Question

The members of a chess club took part in a round robin competition in which each player plays with other once. All members scored the same number of points, except four juniors whose total score were $17.5$. How many members were there in the club? Assume that for each win a player scores $1$ point : $1/2$ for draw point and zero for losing.

Answer 1

Given that each player plays against each other. So, total no. of matches $={}^n{C}_2.$

Each match will result in $1$ point ( with draw or otherwise).

Total points $={(}^n{C}_2)\times 1=17.5+(n-4)x$

Where $x$ is the point achieved by all players except $4$ juniors.

$\Rightarrow \frac{n(n-1)}{2}=17.5+(n-4)x$

$\Rightarrow \frac{n(n-1)-35}{n-4}=2x$

WKT $x$ is always a multiple of $0.5$

$(x$ can be $0.5,1,\mathrm{1.5.....}$ soon $)$

$2x$ is an integer.

$\frac{n(n-1)-35}{n-4}$ must be an integer

$\Rightarrow \frac{(n+3)(n-4)-23}{n-4}=2x$

$=(n+3)-\frac{23}{n-4}$

$\frac{23}{n-4}$ must be integer

$\Rightarrow n-4=23$

$\Rightarrow n=27.$

Hence, the answer is $27.$