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GMAT Instructor
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A certain league has four divisions. The respective [#permalink]
 16 Jul 2003, 04:59
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A certain league has four divisions. The respective divisions had 9, 10, 11, and 12 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?
(A) 79
(B) 83
(C) 85
(D) 87
(E) 88
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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SVP
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A hard nut...
As for divisional games: a minus is given for losing a game.
the 9-team division) lets count minuses; 8 teams can have 2 minuses and one (a champion) can have 1 minus. Total 17 minuses.
the 10-team division) lets count minuses; 9 teams can have 2 minuses and one (a champion) can have 1 minus. Total 19 minuses.
the 11-team division) lets count minuses; 10 teams can have 2 minuses and one (a champion) can have 1 minus. Total 21 minuses.
the 12-team division) lets count minuses; 11 teams can have 2 minuses and one (a champion) can have 1 minus. Total 23 minuses.
After that, four teams remained. I assume that when 3 teams can have 1 minus and the overall champion has none. Total 3 minuses.
Overall, there can be 83 minuses. Tus, it is B.
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My answer A = 79.
I generalised the solution again...
Consider 4 teams.
Max possible games if a team is eliminated after 2 loses = 3 + 2 = 5
Similarly, if 5 teams are there,
Max possible games = 4 + 3 = 7.
So going forth we can play 15,17,19,21 max possible games for 9,10,11,12 teams respectively.
Adding them up gives 72. Now there are two teams left in each group still. So four more games to give a winner in each team.
So 72+ 4 = 76.
Now of the four teams we can have max 3 games if its a knock out round.
So 76 + 3 = 79.
Please confirm the answer...
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GMAT Instructor
Joined: 07 Jul 2003
Posts: 773
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 5
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stolyar wrote: A hard nut...
As for divisional games: a minus is given for losing a game.
the 9-team division) lets count minuses; 8 teams can have 2 minuses and one (a champion) can have 1 minus. Total 17 minuses.
the 10-team division) lets count minuses; 9 teams can have 2 minuses and one (a champion) can have 1 minus. Total 19 minuses.
the 11-team division) lets count minuses; 10 teams can have 2 minuses and one (a champion) can have 1 minus. Total 21 minuses.
the 12-team division) lets count minuses; 11 teams can have 2 minuses and one (a champion) can have 1 minus. Total 23 minuses.
After that, four teams remained. I assume that when 3 teams can have 1 minus and the overall champion has none. Total 3 minuses.
Overall, there can be 83 minuses. Tus, it is B.
Very nicely done. This is correct and a fine approach to the problem.
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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