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A club with a total membership of 30 has formed 3 committees [#permalink]
02 Sep 2010, 20:52

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Difficulty:

35% (medium)

Question Stats:

57% (02:12) correct
42% (01:12) wrong based on 107 sessions

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 member 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 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 member 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

please answer with explanation

thank you so much

Given no member of committee M is on either of the other 2 committees -- hence (M n R) and (M n S) and (M n R n S) is zero.

Hence only M is 8. Now we need to consider only S, (S n R) and R.

(M U R U S) = M + R + S - (M n R) - (M n S) - (R n S) + (M n R n S) + Neither.

Now the max value of x could be 5 and the min value of x could be 0.

When x is 5 (max), Neither is 10. When x is 0 (min), Neither is 5. We need maximum no of people who do not belong to any group. Hence max value of neither is 10.

Answer 10 (D).
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For a less mathematical approach (I still haven't gotten a good handle on permutations and combinations):

8 members of the club are on committee M. None of those 8 are on the other committees, so the 12 members of committee S are all different members (you can ignore R since there might be overlap with S and it's a smaller number than S). 12+8=20 members minimum of the club are on committees. 30-20=10 members maximum are not on any committees.

I don't know if this would work on more complex problems, though, so it's probably good to learn and understand the mathematical approach (which I need to do as well).

I solved it correct but made a silly mistake acc to me 12 + 8 =22 hence answer is 30 - 22 = 8 what a mess I am into anyways

Answer should be 10 that all the members of the R are also members of the S committee hence 12 +8 =20 30 -20 = 10 Answer D
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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 member 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

please answer with explanation

thank you so much

Given no member of committee M is on either of the other 2 committees -- hence (M n R) and (M n S) and (M n R n S) is zero.

Hence only M is 8. Now we need to consider only S, (S n R) and R.

(M U R U S) = M + R + S - (M n R) - (M n S) - (R n S) + (M n R n S) + Neither.

Now the max value of x could be 5 and the min value of x could be 0.

When x is 5 (max), Neither is 10. When x is 0 (min), Neither is 5. We need maximum no of people who do not belong to any group. Hence max value of neither is 10.

Answer 10 (D).

first of all, thanks your great answer. However, I just have one more question

HOw can you draw the equation

(M U R U S) = M + R + S - (M n R) - (M n S) - (R n S) + (M n R n S) + Neither.

help me explain it to me. I try to figure out but i can't

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 member 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

please answer with explanation

thank you so much

Given no member of committee M is on either of the other 2 committees -- hence (M n R) and (M n S) and (M n R n S) is zero.

Hence only M is 8. Now we need to consider only S, (S n R) and R.

(M U R U S) = M + R + S - (M n R) - (M n S) - (R n S) + (M n R n S) + Neither.

Now the max value of x could be 5 and the min value of x could be 0.

When x is 5 (max), Neither is 10. When x is 0 (min), Neither is 5. We need maximum no of people who do not belong to any group. Hence max value of neither is 10.

Answer 10 (D).

first of all, thanks your great answer. However, I just have one more question

HOw can you draw the equation

(M U R U S) = M + R + S - (M n R) - (M n S) - (R n S) + (M n R n S) + Neither.

help me explain it to me. I try to figure out but i can't

I had setup the Venn diagram to solve this question. Also I have incorrectly setup the equation but when I substituted the values I did use the equation correctly because I referred from the Venn diagram.

(M U R U S) = M + R + S [highlight]-[/highlight] (M n R) [highlight]-[/highlight] (M n S) [highlight]-[/highlight] (R n S) [highlight]+[/highlight] (M n R n S) + Neither.

In fact, I can't figure out why (M U R U S) = 30. because ( M U R U S ) means that elements of M, R, and S. In fact, there are some elements doesn't belong to these sets. Therefore, I believe there should be less than 30.

In fact, I can't figure out why (M U R U S) = 30. because ( M U R U S ) means that elements of M, R, and S. In fact, there are some elements doesn't belong to these sets. Therefore, I believe there should be less than 30.

Could you help me clear this

This information is given in the question itself -- "A club with a total membership of 30 has formed 3 committees, M, S and R".

The club's total membership consists of members in the committees (M, S and R) and also members not in any of the committees.
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any one can help me with full explanation with this question.

I have spent two days to try to understand it and understand the explanation of others. But I haven't figured out them yet.

I think this question is so important. everyone helps me????

thank you in advance

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 member 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.

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 member 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

please answer with explanation

thank you so much

Solved this one pretty quickly. 30 people/ 3 committees. M,S,R with 8,12,5 respectively.

--> If the 8 people in M can not be in any other committee, that leaves 22 people left to be in the two committees remaining, or not be in a committee at all. Out of these 22, 12 must be in group S. Of the 12 in group S, 5/12 could serve in both committee S and committee R. You would not need to draw any more people out of the remaining to form group R since this group is represented by individuals from group S that are serving double duty. Hence, the 8 from M, plus the 12 from S (including the 5 that serve on R as well) is 20 people out of the 30. 30-20=10, or the maximum number of people possible in the club that do jack shit!

This was just what quickly went through my head when I saw this problem. I'm not sure this is even a question to deal with probabilities. I thought it was gonna be a permu/combo question, but that wasn't necessary.

The easiest way to solve this question is to just think about what conditions need to be true in order to maximize the number of people who are not in any of the three groups. You know that M is completely self-contained with 8 people, so you don't even need to think about them. Just consider a group of 22 people split into R, S, both, or neither. You know how many people are in R (12) and how many are in S (5). You don't know how many people are in both. So what number of people do you need to have in both groups to maximize the number of people who are in neither?

Think about it like this: if NOBODY was in both R and S, you would have 17 people in the groups, leaving 5 people in neither group. What if EVERYONE in group S was also in group R? Then you would have 12 people in the groups, leaving 10 in neither group. So 10 is the answer.

I don't agree with (or can't find) the official answer. Without coming up with any equation, I find that the answer is 5. My line of reasoning: since there is no overlapping between any two other member groups, to maximize the non-group members, the number of members belonging to all the 3 groups must zero 30 = (12 + 8 + 5 ) + 5 Also with Equations: Group M :8 Group S : 12 Group R: 5 Non-Group: y Total : 30 Let's x = M&S&R, a = M&S, b = M&R and c = S&R. So 30 = (8-a-b-x) + (12-a-c-x) + (5-c-b-x) + x + y a, b and c are said to be 0 30 = 25 -2x + y ==> y = 5 + 2x. To minimize y x got to be 0, thus y = 5. What's wrong w/t my line of reasoning Please help out. Brother Karamazov

Re: A club with a total membership of 30 has formed 3 committees [#permalink]
17 Oct 2013, 12:17

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