The sequence a 1 , a 2 , a 3 ,... , an,... is such that an = an−1 + 7, for all n > 1.
QUANTITY A QUANTITY B- a 3 a 1 + 15 A B C D
Exercise Answers
Discrete Quantitative Questions
- 8 Plug in values: Let the original edge of the cube = 2. The original volume
of the cube is thus 2^3 = 8. If the original edge = 2, then the new edge = 4. The
new volume of the cube is thus (4^3 ) = 64. The new volume is thus 8 times the
original volume. - C Plug in values: Let t = 1 for Gerald and Harry. Let d = 2 for Harry and let d
= 4 for Gerald. Harry’s efficiency is thus 3(1 223 ) =^34. Gerald’s efficiency is 3(1 423 ) = 163.
3
4 is 4 times greater than
3
16.- 21 The formula specifies to triple the value in front of the star, double the
value after the star, and then add the resulting terms. Substitute 5 for a in the
original formula and substitute 3 for b to arrive at 3(5) + 2(3) = 21. - E Solution: When working with compound functions, solve for the inside
value first: g(2) = 1 + 2^1 =^13. f(g(2)) = f(^13 ) =^11
3
= 3.
- B Substitute 3 for x in the original function:
a(3 33 ) =^273 a
9 a = 45
a = 5 - C When working with compound functions, first obtain the value of the
function within the parentheses. 5 $ 8 = (5)(8) 2 = 20. Substitute 20 for (5 $ 8) and
arrive at a value for: 3 $ 20. Repeat the formula: (3)(20) 2 = 30. - C Substitute (x − 3) for x:
3(x − 3)^2 = 12
(x − 3)^2 = 4
(x − 3) = +2 or −2
If (x − 3) = 2, then x = 5. If (x – 3) = –2, then x = 1. Only 5 appears in the
choices. - B and C Substitute values for a and b. Let a = 2 and b = 3. Now identify for
which of the choices 2 ★ 3 = 3 ★ 2.
A: 3 ★ 2 = 2(3 − 2) = 2; 2 ★ 3 = 3(2 − 3) = −3. → Eliminate Choice A.
B: 3 ★ 2 = 3(3 + 2) = 3(5) = 15; 2 ★ 3 = 3(2 + 3) = 3(5) = 15. The outputs are
equal. → B is an answer.
C: 3 ★ 2 = (3 × 2)^3 = 6^3 = 216; 2 ★ 3 = (2 × 3)^3 = 6^3 = 216. The outputs are
equal. → C is an answer.
CHAPTER 11 ■ ALGEBRA 29903-GRE-Test-2018_173-312.indd 299 12/05/17 11:56 am