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

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Re: DS Weighted Average [#permalink] New post 24 Jan 2013, 03:48
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kingston wrote:
In a work force, the employees are either managers or directors. What is the percentage of directors?
(1) the average salary for manager is $5,000 less than the total average salary.
(2) the average salary for directors is $15,000 more than the total average salary.


c


This is actually a direct application of weighted averages. If you recall the scale method (explained here: http://www.veritasprep.com/blog/2011/03 ... -averages/ ), this is what the number line will look like

AVG-5000 _____________ AVG _________________________AVG+15000

Number of managers/Number of directors = [(AVG+15000) - AVG]/[AVG - (AVG - 5000)] = 3/1

Percentage of directors = 1/(1+3) = 1/4 = 25%
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Re: Each employee of a certain task force is either a manager or [#permalink] New post 18 Feb 2013, 09:09
I would prefer method of allegations in this problem.
achan wrote:
Each employee of a certain task force is either a manager or a director. What percent of the employees on the task force are directors?

(1) the average ( Arithemetic mean) salary of the managers on the task force is 5000 less than the average salary of all the employees on the task force.
(2) the average ( Arithemetic mean) salary of the directors on the task force is 15000 greater than the average salary of all the employees on the task force.


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Re: Each employee of a certain task force is either a manager or [#permalink] New post 25 Jan 2014, 08:18
Or one can solve this by using differentials

(1) and (2) together

We have that M = X-5000, and D = X+15,000

X being the total average so

-5000M+15000D = 0

The weights M,D are in a relation 1:3

So we know that 25% of the employers are directors

Hope it helps!
Cheers!
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Re: Each employee of a certain task force is either a manager or [#permalink] New post 28 Aug 2014, 10:26
ConnectTheDots wrote:
This can be solved in allegations method.
The ratio of the components is inverse of the ratio of their differences from the average.

#Managers:#Directors = Difference between salary between combined avg and directors : Difference between salary between combined avg and managers

M:D = 15000:500 = 3:1

D:M = 1:3

D:(M+D) = 1: (1+3) = 1:4

this way you can solve under 30 secs.


Can you really do this here or was this pure luck? Doesn't the -5000 vs. +15,000 actually give us a ratio of 20,000 to 5,000, therefore, D/M=1/4?
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Re: Each employee of a certain task force is either a manager or [#permalink] New post 30 Aug 2014, 08:58
Bunuel wrote:
achan wrote:
Each employee of a certain task force is either a manager or a director. What percent of the employees on the task force are directors?
1) the average ( Arithemetic mean) salary of the managers on the task force is 5000 less than the average salary of all the employees on the task force.
2) the average ( Arithemetic mean) salary of the directors on the task force is 15000 greater than the average salary of all the employees on the task force.


\(S_a\) - Average salary of all employees
\(S_m\) - Average salary for manager
\(S_d\) - Average salary of directors
\(d\) - # of directors;
\(m\) - # of managers.
Question \(\frac{d}{m+d}=?\)

(1) \(S_m=S_a-5000\) --> Not sufficient to calculate ratio.
(2) \(S_d=S_a+15000\) --> Not sufficient to calculate ratio.

(1)+(2) \(S_a=\frac{S_m*m+S_d*d}{d+m}\) --> substitute \(S_m\) and \(S_d\) --> \(S_a=\frac{(S_a-5000)*m+(S_a+15000)*d}{d+m}\) --> \(S_a*d+S_a*m=S_a*m-5000*m+S_a*d+15000*d\) --> \(S_a*d\) and \(S_a*m\) cancel out --> \(m=3d\) --> \(\frac{d}{m+d}=\frac{d}{3d+d}=\frac{1}{4}\). Sufficient.

Answer: C.

Or for (1)+(2): if we say that the fraction of the directors is \(x\) (\(x=\frac{d}{d+m}\)) then the fraction of the managers will be \((1-x)\) (\(1-x=\frac{m}{d+m}\)) --> \(S_a=x(S_a+15000)+(1-x)(S_a-5000)\) --> \(S_a=x*S_a+15000x+S_a-5000-x*S_a+5000x\) --> \(x=\frac{1}{4}\).


Hi Bunuel,

I utilized a similar approach but ended up getting the wrong answer. I set up the two equations just like you have but instead of Setting up the weighted average formula, I plugged the two equated directly into d/(d+m) = which made (Sa) to be 5,000 == when I plugged this back in, I got SM = 0 which made everything screwy.

What did I do wrong here? Was it wrong because I plugged the two equations directly into d/(d+m)? Why is that wrong?
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Re: Each employee of a certain task force is either a manager or [#permalink] New post 31 Aug 2014, 23:37
Expert's post
russ9 wrote:
Bunuel wrote:
achan wrote:
Each employee of a certain task force is either a manager or a director. What percent of the employees on the task force are directors?
1) the average ( Arithemetic mean) salary of the managers on the task force is 5000 less than the average salary of all the employees on the task force.
2) the average ( Arithemetic mean) salary of the directors on the task force is 15000 greater than the average salary of all the employees on the task force.


\(S_a\) - Average salary of all employees
\(S_m\) - Average salary for manager
\(S_d\) - Average salary of directors
\(d\) - # of directors;
\(m\) - # of managers.
Question \(\frac{d}{m+d}=?\)

(1) \(S_m=S_a-5000\) --> Not sufficient to calculate ratio.
(2) \(S_d=S_a+15000\) --> Not sufficient to calculate ratio.

(1)+(2) \(S_a=\frac{S_m*m+S_d*d}{d+m}\) --> substitute \(S_m\) and \(S_d\) --> \(S_a=\frac{(S_a-5000)*m+(S_a+15000)*d}{d+m}\) --> \(S_a*d+S_a*m=S_a*m-5000*m+S_a*d+15000*d\) --> \(S_a*d\) and \(S_a*m\) cancel out --> \(m=3d\) --> \(\frac{d}{m+d}=\frac{d}{3d+d}=\frac{1}{4}\). Sufficient.

Answer: C.

Or for (1)+(2): if we say that the fraction of the directors is \(x\) (\(x=\frac{d}{d+m}\)) then the fraction of the managers will be \((1-x)\) (\(1-x=\frac{m}{d+m}\)) --> \(S_a=x(S_a+15000)+(1-x)(S_a-5000)\) --> \(S_a=x*S_a+15000x+S_a-5000-x*S_a+5000x\) --> \(x=\frac{1}{4}\).


Hi Bunuel,

I utilized a similar approach but ended up getting the wrong answer. I set up the two equations just like you have but instead of Setting up the weighted average formula, I plugged the two equated directly into d/(d+m) = which made (Sa) to be 5,000 == when I plugged this back in, I got SM = 0 which made everything screwy.

What did I do wrong here? Was it wrong because I plugged the two equations directly into d/(d+m)? Why is that wrong?


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Re: Each employee of a certain task force is either a manager or   [#permalink] 31 Aug 2014, 23:37

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