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
    Answer (A)

    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
    Answer (A)

    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
    Answer (B)

    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
    Answer (C)

    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.
    Answer (D)

    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.
    Answer (D)

    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
    Answer (B)

    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.
    Answer (B)

    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
    Answer (A)

    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).
    Answer (A)


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