mukgera wrote:Please help me with this problem.
Each employee on a task force is either a manager or a director.What percentage of the employee on the task force are directors?
1) Arithmatic mean salary of the managers on the task force is $5000 less than the average salary of all employees on the task force.
2) Arithmatic mean salary of the directors on the task force is $15000 more than the average
salary of all employees on the task force.
OA after some discussion.
This is a weighted average question.
The managers are being combined with the directors to form a mixture of managers and directors.
We can use alligation, which dictates the following:
The proportion of each element in the mixture is equal to the distance between the average attributed to the other element in the mixture and the average attributed to the entire mixture.
Directors:
The proportion of directors in the mixture = the distance between the average salary of the MANAGERS (the other element in the mixture) and the average salary of all the employees (the entire mixture).
This information is provided by Statement 1:
The average salary of the managers on the task force is $5000 less than the average salary of all the employees on the task force.
Managers:
The proportion of managers in the mixture = the distance between the average salary of the DIRECTORS (the other element in the mixture) and the average salary of all the employees (the entire mixture).
This information is provided by Statement 2:
The average salary of the directors on the task force is $15000 more than the average salary of all the employees on the task force.
Combining the statements:
Proportion of directors : proportion of managers = 5000:15000 = 1:3.
Thus, of every 4 employees, 1 is a director:
1/4 = 25%.
Sufficient.
The correct answer is
C.
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