...
The twentieth term of the sequence is 20 × 20 = 400
...
c
...
The 35th term is 35 × 35 =
1,225; 1,225 − 1,156 = 69
...
The
13th term is 13 × 13 = 169; 169 − 100 = 69
...
b
...
1 × 109 > 1 × 106 since 109 > 106
...
d
...
The fiftieth percentile of a
data set is the same as the median of the set
...
For example, in the data set 2, 3, 4, the median, or
fiftieth percentile, is 3
...
In the data set 1, 5,
8, 9, the median, or fiftieth percentile is the average of 5 and 8
...
5
...
75, so the median is greater
...
c
...
479
...
The 25th percentile is, by definition, the same as the first
quartile
...
b
...
Finding the sum of these
143
501 Quantitative Comparison Questions
squares, dividing by the number of elements in the set, then
finding the square root of this quantity
...
The sum of these square differences is
10
4 + 1 + 0 + 1 + 4 = 10; ᎏ5ᎏ = 2
...
ෆ
16
For set Q: 2 + 3 + 6 + 5 = 16; ᎏ4ᎏ = 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
...
5
...
5 ͙2
...
ෆ; ෆ
ෆ
481
...
The mean of set P is 3
...
482
...
1 kilometer is equal to 1,000 meters
...
c
...
46 × 109 means the decimal point in
3
...
This would yield 3,460,000,000
...
6 × 108, the decimal point in 34
...
484
...
1
...
0000015;
...
00000015
...
a
...
In 1990, the
profit was approximately $4,800
...
The profit in 1987 was approximately
$3,250
...
The
profit increase was, therefore, about $350
...
486
...
The profit in 1990 was approximately $4,800
...
The profit decrease was,
therefore, about $200
...
In 1989, the profit was approximately $3600
...
$500 > $200
...
a
...
In 1993, the
profit was approximately $5,250
...
$2,000 > $1,500
...
b
...
So, the
first term, 1, times 2 = 2, the second term
...
The third term, 6, times 4 = 24, the
fourth term
...
150 > 120
...
c
...
490
...
The number of each term in the sequence can be determined by
examining its tens and hundreds places
...
The second term is 12 and there
is a 1 in the 10s place
...
there is a 6 in
the 10s place
...
Note that the eleventh term should
therefore be 102 and the twelfth term should be 112
...
491
...
Rewriting the series in backward order beneath the original
series, then adding vertically, gives 14 consistent sums of 134
...
492
...
4
...
0000423;
...
0000423 as well
...
b
...
Therefore, 234 milliliters, divided
1,000
by 1,000, =
...
34 >
...
c
...
45 centigrams, divided by 100, is
100
...
495
...
Each square yard measures 3 feet long by 3 feet wide and so, is
9 square feet
...
496
...
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
...
c
...
498
...
The answer 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
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
...
a
...
500
...
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
...
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
...
a
...
146
Title: quantitative comparision questions.
Description: Best ever Quant comparision NoteS.
Description: Best ever Quant comparision NoteS.