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# A certain league has four divisions. The respective

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16 Jul 2003, 03:59
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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
[Reveal] Spoiler: OA

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16 Jul 2003, 05:36
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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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16 Jul 2003, 07:14

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.

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16 Jul 2003, 09:30
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.
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Re: A certain league has four divisions. The respective [#permalink]

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Re: A certain league has four divisions. The respective [#permalink]

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27 Feb 2014, 13:30
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AkamaiBrah wrote:
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

Let's name the teams in group 1 as 1,2,3,4,5,6,7,8,9.

Case 1; team1 played with every other team and won all of its matches.
so total number of matchs =8
case 2: team2 , played with team 3,4,5,6,7,8,9 and won all of its matches.
total number of matches =7
after case 1 and case 2 we have only two teams remaining in the group1 which are team 1 and team 2. Now since question asks us for the maximum no. of matches. Therefore we must include the extra case in which team 2 defeated team 1. Now both team 2 and team 1 have 1 loss each. Now in the final match, we will found out about the eventual winner in group 1.
maximum no. of matches in group 1 are 8+7+1(in which team2 defeated team1) + 1 ( final) =17

Similarly in group 2 we have 9 + 8 +1 +1 =19
group 3 = 10+9+1+1 =21
group 4 = 11+10+1+1=23

After this we will have 4 winner from each group. lets name them as w1,w2,w3,w4
Let's assume w1 won all of its matches from the remaining three teams and eventually emerged as a winner. Therefore total matches among four winners=3

Therefore maximum total no. of matches played are 17+19+21+23+3=83
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Re: A certain league has four divisions. The respective [#permalink]

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03 Mar 2014, 20:48
manpreetsingh86 wrote:
AkamaiBrah wrote:
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

Let's name the teams in group 1 as 1,2,3,4,5,6,7,8,9.

Case 1; team1 played with every other team and won all of its matches.
so total number of matchs =8
case 2: team2 , played with team 3,4,5,6,7,8,9 and won all of its matches.
total number of matches =7
after case 1 and case 2 we have only two teams remaining in the group1 which are team 1 and team 2. Now since question asks us for the maximum no. of matches. Therefore we must include the extra case in which team 2 defeated team 1. Now both team 2 and team 1 have 1 loss each. Now in the final match, we will found out about the eventual winner in group 1.
maximum no. of matches in group 1 are 8+7+1(in which team2 defeated team1) + 1 ( final) =17

Similarly in group 2 we have 9 + 8 +1 +1 =19
group 3 = 10+9+1+1 =21
group 4 = 11+10+1+1=23

After this we will have 4 winner from each group. lets name them as w1,w2,w3,w4
Let's assume w1 won all of its matches from the remaining three teams and eventually emerged as a winner. Therefore total matches among four winners=3

Therefore maximum total no. of matches played are 17+19+21+23+3=83

How long it took to solve this? Its taking a lot of time for me
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03 Mar 2014, 21:18
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PareshGmat wrote:
manpreetsingh86 wrote:
AkamaiBrah wrote:
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

Let's name the teams in group 1 as 1,2,3,4,5,6,7,8,9.

Case 1; team1 played with every other team and won all of its matches.
so total number of matchs =8
case 2: team2 , played with team 3,4,5,6,7,8,9 and won all of its matches.
total number of matches =7
after case 1 and case 2 we have only two teams remaining in the group1 which are team 1 and team 2. Now since question asks us for the maximum no. of matches. Therefore we must include the extra case in which team 2 defeated team 1. Now both team 2 and team 1 have 1 loss each. Now in the final match, we will found out about the eventual winner in group 1.
maximum no. of matches in group 1 are 8+7+1(in which team2 defeated team1) + 1 ( final) =17

Similarly in group 2 we have 9 + 8 +1 +1 =19
group 3 = 10+9+1+1 =21
group 4 = 11+10+1+1=23

After this we will have 4 winner from each group. lets name them as w1,w2,w3,w4
Let's assume w1 won all of its matches from the remaining three teams and eventually emerged as a winner. Therefore total matches among four winners=3

Therefore maximum total no. of matches played are 17+19+21+23+3=83

How long it took to solve this? Its taking a lot of time for me

Actually solving the problem doesn't take very long. Think of it this way:

We need to keep track of losses. Let's focus on those and forget about the wins.
Every time a game is played, someone loses. You can give at most 2 losses to a team since after that it is out of the tournament.
Consider the division which has 9 teams. What happens when 18 games are played? There are 18 losses and each team gets 2 losses (you cant give more than 2 to a team since it gets kicked out after 2 losses) so all are out of the tournament. But we need a winner so we play only 17 games so that the winning team get only 1 loss.

