DS

Question

Please, explain your answer.

ashkrs · Answer

Well I could not solve this within 5 mins. And so get E .
But down somewhere in my heart I feel its C . 
Just wish somebody could prove this!

TriColor · Answer

askrsh, I also chose E because I could not solve the simulataneous equation quickly. But OA is C. Who can prove this???

IrinaOK · Answer

C,

Lets denote 
Am- average salary of managers
Ad- average salary of directors
M- number of managers
D- number of directors
A- average salary of all employees

Then stem is asking what is M/(M+D), just for sake of brevity, let`s say M/(M+D) is X, it is equal to 1-  or (1-X), .

1) Am+5000=A
2) Ad-15000=A

thus Am=Ad-20000

A=total salary/ (M+D)=  / (M+D) = Am   + Ad   = Am.X+Ad (1-X)=Am+5000-----> (Ad-20000).X+Ad(1-X)=Am+5000=Ad-20000+5000

So X=3/4

johnrb · Answer

The fastest way I can see is by picking numbers: say the managers are at 20,000, the employee's average is 25,000, and the directors are at 40,000. Then, if M and D are the numbers of managers and directors:

(20,000 M + 40,000 D)/ (M + D) = 25,000 (formula for employee average)

20, 000M + 40, 000D = 25,000M + 25,000D

15,000D = 5,000M

D/M = 1/3; since all the employees are either D or M, D/employees = 1/4 or 25%.

Different values don't change the percentage, so pick C.

The long way (too long to be realistic, I think, but maybe someone else sees a shortcut):

Am is the average for managers, Ad is the average for directors, M is the number of managers, and D is the number of directors.

a. Am + 5,000 = (M*Am + D*Ad)/(M + D) (from 1)
b. Am + 20,000 = D*Ad/D (from 2)
c. (M + D)(Am + 5000) = M*Am + D*Ad (from a)
d. M*Am +D*Am + 5000M + 5000D = M*Am +D*Ad (from c)
e. D*Am + 5000M + 5000D = D*Ad (from d)
f. D(Am + 20,000) = D*Ad (from b)
g. D*Am + 20,000D = D*Ad (from f)
h. D*Am + 5000M + 5000D = D*Am + 20,000D (from e, g)
i. 5000M + 5000D = 20,000D (from h)
j. 5000M = 15,000D
k. M/D = 3, so D/M = 1/3, so as above D = 25% of total employees.

Whew.

johnrb · Answer

Of course, once you know the general point that knowing the average value of different parts and the average value of the whole gives you the ratios of the parts to each other or the parts to the whole, you don't need to go through such calculations.

TriColor · Answer

Thank you everybody! I learned something new today.

DS

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