McGraw-Hill Education GRE 2019

(singke) #1
value, the spread between the data points will not change. Thus the standard
deviation will not change. Since the standard deviation of the original set was
a, the standard deviation of the new set is also a.


  1. D Since the frequency of the data points differs, this is a weighted average.
    Let’s use this formula: weighted average = (frequency of data point 1/total
    number of data points) × data point 1 + (frequency of data point 2/total
    number of data points) × data point 2. In the question, the data points are
    the average ages for the two groups. Let the average age for the directors be
    data point 1 and the average age for the managers be data point 2. Thus the
    weighted average = (^300500 )42 + (^200500 )38 = (^35 )42 + (^25 )38 = 25.2 + 15.2 = 40.4.

  2. C Since the choices have variables, you can plug in numbers.
    Step 1: Let x = 100.
    Step 2: Answer the question when x = 100. Originally, the sum of the scores
    is thus A × N = 100 × 10 = 1,000. After the corrections, one of the scores
    will decrease by 10 points, and one of the scores will increase by 20 points.
    The net change is thus (20 –10) = 10. The new sum is 1,010. The new average
    is thus NS = 1,010/10 = 101. The goal is 101.
    Step 3: Substitute 100 for x in the choices, and identify which choice
    matches the target of 101.

  3. C The sum of the terms is n + (n + 1) + (n + 7) + (n + 9) + (n + 13) = 5n + 30.
    Since there are 5 terms, the average of the set is^5 n + 30 5 = n + 6.


Quantitative Comparison Questions



  1. A The average for a set must fall between the average of each subset. Thus the
    average annual salary must be greater than $52,000.

  2. A Don’t calculate! Look at how the average of 6, 6, 8, and 10 compares to
    7. The average of those numbers is 6 + 6 + 8 + 10 4 =^304 > 7. Quantity A has four
    numbers with an average of 7 (7, 7, 7, and 7) and four numbers with an average
    greater than 7. Its average must therefore be greater than 7.

  3. D Without knowing where the average age for the entire school falls in
    relation to 16.3 and 16.8, it is not possible to determine the ratio of boys
    to girls.

  4. A Express Quantities A and B algebraically:
    Quantity A →^ A = NS =^2 a + 3 b
    Quantity B → A = NS =^3 a + 2 5 b


338 PART 4 ■ MATH REVIEW

04-GRE-Test-2018_313-462.indd 338 12/05/17 12:03 pm

Free download pdf