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

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Manager
Joined: 28 Aug 2004
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Each employee on a certain task force is either a manager or [#permalink]  22 Jul 2005, 22:07
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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
Director
Joined: 13 Nov 2003
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Location: BULGARIA
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Kudos [?]: 27 [0], given: 0

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) Senior Manager Joined: 29 Nov 2004 Posts: 486 Location: Chicago Followers: 1 Kudos [?]: 9 [0], given: 0 Re: 'mean' DS [#permalink] 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 _________________ Fear Mediocrity, Respect Ignorance Director Joined: 27 Dec 2004 Posts: 908 Followers: 1 Kudos [?]: 19 [0], given: 0 Re: 'mean' DS [#permalink] 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 $15,000 greater than the average salary of all employees on the task force My answer is C, Obviously both conditions alone are insufficient.. So let us take a situation that combines both condition Group avg of 10000 M for no of managers (Avg salary of 5000) D for no of directors (Avg salary of 25000) Now according to the conditons (M*5000 + D*25000)/(M+D) = 10000 solving this D/M = 1/3, so directors are 25% of the group. If somethings wrong, Pl.let me know Ranga, do you care to explai how you got group average = 10000? Manager Joined: 28 Aug 2004 Posts: 205 Followers: 1 Kudos [?]: 1 [0], given: 0 [#permalink] 24 Jul 2005, 22:36 OA is C but I don't get other than E. Director Joined: 13 Nov 2003 Posts: 793 Location: BULGARIA Followers: 1 Kudos [?]: 27 [0], given: 0 [#permalink] 25 Jul 2005, 00:34 The sum of both averages is AvM+AvD=2AvAll+10000$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
Senior Manager
Joined: 29 Jun 2005
Posts: 403
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Kudos [?]: 15 [0], given: 0

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

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:)
Director
Joined: 13 Nov 2003
Posts: 793
Location: BULGARIA
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Kudos [?]: 27 [0], given: 0

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.
Senior Manager
Joined: 29 Jun 2005
Posts: 403
Followers: 1

Kudos [?]: 15 [0], given: 0

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,
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.
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