Question

The members of a chess club took part in a round robin competition in which each player plays with the other once. All members scored the same number of points, except four juniors whose total score was 17.5. How many members were there in the club? Assume that for each win a player scores 1 point, for a draw 0.5 points and 0 points for a loss.

Answer 1

Hint: In this question the concept of permutations and combinations will be used. We need to apply logic and general algebra to proceed through the problem. We will first find the total number of points in all the games by all the players. Then we will equate them to the sum of the points scored by juniors and seniors. In order to select r unique things from a total number of n things, the number of ways are-
${}_{}^{\text{n}}{\text{C}}_{\text{r}}^{} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!{\text{r}}!}}$

Complete step-by-step answer:
In every game that is played, either one of the players wins and gets 1 point or there is a draw and both of the players get half a point each. So, the number of points awarded per game will be 1. Let the number of players be n, and let the senior members score x points each.

The number of games can be obtained by selecting two members out of the total n number of members. This can be done by
 ${}_{}^{\text{n}}{\text{C}}_2^{} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n}} - 2} \right)!2!}} = \dfrac{{{\text{n}}\left( {{\text{n}} - 1} \right)}}{2}$
The total number of points in each game is 1, so the total number of points obtained will also be the same.

Also, the total number of points by the four junior members are 17.5
The number of senior members are n - 4, and each of them scored x points so-
The total number of points scored by senior members are x(n - 4).

Hence, we can say that-
Total points = points by seniors + points by juniors
$\dfrac{{{\text{n}}\left( {{\text{n}} - 1} \right)}}{2} = \left( {{\text{n}} - 4} \right){\text{x}} + 17.5$
$\dfrac{{{\text{n}}\left( {{\text{n}} - 1} \right)}}{2} - 17.5 = \left( {{\text{n}} - 4} \right){\text{x}}$
${\text{n}}\left( {{\text{n}} - 1} \right) - 35 = 2\left( {{\text{n}} - 4} \right){\text{x}}$
We know that x will always be a multiple of 0.5, because a player can obtain either 0.5 or 1 point in a match. So, we can say that 2x will always be an integer. For example, if x=1.5, then 2x = 3, which is an integer.
$\dfrac{{{\text{n}}\left( {{\text{n}} - 1} \right) - 35}}{{{\text{n}} - 4}} = 2{\text{x}}\left( {integer} \right)$
$\dfrac{{{{\text{n}}^2} - {\text{n}} - 35}}{{{\text{n}} - 4}} = {\text{I}}$
$\dfrac{{{\text{n}}\left( {{\text{n}} - 4} \right) + 3\left( {{\text{n}} - 4} \right) - 23}}{{{\text{n}} - 4}} = {\text{I}}$
$\left( {{\text{n}} + 3} \right) - \dfrac{{23}}{{{\text{n}} - 4}} = {\text{I}}$
We will use the concept that the sum of two integers is always an integer. We can see that (n + 3) is an integer. It is adding to another term which gives an integer as the sum, so the second term should also be an integer. So,
$\dfrac{{23}}{{{\text{n}} - 4}} = {\text{I}}$
For this term to be an integer, then n - 4 should be a factor of 23. So, we can write that-
$n - 4 = 23$ or $n - 4 = 1$
$n = 27$ or $n = 5$

When we substitute n = 5 in the given equation-
$(5 + 3) - 23 = 2x$
$2x = -15$
$x = -7.5$
But the points scored cannot be negative, so the correct option is n = 27.

Hence, the number of members in the club are 27.


Note: There is no direct method to solve this problem. We need to form the equations according to the question, and simplify it using general algebra. One common mistake is that students often forget the formula for combinations and permutations, which should be remembered. One should remember the property that the ratio of two numbers is an integer if and only if the denominator is a factor of the numerator. Here we used the property as- (n - 4) is a factor of 23