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Inequalities 241

15. If x – y equaled 0, the inequality (x – y)3 > (x – y)2 would not be valid. Hence, x – y is nonzero. Since the square of a nonzero number is positive, (x – y)2 is positive. Dividing both sides of the inequality (x – y)3 > (x – y)2 by the positive term (x – y)2 yields x – y > 1. Adding y to both sides yields x > y + 1. This inequality says that x is at least one unit larger than y, so x > y. The relation can be carried to the cubes as x3 > y3, so the answer is (E). The remaining choices need not be true.

16.Combining the given inequalities x > 2 and x < 3 yields 2 < x < 3. So, x lies between 2 and 3. Hence,

x– 2 is positive and x – 3 is negative. Hence, the product (x – 2)(x – 3) is negative. I is false.

2 – x is negative and x – 3 is negative. Hence, the product (x – 2)(x – 3) is positive. II is true.

2 – x is negative and 3 – x is positive. Hence, the product (2 – x)(3 – x) is negative. III is false.

Hence, the answer is (B), II only is correct.

17. We are given the inequality x > y and that x is negative. Since x > y, y must also be negative. Hence, xy, the product of two negative numbers, must be positive. Dividing the inequality by the positive expression

xy yields

, or

. Rearranging yields

. Hence, I is true.

Since x is negative, x – 1 is also negative. Similarly, since y is negative, y – 1 is also negative. Hence, the product of the two, (x – 1)(y – 1), must be positive. Subtracting –1 from both sides of the given inequality

x > y yields x – 1 > y – 1. Dividing the inequality by the positive value (x – 1)(y – 1) yields

y −1

x −1

Rearranging the inequality yields

. Hence, II must be true.

x −1

y −1

Though x is negative, it is possible that x + 1 is positive while y + 1 is still negative. Here,

x +1

y +1

false because the left-hand side is positive while the right-hand side is negative. Hence, III need not be true.

Hence, the answer is (D), I and II must be true.

18. Factoring out the common factor a on the left-hand side of the inequality a2 + 7a < 0 yields a(a + 7) < 0. The product of two numbers (here, a and a + 7) is negative when one is negative and the other is positive. Hence, we have two cases:

1)a < 0 and a + 7 > 0

2)a > 0 and a + 7 < 0

Case 2) is impossible since if a is positive then a + 7 cannot be negative.

Case 1) is valid for all values of a between 0 and –7. Any number in the range is less than 0. So, Column A is less than Column B, and the answer is (B).

Hard

19. The payments to Worker A for the 4 days are the four integers 4, 6, 8, and 10. The sum of the payments is 4 + 6 + 8 + 10 = 28.

The payments to Worker B for the 4 days are the four integers 3, 5, 7, and 9. The sum of the payments is 3 + 5 + 7 + 9 = 24.

The payments to Worker C for the 4 days are 1, 1 + 2 = 3, 3 + 3 = 6, and 6 + 5 = 11. The sum of the payments is 1 + 3 + 6 + 11 = 21.

From the calculations, A > B > C. The answer is (A).

242 GRE Math Bible

20. The choices given are positive. Multiplying both sides of the given inequality x + y > x – y by the positive value x – y yields

(x + y)(x − y) > (x − y)2

x2 − y2 > (x − y)2

16 > (x − y)2

16 > x − y

4 > x − y

Since choice (A) is one such suitable choice, the answer is (A).

21.Combining the two given inequalities 1/55 < x < 1/22 and 1/33 < x < 1/11 yields 1/33 < x < 1/22. Since among the three positive numbers 12, 23, and 54, the number 23 is the only one in the positive range between 22 and 33, only the number 1/23 lies between 1/33 and 1/22 and the numbers 1/54, and 1/12 do not. Hence, x = 1/23. The answer is (B), II only.

22.The possible solutions of the equation a2 = 9 are a = 3 and a = –3.

Since the square of any nonzero number is positive, b2 must be positive. We are given that ab2 = –8, a negative number. Since the product of a and b2 is negative, a must be negative. The negative solution of a is a = –3.

Substituting this value of a in the given inequality x/a > 4 yields

−x3 > 4

x < –12 by multiplying both sides by the negative number –3 and flipping the direction of the inequality

Hence, Column A is less than –12.

