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 Post subject: Help with questionPosted: Wed Jun 25, 2008 6:41 pm

Joined: Sun Mar 16, 2008 5:15 pm
Posts: 253
Location: Birmingham
Please can anyone advise how to approach this question and is there a formula which can be used to solve it,please?

The cost of 2 full fares and 1 half fare is Â£55. Find the full fare and the half fare.

Thanks for any help.

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 Post subject: Posted: Wed Jun 25, 2008 6:58 pm

Joined: Mon Feb 20, 2006 1:29 pm
Posts: 1805
Location: Berkshire
2 full fares = four 1/2 fare's

add other 1/2 fare and you have 5..... 1/2 fares.

(still with me? )

55/5(half fares) =11

therefore Â£11= one half fare
and Â£22= one full fare

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 Post subject: Posted: Wed Jun 25, 2008 9:27 pm

Joined: Sun Mar 16, 2008 5:15 pm
Posts: 253
Location: Birmingham
Thanks for the reply... appears so simple when explained!

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 Post subject: help required with maths questionPosted: Wed Jul 09, 2008 10:38 am

Joined: Mon May 14, 2007 12:02 pm
Posts: 26
7 boxes each contain 16 coins
7 boxes each contain 14 coins
10 boxes each contain 9 coins

so total no of coins is 300

share these coins equally between 3 people without opening the boxes

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 Post subject: Posted: Wed Jul 09, 2008 12:05 pm

Joined: Wed May 09, 2007 2:09 pm
Posts: 875
Location: Solihull, West Midlands
Not sure if there is a clever way to do this, but first note that 16+9=25 so give one person 4 boxes of 16 and 4 boxes of 9 = 4 X 25 = 100

Then have 3 boxes of 16 left - so possible multiples are 16, 32, 48
All 7 boxes of 14 left, multiples 14, 28, 42, 56, 70, 84

AHA !! = notice that 16 plus 84 = 100

so 2nd person has 1 box of 16 and 6 boxes of 14

3rd person has remaining 2 boxes of 16, 1 box of 14 and 6 boxes of 9, check that works 32+14+54 = 100

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 Post subject: Re: help required with maths questionPosted: Wed Jul 09, 2008 1:00 pm

Joined: Thu Jan 03, 2008 8:26 am
Posts: 1326
Location: Watford, Herts
fusspot99 wrote:
7 boxes each contain 16 coins
7 boxes each contain 14 coins
10 boxes each contain 9 coins

so total no of coins is 300

share these coins equally between 3 people without opening the boxes

100 coins each, so each person must receive an even number of boxes of 9s, with the rest of their 100 made up of 14s and 16s. (They could just as well have said 5 boxes of 18s.) However not all even numbers can be made up from 14s and 16s, and those up to 100 that can can be done in only one way. So consider the different numbers of 9s and how the rest might be made up:

100 - 0x9 = 100 = 6x14 + 16
100 - 2x9 = 82, which can't be done
100 - 4x9 = 64 = 4x16
100 - 6x9 = 46 = 14 + 2x16
100 - 8x9 = 28 = 2x14
100 - 10x9 = 10, which can't be done

Eight 9s are impossible because two 9s can't be done. So the only possibility is to allocate 0, 4 and 6 nines, with the rest allocated as above.

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 Post subject: Posted: Thu Jul 10, 2008 10:13 am

Joined: Mon May 14, 2007 12:02 pm
Posts: 26

It took me a very long time to work it out and even longer to work out how to explain it.

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