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# Each employ on a certain task is either manager or director.

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Manager
Joined: 13 Mar 2006
Posts: 76
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Each employ on a certain task is either manager or director. [#permalink]  04 May 2006, 21:34
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Each employ on a certain task is either manager or director. What % of the employees on the task force are directors?
(1) The average salary of managers on the TF is \$5,000 less that the av. sal. of all employees.
(2) The av. sal. of directors on the TF is \$15,000 greater that the av. sal. of all employees.
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VP
Joined: 06 Jun 2004
Posts: 1061
Location: CA
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Kudos [?]: 36 [0], given: 0

[#permalink]  04 May 2006, 23:09
C

Let x be the average, m = manager, d = director. We are looking for d/(d+m)?

(1) x - 5000 ==> Not sufficient
(2) x = 15,000 ==> Not sufficient

(1) + (2)

m(x-5000) + d(x+15000)/(m+d) = x ==> After simplifying we get
3d = m and we can find out d/(d+m) ==> 1/4 ==> 25%
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VP
Joined: 29 Apr 2003
Posts: 1405
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Kudos [?]: 17 [0], given: 0

[#permalink]  05 May 2006, 17:59
I also get E..

It says Average Salaray of ALL EMPLOYEEs. Not just the ones in the Task Force!
Manager
Joined: 27 Mar 2006
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[#permalink]  05 May 2006, 21:49
I say E
Director
Joined: 16 Aug 2005
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Location: France
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[#permalink]  10 May 2006, 11:13
TeHCM your method of solving the problem I agree with, except those who are saying E has a point that (1) and (2) says "av. sal. of all employees". The question could be clear if it mentioned "av. sal. of all employees in the Task Force"

What is the source of this question?
Intern
Joined: 23 Feb 2006
Posts: 40
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[#permalink]  11 May 2006, 05:05
Go for C.
here is the solution.
let m - manager salary, d-director salary.
n-number of manager,m-number of directors
From 1:
(m1+m2+...mn/n)+5000=(m1+...+mn+d1+...+dm)/(m+n) - nothing we can get since don't know relationship between m and n
From 2:
(d1+d2+...dm/m)-15000=(m1+...+mn+d1+...+dm)/(m+n) - nothing we can get again
Combying both we can solve equation.
[#permalink] 11 May 2006, 05:05
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# Each employ on a certain task is either manager or director.

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