GRE_Math_Bible_eBook

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Canceling the 2 on the left side yields

Canceling the 2 on the left side yields

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Fractions & Decimals 251

20.Which one of the following is closest to 1?

(A)		3
(A)	3	+ 0.3
	3	+ 0.3
(B)		3
(B)	3	+ 0.32
	3	+ 0.32
(C)		3
(C)	3	− 0.3
	3	− 0.3
(D)		3
(D)	3	− 0.32
	3	− 0.32
(E)		3
(E)	3	+ 0.33
	3	+ 0.33

21.Jane gave three-fifths of the amount of money she had to Jack. Jane now has 200 dollars. How much did she give to Jack?

(A)$80

(B)$120

(C)$200

(D)$300

(E)$500

Very Hard

22.In a country, 60% of the male citizen and 70% of the female citizen are eligible to vote. 70% of male citizens eligible to vote voted, and 60% of female citizens eligible to vote voted. What fraction of the citizens voted during the election?

(A)0.42

(B)0.48

(D)0.54

(E)0.60

252GRE Math Bible

Answers and Solutions to Problem Set N

Easy

1. First, cancel (subtract) the common term 3/100 from both columns:

	2	+		4	+	5		4	+	2
10		+	1000		+	10000	10		+	1000
10			1000			10000	10			1000
Next, multiply both columns by 10000 to clear the fractions:
		2000 + 40 + 5					4000 + 20
Finally, add the numbers:
			2045						4020

The answer is (B).

Method II

Column A equals 2/10 + 3/100 + 4/1000 + 5/10000 = 0.2345.

Column B equals 4/10 + 3/100 + 2/1000 = 0.432.

Since 0.432 is greater than 0.2345, Column B is greater. The answer is (B).

2. The least common multiple of the denominators of all the fractions is 60. Multiplying both columns by 60 to clear the fractions yields

40 – 45

45 – 48

Subtracting the numbers yields

–5

–3

Since –5 < –3, Column B is greater than Column A and the answer is (B).

3. The dominant term 101 appears in both columns, but has more weight (5) in Column B. Hence, Column B is greater. Let's still evaluate the expressions:

Column A = 2 × 101 + 3 × 100 + 4 × 10–1 + 5 × 10–2 = 20 + 3 + 0.4 + 0.05 = 23.45.

Column B = 1 × 10–3 + 2 × 10–2 + 3 × 10–1 + 4 × 100 + 5 × 101 = 0.001 + 0.02 + 0.3 + 4 + 50 = 54.321.

Hence, Column B is greater than Column A. The answer is (B).

4. Column A:	30	of	31 =	30	31	=	30	. Now,	30	<	30	because the fractions have the same numerators
	31		32	31	32		32		32		31
and the denominator of 30				is larger than the denominator of									30	. Hence, Column B is larger than Column
			32										31

A, and the answer is (B).

5. The given decimal 0.313233 rounded to first, second, third, fourth and fifth digits after the decimal respectively equal 0.3 [= Choice (A)], 0.31 [= Choice (B)], 0.313 [= Choice (C)], 0.3132 [= Choice (D)], and 0.31323 [= Choice (E)]. Accuracy can be maintained by rounding the decimals only to later digits. So, choice (E) is the most accurate and hence the nearest. The answer is (E).

Fractions & Decimals 253

6. Substituting 1/y for x in Column A yields

1y +1+ 11 =

1y +1+ y =

Column B

The answer is (C).

7. Let’s multiply both columns by 16 to clear the fractions. (Remember, this can only be done if the number you are multiplying by is positive.)

8 + 4 + 2 + 1	16
15	16

Hence, Column A is less than Column B, and the answer is (B).

Medium

8. Let the number be x. Now, 3/8 of the number is 3x/8, and 2 times the number is 2x. Forming the

	3	x		3		3	1		3
fraction yields	8		=	8	=			=		. The answer is (A).
	2x			2		8			16
							2

9.Adding the fractions yields

1	+	1	=	p + q	=	12	we are given that p + q = 12, and pq = 35
p				pq
		q				35

The answer is (D).

