At a certain hedge fund, there are only CFAs and MBAs. The average num

Question

At a certain hedge fund, there are only CFAs and MBAs. The average number of vacation days taken by the CFAs is 5 fewer than the average number of vacation days taken per employee at the hedge fund. The average number of vacation days taken by the MBAs is 10 more than the average number of vacations days taken per employee at the hedge fund. What fraction of the hedge fund employees are CFAs?

A)\(\frac{1}{8}\)

B)\(\frac{1}{4}\)

C)\(\frac{1}{2}\)

D)\(\frac{2}{3}\)

E)\(\frac{3}{4}\)

A) \frac{1}{8}

B) \frac{1}{4}

C) \frac{1}{2}

D) \frac{2}{3}

E) \frac{3}{4}

laulau · Accepted Answer

This problem can be solved by treating it like a weighted average prooblem.

Let the average number of holidays/employee be x (the weighted average).
From the stem we know that the average number of holidays/CFA = x-5 and the average number of holidays/MBA = x+10.

Apppying the teeter totter method (from MGMAT): Thus, the the total distance between x-5 and x+10 = 15.

x-5--------------------x-----------------------x+10
distance from x=5 , distance from x=10

As we want to determine the number of CFAs, we can use the above teeter totter to conclude CFAs make up \(\frac{10}{15}=\frac{2}{3}\)of the total number of employees.

Using the teeter totter is a great way to simplify and visualise. Moreover, this approach is quite fast and took me around 1.5 minutes.

Please hit Kudos if you liked this approach

chetan2u · Answer

Use weighted average method ..
C... Average....M
x-5.....x.......x+10
So C/total =  \frac{(x+10)-x}{x+10-(x-5)} = \frac{10}{15}=\frac{2}{3} 
D

Or

Total number of leave by CFA= C*(x-5)
Total number of leave by MBA= M*(x+10)
Total number of leave = T*x
So T*x=Cx-5C+Mx+10M=(C+M)*X.......5C=10M....C=2M
Total =C+M=2M+M=3M
So C/T=2M/3M=2/3
D

Salsanousi · Answer

If we let X be the number of CFAs and Y the number of MBAs.

A be the sum of leaves taken by CFAs
B be the sum of leaves taken by MBAs

Let the average of leaves for all employees be z

We want to know x/(x+y)

We are given a/x = z - 5

a = x(z-5)

b/y = z + 10

b = y (z + 10)

Average of all = (a+b)/(x+y) = z

we get x = 2y

x/x+y = 2y/(2y+y) = 2/3 answer choice D

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Keismith · Answer

You can think about this question logically
10 more vacation than average has a bigger impact than 5 fewer days
So 10 more vacations (MBA) definitely represents a smaller portion of employees than CFAs.
Thus, the question becomes a choice between D and E
You can then set up an equation and test it out: 2/3(v-5) + 1/3(v+10) = v

Rishab2409 · Answer

Let’s picture this using a simple teeter-totter (seesaw).

The overall average of the company sits right in the middle as the balance point (the fulcrum).
The CFAs are lighter on vacation days. They sit 5 steps below the average.
The MBAs are heavier on vacation days. They sit 10 steps above the average.

Because the MBAs are twice as far away from the center (10 steps vs 5 steps), they exert twice as much leverage. To keep the seesaw perfectly balanced, you need twice as many CFAs on the other side to counter them.

For every 1 MBA, you need 2 CFAs to keep the average perfectly balanced.

If you look at the whole company as a group:2 parts are CFAs1 part is an MBATotal parts = 3Therefore, the CFAs make up 2/3 of the entire hedge fund.

At a certain hedge fund, there are only CFAs and MBAs. The average num

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At a certain hedge fund, there are only CFAs and MBAs. The average num

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