Now, replacing a with –3 in the inequality y/a < –6 yields

−y3 < −6

y > 18 by multiplying both sides by –3 and flipping the direction of the inequality

Hence, y is positive and therefore greater than x. Hence, Column B is greater, and the answer is (B).

Fractions & Decimals

Fractions

A fraction consists of two parts: a numerator and a denominator. numerator

denominator

If the numerator is smaller than the denominator, the fraction is called proper and is less than one. For

example: 2 , 5 , and

are all proper fractions and therefore less than 1.

If the numerator is larger than the denominator, the fraction is called improper and is greater than 1.

3 5

For example:

2 , 4 , and 3 are all improper fractions and therefore greater than 1.

An improper fraction can be converted into a mixed fraction by dividing its denominator into its

numerator. For example, since 2 divides into 7 three times with a remainder of 1, we get

= 3

To convert a mixed fraction into an improper fraction, multiply the denominator and the integer and

then add the numerator. Then, write the result over the denominator. For example, 5

2 =

3 5 + 2

17 .

In a negative fraction, the negative symbol can be written on the top, in the middle, or on the bottom;

however, when a negative symbol appears on the bottom, it is usually moved to the top or the middle:

−5

= −

3 . If both terms in the denominator of a fraction are negative, the negative symbol is often

−3

−1

factored out and moved to the top or middle of the fraction:

= −

− x − 2

−(x + 2)

x + 2

To compare two fractions, cross-multiply. The larger number will be on the same side as the larger fraction.

Strategy

Example: Which of the following fractions is larger?
	9	10
10		11

Cross-multiplying gives 9 11 versus 10 10 , which reduces to 99 versus 100. Now, 100 is greater than 99.

10 9

Hence, 11 is greater than 10 .

243

244 GRE Math Bible

Always reduce a fraction to its lowest terms.

Strategy
Example:	If x ≠ –1, then	2 x2 + 4 x + 2		=
Example:	If x ≠ –1, then	(x + 1)2		=
	(A) 0	(B) 1	(C) 2		(D) 4	(E) 6

Factor out the 2 in the expression:

2(x2 + 2x + 1) (x + 1)2

Factor the quadratic expressions:

2( x +1)( x +1) (x + 1)(x + 1)

Finally, canceling the (x + 1)’s gives 2. The answer is (C).

To solve a fractional equation, multiply both sides by the LCD (lowest common denominator) to clear fractions.

Strategy

Example:	If	x + 3	= y , what is the value of x in terms of y?
		x − 3
		(A) 3 – y		(B)	3	(C)
		(A) 3 – y		(B)	3	(C)	y +12
					y

First, multiply both sides of the equation by x – 3:

Cancel the (x – 3)'s on the left side of the equation: Distribute the y:

Subtract xy and 3 from both sides:

Factor out the x on the left side of the equation:

Finally, divide both sides of the equation by 1 – y:

Hence, the answer is (D).

(D) −3y − 3	(E) 3y2
1− y

(x − 3) xx +− 33 = (x − 3)y

x + 3 = (x – 3)y x + 3 = xy – 3y x – xy = –3y – 3

x(1 – y)= –3y – 3 x = −3y − 3

1 − y

	Complex Fractions: When dividing a fraction by a whole number (or vice versa), you must
Note!	keep track of the main division bar:

1− 1

Example: 32 =

	(A) 6
1−		1			2	−	1
				2
Solution:		2	=				2	=
	3					3

= a

ac .

But

	(B) 3							(C) 1/3	(D) 1/6	(E) 1/8
2−1		1		1	1		1
2	=	2	=			=		. The answer is (D).
3	=	3	=			=		. The answer is (D).
3		3		2	3		6

Fractions & Decimals 245

Example: If z ≠ 0 and yz ≠ 1, then

1 =

y −

yz − z

y − z

(A)

zy − 1

(B)

(C)

z −1

(D) zy − 1

(E) zy − 1

= 1

Solution:

1 =

zy − 1

= zy − 1 . The answer is (D).

y − z

Multiplying fractions is routine: merely multiply the numerators and multiply the

Note!

denominators:

c =

. For example,

1 3

= 3 .

