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In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]
01 May 2012, 20:28

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Difficulty:

45% (medium)

Question Stats:

59% (02:44) correct
41% (02:16) wrong based on 200 sessions

In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football. 7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football. If 18 students do not play any of these given sports, how many students play exactly two of these sports?

Re: In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]
01 May 2012, 22:27

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Expert's post

gmihir wrote:

In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football. 7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football. If 18 students do not play any of these given sports, how many students play exactly two of these sports?

A. 12 B. 10 C. 11 D. 15 E. 14

Notice that "7 play both Hockey and Cricket" does not mean that out of those 7, some does not play Football too. The same for Cricket/Football and Hockey/Football.

Those who play ONLY Hockey and Cricket are 7-2=5; Those who play ONLY Cricket and Football are 4-2=2; Those who play ONLY Hockey and Football are 5-2=3;

Hence, 5+2+3=10 students play exactly two of these sports.

Re: In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]
09 Oct 2012, 10:39

I don't agree with the answer given If the question is correctly worded the answer ought to be 7 + 5 + 4 = 16 But if the question is: How many students play exactly one of these sports then the answer could be 10 (I haven't checked) Brother Karamazov

Re: In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]
07 Jul 2013, 22:02

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Expert's post

josemnz83 wrote:

This question asks for the number of students who played exactly two sports? Why does the second formula not work here?

Notice that "7 play both Hockey and Cricket..." does NOT mean that these 7 students play ONLY Hockey and Cricket, some might play Football too. The same for "4 play Cricket and Football and 5 play Hockey and football". So, we cannot use the second formula directly. Also notice that we don't know the number of students who play all three sports.

But we CAN use the first formula, find the number of students who play all three and then find the number of students who play exactly two of the sports.

Re: In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]
18 Oct 2013, 22:00

Bunuel wrote:

josemnz83 wrote:

This question asks for the number of students who played exactly two sports? Why does the second formula not work here?

Notice that "7 play both Hockey and Cricket..." does NOT mean that these 7 students play ONLY Hockey and Cricket, some might play Football too. The same for "4 play Cricket and Football and 5 play Hockey and football". So, we cannot use the second formula directly. Also notice that we don't know the number of students who play all three sports.

But we CAN use the first formula, find the number of students who play all three and then find the number of students who play exactly two of the sports.

Hope it's clear.

hey bunuel, if we had to find exactly one sport then how are we suppose to go ahead ?

Re: In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]
04 Nov 2013, 14:29

Bunuel wrote:

josemnz83 wrote:

This question asks for the number of students who played exactly two sports? Why does the second formula not work here?

Notice that "7 play both Hockey and Cricket..." does NOT mean that these 7 students play ONLY Hockey and Cricket, some might play Football too. The same for "4 play Cricket and Football and 5 play Hockey and football". So, we cannot use the second formula directly. Also notice that we don't know the number of students who play all three sports.

But we CAN use the first formula, find the number of students who play all three and then find the number of students who play exactly two of the sports.

Hope it's clear.

Hi All,

That's exactly what I did, started with formula 1 and then used #2 once had "g" (all three):

Formula #1: 50=20+15+11-(7+4+5)+18 --> 2 (all three or "g") Formula #2: 50=20+15+11-x-(2*2)+18 which leads to 50=60-x ; x=10