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

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Each employee in a certain task force is either a manager or [#permalink]  21 Apr 2008, 23:37
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Each employee in a certain task force is either a manager or a director. What percent of the employee on the task force are directors ?

1.The Average salary of the Managers on the task force is $5000 less than the avg salary of all Employees on the task force. 2.The avg salary of the directors on the task force is$15000 greater than the avg salary of all employees on the task force.

a)
b)
c)
d)
e)
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Re: DS- Employee [#permalink]  22 Apr 2008, 05:44
patrannn wrote:
Each employee in a certain task force is either a manager or a director. What percent of the employee on the task force are directors ?

1.The Average salary of the Managers on the task force is $5000 less than the avg salary of all Employees on the task force. 2.The avg salary of the directors on the task force is$15000 greater than the avg salary of all employees on the task force.

a)
b)
c)
d)
e)

I got C

Set S = average salary of all
Sm = average salary of managers
Sd = average salary of directors
d = number of directors
m = number of managers

We are looking for:
d / (m+d) = ?

The equation of average salary is the following:
S = (m*Sm + d*Sd) / (m+d)

In (1), we know that Sm = S-5000, and plug it back in, you get no where.
In (2), same problem as (1).
If using both conditions, we have:
S = (m*(S-5000) + d*(S+15000)) / (m+d) = (m*S - 5000m + d*S + 15000d) / (m+d)
S = (S*(m+d) - 5000m + 15000d) / (m+d) = S +(15000d - 5000m)/(m+d)
0 = (15000d - 5000m)/(m+d)
15000d = 5000m
3d = m
And so, d / (m+d) = d / (3d + d) = 1/4
SUFFICIENT
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Re: DS- Employee [#permalink]  22 Apr 2008, 06:56
NICE!!!

i had guessed it would be C..givien it seemed like a weighted avg..but as usual setting up the equation took..way tooo long..

any short cuts to this??
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Re: DS- Employee [#permalink]  24 Apr 2008, 01:38
fresinha12 wrote:
NICE!!!

i had guessed it would be C..givien it seemed like a weighted avg..but as usual setting up the equation took..way tooo long..

any short cuts to this??

Just a general understanding.. Difference between average director's salary from average employer's is +15.000, difference of manager's salary -5.000. So, every single director's salary should be "compensated" by 3 salaries of managers to get to average employer's one.
Let's say average salary of employer is x, then

x+15.000 + (x-5.000) + (x-5.000) + (x-5.000)... gives us an average of "x"
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Re: DS- Employee [#permalink]  24 Apr 2008, 05:22
NICE...thats exactly what i had in mind..but couldnt somehow put it into a equation...

Sunchaser20 wrote:
fresinha12 wrote:
NICE!!!

i had guessed it would be C..givien it seemed like a weighted avg..but as usual setting up the equation took..way tooo long..

any short cuts to this??

Just a general understanding.. Difference between average director's salary from average employer's is +15.000, difference of manager's salary -5.000. So, every single director's salary should be "compensated" by 3 salaries of managers to get to average employer's one.
Let's say average salary of employer is x, then

x+15.000 + (x-5.000) + (x-5.000) + (x-5.000)... gives us an average of "x"
Re: DS- Employee   [#permalink] 24 Apr 2008, 05:22
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