MHB What Could Be the Average Salary for All Employees in the Company?

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The discussion revolves around calculating the average salary for all employees in a company with two departments, X and Y, where Department X has more than twice the number of employees as Department Y. The average salary in Department X is $25,000, while in Department Y it is $35,000. The derived inequality indicates that the overall average salary must be between $25,000 and approximately $28,333.33. Consequently, the only viable average salary options from the provided list are $26,000 and $28,000. Thus, the average salary for all employees could only be $26,000 or $28,000.
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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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Congratulations to the following members for their correct solutions:

1) Sudharaka

Solution (from Sudharaka):

[sp]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.[/sp]
 

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