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GRE GRE Test Exam Questions
Graduate Record Examination Test: Verbal, Quantitative, Analytical Writing (Page 10 )

Updated On: 9-Mar-2026
Page 10 of 119
PDF with 793 Questions

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

  1. x < x2 < x3
  2. x < x3 < x2
  3. x2 < x < x3
  4. x3 < x < x2
  5. 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.



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

  1. 0
  2. 5
  3. 8
  4. 12
  5. 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



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

  1. -2 < x < 2
  2. -2 < x < 6
  3. -4 < x < 4
  4. -6 < x < 2
  5. -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



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?

  1. 18k
  2. 6k + 6
  3. k + 18
  4. k + 6
  5. 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.



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

  1. x + y
  2. x - y + l
  3. 2x + y + 1
  4. 2y + x + 1
  5. 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.



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