Help with question

11 Plus Maths – Preparation and Information

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clarendon
Posts: 253
Joined: Sun Mar 16, 2008 6:15 pm
Location: Birmingham

Help with question

Post by clarendon »

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 £55. Find the full fare and the half fare.

Thanks for any help.
Bewildered
Posts: 1806
Joined: Mon Feb 20, 2006 2:29 pm
Location: Berkshire

Post by Bewildered »

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

:D
clarendon
Posts: 253
Joined: Sun Mar 16, 2008 6:15 pm
Location: Birmingham

Post by clarendon »

Thanks for the reply... appears so simple when explained!
fusspot99
Posts: 26
Joined: Mon May 14, 2007 1:02 pm

help required with maths question

Post by fusspot99 »

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
solimum
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Location: Solihull, West Midlands

Post by solimum »

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
WP
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Location: Watford, Herts

Re: help required with maths question

Post by WP »

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.
fusspot99
Posts: 26
Joined: Mon May 14, 2007 1:02 pm

Post by fusspot99 »

Thank yoo for your help

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

:lol:
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