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by sparkle6 » Thu Sep 22, 2011 10:53 pm
Each employee on a certain task force is either a manager or director. What percent of the employees on the task force are directors?

1) The average salary of the managers on the task force is $5000 less than the average of all employees on the task force
2) The average salary of the directors on the task force is $15,000 greater than the average salary of all employees on the task force
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Source: — Data Sufficiency |

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by Anurag@Gurome » Thu Sep 22, 2011 11:06 pm
sparkle6 wrote:Each employee on a certain task force is either a manager or director. What percent of the employees on the task force are directors?

1) The average salary of the managers on the task force is $5000 less than the average of all employees on the task force
2) The average salary of the directors on the task force is $15,000 greater than the average salary of all employees on the task force

Let the average salary of managers of the task force = S(m), the average salary of the directors on the task force = S(d), and the average salary of all the employee on the task force = S(e).
Let the no. of managers = m and no. of directors = d. We have to find d/(m + d).

(1) S(m) = S(e) - 5000, which ALONE is NOT SUFFICIENT.

(2) S(d) = S(e) + 15000, which ALONE is NOT SUFFICIENT.

Combining (1) and (2), we know that S(e) = {m * S(m) + d * S(d)}/(m + d)
So, S(e) = {m * [S(e) - 5000] + d * [S(e) + 15000]}/(m + d)
m * S(e) + d * S(e) = m * S(e) - 5000m + d * S(e) + 15000d
15000d = 5000m
3d = m
So, d/(m + d) = d/4d = 1/4, which is SUFFICIENT.

The correct answer is C.
Anurag Mairal, Ph.D., MBA
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