How to solve this maths problem?

[crisscrosscranky]

I've worked out an answer based on trial and error but how would an 11year old be expected to work this out 'properly'?

Algerbra.

(C+2) +P = 144 as a starting point.

I'd do my 17 times table...
1710. Is. 170
179. Is. 153. No good.
17*8. Is. 138....remainder. 6p
Each coloured pencil is 2p. So 3 coloured pencils were brought.

The answer I get is 13 plain pencils and 4 coloured pencils.

Divide £1.44 by 17 to give you slightly over 8p a pencil. You can’t have parts of a penny, so use 8p as the cost of 1 plain pencil.

17plain pencils will come to a total of £1.36, which leaves 8p left over.

Divide that cost by the extra cost of a coloured pencil (2p) gives the number of coloured pencils.

It’s a slightly more complicated version of the sort of problems that usually involve 2 legged and 4legged animals or wheels in younger age groups.

At 11 trial and error is likely to be an equally acceptable method though.

178 is 136 though, mummy*, but otherwise that method works.

I think Mummy has done it a good way, and that her method of logic / trial and error would be the way to go.

but 17x8 = 136

13 plain @ 8p ( £1.04) and 4 coloured at 10p (40p)

Yes .. my bad. 144- 136 is 8p not 6p....
8 divided by 2 is colour coloured...

rafals you've assumed the plain is 8p. What if the plain is lower?

Good question Bernadette

Say the plain was 7p. 7x17 = £1.19 so that leaves an odd number

6x17 = £1.02 with 42p left over and that doesn't work either

I did it the other way...

144/17 = 8.4 and some stuff.
17*8 = 136
144 - 136 = 8

So 4 coloured pencils

13 * 8p = 104p 4 * 10 = 40p

144p
.

No you have divide the amount by the number purchased....
We know there are 17 pencils.
Find the remainder.....8p
We know for each extra 2p. A colour pencils was purchased....
So 8p for 17 pencils...
The extra means 4 colour pencils ....

I cheated...
nrich.maths.org/595/solution

Well, then you would have to use trial and error. But dividing 1.44 by 17 will at least have saved you a bit of time ar you’ll know that 8p is the maximum it can cost.

My guess is that most of the time that won’t be an issue though. Although whether there’p could be more than one correct answer might be an interesting discussion.

What's worrying is that you think that no 11 year old would even know how to attempt it. And the impression I get is that it would be surprising to you if he did. It really isn't difficult. And kids that age, (barring special circumstances), certainly have the brainpower to do problems like this.

Teachers have taught him how to approach these types of questions so he should know. Maths shouldn't be seen as something that is "difficult" any more than English should. Would you ask how an 11 year old could be expected to read the front page of a newspaper?
We need to have more faith in our young people - and in the teachers who work with them.

I could be wrong, but I think you’ve misread and/or misunderstood the OP’s question.

She’s not saying that 11 year olds can’t be expected to do it, she’s asking what methods they would have been shown.

That's a beautiful little question. So simple and yet not immediately obvious. Trial and error is fine. 11 year olds might also notice that you can turn it into algebra but not get an immediate solution. But you can notice some patterns, the maximum possible etc. as others have pointed out. You can even draw out a table of possible solutions and see which actually works.

A great introduction to Diophantine equations. Bravo to their teacher.

Thanks all.

@CellularBlanket I'm not sure what you're on about!

Ok Sorry OP, I wasn't clear and I obviously misunderstood your post.

I was making a point about the general attitude that I regularly come across that underestimates our children. It manifests in helicopter parenting, complaints to teachers and even employers(!) and a belief that everything is too difficult unless it is spoonfed. A view that our kids should not fail - which does not help them in the long run.

I am often surprised at what kids can do if you let them. (But clearly wrong thread -