In Dives Corporation, \(75\%\) of the employees are customer service representatives, \(15\%\) are programmers, and

Given that:
Customer service representation = 75% = 0.75
Programmers = 15% = 0.15
Managers = 10% = 0.1
Average salary of customer service rep = $60,000
Average salary of programmers = $100,000
Target question: what is the average salary of the managers?
Using weighted average,
$$Weighted\ average=a_1b_1+a_2b_2+...+a_nb_n$$
Where a = no. of employees and b = average salary of employee.
Let the average salary of managers = x
Weighted average = total employees average = $100,000
Therefore;
$100,000 = (0.75 * $60,000) + (0.15 * $100,000) + (0.1 * x)
$100,000 = $45,000 + $15,000 + 0.1x
$100,000 - $60,000 = 0.1x
$40,000 = 0.1x
$$x=\frac{40,000}{0.1}=$400,000\ \ \ \ \ \left(Option\ D\right)$$

$100,000 = $45,000 + $15,000 + 0.1x
0.1x = $40,000

Scott@TargetTestPrep wrote: ↑

Mon Oct 12, 2020 4:22 am

Gmat_mission wrote: ↑

Thu Oct 08, 2020 3:53 am

In Dives Corporation, \(75\%\) of the employees are customer service representatives, \(15\%\) are programmers, and \(10\%\) are managers. The customer service representatives have an average salary of \(\$60,000,\) and the programmers have an average salary of \(\$100,000.\) If the average salary of all employees is also \(\$100,000,\) what is the average salary of the managers?

A. \(\$140,000\)
B. \(\$180,000\)
C. \(\$260,000\)
D. \(\$400,000\)
E. \(\$560,000\)

Answer: D

Solution:

We can let n = the average salary of the managers and create the weighted average equation:

0.75(60,000) + 0.15(100,000) + 0.1n = 100,000

45,000 + 15,000 + 0.1n = 100,000

0.1n = 40,000

n = 400,000

Answer: D

Great!