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by srcc25anu » Wed Mar 09, 2011 4:46 pm
There are two women participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves proved to exceed by 66 the number of games that the men played with the women.The number of participants is

A. 6
B. 11
C. 13
D. 9
E. 17

Source: gmatmaths.com
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by Night reader » Wed Mar 09, 2011 5:35 pm
the number of men playing between each other can be defined as m(m-1) --> check point 5 by 5; 2 women played with all men are less by 66 --> 4m+66; m(m-1)=4m+66, m^2-5m-66=0, only +ve integers count m=(5+17)/2=11. 11 men + 2 women makes 13 participants.

choice C.
srcc25anu wrote:There are two women participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves proved to exceed by 66 the number of games that the men played with the women.The number of participants is

A. 6
B. 11
C. 13
D. 9
E. 17

Source: gmatmaths.com
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by anshumishra » Wed Mar 09, 2011 5:41 pm
srcc25anu wrote:There are two women participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves proved to exceed by 66 the number of games that the men played with the women.The number of participants is

A. 6
B. 11
C. 13
D. 9
E. 17

Source: gmatmaths.com
n-> men
2-> women

# of games among men = 2*nC2 = n(n-1)
# of games among men and women = 2*(2n) = 4n

So, n(n-1) - 4n = 66
=> n^2-5n - 66 = 0
=> (n-11)(n+6) = 0
=> n = 11

Total no. of participants = no. of men + no. of women = n+2 = 13 C
Thanks
Anshu

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by gmat7202011 » Fri Mar 11, 2011 6:32 am
Hi Anshu,

Can you please explain a little more on how you arrived at

# of games among men = 2*nC2 = n(n-1)
# of games among men and women = 2*(2n) = 4n

Also, i am starting with combinations, what can be a good source for getting started from the basics.

Thank You for your help,
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by GMATGuruNY » Fri Mar 11, 2011 7:10 am
srcc25anu wrote:There are two women participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves proved to exceed by 66 the number of games that the men played with the women.The number of participants is

A. 6
B. 11
C. 13
D. 9
E. 17

Source: gmatmaths.com
Each game is played by a pair of participants. Each pair must play twice.
We can plug in the answer choices, which represent the total number of participants.

Answer choice C: 13 total participants
13 total = 11 men + 2 women
Number of pairs that can be formed from 11 men = 11C2 = 55.
Since each pair plays twice, number of games played among only the men = 2*55 = 110.
Number of possible pairs that include 1 man and 1 woman = 11*2 = 22.
Since each pair plays twice, number of games played between the men and the women = 2*22 = 44.
110-44 = 66. Success!

The correct answer is C.
Last edited by GMATGuruNY on Fri Mar 11, 2011 7:37 am, edited 1 time in total.
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by anshumishra » Fri Mar 11, 2011 7:26 am
gmat7202011 wrote:Hi Anshu,

Can you please explain a little more on how you arrived at

# of games among men = 2*nC2 = n(n-1)
# of games among men and women = 2*(2n) = 4n

Also, i am starting with combinations, what can be a good source for getting started from the basics.

Thank You for your help,
Sure.
There are "n" men. To play a game among men, we need to find 2 men. So, no. of ways to find 2 men out of n = nC2.
Hence, there could be nC2 combination (team/participant) of the match. Also, given that each of them play 2 matches among themselves, so it is 2*nC2.

Similarly : for matches among men and women. (Select one man out of n men)*(select one women out of 2) = n*2C1 = 2n
Again, since they have to play 2 matches among themselves, no. of matches among women and men = 2*2n

Hope that helps.
I guess, MGMAT book should give you a good start. After that, practice and try to learn different ways to solve the same problem.
All the best !
Thanks
Anshu

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