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A club with a total membership of 30 has formed 3 committees [#permalink]
14 Nov 2010, 22:42

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A club with a total membership of 30 has formed 3 committees, M, S and R, which have 8, 12 and 5 members respectively. If no members of committee M is on either of the other 2 committees, what is the greatest possible number of members in the club who are on none of the committees?

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Re: A club with a total membership of 30 [#permalink]
15 Nov 2010, 00:48

Expert's post

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monirjewel wrote:

A club with a total membership of 30 has formed 3 committees, M, S and R, which have 8, 12 and 5 members respectively. If no members of committee M is on either of the other 2 committees, what is the greatest possible number of members in the club who are on none of the committees? (A) 5 (B) 7 (C) 8 (D) 10 (E) 12

As "no member of committee M is on either of the other 2 committees" then 30-M=30-8=22 people are on committee S, committee R or on none of the committee. We want to maximize the last group: members in the club who are on none of the committees

General rule for such kind of problems: to maximize one quantity, minimize the others; to minimize one quantity, maximize the others.

So we should minimize total # of people who are on committee S and committee R. Now if ALL 5 people who are the members of committee R are also the members of committee S (if R is subset of S) then total # members of committee S and committee R would be minimized and equal to 12. Which means that 22-12=10 is the greatest possible number of members in the club who are on none of the committees.

Re: A club with a total membership of 30 [#permalink]
15 Nov 2010, 00:54

monirjewel wrote:

A club with a total membership of 30 has formed 3 committees, M, S and R, which have 8, 12 and 5 members respectively. If no members of committee M is on either of the other 2 committees, what is the greatest possible number of members in the club who are on none of the committees? (A) 5 (B) 7 (C) 8 (D) 10 (E) 12

IMO D: 10

Total members to be considered for committees S and R = 30 - 8 = 22

Greatest possible members on none of the committees would be a situation when all the members in R are from S, leading to answer as 22 - 12 = 10 members on none of the committees.

Re: A club with a total membership of 30 [#permalink]
16 Nov 2010, 18:00

I used a venn diagram to solve this, usually find these problems easier to solve that way.

So we have a total of 8+12+5=25 members in the 3 groups M,S,R. But we have 30 members in total. This tells you that 5 members could be those that participate in none of the groups. Further since the members in M are not part of any of the other committees, only a total of 17 members are possible that remain. Out of these 5 of group R are also part of S since you need to minimize the number of committee participants hence 7 are only in committee R. This leave 17-7=10 that can potentially be the maximum number not participating in any of the committees.

Re: A club with a total membership of 30 [#permalink]
16 Nov 2010, 18:44

Expert's post

gettinit wrote:

I used a venn diagram to solve this, usually find these problems easier to solve that way.

So we have a total of 8+12+5=25 members in the 3 groups M,S,R. But we have 30 members in total. This tells you that 5 members could be those that participate in none of the groups. Further since the members in M are not part of any of the other committees, only a total of 17 members are possible that remain. Out of these 5 of group R are also part of S since you need to minimize the number of committee participants hence 7 are only in committee R. This leave 17-7=10 that can potentially be the maximum number not participating in any of the committees.

You are right gettinit. Generally venn diagrams work the best for these kind of questions. One good thing to note here is that M is disjoint from the other two since no member of M can be a member of either of the other two sets. Therefore, out of 30 members, 8 are already out. Out of the other 22, we have to give 12 to S and 5 to R. Once we give 12 to S, just put the circle of R inside S (5 of the members of S become members of R too) so that you have 10 left outside who needn't be in any committee. _________________

Re: A club with a total membership of 30 [#permalink]
30 Nov 2010, 10:56

VeritasPrepKarishma wrote:

gettinit wrote:

I used a venn diagram to solve this, usually find these problems easier to solve that way.

So we have a total of 8+12+5=25 members in the 3 groups M,S,R. But we have 30 members in total. This tells you that 5 members could be those that participate in none of the groups. Further since the members in M are not part of any of the other committees, only a total of 17 members are possible that remain. Out of these 5 of group R are also part of S since you need to minimize the number of committee participants hence 7 are only in committee R. This leave 17-7=10 that can potentially be the maximum number not participating in any of the committees.

You are right gettinit. Generally venn diagrams work the best for these kind of questions. One good thing to note here is that M is disjoint from the other two since no member of M can be a member of either of the other two sets. Therefore, out of 30 members, 8 are already out. Out of the other 22, we have to give 12 to S and 5 to R. Once we give 12 to S, just put the circle of R inside S (5 of the members of S become members of R too) so that you have 10 left outside who needn't be in any committee.

Can someone show how to solve this with the image of ven diagram.. i know its cumbersome to draw and all.. but that will be a great help

Re: A club with a total membership of 30 [#permalink]
17 Mar 2011, 19:10

Bunuel wrote:

monirjewel wrote:

A club with a total membership of 30 has formed 3 committees, M, S and R, which have 8, 12 and 5 members respectively. If no members of committee M is on either of the other 2 committees, what is the greatest possible number of members in the club who are on none of the committees? (A) 5 (B) 7 (C) 8 (D) 10 (E) 12

As "no member of committee M is on either of the other 2 committees" then 30-M=30-8=22 people are on committee S, committee R or on none of the committee. We want to maximize the last group: members in the club who are on none of the committees

General rule for such kind of problems: to maximize one quantity, minimize the others; to minimize one quantity, maximize the others.

So we should minimize total # of people who are on committee S and committee R. Now if ALL 5 people who are the members of committee R are also the members of committee S (if R is subset of S) then total # members of committee S and committee R would be minimized and equal to 12. Which means that 22-12=10 is the greatest possible number of members in the club who are on none of the committees.

Answer: D.

Hope it's clear.

Bunuel, This is definitely perfect. But I have a question. Can this be solved with the "Exactly two" set formula. Bcos I tried and it also gives me the right answer, here it is:

Formula: Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + Neither

30=8+12+5-(5)-2*0+N - ( here intersection of all three=0; and to maximize the result, I took 2 group overlap as 5 -between 12 and 5) => N = 30-20=10

Please confirm if this is the right approach as well. _________________

Consider giving Kudos if my post helps in some way

Re: A club with a total membership of 30 [#permalink]
19 Mar 2011, 00:38

Because 8 members from committee M are not common, so to minimize non-members we havt to "commonize" S and R, who can have 5 common (and total # of members = 12 including B and C). So total number of members = 12+8 = 20

=> Non-members = 30-20 = 10 _________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Re: A club with a total membership of 30 has formed 3 committees [#permalink]
01 Mar 2015, 19:11

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Would one be wrong to assume that since you have the total groups that sum to 25, 5 ppl are unaccounted for, and in order to maximize members you essentially have an additional 5 spaces to bring you to 30 members... 5+5=10....thus answer D? Or did I just get lucky?

Re: A club with a total membership of 30 has formed 3 committees [#permalink]
17 Apr 2015, 09:57

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Expert's post

Hi FTS185,

The method you've described is not perfectly clear, so it's tough to say if it's logical or lucky.

In this question, to MAXIMIZE the number of people who are NOT on a committee, we have to "overlap" as many people as possible (put as many of them onto MORE than one committee as possible). We're told that the 8 members of committee M are NOT on any other committee, so we can't do anything with them. However, the members of the other 2 committees COULD overlap (the 5 members of committee R COULD be on committee S). This means that we COULD be dealing with just 12 members accounting for everyone on those 2 committees. With the other 8 members from committee M, we have 12 + 8 = 20 members. THAT leaves 10 members that are on NO committee.

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