Maths help please!

[Land0r]

I have worked this out using trial and error (I think!) so I know the answer (I think!) but how should it be written out to 'show working' please? I know it's algebra but can't remember how to do it!

There are 72 children in the library and classroom altogether. After 12 pupils leave the library and enter the classroom, there are 3 times as many children in the classroom as the library. How many children were there in each room to start with?

This is part of DC2's homework - strong at maths but they haven't done algebra yet! The sheet says to show working not use trial and error (like me!).

I wish I'd been taught that in school. It makes so much more sense to me.

Smartiepants shows the silliness of the sentence construction, by giving the correct answer to the wrong question.

Smartiepants79 She's 9, but youngest in the year, yr5 at private school.

Yes, the bar modelling (in the link) is what she did in the lesson today, but she didn't call it that!

whole piece is 90cm
you want 2 bits
one bit is three times the size of the otther (give or take 2cm!)
so you're dealing with 4 bits - 1 for one piece of ribbon, 3 for the other
90 / 4 = 22.5cm
so piece 2 = 22.5cm + 2cm = 24.5cm
piece 2 - 22.5 x 3 = 67.5 - 2cm
so final answer = 65.5cm

Is what I did - I think! You know when I said I had .5 in my answers. But that's a different answer to others who said 67cm rather than 65.5cm.

So it sounds like bar modelling’s main advantage is that it gives you the wrong answer. Read the question. Now, 24.5 x 3 = 73.5. 73.5-65.5=8. So clearly wrong.

Whether the mistake is converting the question to symbols or in this “bar method” stuff, but whichever, it’s interesting that the bar method is being enthused about by people who have the wrong answer, while algebra is giving the right answer...

Can anyone explain their answer without using algebra??

If there is a problem with the bar modelling then it is that I'm not using it correctly! I've only just started learning how it works. It has been used to great effect across the world to support children with their problem solving.
I am not a mathematician so apologies if I appear slow ReflectentMonatomism but I'm don't understand where your figures in the last post came from?
Can you explain your answer without algebra in a way that a 9 year old might understand?

I have no idea about bar modelling! Had never heard of it until now. I am certainly not a mathematician! Grateful to you all for your help, thank you.

DD1 did a lot of algebra in yr4 at the same school but with a different teacher.

t I'm don't understand where your figures in the last post came

They are the answers you gave. Your answers were 65.5 for the large piece, 24.5 for the smaller. I just multiplied 24.5 by three, as per the question, and showed the difference between the two parts is not 2, as per the question.

Primary maths teaching appears to be full of complex, error-prone methods to desperately avoid simple algebra, when it would be simpler and easier to just use simple algebra if the problems need it. As the problems are pointless, it would be even better to use neither.

To update: DD2 didn't use the bar modelling method! She saw that to get 3 times the amount of something you should divide into 4 equal parts, then do the other bits.

I 'd not heard of bar modelling before but I think you could do a bar modelling-esque calculation which is perhaps a little more visual but it's really just algebra in disguise with a bar drawing instead of an 'x'.

ribbon = |\\\\\\|::::::::::|

piece1 = |\\\\\\| = |::::::::::|::::::::::|::::::|(-2)|

piece2 = |::::::::::|

ribbon = piece1 + piece2 = 90 = |::::::::::|::::::::::|::::::::::|::::::|(-2)| =
|::::::::::|::::::::::|::::::::::|::::::::::| -2

So |::::::::::| = 23 and |\\\\\\| = 67

My ribbon bit only works because of the large difference in the numbers, an assumption about whole number answers, and is probably close to trial and error!

I split the 90cm into 4, and ended up with the 22.5.
Then "knew" the answer was going to be a whole number, so tried it with 22 and 23cm, to come up with 23 and 67 as the solution.

You xant take 2cm off the 22.5, because then the whole thing wont add up to 90 at the end.
Maybe you need to say the 2cm would be split evenly between the 2cm, so is worth 0.5 per quarter piece?
Will think more when it's not quite as early!

The first question is OK for bar modelling.

The second one really, really doesn't lend itself to bar modelling at all.

Trial and error, with some feeling for quantities, might help but ... all in all, DadDadDad's substitution is probably the best way to approach it.

Bar modelling is a step into algebra, via concretes and proportionality. It can be really effective. But that second question - I can't see what the teacher is aiming for there.

Very odd.

Oh. Just seen outwith's post.

I stand corrected!

*But that second question - I can't see what the teacher is aiming at"

I was always puzzled by primary maths problems like this. The only reason they exist is as a contorted way to say solve 4x-2=90. So if you want to argue that algebra shouldn't be in primary (and I probably agree) then why set the problems? They're not useful in and of themselves.