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A club with a total membership of 30 has formed 3 committees

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A club with a total membership of 30 has formed 3 committees [#permalink] New post 02 Sep 2010, 20:52
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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
[Reveal] Spoiler: OA
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Re: hard question [#permalink] New post 02 Sep 2010, 21:09
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bupbebeo 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 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.

30 = 8 + (12 - x) + (5 - x) + 0 + 0 + x + 0 + Neither.

22 = (12-x) + (5-x) + x + 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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Re: hard question [#permalink] New post 02 Sep 2010, 22:01
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).
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Re: hard question [#permalink] New post 02 Sep 2010, 23:40
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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Re: hard question [#permalink] New post 03 Sep 2010, 00:04
ezhilkumarank wrote:
bupbebeo 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 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.

30 = 8 + (12 - x) + (5 - x) + 0 + 0 + x + 0 + Neither.

22 = (12-x) + (5-x) + x + 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
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Re: hard question [#permalink] New post 03 Sep 2010, 14:34
bupbebeo wrote:
ezhilkumarank wrote:
bupbebeo 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 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.

30 = 8 + (12 - x) + (5 - x) + 0 + 0 + x + 0 + Neither.

22 = (12-x) + (5-x) + x + 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.

30 = 8 + (12 - x) + (5 - x) [highlight]+[/highlight] 0 [highlight]+[/highlight] 0 [highlight]+[/highlight] x [highlight]+[/highlight] 0 + Neither.

Please refer the diagram below.

Attachment:
Venn-Diagram.jpg
Venn-Diagram.jpg [ 14.13 KiB | Viewed 2289 times ]

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Re: hard question [#permalink] New post 03 Sep 2010, 18:49
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
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Re: hard question [#permalink] New post 03 Sep 2010, 18:56
bupbebeo wrote:
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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Re: hard question [#permalink] New post 03 Sep 2010, 22:24
as you say: " The club's total membership consists of members in the committees (M, S and R) and also members not in any of the committees."

Therefore, I guess 30 = ( M U R U S ) + member not belong to any these sets.

(M U R U S) alone cannot be 30

do you think so
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Re: hard question [#permalink] New post 04 Sep 2010, 22:58
eazy question.... just make the smaller set a subset of bigger one.
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Re: hard question [#permalink] New post 05 Sep 2010, 06:56
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
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Re: hard question [#permalink] New post 05 Sep 2010, 07:13
Expert's post
bupbebeo wrote:
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.

Answer: D.

Hope it's clear.
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Re: hard question [#permalink] New post 05 Sep 2010, 08:07
bupbebeo 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 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.
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Re: hard question [#permalink] New post 06 Sep 2010, 23:46
Can anyone help me, why we have this formula

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

I really appreciate who can help me clear this.
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Re: hard question [#permalink] New post 07 Sep 2010, 04:36
Expert's post
bupbebeo wrote:
Can anyone help me, why we have this formula

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

I really appreciate who can help me clear this.


I don't think that Venn diagram is the best way to solve this question. Moreover the formula you posted has typos in it. Anyway check below link for formulas of 3 overlapping sets:
formulae-for-3-overlapping-sets-69014.html?hilit=formula%20exactly

Hope it helps.
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Re: hard question [#permalink] New post 07 Sep 2010, 06:34
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.
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Re: hard question [#permalink] New post 05 Oct 2012, 23:33
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.
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Re: hard question [#permalink] New post 05 Oct 2012, 23:43
I am sorry for my last post. I read the stem incorrectly
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Re: A club with a total membership of 30 has formed 3 committees [#permalink] New post 17 Oct 2013, 12:17
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