Question

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

Hint:

In weighted average problems with inequalities, first find the value at the boundary of the inequality (e.g., using equality). Then, reason about which direction the average will shift. If the group with the lower value becomes larger, the average moves down; if the group with the higher value becomes larger, the average moves up.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a weighted average problem. The overall average salary is influenced more by the department with more employees. We are given the average salaries of two groups and an inequality describing the relative sizes of the groups.
Step 2: Key Formula or Approach:
Let \(N_X\) and \(N_Y\) be the number of employees in departments X and Y, respectively. Let \(S_X = $25,000\) and \(S_Y = $35,000\).
The overall average salary, \(S_{avg}\), is calculated as: \[ S_{avg} = \frac{N_X \cdot S_X + N_Y \cdot S_Y}{N_X + N_Y} \] We are given the condition \(N_X>2N_Y\). We can find the boundary for the average salary by considering the case where \(N_X = 2N_Y\).
Step 3: Detailed Explanation:
The overall average salary must be between the two individual averages, so \($25,000<S_{avg}<$35,000\).
Let's calculate the average salary for the boundary case where there are exactly twice as many employees in X as in Y, i.e., \(N_X = 2N_Y\). We can substitute \(N_X = 2\) and \(N_Y = 1\) into the formula.
\[ S_{boundary} = \frac{2 \cdot ($25,000) + 1 \cdot ($35,000)}{2 + 1} \] \[ S_{boundary} = \frac{$50,000 + $35,000}{3} = \frac{$85,000}{3} \approx $28,333.33 \] This value, $28,333.33, is the average salary when the ratio of employees in X to Y is exactly 2 to 1. The problem states that there are more than twice as many employees in Department X (\(N_X>2N_Y\)).
Since Department X has the lower salary ($25,000), increasing its proportion of employees beyond the 2:1 ratio will pull the overall average closer to $25,000. Therefore, the actual average salary must be less than the boundary value we calculated.
So, the possible range for the average salary is \( $25,000<S_{avg}<$28,333.33 \).
Now let's check the options against this range:
- (A) $26,000: This is in the range. Correct.
- (B) $28,000: This is in the range. Correct.
- (C) $29,000: This is greater than $28,333.33. Incorrect.
- (D) $30,000: Incorrect.
- (E) $31,000: Incorrect.
- (F) $32,000: Incorrect.
- (G) $34,000: Incorrect.
Step 4: Final Answer:
The amounts that could be the average salary for all employees are those between $25,000 and $28,333.33. From the list, these are $26,000 and $28,000. The correct answers are (A) and (B).