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 Post subject: coin problemPosted: Wed Aug 28, 2013 10:05 pm

Joined: Wed Aug 28, 2013 9:53 pm
Posts: 3
Hi

Please can anyone help me with this problem. i think it is a ratio problem which for the life of me i cannot work out.

A cash box contains some coins to the value of £5.25.
There are twice as many 5p coins as 2p coins, and twice as many 2p coins as 1p coins.

This means there are:
1. how many 5p coins
2. how many 2p coins
3. how many 1p coins

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 Post subject: Re: MathsPosted: Wed Aug 28, 2013 10:17 pm

Joined: Fri Oct 12, 2007 12:42 pm
Posts: 3826
Think how much money is made from one set of the coins.

1p + 4p (2, 2p - 2 x no. of 1p) + 20p (4, 5p - 2 x no. of 2p coins) = 25p

How many 25p in £5.25? 21

So 21 1p coins is 21p
42 2p coins is 84p and
84 5p coins is 420p

Check
21 + 84 + 420 = 525

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 Post subject: Re: MathsPosted: Wed Aug 28, 2013 10:30 pm

Joined: Wed Aug 28, 2013 9:53 pm
Posts: 3
Thanks, however i am trying to teach this to a 10 year old and really need to break it down to him. Please if possible could you break it down further for me as i am still very confused. I am really useless at maths. sorry

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 Post subject: Re: MathsPosted: Wed Aug 28, 2013 10:45 pm

Joined: Fri Oct 12, 2007 12:42 pm
Posts: 3826
I taught ratio by drawing beads.

GRR

GRRBBBB
1+2+2+5+5+5+5

The cost of these beads is: 1p + 4p + 20p = 25p

How many sets of GRRBBBB can you buy for £5.25. The answer is 21 (525/25)
21 G
42 R and
84 B.

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 Post subject: Re: MathsPosted: Wed Aug 28, 2013 11:13 pm

Joined: Sat Jan 12, 2013 10:02 pm
Posts: 88
Hi

Please can anyone help me with this problem. i think it is a ratio problem which for the life of me i cannot work out.

A cash box contains some coins to the value of £5.25.
There are twice as many 5p coins as 2p coins, and twice as many 2p coins as 1p coins.

This means there are:
1. how many 5p coins
2. how many 2p coins
3. how many 1p coins

*If there are two 5p coins then there is one 2p coin (the ratio is 2:1)
*If there are two 2p coins then there is one 1p coin (the ratio is 2:1)

Which can also be rewritten as:

*If there are four 5p coins, then there are two 2p coins and one 1p coin (the ratio is 4:2:1, we are taking all the coins into account to form a ratio)

*If we write this in the form of an equation it will be like this:

4(5p) + 2(2p) + 1(1p) = 20p + 4p + 1p = 25p

*Now that you know that this gives you 25p you have to do:

5.25/0.25 = 21

*This means the following equation occurs 21 times:

4(5p) + 2(2p) + 1(1p) = 20p + 4p + 1p = 25p

*Therefore there are:

21 x 4(5p) = 84 x 5p coins
21 x 2(2p) = 42 x 2p coins
21 x 1(1p) = 21 x 1p coins

(84 x 5) + (42 x 2) + (21 x 2) = 525p

Last edited by Rainbow Petals on Thu Aug 29, 2013 10:28 am, edited 1 time in total.

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 Post subject: Re: MathsPosted: Wed Aug 28, 2013 11:29 pm

Joined: Wed Aug 28, 2013 9:53 pm
Posts: 3
Thanks so much, got it. Much appreciated.

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