Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

47% (02:30) correct
53% (03:27) wrong based on 56 sessions

HideShow timer Statistics

This game season, five divisions are going to play. Out of all the teams in each division, 6, 9, 12, 13 and 14 teams have qualified from the respective divisions. Each division will hold its own tournament – where a team is eliminated from the tournament upon losing two games – in order to determine its champion. The five division champions will then play in a knock-off tournament – a team is eliminated as soon as it loses a game – in order to determine the overall 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?

Re: This game season, five divisions are going to play. Out of all the tea [#permalink]

Show Tags

22 Apr 2015, 19:01

# of games in division I: Total teams qualified \(N\,=\,6\) number of games to determine top 2 teams \(=\,(total\,teams\,-2)\,*\,2_{losses}\,=\,8\,\); each team has to lose \(2\) games number of games to pick winner among top two teams \(=\,two\,losses\,+\,one\,win\,=\,3\,\); each team lose one game & a win so total games \(=\,(N-2)*2\,+\,3\,=\,(6-2)*2\,+\,3\,=\,11\)

applying the above method Division II: Total games \(=\,(N-2)*2\,+\,3\,=\,(9-2)*2\,+\,3\,=\,17\) Division III: Total games \(=\,(N-2)*2\,+\,3\,=\,(12-2)*2\,+\,3\,=\,23\) Division IV: Total games \(=\,(N-2)*2\,+\,3\,=\,(13-2)*2\,+\,3\,=\,25\) Division V: Total games \(=\,(N-2)*2\,+\,3\,=\,(14-2)*2\,+\,3\,=\,27\)

Total \(=\,11\,+\,17\,+\,23\,+\,25\,+27\,=\,103\)

League Championship: # of teams qualified \(=\,5\) # of games \(=\,4\); knock-off system

Maximum Games \(=\,103\,+\,4\,=\,107\)

or

Division I: 6 * 2 losses but one [(6*2) - 1] will give the winner; total = \([(6*2) - 1] = 11\) Division II: 9 * 2 losses but one [(9*2) - 1] will give the winner; total = \([(9*2) - 1] = 17\) Division III: 12 * 2 losses but one [(12*2) - 1] will give the winner; total = \([(12*2) - 1] = 23\) Division IV: 13 * 2 losses but one [(13*2) - 1] will give the winner; total = \([(13*2) - 1] = 25\) Division V: 14 * 2 losses but one [(14*2) - 1] will give the winner; total = \([(14*2) - 1] = 27\)

League Championship: \(4\) losses for qualified five teams will determine the winner

Maximum Games \(=\,11\,+\,17\,+\,23\,+\,25\,+27\,+\,4\,=\,107\)

Re: This game season, five divisions are going to play. Out of all the tea [#permalink]

Show Tags

23 Apr 2015, 15:55

1

This post received KUDOS

Since we aren't told about how the games are played, lets just imagine that every division's group represents a linked N-angle polygon. When you go by the edges of it, like on the picture below, you can easily determine that for N teams you have to play max 2*N - 1 games to determine a winner and I honestly can't think of an approach to play more games (although you can play 1 less game if you connect all non-adjacent vertexes and call them "games", this way you will get 2*N - 2 games, but its irrelevant I guess) Numbers are games, vertexes are teams

For the last part you can also use that polygon approach which provided 1 loss is elimination easily gives us answer N - 1 where N - amount of teams (4 as a result). Resulting answer ends up being 2(6 + 9 + 12 + 13 + 14) - 5 + 4 = 107 which corresponds to answer D

This game season, five divisions are going to play. Out of all the teams in each division, 6, 9, 12, 13 and 14 teams have qualified from the respective divisions. Each division will hold its own tournament – where a team is eliminated from the tournament upon losing two games – in order to determine its champion. The five division champions will then play in a knock-off tournament – a team is eliminated as soon as it loses a game – in order to determine the overall 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?

Is it a max-min problem? Perhaps, but which guiding principle of max-min will we use to solve this problem? First think on your own how you will solve this problem.

Will you focus on the method or the result i.e. will you worry about who plays against whom or just focus on each result which gives one loss and one win? If you don’t worry about the method and just focus on the result, you can use a concept of mixtures here. In mixture questions, we focus on one component and how it changes. Here, we need to keep track of losses. Let’s focus on those and forget about the wins. As given, there were no ties so every loss has a win on the other side.

Every time a game is played, someone loses. We can give at most 2 losses to a team since after that it is out of the tournament. Don’t worry about against who it plays those two games. Whenever a team loses 2 games, it is out. The team could have won many games but we are not counting the wins and hence, are not concerned about its wins. As discussed, we are counting the losses so each win of that team will be counted on the other side i.e. as a loss for the other team.

Consider the division which has 6 teams – what happens when 12 games are played? There are 12 losses and each team gets 2 losses (we can’t give more than 2 to a team since the team gets kicked out after 2 losses), so all are out of the tournament. But we need a winner so we play only 11 games so that the winning team gets only 1 loss. We want to maximize the losses (and hence the number of games), therefore the winning team must be given a loss too.

So maximum number of games that can be played by the district in its own tournament = 2*6 – 1 = 11

Similarly, the division with 9 teams can play at most 2*9 – 1 = 17 games.

The division with 12 teams can play at most 2*12 – 1 = 23 games.

The division with 13 teams can play at most 2*13 – 1 = 25 games.

The division with 14 teams can play at most 2*14 – 1 = 27 games.

This totals up to 11 + 17 + 23 + 25 + 27 = 103 games

Now we come to the games between the district champions.

We have 5 teams. 1 loss gets a team kicked out. If the teams play 4 games, there are 4 losses and 4 teams get kicked out. We have a final winner!

Hence the total number of games = 103 + 4 = 107

There are a lot of variations you can consider for this question.

Say, if we need to minimize the number of games, how many total games would have been played?

Notice that the only games you can avoid are the ones in which the 5 district champions lost. You do still need 2 losses for each team to get the district champion and one loss each for four district champions to get the winner. Hence, at least 107 – 5 = 102 games need to be played.

Look at it in another way:

To kick out a team, it needs to have 2 losses so if the district had 6 teams, there would be 5*2 = 10 games played.

Similarly, the division with 9 teams will play at least 2*8 = 16 games.

The division with 12 teams will play at least 2*11 = 22 games.

The division with 13 teams will play at least 2*12 = 24 games.

The division with 14 teams will play at least 2*13 = 26 games.

This totals up to 10 + 16 + 22 + 24 + 26 = 98 games

Now we have 5 champions and they will need to play at least 4 games to pick a winner.

Therefore, at least 98 + 4 = 102 games need to be played.

You can try other similar variations – what happens when a team is kicked out after it loses 3 games instead of 2? What happens if you don’t have the knock-off tournament and instead need each district champion to lose 2 games to get knocked out?
_________________

Re: This game season, five divisions are going to play. Out of all the tea [#permalink]

Show Tags

04 Sep 2017, 03:35

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________