10. 0 , p , q , 1
Column A
Column B
p
q
p
q
+
+
1
1
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (B). Remember three rules, which apply to any positive
numerator p and denominator q:
If p = q, the fraction’s value remains unchanged (and is always 1).
If p . q, the fraction’s value will decrease.
If p , q, the fraction’s value will increase (as given in the example).
It makes no difference that p and q are fractional values less than 1 rather than
larger numbers.As long as p and q are both positive, the above three rules always
apply. So in the example, if the centered inequality were 1 , p , q or 0 , p , 1
, q, it wouldn’t make any difference—quantity B would still be greater than
quantity A. (If you’re not convinced, try plugging in some sample values for p
and q.)
A GRE question might ask you to alter a ratio by adding or subtracting from one or
both terms in the ratio. The rules for altering ratios are the same as for altering
fractions. In either case, set up a proportion and solve algebraically for the unknown
term.
11. A drawer contains exactly half as many white shirts as blue shirts. If four
more shirts of each color were to be added to the drawer, the ratio of white
to blue shirts would be 5:8. How many blue shirts did the drawer
originally contain?
(A) 14
(B) 12
(C) 11
(D) 10
(E) 9
The correct answer is (B). Represent the original ratio of white to blue shirts
by the fraction
x
2x
, where x is the number of white shirts, then add 4 to both the
numerator and denominator. Set this fraction equal to
5
8
(the ratio after adding
shirts). Cross-multiply to solve for x:
Chapter 9: Math Review: Number Forms, Relationships, and Sets 203
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x 1 4
2x 1 4
5
5
8
8x 1 32 5 10x 1 20
12 5 2x
x 5 6
The original denominator is 2x,or12.
Ratios with More Than Two Quantities
You approach ratio problems involving three or more quantities the same way as
those involving only two quantities. The only difference is that there are more “parts”
that make up the “whole.”
12. Three lottery winners—X, Y, and Z—are sharing a lottery jackpot. X’s
share is
1
5
of Y’s share and
1
7
of Z’s share. If the total jackpot is $195,000,
what is the dollar amount of Z’s share?
(A) $15,000
(B) $35,000
(C) $75,000
(D) $105,000
(E) $115,000
The correct answer is (D). At first glance, this problem doesn’t appear to
involve ratios. (Where’s the colon?) But it does. The ratio of X’s share to Y’s share
is 1:5, and the ratio of X’s share to Z’s share is 1:7. So you can set up the following
triple ratio:
X:Y:Z 5 1:5:7
X’s winnings account for 1 of 13 equal parts (1 1 5 1 7) of the total jackpot.
1
13
of
$195,000 is $15,000. Accordingly, Y’s share is 5 times that amount, or $75,000,
and Z’s share is 7 times that amount, or $105,000.
In handling word problems involving ratios, think of a whole as the sum of its
fractional parts, as in the method used to solve the preceding problem:
1
13
(X’s share)
1
5
13
(Y’s share) 1
7
13
(Z’s share) 5 1 (the whole jackpot).
PART IV: Quantitative Reasoning204
TIP
Remember: When you add or
subtract the same number
from both the numerator and
denominator of a fraction—or
from each term in a ratio—you
alter the fraction or ratio,
unless the original ratio was 1:1
(in which case the ratio i s
unchanged).
www.petersons.com
Proportion Problems with Variables
A GRE proportion question might use letters instead of numbers, to focus on the
process rather than the result. You can solve these problems algebraically or by using
the plug-in strategy.
13. A candy store sells candy only in half-pound boxes. At c cents per box,
which of the following is the cost of a ounces of candy? [1 pound 5 16
ounces]
(A)
c
a
(B)
a
16c
(C) ac
(D)
ac
8
(E)
8c
a
The correct answer is (D). This question is asking: “c cents is to one box
as how many cents are to a ounces?” Set up a proportion, letting x equal
the cost of a ounces. Because the question asks for the cost of ounces,
convert 1 box to 8 ounces (a half pound). Use the cross-product method to
solve quickly:
c
8
5
x
a
8x 5 ac
x 5
ac
8
You can also use the plug-in strategy for this question, either instead of algebra
or, better yet, to check the answer you chose using algebra. Pick easy numbers to
work with, such as 100 for c and 16 for a. At 100 cents per 8-ounce box, 16 ounces
of candy cost 200 cents. Plug your numbers for a and c into each answer choice.
Only choice (D) gives you the number 200 you’re looking for.
