I have changed the question stem slightly, because in the original wording it would be impossible to each of two teams win all its games (one played the other)!

**Quote:**

Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If each of 5 teams lost exactly 5 games, each of 4 teams won exactly 3 games, and each of the remaining teams lost exactly 1 game, what is the total number of games played during the tournament?

A) 36

B) 45

C) 55

D) 66

E) 78

Very nice problem, Brent. Congrats! (kudos!)

Let´s suppose there are (N+1) teams. Hence each one will play the N other teams, so that

\(? = {{\left( {N + 1} \right) \cdot N} \over 2}\)

Each of 5 teams lost 5 games and won (N-5) games.

Each of 4 teams lost (N-3) games and won 3 games.

Each of the remaining (N+1-9) teams lost 1 game and won N-1 games. (*)

(*) If there were at least one SINGLE team who lost exactly 5 games AND won exactly 3 games, we would have N=8 (hence 9 teams), impossible.

(No team would have lost exactly one game, contradicting the question stem.)

There were no ties, therefore the number of games lost must equal the number of games won, hence:

\(5 \cdot 5 + 4\left( {N - 3} \right) + \left[ {\left( {N + 1} \right) - 9} \right] \cdot 1 = 5\left( {N - 5} \right) + 4 \cdot 3 + \left[ {\left( {N + 1} \right) - 9} \right] \cdot \left( {N - 1} \right)\)

\({N^2} - 9N - 10 = 0\,\,\,\,\mathop \Rightarrow \limits^{{\rm{sum}} = 9\,\,,\,\,\,{\rm{product}} = - 10} \,\,\,\,\,N = 10\,\,\,\,{\rm{or}}\,\,\,\,N = - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,N = 10\)

\(? = {{\left( {N + 1} \right) \cdot N} \over 2} = 55\)

This solution follows the notations and rationale taught in the GMATH method.

Regards,

Fabio.

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