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

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.

GMAT assassins aren't born, they're made,
Rich

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.

Find the image below.Hope you like it.

It is but only when you have to draw on a laptop! On paper, it is the easiest tool to get the answer.
There you go:

To answer your question - Yes. They would still continue to be members of R but now they would also be members of S and no "new members from the 30 total" would be needed to populate R. This, in turn, would maximize the no. of people who are in neither MSR.

So M is unique and has 8 people who are only in M. Now with the maximizing step, we have 12 members in S ( among these 12 there are 5 who are also in R). So finally only 12 + 8 = 20 are in either of the committees. Hence 10 are such that they are in neither.

Hope it is clear. If you can imagine the solution this question can be solved in a matter of seconds.

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

thanks

Bunuel

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?

Thanks!

Hi Gladiator59

Just a prompt question the problem asks
"what is the greatest possible number of members in the club who are on none of the committees" I know how to get the answer (by merging R to S) thus , in this way only R will be equal to 0 , however even in this way those 5 who were merged with S are still part of committee R right? I mean that they now belong to both R and S but how can we deduce that those 5 have seized to be R members and are just S members after the merging.

Attached is a visual that should help. The basic strategy here is to maximize the overlap, thus minimizing the rest of the numbers, and maximizing the value of the "none" category.

Note that because only two of the groups overlap, we should use the formula for a double Venn diagram instead of the formula for a triple Venn diagram.



Please note that if the question asked for the minimum value, then Choice A (5) would be the correct answer.