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

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#### Jameson

Staff member
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
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#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

1) Sudharaka

Solution (from Sudharaka):

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.

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