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

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

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Updated on: 01 Nov 2013, 00:13
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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 (arithmetic 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 (arithmetic mean) salary of the directors on the task force is 15000 greater than the average salary of all the employees on the task force.

Originally posted by achan on 18 May 2010, 07:01.
Last edited by Bunuel on 01 Nov 2013, 00:13, edited 2 times in total.
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Re: Each employee..  [#permalink]

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18 May 2010, 07:21
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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.

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 $$(1-x)$$ ($$1-x=\frac{m}{d+m}$$) --> $$S_a=x(S_a+15000)+(1-x)(S_a-5000)$$ --> $$S_a=x*S_a+15000x+S_a-5000-x*S_a+5000x$$ --> $$x=\frac{1}{4}$$.
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Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS  [#permalink]

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03 Jul 2010, 23:46
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[quote2raulmaldonadomtz]I don't understand how can you get the number of employees from an average salary. Can someone explain this problem. For me neither of the 2 statements answer the question.[/quote2]

We can't get the number. We can get the ratio of director to total though.

This is a weighted average problem. From (1) and (2) together, we know that the managers are 5,000 smaller than the grand average, and that the directors are 15,000 greater than the grand average, or:

5000----grand average-------------15000
(mgrs)----------------------------(dirctrs)

The managers are way closer to the grand average than are the directors. So, there must be way more managers than directors. In fact, there are 15000/5000 or 3 times as many managers as directors. So, the manager to director ratio is 3:1. Thus, the director to total ratio is 1:4, or 25%.

I discuss weighted average strategy in more detail here: mixtures-96284.html

(For question 2, whiplash's explanation was simply superb).
##### General Discussion
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Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS  [#permalink]

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03 Jul 2010, 21:54
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Question 1:

Okay, let us assume the number of managers are M and the number of directors are D. So the total number of employees = M +D

Average Salary of Manager = x
Average Salary of Director = y

So, we have: Total salary of all employees = Mx + Dy
Average salary of employee = Total salary/Total employees = $$\frac{Mx+Dy}{M+D}$$

Statement 1:

Average salary of manager = Average salary of employee - 5000

$$x = \frac{Mx+Dy}{M+D} - 5000$$

Cross multiplying to the other side we get:

$$x(M+D) = Mx + Dy - 5000$$

Cancelling Mx on both sides and rearranging like terms together we get:

$$D(y-x) = 5000$$ - (1)

However, this doesn't say anything specific to us. So we move on to the second statement.

Statement 1:

Average salary of director = Average salary of employee + 15000

$$y = \frac{Mx+Dy}{M+D} + 15000$$

Cross multiplying to the other side we get:

$$y(M+D) = Mx + Dy + 15000$$

Cancelling Mx on both sides and rearranging like terms together we get:

$$M(y-x) = 15000$$ - (2)

This doesn't say anything by itself either. But when we put the results of the two statements [(1) and (2)] together, we get:

$$D(y-x) = 5000$$
$$M(y-x) = 15000$$

Dividing statement 1 by 2 we get

$$\frac{D}{M} = \frac{5000}{15000} = \frac{1}{3}$$

M = 3D

Total = M+D = 4D

From here, we can say percentage of directors = $$\frac{D}{4D}*100 = 25%$$.

Hence answer is C.
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Re: Each employee..  [#permalink]

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27 Jul 2010, 04:01
Let's call all employee A, manager M, and director D, so A=M+D. Their average salary is Sa, Sm, and Sd for A, M, and D respectively. The percentage of M and D are x% and y% respectively, so x+y=100%. We have the following equation:
Sa=\frac{x*Sm+y*Sd}{x+y} or Sa=\frac{x*Sm+y*Sd}{100}
Now consider the statement 1, it can be written as Sm=Sa-5000, not sufficient to determine x and y.
The same for statement 2, it leads to Sd=Sa+15000, not sufficient to determine x and y.
Nevertheless, using both statement, by eliminating the constant as follows:
3Sm+Sd=3*(Sa-5000)+(Sa+15000)
3Sm+Sd=4Sa
Sa=\frac{3Sm+Sd}{4}=Sa=\frac{75Sm+25Sd}{100}

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Re: Each employee..  [#permalink]

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15 Aug 2010, 11:14
thanks a lot bunuel!! i wonder how you know that someone posted the same question before. i did try to do a search, but couldn't find any...
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Re: Percentage of salary of directors.  [#permalink]

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05 Jan 2011, 18:58
From the given, avg salary = (D(X%) + M(100-X%))/100

Statement 1 doesn't provide any info on D. Not Sufficient
Statement 2 doesn't provide any info on M. Not Sufficient

Now we have 3 unknowns - with 2 equations...

Statement 1 & Statement 2 can come together to provide us with our 3rd equation, which is, on avg, M = D - 20,000

With 3 equations and 3 unknowns - u can find M, D & X. Sufficient

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Re: Percentage of salary of directors.  [#permalink]

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05 Jan 2011, 23:37
5
shan123 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.

(1) : Tells us nothing about how many directors or managers
(2) : Again tells us nothing about how many

(1+2) : Say average salary is x and there be m fraction of managers and hence (1-m) directors

m(x-5000) + (1-m)(x+15000) = x
mx - 5000m + x + 15000 -mx - 15000m = x
15000 - 20000m = 0
m = (3/4)
Hence fraction of directors = (1/4)

Sufficient !
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Re: Each employee..  [#permalink]

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06 Jan 2011, 01:34
Bunuel - is my minimalist solution of just identifying 3 equations satisfactory?

Or do you recommend fully solving it out to see if I net a result?
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Re: Each employee..  [#permalink]

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06 Jan 2011, 02:31
MackyCee wrote:
Bunuel - is my minimalist solution of just identifying 3 equations satisfactory?