2 4

Note!	Two fractions can be added quickly by cross-multiplying:
Example:	1	−	3	=
	2		4

a	±	c	= ad ± bc
b		d
			bd

(A)

–5/4

(B)

–2/3

(C)

–1/4

(D)

–1/2

(E)

–2/3

1 4 − 2 3

4 − 6

−2

Cross multiplying the expression 2 −

yields

2 4

= 8

= − 4 . Hence, the answer is (C).

Example: Which of the following equals the average of x and x ?

2x2 + 1

x + 1

x + 2

x2 + 1

x + 1

(A)

(B)

(C)

(D)

(E) x

x2 + 1

The average of x and

x + x

= x2 + 1

x2 + 1 . Thus, the answer is (B).

Note!

To add three or more fractions with different denominators, you need to form a common

denominator of all the fractions.

For example, to add the fractions in the expression 1

1 , we have to change the denominator of each

fraction into the common denominator 36 (note, 36 is a common denominator because 3, 4, and 18 all divide into it evenly). This is done by multiply the top and bottom of each fraction by an appropriate number (this does not change the value of the expression because any number divided by itself equals 1):

1	12		1	9		1	2		12		9		2		12 + 9 + 2		23
		+			+			=		+		+		=		=
3	12		4	9		18	2		36		36		36		36		36

You may remember from algebra that to find a common denominator of a set of fractions, you prime factor the denominators and then select each factor the greatest number of times it occurs in any of the factorizations. That is too cumbersome, however. A better way is to simply add the largest denominator to itself until all the other denominators divide into it evenly. In the above example, we just add 18 to itself to get the common denominator 36.

cannot be canceled. However, the c’s in the expression

For example, the c’s in the expression

246 GRE Math Bible

Note!

To find a common denominator of a set of fractions, simply add the largest denominator to itself until all the other denominators divide into it evenly.

Fractions often behave in unusual ways: Squaring a fraction makes it smaller, and taking the square root of a fraction makes it larger. (Caution: This is true only for proper fractions,

that is, fractions between 0 and 1.)

Example:		1	2	=	1	and	1	is less than	1	. Also	1	=	1	and	1	is greater than	1	.
		3			9		9		3		4		2		2		4

Note!	You can cancel only over multiplication, not over addition or subtraction.

cx + c c

c + x c

can be canceled as follows: cx + c = c/(x + 1) = x + 1. c c/

Decimals

If a fraction’s denominator is a power of 10, it can be written in a special form called a decimal fraction. Some common decimals are 101 =.1, 1002 =.02, 10003 =.003. Notice that the number of decimal places

corresponds to the number of zeros in the denominator of the fraction. Also, note that the value of the decimal place decreases to the right of the decimal point:

tenths

hundredths thousandths

ten-thousandths

This decimal can be written in expanded form as follows:

.1234 = 101 + 1002 + 10003 + 100004

Sometimes a zero is placed before the decimal point to prevent misreading the decimal as a whole number. The zero has no affect on the value of the decimal. For example, .2 = 0.2.

Fractions can be converted to decimals by dividing the denominator into the numerator. For example,

to convert 8 to a decimal, divide 8 into 5 (note, a decimal point and as many zeros as necessary are added

after the 5):

.625

8)548.000

Fractions & Decimals 247

The procedures for adding, subtracting, multiplying, and dividing decimals are the same as for whole numbers, except for a few small adjustments.

Adding and Subtracting Decimals: To add or subtract decimals, merely align the decimal points and then add or subtract as you would with whole numbers.

1.369	12.45
+ 9.7		− 6.367
11.069	6.083

Multiplying Decimals: Multiply decimals as you would with whole numbers. The answer will have as many decimal places as the sum of the number of decimal places in the numbers being multiplied.

1.232 decimal places

	× 2.4		1 decimal place
	492
	246
	2.952		3 decimal places

Dividing Decimals: Before dividing decimals, move the decimal point of the divisor all the way to the right and move the decimal point of the dividend the same number of spaces to the right (adding zeros if necessary). Then divide as you would with whole numbers.

