How many members play exactly two games

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gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

by gmattesttaker2 » Sat Nov 02, 2013 7:27 pm

Hello,

Can you please assist with the following:

At a certain 65-member gaming club 28 of the members play backgammon, 23 play
chess, and 24 play bridge. If 12 members do not play any game and the number of
members that play the three games is twothirds that of the members that play nothing,
how many members play exactly two games?

OA: 6

Thanks - Sri

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Source: Beat The GMAT — Problem Solving | Sat Nov 02, 2013 7:27 pm

gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

by gmattesttaker2 » Sat Nov 02, 2013 8:04 pm

Hello,

I tried to solve as follows:

Number of players that do not play any game = 12 (Given)

Number of players that play the 3 games = 2/3(12) (Given)
=> Number of members that play the three games = 8

From the diagram,

a + b + d + 8 = 28 => a + b + d = 20 - Eq. 1
a + c + e + 8 = 23 => a + c + e = 15 - Eq. 2
b + c + f + 8 = 24 => b + c + f = 16 - Eq. 3

Also,

a + b + c + d + e + f + 8 = 53 => a + b + c + d + e + f = 45 - Eq. 4

Adding Eqs. 1, 2 and 3:

2a + 2b + 2c + d + e + f = 51
=> a + b + c + d + e + f + (a + b + c) = 51
=> 45 + ( a + b + c ) = 51 (From Eq. 4)
=> a + b + c = 6

Thanks,
Sri

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Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Sat Nov 02, 2013 9:49 pm

gmattesttaker2 wrote:Hello,

Can you please assist with the following:

At a certain 65-member gaming club 28 of the members play backgammon, 23 play
chess, and 24 play bridge. If 12 members do not play any game and the number of
members that play the three games is twothirds that of the members that play nothing,
how many members play exactly two games?

OA: 6

Thanks - Sri

Hi Gmattesttaker2,
There is a formula to find the EXACTLY 2- group Overlaps and here it is,

Total = A+B+C -(Sum of EXACTLY 2-group overlaps) -2*(All three) + Neither

From question we can get to know
A = 28
B = 23
C = 24
Neither = 12
All three = 8(2/3*12)

and let X be Sum of EXACTLY 2-group overlaps
so,
65 =28+23+24 -(X) -2*8 +12
Solve for X you will get X = 6

Note: Above mentioned formula to be used only if they mentioned as EXACTLY 2-group
Else use
Total = A+B+C -(Sum of 2 group overlaps) +(all three) +Neither

Hope it is clear to you.

Regards,
Uva.

Known is a drop Unknown is an Ocean

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pareekbharat86: Master | Next Rank: 500 Posts; Posts: 168; Joined: Thu Nov 01, 2012 7:43 pm; Thanked: 1 times

by pareekbharat86 » Mon Nov 04, 2013 1:44 am

Use the ABC-2-23 rule here.

Total No. of members= n(a)+n(b)+N(c)- n(members playing 2 games)- 2*n(members playing 3 games)+n(members playing no games)

Let, members playing 2 games shall be x

Members playing no games= 12

Members playing 3 games = 2/3*12= 8 (given)

Putting all the values in the above-mentioned equation,

65= 23+24+28-x-2*8+12

x=6.

Thanks,
Bharat.

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