  Question

There are $$2$$ women participating in a chess tournament. Every participant played $$2$$ games with the other participants. The number of games that the men played between themselves exceeded by $$66$$ as compared to the number of games that the men played with the women.The number of participants & the total numbers of games played in the tournament are respectively

A
11,130  B
12,144  C
13,156  D
14,168  Solution

The correct option is B 13,156let the number of men be $$m$$.every participant will play 2 games with other participant.For a game within men, we have to select any two out of $$m$$ men. This can be done in $$^mC_2$$ ways.Since, any pair will play two games between them, total number of games played between men is $$2 \times {^mC_2}=m(m-1)$$....(1)One man can play 2 games with any one woman. So he can play $$4$$ games with two women. Since, there are $$m$$ men, number of games played between men and women is $$4 \times m$$....(2)As per the question,$$m(m-1)=4m+66$$$$m^{2}-m-4m-66=0$$$$(m-11)(m+6)=0$$$$m=11$$ or $$m=-6$$m cannot be negative so m is $$11$$hence the total number of participants are $$11+2=13$$ and total number of games 13 players played are $$13(13-1)=156$$Hence option C is correct.Maths

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