Each employee on a certain task force is either a manager

[a7lee]

Each employee on a certain task force is either a manager or a director. What percent of the employees on the task force are directors.

A) The average salary of the managers on the task force is 5,000 less than the average salary of all employees on the task force.

B) The average salary of the directors on the task force is 15,000 greater than the average salary of all employees on the task force.

Answer is C.

Is there a quick to determine the sufficiency without getting long algebraic equations?

[StaceyKoprince]

Complicated. If you understand really well how weighted averages work, you can do something here without using algebra. Otherwise, this is probably a good one on which to make an educated guess and move on.

I can't just average the two averages, right? The point of a weighted average is to know how much weight to give these two individual groups, the managers and the directors.

Statement 1 tells me how the managers' salaries relate to the all employee average but, since I know how weighted averages work (that is, I know that I need to know something about both groups), this isn't enough info.

Statement 2:
Again, I need to know something about both groups; not enough info.

Statements 1 and 2 together:
Now I really have to understand how weighted averages work. The manager average is 5000 less than the combined average. The director average is 15000 greater than the combined average. The difference between the manager average and the director average is 20000. If there were an equal number of managers and directors, they would each be 10000 off of the combined average - that would be a 50/50 weighting. But the combined average is closer to the manager average, so there are more managers than directors. And I can actually use the above three numbers to know how much: the difference is 20000, and the combined average is three-quarters of the way towards the manager average. Sketch it out:
manager avg ----- combined average ----- ----- ----- director average (each dash represents 1000)

So 3/4 of the employees ar managers and 1/4 are directors. Sufficient.

[Borcho]

Here are the steps I used to confirm C makes sense:
1) The difference between managers's average and group average is 5,000
2) The corresponding figure for directors is 15,000.
3) The ratio between the differences is 1:3. Since the average is a weighted average, the corresponding ratio for the number of managers to number of directors will be 3:1.
4) Do a quick check to make sure this is right. Assume the average is 10, and use 5 for manager's average and 25 for directors. Assume 3 managers and 1 directors. Hence, 3x5 + 1x25 = 15+25 = 40. Divide by number of employees 40/(3+1) = 10.

[Guest]

Hi instructors,

Is it possible to solve this problem using algebra as well, if so how can one go about doing so?

Thanks

[Guest]

In addition to the question above could define by what you mean by weighted average and when to use it.
Thanks

[guest12]

can you pls confirm borocho's method is correct?

stacey, i'm not sure where the 10,000 comes from.

[Guest]

Hi instructors,

Is it possible to solve this problem using algebra as well, if so how can one go about doing so?

Thanks

1) M_sal = All_sal - 5000
2) D_sal = All_sal + 15000

3) We know that All_sal = M_sal*M + D_sal*D / M+D

Substitute 1 and 2 in 3.

[RonPurewal]

can you pls confirm borocho's method is correct?

yeah, you can do that; the ratio of the 'weights' is the inverse of the ratio of the distances between the weighted average and the individual data points.
(there is absolutely no way to express this truth in a way that's actually easy to read; it's very easy to illustrate in a couple minutes of a live session.)

[RonPurewal]

Hi instructors,

Is it possible to solve this problem using algebra as well, if so how can one go about doing so?

Thanks

well, sure, of course it's possible, although the process is somewhat formidable.

here's how you do it:

let M be the number of managers, and D the number of directors; also, let S stand for the average salary.
also, express everything in thousands (so that you don't have to waste time writing ",000" on the end of every single number in the problem)
statement (1) says that the managers' salary is S - 5, and statement (2) says that the directors' salary is S + 15.
so
the sum of everybody's salaries is M(S - 5) + D(S + 15).
but
the sum of everybody's salaries is also given by the average formula: Sum = Average x Number, or (M + D)S.

therefore,
(M + D)S = M(S - 5) + D(S + 15)
MS + DS = MS - 5M + DS + 15D
0 = -5M + 15D
5M = 15D
M = 3D
the ratio of M to D is 3:1
sufficient

[acethegmat]

Ron, would you please explain how/why would one assume Salary of a manager and director to be equal? You have assumed S to be the salary for a manager and a director. I got the following equaltion instead:
M - Manager, x - salary of manager, D - Director, y - salary of director.

x + 5000 = (Mx + Dy)/ (M+D)
y - 15000 = (Mx + Dy)/ (M+D)
x - y = 20,000

It is possible to have equal number of managers and directors, but a difference in their salary could result in the difference. No? How do we assume that it is the number of managers and directors that is different and not their salary figures? Would you please elaborate. Thank you.

