GRE Test Exam Questions and Answers

QUESTION: 41

If -1 < x < 0, which of the following inequalities must be true?

x < x2 < x3
x < x3 < x2
x2 < x < x3
x3 < x < x2
x3 < x2 < x

Answer(s): D

Explanation:

Since x is negative, squaring it makes it positive, so x2 > 0.
Cubing a negative number keeps it negative, but it is closer to 0 than x, so x3 > x (since for negative numbers, cubing makes them smaller in magnitude).
Since x2 is positive and both x and x3 are negative, it follows that x3 < x < x2.

Reveal Solution Next Question

QUESTION: 42

For all integers x, x is defined by x = (x - 1)2 + 1.
What is the value of (-3) - (3)?

0
5
8
12
17

Answer(s): D

Explanation:

Substitute x = -3 into the definition of x:
(-3) = ((-3) - 1)2 + 1 = (-4)2 + 1 = 16 + 1 = 17
Substitute x = 3 into the definition of x:
(3) = (3 - 1)2 + 1 = 22 + 1 = 4 + 1 = 5
Now, subtract the two values:
(-3)- (3) = 17 - 5 = 12

Reveal Solution Next Question

QUESTION: 43

Which of the following inequalities is equivalent to |x - 2| < 4?

-2 < x < 2
-2 < x < 6
-4 < x < 4
-6 < x < 2
-6 < x < 6

Answer(s): B

Explanation:

The inequality |x - 2| < 4 means
-4 < x - 2 < 4
-4 + 2 < x - 2 + 2 < 4 + 2
-2 < x < 6

Reveal Solution Next Question

QUESTION: 44

w + 6, m + 6, s + 6

If k is the standard deviation of the three numbers in the list above, what is the standard deviation of w, m, and s?

18k
6k + 6
k + 18
k + 6
k

Answer(s): E

Explanation:

The standard deviation measures the spread of numbers, and adding the same constant to each number in a dataset does not change the standard deviation.
In this case, each number in the dataset has 6 added to it, but the differences between the numbers remain the same. Since standard deviation is only affected by the spread and not by shifts in values, the standard deviation of w, m, and s remains k.

Reveal Solution Next Question

QUESTION: 45

If -1 < x < 0 and 0 < y < 1, which of the following CANNOT be equal to 0?

x + y
x - y + l
2x + y + 1
2y + x + 1
2y - 3x - 1

Answer(s): D

Explanation:

Option x + y

Since x is negative and y is positive, their sum can be either negative or positive, meaning it could be 0 if x and y are chosen appropriately.
Option x - y + 1

Rewriting: x + 1 - y.
Since x is greater than -1, we know x + 1 is always positive. Subtracting a positive y means this expression can be positive, negative, or 0 depending on the values of x and y.
Option 2x + y + 1

Rewriting: 2x + (y + 1).
Since -1 < x < 0, multiplying by 2 gives -2 < 2x < 0.
Since 0 < y < 1, then 1 < y + 1 < 2.
Adding 2x + (y + 1) means

-2 + 1 < 2x + y + 1 < 0 + 2.
-1 < 2x + y + 1 < 2.
Since this range includes 0, the expression can be 0.
Option 2y + x + 1

Rewriting: x + 1 + 2y.
Since -1 < x < 0, we know x + 1 is positive. Since 2y is also positive, their sum must always be positive.
Thus, this expression cannot be equal to 0.
Option 2y - 3x - 1

Rewriting: 2y - 3x - 1.
Since 0 < y < 1, multiplying by 2 gives 0 < 2y < 2.
Since -1 < x < 0, multiplying by -3 reverses the inequality, giving 0 < -3x < 3.
Since 2y is positive, -3x is positive, and we subtract 1, this can be 0 if chosen appropriately.

Reveal Solution Next Question

QUESTION: 46

If is an integer, which of the following numbers must also be an integer? (Choose all that apply.)

x - 1
x + 1

Answer(s): A,C,D

Explanation:

x - 1 is a multiple of 4.
x - 1 = 4k for some integer k.
Option x - 1
Since x - 1 = 4k, which is a multiple of 4, it must be an integer.
Option x + 1
Rewriting x + 1 in terms of k:

x + 1 = (x - 1) + 2 = 4k + 2
Since 4k is a multiple of 4 but adding 2 does not ensure divisibility by 4, this may not always be an integer.
Others options:

2k and 2k + 1 must be an integer.
The last option is not necessarily an integer.

Reveal Solution Next Question

QUESTION: 47

If x is an integer and the sides of a triangle are x + 3, 2x, and x + 5, respectively, which of the following could NOT be the perimeter of the triangle?

16
20
28
30
32

Answer(s): D

Explanation:

The perimeter of the triangle is:
(x + 3) + (2x) + (x + 5) = 4x + 8.
Since the perimeter is 4x + 8, we check which given perimeter values are possible by solving for x.
For perimeter 16:
4x + 8 = 16
4x = 8
x = 2
For perimeter 20:
4x + 8 = 20
4x = 12
x = 3
For perimeter 28:
4x + 8 = 28
4x = 20
x = 5

For perimeter 30:
4x + 8 = 30
4x = 22
x = 5.5
For perimeter 32:
4x + 8 = 32
4x = 24
x = 6
The only perimeter value that is not possible because it results in a non-integer x is 30.

Reveal Solution Next Question

QUESTION: 48

If P and Q are positive integers that have no common factors other than 1, what is the value of Q?

30
60
120
240
360

Answer(s): B

Explanation:

The denominators of the fractions are 2, 3, 4, 5, and 6. The least common denominator (LCD) of these numbers is the least common multiple (LCM) of 2, 3, 4, 5, and 6.
The prime factorizations are:
2 = 2

3 = 3

4 = 22

5 = 5

6 = 2 × 3

The LCM is the product of the highest powers of all prime factors:
LCM(2, 3, 4, 5, 6) = 22 × 3 × 5 = 60.

The result is already in its simplest form because 23 is a prime number, and it has no common factors with 60.

Reveal Solution Next Question

GRE GRE Test Exam (page: 6) GRE Graduate Record Examination Test: Verbal, Quantitative, Analytical Writing Updated on: 15-Feb-2026

QUESTION: 41

Explanation:

QUESTION: 42

Explanation:

QUESTION: 43

Explanation:

QUESTION: 44

Explanation:

QUESTION: 45

Explanation:

QUESTION: 46

Explanation:

QUESTION: 47

Explanation:

QUESTION: 48

Explanation:

GRE GRE Test Exam (page: 6)
GRE Graduate Record Examination Test: Verbal, Quantitative, Analytical Writing
Updated on: 15-Feb-2026