Question

In a chess tournament where the participants were to play one game with one another, two players fell ill having played 6 games each, without playing among themselves. If the total number of games is $117$ , then the number of participants at the beginning was:

• A. 15
• B. 16
• C. 17
• D. 18

Answer 1

Answer: (A)

Let the no. of participants at the beginning was $n$.
Now, we have $6$ games and each participant will play $2$ games.
$\therefore$ Total no. of games played by $2$ persons
$=6\times 2=12$
Since, two players fell ill having played $6$ games each,
without playing among them selves and total no. of games $=117$
$\therefore \frac{n(n-1)}{2}=117-12$
$\Rightarrow n(n-1)=2(105{)}^2=210$
$\Rightarrow n^2-n-210=0$
$\Rightarrow n^2-15n+14n-210=0$
$\Rightarrow n(n-15)+14(n-15)=0$
$\Rightarrow n=-14,15$
But no. of participants can not be-ve $n=15$