GMATPrep: On July 1 of last year, total employees at company

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On July 1 of last year, total 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 the 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?

A. 90%
B. 99%
C. 100%
D. 101%
E. 110%

OA: B

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by Brent@GMATPrepNow » Sat Oct 21, 2017 6:48 am
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%
Choose some nice values that satisfy the given information.

BEFORE July 1
Let's say that there were 10 employees
And let's say the average salary was $10.
TOTAL payroll = (10)($10) =$100

Number of employees decreases by 10%
Average salary increases by 10%
So....

AFTER July 1
There are 9 employees
Average salary is $11.
TOTAL payroll = (9)($11) =$99

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?

99/100 = [spoiler]99%[/spoiler]

Answer: B
Brent Hanneson - Creator of GMATPrepNow.com
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answer

by [email protected] » Sat Oct 21, 2017 10:29 am
Hi NandishSS

I'm a big fan of Brent's solution (to TEST Values). Here's how you can solve the 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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by Scott@TargetTestPrep » Sun Nov 11, 2018 7:20 pm
NandishSS wrote:On July 1 of last year, total 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 the 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?

A. 90%
B. 99%
C. 100%
D. 101%
E. 110%
We can start by defining a few variables.

n = the number of employees at Company E last year before July 1

x = the average salary of each employee at company E last year before July 1

We are given that on July 1 of last year, the total number of employees at Company E was decreased by 10 percent. Thus, we can represent the remaining number of employees as 0.9n.

We are also given that the average (arithmetic mean) employee salary was 10 percent more after the decrease in number of employees than before the decrease. We can represent this new average salary as 1.1x.

We must determine what percent the total of the combined salaries of all of the employees at Company E after July 1 last year is of that before July 1 last year.

The combined salaries of the employees before July 1 is nx and the combined salaries of the employees after July 1 is 0.9n * 1.1x = 0.99nx. We can create the following expression:

(salaries after July 1)/(salaries before July 1) * 100%

(0.99nx)/(nx) * 100%

0.99 * 100% = 99%

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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