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Percent of salaries

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by Rudy414 » Fri May 24, 2013 4:31 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%

Thanks!
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Source: — Problem Solving |

by srcc25anu » Fri May 24, 2013 6:13 pm
# of employees prior to July 1: X
# of employees after July 1: 0.9X

Avg Salary prior to July1: Y
Avg Salary after July1: 1.1Y

Total salary prior to July 1: XY
Total Salary after July 1: 0.9X * 1.1Y = 0.99XY
Ratio = 0.99XY/XY = 99%
Ans B
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by GMATGuruNY » Fri May 24, 2013 6:21 pm
Rudy414 wrote: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 J N » Fri May 24, 2013 6:31 pm
before

A= Sum /n ----> sum = An

after layoff of 10%

1.1A = (Sum -x)/ .9n -----> sum-x = 1.1A * .9n


want (sum-x) / sum

(1.1A * .9n) /(An) * 100 = 99%
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