Similarly, the division with 10 teams can have at most 2*10 - 1 = 19 games.
The division with 11 teams can have at most 2*11 - 1 = 21 games.
The division with 12 teams can have at most 2*12 - 1 = 23 games.
This totals up to 80 games (note that the average of 17, 19, 21 and 23 will be 20 so the sum will be 4*20 = 80).

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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Re: A certain league has four divisions. The respective [#permalink]

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22 Jul 2014, 05:35

If I went with the same logic, the answer would be 4 lesser (eliminating the number of games by the group winners lost in the Prelims) = 79 Games.

Can I generalize and use the above presented logic for say, a single elimination ( 1 Loss only) or triple elimination (3 Losses) format too?
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Re: A certain league has four divisions. The respective [#permalink]

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22 Jul 2014, 21:14
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SachinWordsmith wrote:

If I went with the same logic, the answer would be 4 lesser (eliminating the number of games by the group winners lost in the Prelims) = 79 Games.

Can I generalize and use the above presented logic for say, a single elimination ( 1 Loss only) or triple elimination (3 Losses) format too?

Yes, you are correct. 4 fewer games is all you can afford.

Whether you can generalize will depend on the question. If the question remains the same:
In case of a single loss elimination, there will be fixed number of games which will be played so there will be no maximum - minimum. Every game will have a loss and will eliminate exactly one team.
In case of 3 loss elimination format, the qualifying team will not suffer any losses if we want to minimize the number of games and it will suffer 2 losses in case we want to maximize the number of games.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 12 Sep 2014 Posts: 172 Concentration: Strategy, Leadership GMAT 1: 740 Q49 V41 GPA: 3.94 Followers: 0 Kudos [?]: 76 [0], given: 103 Re: A certain league has four divisions. The respective [#permalink] ### Show Tags 04 Oct 2014, 12:08 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. Deceptive question and you're right! I double counted at the end making mine 18, 20 and so far to get 87 total instead of 83! Intern Joined: 18 May 2011 Posts: 4 Followers: 0 Kudos [?]: 0 [0], given: 13 Re: A certain league has four divisions. The respective [#permalink] ### Show Tags 01 Apr 2015, 20:42 Hi, I had a crazy thought. Here our goal is to maximize the matches. In Divison A we have 9 teams let them be a,b,c,d,e,f,g,h,i I let team "a" play with all other teams twice and lose every time. No. of matches=16 Similarly for b=14,c=12 etc. So we get a total of 72 matches in Group A alone. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7187 Location: Pune, India Followers: 2171 Kudos [?]: 14039 [0], given: 222 Re: A certain league has four divisions. The respective [#permalink] ### Show Tags 01 Apr 2015, 21:23 rmselva wrote: Hi, I had a crazy thought. Here our goal is to maximize the matches. In Divison A we have 9 teams let them be a,b,c,d,e,f,g,h,i I let team "a" play with all other teams twice and lose every time. No. of matches=16 Similarly for b=14,c=12 etc. So we get a total of 72 matches in Group A alone. Here is what the question says: "Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games --" When a loses its matches against b and c, it will be eliminated. It will not play any other game. It doesn't need to lose 2 matches against the same team for every team. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A certain league has four divisions. The respective [#permalink]

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01 Apr 2015, 22:20
VeritasPrepKarishma wrote:
Actually solving the problem doesn't take very long. Think of it this way:

We need to keep track of losses. Let's focus on those and forget about the wins.
Every time a game is played, someone loses. You can give at most 2 losses to a team since after that it is out of the tournament.
Consider the division which has 9 teams. What happens when 18 games are played? There are 18 losses and each team gets 2 losses (you cant give more than 2 to a team since it gets kicked out after 2 losses) so all are out of the tournament. But we need a winner so we play only 17 games so that the winning team get only 1 loss.

Similarly, the division with 10 teams can have at most 2*10 - 1 = 19 games.
The division with 11 teams can have at most 2*11 - 1 = 21 games.
The division with 12 teams can have at most 2*12 - 1 = 23 games.
This totals up to 80 games (note that the average of 17, 19, 21 and 23 will be 20 so the sum will be 4*20 = 80).

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

Excellent solution .
Thanks VeritasPrepKarishma.
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Re: A certain league has four divisions. The respective   [#permalink] 01 Apr 2015, 22:20

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