achan wrote:

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

1) the average ( Arithemetic mean) salary of the managers on the task force is 5000 less than the average salary of all the employees on the task force.

2) the average ( Arithemetic mean) salary of the directors on the task force is 15000 greater than the average salary of all the employees on the task force.

\(S_a\) - Average salary of all employees

\(S_m\) - Average salary for manager

\(S_d\) - Average salary of directors

\(d\) - # of directors;

\(m\) - # of managers.

Question \(\frac{d}{m+d}=?\)

(1) \(S_m=S_a-5000\) --> Not sufficient to calculate ratio.

(2) \(S_d=S_a+15000\) --> Not sufficient to calculate ratio.

(1)+(2) \(S_a=\frac{S_m*m+S_d*d}{d+m}\) --> substitute \(S_m\) and \(S_d\) --> \(S_a=\frac{(S_a-5000)*m+(S_a+15000)*d}{d+m}\) --> \(S_a*d+S_a*m=S_a*m-5000*m+S_a*d+15000*d\) --> \(S_a*d\) and \(S_a*m\) cancel out --> \(m=3d\) --> \(\frac{d}{m+d}=\frac{d}{3d+d}=\frac{1}{4}\). Sufficient.

Answer: C.

Or for (1)+(2): if we say that the fraction of the directors is \(x\) (\(x=\frac{d}{d+m}\)) then the fraction of the managers will be \((1-x)\) (\(1-x=\frac{m}{d+m}\)) --> \(S_a=x(S_a+15000)+(1-x)(S_a-5000)\) --> \(S_a=x*S_a+15000x+S_a-5000-x*S_a+5000x\) --> \(x=\frac{1}{4}\).

I utilized a similar approach but ended up getting the wrong answer. I set up the two equations just like you have but instead of Setting up the weighted average formula, I plugged the two equated directly into d/(d+m) = which made (Sa) to be 5,000 == when I plugged this back in, I got SM = 0 which made everything screwy.

What did I do wrong here? Was it wrong because I plugged the two equations directly into d/(d+m)? Why is that wrong?