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 Post subject: General Paper question
PostPosted: Mon Oct 11, 2010 8:26 am 
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Joined: Fri Mar 26, 2010 7:38 pm
Posts: 105
Would anyone be able to work this out for me?

I have spent hours trying to figure it out and the whole test (comprising of this and other questions) is meant to be completed in an hour. It must be quite simple logic but I guess my grey cells have taken over. Here goes:

"I have lost the results of the local Water Polo league, but luckily I think I can remember enough information to reconstruct the table of results.

Each team played every other team twice.
The final order was alphabetical.
Only two teams won more than half their games, but only one team lost more than half of its games.
No two teams won the same number of matches.
No team was unbeaten and only Broadstairs failed to draw at least one match.
Ashford won the same number of matches as Elham lost.
Dover lost half of its matches.
Canterbury got 8 points.
No two teams ended up with the same number of points.

Given that a 'Win' scores 2 points, a 'draw' 1 and a loss 0, fill in the table below.

Team Played Won Drew Lost Points
Ashford
Broadstairs
Canterbury
Dover
Elham"

(Sorry, can't seem to get the table starting with the headings Team, Played, Won, Drew, Lost, Points to show up as a table layout on this message - hope you can all decipher - there are 6 columns in all)


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PostPosted: Mon Oct 11, 2010 9:05 am 
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Joined: Fri Jan 18, 2008 3:34 pm
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Location: Watford, Herts
I started with trying to work out the figures for the top 2 teams.

All teams played 8 games.

You know the top two both won 5 or more games and lost at least 1 also Broadstairs didn't draw any games but Ashford drew at least 1. From this you can work their figures.

Alison


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PostPosted: Mon Oct 11, 2010 11:03 am 
I think the answer may be:

team won drew lost points
ashford 6 1 1 13
broad 5 0 3 10
cant 3 2 3 8
dover 2 2 4 6
elham 1 1 6 3


Taking Ashford first, only 2 teams won more than half of their games, so that is going to be Ashford and Broadstairs. Ashford, being on top, either won 7 or 6. It can't be 7 because they had to draw at least one and, because no team was unbeaten, it had to lose one, so it must be 6.
Broadstairs must be 5 and, because it had no draws, the lost must be 3.
Canterbury had 8 points. We know it can't have lost more than half its matches because Elham is the only team to do that. It can't have won 4 to produce the 8 points because it has to draw at least one match so it must have won 3 and drawn 2 to make up the eight, leaving 3 lost.
Dover lost 4 matches, so the other 4 has to be spread amoung win and draw. It has to have at least 1 draw but it can't have 3 wins because no two teams have the same number of wins and Canterbury already has 3 Dover must be 2 wins and 2 draws.
Elham has 6 losses (the same number as Ashford). It can either have 1 win and 1 draw or 2 draws, but the latter would leave you with an uneven number of draws which is not possible, so it must have 1 win and 1 draw.

I think this is correct but possibly there is a flaw somewhere in my logic.

I can't get the numbers to line up properly either.


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PostPosted: Mon Oct 11, 2010 12:32 pm 
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I agree with your answer but quibble slightly with the logic - there are three possible results for Canterbury that produce 8 pts, so I think you have to work out the results for D & E first (must have three wins between them because total wins for the league must = total losses). But I may be missing something that would make it simpler?

Mike


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PostPosted: Mon Oct 11, 2010 1:02 pm 
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Joined: Mon Mar 12, 2007 11:49 am
Posts: 450
What is this from? What is a 'General paper'?


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PostPosted: Mon Oct 11, 2010 1:17 pm 
Mike,

I can only work out 2 possible scenarios for 8 points which is 3 wins and 2 draws or 2 wins and 4 draws because it can't be 1 win because all teams won a different number of matches.

But you are right. I forgot to include in my explanation that I counted the number of losses as 17, then worked out that Canterbury , Dover and Elham's wins had also to add up to 17, so, between them, they must have had 6 wins which then would mean C would have to be 3.


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PostPosted: Mon Oct 11, 2010 3:25 pm 
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Also 1 win, 6 drawn and 1 lost. In fact, I think there is another solution:

C win 2, draw 4, lose 2 = 8 pts
D win 3, draw 1, lose 4 = 7 pts
E win 0, draw 2, lose 6 = 2 pts

Can't see anything wrong with that at the moment but since I'm at work it's not exactly got my full attention.

Mike


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PostPosted: Mon Oct 11, 2010 5:38 pm 
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Joined: Fri Mar 26, 2010 7:38 pm
Posts: 105
Thanks all for your help. I will have to sit and digest all your messages - have just come in from work.

This question is from an Indie entrance exam for Sixth Form - From a General Paper question. Did it take any of you less than an hour to figure it out? The paper itself was made up of about 4 questions - with 2 essays - to be completed in an hour.


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PostPosted: Mon Oct 11, 2010 7:04 pm 
Quote:
Also 1 win, 6 drawn and 1 lost


Mike, I don't think you can have this for C because you would then have 3 more wins to distribute between D and E and, because you can't have 2 and 1 for D and E (because that would be repeating the number of wins that C has had), it would have to be 3 for D and 0 for E. This would leave D drawing once and E drawing twice, making altogether 4 draws among AD and E so with whom was C drawing 6 times?
Your alternative seems fine which makes me wonder whether mine is wrong unless there are 2 answers.

Nigs, it took me about 10 minutes to solve. Once I had worked out what A and B had to be, I just played around with the numbers until I found something that worked for all the conditions. That said, now Mike has another solution I am wondering if mine is right.


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PostPosted: Mon Oct 11, 2010 9:35 pm 
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I think there are two solutions - any solution that distributes three wins between D&E, gives C a unique number of wins and gives an even number of draws would be correct (I think!!!), so I think there are probably two correct solutions. It took me longer than fm but certainly not an hour.

Mike


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