Leaders, Directors and Fundrasiers

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Leaders, Directors and Fundrasiers

by faraz_jeddah » Wed Aug 21, 2013 9:51 pm
A nonprofit group organizes its local fundraisers in teams, with each of its L team leaders responsible for D group directors, and each of those D group directors responsible for F fundraisers. If the only three positions on each local team are team leaders, group directors, and fundraisers, and there are more group directors than team leaders, how many team leaders are on the Dallas team?

(1) There are 81 total members on the Dallas team

(2) There are 5 group directors on the Dallas team

My approach

Given informaton:
Total # of members T = L + D + F
D > L

Target question - what is L?

Statement 1:
T = 81

I used the brute approach of testing values and see if I could get only value of L

Let
L = 1
D = 2

This means each Leader will have 2 directors and each director will have 39 fund raisers
How did I reach that?
81 = 1 + 1 + 1 + F + F
2F = 78
F = 39
Now if I take any value of L > 1 I cannot distribute the fund raisers evenly among the Directors
For example
L = 2
D = 3 ( I cannot take D = 2 as it is stated D>L)
81 = 2+3+3+F+F+F
3F = 73 => Not divisible


This would make L=1 as the only value.

So statement 1 should be sufficient.

OA is NOT A
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by [email protected] » Wed Aug 21, 2013 10:54 pm
Hi faraz_jeddah,

I'm going to give you a hint and have you retry this question:

For Fact 1, there IS a way to get to a total of 81 in which there are 3 team leaders. Can you figure out how?

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by faraz_jeddah » Wed Aug 21, 2013 11:10 pm
Ahh Rich! Just the person I was looking forward to replying. :)

So if I take L = 3

D = 4

81 = 3 + 4 + 4 + 4 + 12F
12 F = 66 (Not divisible)

if D = 5

81 = 3 + 5 + 5 + 5 + 15 F
15 F = 63 (not divisible)

If D = 6

81 = 3 + 6 + 6 + 6 + 18F
18F = 60 (not divisible)

Should I keep going? Even if I do end up finding a suitable value, there must be a more efficient way.
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by GMATGuruNY » Thu Aug 22, 2013 3:34 am
faraz_jeddah wrote:
So if I take L = 3

D = 4

81 = 3 + 4 + 4 + 4 + 12F
You're overly constraining the problem.
It is not necessary that the number of directors for EACH team leader be greater than L.
It is only necessary the the TOTAL number of directors be greater than L.

Let L = the total number of team leaders, D = the total number of directors, and F = the total number of fundraisers.
Thus:
L + D + F = 81.

Let L = 3.
If each of these 3 team leaders oversees 2 directors, then D = 3*2 = 6.
Since the remaining number of members must be fundraisers, F = 81-3-6 = 72.
For there to be 72 fundraisers, each of the 6 directors must oversee 12 fundraisers.
Thus, it's possible that L=3, D=6, and F=72, for a total of 81 members.
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by faraz_jeddah » Thu Aug 22, 2013 6:50 am
GMATGuruNY wrote:
faraz_jeddah wrote:
So if I take L = 3

D = 4

81 = 3 + 4 + 4 + 4 + 12F
You're overly constraining the problem.
It is not necessary that the number of directors for EACH team leader be greater than L.
It is only necessary the the TOTAL number of directors be greater than L.

Let L = the total number of team leaders, D = the total number of directors, and F = the total number of fundraisers.
Thus:
L + D + F = 81.

Let L = 3.
If each of these 3 team leaders oversees 2 directors, then D = 3*2 = 6.
Since the remaining number of members must be fundraisers, F = 81-3-6 = 72.
For there to be 72 fundraisers, each of the 6 directors must oversee 12 fundraisers.
Thus, it's possible that L=3, D=6, and F=72, for a total of 81 members.
Thanks Mitch. I see where I was going wrong.

But is there any algebraic / factoring / inequality method that you may have up your sleeve to solve this problem?
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by GMATGuruNY » Thu Aug 22, 2013 7:48 am
faraz_jeddah wrote:
But is there any algebraic / factoring / inequality method that you may have up your sleeve to solve this problem?
Leaders + Directors + Fundraisers = 81.

Let:
L = the TOTAL number of leaders.
d = the number of directors PER team leader.
f = the number of fundraisers PER director.

Since each of the L leaders oversees d directors, the total number of directors = Ld.
Since each of the Ld directors oversees f fundraisers, the total number of fundraisers = Ldf.
Thus:
L + Ld + Ldf = 81.
L(1 + d + df) = 81.

Thus, L must be a FACTOR of 81:
1, 3, 9...

From here, we can work out one case where L=1 and another where L=3 or L=9.

Another viable case:
Let L = 9.
If each of these 9 leaders oversees 2 directors, then D = 9*2 = 18.
Here, F = 81 - 9 - 18 = 54.
This works:
Since F/D = 54/18 = 3, each director oversees 3 fundraisers.
Last edited by GMATGuruNY on Thu Aug 22, 2013 5:14 pm, edited 2 times in total.
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by [email protected] » Thu Aug 22, 2013 12:39 pm
Hi faraz_jeddah,

This question has some built-in Number Properties that can help you to save time:

1) The total is an odd number, so L MUST be odd (if L were even, then we'd be multiplying by an even number, so the total would have to be even).

2) As Mitch noted, the TOTAL number of Directors has to be greater than L; this doesn't necessarily mean that D > L

If you notice how each variable relates to one-another, then it doesn't take long to come up with a few options:

L = 1
D = 2
F = 39

L = 1
D = 5
F = 15

L = 3
D = 6
F = 4

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by Java_85 » Thu Aug 22, 2013 4:50 pm
My answer is B.

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by faraz_jeddah » Fri Aug 23, 2013 3:48 am
Thanks Mitch & Rich (I can be a rapper now!)

The OA is B for people who are interested.
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