Or do you recommend fully solving it out to see if I net a result?

Generally you can stop solving a DS question at the point you realize a statement is sufficient to get the answer. Note that for this question we don't need to solve for unknowns, we need to get the ratio of directors to total employees. I don't know what 3 equations are you talking about but the statements together are indeed sufficient to get the desired ratio and one can get this even not calculating its exact value.
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Re: Each employee..  [#permalink]

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06 Jan 2011, 09:56
Good question. While you can not figure out the number of manager and directors available, you still can manipulate the two equations to get a ratio or percentage. I didn't do any calculations but knowing that there where two statements that are relative to one another and that has factors that do not cancel one another out when setting up the equation to solve for the ratio is enough to conclude that C is the answer.
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Re: Each employee..  [#permalink]

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15 Mar 2011, 22:31
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.

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 $$(1-x)$$ ($$1-x=\frac{m}{d+m}$$) --> $$S_a=x(S_a+15000)+(1-x)(S_a-5000)$$ --> $$S_a=x*S_a+15000x+S_a-5000-x*S_a+5000x$$ --> $$x=\frac{1}{4}$$.

GREAT EXPLANATION.
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08 Jan 2012, 17:03
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Good one.

To solve it under 2 minutes, I had to guess this one as C. You are provided two relationships: managers-all and directors-all. That should be likely enough to determine the number of managers and directors as note that the differences in average numbers are specific to the number of managers and directors.

To calculate, you can solve two equations such as:
1. Sm/m = (Sm+Sd)/(m+d) - 5000.
2. Sd/d = (Sm+Sd)/(m+d) + 15000.

You need to solve for d/m+d.
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10 Jan 2012, 01:26
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Supergmatgirl wrote:
Can you please elaborate?

One has to understand the concept of weighted averages pretty well to understand my solution.

The point of a weighted average is to know how much weight to give these two individual groups, the managers and the directors.

Statement 1 tells how the managers' salaries relate to employee average but there is no information about how the directors' salaries relate to the employee average.
Insufficient
Eliminate A & D

Statement 2 tells how the directors' salaries relate to employee average but there is no information about how the managers' salaries relate to the employee average.
Insufficient
Eliminate B

Statements 1 and 2 together:
The manager average is 5000 less than the combined average. The director average is 15000 greater than the combined average. The difference between the manager average and the director average is 20000.

If there were an equal number of managers and directors they would each be 10000 off of the combined average - that would be a 50/50 weighting.
But the combined average is closer to the manager average, so there are more managers than directors. Use the above three numbers to know how much: the difference is 20000, and the combined average is three-quarters of the way towards the manager average.

Hence, 3/4 of the employees are managers & 1/4 are directors.
Sufficient.

Hence C.
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10 Jan 2012, 04:25
There should be a minor correction in the above equations - as the salaries are not same.
1. S1m/m = (S1m+S2d)/(m+d) - 5000.
2. S2d/d = (S1m+S2d)/(m+d) + 15000.

Going back to the weighted average method. I would think in this way, say if the average of all employee’s salary is 10K then that of Manager is(10-5) = 5K and of Directors is (10+15) = 25 k . So for each new directors, we need 3 managers to offset the total average from increasing. That’s it - meaning for every 4 employees - 3 are manager and 1 is director.
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Re: Each employee of a certain task force is either a manager or  [#permalink]

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12 May 2012, 18:19
Hey Bunuel,

What clue told you that "average salary of all the employees", your S_alpha was a weighted average?I tried solving the problem thinking it was an arithmetic average. Was the clue that the question did not specify that it was an arithmetic? Just puzzled. I mean it makes sense that its a weighted average but the question did not specify that info.
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Re: Each employee of a certain task force is either a manager or  [#permalink]

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26 Aug 2012, 05:35
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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.
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Re: Each employee of a certain task force is either a manager or  [#permalink]

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26 Aug 2012, 07:56
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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.

This is a question involving weighted average.
Having two quantities $$Q_1$$ and $$Q_2$$ with averages $$a_1$$ and $$a_2$$ respectively, if the combined average is $$a$$, and let assume that $$a_1>a>a_2,$$ then we can write:

$$\frac{a_1Q_1+a_2Q_2}{Q_1+Q_2}=a$$ from which $$a_1Q_1+a_2Q_2=aQ_1+aQ_2$$ or $$(a_1-a)Q_1=(a-a_2)Q_2,$$ which means that the distances from the combined average are inversely proportional to the quantities.
This equality we can also be written as $$\frac{a_1-a}{a-a_2}=\frac{Q_2}{Q_1}.$$

To answer the question it is enough to know the ratio between the two types of employees.
In our case we have a certain number of managers $$Q_1$$ and a certain number of directors $$Q_2.$$
From the above, if we know the two differences between the combined average (average salary of all employees) and each type of average, then in fact we have the ratio between $$Q_1$$ and $$Q_2.$$
Sufficient

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Re: DS Weighted Average  [#permalink]

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17 Jan 2013, 01:07
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Let T be the total average salary.
M be the no of Managers
D be no of Directors.

1st Statement: Aveg salary of Managers is T-5000.
2nd Statement: Avg salary of Directors is T+15000

1st things first. None of the statements provide sufficient information when taken one at a time. So the answer is either C or E
Combine two statements and you get one equation:

(T-5000)*M + (T+15000)*D = T*(M+D)
(Average Salary of M * no of Managers + Average salary of D* no of directors = Total Average Salary T * (M+D)

solving this we get M=3D

So ratio of Directors = D/(D+M) = D/(D + 3D) = 1/4 = 0.25

Hence 25%. So the answer is C

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Re: DS Weighted Average  [#permalink]

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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: DS Weighted Average &nbs [#permalink] 24 Jan 2013, 03:48

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