Bunuel wrote:

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}.

Hi Bunuel,

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?