# Free GRE Quantitative Comparison Questions with Solutions and Explanations Sample 1

Solutions and detailed explanations to questions and problems similar to the quantitative comparison GRE questions in sample 1.
When solving quantitative comparison questions, you asked to compare two quantities – Quantity A and Quantity B – and then determine which of the following statements describes the comparison:
(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.

## Solution to Questions 1

Quantity A

Quantity B

2(3 + 4)2 + 10 2 + (3 + 4)2 + 10
Evaluate expressions in A and B
A: 2(3 + 4)2 + 10 = 2(7)2 + 10 = 2(49) + 10 = 108
B: 2 + (3 + 4)2 + 10 = 2 + 49 + 10 = 61
Quantity A is greater than quantity B

## Solution to Questions 2

Quantity A

Quantity B

2×310 + 2×310 + 2×310 3×310
Simplify expressions in A and B
A: 2×310 + 2×310 + 2×310 = 2× ( 310 + 310 + 310 )
= 2× ( 3×310 ) = 2×311
B: 3×310 = 311
Quantity A is greater than quantity B

## Solution to Questions 3

Quantity A

Quantity B

25% of 300 Three quarters of 200
Evaluate expressions in A and B
A: 25% of 300 = (25/100)×300 = 25 × 300 / 100 = 75
B: Three quarters of 200 = (3/4) × 200 = 150
Quantity B is greater than quantity A

## Solution to Questions 4

ABC is a triangle such that the measure of angle A is 45°. The measure of angle C is twice the measure of angle B.

Quantity A

Quantity B

The measure of angle A The measure of angle B
Quantity A is known. Let us calculate quantity B. The measure of angle C is twice the measure of angle B is translated as follows.
C = 2 B
The sum of the measures of the three angles A, B and C of the triangles is 180°
A + B + C = 180
Substitute A by 45 and C by 2B in the above equation
45 + B + 2B = 180
Solve for B
3 B = 135
B = 135 / 3 = 45°
The measures of angle A and B are equal

## Solution to Questions 5

x is a variable that may take any real value.

Quantity A

Quantity B

x2 + 1 100x + 1
Evaluate expressions in A and B for different values of x.
x = 0
A: x2 + 1 = 02 + 1 = 1
B: 100x + 1 = 100(0) + 1 = 1
x = 1
A: x2 + 1 = 12 + 1 = 2
B: 100x + 1 = 100(1) + 1 = 101
We have tried two values of x 0 and 1 and they gave different conclusions. Therefore the relationship between the quantities in A and B cannot be determined from the information given.

## Solution to Questions 6

Quantity A

Quantity B

Area of rectangle of perimeter 240. Area of square of perimeter 240
Given the perimeter of a rectangle, we cannot calcualte its area and therefore the relationship between the quantities in A and B cannot be determined from the information given.

## Solution to Questions 7

x = - 10-2

Quantity A

Quantity B

x3 x2
Evaluate expressions in A and B for the given value of x
A: x3 = (- 10-2)3 = (-1)3(10-2)3
= - 10-6
B: x2 = (- 10-2)2 = (-1)2(10-2)2
= 10-4
Quantity B is greater than quantity A

## Solution to Questions 8

Quantity A

Quantity B

The volume of a cylindrical tube of radius 1 cm and a length of 100 m The Volume of a cylindrical container of radius 10 cm and a length of 10 m
Before we calculate the volume, we need to convert all given units to one single unit, cm in this case.
A: radius 1 cm , length = 100 m = 100 * 100 cm = 104 cm
B: radius 10 cm , length = 10 m = 10 * 100 cm = 103 cm
We now calculate the volume using the formula for the volume of a cylindrical tube: Pi * radius2 * height.
A: Pi * 12 * 104 = 104 Pi
B: Pi * 102 * 103 = 105 Pi
The volume in B is greater than the volume in A.

## Solution to Questions 9

0 < x < 1

Quantity A

Quantity B

1 / x2 1 / x
Since x is positive, we can mutliply all terms of the given inequality 0 < x < 1 by x without changing the inequality symbols.
0*x < x*x < 1*x
Which simplifies to
0 < x2 < x
Again since x is positive, the inequality x2 < x gives an inequality of the reciprocals as follows
1 / x2 > 1 / x

Quantity A is greater than quantity B

## Solution to Questions 10

A bag contains green, blue and yellow balls. The ratio of green to blue balls is 2:7. The ratio of green to yellow balls is 3:5.

Quantity A

Quantity B

Number of blue balls Number of yellow balls
Let g, b and y be the numbers of green, blue and yellow balls respectively. Using the given ratios, we can write the following fractions
g / b = 2 / 7 and g / y = 3 / 5
The first fraction gives
b / g = 7 / 2
We now evaluate the product of fractions g / y and b / g as follows
(g / y) * (b / g) = (3 / 5) * (7 / 2)
Note that (g / y) * (b / g) simplifies to b / y, hence
b / y = 21 / 10
The above fraction indicates that the number of blue balls (A) is greater than the number of yellow balls (B).