McGraw-Hill Education GRE 2019

Section 6. Quantitative Reasoning

1. B Since a, b, and c are the angles inside a triangle, the sum of their measures
   must be 180. Since each of angles d, e, and f are supplementary to angles a, b,
   and c, respectively, d + e + f = 540 − 180 = 360.


2. D Plug in numbers. First, let a = 2, and b = 2. In this case, the value of
   Quantity A is^12 +^12 = 1. The value of Quantity B is

1

1

2 +

1

2

=^11 = 1. In this case, the

two quantities are equal. Now, let a = 10 and b = 10. In this case, the value of

Quantity A is 101 + 101 = 102 =^15. The value of Quantity B is

1

1

10 +

1

10

=

1

2

10

=

1

1

5

= 5. In this case, Quantity B is greater. A relationship cannot be determined.

3. B Don’t calculate! The number of terms in the set in Quantity A is^982 = 49.
   The number of terms in Quantity B is^982 + 1 = 50. Each term in Quantity A is
   one more than the corresponding term in Quantity B, Thus, up until 98, the
   sum of the terms in Quantity A is 49 greater. However, this is offset by the
   additional 99 in Quantity B. Thus, Quantity B is greater.


4. D Plug in values. First, let x = −1 and let y = 1. In this case, the two quantities
   are equal. The correct answer must be C or D. Next, let x = −1 and let y = 2. In
   this case, the values in the quantities are different.


5. A Answer: Since a is an integer, the greatest possible value for a is 12, since 12^2
   = 144 (note that 13 is too large since 13^2 = 169). The least possible value for a
   is –12, since (–12)^2 = 144 (note that –13 is too small since (–13)^2 = 169. We can
   express a’s range as: –12 ≤ a ≤ 12. In this range, there are 12 negative integers
   (–12 thru –1), 12 positive integers (1–12), and zero. Therefore, there are 25
   possible values for a.


6. D The given information is insufficient to conclude any relationship between
   all male professors and all female professors.


7. C Simplify Quantity A by expressing the numerator as base 4: 16−a = (4^2 )−a = 4−2a.
   Therefore,^16

–a

4 –b =

4 –^2 a

4 –b = 4

(–2a + b) or 4b – 2a. The two quantities are equal.

8. A If (^17 )x of the products were defective, then (^67 )x were not defective.
   Of these nondefective devices, 30% = 103 were not sold. ( 103 )(^67 )x = ( 7018 )x = ( 359 )x.
   ( 359 )x > (^14 )x. Quantity A is greater.


9. C Solution: To isolate a, divide both sides of the equation by 16 to arrive at:
   a =^64

b

16

Next, express the bases in base-2. 64 = 2^6 , so the numerator = (2^6 )b = 2^6 b.

Expressed in base-2, the denominator = 2^4. So the equation simplifies to:

a = (2

6 b)

24.

Since we’re dividing terms with the same base, we’ll keep the base and subtract

the exponents to arrive at: 2(6b – 4).

572 PART 5 ■ GRE PRACTICE TESTS

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