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Each employee on a certain task force is either a manager or

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Each employee on a certain task force is either a manager or [#permalink] New post 23 Jul 2007, 07:22
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A
B
C
D
E

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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 the managers on the task force is $5,000 less than the average salary 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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 [#permalink] New post 23 Jul 2007, 19:34
Ans : E

But are u sure this is the right question?
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 [#permalink] New post 23 Jul 2007, 21:15
Well neither would be sufficient and together they would be insufficient as well.
We cannot associate salaries with % of employees.
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 [#permalink] New post 24 Jul 2007, 03:56
Yeah the answer is actually C.

I figured it out last night.

Let A(t)=the average salary of all employees on the task force, A(m)= the avg. salary of managers, and A(d)=avg salary of directors. Then from (1), we have that A(m)=A(t)-5000. Insufficient. From (2), we have that A(d)=A(t)+15000. Insufficient.

Together, we have that the A(t)=(A(m)*M+A(d)D)/(M+D) where M and D are the amount of directors. So A(t)*M+A(t)*D=A(m)*M+A(d)*D. Substituting in (1) and (2), we get that A(t)M+A(t)D=A(t)M-5000M+A(t)D+15000D. Cancelling out yields 5000M=15000D. So M=3D. So the directors are 75% of the task force.
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 [#permalink] New post 24 Jul 2007, 06:28
briks123 wrote:
Yeah the answer is actually C.

I figured it out last night.

Let A(t)=the average salary of all employees on the task force, A(m)= the avg. salary of managers, and A(d)=avg salary of directors. Then from (1), we have that A(m)=A(t)-5000. Insufficient. From (2), we have that A(d)=A(t)+15000. Insufficient.

Together, we have that the A(t)=(A(m)*M+A(d)D)/(M+D) where M and D are the amount of directors. So A(t)*M+A(t)*D=A(m)*M+A(d)*D. Substituting in (1) and (2), we get that A(t)M+A(t)D=A(t)M-5000M+A(t)D+15000D. Cancelling out yields 5000M=15000D. So M=3D. So the directors are 75% of the task force.


ooh... yeah. Its always a good thing to write down the equations. It gives u a better picture.
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 [#permalink] New post 24 Jul 2007, 11:52
Simpler solution would be:

Let m be the no. of managers and d be the no. of directors and x be the average salary of all employees, then

stm 1 gives: Total salary of all managers= m*(x-5000)
Stm 2 gives: Total salary of all directors= d*(x+15000)

From stm 1 and 2 and the stem: m*(x-5000) + d*(x+15000)=(m+d)*x

which gives: 15000d=5000m, or d/(d+m)=1/4= 25%
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 [#permalink] New post 25 Jul 2007, 07:51
Good one.. i guess i jumped too soon.

using means gives us the answer.
  [#permalink] 25 Jul 2007, 07:51
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