Q. In a work force, the employees are either managers or directors. What is the percentage of directors?
(1) the average salary for manager is $5,000 less than the total average salary.
(2) the average salary for directors is $15,000 more than the total average salary.
IMO - E, am I correct?
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goalevan
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M: number of managers
D: number of directors
The question is asking for the value of D/(M + D).
Statement 1) This gives no information on the absolute or relative quantities of D or of M. There could be 5,000 directors and 1 manager (with about $5,000 less than the average director's pay), or there could be 5,000 managers and 1 director (with an outrageously huge salary that moves the average up $5,000).
Statement 2) The same logic holds on this statement. No information is given on the absolute or relative quantities of D or M.
Combined) This is where it will serve you to know the weighted average formula well. Combined, the statements tell us that the manager average is $5,000 below the weighted average, and that the directors average is $15,000 above.
With A = weighted average salary of all employees
[M(A - 5000) + D(A + 15000)] / M + D = A
MA - 5000M + DA + 15000D = AM + AD
15000D - 5000M = 0
3D = M
This gives us a relationship between the quantities of directors and managers that we can use in the original question: D/(M + D)?
D/(M + D)
D/(3D + D)
D/(4D)
1/4 = 25%
Answer: 25% of employees are directors and 75% are managers.
Note that all of these calculations are not necessary, and in fact, weighted average problems can be visualized fairly easily on a number line, where the weighted average will fall closer to the item with more weight (quantity in this case)
Manager average - - Weighted average - - - - - - - - Director average
0 - - - - 5,000 - - - - - - - - - - - - - - 20,000
Since 5 is 1/4 of 20, 3/4 of the quantity comes from the managers.
C
D: number of directors
The question is asking for the value of D/(M + D).
Statement 1) This gives no information on the absolute or relative quantities of D or of M. There could be 5,000 directors and 1 manager (with about $5,000 less than the average director's pay), or there could be 5,000 managers and 1 director (with an outrageously huge salary that moves the average up $5,000).
Statement 2) The same logic holds on this statement. No information is given on the absolute or relative quantities of D or M.
Combined) This is where it will serve you to know the weighted average formula well. Combined, the statements tell us that the manager average is $5,000 below the weighted average, and that the directors average is $15,000 above.
With A = weighted average salary of all employees
[M(A - 5000) + D(A + 15000)] / M + D = A
MA - 5000M + DA + 15000D = AM + AD
15000D - 5000M = 0
3D = M
This gives us a relationship between the quantities of directors and managers that we can use in the original question: D/(M + D)?
D/(M + D)
D/(3D + D)
D/(4D)
1/4 = 25%
Answer: 25% of employees are directors and 75% are managers.
Note that all of these calculations are not necessary, and in fact, weighted average problems can be visualized fairly easily on a number line, where the weighted average will fall closer to the item with more weight (quantity in this case)
Manager average - - Weighted average - - - - - - - - Director average
0 - - - - 5,000 - - - - - - - - - - - - - - 20,000
Since 5 is 1/4 of 20, 3/4 of the quantity comes from the managers.
C
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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MI3
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Hello Goalevan,goalevan wrote:Combined) This is where it will serve you to know the weighted average formula well. Combined, the statements tell us that the manager average is $5,000 below the weighted average, and that the directors average is $15,000 above.
With A = weighted average salary of all employees
[M(A - 5000) + D(A + 15000)] / M + D = A
Is there is standard formula for weighted averages?
Thanks,
M.
