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GRE Practice - Quantitative Reasoning - Set 13

  1. If N is an integer, and 5N2 is even, then which of the following must be true? Select all that apply.
  1. N + 5 is even
  2. N2–4 is even
  3. N + 8 is even
  4. N/2 is even
  5. N/2 is odd

The only way 5N2 is even is if N2 is even, which means N is even
Clearly N + 5 is odd, eliminate A
Now if N = 6 then N/2 is NOT even eliminate D
Now if N = 4 then N/2 is NOT odd eliminate E
Don’t over think B even if N = 2 then N2 – 4 is equal to 0 and it is even
C is also true.

  1. If a4b3c7 > 0, then which of the following must be true? Select all that apply
  1. bc > 0
  2. abc > 0
  3. ab < 0
  4. b/c > 0

a4b3c7 > 0 can be written as a4.(bc)3.c4 > 0
Now, a4 and c4 must be positive so (bc)3> 0
An odd power of bc is positive so bc is definitely positive. Select A
If bc is positive then b and c are both positive or negative. Select D

  1. x > y; xy ≠ 0
Quantity A
x2
Quantity B
y2
  1. if Quantity A is greater
  2. if Quantity B is greater
  3. if the two quantities are equal
  4. if the relationship between the two quantities cannot be determined

Solution: Don’t jump to option A
Even though x > y, if x and y are negative then
-2 > -3 but (-2)2 < (-3)2
This double case tells us we cannot be sure of the relationship between the two quantities

  1. N is an integer that leaves a remainder 1 when divided by 7
Quantity A
The remainder when N is divided by 3

Quantity B
5

  1. if Quantity A is greater
  2. if Quantity B is greater
  3. if the two quantities are equal
  4. if the relationship between the two quantities cannot be determined

Solution: This is a classic misdirection tactic.
N is an integer that leaves a remainder 1 when divided by 7 so N is of the form 7k + 1
But it’s irrelevant because when you divide N by 3 you can only get remainders of 0, 1, and 2 all of which are smaller than 5

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Q. If the function f(x) is defined as f(x) = 3(x + 1) + 7, then f(a − 2) =

  1. 3a + 5
  2. 3a – 9
  3. 3a + 10
  4. 3a – 4
  5. 3a + 4
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