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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 22 Jul 2005, 22:07
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C
D
E

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Each employee on a certain task force is either a manager or a director. What percent ofthe employees on the task force are directors?

(1) the average (arithematic mean) of the 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 (arithematic mean) of the salary of the directors on the task force is $15,000 greater than the average salary of all employees on the task force
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 [#permalink] New post 23 Jul 2005, 00:15
from stmnt 1) AvM=AvAll-5000$
from stmnt2) AvD=AvAll+15000$
Each of them is by itself insufficient
Taken together when substract 1) from 2) WE GET AvD-AvM=20000$
Think it should be E)
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Re: 'mean' DS [#permalink] New post 23 Jul 2005, 05:38
Dan wrote:
Each employee on a certain task force is either a manager or a director. What percent ofthe employees on the task force are directors?

(1) the average (arithematic mean) of the 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 (arithematic mean) of the salary of the directors on the task force is $15,000 greater than the average salary of all employees on the task force


My answer is C,

Obviously both conditions alone are insufficient..

So let us take a situation that combines both condition

Group avg of 10000
M for no of managers (Avg salary of 5000)
D for no of directors (Avg salary of 25000)

Now according to the conditons

(M*5000 + D*25000)/(M+D) = 10000

solving this

D/M = 1/3, so directors are 25% of the group.

If somethings wrong, Pl.let me know
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Re: 'mean' DS [#permalink] New post 23 Jul 2005, 08:01
ranga41 wrote:
Dan wrote:
Each employee on a certain task force is either a manager or a director. What percent ofthe employees on the task force are directors?

(1) the average (arithematic mean) of the 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 (arithematic mean) of the salary of the directors on the task force is $15,000 greater than the average salary of all employees on the task force


My answer is C,

Obviously both conditions alone are insufficient..

So let us take a situation that combines both condition

Group avg of 10000
M for no of managers (Avg salary of 5000)
D for no of directors (Avg salary of 25000)

Now according to the conditons

(M*5000 + D*25000)/(M+D) = 10000

solving this

D/M = 1/3, so directors are 25% of the group.

If somethings wrong, Pl.let me know


Ranga, do you care to explai how you got group average = 10000?
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 [#permalink] New post 24 Jul 2005, 22:36
OA is C but I don't get other than E.
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 [#permalink] New post 25 Jul 2005, 00:34
The sum of both averages is AvM+AvD=2AvAll+10000$I assume that the ratio of M/D=1 and in this case need to devide the equation by2 to find the Av of the group.When M/D=1 then we have 1/2AvM=(AvAll-5000$)/2 and 1/2AvD=(AvAll+15000$)/2 and we get AvM+AvD=AvAll+5000$ or the ratio is 50:50
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 [#permalink] New post 25 Jul 2005, 02:34
Lets solve this question "hard" way, WITHOUT any assumptions and picking numbers:

M=Num. of Managers
D=Num. of Directors
S1=Salary of Managers
S2=Salary of Directors.

S1+5000 = (MS1+DS2)/(M+D)
S2-15000 = (MS1+DS2)/(M+D)

E.g., S1+5000=S2-15000 -> S2-S1=20000

Lets return to our equation.
Combine both equations and get:
S1 + S2 - 10000 = 2(MS1 + DS2) / (M + D)
S1 + S2 - 2(MS1 + DS2)/(M + D)) = 10000
(M - D) * (S2 - S1)/(M + D) = 10000
S2 - S1 = 20000
20000 * (M - D) = 10000 * (M + D)
M = 3D
1/(1 + 3) = 0.25 - > 25%

P.S.: Don't correct me, because I know that I'm right:)
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 [#permalink] New post 25 Jul 2005, 03:39
Can you please elaborate on this part:
S1 + S2 - 2(MS1 + DS2)/(M + D)) = 10000
(M - D) * (S2 - S1)/(M + D) = 10000
S2 - S1 = 20000

i can't get it clearly
thanks.
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 [#permalink] New post 25 Jul 2005, 04:20
BG,
We have 2 equations

S1+5000 = (MS1+DS2)/(M+D)
S2-15000 =(MS1+DS2)/(M+D)

Substract first equation from the second one, and get
S1-S2+20000=0 -> S2-S1=20000(!)Remeber this equation, you will use it later.

then,
Add both equations, and get:
S1+S2-10000=2*(MS1+DS2)/(M+D),
simplify and get:

(M - D) * (S2 - S1)/(M + D) = 10000

Now, as we now that S2-S1=20000(!), we can use it here,
(M-D)*20000/(M+D)=10000
Multiply both sides by M+D and get
20000(M-D)=10000(M+D) -> 2M-2D=M+D -> M=3D

Hope it helps.
  [#permalink] 25 Jul 2005, 04:20
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