Question

In a chess tournment, where the participants were to play one game with another. Two chess players fell ill, having played $3$ games each but not with each other. If the total number of games played is $84$, the number of participants at the begining was

• A. $15$
• B. $16$
• C. $20$
• D. $21$

Answer 1

Answer: (A)

$15$

Suppose the two players did not play at all so that the
remaining ( n-2) players played $(\frac{n-2}{2})$ matches
Since these two players 3 matches each
Hence , the total number of matches is
$(\frac{n-2}{2})+3+3=84$
$\frac{(n-1)(n-2)}{1.2}=78$
$n^2-5n+6=156$
$n^2-5n-150=0$
On solving this , we get
n = 15
Hence , the number of participants at the beginning was
= 15