ARITHMETIC MEAN, MEDIAN, MODE, AND RANGE
Arithmetic mean (simple average), median, mode, and range are four different ways
to describe a set of terms quantitatively. Here’s the definition of each one:
• Arithmetic mean (average): In a set of n measurements, the sum of the
measurements divided by n.
• Median: The middlemeasurement after themeasurements are orderedby size (or
the average of the two middle measurements if the number of measurements is
even).
• Mode: The measurement that appears most frequently in a set.
• Range: The difference between the greatest measurement and the least
measurement.
Chapter 9: Math Review: Number Forms, Relationships, and Sets 205
www.petersons.com
For example, given a set of six measurements, {8,24,8,3,2,7}:
Mean 5 4(82 4 1 8 1 3 1 2 1 7) 4 6 5 24 4 6 5 4
Median 5 5 The average of 3 and 7—the two middle measurements in the set
ordered in this way: {24,2,3,7,8,8}
Mode 5 8 8 appears twice (more frequently than any other measurement)
Range 5 12 The difference between 8 and 24
For the same set of values, the mean (simple average) and the median can be, but are
not necessarily, the same. For example: {3,4,5,6,7} has both a mean and median of 5.
However, the set {22,0,5,8,9} has a mean of 4 but a median of 5.
For any set of terms, the arithmetic mean (AM), also called the simple average,isthe
sum of the terms (a 1 b 1 c 1 . . .) divided by the number of terms (n) in the set:
AM 5
~a 1 b 1 c 1 !
n
On the GRE, easier questions involving simple average might ask you to add numbers
together and divide a sum. In finding a simple average, first be sure the numbers
being added are all of the same form or in terms of the same units.
14. What is the average of
1
5
, 25%, and 0.09?
(A) 0.18
(B) 20%
(C)
1
4
(D) 0.32
(E)
1
3
The correct answer is (A). Since the answer choices are not all expressed in the
same form, first rewrite numbers in whichever form you think would be easiest to
work with when you add the numbers together. In this case, the easiest form to
work with is probably the decimal form. So rewrite the first two numbers as
decimals, and then find the sum of the three numbers: 0.20 1 0.25 1 0.09 5 0.54.
Finally, divide by 3 to find the average: 0.54 4 3 5 0.18.
To find a missing number when the average of all the numbers in a set is given, plug
into the arithmetic-mean formula all the numbers you know—which include the
average, the sum of the other numbers, and the number of terms. Then, use algebra to
find the missing number. Or, you can try out each answer choice, in turn, as the
missing number until you find one that results in the average given.
PART IV: Quantitative Reasoning206
ALERT!
The mean and median of the
same set of numbers can be
the same—but this is not
necessarily the case.
www.petersons.com
15. The average of five numbers is 26. Four of the numbers are 212, 90, 226,
and 10. What is the fifth number?
(A) 16
(B) 42
(C) 44
(D) 68
(E) 84
The correct answer is (D). To solve the problem algebraically, let x 5 the
missing number. Set up the arithmetic-mean formula, then solve for x:
26 5
~90 1 10 2 12 2 26!1x
5
26 5
62 1 x
5
130 5 62 1 x
68 5 x
Or you can try out each answer choice in turn. Start with choice (C), which
supplies the middle value, 44. The sum of 44 and the other four numbers is 106.
Dividing this sum by 5 gives you 21.2—a number less than the average of 26 that
you’re aiming for. So you know the fifth number is greater than 44—and that
leaves choices (D) and (E). Try out the number 68, choice (D), and you’ll obtain
the average of 26.
If the numbers are easy to work with, you might be able to determine a missing term,
given the simple average of a set of numbers, without resorting to algebra. Simply
apply logic.
16. If the average of six consecutive multiples of 4 is 22, what is the greatest
of these integers?
(A) 22
(B) 24
(C) 26
(D) 28
(E) 32
The correct answer is (E). You can answer this question with common
sense—no algebra required. Consecutive multiples of 4 are 4, 8, 12, 16, etc. Given
that the average of six such numbers is 22, the two middle terms (the third and
fourth terms) must be 20 and 24. (Their average is 22.) Accordingly, the fifth term
is 28, and the sixth and greatest term is 32.
Simple Average and Median
Atougherquestionmight ask you tofind the value ofa number that changes an average
from one number to another.
Chapter 9: Math Review: Number Forms, Relationships, and Sets 207
TIP
Numerical answer choices are
listed either in ascending or
descending order of value, so
if you’re working backward
from the choices, start with
choice (C), the median value.
