can someone please explain how to solve the following...
Each employee on a task force is either a manager or a director. What percentage of the employees on the task force are directors?
(1) Arithmetic mean salary of the managers on the task force is $5000 less than the average salary of all employees on the task force.
(2) Arithmetic mean salary of the directors on the task force is $15000 more than the average salary of all employees on the task force.
OA is C
arithmetic means
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Neither statement alone is sufficient.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 certain number of $5000 salaries (on average) are COMBINED with a certain number of $15,000 salaries (on average) to yield a task force that reflects the average salary of ALL of the employees.
To determine the ratio of $5000 salaries to $15,000 salaries in the task force, 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 in the mixture 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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Heyo Topspin,
GMATGuru's explanation here is Grade A. But just to make sure the science is clear here, this concept is applied whenever you have a weighted average question.
As a basic example:
The boys in a class weigh 50 pounds. The girls weigh 10 pounds. The average weight of a student in the class is 45 pounds.
The first thing you know is that there must be more boys than girls. Why? Because the class average is so much closer to the boys average than the girls average. That knowledge alone is enough to get you through a surprising number of weighted average questions.
But if you needed to know the exact ratio of boys to girls, you can use Guru's method:
If there were an equal number of boys and girls in the class, the girls are 35 pounds away from the class average, and the boys are 5 pounds away.
That's a ratio of 5:35, or 1:7. That means there are 7 times as many boys as girls. (The reason that Guru mentioned the reciprocal is that, whichever group is most distant from the mean actually has fewest members.)
Good luck!
-t
GMATGuru's explanation here is Grade A. But just to make sure the science is clear here, this concept is applied whenever you have a weighted average question.
As a basic example:
The boys in a class weigh 50 pounds. The girls weigh 10 pounds. The average weight of a student in the class is 45 pounds.
The first thing you know is that there must be more boys than girls. Why? Because the class average is so much closer to the boys average than the girls average. That knowledge alone is enough to get you through a surprising number of weighted average questions.
But if you needed to know the exact ratio of boys to girls, you can use Guru's method:
If there were an equal number of boys and girls in the class, the girls are 35 pounds away from the class average, and the boys are 5 pounds away.
That's a ratio of 5:35, or 1:7. That means there are 7 times as many boys as girls. (The reason that Guru mentioned the reciprocal is that, whichever group is most distant from the mean actually has fewest members.)
Good luck!
-t
Tommy Wallach, Company Expert
ManhattanGMAT
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