ratio question
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ratio question
Dear Maths man,
Please can you explain the following question's answer in detail.
A bag contains One pence and Two pence coins to the value of £5.28. There are Five times as many Two pence coins as One pence coins. How many 1 pence coins are there in the bag?
Please can you explain the following question's answer in detail.
A bag contains One pence and Two pence coins to the value of £5.28. There are Five times as many Two pence coins as One pence coins. How many 1 pence coins are there in the bag?
Re: ratio question
5 x 2p coin + 1 x 1p coin = 11p
i.e. for every 11p value there will be one 1p coin.
Divide £5.28 by 11 gives 48.
Therefore 48 x 1p coins
i.e. for every 11p value there will be one 1p coin.
Divide £5.28 by 11 gives 48.
Therefore 48 x 1p coins
Re: ratio question
Maths lady here!
If there are 5 x the number of 2p then the 2p conis make a multiple of 10p. This means the '8' at the end comes from the 1p coins and these must therefore be a multiple of 8.
Trial and improvement
8 1p 40 2p we know this isn't enought to make £5 plus
16 1p 80 2p ditto
48 1p 240 2p will give £5.28
No need for algebra because you know most of the money is 2p coins and the number has to be a multiple of 40 ...
If there are 5 x the number of 2p then the 2p conis make a multiple of 10p. This means the '8' at the end comes from the 1p coins and these must therefore be a multiple of 8.
Trial and improvement
8 1p 40 2p we know this isn't enought to make £5 plus
16 1p 80 2p ditto
48 1p 240 2p will give £5.28
No need for algebra because you know most of the money is 2p coins and the number has to be a multiple of 40 ...
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aliportico
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Re: ratio question
Doesn't the 8 at the end mean the 1p's must come in an amount that ends in 8? 8, 18, 28, etc? 16 Pennies would give you an amount ending in 6.
If I were explaining that question to someone, I would use Okanagan's method I think, as it seems easier to me to visualise what's going on - imagine one penny, so that must have five 2p's with it. So how much in that pile? And then how many piles would you need?
If I were explaining that question to someone, I would use Okanagan's method I think, as it seems easier to me to visualise what's going on - imagine one penny, so that must have five 2p's with it. So how much in that pile? And then how many piles would you need?
