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vcbabu
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venn diagram 29 Jan 2009, 11:20
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Q: THERE are 30 students. 12 play basket ball . 15 play soccer and 19 play volleyball. if 7 play all 3 games and exactly 6 play 2 games , how many of them play none of 3 games?

0 2 4 6 8



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venn diagram 29 Jan 2009, 11:34
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vcbabu
Q: THERE are 30 students. 12 play basket ball . 15 play soccer and 19 play volleyball. if 7 play all 3 games and exactly 6 play 2 games , how many of them play none of 3 games?

0 2 4 6 8
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No of students who play games = (15+19+12) - 2(7)-6 = 46-20=26

Ans= 30-26= 4
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venn diagram 01 Feb 2009, 07:30
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Hi x2suresh,

I could solve this via Venn Diagram but in your solution I couldnt understand -2(7)-6 !!

Normally when we have 2 entities like X and Y , we can do it the following way:

X + Y + Neither - Both = Total

In case of >2 entities I am not sure how to go about..only Venn diagram seems to be the way to go...Does anyone know any other shortcut?


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vcbabu
Q: THERE are 30 students. 12 play basket ball . 15 play soccer and 19 play volleyball. if 7 play all 3 games and exactly 6 play 2 games , how many of them play none of 3 games?

0 2 4 6 8

No of students who play games = (15+19+12) - 2(7)-6 = 46-20=26

Ans= 30-26= 4
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venn diagram 01 Feb 2009, 11:20
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Can the formula used be elaborated pls
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venn diagram 01 Feb 2009, 11:41
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vcbabu
Q: THERE are 30 students. 12 play basket ball . 15 play soccer and 19 play volleyball. if 7 play all 3 games and exactly 6 play 2 games , how many of them play none of 3 games?

0 2 4 6 8
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B + S+ V - (a+b+c)-2d+n=30
12+15+19-6-2*3+n=30
n=4
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venn diagram 17 Mar 2009, 17:36
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Can someone demonstrate how to solve this using Venn diagrams ?
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venn diagram Updated on: 18 Mar 2009, 13:00
Originally posted by peterssharp on 18 Mar 2009, 12:57.
Last edited by peterssharp on 18 Mar 2009, 13:00, edited 1 time in total.
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Hi pmenon,

This is how I solved the question using Venn Diagram. I don't know if it is the best aproach, but it worked.



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venn diagram 18 Mar 2009, 12:59
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Peter, thanks a million ! +1 for you !
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venn diagram 18 Mar 2009, 13:02
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Thank you for posting this question. I uploaded a better version.

Peter
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venn diagram 18 Mar 2009, 17:27
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When 3 groups are given(And I hope, GMAT does not go beyond this limit) we have a formula worth remembering -

Grp1 + Grp2 + grp3 - Grp(1&2) - Grp(2&3) - Grp(1&3) + Grp(1&2&3) + None = Total.

Now coming back to the problem, we can solve it as -

12 + 15 + 19 - (a +7) -(b + 7) - (c+7) + 7 + None = 30. (Where a,b,c represent ONLY 2 games count)

It is given that a+b+c = 6.

Substitute (a+b+c) in the main eq, we get 4 as the ans.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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