If choice (C) is too great or
too little, you’ve narrowed
down the options either to
choices (A) and (B) or to
choices (D) and (E).
www.petersons.com
When an additional number is added to a set, and the average of the numbers in the
set changes as a result, you can determine the value of the number that’s added by
applying the arithmetic-mean formula twice.
17. The average of three numbers is 24. If a fourth number is added, the
arithmetic mean of all four numbers is 21. What is the fourth number?
(A) 210
(B) 2
(C) 8
(D) 10
(E) 16
The correct answer is (C). To solve the problem algebraically, first determine
the sum of the three original numbers by the arithmetic-mean formula:
24 5
a 1 b 1 c
3
212 5 a + b + c
Then, apply the formula again accounting for the additional (fourth) number. The
new average is 21, the sum of the other three numbers is 212, and the number
of terms is 4. Solve for the missing number (x):
21 5
212 1 x
4
24 5212 1 x
8 5 x
You approach arithmetic-mean problems that involve variables instead of (or in
addition to) numbers in the same way as those involving only numbers. Just plug the
information you’re given into the arithmetic-mean formula, and then solve the
problem algebraically.
18. If A is the average of P, Q, and another number, which of the following
represents the missing number?
(A) 3A 2 P 2 Q
(B) A 1 P 1 Q
(C) A 1 P 2 Q
(D) A 2 P 1 Q
(E) 3A 2 P 1 Q
The correct answer is (A). Let x 5 the missing number. Solve for x by the
arithmetic-mean formula:
A 5
P 1 Q 1 x
3
3A 5 P 1 Q 1 x
3A 2 P 2 Q 5 x
PART IV: Quantitative Reasoning208
ALERT!
Don’t try the plug-in strategy
(working backward from the
answer choices) to solve
problems like the one here; it’s
too complex. Be flexible and
use shortcuts wherever you
can—but recognize their
limitations.
www.petersons.com
A GRE question involving arithmetic mean might also involve the concept of median.
If the question deals solely with numbers, simply do the math to calculate the average
and the median.
19. Column A
Column B
The median value of
1
2
,
2
3
,
3
4
, and
5
6
The arithmetic mean (average) of
1
2
,
2
3
,
3
4
, and
5
6
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (A). First, calculate Quantity A. Both columns list the
same four numbers from smallest to greatest in value. So Quantity A (the
median) is the average of
2
3
and
3
4
:
2
3
3
4
2
8
12
9
12
2
17
12
2
17
24
+
=
+
==
Next, calculate Quantity B, using the common denominator 12:
1
2
2
3
3
4
5
6
68910
12
33
12
+++=
+++
=
The mean of the four numbers is one fourth of
33
12
,or
33
48
. To make the com-
parison, convert Quantity A to a fraction with the denominator 48:
17
24
34
48
=
.
Since the numerator in QuantityA(34) is greater than the one in Quantity B (33),
Quantity A is greater than Quantity B.
A more challenging question might involve variables instead of (or in addition to)
numbers. Here’s a good example:
20. Distribution K contains four terms, represented by the four
expressions p, q, p + q, and q 2 p, where 0 , q , p.
Column A
Column B
The median of
Distribution K
The arithmetic mean (simple average)
of Distribution K
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (A). First, express the median of the four terms
(Quantity A). Given that q , p and that p and q are both positive, (q 2 p) must be
negative and hence is least in value among the four terms, while (q + p) must be
Chapter 9: Math Review: Number Forms, Relationships, and Sets 209
www.petersons.com
greatest in value among the four terms. Here are the four terms, ranked from
least to greatest in value:
(q 2 p), q, p, (p + q)
The median value, then, is the average (arithmetic mean) of the two middle terms
p and q:
pq+
2
Next, express the arithmetic mean (AM) of the four terms (Quantity B):
AM =
++ +
(
)
+−
(
)
=
+
qp pq qp
qp
4
3
4
Since both quantities are positive, you can multiply by 4 across columns to
eliminate fractions, leaving the following simpler comparison:
Quantity A: 2p +2q
Quantity B: 3q + p
Subtract p and 2q from both columns to further simplify the comparison:
Quantity A: p
Quantity B: q
As you can easily see now, Quantity A is greater than Quantity B.
STANDARD DEVIATION
Standard deviation is a measure of dispersion among members of a set. Computing
standard deviation involves these five steps:
Compute the arithmetic mean (simple average) of all terms in the set.
Compute the difference between the mean and each term.
Square each difference you computed in step 2.
