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# why I am not getting what OA says after many trials

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Manager
Joined: 28 Aug 2006
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Location: Albuquerque, NM
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why I am not getting what OA says after many trials [#permalink]  06 Jan 2007, 18:25
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why I am not getting what OA says after many trials
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Director
Joined: 28 Dec 2005
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[#permalink]  06 Jan 2007, 19:13
first off, I must admit this is not the first time I am seeing this question. I got it wrong on my GmatPrep and struggled with it too.

Here goes:

x --> average salary of managers
m --> number of managers

y --> average salary of directors
d --> number of directors

sum of managers' salaries --> xm
sum of directors salaries --> dy
From 1)
x = (xm+yd)/(m+d) - 5000

=> xm+yd - 5000(m+d) = x(m+d)
=> xm+yd - 5000m - 5000d = xm +xd

=> d(y-x) = 5000(m+d) --> INSUFF

From 2)

y = (xm+yd)/m+d + 15000

=> ym+yd = xm+yd + 15000(m+d)

=> m(y-x) = 15000(m+d) --> insuff

Together:

equating y-x,

5000(m+d)/d = 15000 (m+d)/m

d/m = 1/3

d/(m+d) = 1/(3+1) = 1/4
Manager
Joined: 20 Dec 2004
Posts: 180
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[#permalink]  07 Jan 2007, 01:07
you can also use the following.

Let M = no of managers
Let D = no of directors
Let AE = Average salary of all employees

Avg sal of all directors = AE + 15000
Avg sal of all managers = AE - 5000

Now equate sum of salary of all directors = salary of all employees - salary of all managers.

D*(AE + 15000)= (D + M)*AE - M*(AE - 5000)

-->D*AE + 15000*D = D*AE + M*AE - M*AE + 5000M
-->15000D=5000M
--> D/M = 5000/15000 = 1/3
--> D/(D+M) = 1/4

Hence answer is C. I need not explain why it can't be A or B.
_________________

Regards

Subhen

[#permalink] 07 Jan 2007, 01:07
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# why I am not getting what OA says after many trials

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