First, acknowledge there are two groups Ms (managers) and Ds (Directors). I will use the variable (x) to represent the % of employees that are directors. Since there are only 2 categories, (1-x) will represent the % of employees that are managers).
Statement 1: The average salary of the managers on the task force is $5,000 less than the average salary of all of the employees on the task force.
If the average salary is $50,000, the manager's average salary would be $45,000. The director's average salary could be $55,000.
55x + (1-x)45 = 50
55x + 45 -45x = 50
10x = 5
x = 5/10
x = 50%
However, what if the average director's salary is $60,000 and the average salary of all employees is still $50,000.
60x + (1-x)45 = 50
60x + 45 - 45x = 50
15x = 5
x = 5/15
x = 33.3%
So the information in statement one does not give us enough to find a single answer.
Statement 2: The average salary of the directors is $15,000 more than the average salary of all of the employees on the task force.
Statement two gives us the same amount of information for the other side. We could prove statement 2 to be insufficient by going through the steps in statement one (but during the exam we obviously would not want to)
Together:
Using both statements we know that the directors on average make 15k more than the average, and that on average managers make 5k less than the average. This gives us enough information, since there are only two groups. Eyeballing it, you may be able to see that it's a it takes 3 managers on average to lower the average salary of a director to the average salary of all employees. (15:-5) = (3:-1), of that 1 of 4 employees is a director. But just to illustrate that this works regardless of the average salary, below are examples with (1) the average salary of all employees = $100,000 and (2) = $50,000.
115x + (1-x) 95 = 100
115x + 95 - 95x = 100
20x = 5
x = 5/20
x = 25%
65x + (1-x) 45 = 50
65x + 45 - 45x = 50
20x = 5
x = 25%
Answer: C
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