McGraw-Hill Education GRE 2019

6. C Make the columns comparable by manipulating Quantity A to be a fraction.
   0.25 =^14. 15% = 10015 = 203. The value in Quantity A is thus^14 × 203 = 803. The two
   quantities are equal.


7. D Manipulate the quantities:

A B

Multiply by C : 5C^2

Divide by 5 : C^2

5

1

5

5 C C

C > 0

Note that you were able to multiply by c because you were told that c > 0.

The comparison is c^2 versus 1. If c > 1, then Quantity A is greater. If c is a

fraction, then Quantity B is greater.

8. D Choose values for x. If x = −1, then the two quantities are equal. If x = −2,
   then Quantity A = −8, and Quantity B = −32, in which case Quantity A is
   greater. Since a relationship cannot be determined, the correct answer is D.


9. C First, cross-multiply the given proportion: ab =^31 → a = 3b. Next, to make
   the quantities comparable, substitute 3b for a in Quantity A. Thus Quantity
   A = 3(3b) = 9b. The values in the quantities are equal.


10. B Don’t calculate! If the diameter of the circle in Quantity B is 40, then the
    radius of that circle is^12 (40) = 20. Since the radius of the circle in Quantity B
    is greater than the radius of the circle in Quantity A, the area of the circle in
    Quantity B must be greater.


11. A Both quantities concern the same set. Within any finite set of numbers,
    there are fewer multiples of 4 than there are multiples of 3. For example,
    from 1–12, inclusive, there are four multiples of 3 (3, 6, 9, 12), but only three
    multiples of 4 (4, 8, 12). Thus Quantity A is greater.


12. A Since you are not given a value for the radius, assign a variable. Let the
    radius = r. In terms of r, Quantity A = πr2. In terms of r, Quantity B = 2πr.
    Now manipulate the columns to simplify the comparison:
    QUANTITY A QUANTITY B
    Divide by π: πr^2 2πr
    ↓ ↓
    r^2 2 r
    ↓ ↓
    Divide by r: r 2
    Since you are told that the radius is greater than 2, Quantity A must be greater.


13. A Substitute values for x and y that satisfy the given information. Let x = 13
    and y = 1. In this case, both are integers, x > y > 0, and their sum is even. Now,
    substitute these values into Quantity A: 13 – 1 = 12. In this case, Quantity A
    is greater. Choose new values for x and y that satisfy the given information. To
    prove Choice D, you want to make the difference between x and y as small as
    possible. Let x = 3 and y = 1. In this case, all the restrictions are satisfied.
    3 – 1 = 2. Thus Quantity A is still greater.

CHAPTER 8 ■ QUANTITATIVE COMPARISON STRATEGIES 171

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