14. B Your first step should be to set up a proportion, where s represents the
number of students and t represents the number of teachers. However, note
that the proportion will be an inequality since the ratio is greater than 3:1:
s
t >3
1
There are 500 students, so^500 t >^31. Cross-multiply to arrive at a range for
t: 500 > 3t. Solve for t:^5003 > t. Since t must be an integer, the greatest value for t
will be the greatest integer less than^5003.^5003 = 166.6, so the greatest value for t
is 166. The correct answer is B.- C The best approach here is to plug in values. Choose a value for a and let that
value determine b. If a = 6, then 42 = 6b, and b = 7.^17 (a) =^17 (6) =^67 , and^16 (b) =
(^16 )(7) =^76. Now find the ratio of^67 to^76 :6
7
7
6=^67 × 67 =^3649. The correct answer is C.Quantitative Comparison Questions
- B Choose values that satisfy the constraints in the given information: let’s
say there are 3 white marbles and 7 black marbles. In this case, the ratio of
white marbles to all marbles is 103. 103 <^13 , so for the given values, Quantity B
is greater. Choose a “weird” case: 7 is the smallest possible number of black
marbles when there are 3 white marbles. Let’s choose a larger value for the
number of black marbles. Let there be 3 white marbles and 100 black marbles.
In this case, the ratio of white marbles to all marbles is 1033. 1033 <^13 , so Quantity
B is still greater. - D Since there are more boys than girls, the current ratio of boys to girls must
be greater than 1. The impact of adding 10 girls and 20 boys on the ratio will
depend on the original ratio. For example, let’s say there were originally 100
boys and 50 girls. In this case, the original ratio is^10050 =^21. When 20 boys and
10 girls are added, the new ratio is^12060 = 2 to 1. In this case, the quantities are
equal. Now let’s say the original number of boys was 100 and the original
number of girls was 90. In this case, the original ratio is^10090 =^109. The new ratio
will be^120100 =^65. In this case, Quantity B is greater. Since there is more than one
relationship between the quantities, the correct answer is Choice D. - D Set the ratios up as fractions:^12 x = x 3. Next, cross-multiply: x^2 = 36. Since
the exponent on x is even, x = 6 or –6. The relationship between the columns
cannot be determined. - C Let the revenue = 5x and the expenses = 4x. Use the profit formula: profit =
revenue – expenses. Thus profit = 5x – 4x = x. The ratio of profit to expenses is
thus 4 xx =^14. The two quantities are equal. - C When you are given two ratios with a common element, you should
manipulate the ratios so that the common element has the same value in
both ratios. In this case, the common element is b. In the first ratio, ab = 5/12,
and in the second ratio, bc =^47. Manipulate the second ratio to make b = 12:
b
c =
4 × 3
7 × 3 =12
21. Thus the ratio of a:b:c = 5:12:21. Since c < 25 and is an integer,
c must equal 21. If c = 21, then a = 5.CHAPTER 10 ■ PART-TO-WHOLE RELATIONSHIPS 25503-GRE-Test-2018_173-312.indd 255 12/05/17 11:53 am