Method II:

Solving the given equation pq = 35 for q yields q = 35/p. Plugging this into the equation p + q = 12 yields

p + 35/p = 12
p2	+ 35 = 12p	by multiplying both sides by p
p2	– 12p + 35 = 0	by subtracting 12p from both sides

(p – 5)(p – 7) = 0

p – 5 = 0 or p – 7 = 0 p = 5 or p = 7

When p = 5, the equation p + q = 12 shows that q = 7. Similarly, when p = 7, q equals 5. In either case,

1	+	1	=	1	+	1	=	7+ 5	=	12	. The answer is (D).
p								35
		q		5		7				35

254 GRE Math Bible

10. Substituting 1/y for x in Column A yields

1+ y2

1+ y

Column B

The answer is (C).

11.

Since the numerator of the fraction

does not contain a variable, it can never equal 0. Hence, the

x −1

fraction can never equal 0. The answer is (A).

−

, and the term 2x+1 equals 2x 2. Hence, the given expression

2x + 2x−1

12.

The term 2x 1 equals

2x+1 −

becomes

2x +

2x 2 − 2x

by factoring out 2x from both numerator and denominator

2x (2 −1)

2 =

by canceling 2x from both numerator and denominator

2 −1

32 =

The answer is (B).

			Fractions & Decimals 255
13. If x = 1 and y = 2, then the columns become
1	+ 1/1		2 + 1/2
	2		2 1/2
In this case, Column B is greater.
If x = 1/2 and y = 1, the columns become
1	+	1	1+	1
1	+	1	1+
2		1	1
		2
	2	1	2
		2

In this case, Column A is greater.

Hence, we have a double case, and the answer is (D).

14.We are given that Kate ate 1/3 of the cake. So, the uneaten part of the cake is 1 – 1/3 = 2/3. Hence, regardless of how much Fritz ate, Emily could not have eaten more than 2/3 of the cake. Hence, Column A is less than 2/3; and since 2/3 < 5/7, Column B is larger. The answer is (B).

15.We are given that the product of x and y is twice the sum of x and y. Hence, we have xy = 2(x + y).

Now, the sum of the reciprocals of x and y is

1x + 1y = yxy+ x =

x + y =

2(x + y) 1 2

The answer is (C).

256GRE Math Bible

16. Column A =

x2 + x + 2 = x

x2 + x + 2 = x x x

x +1+ 2x

Now, substituting 1/y for x yields

1y +1+ 12 =

1y +1+ 2y =

1+ y + 2y2 = y

Column B

The answer is (C).

Hard

17. Let T be the total number of balls, R the number of balls having red color, G the number having green color, and B the number having both colors.

So, the number of balls having only red is R – B, the number having only green is G – B, and the number having both is B. Now, the total number of balls is T = (R – B) + (G – B) + B = R + G – B.

We are given that 2/7 of the balls having red color have green also. This implies that B = 2R/7. Also, we are given that 3/7 of the green balls have red color. This implies that B = 3G/7. Solving for R and G in these two equations yields R = 7B/2 and G = 7B/3. Substituting this into the equation T = R + G – B yields T = 7B/2 + 7B/3 – B. Solving for B yields B = 6T/29. Hence, 6/29 of all the balls in the jar have both colors. The answer is (D). Note that we did not use the information: “There are 87 balls.” Sometimes, not all information in a problem is needed.

18. Let n be an integer. Then its reciprocal is 1/n, and the sum of the two is n +	1	=	n2 +1	. The resultant
	n		n

fraction has a numerator 1 unit greater than the square of the denominator. Now, choose the answer-choice that has the fraction in this format.

Choice (A): Denominator is 8. Expected numerator is 82 + 1 = 65 ≠ 15, numerator in the choice. Hence, reject.

