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# venn diagram

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Manager
Joined: 04 Sep 2006
Posts: 113

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29 Jan 2009, 11:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York

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29 Jan 2009, 11:34
vcbabu wrote:
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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Director
Joined: 01 Apr 2008
Posts: 848
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

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01 Feb 2009, 07:30
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?

x2suresh wrote:
vcbabu wrote:
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
Manager
Joined: 04 Sep 2006
Posts: 113

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01 Feb 2009, 11:20
Can the formula used be elaborated pls
Director
Joined: 29 Aug 2005
Posts: 836

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01 Feb 2009, 11:41
vcbabu wrote:
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

B + S+ V - (a+b+c)-2d+n=30
12+15+19-6-2*3+n=30
n=4
SVP
Joined: 28 Dec 2005
Posts: 1500

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17 Mar 2009, 17:36
Can someone demonstrate how to solve this using Venn diagrams ?
Intern
Joined: 05 Mar 2009
Posts: 2

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18 Mar 2009, 12:57
1
KUDOS
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.

Peter

Last edited by peterssharp on 18 Mar 2009, 13:00, edited 1 time in total.
SVP
Joined: 28 Dec 2005
Posts: 1500

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18 Mar 2009, 12:59
Peter, thanks a million ! +1 for you !
Intern
Joined: 05 Mar 2009
Posts: 2

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18 Mar 2009, 13:02
Thank you for posting this question. I uploaded a better version.

Peter
Senior Manager
Joined: 30 Nov 2008
Posts: 482
Schools: Fuqua

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18 Mar 2009, 17:27
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.
Re: venn diagram   [#permalink] 18 Mar 2009, 17:27
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# venn diagram

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