Section 6. Quantitative Reasoning
- A Simplify the given inequality. 6x + 8 y = 2(3x + 4 y). Thus: 2(3x + 4y) > 36 →
3 x +4y > 18. Quantity A is greater. - D Rewrite quantity A as (dc)(ab). Next, divide both quantities by (cd). The new
relationship is thus: ab vs 1. Without knowing the relationship between a and b,
we cannot determine a relationship. - A In the set of integers from 1–98, inclusive, half of the numbers are even
and half are odd. Therefore, in the set of integers from 1–99, inclusive, there
are more odd integers than even integers. The probability of choosing an odd
integer is thus greater than the probability of choosing an even integer. - D Since the equation x^2 = x is quadratic, set it equal to zero, and determine the
possible values for x: x^2 − x = 0 → x (x − 1) = 0 → x = 0 or (x − 1) = 0 → x= 0
or x = 1. The relationship cannot be determined. - C The sum of the exterior angles of any regular polygon is 360 degrees.
- D Without any information about the overlap between the percentage of
students that are boys and the percentage of students that have brown hair, a
relationship cannot be determined. - C The number of ways to select 1 person from 5 people is equivalent to the
number of ways of selecting 4 people from a group of 5. - A Use the RTD formula: 125 meterssecond × (# of seconds traveled) > 2 kilometers.
Convert kilometers to meters: 125 meterssecond × (# of seconds traveled) > 2,000
meters → # of seconds traveled >^2000125 → # of seconds traveled > 16.16
seconds >^14 of a minute. Thus, Quantity A is greater. - 163 Since multiple events must be satisfied, we should multiply the probabilities
of each individual event. The probability of rain on Tuesday is^14. The
probability of no rain on Wednesday is^34. (^14 ) × (^34 ) = 163. - B Plug in values for x. First, choose an odd value, then choose an even value.
The only choice that yields an even for both conditions is choice B. - E To achieve a maximum profit, the cost of an item should be as low as
possible and the revenue on the item should be as high as possible. The least
possible cost of the car is $8,000, and the greatest possible selling price of
the car is $18,000. The greatest profit on one car is thus $18,000 − $8,000 =
$10,000. The greatest profit on 10 cars is thus $10,000 × 10 = $100,000. - A The combined rate of the two machines is 400 sheets/5 hours = 80 sheets/
hour. The rate of the first machine is^2003. Let the rate of the second machine be
y. Thus,^2003 + y = 80. Solve for y: y = 80 −^2003 → y =^2403 −^2003 → y =^403 - C Plug in values that satisfy the given constraints. Let a = 50 and b = 25. If 50
is 25% of z, then z = 200. If z = 200, then y = 100. Now, plug in 50 for a and 25
for b into the choices, and see which choice yields a value of 100.
CHAPTER 15 ■ PRACTICE TEST 1 51505-GRE-Test-2018_463-582.indd 515 12/05/17 12:14 pm