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

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Re: DS Weighted Average [#permalink]

17 Jan 2013, 04:28

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

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Re: Each employee.. [#permalink]

23 Jan 2013, 06:58

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}.

Do you have more example of this question type?

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Re: DS Weighted Average [#permalink]

24 Jan 2013, 04:48

kingston wrote:

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]

18 Feb 2013, 10:09

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] 18 Feb 2013, 10:09

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