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3/8 of all students at Social High are in all three of the

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3/8 of all students at Social High are in all three of the [#permalink]

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3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?

(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 5/8
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Re: The Quest for 700: Weekly GMAT Challenge [#permalink]

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New post 21 Aug 2010, 05:33
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zisis wrote:
3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?

(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 5/8


Let's # of students at Social High be 8 (I picked 8 as in this case 3/8 of total and 5/8 of total will be an integer).

3/8 of all students at Social High are in all three clubs --> 3/8*8=3 people are in exactly 3 clubs;

1/2 of all students are in Albanian club --> 1/2*8=4 people are in Albanian club;
5/8 of all students are in Bardic club --> 5/8*8=5 people are in Bardic club;
3/4 of all students are in Checkmate club --> 3/4*8=6 people are in Checkmate club;

Also as every student is in at least one club then # of students in neither of clubs is 0;

Total=A+B+C-{# of students in exactly 2 clubs}-2*{# of students in exactly 3 clubs}+{# of students in neither of clubs};

8=4+5+6-{# of students in exactly 2 clubs}-2*3+0 --> {# of students in exactly 2 clubs}=1, so fraction is 1/8.

Answer: A.

For more about the formula used check my post at: formulae-for-3-overlapping-sets-69014.html?hilit=exactly%20groups or overlapping-sets-problems-87628.html?hilit=exactly%20members

Hope it helps.
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Re: The Quest for 700: Weekly GMAT Challenge [#permalink]

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New post 20 Sep 2010, 21:30
If you plot this data on a venn diagram, the problem would be much clearer .
General formula for 3 intersecting sets of data:
n(A u B u C) = n(A) + n(B) + n(C) - n(A int B) - n(B int C) - n(A int C) + n(A int B int C).
Assuming all students were part of atleast 1.
My solution is giving me an absurd result and it was definitely not 1/8. Please explain, I might have missed the point.
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Re: The Quest for 700: Weekly GMAT Challenge [#permalink]

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New post 20 Sep 2010, 21:43
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vicksikand wrote:
If you plot this data on a venn diagram, the problem would be much clearer .
General formula for 3 intersecting sets of data:
n(A u B u C) = n(A) + n(B) + n(C) - n(A int B) - n(B int C) - n(A int C) + n(A int B int C).
Assuming all students were part of atleast 1.
My solution is giving me an absurd result and it was definitely not 1/8. Please explain, I might have missed the point.


It's not clear how you applied the formula you wrote and what answer did you get, so it's hard to tell where you went wrong.

But the problem might be in formula you apply. Actually there are 2 formulas for 3 overlapping sets and the second one is better for this particular problem. Check the solution above and also this link about the mentioned 2 formulas: formulae-for-3-overlapping-sets-69014.html?hilit=exactly%20groups

Hope it helps.
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Re: The Quest for 700: Weekly GMAT Challenge [#permalink]

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New post 20 Sep 2010, 21:57
1. For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
Thats the one I have used all along and has worked fine for me. I would think that this formula would apply to this problem.
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New post 20 Sep 2010, 22:02
vicksikand wrote:
1. For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
Thats the one I have used all along and has worked fine for me. I would think that this formula would apply to this problem.


It depends how you apply the formula. Again it's not the only formula for 3-overlapping sets, there is a second one and it fits this problem better. You can check the link I gave if you are interested to know the difference and to study how to apply either of them.
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New post 20 Sep 2010, 22:08
5. To determine the No of persons in two or more sets (at least 2 sets) : P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)

3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?

It doesnt say 1/2 of students are in Albanian only - In my opinion 1/2 includes the number that are common with Bardic and Checkmate. Formula 2 is for finding no. of persons in one set. I kind of see how the author arrived at that formula, but I wouldnt apply it to our case.

We have to find (P(A n B) + P(A n C) + P(B n C))/2
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Re: The Quest for 700: Weekly GMAT Challenge [#permalink]

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New post 20 Sep 2010, 22:23
vicksikand wrote:
5. To determine the No of persons in two or more sets (at least 2 sets) : P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)

3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?

It doesnt say 1/2 of students are in Albanian only - In my opinion 1/2 includes the number that are common with Bardic and Checkmate. Formula 2 is for finding no. of persons in one set. I kind of see how the author arrived at that formula, but I wouldnt apply it to our case.

We have to find (P(A n B) + P(A n C) + P(B n C))/2


You are right it doesn't say that 1/2 of students are in Albanian only and nowhere in the solution above it's mentioned. Also this is not the formula I meant.

Refer to my post at: formulae-for-3-overlapping-sets-69014.html?hilit=exactly%20groups

And refer to the solution above to see how the second formula is applied to this problem.
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New post 20 Sep 2010, 23:07
I agree. Good explanation on the proof.
(a n b n c) was bothering me and you did a good job explaining why we need to subtract it twice and let that 1 remain so that it is counted at least once.

I remember having seen 1 question based on this in the MGMAT test series.

Thanks!
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Re: 3/8 of all students at Social High are in all three of the [#permalink]

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Re: 3/8 of all students at Social High are in all three of the [#permalink]

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New post 30 Oct 2014, 08:58
Why did you multiply {# of students in exactly 3 clubs} by 2?
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Re: 3/8 of all students at Social High are in all three of the [#permalink]

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3/8 of all students at Social High are in all three of the [#permalink]

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New post 01 Nov 2014, 21:40
Basic formula is:

Total=A+B+C - exactly two - 2*all three, so

Let pick 16 as total number of students

16=8+10+12 - exactly two - 2*6
16=30-x-12
x=2

2/16=1/8

with fractions:

8/8=1/2+5/8+3/4 - x - 3/4
8/8=15/8-x-3/4
x=1/8

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Re: 3/8 of all students at Social High are in all three of the [#permalink]

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