2.5

.24).6 = 24)60.0

120

1
Example: 5	of .1 percent equals:
(A) 2		(B) .2	(C) .02	(D) .002

Recall that percent means to divide by 100. So .1 percent equals 100 divide 5 into 1:

.2 5)1.0

(E) .0002
= .001	1	to a decimal,
	. To convert 5

In percent problems, “of” means multiplication. So multiplying .2 and .001 yields

.001 × .2

.0002

Hence, the answer is (E). Note, you may be surprised to learn that the GRE would consider this to be a hard problem.

248GRE Math Bible

Example: The decimal .1 is how many times greater than the decimal (.001)3 ?

(A)10

(B)102

(D)108

(E)101 0

1			1		, or 10	−3
Converting .001 to a fraction gives		.	This fraction, in turn, can be written as				. Cubing
	1000			103

this expression yields (.001)3

(

)

= 10−3

3 = 10−9 . Now, dividing the larger number, .1, by the smaller

number, (.001)3 , yields

10−1

−1− −9

−

= 10

( )

= 10

9 = 10

(.001)3

10−9

Hence, .1 is 108 times as large as (.001)3 . The answer is (D).

Example: Let x = .99, y = .99 , and z = (.99)2. Then which of the following is true?

(A)x < z < y

(B)z < y < x

(C)z < x < y

(D)y < x < z

(E)y < z < x

99 99

Converting .99 into a fraction gives 100 . Since 100 is between 0 and 1, squaring it will make it smaller

and taking its square root will make it larger. Hence, (.99)2 < .99 < .99 . The answer is (C). Note, this property holds for all proper decimals (decimals between 0 and 1) just as it does for all proper fractions.

Problem Set N:

Easy
1.			Column A								Column B
	2	+	3	+	4		+	5		4	+	3	+	2
10		+	100	+	1000		+	10000	10		+		+	1000
10			100		1000			10000	10		100			1000
2.			Column A								Column B
			2/3 – 3/4								3/4 – 4/5
3.							Column A		Column B
		2 × 101 + 3 × 100 + 4 × 10–1 + 5 × 10–2							1 × 10–3 + 2 × 10–2 + 3 × 10–1 + 4 × 100 + 5 × 101
4.			Column A								Column B
			30		of	31					30
			31			32					31

Fractions & Decimals 249

5.Which one of the following is nearest to 0.313233 ?

(A)3/10

(B)31/100

(D)3132/10000

(E)31323/10000

6.	Column A	x = 1/y	Column B
	x + 1 + 1/x		y + 1 + 1/y
7.	Column A		Column B
	1/2 + 1/4 + 1/8 + 1/16		1

Medium

8.3/8 of a number is what fraction of 2 times the number?

(A)3/16

(B)3/8

(D)4/6

(E)3/4

9.If p + q = 12 and pq = 35, then p + q =

(A)1/5

(B)1/7

(D)12/35

(E)23/35

10.	Column A		x = 1/y	Column B
		x2 +1			y2 +1
		x			y

11. If x is not equal to 1 and y = x −1, then which one of the following cannot be the value of y ?

(A) 0

(B) 1

(D) 3

(E) 4

2x + 2x−1

12. Which one of the following does the expression 2x+1 − 2x equal?

(A) 1

(B) 3/2

(D) 5/2

(E) 3

250 GRE Math Bible

13.	Column A	0 < x < y	Column B
	x + 1/x		y + 1/y
14.	Column A	Kate ate 1/3 of a cake, Fritz ate	Column B
		1/5 of the remaining cake, and
	Fraction of the cake eaten by	what was left was eaten by Emily.	5/7
	Fraction of the cake eaten by		5/7
	Emily

15.The product of two numbers x and y is twice the sum of the numbers. What is the sum of the reciprocals of x and y ?

(A)1/8

(B)1/4

(D)2

(E)4

16.	Column A		x = 1/y	Column B
		x2 + x + 2		2y2 + y +1
		x		y

Hard

17.There are 87 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What fraction of the balls in the jar have both red and green colors?

(A)6/14

(B)2/7

(D)6/29

(E)6/42

18.Which one of the following fractions is a result of the sum of an integer and its reciprocal?

(A)15/8

(B)17/5

(D)37/5

(E)65/8

19.	Column A		x < 1/y, and x and y are positive	Column B
		2+ x − x2		2y2 + y −1
		x		y

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