Exercise Answers
Discrete Quantitative Questions
- A To simplify the inequality, combine like terms. Add x to both sides:
y > y + 3x. Subtract y from both sides: 0 > 3x. Divide both sides by 3: 0 > x. - E To determine the maximum value of xy, you should first find the solutions
for x and y. Solve for x:
(x + 3) = 12 or (x + 3) = −12
x = 9 x = −15
Solve for y:
(y + 2) = 9 or (y + 2) = −9
y = 7 y = −11
The maximum value for xy will be (−15)( −11) = 165. - A and C When solving for an absolute value, you must isolate the term
or expression inside the absolute value. Here, subtract 3 from both sides:
9|x + 3| = 45. Divide both sides by 9: |x + 3| = 5. Thus
(x + 3) = 5 or (x + 3) = −5
x = 2 x = −8 - E Back-solving is a good approach here. Start with B: (−2)^2 = 4. 4 is not
greater than 16, so eliminate B. Since B is too small, any choice with an
absolute value smaller than 2 will be too small. Thus C is also too small. Now
look at 3: 3^2 = 9. 9 < 16. So you can eliminate D. 3^2 a n d (−3)^2 have the same
value, so A is also out. - A To maximize the value of a, you should first maximize the value of
a + b + c. Since a + b + c < 27 and all the variables are integers, the maximum
value of a + b + c = 26. Next, you should minimize the values of b and c. Since
they are both positive integers and b > c, the minimum value for c = 1 and the
minimum value for b = 2. Plug these values into the equation: a + 2 + 1 = 26.
Solve for a: a = 23. - B, C, D, and E Use properties of positives and negatives to manipulate the
choices:
A: Divide both sides by y: x > 1 (remember to flip the sign!). You know that
x < 0, so x cannot be greater than 1. → Eliminate Choice A.
B: The sum of two negatives is negative. → Choice B is true.
C: negative/negative > 0. → Choice C is true.
D: If x is more negative than y, then x is further from zero than y is.
→ Choice D is true.
E: Multiply both sides by y: x < y (remember to flip the sign!). → Choice E
is true.
310 PART 4 ■ MATH REVIEW03-GRE-Test-2018_173-312.indd 310 12/05/17 11:57 am