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by selango » Mon Jul 05, 2010 5:43 am
Each employee on a certain task force is either a manager or a director. What percent of employees on the task force are directors?
1.The average salary of managers on the task force is $5000 less than the avg salary of all the employees
2.The average salary of directors on the task force is $15000 more than the avg salary of all the employees

I solved this using algebra.Any quick method appreciated.

OA C
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Source: — Data Sufficiency |
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by papgust » Mon Jul 05, 2010 5:58 am
Did you do a search on this question here? This question has been discussed a million times.
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by Rahul@gurome » Mon Jul 05, 2010 5:58 am
Let D = no. of directors, M = no. of managers, S(M) = Avg salary of managers, S(D) = Avg salary of directors and A = avg salary of all employees.
(1) S(M) = A - 5000, which is clearly NOT SUFFICIENT.
(2) S(D) = 15000 + A, which is clearly NOT SUFFICIENT.

Combining (1) and (2), A = {S(M) × M} + {S(D) × D} / {M + D} = [(A - 5000)M + (15000 + A)D] / {M + D}
AM + AD = AM + AD - 5000M + 15000D implies 15000D = 5000M or M = 3D
So, % of employees who directors are [D/(M + D)] * 100 = (D/ 4D)*100 = 25%

The correct answer is (C).
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