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A certain league has four divisions.

This topic has 3 expert replies and 1 member reply

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Vincen Legendary Member
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A certain league has four divisions.

Mon Sep 25, 2017 6:13 pm
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

The OA is B.

Uff. There are so much conditions. I got confused. I need an expert to explain this PS question, please.

GMAT/MBA Expert

Scott@TargetTestPrep GMAT Instructor
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Fri Dec 15, 2017 10:21 am
Vincen 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
Letâ€™s call the four divisions A, B, C, and D, with 9, 10, 11, and 12 teams, respectively. Now letâ€™s analyze the maximum number of games that can be played in each division.

Letâ€™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 itâ€™s a double-elimination tournament. The winning team can still lose once (but not twice; otherwise itâ€™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 single-elimination is 17 + 19 + 21 + 23 = 80.

In the single-elimination, only 3 games will be played since there will be 2 semi-finals and 1 final. (For example, in the semi-final 1, Aâ€™s champion vs. Bâ€™s champion; in the semi-final 2, Câ€™s champion vs. Dâ€™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.

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Matt@VeritasPrep GMAT Instructor
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Tue Sep 26, 2017 4:46 pm
An easy way to think of this is to think of the number of losses there would have to be to eliminate each team. Since only one team loses each game, the number of losses / 2 = the number of eliminations.

We need to eliminate 8, 9, 10, and 11 teams during the divisional rounds, a process that will require at least 16, 18, 20, and 22 games, respectively. Since we want to maximize the number of games played, we want each of division games to lose one game too, adding one to each of our tallies above and leaving us with 17, 19, 21, and 23 games.

We then need three teams to lose ONCE in the playoffs, adding three more losses.

In all, we've got 17 + 19 + 21 + 23 + 3 => 83 games.

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pannalal Junior | Next Rank: 30 Posts
Joined
20 Sep 2017
Posted:
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Tue Sep 26, 2017 5:15 am
Vincen 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

The OA is B.

Uff. There are so much conditions. I got confused. I need an expert to explain this PS question, please.
The maximum number of games in a double-elimination tournament is one less than twice the number of teams participating (e.g., 8 teams - 15 games). The minimum number is two less than twice the number of teams (e.g., 8 teams - 14 games).

Let us calculate maximum number of games for Each Division:
(1) Division with 9 teams = 9*2 - 1 = 17 games
(2) Division with 10 teams = 10*2 - 1 = 19 games
(3) Division with 11 teams = 11*2 - 1 = 21 games
(4) Division with 12 teams = 12*2 - 1 = 23 games

On adding these, we get 80 games. Thus, after maximum of 80 games, we get 4 teams.

Now, three games are sufficient to find winner. Two semi-final and one final game.

So, the total maximum number of games required are 80 + 3 = 83 games.

GMAT/MBA Expert

Rich.C@EMPOWERgmat.com Elite Legendary Member
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Tue Sep 26, 2017 11:25 am
Hi Vincen,

While this question is 'wordy', it's a relatively straight-forward 'concept' question that doesn't require any difficult math to solve.

In a double-elimination 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.

Double-elimination 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 single-elimination 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

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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