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by lukaswelker » Tue Apr 08, 2014 10:23 am
Hey guys, here is the question.

On July 1st of last year, the total number of employees at Company E was decreased by 10 percent. Without any change in the salaries of the remaining employees, the average (arithmetic mean) employee salary was 10 percent more after the decrease in number of employees than before the decrease. The total of the combined salaries of all of the employees at Company E after July 1 last year was what percent of that before July 1 last year?

90%; 99%;100%; 101%; 110%

I know it has to be under a 100% so the last three option are not possible. But I can't figure out why it's 99% and not 90%.

Any suggestions?
Many thanks
Lukas
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by GMATGuruNY » Tue Apr 08, 2014 12:13 pm
On July 1 of last year, the total number of employees at Company E was decreased by 10 percent. Without any change in the salaries of the remaining employees, the average employee salary was 10 percent more after the decrease in number of employees than before the decrease. The total of the combined salaries of all the employees at Company E after July 1 last year was what percent of that before July 1 last year?

90%
99%
100%
101%
110%
Before July 1:
Let the number of employees = 10.
Let the average salary = 10.
Old total payroll = 10*10 = 100.

After July 1:
10% fewer employees = 10 - .1(10) = 9.
Average salary increased by 10% = 10 + .1(10) = 11.
New total payroll = 9*11 = 99.

Thus:
(new total payroll)/(old total payroll) = 99/100 = 99%.

The correct answer is B.
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by [email protected] » Tue Apr 08, 2014 3:59 pm
Hi lukaswelker,

I'm a big fan of Mitch's solution (to TEST Values). Here's how you can solve with problem with algebra:

N = Number of Employees
S = Average Salary of Employees

Original Formula:

(Sum of Salaries)/N = S

Sum of Salaries = N(S)

On July 1st, the Formula changes:
10% fewer employees
10% increase to average salary

(Sum of Salaries)/(.9N) = 1.1(S)

Sum of Salaries = (.9(N)(1.1)(S)
Sum of Salaries = .99(N)(S)

The original sum of salaries = NS
The new sum of salaries = .99NS

Final Answer: B

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