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# The 39 parents participating in the Smithville PTA have been

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The 39 parents participating in the Smithville PTA have been [#permalink]

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27 Sep 2013, 07:33
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66% (02:43) correct 34% (02:04) wrong based on 136 sessions

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The 39 parents participating in the Smithville PTA have been assigned to at least 1 of 3 committees: festival planning, classroom aid, and teacher relations. 21 parents are assigned to the festival planning committee, 18 parents are assigned to the classroom aid committee, and 19 parents are assigned to the teacher relations committee. If 5 parents are assigned to all 3 committees, how many parents are assigned to exactly 2 committees?

A. 4
B. 6
C. 8
D. 9
E. 10
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Sep 2013, 08:31, edited 1 time in total.
Edited the question.
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Re: The 39 parents participating in the Smithville PTA [#permalink]

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27 Sep 2013, 08:30
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violetsplash wrote:
The 39 parents participating in the Smithville PTA have been assigned to at least 1 of 3 committees: festival planning, classroom aid, and teacher relations. 21 parents are assigned to the festival planning committee, 18 parents are assigned to the classroom aid committee, and 19 parents are assigned to the teacher relations committee. If 5 parents are assigned to all 3 committees, how many parents are assigned to exactly 2 committees?

A. 4
B. 6
C. 8
D. 9
E. 10

$$Total = A + B + C - (sum \ of \ EXACTLY \ 2-group \ overlaps) - 2*(all \ three) + Neither$$.

Since each parent is assigned to at least 1 of 3 committees, then the # of parents assigned to neither of them is 0.

$$39=21+18+19-x-2*5+0$$ --> $$x=9$$.

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Re: The 39 parents participating in the Smithville PTA have been [#permalink]

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27 Sep 2013, 11:09
The formula is Total = A+B+C - sum of exactly two + 2*all three + neither
21+18+19-x-2*5=39
solving for x you get 9
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Re: The 39 parents participating in the Smithville PTA have been [#permalink]

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29 Aug 2015, 01:11
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Re: The 39 parents participating in the Smithville PTA have been [#permalink]

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27 Oct 2015, 07:15
Could someone explain to me why it is necessary to double the amount of parents in all groups?

Thanks
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The 39 parents participating in the Smithville PTA have been [#permalink]

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29 Oct 2015, 23:43
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Condon wrote:
Could someone explain to me why it is necessary to double the amount of parents in all groups?

Thanks

This post tells you how you get the formula for n(Exactly two sets) and others:
http://www.veritasprep.com/blog/2015/10 ... ping-sets/

This is our standard formula:
Total = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) + n(No Set)
39 = 21 + 18 + 19 - (n(A and B) + n(B and C) + n(C and A)) + 5 + 0
(n(A and B) + n(B and C) + n(C and A)) = 24

Now let’s see how we can calculate the number of people in exactly two sets.
n(A and B) includes people who are in both A and B and it also includes people who are in A, B and C. Because of this, we should remove n(A and B and C) from n(A and B) to get n(A and B only). Similarly, you get n(B and C only) and n(C and A only), so adding all these three will give us number of people in exactly 2 sets.
n(Exactly two sets) = n(A and B) – n(A and B and C) + n(B and C) – n(A and B and C) + n(C and A) – n(A and B and C).

Therefore:
n(Exactly two sets) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C)
n(Exactly two sets) = 24 - 3*5 = 9

In other words, you can substitute from formula 2 into formula 1 to get
n(A and B) + n(B and C) + n(C and A) = n(Exactly two sets) + 3*n(A and B and C)

Put in formula 1:
Total = n(A) + n(B) + n(C) – (n(Exactly two sets) + 3*n(A and B and C)) + n(A and B and C) + n(No Set)
Total = n(A) + n(B) + n(C) – n(Exactly two sets) - 2*n(A and B and C)) + n(No Set)
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The 39 parents participating in the Smithville PTA have been   [#permalink] 29 Oct 2015, 23:43
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