Question

There were 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 the number of games that the men played with the women by $66$. The number of participants are

• A. $6$
• B. $11$
• C. $13$
• D. $10$

Answer 1

The correct option is C $13$

Let number of men be $n$
$\therefore$ No of games between males $=2\left({}^n{C}_2\right)$
Number of games between players of opposite sex $=2(n\times 2)=4n$
$\therefore$ By the given condition:
${2.}^n{C}_2-4n=66$

$\Rightarrow n(n-1)-4n=66$
$\Rightarrow n^2-5n-66=0$

$\Rightarrow n=11$
$(\because n=-6$ is impossible$)$
$\therefore$ number of participants $=11+2=13$