Maths Question

[Aroundtheworldin80moves]

I know the answer. My brother, with his Master's in Maths, has confirmed my alternative answer (well DDs answer!) Is also correct. But I'm intrigued to know what everyone else gets...
It's Yr5 White Rose Maths Equivalent Fractions.

With my algebra loving brain I jumped straight into simultaneous equations and quadratics and got C = 28 or 24.

Then I looked more carefully at your post! It is a Y5 question.

Your DD, using trial and error, has exactly the right approach.

It is more of a ratio question
3 7
6 14
12 28

Did the question ask ‘what could C be’ ? In which case either answer is correct.

I think 28 is perhaps the expected answer if they assume that children will use logic of factors to help. The 14 is the only number on the bottom, and suggests that they must have had to multiply the first denominator by something to get that - ie factors of 14, so 2 or 7. So 7 works, and then it's easy to show that they get 6 for the numerator of the second fraction, which does make 13. Then they can use x4 to work out the final one and get C. Of course it isn't the only solution, but it might be the likely way they want the children to approach it, rather than random trial and error.

3 is the smallest number and it is a prime number and that goes at the top of one column.

It would be nice if the question had asked for more than one possible answer for C. I really wouldn’t want to see either marked wrong.

I also used the quadratic formula to get 24 or 28 but sensible trial and error is how I would expect a year 5 to do it.

I got 28 but 24 seems the far more logical (easier/obvious) answer.
I'm a teacher btw and I hate WR. There are always so many mistakes in their resources. The students are so used to it now.

Ds also had this question earlier this week (or possibly last, it's all blurring in). Without looking at anyone else's answers, he made a=6 and b=7, making c=24 which his teacher marked wrong. I queried it but she never has come back about it.

@AllThatFancyPaintsAsFair

You've written out the question, how does it appear in the original?

Without any other info I think most people would assume A = 6 just because the maths is easier

This is the question as set out on the pdf.

I bloody hate white rose sometimes. “The value for C”.

@Pascha

Ds also had this question earlier this week (or possibly last, it's all blurring in). Without looking at anyone else's answers, he made a=6 and b=7, making c=24 which his teacher marked wrong. I queried it but she never has come back about it.

In the absence of any other information, it seems that the teacher got it wrong. I'm surprised: it's not difficult to verify.

There must be some very confused children out there at the moment if this is the level of debate in adults...

(DD is perfectly happy, as in her mind her answer is correct. She's about average, maybe a bit below average ability in Maths (her times tables let her down, but my observation is her logic is sound)

I got 24 and worked it out by trial and error. Clearly I have the maths ability of a year 5 child
I've just sent it to year 11 DS1 who has a maths brain. Let's see what he says

I got 24 and worked it out by trial and error. Clearly I have the maths ability of a year 5 child.

There isn’t just one way to solve the problem. There’s nothing wrong with using trial and error.

I got 28 first because I figured A and B were 6 or 7, then picked the 7 for A because it goes into 14.
3/7 = 6/14 you get by doubling top and bottom. Then doubling 6/14 gives 12/28 for the last one.

My instinct was to get one fraction from the other rather than simplifying.

C=28

Assuming all integers
Simply: You know that as 3 is a prime number (ie can't be factorised any more) then A must be a factor of 14. So 1, 2, 7 or 14
3/A = B/14
Equivalent fractions
A =1, B=42
A=2 B=21
A=7, B =6
A =14, B = 3

We also know A+B=13, so it must be A=7, B=6

To find C, we again look at equivalent fractions:
3/7=12/C
3x4=12
So 7x4=C
Therefore C=28

But @Witchend it also works if A=6, B=7 and C=24.

@PurpleDaisies

I got 24 and worked it out by trial and error. Clearly I have the maths ability of a year 5 child.

There isn’t just one way to solve the problem. There’s nothing wrong with using trial and error.

Oh I know. It was a tongue in cheek comment. I've always used trial and error for maths. Which was a problem when teachers asked where my working out was. There wasn't any really. Just process of elimination

DS2 (year 9 and used to be so good at maths he could practically do it in his sleep) said c=18. Which makes no sense.

@HappyFlamingo

But *@Witchend* it also works if A=6, B=7 and C=24.

Often year 5’s are used to seeing fractions equivalent to 1/2 so might spot 3/6 = 7/14 = 12/24 first.

DS2 (year 9 and used to be so good at maths he could practically do it in his sleep) said c=18. Which makes no sense.

What was their strategy/reasoning?

Which was a problem when teachers asked where my working out was. There wasn't any really. Just process of elimination.

That is a valid strategy! It’s not always very efficient or elegant but it gets you the marks if it’s clear (assuming you’re not told to use a specific method in the the question).

@HappyFlamingo

I know, but I was trying to explain the maths behind 28.
24 you can spot, but 28 is more of reasoning out.

I'd do it by simultaneous equations to get both answers.

There's a method for getting both answers without relying on trial and error:

Notice that: B = 13 - A.
Substitute for B to get:
3/A = (13 - A)/14
Multiply out: 42 = 13A - A2 (A2 means 'A squared')
Rearrange: A^2 - 13A + 42 = 0
Rewrite this by Factorising to get the equivalent equation:
(A - 7)(A - 6) = 0.
This clearly has precisely two solutions: A = 7 and A = 6.
Solving for C gives the two solutions.

@yossell

There's a method for getting both answers without relying on trial and error:

Notice that: B = 13 - A.
Substitute for B to get:
3/A = (13 - A)/14
Multiply out: 42 = 13A - A2 (A2 means 'A squared')
Rearrange: A^2 - 13A + 42 = 0
Rewrite this by Factorising to get the equivalent equation:
(A - 7)(A - 6) = 0.
This clearly has precisely two solutions: A = 7 and A = 6.
Solving for C gives the two solutions.

Not for year 5.

A is 7, B is 6 and C is 28.