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There are 230 students. 80 play football, 42 play soccer and [#permalink]
21 Aug 2007, 19:02

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There are 230 students. 80 play football, 42 play soccer and 12 play rugby. 32 play exactly 2 sports and 4 play all three. How many students play none?

Re: PS- Venn diagram question [#permalink]
21 Aug 2007, 19:18

r019h wrote:

There are 230 students. 80 play football, 42 play soccer and 12 play rugby. 32 play exactly 2 sports and 4 play all three. How many students play none?

Re: PS- Venn diagram question [#permalink]
21 Aug 2007, 19:40

bkk145 wrote:

r019h wrote:

There are 230 students. 80 play football, 42 play soccer and 12 play rugby. 32 play exactly 2 sports and 4 play all three. How many students play none?

Re: PS- Venn diagram question [#permalink]
21 Aug 2007, 19:52

abhi_ferrari wrote:

bkk145 wrote:

r019h wrote:

There are 230 students. 80 play football, 42 play soccer and 12 play rugby. 32 play exactly 2 sports and 4 play all three. How many students play none?

I am getting a different answer. Just applying the set theory formula:

AUBUC = A + B + C - {A n B + B n C + C n A} + A n B n C

230 = 80 + 42 + 12 - 32 + 4 + none

So, none = 124.

bkk, can you please explain why did you had a -ive 2*4 ? May be I am missing something. Thanks.

Basically, you count students who play "all three sports" twice because question said "exactly two sports". If the question ask "two sports and more", then your formula is correct since it is already count "all three sports" once.

Sumithra, what is this MGMAT method? Could you elaborate? I get the two-group Venn Diagram questions but get confused when there are three groups. Thanks, would appreciate ur help.

Why would you count the number of people who play exactly three sports twice?

To put it numerically, say you have the 4 students who each play 1 sport, so you have:

4 in football
4 in soccer
4 in rugby

However, if we discover that we actually have 4 students each playing all 3 sports, then the number of students represented has been exagerated. We really have 4 students represented, not 12 which is why we subtract 2*4 or 8 from 12....

I guess what I wasn't seeing is that the 32 comes from the number of people who are playing exactly 2 sports. However, the formula reads...

|A n B n C| = |A| + |B| + |C| - |A U B| - |A U C| - |B U C| + |AnBnC|

but |AUB| + |AUC| + |BUC| is NOT 32. It is larger than that because it it also counts the number of people who play all 3 sports, each of whom are in the unions of any two of the sets.

So, to use the equation it should be...

|A n B n C| = 80 + 42 + 12 - (32 + 4 + 4 + 4) + 4 = 134 - 44 + 4 = 94