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GRE GRE Test Exam (page: 2)
GRE Graduate Record Examination Test: Verbal, Quantitative, Analytical Writing
Updated on: 31-Mar-2026

Viewing Page 2 of 75
GRE Test PDF 793 Questions



In the xy-plane, line segment, RS is one of the sides of square RSTU (not shown).

Quantity A
The area of square RSTU

Quantity B
13

  1. Quantity A is greater.
  2. Quantity B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given.

Answer(s): C

Explanation:

The given points are:
· R (0, 2)
· S (3, 0)
Since RS is one side of square RSTU, the side length of the square is d (the result of the distance formula):



Area = side length2



Quantity A



Quantity B

  1. Quantity A is greater.
  2. Quantity B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given.

Answer(s): B

Explanation:





Seventy-five boxes of light bulbs were examined during a quality control check. The number of defective light bulbs per box and the corresponding frequency are given in the table.

Quantity A
The average (arithmetic mean) number of defective light bulbs per box for the 75 boxes

Quantity B
The median number of defective light bulbs per box for the 75 boxes

  1. Quantity A is greater.
  2. Quantity B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given.

Answer(s): B

Explanation:

To calculate the average number of defective bulbs, we first compute the total number of defective bulbs in all 75 boxes:
For 0 defective bulbs: 35 × 0 = 0

For 1 defective bulb: 20 × 1 = 20

For 2 defective bulbs: 12 × 2 = 24

For 3 defective bulbs: 2 × 3 = 6

For 4 or more defective bulbs: 6 × 4 (using 4 as an estimate) = 24


Now, add all these values together:
0 + 20 + 24 + 6 + 24 = 74
The total number of defective bulbs is 74, and the total number of boxes is 75. So, the average number of defective bulbs per box is:



The median is the middle value in the data set when ordered. To find the median, we first determine the cumulative frequency:
Cumulative frequency for 0 defective bulbs: 35

Cumulative frequency for 1 defective bulb: 35 + 20 = 55

Cumulative frequency for 2 defective bulbs: 55 + 12 = 67

Cumulative frequency for 3 defective bulbs: 67 + 2 = 69

Cumulative frequency for 4 or more defective bulbs: 69 + 6 = 75

There are 75 boxes, so the median is the 38th value. From the cumulative frequency, we see that the 38th value falls in the range where there are 1 defective bulb, since the cumulative frequency for 0 defective bulbs is 35 and for 1 defective bulb is 55.
Thus, the median number of defective bulbs is 1.
Compare Quantity A and Quantity B
Quantity A: The average number of defective bulbs per box is approximately 0.987.

Quantity B: The median number of defective bulbs per box is 1.

Since 0.987 is less than 1, Quantity B is greater.



Quantity A
The number of different prime factors of 500

Quantity B
The number of different prime factors of 360

  1. Quantity A is greater.
  2. Quantity B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given.

Answer(s): B

Explanation:

To compare the number of different prime factors of 500 and 360, we first need to find the prime factorization of each number.
For 500, start by dividing by 2:
500 ÷ 2 = 250
250 ÷ 2 = 125
Next, divide by 5:

125 ÷ 5 = 25
25 ÷ 5 = 5
5 ÷ 5 = 1
So, the prime factorization of 500 is:
500 = 22 × 53
The distinct prime factors are 2 and 5, meaning there are 2 different prime factors.
For 360, start by dividing by 2:
360 ÷ 2 = 180
180 ÷ 2 = 90
90 ÷ 2 = 45
Next, divide by 3:
45 ÷ 3 = 15
15 ÷ 3 = 5
Finally, divide by 5:
5 ÷ 5 = 1
The prime factorization of 360 is:
360 = 23 × 32 × 5.
The distinct prime factors are 2, 3, and 5, meaning there are 3 different prime factors.
Now, comparing the two quantities:
The number of different prime factors of 500 is 2.

The number of different prime factors of 360 is 3.

Since 2 is less than 3, the number of different prime factors of 360 is greater.





Point P is shown in the rectangular coordinate system.

Quantity A
The ratio of the x-coordinate of P to the y-coordinate of P

Quantity B
-1

  1. Quantity A is greater.
  2. Quantity B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given.

Answer(s): C

Explanation:

Point P is in the second quadrant, and it is located along a dashed line that forms a 45-degree angle with the negative x-axis. In this case, the coordinates of P can be represented as (x, y).
Since the angle between the line and the negative x-axis is 45 degrees, the coordinates of P must satisfy the property that the absolute values of x and y are equal. Given that the point is in the second quadrant, the x- coordinate is negative and the y-coordinate is positive. This means x = -a and y = a for some positive value a.
Now, calculating the ratio of the x-coordinate to the y-coordinate: x divided by y = (-a) divided by a = -1.
Thus, Quantity A is -1. Quantity B is also given as -1.



1 kilogram is approximately equal to 2.2 pounds.

x < y

Quantity A
The number of pounds equal to x kilograms

Quantity B
The number of kilograms equal to y pounds

  1. Quantity A is greater.
  2. Quantity B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given.

Answer(s): D

Explanation:

The number of pounds equal to xxx kilograms can be found by multiplying x by 2.2:
Quantity A = 2.2x
To convert y pounds to kilograms, we divide y by 2.2:



Since we are only given x < y, it does not directly establish a clear comparison between quantities because of the conversion factor. The inequality x < y does not necessarily imply that the converted values will have a strict relationship.



A box contains 10 red balls, 5 blue balls, and no other balls. Ann will randomly select, without replacement, 2 balls from the box.

Quantity A
The probability that Ann will select 2 red balls

Quantity B

  1. Quantity A is greater.
  2. Quantity B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given.

Answer(s): C

Explanation:

The total number of balls in the box is:
10(red balls) + 5(blue balls) = 15(total balls)
The total number of ways to select 2 balls from 15 is given by the combination formula:



Now, the number of ways to select 2 red balls from the 10 red balls is:



Thus, the probability that Ann will select 2 red balls is:



Since both quantities are equal, we conclude that the two quantities are the same.



x is the units digit of the number 1114 - 1.

Quantity A
x

Quantity B
1

  1. Quantity A is greater.
  2. Quantity B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined from the information given.

Answer(s): B

Explanation:

The units digit of powers of 11 repeats in a cycle. Let's examine the first few powers of 11:
111 = 11, so the units digit is 1.

112 = 121, so the units digit is 1.

113 = 1331, so the units digit is 1.

114 = 14641, so the units digit is 1.

We observe that the units digit of 11n for any positive integer n is always 1.
Since the units digit of 1114 is 1, the units digit of 1114 - 1 is:
1 - 1 = 0
Quantity A: The units digit of 1114 - 1, which is 0.

Quantity B: 1.

Since 0 < 1, Quantity B is greater.



Viewing Page 2 of 75



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