Two women and some men participated in a ches | vedclass

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(C) Let the number of men be $n$ and the number of women be $2$. The total number of participants is $n+2$.Each participant plays $2$ games with every

Vedclass · Accepted Answer

$(11, 13]$

Vedclass · Answer

(C) Let the number of men be $n$ and the number of women be $2$. The total number of participants is $n+2$.Each participant plays $2$ games with every other participant.The number of games played between men themselves:Each man plays $2$ games with every other $(n-1)$ man. The total number of games played by $n$ men is $\frac{n 	imes 2(n-1)}{2} = n(n-1)$.The number of games played between men and women:Each man plays $2$ games with each of the $2$ women,which is $2 	imes 2 = 4$ games per man. For $n$ men,the total number of games played with women is $4n$.According to the problem,the difference between these two is $66$:$n(n-1) - 4n = 66$$n^2 - n - 4n = 66$$n^2 - 5n - 66 = 0$$(n - 11)(n + 6) = 0$Since the number of men cannot be negative,$n = 11$.The value $n = 11$ lies in the interval $(11, 13]$. Thus,the correct option is $C$.

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