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

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25 Mar 2009, 11:07

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A

B

C

D

E

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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.

it is a weighted average type problem, we know that Director income is 15000 more than the avg and we know that manager income is 5000 less than the avg..

so if you look at it, the avg is closer to the manager salary than it is to the director salary clearly there are more managers than directors!

it is a weighted average type problem, we know that Director income is 15000 more than the avg and we know that manager income is 5000 less than the avg..

so if you look at it, the avg is closer to the manager salary than it is to the director salary clearly there are more managers than directors!

Well that still doesn't give the percentage of directors in the group. I'm still not convinced. Sure there are more managers than directors - but to find the percentage of directors, won't we need the number of directors ? I may be missing something very basic here. Any help will be appreciated.

- pradeep _________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

I tried this way and agree that A & B are not the answers...

Total avg of Manager and Director = x

For Managers , Salary Avg = m = (x-5000) Managers Count = M

For Directors , Salary Avg = d = (x+15000) Directors Count = D

We have a clue in stmt that they are talking abt averages so lets substitute in the formula

Avg (x) = {M(x-5000) + D(x+15000) } / (M+D) After solving the above eqn you will get a relationship between M and D and it is M/D =1/3 and hence the answer is C.

Each employee on a certain task force is either a manager or a director. What percentage of the employees are directors:

1) Average salary for manager is $5,000 less than average of all employees. 2) Average salary of directors is $15,000 greater than average salary of all employees

Please explain your answers.

Suppose: no. of managers = m managers' avg. salary = s1 no. of directors = d directors avg salary = s2

From st 1: s1 + 5000 = (s1 m + s2 d) / (m+d) s1 (m+d) + 5000 (m+d) = (s1 m + s2 d) s1 m + s1 d + 5000m + 5000d = s1 m + s2 d s1 d + 5000m + 5000d = s2 d s2 d - s1 d = 5000 (m + d) d (s2 - s1) = 5000 (m + d) s2 - s1 = 5000 (m + d)/d

From st 2: s2 - 15000 = (s1 m + s2 d) / (m+d) s2 - s1 = 15000 (m+d)/m

From 1 and 2: 5000 (m+d)/d = 15000 (m+d)/m m = 3d d/(m+d) = d/(3d+d) = 1/4

Let m = no of managers and M be the avg salaray of a manager d = no of directors and D be the avg salary of the director. T be the avg salary of the total group.

We can say that Mm + Dd = T(m + d) ----> Eq 1

Question asked is what is the Percentage of directors ie (d / (m+d)) = ?

For simplicity lets consider 5 and 15 instead of 5000 or 15000.

From stmt1, avg salary of managers is 5 less than total avg

==> Mm = (T-5)m. ---> Eq 2. We don't know any thing abt d or D. Cannot simplfy this further to find out what is d / (m+d). Hence Insufficient.

From stmt2, avg salary of directors is 15 more than total avg

==> Dd = Td + 15d. --> Eq 3. We don't know anything abt M or m. Cannot simplfy this further to find out what is d / (m+d). Hence Insufficient.

Combine stmt 1 and 2 for Mm and Dd in Eq 1, we get,

Tm - 5m + Td + 15d = Tm + Td. ==> 3d = m.

Now we can solve for d / (m+d) which is d / 4d = 1/4. Hence sufficient.

Each employee on a certain task force is either a manager or a director. What percentage of the employees are directors:

1) Average salary for manager is $5,000 less than average of all employees. 2) Average salary of directors is $15,000 greater than average salary of all employees

Please explain your answers.

This question is testing weighted average. D - % of Directors M- % of Managers D+M=1 D? ____________ Let "a" be total average Stmt1. a-5k=m (i.e. average salary for manager)

Stmt2. a+15k=d (i.e. average salary for director)

None is suff by itself.

Stmt 1 &2. Formula for weighted average: M*m+D*d=a M(a-5k)+D(a+15k)=a Ma-M5k+Da+D15k=a a(M+D)+D15k-M5k=a As M+D=1, we are left with D15k-M5k=0 Also, M=1-D => D15k-5k+D5k=0 D=1/4 or 25%

Yes this probelm is a bit time consuming. I liked GMAT tigers equation approach. It was more understandable to me.

Have seen this problem in some other forum some time back and i remember they put the below explaination. It will be good if someone can expalin. Just wanted to share with u all.

Concept of weighted averages

5000-------- Av ------------------150000

salarys are the ratio M/D = 5000 / 15000 = 1/3

the number of mangers and directors will be in the inverse ratio M / D = 3/1 .

Solving this ques. by equations will take a lot of time.

Alternative approach.

Manager manager manager............x managers ................Average..............director director director.....y director instaed of 5000 and 15000 lets deal with 5 and 15.

