Data Sufficiency

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?

(1) The average (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) The average (arithmetic mean) salary of the directors on the task force is $15000 greater than the average salary of all employees on the task force.

OA: C

lightbulb wrote: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?

(1) The average (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) The average (arithmetic mean) salary of the directors on the task force is $15000 greater than the average salary of all employees on the task force.

OA: C

Lets say there are 2 girls who are 20 years old and 4 boys who are 10 years old.

Girls average age : 20

Boys average age: 10

Overall average : [(2x20) + (4x10)] / 6 = 13.333

10 13.3 20

so our question gives us the same information bu we clearly have no information regarding to the numbers of managers and directors. But the distance between the averages to overall average gives us an idea about the ratio of two sides, and this is exactly what the question is asking!

If you pay attention to the girls and boys of there is same number of boys and girls:

10 15 20 and the average will be equal distance to the both end. But when the number of one part goes increases the average moves close to the that number and actually the ratio of the distance gives us the answer!

In our question

Managers ---5000 ----AVERAGE---------15000---------Directors

SO M/D=5000/15000 = 1/3

Hence, C

Thanks for your reply. Here's the math in case anybody is interested:

(SumD + SumM)/(D + M) - SumM/M = 5000 ...(1)
(SumD + SumM)/(D + M) - SumD/D = - 15000 ...(2)

From 1:
M*SumD + M*SumM - D*SumM - M*SumM = 5000*M*(D + M)
M*SumD - D*SumM = 5000*M* (D + M) ...(3)

From 2:
D*SumD + D*SumM - D*SumD - M*SumD = -15000*D*(D + M)
D*SumM - M*SumD = -15000*D*(D + M) ...(4)

Dividing (3) by (4):

=> -1 = -(1/3) * (M/D)
=> M/D = 1/3
=> (M + D)/D = 4/3
=> D/( M+ D) = 3/4

Nice solution.

(3) + (4)

lightbulb wrote:Thanks for your reply. Here's the math in case anybody is interested:

(SumD + SumM)/(D + M) - SumM/M = 5000 ...(1)
(SumD + SumM)/(D + M) - SumD/D = - 15000 ...(2)

From 1:
M*SumD + M*SumM - D*SumM - M*SumM = 5000*M*(D + M)
M*SumD - D*SumM = 5000*M* (D + M) ...(3)

From 2:
D*SumD + D*SumM - D*SumD - M*SumD = -15000*D*(D + M)
D*SumM - M*SumD = -15000*D*(D + M) ...(4)

Dividing (3) by (4):

=> -1 = -(1/3) * (M/D)
=> M/D = 1/3
=> (M + D)/D = 4/3
=> D/( M+ D) = 3/4

Ian,

I remember you giving a great tip on this kind of problems.

When we know 3 out of 4 things, we can find out the fourth.

Could you please implement the same in this problem and show us the way?

Also,

(SumD + SumM)/(D + M) - SumM/M = 5000 ...(1)
- (SumD + SumM)/(D + M) +SumD/D = 15000 ...(2)

I added (1) + (2) and ended up with

SumD/D + SumM/M = 20000

I don't know how to head further. Please help.

since the distance from the average should sum to zero, then the distance from the total average of one group plus the distance from the average of the second group should be equal to zero! So, a quick formula would be:
(-5000)x(number of managers) + (15000)(number of directors) = 0
So: number of managers/number of directors = 15000/5000 = 3/1

Out of the total number, the managers would be 3 times the number of directors.

so, D = 1/4 of total
and M = 3/4 of total

Hope that helps

Seems like some cool way to do. If you don't mind, could you please elaborate a bit? Does it hold in all situations?

I'd appreciate your help.

Thanks,
Raj Peddisetty.

Understood myself. Thanks.

this approach might be similar to the earlier one, assume M-Managers, M1- Average salary of managers,
D- Directors,D1- Average salary of directors,

we need to find out % of directors = D/(M+D)

State1. M1 = MM1+DD1/(M+D) - 15,000
State2. D1 = MM1+DD1/(M+D) + 15000
Solving these 2, we will reach (D-M)(M1-D1) = 0, now M1 cant be equal to D1 as 2 statements say, both are +, - 15,000 and hence D = M or % = 50%

lightbulb wrote: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?

(1) The average (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) The average (arithmetic mean) salary of the directors on the task force is $15000 greater than the average salary of all employees on the task force.

OA: C

If you understand how weighted averages work, almost NO MATH is needed for this problem.
I posted an explanation here:

https://www.beatthegmat.com/manager-or-d ... 90558.html

Anurag@Gurome wrote:

lightbulb wrote: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?

(1) The average (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) The average (arithmetic mean) salary of the directors on the task force is $15000 greater than the average salary of all employees on the task force.

OA: C

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).
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.

Hi,

I kind of disagree with what you have written. The S(e) formula applies only when your S(d) is the sum of salaries of Directors. In your notation you have used S(d) as the average of directors

The formula generally is = m(sum of sal of managers) + d(sum of sal of dir) / (m + d)

Correct me if I am wrong.

--Praveen

Hi anurag,

I kind of disagree with what you have written. The S(e) formula applies only when your S(d) is the sum of salaries of Directors. In your notation you have used S(d) as the average of directors

The formula generally is = m(sum of sal of managers) + d(sum of sal of dir) / (m + d)

Correct me if I am wrong.

--Praveen

Hello Guru's:

Shouldn't the question clearly say that all Managers have same salary and so does all Directors? If this was a real GMAT question, would this additional statement be mentioned?

Thanks.