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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] New post 03 Jun 2013, 20:17
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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] New post 28 Nov 2014, 11:38
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Re: In the Mundane Goblet competition, 6 teams compete in a [#permalink] 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
Re: In the Mundane Goblet competition, 6 teams compete in a   [#permalink] 18 Jan 2015, 20:41

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In the Mundane Goblet competition, 6 teams compete in a

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