So managers are trying to bring down the average and directors are trying to bring up the average. however the average is fixed.

1 director can raise the average by 15 so to nullify it we need 3managers. so total employees = 4.Hence ratio of manager to director = 1/4.

Yes this probelm is a bit time consuming. I liked GMAT tigers equation approach. It was more understandable to me.

Have seen this problem in some other forum some time back and i remember they put the below explaination. It will be good if someone can expalin. Just wanted to share with u all.

Concept of weighted averages

5000-------- Av ------------------150000

salarys are the ratio M/D = 5000 / 15000 = 1/3

the number of mangers and directors will be in the inverse ratio M / D = 3/1 .

This is the best approach I've ever seen for averages. I might never have figured this out myself had i not seen this.

Actually the above diagram can be better represented as

Avg M-----(5000)-----Avg Tot-----------------(15000)--------------Avg D

ie, Avg of managers is at distance of 5000 from total avg, and avg of directors is at a distance for 15000 from tot avg.

Now concept of weighted avg is that The total avg will be most affected by the heaviest weight, ie, more the number of managers, closer will be tot avg to their avg salary, more the number of directors, closer will be the tot avg to their avg.

Now Avg tot is most affected by avg M, that means managers are more. How many more relative to directors?

Avg D is 3 times as far from tot avg as is Avg M.

that implies that weight of AvgM is 3 times greater than weight of AvgD. weight here is nothing but number of managers.

Using pen and paper, it takes arround 2-3 minuets.

reply2spg wrote:

btw....How much time will it take to answer the question?????

GMAT TIGER wrote:

Accountant wrote:

Each employee on a certain task force is either a manager or a director. What percentage of the employees are directors:

1) Average salary for manager is $5,000 less than average of all employees. 2) Average salary of directors is $15,000 greater than average salary of all employees

Please explain your answers.

Suppose: no. of managers = m managers' avg. salary = s1 no. of directors = d directors avg salary = s2

From st 1: s1 + 5000 = (s1 m + s2 d) / (m+d) s1 (m+d) + 5000 (m+d) = (s1 m + s2 d) s1 m + s1 d + 5000m + 5000d = s1 m + s2 d s1 d + 5000m + 5000d = s2 d s2 d - s1 d = 5000 (m + d) d (s2 - s1) = 5000 (m + d) s2 - s1 = 5000 (m + d)/d

From st 2: s2 - 15000 = (s1 m + s2 d) / (m+d) s2 - s1 = 15000 (m+d)/m

From 1 and 2: 5000 (m+d)/d = 15000 (m+d)/m m = 3d d/(m+d) = d/(3d+d) = 1/4

since the distance from the average should sum to zero, then the distance from the total average of one group plus the distance from the average of the second group should be equal to zero! So, a quick formula would be: (-5000)x(number of managers) + (15000)(number of directors) = 0 So: number of managers/number of directors = 15000/5000 = 3/1

Out of the total number, the managers would be 3 times the number of directors.

as this is DS problem and understanding the problem's concept and what is solution for this is important though, it does not need actual solution. As above mentioned in various post, weight average concept should yield percentage of director as relation 1) between manager Avg and total avg is available 2) between director avg and total avg is available Hence answer is C.

Each employee on a certain task force is either a manager or a director. What percentage of the employees are directors:

1) Average salary for manager is $5,000 less than average of all employees. 2) Average salary of directors is $15,000 greater than average salary of all employees

Statement 1 Let average of all employees be x. Avg salary for manager: x - 5000 Insufficient.

Statement 2 Avg salary of directors (D): x + 15000 Insufficient.

Statement 1 + 2

\(\frac{M(x-5000)+D(x+15000)}{M+D}= x\)

\(\frac{Mx-5000M+Dx+15000D}{M+D} = x\)

\(\frac{x(M+D)+15000D-5000M}{M+D} = x\)

\(\frac{15000D-5000M}{M+D}=0\)

\(5000(3D-M)=0\)

\(3D=M\) (from here you can actually derive the required percentage)

Sufficient!

Answer: C

The below by karimsafi is a really quick method though! +1 Kudos!

Quote:

Concept of weighted averages

(-5000)............x............................(+15000) M-------- Average ------------------D

since the distance from the average should sum to zero, then the distance from the total average of one group plus the distance from the average of the second group should be equal to zero! So, a quick formula would be: (-5000)x(number of managers) + (15000)(number of directors) = 0 So: number of managers/number of directors = 15000/5000 = 3/1

Out of the total number, the managers would be 3 times the number of directors.

so, D = 1/4 of total and M = 3/4 of total

_________________

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

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31 Oct 2013, 22:19

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