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This game season, five divisions are going to play. Out of all the tea

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

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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?

(A) 89
(B) 100
(C) 102
(D) 107
(E) 112


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[Reveal] Spoiler: OA

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

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New post 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 \(=\,11\,+\,17\,+\,23\,+\,25\,+27\,+\,4\,=\,107\)

Answer D

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

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New post 23 Apr 2015, 15:55
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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
Image
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

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

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New post 27 Apr 2015, 03:33
Bunuel wrote:
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?

(A) 89
(B) 100
(C) 102
(D) 107
(E) 112


Kudos for a correct solution.


VERITAS PRGEP OFFICIAL SOLUTION:

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?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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Re: This game season, five divisions are going to play. Out of all the tea   [#permalink] 04 Sep 2017, 03:35
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