Choice (B): Denominator is 5. Expected numerator is 52 + 1 = 26 ≠ 17, numerator in the choice. Hence, reject.

Choice (C): Denominator is 7. Expected numerator is 72 + 1 = 50 ≠ 36, numerator in the choice. Hence, reject.

Choice (D): Denominator is 5. Expected numerator is 52 + 1 = 26 ≠ 37, numerator in the choice. Hence, reject.

Choice (E): Denominator is 8. Expected numerator is 82 + 1 = 65, numerator in the choice. Correct.

The answer is (E).

Fractions & Decimals 257

19. Breaking up the fractions in both columns yields

Column A:

2+ x − x2

−

+1− x

Column B:

2y2

+ y −1

2y2

−

= 2y +1

−

Now, multiplying both sides of the given inequality x < 1/y by –1 and flipping the direction of the inequality yields

–x > –1/y

…(1)

Since x and 1/y are both positive (since y is positive, so is its reciprocal 1/y), we can safely invert both sides of the inequality x < 1/y and flip the direction of the inequality to yield 1/x > y. Multiplying both sides of this inequality by 2 yields

2/x > 2y

…(2)

Now, adding inequalities (1) and (2) yields

2x − x > 2y − 1y

Adding 1 to both sides of this inequality yields

	2	+1− x > 2y +1−	1
	x		y

Column A > Column B				from the known results

Hence, the answer is (A).

20. Let’s subtract 1 from each answer-choice. The answer-choice that has the lowest positive value should be closest to 1.

Choice (A):

Choice (B):

−1 =

3− (3+ 0.3)

−0.3

−1

. Hence, Choice (A) is 1/11 units away from 1.

3+ 0.3

3.3

−1

3− (3+ 0.32)

−0.32

−0.09

−9

. Hence, Choice (B) is 9/309

+ 0.32

3+ 0.32

0.32

0.09

3.09

309

units away from 1. Since 9/309 is less than 1/11, Choice (B) is closer than Choice (A). Eliminate choice

(A).

Choice (C):

−1=

3− (3− 0.3)

0.3

= 1 . Hence, Choice (C) is 1/9 units away from 1. Clearly, this

3− 0.3

2.7

3− 0.3

is greater than 9/309. Hence, Choice (B) is closer than Choice (C). Hence, eliminate choice (C).

Choice (D):

−1

3− (3− 0.32)

0.32

0.09

. Hence, Choice (D) is 9/291 units away

− 0.32

3− 0.32

2.91

291

from 1. This is greater than 9/309. Hence, Choice (B) is closer than Choice (D). Hence, eliminate choice

(D).

Choice (E):	3	−1 =	3− (3+ 0.33)	=	−0.33	=	−33	. So, Choice (E) is 33/333 units away from 1. This is
	+ 0.33
3		3+ 0.33		3.33 333

greater than 9/309. Hence, Choice (B) is closer than Choice (E). Hence, eliminate choice (E).

The answer is (B).

258 GRE Math Bible

21. Let the original amount of money Jane had be x. Since she gave 3/5 of her money to Jack, she now has 1 – 3/5 = 2/5 of the original amount. We are given that this 2/5 part equals 200 dollars. Hence, we have the

equation 25 x = 200. Solving for x yields x = 500. Since she gave 3/5 of this amount to Jack, she gave him $300 ( = 35 500). The answer is (D).

Very Hard

22. Let the number of male and female citizens in the country be m and f, respectively.

Now, 60% of the male citizens are eligible to vote, and 60% of m is 60m/100. 70% of female citizens are eligible to vote, and 70% of f is 70f/100.

We are given that 70% of male citizens eligible to vote voted:

70% of 60m/100 is	70	×	60m	=	70× 60m	= 0.42m
	100
			100		10,000

We are also given that 60% of female citizens eligible to vote voted:

60% of 70f/100 is	60	×	70 f	=	60× 70 f	= 0.42 f
	100
			100		10,000

So, out of the total m + f citizens, the total number of voters who voted is

0.42m + 0.42f = 0.42(m + f)

Hence, the required fraction is

0.42(m + f ) = 0.42 m + f

The answer is (A).

