Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss!
Author 
Message 
TAGS:

Hide Tags

GMAT Instructor
Joined: 07 Jul 2003
Posts: 654
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
16 Jul 2003, 04:59
Question Stats:
37% (02:27) correct 63% (02:57) wrong based on 502 sessions
HideShow timer Statistics
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 doubleelimination 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 singleelimination 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
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




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
03 Mar 2014, 22:18
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 doubleelimination 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 singleelimination 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
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >




VP
Joined: 03 Feb 2003
Posts: 1400

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
16 Jul 2003, 06:36
A hard nut...
As for divisional games: a minus is given for losing a game.
the 9team division) lets count minuses; 8 teams can have 2 minuses and one (a champion) can have 1 minus. Total 17 minuses.
the 10team division) lets count minuses; 9 teams can have 2 minuses and one (a champion) can have 1 minus. Total 19 minuses.
the 11team division) lets count minuses; 10 teams can have 2 minuses and one (a champion) can have 1 minus. Total 21 minuses.
the 12team 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.




Senior Manager
Joined: 13 Jun 2013
Posts: 266

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
27 Feb 2014, 14:30
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 doubleelimination 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 singleelimination 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



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1749
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
03 Mar 2014, 21:48
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 doubleelimination 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 singleelimination 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
_________________
Kindly press "+1 Kudos" to appreciate



Intern
Joined: 18 Sep 2013
Posts: 7

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
10 Mar 2014, 11:09
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?
Thanks & regards



Senior Manager
Joined: 13 Jun 2013
Posts: 266

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
10 Mar 2014, 13:34
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. i hope it is clear now.



Director
Joined: 17 Dec 2012
Posts: 626
Location: India

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
10 Mar 2014, 22:32
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 Total = 80+3=83
_________________
Srinivasan Vaidyaraman Sravna Test Prep http://www.sravnatestprep.comHolistic and Systematic Approach



Intern
Joined: 03 Oct 2012
Posts: 26
Concentration: Strategy, General Management

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
22 Jul 2014, 04:17
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?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
22 Jul 2014, 06:25
SachinWordsmith wrote: 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? 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?
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 03 Oct 2012
Posts: 26
Concentration: Strategy, General Management

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
22 Jul 2014, 06:35
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?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
22 Jul 2014, 22:14
SachinWordsmith wrote: 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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 18 May 2011
Posts: 4

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
01 Apr 2015, 21:42
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: 9706
Location: Pune, India

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
01 Apr 2015, 22:23
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 doubleelimination 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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
21 Nov 2016, 08:38
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
Responding to a pm: Quote: Hey Karishma,
why have we taken 17 losses for the first division. cant one of the 9 teams win 2 and rest lose 2 each i.e 16 games and so on for other divisions
Note that every time a team wins, another team loses. So if one of the 9 teams wins 2 matches, someone loses those 2 matches. So 2 losses of the rest of the 8 teams are already accounted for. Say one team loses those two matches. So there can be only 7 teams losing 2 matches each so that we have 14 more losses which gives us a total of 16 games played. Instead, let's make the winning team lose a match too. It won't be kicked out but another match would be played.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Director
Joined: 26 Oct 2016
Posts: 612
Location: United States
Concentration: Marketing, International Business
GPA: 4
WE: Education (Education)

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
04 Jan 2017, 06:33
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 doubleelimination 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 nteam doubleelimination 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 9team 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 10team 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 11team 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 12team 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.
_________________
Thanks & Regards, Anaira Mitch



Senior Manager
Joined: 03 Apr 2013
Posts: 264
Location: India
Concentration: Marketing, Finance
GPA: 3

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
18 Jun 2017, 01:51
Hi VeritasPrepKarishmaLooks very easy after one sees the solution. I actually got lost on making sequences for players in group 1(9 players), such that any player loses the second match as late as possible, so that he plays more games before getting eliminated. In this way, I got completely lost in this question. Please help me understand these types of questions. Please suggest somewhere to read so I understand every bit of these types of problems.
_________________
Spread some love..Like = +1 Kudos



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
19 Jun 2017, 01:52
ShashankDave wrote: Hi VeritasPrepKarishmaLooks very easy after one sees the solution. I actually got lost on making sequences for players in group 1(9 players), such that any player loses the second match as late as possible, so that he plays more games before getting eliminated. In this way, I got completely lost in this question. Please help me understand these types of questions. Please suggest somewhere to read so I understand every bit of these types of problems. What you have to remember is that when you are trying to make a team win, the other team is losing. So somewhere the losses are getting stacked up and those teams are getting eliminated. To you, all the teams are the same. So you should focus only on the number of losses. Each loss represents one game. I would suggest you to not worry too much about this question. It is a tricky 700 level combinations problem. You would see at most one such question and you will need to apply a different logic to each. That is the thing about 700 level questions  you need to "figure out" each one at run time. Your best bet is to practice as many high level questions as possible to make this more intuitive.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2815

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
20 Jun 2017, 18:54
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 doubleelimination 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 singleelimination 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 call the four divisions A, B, C, and D, with 9, 10, 11, and 12 teams, respectively. Now let’s analyze the maximum number of games that can be played in each division. Let’s take division A, for example. It has 9 teams and only 1 is the winning team. So, there must be 8 losing teams, and each of these teams must lose twice since it’s a doubleelimination tournament. The winning team can still lose once (but not twice; otherwise it’s out of the tournament). Therefore, the maximum number of games played in division A is 8 x 2 + 1 = 17. Notice that 8 = 9  1; thus, using the same argument, the maximum number of games played in each remaining division is: Division B: (10  1) x 2 + 1 = 18 + 1 = 19 Division C: (11  1) x 2 + 1 = 20 + 1 = 21 Division D: (12  1) x 2 + 1 = 22 + 1 = 23 Thus, the maximum total number of games played in the 4 divisions before the singleelimination is 17 + 19 + 21 + 23 = 80. In the singleelimination, only 3 games will be played since there will be 2 semifinals and 1 final. (For example, in the semifinal 1, A’s champion vs. B’s champion; in the semifinal 2, C’s champion vs. D’s champion; and in the final, the winner between A and B takes on the winner between C and D to determine the overall league champion.) Thus, the maximum total number of games played to determine the overall league champion is 80 + 3 = 83. Answer: B
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15271
Location: United States (CA)

Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
Show Tags
04 Dec 2017, 14:33
Hi All, While this question is 'wordy', it's a relatively straightforward 'concept' question that doesn't require any difficult math to solve. In a doubleelimination tournament, a team that loses 2 times is eliminated, so every team EXCEPT for the 'champion' will lose twice (and the champion will lose either 0 or 1 times). We're asked to MAXIMIZE the number of games played, so we'll need each champion to lose 1 time. Doubleelimination Division games: 9 teams = (8 teams)(2 losses each) + 1 loss for the champ = 17 games 10 teams = (9 teams)(2 losses each) + 1 loss for the champ = 19 games 11 teams = (10 teams)(2 losses each) + 1 loss for the champ = 21 games 12 teams = (11 teams)(2 losses each) + 1 loss for the champ = 23 games In the singleelimination tournament for the 4 champions, 3 of the teams will lose once and the winner will lose 0 times. That will require 3 more games total. Maximum total games: 17 + 19 + 21 + 23 + 3 = 83 total games Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★




Re: A certain league has four divisions. The respective divisions had 9, 1
[#permalink]
04 Dec 2017, 14:33



Go to page
1 2
Next
[ 21 posts ]



