Brent@GMATPrepNow 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 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?

A) 36
B) 45
C) 55
D) 66
E) 78

Source: www.gmatprepnow.com

Very nice problem, Brent. Congrats!

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 4 teams lost 5 games and won (N-5) games.
Each of 5 teams lost (N-3) games and won 3 games.
Each of the remaining (N+1-9) teams lost 0 games and won N 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 teams would have won all its games, contradicting the question stem.)

There were no ties, therefore the number of games lost must equal the number of games won, hence:
$$4 \cdot 5 + 5 \cdot \left( {N - 3} \right) + \left( {N + 1 - 9} \right) \cdot 0 = 4\left( {N - 5} \right) + 5 \cdot 3 + \left( {N + 1 - 9} \right) \cdot N$$
$${N^2} - 9N - 10 = 0\,\,\,\,\mathop \Rightarrow \limits^{{\rm{sum}} = 9\,\,,\,\,\,{\rm{product}} = - 10} \,\,\,\,\,N = 10\,\,\,\,{\rm{or}}\,\,\,\,N = - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,N = 10$$
$$? = {{\left( {N + 1} \right) \cdot N} \over 2} = 55$$

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

Regards,
Fabio.

fskilnik@GMATH wrote:

Brent@GMATPrepNow 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 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?

A) 36
B) 45
C) 55
D) 66
E) 78

Source: www.gmatprepnow.com

Very nice problem, Brent. Congrats!

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 4 teams lost 5 games and won (N-5) games.
Each of 5 teams lost (N-3) games and won 3 games.
Each of the remaining (N+1-9) teams lost 0 games and won N 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 teams would have won all its games, contradicting the question stem.)

There were no ties, therefore the number of games lost must equal the number of games won, hence:
$$4 \cdot 5 + 5 \cdot \left( {N - 3} \right) + \left( {N + 1 - 9} \right) \cdot 0 = 4\left( {N - 5} \right) + 5 \cdot 3 + \left( {N + 1 - 9} \right) \cdot N$$
$${N^2} - 9N - 10 = 0\,\,\,\,\mathop \Rightarrow \limits^{{\rm{sum}} = 9\,\,,\,\,\,{\rm{product}} = - 10} \,\,\,\,\,N = 10\,\,\,\,{\rm{or}}\,\,\,\,N = - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,N = 10$$
$$? = {{\left( {N + 1} \right) \cdot N} \over 2} = 55$$

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

Regards,
Fabio.

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 ?

Let's start with the option C.
If the total number of games = 55, then the number of team = n*(n-1)/2 ie., N = 11.

If total team is 11 : 4 teams lost 5 games ie., they won 5=10-5 games since (win = Total - Loss). So, total wins = 4*5 = 20 wins
similarly 5 teams won 3 so total wins = 15 wins and rest of the team (11-5-4) = 2 won all games (10 games) =2*10 = 20 wins
Total = 20+20+15 = 55.

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

Brent@GMATPrepNow wrote: I'll have to rework the numbers and create a question that's actually valid!!

Hi, Brent!

I did that above. The problem is really nice, by the way.
Feel free to substitute the original one with the one I presented.

Kind Regards,
Fabio.