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In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]

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01 May 2012, 20:28

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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?

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.

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]

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04 Nov 2013, 14:29

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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

In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]

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29 Jun 2014, 13:12

PareshGmat wrote:

Answer = 10 Using Venn Diagram Bunuel, can you please tell if this method is correct?. Got x -ve in this case

I do not agree with given explanation for the next reason.

If we apply the same logic as we did in previous questions, we get the next....

H=20 C=15 F=11

32 students in game

H and C max 7 C and F max 4 H and F max 5

so, H=20-7-5=8 max C=15-7-4= 4 max F=11-4-5=2 max

at this moment we have the max number of students= 7+4+5+8+4+2=30...2 less than the number of students who participate in any class. If we put any of these students into all three group we will reduce all three numbers of every two classes and can never get the total of 32.

So the formula might be applied well, but this is wrong answer, or even more possible wrong figures in task.

I was surprised that Bunuel did not notice this mistake.

Re: In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]

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17 Aug 2014, 05:21

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

Bunnel,

Will you please explain, why you have chosen this formula against exactly 2 formulae.. LS
_________________

Re: In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]

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15 Sep 2014, 03:26

lastshot wrote:

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

Bunnel,

Will you please explain, why you have chosen this formula against exactly 2 formulae.. LS

Read the above posts from Bunuel. He has used both the formulas in this problem. Formula no.1 to calculate the no. of students playing all the 3 sports, and then Formula no.2 to calculate the no. of students who play exactly two sports....Hope this helps!

Re: In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]

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14 Jun 2015, 04:50

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I've solved this one with the second formula Let X be the area where all 3 overlap and Exactly 2-Groups overlaps = 2-Group ovelaps - 3*X --> 50=20+15+11 -(7+4+5-3X) - 2X+18 X=2 and 16-3X=10
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Re: In a class of 50 students, 20 play Hockey, 15 play Cricket [#permalink]

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28 Jan 2017, 09:16

MJ23 wrote:

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

Hope this helps.

Cheers, MJ

Dense, but helpful. Somehow, I only noticed my mistake with this explanation and not the one from Bunuel (which is literally the same) thanks