Each employee on a certain task force is either a manager or a director. What percent of the employees on the task force
are directors?
(1) The average (arithmetic mean) salary of the managers on the task force is $5,000 less than the average salary of all
employees on the task force.
(2) The average (arithmetic mean) salary of the directors on the task force is $15,000 greater than the average salary of all
employees on the task force.
Clearly, neither statement on its own is sufficient.
The two statements combined constitute a MIXTURE problem: two ingredients (managers and directors) are combined to form a mixture (all of the employees on the task force).
To evaluate the two statements combined, use ALLIGATION -- a great way to handle mixture problems.
Let M = managers, D = directors, and T = the entire task force.
Step 1: Draw a number line, with the two ingredients (M and D) on the ends and the mixture (T) in the middle:
M---------------T---------------D
Step 2: Calculate the distances between the averages.
Since the managers' average is 5000 less than the average of the entire task force, and the directors' average is 15,000 more than the average of the entire task force, we get the following distances between the averages:
M-----
5000------T----
15000------D
Step 3: Determine the ratio in the mixture.
The ratio of M to D in the mixture is the RECIPROCAL of the distances in red.
M : D = 15000:5000 = 3:1.
Since M : D = 3:1, there are 3 managers for every 1 director.
Thus, of every 4 employees, 1 is a director, implying that directors/total = 1/4 = 25%.
SUFFICIENT.
The correct answer is
C.
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