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Percent Problem

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Percent Problem

by mpaudena » Tue Oct 20, 2009 4:25 pm
Each employee on 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 (arithmetic mean) salary of the managers 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 of the task force is $15,000 greater than the average salary of all the employees on the task force.

Answer and explanation please. Thanks in advance.
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Source: — Data Sufficiency |
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by sanjana » Wed Oct 21, 2009 4:44 am
IMO : C

Let the number of directors be : D
Let the number of managers be : M

We need D/D+M

This is a weighted average question. Whenever there are 2 groups you need to find the ratio of number of ppl in the individual groups you need to know either :

1)The avg of the individual groups and the overall average of the whole group
2)the avg of the individual groups and their respective differences from the overall avg(this is what we will use in this question)

Statement1
----------

Given :
Manager(avg sal) = overall(avg sal)-5000
Not sufficient as we dont have any info on the directors avg(sal)

Statement2
----------
Given
Director(avg sal)= overall(avg sal)+15000
Again insufficient as we dont have any info on Managers avg(sal)

Combining 1 and 2,

Mang sal(avg)-----overall sal(avg)-----Dir avg(sal)
5000 15000

Hence M/D = 15000/5000 = 3/1
Hence D:M = 1:3
Hence,D/D+M = 1/4 or 25% employees are directors.
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by mpaudena » Wed Oct 21, 2009 6:23 am
sanjana wrote:Mang sal(avg)-----overall sal(avg)-----Dir avg(sal)
5000 15000
Hence M/D = 15000/5000 = 3/1
How are you getting the number of employees from the average? Average = Sum sal/Number of Employees. I don't see how Mang sal(avg) = overal sal(avg) - 5000 gives you the sum of the salaries or the number of employees(same for Dir avg(sal). So how do you get M/D = 15000/5000? Please explain.

Also can you explain how this may work for me too:
1)The avg of the individual groups and the overall average of the whole group
Thanks in advance.
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by life is a test » Wed Oct 21, 2009 7:52 am
mpaudena wrote:
sanjana wrote:Mang sal(avg)-----overall sal(avg)-----Dir avg(sal)
5000 15000
Hence M/D = 15000/5000 = 3/1
How are you getting the number of employees from the average? Average = Sum sal/Number of Employees. I don't see how Mang sal(avg) = overal sal(avg) - 5000 gives you the sum of the salaries or the number of employees(same for Dir avg(sal). So how do you get M/D = 15000/5000? Please explain.

Also can you explain how this may work for me too:
1)The avg of the individual groups and the overall average of the whole group
Thanks in advance.
good explanation for this problem at:
https://www.manhattangmat.com/forums/eac ... -t772.html
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by mpaudena » Wed Oct 21, 2009 8:19 am
life is a test wrote:
mpaudena wrote:
sanjana wrote:Mang sal(avg)-----overall sal(avg)-----Dir avg(sal)
5000 15000
Hence M/D = 15000/5000 = 3/1
How are you getting the number of employees from the average? Average = Sum sal/Number of Employees. I don't see how Mang sal(avg) = overal sal(avg) - 5000 gives you the sum of the salaries or the number of employees(same for Dir avg(sal). So how do you get M/D = 15000/5000? Please explain.

Also can you explain how this may work for me too:
1)The avg of the individual groups and the overall average of the whole group
Thanks in advance.
good explanation for this problem at:
https://www.manhattangmat.com/forums/eac ... -t772.html
That didn't do it for me either. I want to understand the mechanics so that when the problem changes a bit I'm able to do it. I thought I understood weighted averages but it's just not clicking for me.

First I don't understand why the 20,000 difference matters. It seems to me that in order to find the number of employees we still need to know either the individual salary sums or the combine salary sums. This seems to suggest that the sums are equal and the weights is the only difference. Now all we have is the difference between the averages of group 1 and 2.
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by sanjana » Wed Oct 21, 2009 9:37 am
Remember this general Rule

Given
2 groups A and B,
The avg of the 2 groups A and B
and the overall Avg of the whole group(A+B)

Let Overall avg - Avg A = X
and Overall Avg - Avg b = Y

Then A/B = Y/X

Eg
A class writes a math test and the overall avg is 75%.The girls avg 85% and the boys avg 70%,what fraction of the class are boys?

Overall avg - Avg boys = X = 75-70=5

Overall avg - Avg Girls = Y = 85-75=10

Hence ,B/G = 10/5 = 2:1
Hence fraction of the class that are boys : 2/3

Hope this clears.
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by rohan_vus » Wed Oct 21, 2009 9:58 am
In other words , one can simply deduce such rules easily.
Let D be no of directors and M be no of managers.

Let say avg salary of Directors = Sd and Avg salary of managers be Sm .. and the Overall salary be So

so one can write Sd * D + Sm*M = So* (D + M) ..I hope this is atleast makes sense to you.

Given Sd = So + 15000 and Sm = So - 5000.

substitue this into the above eqn..

(So + 15000) * D + (So - 5000) * M = So* D + So*M
==> So*D + So*M + 15000*D - 5000*M = So*D + So*M
==> 15000*D = 5000*M
==> D/M = 1/3.. Hope this helps
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by mpaudena » Wed Oct 21, 2009 10:16 am
Thanks for the example and formula. I was able to work out why A/B = x/y with that example.
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