Question

In a chess tournament where the participants were to play one game with another$,$ two players fell ill having played $3$ games each$.$ If the total number of games played is $84,$ the number of participants at the beginning was

• A. $13$
• B. $14$
• C. $15$
• D. $10$

Answer 1

Answer: (C)

$15$

Let the number of participants be x

If no participant fell ill, then number of games played $={}^n{C}_2$

Each player will play $n-1$ games in the tournament

There $n-1$ games most include $1$ game which should be played by the player with other.

No. of games played $={}^n{C}_2-\left[\right(n-1)+(n-1)-1]+6=84$

$\Rightarrow \frac{(n-1)n}{2}-[n-1+n-1-1]+6=84$

$\Rightarrow \frac{(n-1)n}{2}-2n+3+6=84$

$\Rightarrow \frac{(n-1)n}{2}-2n+9=84$

$\Rightarrow \frac{(n^2-n)}{2}-2n+9=84$

$\Rightarrow n^2-5n-150=0$

$\Rightarrow n^2+10n+15n-150=0$

$\Rightarrow n(n+10)-15(n+10)=0$

$\Rightarrow (n+10)(n-15)=0$

$n=10$ [Not possible]

$n=15$.