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collection of tough problems from G PREP - 7

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collection of tough problems from G PREP - 7

by abhasjha » Sat Feb 06, 2010 5:23 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 ?

A) The average salary of the managers on the task force is 5,000 less than the average salary of all employees on the task force.

B) The average salary of the directors on the task force is 15,000 greater than the average salary of all employees on the task force.
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by harsh.champ » Sat Feb 06, 2010 6:23 am
abhasjha wrote: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 ?

A) The average salary of the managers on the task force is 5,000 less than the average salary of all employees on the task force.

B) The average salary of the directors on the task force is 15,000 greater than the average salary of all employees on the task force.

Solve,using x=no. of managers and y= no. of directors.
a=avg. salary of a manager,b=avg. salary of a director

We have to calculate y/(x+ y)

Statement 1 alone:- a = [(ax+by)/(x+y)] - 5000 ( 4 variables,1 equation - insufficient)
=> 5000x = y(b - 5000 - a) -(1)

Statement 2 alone:- b = [(ax+by)/(x+y)] + 15,000 (4 variables, 1 equation - insufficient)
=> x(b - 15,000 - a) = 15000y -(2)

Comparing (1) and (2),
we get a quadratic relation between a and b.(suppose x = (&)y [& being the relationship constant])


We find that no soln. emerges out


Ans is E.
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by ajith » Sat Feb 06, 2010 7:53 am
abhasjha wrote: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 ?

A) The average salary of the managers on the task force is 5,000 less than the average salary of all employees on the task force.

B) The average salary of the directors on the task force is 15,000 greater than the average salary of all employees on the task force.
Say - Avg salary of managers, Sm and let no of Managers be m
and - Avg salary of directors , Sd and let no of directors be d

1. (Sd*n +Sm*m)/(m+n) - Sm = 5000

Insufficient

2. Sd- (Sd*n +Sm*m)/(m+n) = 15000

Insufficient

Combining

Sd - Sm = 20000

we have 4 variables and 2 equations (3 variables if we consider m/n as one variable)

I think E
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by abhasjha » Mon Feb 08, 2010 12:12 am
OA is C

possible explanation :

let M be the number of managers, and D the number of directors; also, let S stand for the average salary.
also, express everything in thousands (so that you don't have to waste time writing ",000" on the end of every single number in the problem)
statement (1) says that the managers' salary is S - 5, and statement (2) says that the directors' salary is S + 15.
so
the sum of everybody's salaries is M(S - 5) + D(S + 15).
but
the sum of everybody's salaries is also given by the average formula: Sum = Average x Number, or (M + D)S.

therefore,
(M + D)S = M(S - 5) + D(S + 15)
MS + DS = MS - 5M + DS + 15D
0 = -5M + 15D
5M = 15D
M = 3D
the ratio of M to D is 3:1

sufficient
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by shashank.ism » Mon Feb 08, 2010 3:06 am
abhasjha wrote: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 ?

A) The average salary of the managers on the task force is 5,000 less than the average salary of all employees on the task force.

B) The average salary of the directors on the task force is 15,000 greater than the average salary of all employees on the task force.
I think the ans is C

Let the average salary of total employee = X
Let Y be the percentage of directors. So percentage of managers = 100-Y.
So average salary of managers = X-5000, average salary of directors = X+15000
So, [(100-Y)(X-5000)+ Y(X-15000)]/100 = X
100X-XY-500000 + XY -15000Y =100X --> Y = 500/15 = 100/3 = 33.33% ans.

So Ans is C i.e. combined I and II statement is sufficient.
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