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$
$\Rightarrow n ^{2}-15 n +14 n -210=0$
$ \Rightarrow n ( n -15)+14( n -15)=0$
$\Rightarrow n=-14,15$
But no. of participants can not be-ve $n =15$