JEE Main 12 Jan 2019 Evening Question 21
Question: There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 11
B) 9
C) 12
D) 7
Show Answer
Answer:
Correct Answer: C
Solution:
- There are m men and 2 women.
So, number of games played by the men between themselves $ {{=}^{m}}C _2\times 2 $
And number of games played between men and women $ {{=}^{m}}C _1^{2}C _1\times 2 $
$ \therefore $ $ ^{m}C _2\times 2{{=}^{m}}C _1^{2}C _1\times 2+84 $
$ \Rightarrow $ $ m(m-1)=m\times 2\times 2+84 $
$ \Rightarrow $ $ m^{2}-5m-84=0 $
$ \Rightarrow $ $ (m-12)(m+7)=0 $
$ \Rightarrow $ $ m=12 $ or $ -7 $ $ \because $ m cant be negative, so m = 12.