Equations

When simplifying algebraic expressions, we perform operations within parentheses first and then exponents and then multiplication and then division and then addition and lastly subtraction. This can be remembered by the mnemonic:

PEMDAS

Please Excuse My Dear Aunt Sally

When solving equations, however, we apply the mnemonic in reverse order: SADMEP. This is often expressed as follows: inverse operations in inverse order. The goal in solving an equation is to isolate the variable on one side of the equal sign (usually the left side). This is done by identifying the main operation—addition, multiplication, etc.—and then performing the opposite operation.

Example: Solve the following equation for x: 2x + y = 5

Solution: The main operation is addition (remember addition now comes before multiplication, SADMEP), so subtracting y from both sides yields

	2x + y – y = 5 – y
Simplifying yields	2x = 5 – y

The only operation remaining on the left side is multiplication. Undoing the multiplication by dividing both sides by 2 yields

Example: Solve the following equation for x: 3x – 4 = 2(x – 5)

Solution: Here x appears on both sides of the equal sign, so let’s move the x on the right side to the left side. But the x is trapped inside the parentheses. To release it, distribute the 2:

3x – 4 = 2x – 10

Now, subtracting 2x from both sides yields*

x – 4 = –10

Finally, adding 4 to both sides yields

x = –6

We often manipulate equations without thinking about what the equations actually say. The GRE likes to test this oversight. Equations are packed with information. Take for example the simple equation 3x + 2 = 5. Since 5 is positive, the expression 3x + 2 must be positive as well. An equation means that the terms on either side of the equal sign are equal in every way. Hence, any property one side of an equation has the

* Note, students often mistakenly add 2x to both sides of this equation because of the minus symbol between 2x and 10. But 2x is positive, so we subtract it. This can be seen more clearly by rewriting the right side of the equation as –10 + 2x.

259

260 GRE Math Bible

other side will have as well. Following are some immediate deductions that can be made from simple equations.

Equation				Deduction
y – x = 1				y > x
	y2 = x2			y = ± x, or	y	=	x	. That is,	x and y can differ only
				in sign.
	y3 = x3
				y = x
	y = x 2			y 0
	y	= 1		y > 0
	2
	x
	y	= 2		Both x and y are positive					or both x and y are
	x3			negative.
	x2 + y2 = 0			y = x = 0
3y = 4x and x > 0				y > x and y is positive.
3y = 4x and x < 0				y < x and y is negative.
	y =		x + 2	y 0 and x –2
y = 2x				y is even
y = 2x + 1				y is odd
yx = 0				y = 0 or x = 0, or both

	In Algebra, you solve an equation for, say, y by isolating y on one side of the equality
Note!	symbol. On the GRE, however, you are often asked to solve for an entire term, say, 3 – y
	by isolating it on one side.

Example:	If a + 3a is 4 less than b + 3b, then a – b =
	(A) –4	(B) –1	(C) 1/5	(D)	1/3	(E) 2
Translating the sentence into an equation gives					a + 3a = b + 3b – 4
Combining like terms gives					4a = 4b – 4
Subtracting 4b from both sides gives					4a – 4b = –4
Finally, dividing by 4 gives					a – b = –1

Hence, the answer is (B).

Sometimes on the GRE, a system of 3 equations will be written as one long “triple”

Note!

equation. For example, the three equations x

= y , y

= z, x = z, can be written more

compactly as x = y = z.

Example:

If w 0 and w = 2x =

2y , what is the value of w – x in terms of y ?

(A) 2y

(B)

(C)

(D)

y (E) y

The equation w = 2x =

2y stands for three equations: w = 2x, 2x =

2y , and w = 2y . From the last

equation, we get w =

2y ; and from the second equation, we get x =

2 y . Hence,

w x = 2y

y =

2 2y 2y =

Hence, the answer is (B).

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