Problem of the week #28 - October 8th, 2012

[Jameson]

Each employee of a certain company is in either Department X or Department Y, and there are more than twice as many employees in Department X as in Department Y. The average (arithmetic mean) salary is \$25,000 for the employees in Department X and \$35,000 for the employees in Department Y. Which of the following amounts could be the average salary for all of the employees of the company?

Indicate all such amounts.

A. \$26,000
B. \$28,000
C. \$29,000
D. \$30,000
E. \$31,000
F. \$32,000
G. \$34,000
--------------------
Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) Sudharaka

Solution (from Sudharaka):

Spoiler

Let \(x\) be the number of employees in department \(X\) and \(y\) be the number of employees in department \(Y\). We know that, \(2y<x\Rightarrow \frac{x}{y}>2\).

Total number of money paid for all the employees = \(25000x+35000y\)

Total number of employees = \(x+y\)

Average salary of an employee in the company = \(\dfrac{25000x+35000y}{x+y}\)

Let \(c\) be the average salary of an employee in the company. That is,

\[c=\frac{25000x+35000y}{x+y}\]

\[\Rightarrow \frac{x}{y}=-\frac{c-35000}{c-25000}\]

Since, \(\dfrac{x}{y}>2\) we have,

\[-\frac{c-35000}{c-25000}>2\]

\[\Rightarrow \frac{3c-85000}{c-25000}<0\]

This inequality is only satisfied when, \(25000<c<\dfrac{85000}{3}\approx 28333.33\).

Hence the only possible answers are A and B.