McGraw-Hill Education GRE 2019

Section 5. Quantitative Reasoning

1. A Backsolve. If x = 7, then x^2 = 49 < 50. 7 is too small of a value for x. Thus,
   Quantity A must be greater.


2. A Plug in values. Assume that Bob worked 100 hours in November and made
   $10/hr. In this case, Bob earned 100 × $10 = $1,000 in November. In December,
   he thus worked 90 hours and earned $11/hr. In this case, he earned 90 × 11 =
   $990 in December. Since he earned more in November, Quantity A is greater.


3. C To determine a value for Quantity A, use the formula for evenly spaced sets.
   The smallest multiple of 9 in the set is 18, and the last multiple of 9 in the set is
   297. Plug these values in:
   297 − 18
   9 + 1 = 32


4. C The circumference of the circle is π × d. πdd = π. Thus, the quantities are
   equal.


5. A Based on the prompt, z must be between 8 and 10 units away from zero.
   Thus, the positive values of z can equal 8, 9, or 10, and the negative values of
   z can equal −8, −9, and −10. There are 6 possible values for z. Quantity A is
   greater.


6. B Since the fractions we’re comparing have the same numerator, the fraction
   with the smaller denominator will be larger.
   The comparison is thus:

√(1 + √^13 ) versus √(1 + √^14 )

To simplify the comparison, square both quantities to arrive at:

1 + √

1

3 versus 1 + √

1

4

Simplify further by subtracting 1 from both quantities:

√

1

3 versus √

1

4

Simplify further by squaring both quantities:

1

3 versus

1

4

Since^13 >^14 , the denominator in Quantity A is greater, which means the

fraction in Quantity A is smaller.

7. C If two lines are perpendicular, then their slopes must be negative
   reciprocals, meaning their product is −1.


8. A Plug in values: Let a = 4, b = 2, and k = 1. In this case, Quantity A =^42 = 2 ,
   and Quantity B =^53. With these values, Quantity A is greater. Now, choose new
   values: Let a = 100, b = 2, and k = 1. In this case, Quantity A =^1002 = 50 ,
   and Quantity B =^1013. Quantity A is still greater.


9. A Square-root both sides: (x + 3) = 5 or (x + 3) = −5. Solve for x in each
   equation:
   x = 2 or x = − 8

CHAPTER 15 ■ PRACTICE TEST 1 513

05-GRE-Test-2018_463-582.indd 513 12/05/17 12:14 pm