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

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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]

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New post 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]

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New post 25 Jan 2014, 09:18
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!
J :)
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Re: Each employee of a certain task force is either a manager or [#permalink]

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New post 28 Aug 2014, 11:26
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]

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New post 30 Aug 2014, 09:58
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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New post 01 Sep 2014, 00:37
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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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New post 19 Oct 2015, 10:21
Each employee on a certain task force is either a manager or director. What percent of the employees on the task force are director?
1. The average (Arithmetic Mean) salary of the mangers on the task force is $5,000 less than the average salary of all employees on the task force.
2. The average (Arithmetic Mean) salary of the directors on the task force is $15,000 greater than the average salary of all employees on the task force.
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New post 19 Oct 2015, 11:32
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Hi kumarpritam,

You'll likely get more of a response if you post this in the DS Forum here:

gmat-data-sufficiency-ds-141/

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Re: Data Sufficiency Question from GMAT Prep [#permalink]

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kumarpritam wrote:
Each employee on a certain task force is either a manager or director. What percent of the employees on the task force are director?
1. The average (Arithmetic Mean) salary of the mangers on the task force is $5,000 less than the average salary of all employees on the task force.
2. The average (Arithmetic Mean) salary of the directors on the task force is $15,000 greater than the average salary of all employees on the task force.



As Rich said, the right forum for this question is the DS forum.

The employees are a mix of managers and directors. We need to find the percentage of directors.

No statement alone gives you information on both managers and directors. We need to find whether both statements together are sufficient.

Use the scale method of weighted averages here.

w1/w2 = (A2 - Aavg)/(Aavg - A1) = 15000/5000 = 3:1
So for every 3 managers, there is one director. Hence, directors are 25% of the employees task force.
Answer (C)

Check this post for details of the scale method: http://www.veritasprep.com/blog/2011/03 ... -averages/
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Re: Data Sufficiency Question from GMAT Prep [#permalink]

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New post 19 Oct 2015, 22:24
Hi Kumarpritam,

Let the average salary of all the employees be $x.
Statement 1. the avg salary of the employees is $(x-5000) {Insufficient}
Statement 2. the avg salary of the directors is $(x+15000) {Insufficient}

Lets take S1 and S2 together. We can see that whatever may be the value of x, three employees salary would average out the salary of 1 director to have the average as x. [3*-5000 + 1*15000 = 0]
Hence, we can say that 25% are directors [1/(3+1)*100=1/4*100=25%] and rest 75% will be employees.

Answer C.

Let me know for any further clarification. Thanks.

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Re: Each employee of a certain task force is either a manager or [#permalink]

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New post 19 Oct 2015, 22:24
Instead of going for such long calculations, I think the best way to get on with this question is as follows:

Let the average salary of all the employees be $x.
Statement 1. the avg salary of the employees is $(x-5000) {Insufficient}
Statement 2. the avg salary of the directors is $(x+15000) {Insufficient}

Lets take S1 and S2 together. We can see that whatever may be the value of x, three employees salary would average out the salary of 1 director to have the average as x. [3*-5000 + 1*15000 = 0]
Hence, we can say that 25% are directors [1/(3+1)*100=1/4*100=25%] and rest 75% will be employees.

Answer C.

Let me know for any further clarification. Thanks.

P.S. Be kind and give Kudos. :D
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Each employee of a certain task force is either a manager or [#permalink]

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kumarpritam wrote:
Each employee on a certain task force is either a manager or director. What percent of the employees on the task force are director?
1. The average (Arithmetic Mean) salary of the mangers on the task force is $5,000 less than the average salary of all employees on the task force.
2. The average (Arithmetic Mean) salary of the directors on the task force is $15,000 greater than the average salary of all employees on the task force.


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