Quote:
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
This is my approach...
Statement 1Let average of all employees be x.
Avg salary for manager: x - 5000
Insufficient.
Statement 2Avg 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}=05000(3D-M)=03D=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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