Bunuel
The members of a chess club took part in a round robin competition in which each plays every one else once. All members scored the same number of point, 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, for draw ½ point and zero for losing.
A. 9
B. 15
C. 18
D. 27
E. 36
Are You Up For the Challenge: 700 Level Questions: 700 Level QuestionsLet the total members be n. So, the total number of matches that can be played is nC2. And, since each match carries 1 point, the total number of points will also be nC2.
Let the total points of the members other than the 4 juniors be x.
Therefore, nC2-17.5 = (n-4)x
\frac{n(n-1)}{2} - 17.5 = (n-4)x
\frac{n(n-1)-35 }{ (n-4)} = 2x
Since x is a multiple of 0.5, 2x will be an integer.
Hence, the expression \frac{n(n-1) - 35}{(n-4)} will be an integer.
\frac{n^2-4n+3n-12-23}{(n-4)} = 2x
\frac{(n+3)(n-4)-23}{(n-4)} = 2x
(n+3) - \frac{23}{(n-4)} = 2x
Therefore, \frac{23}{(n-4)} must be an integer. So, n must be 27.
Answer will be D.