GRE Problem Solving Select Many Sample Questions

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1. Question:

Which of the following is the factorization for 16(x-y)^2-9(x+y)^2? Indicate all correct choices.

A. (x-5y)(5x-y)

B. (5y-x)(5x-y)

C. (x-7y)(7x-y)

D. (7x-y)(x-7y)

E. (x-7y)(y-7x)

Correct Answer:

C and D

Explanation:

16(x-y)^2-9(x+y)^2

=[4(x-y) +3(x+y)][4(x-y)-3(x+y)]

=(4x-4y+3x+3y)(4x-4y-3x-3y)

=(7x-y)(x-7y)

Options C and D are true

2. Question:

4x+6/y=15 and 6x-8/y=14. Which of the following is true? Indicate all correct options.

A. x=2

B. y=3

C. x=1/y

D. y=2

E. x=3

Correct Answer:

D and E

Explanation:

4x+6/y=15 …(1)

6x-8/y=14 …(2)

Multiply (1) by 8 and (2) by 6 and add the two equations

32x+48/y=120

36x-48/y=84

32x+48/y+36x-48/y=120+84

68x=204

x=204/68=3

Put x=3 in (1), we get

4*3+6/y=15

12+6/y=15

6/y=3

y=6/3=2

x=3 and y=2

Options D and E are true.

3. Question:

Three numbers are in AP such that when 1 is subtracted from the first two and 1 is added to the third, the new numbers are in GP. The sum of the
three numbers in GP is 14. Which of the following are the numbers? Indicate all such choices.

A. 2

B. 3

C. 5

D. 8

E. 7

Correct Answer:

B, C and E

Explanation:

Let the numbers be a – d, a and a + d

(a ‘ d ‘ 1), (a ‘ 1) and (a + d + 1) are in GP

a ‘ d ‘ 1 + a ‘ 1 + a + d + 1=14

3a=15

a=5

The numbers are (4-d), 4 and (6+d)

16=(4-d)(6+d)=24 -6d +4d – d^2

d^2 + 2d – 8=0

d^2 + 4d – 2d – 8=0

d(d+4) – 2(d+4)=0

d=2, -4

The numbers are 3, 5 and 7 or 9, 5 and 1.

Options (B), (C) and (E) are true.

[d^2=d*d]

GRE Problem Solving Select Many Sample Questions

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