Several teams are competing in a basketball tournament, and each team

Question

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

*kudos for all correct solutions

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

*kudos for all correct  solutions

Adityasekar · Answer

Start with the option C. 
if the total number of games = 55 then no 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

chetan2u · Answer

Hi..
Brent, shifted it to PS

chetan2u · Answer

Another way..
Number of matches lost = number of matches won

Number of matches lost 
1) 4 teams lost 5 matches so 4*5=20
2) 5 teams won 3 so lost (n-1-(3))=n-4, thus 5*(n-4)
3) all remaining teams won all matches. So 0
Total matches lost = 20+5*(n-4)=20+5n-20=5n

Number of matches won
1) 4 teams lost 5 matches so they won (n-1-(5))=n-6, thus 4*(n-6)
2) 5 teams won 3 so 5*3=15
3) all remaining teams won all matches. Each team plays n-1 matches a there are n teams. Now remaining teams will be n-(4+5)=n-9.. so (n-9)(n-1)
Total matches won = 4*(n-6)+15+(n-9)(n-1)=4n-24+15+n^2-10n+9=n^2-6n

Thus won=lost....5n=n^2-6n......n^2-11n=0....n(n-11)=0
So n 0 or 11, but there are more than 9 teams so n=11

Matches =11C2=11*10/2=55

C

fskilnik · Answer

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)!

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

bumpbot · Answer

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Several teams are competing in a basketball tournament, and each team

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