Question

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 than the number of games that the men played with the women. If the number of participants is $k,$ then $\frac{k}{13}$ is equal to

Answer 1

Let the number of men $=n$
Number of women $=2$
$\therefore$ Total participants $=(n+2)$
Now number of games men played by men among themselves $=2{(}^n{C}_2)$
Number of games men played with women $=4n$
According to question, $2{(}^n{C}_2)=4n+66$
$\Rightarrow n(n-1)=4n+66$
$\Rightarrow n^2+5n-66=0$
$\Rightarrow (n-11)(n+6)=0$
$\Rightarrow n=11$ or $n=-6$
$\therefore$ Total number of participants $=11+2=13$
Hence, $\frac{k}{13}=1$