Help with word problem please

[sallyk57]

In a chess tournament, each competitor has played exactly one game with every other competitor. Five of them lost two games each. All the others won two games a piece. There were no draws. How many competitors were there?

The answer I came up with is 5 competitors. Is that right?

Thanks

 

[daon]

Seems right to me.

The total number of games is n(n-1), and you know that 5 of them lost 2 games each, so that means that (n-5) have won to games each. Also, you know that "all the rest" have won 2 games each, so (n-5) have won 2 games each, and from that you know that (n-(n-5)) have lost 2 games each.

Putting this into an equation:

n(n-1) = 5(2) + (n-5)2 + (n-5)2 + (n-(n-5))2
simplifying,
n^2 - n = 20 + 4(n-5)
n^2 - n = 4n
n^2 -5n = 0

When you factor this, n can only be zero or 5, so 5 is the right answer.

 

[pka]

As described the tournament can be modeled with five players in a directional graph.

However, as you see the number of games is (n)(n−1)/2 or ten.
A plays four other people: wins over B & C but looses to D & E.
In the graph each wins two and each looses two.