[adiagr]

Ron, would you please explain how/why would one assume Salary of a manager and director to be equal? You have assumed S to be the salary for a manager and a director. I got the following equaltion instead:
M - Manager, x - salary of manager, D - Director, y - salary of director.

x + 5000 = (Mx + Dy)/ (M+D)
y - 15000 = (Mx + Dy)/ (M+D)
x - y = 20,000

It is possible to have equal number of managers and directors, but a difference in their salary could result in the difference. No? How do we assume that it is the number of managers and directors that is different and not their salary figures? Would you please elaborate. Thank you.

Hi Ace,


S is the average salary of the entire group. Ron has not assumed it to be the salary of a manager or salary of a director.

Infact if you go through his posts he says:

managers' salary is S - 5---> from statement 1

directors' salary is S + 15---> from statement 2


you have formed correct equations, but just see

(Mx + Dy)/ (M+D) ==>> is common in both equations. let us take it equal to A:

then rewriting your equations:

x + 5000 = A
y - 15000 = A

=> x = A-5000; y = A+15000

Now what is A:

A = [(Mx + Dy)/ (M+D)]..............eqn (A)
replace 'x' and 'y' with (A-5000) and (A+15000) in eqn A, respectively.

You will get:

(D/M) = (1/3)

Now we have to find % of Directors

i.e. [D/(D+M)]

from above it works out as 25%.

[acethegmat]

Thanks..adiagr

[RonPurewal]

Thanks..adiagr

adiagr did a nice job of cleaning this one up.

by the way, it's really, really important to use as few variables as you possibly can when you set these problems up. each new variable that you introduce will cost you more time, since you're going to have to eliminate more variables to solve the problem.

in general, in any instance in which you have a SIMPLE RELATIONSHIP, you should never use more than ONE VARIABLE.
in this case, the average salary, the directors' salary, and the managers' salary are all simply related, so you should simply use expressions in one variable (S, S - 5, S + 15) to define them.
if you use three variables here, there's just no way you're going to finish this problem in time, unless you are unbelievably lightning-fast at algebra.

[james.jt.wu]

I got an even easier way to solve this question. Use the Avg Residual Rule, which states that if you take the delta of each member of a set from the overall average, they must sum to zero.

So, if each Manager is -5,000 from overall average, and each Director is +15,000 from overall average, to get all the delta from the average means to sum to zero, for every Director, you must have three Managers. Since the ratio of Manager to Director is 3 to 1, Director is 25% of the overall set, and Manager is 75% of the overall set.

However, I really want to understand Stacey's explanation. Can someone draw out the WT average line to show the weights? I don't know where Stacey is getting the Combined Average from, as that is not given in the question.

thanks!

[RonPurewal]

I got an even easier way to solve this question. Use the Avg Residual Rule, which states that if you take the delta of each member of a set from the overall average, they must sum to zero.

So, if each Manager is -5,000 from overall average, and each Director is +15,000 from overall average, to get all the delta from the average means to sum to zero, for every Director, you must have three Managers. Since the ratio of Manager to Director is 3 to 1, Director is 25% of the overall set, and Manager is 75% of the overall set.

However, I really want to understand Stacey's explanation. Can someone draw out the WT average line to show the weights? I don't know where Stacey is getting the Combined Average from, as that is not given in the question.

thanks!

for details on the number line shortcut, watch the december 17 study hall at this link:
http://www.manhattangmat.com/thursdays-with-ron.cfm

if you still have questions after watching that, post back. thanks.