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In the Mundane Goblet competition, 6 teams compete in a

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Re: MBAmission: The Quest for 700 [#permalink]

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New post 03 Jun 2013, 20:17
cumulonimbus wrote:
Hi Karishma,

Can you please explain why did we add = 5+4+3+2+1 here?


We need to find the total number of points that can be gained by all teams together. For this, we need the total number of matches played since each match can give either (3 + 0) points to the two teams or (1 + 1) points to the two teams.

We say there are 6 teams - A, B, C, D, E and F

A plays a match with all 5 so 5 matches have been played.
Then B plays a match with C, D, E and F so another 4 matches were played (note that B has already played with A above).
Then C played with D, E and F so another 3 matches were played.

Like this, we get that the total number of matches played = 5 + 4 + 3 + 2 + 1 = 15

The 15 matches could have contributed a maximum of 3*15 = 45 points and a minimum of 2*15 = 30 points (since each match would have a winner-loser or it would be a draw).
Required difference = 45 - 30 = 15
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Re: In the Mundane Goblet competition, 6 teams compete in a [#permalink]

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New post 18 Jan 2015, 20:41
when two teams compete there are 2 results :
1) either team win: 3 points to one team and none to other team3
2) Match Drawn : 1 point to each team that is 2 points
so for maximum points maximum wins
and for minimum points all matches should draw.
It's a round robin process with 6 teams
that means each team will play with 5 other teams exactly once
so, Table will look like this
A 5w 0L 0D 15pts.
B 4w 1L 0D 12pts.
C 3w 2L 0D 9pts.
D 2w 3L 0D 6pts.
E 1w 4L 0D 3pts.
F 0w 5L 0D 0pts.

total points 45
for all draws total points will be 30
so difference will be 15

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Re: In the Mundane Goblet competition, 6 teams compete in a [#permalink]

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New post 11 Oct 2016, 14:33
The first step is to see how many games can be played. It helps to draw a grid. So if there are 6 teams the first team cannot play itself so it goes 5 games+4 games... you get 15 games.

Then to maximize it we can have each team win every game they play. Lets call the teams A-F. So A wins all the games it plays (5 games) 3*5=15, B wins all the games it plays 3*4, etc etc 45 total points.

Now to minimize the number of points we can have every game tie. If there are 15 games and all tie then each team gets a point--30 points.

45-30=15. A.

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Re: In the Mundane Goblet competition, 6 teams compete in a [#permalink]

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New post 18 Nov 2017, 04:10
Bunuel wrote:
zisis wrote:
In the Mundane Goblet competition, 6 teams compete in a “round robin” format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition?

(A) 15

(B) 30

(C) 45

(D) 60

(E) 75


There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).

Now, in one game max points (3 points) will be obtained if one team wins and another looses and min points (1+1=2 points) will be obtained if there will be a tie. Hence, maximum points that can be gained by all teams will be 15 games * 3 points=45 and the minimum points that can be gained by all teams will be 15 games * 2 points=30, difference is 45-30=15.

Answer: A.

Hope it helps.


Hi Bunuel

Cant it be 6*5 = 30 games, then one team won and another lost?

so 15 games won so total max points is 45

Total games 30 ties 15 points to each team

Therefore 45-30=15

Pls help

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Re: In the Mundane Goblet competition, 6 teams compete in a [#permalink]

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New post 19 Nov 2017, 02:34
zanaik89 wrote:
Bunuel wrote:
zisis wrote:
In the Mundane Goblet competition, 6 teams compete in a “round robin” format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition?

(A) 15

(B) 30

(C) 45

(D) 60

(E) 75


There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).

Now, in one game max points (3 points) will be obtained if one team wins and another looses and min points (1+1=2 points) will be obtained if there will be a tie. Hence, maximum points that can be gained by all teams will be 15 games * 3 points=45 and the minimum points that can be gained by all teams will be 15 games * 2 points=30, difference is 45-30=15.

Answer: A.

Hope it helps.


Hi Bunuel

Cant it be 6*5 = 30 games, then one team won and another lost?

so 15 games won so total max points is 45

Total games 30 ties 15 points to each team

Therefore 45-30=15

Pls help


6 teams = 15 games. There are 15 different pairs possible from 6 teams. You could check with smaller numbers, for example, list number of games for 3 teams. Is it 3C2 = 3 or 3*2 = 6?
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Re: In the Mundane Goblet competition, 6 teams compete in a   [#permalink] 19 Nov 2017, 02:34

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