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Managers and Directors

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by dumb.doofus » Sun Jun 14, 2009 11:17 pm
Each employee on a certain task force is either a manager or a director. What percent of employees on the task force are directors?
a] The average salary of managers on the task force is $5000 less than the avg salary of all the employees
b] The average salary of directors on the task force is $15000 more than the avg salary of all the employees
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Source: — Data Sufficiency |
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by bluementor » Mon Jun 15, 2009 1:08 am
No. of directors = x
No. of managers = y

x/(x+y) ?

Statements 1 and 2 are clearly insufficient on their own. There is no link between the average salaries and the number of managers or directors.

Both statements together:

Let's say the average salary of ALL employees = z.

Using the concept of weighted average, we can setup the following:

(z + 15000)x + (z - 5000) y = z(x + y)
zx + 15000x + zy - 5000y = zx + zy
15000x = 5000y
3x = y

we can now use this equation to solve for x/(x+y). sufficient.

Choose C.

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by pg850 » Mon Jun 15, 2009 1:29 am
Answer - C (Both are required)

I feel my solution wasnt the optimal one as it took about 5 mins to solve the question, if someone has a better method please let me know.

Soln:

Total no. of managers = M
Total no. of directors = D
Total no. of employees = E = M+D
Average salary of managers = x
Average salary of directors = y
Avg. salary of all employees = {Dy +(E-D)x}/E}

Therefore, answer = D/E

=> x = {Dy +(E-D)x}/E} - 5000 ---- 1
=> y = {Dy +(E-D)x}/E}+15000 ---- 2

From the two equations we can get the value of D/E as there are two unknowns (x and y)..since we are looking for D/E we do not count D and E as unknowns.

But, if you want to be sure, you can solve it:

solving 1 gives, y-x = (5000/D)*E
solving 2 gives, E(y-x-15000) = D(y-x)

Substitute y-x from 1,
which will give you D/E = 1/4 or 25%

Again, this approach looks a little too long for a GMAT question and if someone has a better solution please post it up.

-PG
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by pg850 » Mon Jun 15, 2009 1:35 am
bluementor's solution is much simpler!
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by dumb.doofus » Mon Jun 15, 2009 7:07 pm
C is right. I was stuck in a similar approach as pg850. Thanks BM.
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by rdchandvadkar » Thu Feb 11, 2010 11:13 am
can someone please explain this rite down to the basics.. just not getting this question..

thanks.
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by ajith » Thu Feb 11, 2010 11:27 am
rdchandvadkar wrote:can someone please explain this rite down to the basics.. just not getting this question..

thanks.
https://www.beatthegmat.com/avg-salary-a ... 23038.html
https://www.beatthegmat.com/need-help-on ... t1225.html

Try these links and if your doubt persists, please post the specific doubt and I am sure people will be ready to help you
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by rdchandvadkar » Thu Feb 11, 2010 11:31 am
got it myself...

but let me know if this approach is correct

no. of directors= x
no. of managers=y
total no of employees =x+y
average salary of all employees = z
total salary to all employees= z(x+y)
average of salary paid to directors =a
average of salary paid to managers =b

using weighted average formula;
z= (xa+yb)/x+y
z(x+y)= xa+xb............................eq1

from st(1)

z-b=5000; b=z-5000 not suff

from st2
a-z=15000; a=15000+z not suff

st(1) and (2) together
replacing in equation 1 values of b and a

z(x+y)= x(15000+z)+y(z-5000)
solving this we get; x/y=1/3
therefore using dividendo we get x/x+y=1/(3+1)=1/4=25%

is this ok??
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by ajith » Thu Feb 11, 2010 11:39 am
rdchandvadkar wrote: therefore using dividendo we get x/x+y=1/(3+1)=1/4=25%

is this ok??
The solution is perfect. I would say it is a brilliant answer if it were a maths exam. If you are comfortable with the lengthy steps and time taken for the solution, I would say it is good in GMAT also.
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by chaya009 » Thu Feb 11, 2010 3:03 pm
That was a nice and simple explanation bluementor.

i knew that both the statements are together sufficient, but ran out of time in solving it.

Thanks.
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