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

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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, 08:01.
Last edited by Bunuel on 13 Jun 2019, 03:22, edited 3 times in total.
Renamed the topic and edited the question.
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New post 18 May 2010, 08:21
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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 (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.


\(S_a\) - 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 \((1-x)\) (\(1-x=\frac{m}{d+m}\)) --> \(S_a=x(S_a+15000)+(1-x)(S_a-5000)\) --> \(S_a=x*S_a+15000x+S_a-5000-x*S_a+5000x\) --> \(x=\frac{1}{4}\).
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Re: Each employee of a certain task force is either a manager or a directo  [#permalink]

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New post 04 Jul 2010, 00:46
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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).
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New post 03 Jul 2010, 22:54
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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 of a certain task force is either a manager or a directo  [#permalink]

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New post 06 Jan 2011, 00:37
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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 !
Answer is (c)
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Re: Each employee of a certain task force is either a manager or a directo  [#permalink]

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New post 06 Jan 2011, 02:34
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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New post 06 Jan 2011, 03:31
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 of a certain task force is either a manager or a directo  [#permalink]

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New post 08 Jan 2012, 18:03
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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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New post 10 Jan 2012, 02:26
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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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Re: Each employee of a certain task force is either a manager or a directo  [#permalink]

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New post 26 Aug 2012, 06:35
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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 a directo  [#permalink]

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New post 26 Aug 2012, 08:56
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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

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

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New post 17 Jan 2013, 02:07
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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: Each employee of a certain task force is either a manager or a directo  [#permalink]

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New post 24 Jan 2013, 04:48
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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 a directo  [#permalink]

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New post 11 Sep 2016, 05:07
VeritasPrepKarishma wrote:
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/


Hi Karishma, I also tried the same method for this but the ratio of 3:1 s the ration of average salary. How can we say that it would be the ratio of Managers and Directors?
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Re: Each employee of a certain task force is either a manager or a directo  [#permalink]

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New post 11 Sep 2016, 22:49
rakaisraka wrote:
VeritasPrepKarishma wrote:
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/


Hi Karishma, I also tried the same method for this but the ratio of 3:1 s the ration of average salary. How can we say that it would be the ratio of Managers and Directors?
Thanks


How would you find the average salary of both - Managers + Directors

Average salary of group = (Avg sal of Managers * No of managers + Avg sal of Directors * No of Directors) / (No of Managers + No of Directors)

So note that the weights in weighted average formula are 'no of managers' and 'no of directors'.

The formula w1/w2 = (A2 - Aavg)/(Aavg - A1) is just an arrangement of the above formula.

w1 and w2 are the weights. A2 is the average salary of Directors, Aavg is the average salary of the group and A1 is the average salary of Managers.

Stmt. 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.
This means Aavg - A1 = 5,000

Stmt 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.
This means A2 - Aavg = 15,000
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Re: Each employee of a certain task force is either a manager or a directo  [#permalink]

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New post 30 Jan 2017, 18:03
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Apply weighted Average formula:

N1/N2 = (M2-M)/(M-M1)

Let N1 = Managers
N2 = Directors

Statement 1: M-M1 = 5000
NOT SUFFICIENT

Statement 2: M2-M = 15000
NOT SUFFICIENT

Combining...

N1/N2 = 15000/5000 = 3:1

SUFFICIENT, SO [C]
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Re: Each employee of a certain task force is either a manager or a directo  [#permalink]

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New post 25 Sep 2019, 03:00
Bunuel - thank you for the explanation above.

May I ask if say there are 3 groups of people that comprise the company's employees (instead of 2) e.g. Manager, Directors, and Executives, each with a different average pay and relative distance to the total employee average, would the above method still apply?
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Re: Each employee of a certain task force is either a manager or a directo   [#permalink] 25 Sep 2019, 03:00
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