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The 39 parents participating in the Smithville PTA have been
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Updated on: 27 Sep 2013, 08:31
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
A. 4
B. 6
C. 8
D. 9
E. 10
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.
Last edited by Bunuel on 27 Sep 2013, 08:31, edited 1 time in total.
Edited the question.
Re: The 39 parents participating in the Smithville PTA
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27 Sep 2013, 08:30
27 Sep 2013, 08:30
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Expert Reply
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
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
The 39 parents participating in the Smithville PTA have been
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29 Oct 2015, 23:43
29 Oct 2015, 23:43
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Expert Reply
Condon wrote:
Could someone explain to me why it is necessary to double the amount of parents in all groups?
Thanks
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)
