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The 39 parents participating in the Smithville PTA have been

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violetsplash
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The 39 parents participating in the Smithville PTA have been [#permalink] New post   Updated on: 27 Sep 2013, 08:31
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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
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D

Originally posted by violetsplash on 27 Sep 2013, 07:33.
Last edited by Bunuel on 27 Sep 2013, 08:31, edited 1 time in total.
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Re: The 39 parents participating in the Smithville PTA [#permalink] New post   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.

\(39=21+18+19-x-2*5+0\) --> \(x=9\).

Answer: D.

FOR MORE CHECK ADVANCED OVERLAPPING SETS PROBLEMS: advanced-overlapping-sets-problems-144260.html
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The 39 parents participating in the Smithville PTA have been [#permalink] New post   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:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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

Answer (D)

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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Re: The 39 parents participating in the Smithville PTA have been [#permalink] New post   27 Oct 2015, 07:15
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Could someone explain to me why it is necessary to double the amount of parents in all groups?

Thanks
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Re: The 39 parents participating in the Smithville PTA have been [#permalink] New post   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
Answer D
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Re: The 39 parents participating in the Smithville PTA have been [#permalink] New post   24 Mar 2017, 10:59
A+B+C - sum of exactly two - 2*all three
ans: D
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Re: The 39 parents participating in the Smithville PTA have been [#permalink] New post   21 Jul 2018, 12:17
Formula for three sets is 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)
Which implies 39 = 21+18+19 – n(A and B) – n(B and C) – n(C and A) + 5
Therefore n(A and B) + n(B and C) + n(C and A) = 24
Between A & B there are 5 parents participating in C & similarly the same thing is happening between A & C and B & C.
Therefore correct answer is 24-15=9
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Re: The 39 parents participating in the Smithville PTA have been [#permalink] New post   10 Sep 2018, 10:58
Let exactly one= A
Let exactly two= B
Let exactly three=C

So, A+B+C=39-----------I
A+2B+3C=58----------II

from I and II
B+2C=19
B=19-10 [C is given 5]
B=9
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Re: The 39 parents participating in the Smithville PTA have been [#permalink] New post   30 Oct 2023, 09:51
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Re: The 39 parents participating in the Smithville PTA have been [#permalink]
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