Each employee on a task force is either a manager or a director. What percentage of the task force are directors?
(1) The average manager salary is $5,000 less than the average of all employees.
(2) The average director salary is $15,000 more than the average of all employees.
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Let the average salary of managers of the task force = S(m), the average salary of the directors on the task force = S(d), and the average salary of all the employee on the task force = S(e).Mjkourtis wrote:Each employee on a task force is either a manager or a director. What percentage of the task force are directors?
(1) The average manager salary is $5,000 less than the average of all employees.
(2) The average director salary is $15,000 more than the average of all employees.
Let the no. of managers = m and no. of directors = d. We have to find d/(m + d).
(1) S(m) = S(e) - 5000, which ALONE is NOT SUFFICIENT.
(2) S(d) = S(e) + 15000, which ALONE is NOT SUFFICIENT.
Combining (1) and (2), we know that S(e) = {m * S(m) + d * S(d)}/(m + d)
So, S(e) = {m * [S(e) - 5000] + d * [S(e) + 15000]}/(m + d)
m * S(e) + d * S(e) = m * S(e) - 5000m + d * S(e) + 15000d
15000d = 5000m
3d = m
So, d/(m + d) = d/4d = 1/4, which is SUFFICIENT.
The correct answer is C.
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Neither statement alone is sufficient.Mjkourtis wrote:Each employee on a task force is either a manager or a director. What percentage of the task force are directors?
(1) The average manager salary is $5,000 less than the average of all employees.
(2) The average director salary is $15,000 more than the average of all employees.
The two statements combined constitute a MIXTURE problem.
A $5000 average salary is being combined with a $15,000 average salary to yield a COMBINED average salary for ALL of the employees.
We can use ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.
Step 1: Draw a number line, with the managers and directors on the ends and the average salary of all of the employees in the middle.
Managers----------All-------------------Directors
Step 2: Enter the distance between the MANAGERS' average and the average salary of ALL of the employees and the distance between the DIRECTORS' average and the average salary of ALL of the employees:
Managers-----5000----All---------15,000---------Directors
Step 3: Determine the ratio in the mixture.
The ratio of managers to directors is equal to the RECIPROCAL of the distances in red.
Managers : Directors = 15,000:5000 = 3:1.
Thus, of every 4 employees, 3 are managers and 1 is a director, implying that 1/4 = 25% of the employees are directors.
The correct answer is C.
For another alligation problem, check here:
https://www.beatthegmat.com/pls-help-wit ... 24884.html
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
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