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can this be a GMAT one?

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can this be a GMAT one?

by sanju09 » Fri May 21, 2010 1:24 am
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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by liferocks » Fri May 21, 2010 1:51 am
Not sure if it can be a GMAT one..but it took more that 2 minutes for me

here is what I got,
to maximize the number of matches ,for each division double-elimination tournament every team has to loose one match in first round
taking this in account
Division with 9 team
round---matches--team Left
--1---------9------------9
--2---------4------------5
--3---------2------------3
--4---------1------------2
--5---------1------------1
total 9+4+2+1+1=17

For Division with 10 team first and second round each will have 1 match extra..so total 17+2=19
similarly Division with 11 team will play total 19+2=21
and Division with 12 team will play total 21+2=23
total 17+19+21+23=80
After that among the 4 teams 2+1=3 matches will be played
so total [spoiler]80+3=83[/spoiler]
Ans option B
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by akdayal » Sat May 22, 2010 6:47 am
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
for each division its double elimination
so for example division 1 no. of team is 9
for team to eliminate from division they have to lose twice hence Maximum number of matches = 2*9 -1 (because one will be champion of div and he can lose 1 match) = 17
similarly for div2 no. of team is 10 ==> max. number of matches = 2*10 -1 = 19
for div3 no. of team 11 => max. no. of matches = 2*11 -1 = 21
for div4 no. of team 12 => max. no. of matches = 2*12 - 1 = 23
Second round 4 teams and elimination criteria lose 1 match
Hence max. no. of match in 2nd round = 3
total nu. of matches = 17 + 19 +21 + 23 +3 = 83
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by Anaira Mitch » Wed Jan 04, 2017 5:35 am
There are two different approaches to solving this problem. The first employs a purely algebraic approach, as follows: Let us assume there are n teams in a double-elimination tournament. In order to crown a champion, n - 1 teams must be eliminated, each losing exactly two games. Thus, the minimum number of games played in order to eliminate all but one of the teams is 2(n - 1). At the time when the (n - 1)th team is eliminated, the surviving team (the division champion) either has one loss or no losses, adding at most one more game to the total played. Thus, the maximum number of games that can be played in an n-team double-elimination tournament is 2(n - 1) + 1. There were four divisions with 9, 10, 11, and 12 teams each. The maximum number of games that could have been played in order to determine the four division champions was (2(8) + 1) + (2(9) + 1) + (2(10) + 1) + (2(11) + 1) = 17 + 19 + 21 + 23 = 80.

The 9-team division) lets count minuses; 8 teams can have 2 minuses and one (a champion) can have 1 minus. Total 17 minuses((2(8) + 1)).

The 10-team division) lets count minuses; 9 teams can have 2 minuses and one (a champion) can have 1 minus. Total 19 minuses((2(9) + 1)).

the 11-team division) lets count minuses; 10 teams can have 2 minuses and one (a champion) can have 1 minus. Total 21 minuses((2(10) + 1)).

the 12-team division) lets count minuses; 11 teams can have 2 minuses and one (a champion) can have 1 minus. Total 23 minuses((2(11) + 1)).

Now you have 4 teams. 1 loss gets a team kicked out. If you have 3 games, there are 3 losses and 3 teams are kicked out. You have a final winner!
Hence the total number of games = 80 + 3 = 83.
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