In a certain class 1/3 of students are honors 1/4 of

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In a certain class 1/3 of students are honors 1/4 of [#permalink]

24 Aug 2004, 21:22

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In a certain class 1/3 of students are honors 1/4 of students play varisity sports. If 12 students are honors and play varisty sports , what is the least number of students in class?

20

22

36

48

144

Please explain your solution..

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24 Aug 2004, 22:34

Backsolve.
Using 36, 1/3(36)=12
1/4(36)=9
Impossible, since 12 are both.

Using 48, 1/3(48) = 16
1/4(48) = 12

Possible, since we have 12 playing sports now. We are asked for least possible number, so this has to be the one, since 144 will satisfy the condition given but have an excess of students who are only playing one role.

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Re: PS: varisty sports [#permalink]

25 Aug 2004, 04:13

The least number comes when both the categories have the highest overlap. Thus, the maximum overlap is all the students that play varsity sports are honors. => x/4 = 12. i.e., x = 48 which is the minimum.

afife76 wrote:

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25 Aug 2004, 08:34

My answer is 144

Total=Group1+group2-both+neither
144=48+36-12+60
144=144.

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25 Aug 2004, 08:35

My answer is 144

Total=Group1+group2-both+neither
144=48+36-12+60
144=144.

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25 Aug 2004, 08:37

I got the math wrong, please ignore last post.

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Re: PS: varisty sports [#permalink]

25 Aug 2004, 12:14

mallelac wrote:

Mallelac, your approach is very cool!!

Thank you to both!! :-)

Re: PS: varisty sports [#permalink] 25 Aug 2004, 12:14

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In a certain class 1/3 of students are honors 1/4 of

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