Question

The members of a chess club took part in a round robin competition in which each player plays with another 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 : 1/2 for draw point and zero for losing.

Answer 1

Hint: Start with assuming the total number of members to be a variable.Each match will bestow 1 point within the participants of the match because either one team will win (taking home 1 point) or the match will be drawn (each of them will share 1/2 points). Apply the formula of combination to find the total number of matches and use the conditions given in the question to find the possible value of the number of members.

Complete step-by-step answer:
Let us assume that the total members in a club are n.
As we know 2 members are required to play a match.
So, the total number of matches will be ${}^n{C_2}$ because we had to choose 2 members out of n.
So, as we know that total point per match is 1. So, the total number of points for ${}^n{C_2}$ matches will be equal to ${}^n{C_2}$.
Let us assume that the score of all members having similar scores is equal to x.
Now as we are given in the question that the total score of 4 junior members in the match is equal to 17.5
So, the remaining members having similar score is n – 4
So, the total score of n – 4 members will be equal to x(n - 4).
And, the remaining score of all other members (i.e. similar score players) must be equal to ${}^n{C_2}$ - 17.5 = x (n – 4) --------------(1)
Now as we know that according to the combination formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
So, the value of ${}^n{C_2}$ must be equal to ${}^n{C_2} = \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = \dfrac{{n!\left( {n - 1} \right)\left( n \right)}}{{2!\left( {n - 2} \right)!\left( {n - 1} \right)\left( n \right)}} = \dfrac{{n!\left( {n - 1} \right)\left( n \right)}}{{2!\left( n \right)!}} = \dfrac{{\left( {n - 1} \right)\left( n \right)}}{2}$
Now putting the value of ${}^n{C_2}$ in equation 1. We get,
$\dfrac{{\left( {n - 1} \right)\left( n \right)}}{2} - 17.5 = x\left( {n - 4} \right)$
Solving LHS of the above equation and then cross multiplying. We get,

$$\dfrac{{\left( {n - 1} \right)\left( n \right) - 35}}{{\left( {n - 4} \right)}} = 2x \\\ \dfrac{{{n^2} - n - 35}}{{\left( {n - 4} \right)}} = 2x \\\$$

Now making factors of LHS of the above equation. We get,

$$\dfrac{{{n^2} - 4n + 3n - 35}}{{\left( {n - 4} \right)}} = 2x \\\ \dfrac{{n\left( {n - 4} \right) + 3\left( {n - 4} \right) - 23}}{{\left( {n - 4} \right)}} = 2x \\\$$

$\left( {n + 3} \right) - \dfrac{{23}}{{\left( {n - 4} \right)}} = 2x$ --------- (2)
As we know that value of x must integer or integer + $\dfrac{1}{2}$ because for every match either there is 0 point, 1 point or $\dfrac{1}{2}$ point.
So, the value of 2x must always be an integer.
Now as we can see that in the LHS of the equation 2 (n + 3) is already an integer and the RHS of the equation is also an integer. So, $\dfrac{{23}}{{\left( {n - 4} \right)}}$ must also be an integer.
Now for $\dfrac{{23}}{{\left( {n - 4} \right)}}$ to be an integer (n – 4) must be a factor of 23. But as we know that 23 has only two factors and that is 1 and 23.
So, (n – 4) = 1 or (n – 4) = 23
n = 5 or 27
Now ${}^n{C_2}$ must be greater than 17.5 because ${}^n{C_2}$ is the total score but when we put n = 5 then the value of ${}^n{C_2}$ will be 10 which is less than 17.5 (i.e. not possible)
Now when we put n = 27 then the value of ${}^n{C_2}$ becomes $\dfrac{{\left( {27 - 1} \right)\left( {27} \right)}}{2} = 351$ which is greater than 17.5 (i.e. possible)
Hence the correct value of n that is the total number of members must be equal to 27.

Note:- Whenever we come up with this type of problem then first, we have to find the total number of matches that will be played and that will also be equal to the total score of all matches because the score for each match is 1. Now we assume that the number of players having similar scores is x (which will be 1/2 or 1 or 0) and then we can find the total score of these similar plays by subtracting 17.5 from the total scores and on solving that equation we will get that the fraction part of the equation i.e. $\dfrac{{23}}{{\left( {n - 4} \right)}}$ should also be an integer because 2x is an integer. Now on comparing (n – 4) with the factors of 23 we will get the required value of n. This will be the easiest and efficient way to find the solution of the problem.