Compute the mean of the squares you computed in step 3.
Compute the non-negative square root of the mean you computed in step 4.
For example, here’s how you’d determine the standard deviation of Distribution A:
{21, 2, 3, 4}:
Arithmetic mean 5
21 1 2 1 3 1 4
4
5
8
4
5 2
The difference between the mean (2) and each term:
2 2 (21) 5 3; 2 2 2 5 0; 3 2 2 5 1; 4 2 2 5 2
The square of each difference: {3
2
,0
2
,1
2
,2
2
} 5 {9,0,1,4}
The mean of the squares:
9 1 0 1 1 1 4
4
5
14
4
5
7
2
The standard deviation of Distribution A 5
Î
7
2
PART IV: Quantitative Reasoning210
www.petersons.com
A GRE question might ask you to calculate standard deviation (as in the preceding
example). Or, a question might ask you to compare standard deviations. You might be
able to make the comparison without precise calculations—by remembering to follow
this general rule: The greater the data are spread away from the mean, the greater the
standard deviation.
21. Distribution A: {1, 2.5, 4, 5.5, 7}
Distribution B: {1, 3, 4, 5, 7}
Column A
Column B
The standard deviation
of Distribution A
The standard deviation of
Distribution B
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (A). In both sets, the mean and median is 4, and the
range is 6. But the standard deviation of A is greater than that of B, because 2.5
and 5.5 are further away than 3 and 5 from the mean.
22. Which of the following distributions has the greatest standard deviation?
(A) {21, 1, 3}
(B) {1,3,5}
(C) {0,4,5}
(D) {23, 21, 2}
(E) {2,4,6}
The correct answer is (C). Notice that in each of the choices (A), (B), and (E),
the distribution’s range is 4. But in choice (C) and choice (D), the range is 5. So
the correct answer must be either choice (C) or (D). Notice that the middle term
in choice (C), 4, is skewed farther away from the mean than the middle term in
choice (D). That means that choice (C) provides the distribution having the
greatest standard deviation.
ARITHMETIC SEQUENCES
In an arithmetic sequences of numbers, there is a constant (unchanging) difference
between successive numbers in the sequence. In other words, all numbers in an
arithmetic series are evenly spaced. All of the following are examples of an arithmetic
sequence:
• Successive integers
• Successive even integers
• Successive odd integers
Chapter 9: Math Review: Number Forms, Relationships, and Sets 211
www.petersons.com
• Successive multiples of the same number
• Successive integers ending in the same digit
On the GRE, questions involving an arithmetic sequence might ask for the average or
the sum of a sequence. When the numbers to be averaged form an arithmetic (evenly
spaced) sequence, the average is simply the median (the middle number or the
average of the two middle numbers if the number of terms is even). In other words,
the mean and median of the set of numbers are the same. Faced with calculating the
average of a long sequence of evenly-spaced integers, you can shortcut the addition.
23. What is the average of the first twenty positive integers?
Express your answer to the nearest tenth.
Enter a number in the box.
The correct answer is (10.5). Since the terms are evenly spaced, the
average is halfway between the 10th and 11th terms—which happen to be
the integers 10 and 11. So the average is 10.5. (This number is also the
median.) If you take the average of the first term (1) and the last term
(20), you get the same result:
1 1 20
2
5
21
2
or 10
1
2
Finding the sum (rather than the average) of an arithmetic (evenly spaced) sequence
of numbers requires only one additional step: multiplying the average (which is also
the median) by the number of terms in the sequence. The trickiest aspect of this type
of question is determining the number of terms in the sequence.
24. What is the sum of all odd integers between 10 and 40?
(A) 250
(B) 325
(C) 375
(D) 400
(E) 450
The correct answer is (C). The average of the described numbers is
25—halfway between 10 and 40 (in other words, half the sum of 10 and 40). The
number of terms in the series is 15. (The first term is 11, and the last term is 39.)
The sum of the described series of integers 5 25 3 15 5 375.
When you calculate the average or sum of a sequence of evenly spaced numbers, be
careful counting the number of terms in the sequence. For instance, the number of
positive odd integers less than 50 is 25, but the number of positive even integers less
than 50 is only 24.
PART IV: Quantitative Reasoning212
www.petersons.com
. of all the numbers in a set is given, plug
into the arithmetic-mean formula all the numbers you know—which include the
average, the sum of the other numbers,. c
Then, apply the formula again accounting for the additional (fourth) number. The
new average is 21, the sum of the other three numbers is 212, and the