CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
12,144
loader
C
13,156
loader
D
14,168
loader

Solution

The correct option is B 13,156
let 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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image