regor60 wrote:
I followed the same approach, but the solution implies 2 teams won all of their games. I must be missing something... didn't they play each other ?
You are right, regor60. The unique potential solution is not viable.
I could fix the question stem with small modifications:
fskilnik@GMATH wrote:
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
Adapting my solution accordingly (same arguments), the equation we get is the following:
$$5 \cdot 5 + 4\left( {N - 3} \right) + \left[ {\left( {N + 1} \right) - 9} \right] \cdot 1 = 5\left( {N - 5} \right) + 4 \cdot 3 + \left[ {\left( {N + 1} \right) - 9} \right] \cdot \left( {N - 1} \right)$$
$${N^2} - 9N - 10 = 0\,\,\,\, \Rightarrow \,\,\,\,N = - 1\,\,{\rm{or}}\,\,N = 10\,\, \ldots $$
The answer is still (C) but now the 11 teams (playing 10 games each) are in a viable situation (*).
Regards,
Fabio.
(*) P.S.: this may be proved explicitly. For instance:
Team 1 - lost to 7, 8, 9, 10, 11
Team 2 - lost to 1, 8, 9, 10, 11
Team 3 - lost to 1, 2, 9, 10, 11
Team 4 - lost to 1, 2, 3, 10, 11
Team 5 - lost to 1, 2, 3, 4, 9, 10, 11
Team 6 - lost to 1, 2, 3, 4, 5, 10, 11
Team 7 - lost to 2, 3, 4, 5, 6, 10, 11
Team 8 - lost to 3, 4, 5, 6, 7, 10, 11
Team 9 - lost to 4, 6, 7, 8, 10
Team 10 - lost to 11
Team 11 - lost to 9
Conclusion:
Teams 1, 2, 3, 4 and 9 each lost 5 games and won 5 games
Teams 5, 6, 7, 8 each lost 7 games and won 3 games
Teams 10 and 11 each lost 1 game and won 9 games
