- E You have a range for x, and you have ranges for y and z in terms of x.
Manipulate to get ranges for y and z. If x > 8, then 2x > 16. If y > 2x, then
y > 16. If z > x and x > 8, then z > 8. Choosing extremes, you know that if
y > 16 and z > 8, then y + z > 24. All of the values in the choices are greater
than 24 except for E. - 14 4 To solve for the maximum value of ab, first get the ranges of a and b
individually. If |a + 3| ≤ 6, then a + 3 must be at most 6 units away from zero.
Expressed algebraically, this means that −6 ≤ a + 3 ≤ 6. Solve for a: −9 ≤ a ≤ 3.
Go through the same process to determine a range for b. Using this reasoning,
you can determine that −12 ≤ b + 4 ≤ 12. Solve for b: −16 ≤ b ≤ 8. It might be
tempting to choose 24 as the maximum value of ab, but notice that (−16)(−9)
will yield 144. - C Solution: Rewrite the inequality as (x/c)(b/y) > x/c. Divide both sides by
(x/c): b/y > 1. Multiply both sides by y: b > y. Note that you were able to divide
by these variables because you are told that all the variables are positive. - E Think of absolute value as distance from zero. If |a| is between 2 and 9,
then a is between 2 and 9 units away from zero. Looking to the right of zero,
this means that a can be any integer from 3 through 8, inclusive. Looking
to the left of zero, this means that a can be any integer from −8 through −3,
inclusive. There are 6 integers from 3 through 8, inclusive, and 6 integers from
−8 through −3, inclusive. Thus there are 12 possible values for a. The correct
answer is E.
Quantitative Comparison Questions
- A Since you are comparing an unknown to a value, let’s determine how
the possible values of that unknown relate to 1. You have ranges for x and y,
so choose extremes. If x =^78 and y =^17 , then sum will be^5756 , which is greater
than 1. If you choose larger values for x and y, the sum will only get larger.
Therefore, x + y will always be greater than 1. - D The best approach here is to choose values. Since the columns deal with
absolute values, you should consider positive and negative cases that
satisfy x > y.
Case 1: x = 3 and y = 2. In this case, |3| is greater than |2| and Column A
is greater.
Case 2: x = 3 and y = −5. In this case, |3| is less than |−5|, and Column B is
greater. A relationship cannot be determined. - B It might be tempting to multiply the minimum value for a and the
minimum value for b to arrive at −6. However, note that when you multiply a
negative by a positive, the larger the positive number is, the more negative the
product is. Thus to minimize ab, you should multiply −2 by the largest value
for b: 5. −2(5) = −10, which is less than −6. - A When comparing an unknown to zero, your concern should be the sign of
that unknown. You want to know the sign of a + b, so you should manipulate
the given information to isolate a + b. Add 3b to both sides and arrive at 3a +
3 b > 0. Factor: 3(a + b) > 0. Divide both sides by 3: a + b > 0.
CHAPTER 11 ■ ALGEBRA 31103-GRE-Test-2018_173-312.indd 311 12/05/17 11:57 am