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

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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?

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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

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

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

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26 Feb 2014, 03:04

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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

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
_________________

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
_________________

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

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10 Mar 2014, 10:09

Dear Manpreet

I was applying the same trick but got confused in between.

Since, we allowed 3 matches between team 1 and team 2 in the first group, are you assuming that more than one match can be played between two teams as long as the team does not lose 2 matches in total?

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

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10 Mar 2014, 12:34

ShantnuMathuria wrote:

Dear Manpreet

I was applying the same trick but got confused in between.

Since, we allowed 3 matches between team 1 and team 2 in the first group, are you assuming that more than one match can be played between two teams as long as the team does not lose 2 matches in total?

Thanks & regards

yes, you are right. see the question asks us to find the maximum number of games that could have been played to determine the overall league champion. hence we go for the additional matches.

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

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10 Mar 2014, 21:32

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The intuition is each team is eliminated after 2 losses. It has to lose 2 games and it can lose only 2 games. The winner loses only once or may not lose at all. But since maximum number of games is asked we will assume that the winner loses once.

Each loss corresponds to 1 game played.

For the 4 divisions the number of games played will be 2*8 + 1, 2*9 +1 , 2*10 +1 and 2*11+1= 17,19,21,23 resp= 80 total

Applying the same logic to the finals, the maximum number of games played will be 1*3 + 0=3

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

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22 Jul 2014, 03:17

Hi,

Could someone comment on how they would solve/approach it if the question were : What's the minimum number of games that could be played before a winner is determined?

Could someone comment on how they would solve/approach it if the question were : What's the minimum number of games that could be played before a winner is determined?

Use the same logic as that used for maximum number of games. You know that each team must lose in at least 2 games (and at most 2 games). The only difference will be that the qualifying team will not lose any match to minimize the number of matches.

Now can you come up with the answer?
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Thanks for the Reply Karishma. I follow your posts and your explanations are top notch.

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

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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.

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.
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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

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