yellowho wrote:Each employee on a certain task force is either a manager or a director. What percent of the employees on the task force are directors.

A) The average salary of the managers on the task force is 5,000 less than the average salary of all employees on the task force.

B) The average salary of the directors on the task force is 15,000 greater than the average salary of all employees on the task force.

Whats the algebra approach to this? I did number line/ weighted average conceptualization. Just wondering the proof for algebra.

I would use alligation.

Let A = average salary of all the employees, M = average salary of the managers, D = average salary of the directors.

Plug in A = 10,000.
M = 10,000 - 5000 = 5000
D = 10,000 + 15,000 = 25,000.

The proportion needed of each starting average is the positive difference between the other two averages.

Proportion needed of M = 25,000-10,000 = 15,000.
Proportion needed of D = 10,000-5,000 = 5000.
M : D = 15,000 : 5000 = 3:1.
Thus, D = 1/4 = 25%.

Here an algebraic approach:
Sum of all salaries = number of employees * average = (M+D)*A
Sum of manager salaries = number of managers * average = M*(A-5000)
Sum of director salaries = number of directors * average = D*(A+15,000)
Thus:
(M+D)*A = M*(A-5000) + D(A+15,000)
MA + DA = MA - 5000M + DA + 15,000D
5000M = 15,000D
M/D = 15000/5000 = 3/1 = 3:1.
Thus, D = 1/4 = 25%.