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Managers and Directors

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Managers and Directors

by f2001290 » Sun Jun 17, 2007 8:52 am
13. 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.

Please provide a details explanations. OA after few explanations
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Source: — Data Sufficiency |
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by jayhawk2001 » Sun Jun 17, 2007 2:07 pm
Is it C ?

Let m be total num of mangers, d be total num of directors and 'a' be
the average salary.

1 or 2 by itself is not sufficient. Using both,

m (a-5000) + d (a+5000) = a*(m+a)
m = d

So, % of d = 50%

Now, on a related note, if half the employees are directors and half
are managers, who actually does the work ;-)
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by f2001290 » Sun Jun 17, 2007 3:55 pm
jayhawk2001 wrote:Is it C ?

Let m be total num of mangers, d be total num of directors and 'a' be
the average salary.

1 or 2 by itself is not sufficient. Using both,

m (a-5000) + d (a+5000) = a*(m+a)
m = d

So, % of d = 50%

Now, on a related note, if half the employees are directors and half
are managers, who actually does the work ;-)
Jay - The equation should be modified;
m (a-5000) + d (a+15000) = a*(m+d)
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by jayhawk2001 » Sun Jun 17, 2007 5:33 pm
f2001290 wrote:
Jay - The equation should be modified;
m (a-5000) + d (a+15000) = a*(m+d)
Ah, yes, sorry for the typo.
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by Jhyphi » Wed Jun 27, 2007 2:41 am
There's actually 3 times as many managers as directors.
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by SaraLotfy » Thu Sep 19, 2013 12:44 am
I got the answer C

Could experts please confirm?
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by theCodeToGMAT » Thu Sep 19, 2013 1:00 am
The ANSWER should be [spoiler][C][/spoiler]

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 [spoiler][C][/spoiler]
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by GMATGuruNY » Thu Sep 19, 2013 1:27 am
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 on the task force is $15,000 greater than the average salary of all
employees on the task force.
Clearly, neither statement on its own is sufficient.
The two statements combined constitute a MIXTURE problem: two ingredients (managers and directors) are combined to form a mixture (all of the employees on the task force).
To evaluate the two statements combined, use ALLIGATION -- a great way to handle mixture problems.

Let M = managers, D = directors, and T = the entire task force.

Step 1: Draw a number line, with the two ingredients (M and D) on the ends and the mixture (T) in the middle:
M---------------T---------------D

Step 2: Calculate the distances between the averages.
Since the managers' average is 5000 less than the average of the entire task force, and the directors' average is 15,000 more than the average of the entire task force, we get the following distances between the averages:
M-----5000------T----15000------D

Step 3: Determine the ratio in the mixture.
The ratio of M to D in the mixture is the RECIPROCAL of the distances in red.
M : D = 15000:5000 = 3:1.

Since M : D = 3:1, there are 3 managers for every 1 director.
Thus, of every 4 employees, 1 is a director, implying that directors/total = 1/4 = 25%.
SUFFICIENT.

The correct answer is C.

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