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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