# 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.

1. Quantity A

Quantity B

2(3 + 4)2 + 10 2 + (3 + 4)2 + 10

Solution

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

2. Quantity A

Quantity B

2×310 + 2×310 + 2×310 3×310

Solution

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

3. Quantity A

Quantity B

25% of 300 Three quarters of 200

Solution

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

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

Solution

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

5. x is a variable that may take any real value.

Quantity A

Quantity B

x2 + 1 100x + 1

Solution

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.

6. Quantity A

Quantity B

Area of rectangle of perimeter 240. Area of square of perimeter 240

Solution

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.

7. x = - 10-2

Quantity A

Quantity B

x3 x2

Solution

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

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

Solution

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.

9. 0 < x < 1

Quantity A

Quantity B

1 / x2 1 / x

Solution

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

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

Solution

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).

More Math Practice Tests

Free GRE Quantitative for Practice

Free Practice for GAMT Math tests

Free Compass Math tests Practice

Free Practice for SAT, ACT Math tests

Free AP Calculus Questions (AB and BC) with Answers