From my GMAT Prep CAT:
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 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 salary of the directors on the task force is $15,000 greater than the average salary of all employees on the task force.
Thanks,
Cappy
What percent of employees?
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IMO C..
tough question for someone who doesnt understand the concept of weighted averages...
Statement 1.
The manager average is 5000 less than the combined average...insufficient
Statement 2.
director average is 15000 greater than the combined average... insuffiicient..
taking both...
we know that there is a difference of 20000 between the managers and directors and average is closer to the managers ..hence there will be more managers than directors..
hence there are 3/4 managers and 1/4 directors..
please let me know if u have doubts...
tough question for someone who doesnt understand the concept of weighted averages...
Statement 1.
The manager average is 5000 less than the combined average...insufficient
Statement 2.
director average is 15000 greater than the combined average... insuffiicient..
taking both...
we know that there is a difference of 20000 between the managers and directors and average is closer to the managers ..hence there will be more managers than directors..
hence there are 3/4 managers and 1/4 directors..
please let me know if u have doubts...
hi sudhir,sudhir3127 wrote:IMO C..
hence there are 3/4 managers and 1/4 directors..
please let me know if u have doubts...
I cld solve this problem by solving the complete equation...but is there any quick way to reach above fractions, considering this is weighted average. Pls let me know
Ans: C
Average: x dollars
Manager made: x-5000 dollars
Director made: x+15000 dollars
Assume there are "m" managers and "d" directors.
[(m(x-5000)+d(x+15000)] / (m+d) = x
i.e. average salary of manager times number of managers
PLUS
average salary of director times number of directors
DIVIDED by the total number of employees (m managers + d directors)
EQUALS to the AVERAGE OF ALL EMPLOYEES
[(m(x-5000)+d(x+15000)] / (m+d) = x
=> mx - 5000m + dx + 15000d = x(m+d)
=> mx - 5000m + dx + 15000d = mx + dx
=> -5000m + 15000d = 0
=> 15000d = 5000m
=> 3d = m
Means: Every 3 directors, there is 1 manager. Thus we can calculate the percentage of employees on the task force are directors. 75%
Average: x dollars
Manager made: x-5000 dollars
Director made: x+15000 dollars
Assume there are "m" managers and "d" directors.
[(m(x-5000)+d(x+15000)] / (m+d) = x
i.e. average salary of manager times number of managers
PLUS
average salary of director times number of directors
DIVIDED by the total number of employees (m managers + d directors)
EQUALS to the AVERAGE OF ALL EMPLOYEES
[(m(x-5000)+d(x+15000)] / (m+d) = x
=> mx - 5000m + dx + 15000d = x(m+d)
=> mx - 5000m + dx + 15000d = mx + dx
=> -5000m + 15000d = 0
=> 15000d = 5000m
=> 3d = m
Means: Every 3 directors, there is 1 manager. Thus we can calculate the percentage of employees on the task force are directors. 75%