3 women and a few men participated in a chess tournament. Each player played two matches with each of the other players. If the number of matches that men played among themselves is 78 more than those they played with the women, how many more men than women participated in the tournament?
Correct Answer : 10. Choice D
Let the number of men who participated in the tournament be ‘n’
Each player played two matches against all the other players.
So, the n men would have played 2 * 3 * n = 6n matches against the women.
Each of the n men would have played two matches among themselves.
Each man plays with each of the other (n – 1) players.
So, the number of matches among the men = n(n -1)
We know that the men have played 78 more matches among themselves than against the women.
i.e., n(n -1) = 78 + 6n
or n2 – n – 6n – 78 = 0
or n2 – 7n – 78 = 0
Factorizing, we get (n – 13)(n + 6) = 0
or n = 13 or n = -6.
n cannot be negative. Hence, n = 13.
Number of men = 13. Number of women = 3.
So, there are 10 more men than the number of women in the tournament.