Can someone do this Y4 maths question for me please?

[cantdomaths]

Brain is muddled...

There are 170 people in a hall.
There are 6 times as many girls as boys.
There are 6 more boys than teachers.
How many teachers are there?

I have Maths A level...but have also been drinking wine....

I think this kind of question is good indicator if the child understand maths well or not. Once they learn the algebra, it's easier, but this kind of thinking process is really important for primary children, and it will stretch children's maths ability.

@Arkadia this is really hard! I'm really interested in there being another method that removes trial and error though.

How about this? Brown is what's left after ticket is bought. Ticket money I've split into 3 pots for each boy. Comparing James's bar to one of John's, we can cancel out a brown box and a blue one to show that a brown box is equivalent to 2 't' boxes. Substituting that into James's bar, each 't' box is one fifth of 25, hence the ticket cost £15.

I seem to be making heavy weather of it compared with the harder sounding original problem! But am I along tthe right lines?

Algebra is not allowed ;) this is supposed to be solved by a year 3-4 child, just like your one

Interesting, I have a child going into year 3, never heard of the bar method (but that doesn't mean she hasn't seen it, but I don't think so) could answer the questions but didn't do it with any sort of visual problem, just by what was essentially algebra, but without any formal writing.

I suspect as we've seen before on these maths threads my unothordox approach to these maths problems may have ended up with her.

Use a bar model. 6 sections for girls, one for boys and one for teachers = 8 sections. Divide 170 by 8 = 22 r. 2. So each section = 22 but the teachers' one is smaller (the remainder) . 22x6 = 132 so 132 girls. Boys = 22. 22-6=16 so 16 teachers. To check 132+22+16=170.

My sons only just finished year one.

Ok, here it goes.
The biggest problem (for me) here is to understand how to draw the problem (see figure 1). As I don't know the price of the ticket they bought I am not sure how to draw that box. Hence it is not immediate (to me) that boxes A and B are identical. To see this forget the ticket which is a complete red herring and consider what is left. What I now know is that one has 6 times the amount of money of the other. Note that I do not know any longer how much each of them have, so I cannot draw the boxes to scale (something that was hindering me earlier).
What do I know one, though? I still know the difference between the two amounts (figure 2) which has remained the same, so it is a simple matter of dividing 50 by 5 which gives me 10.
Now I can go back to figure 1 (even if it has been drawn wrong), do 25-10=15 and that is the price of the ticket.
As you see, you can have any number you like, big or small, but the problem is solved in 30 secs once you know the logic behind it.
For example, you can now solve it if one is left with 2/3 of the money of the other.

The takeaways are:

1. the ticket is a smoke screen to make the problem seem more difficult. Instead it is simply: John has some money, Mary has X times that amount. The difference is Y. How much do they each have? Easy as pie.
2. drawing it right is key, but in order to do so you need some background info, i.e. understanding point 1.

Many got the answer right through alternative methods and kudos to you, but I am sure we all agree this method is much more elegant and powerful. Once you know the technique is like black magic to the uninitiated.

I find that bar modelling is ideal for smaller kids because it allows them to abstract the information given in a problem and understand the same at a higher level than before. Going back to the OP, the teacher, I fear, did not have much of an understanding of it. To give such a problem from cold makes no sense. Instead spending a year (or 2 or more) doing them, makes PERFECT sense ;)

Figures ;)

I've been puzzling through your reply since yesterday @Arkadia and I think it's finally clicked. It's using the "b"s to derive the "a" which feels very weird somehow when it doesn't seem obvious that they'd match. Perhaps that leap is trivially obvious to DS and DH, and is why they both instantly said 15 without knowing why.

Thanks so much for taking the time to share this, and for your patience.

This has sparked a lot of conversation in this house about how best to support DS in his maths. I'm quite impressed by his 2s and 7s approach.

Hattie - they are making the teacher group a better number to work with. If you had the same number of boys and teachers, there would be 176 people in the room

Thank you. 😊

@Arkadia

I got DS2 to do the original problem. He did it quite quickly through trial and error to start with, so it clearly wasn't as daunting as I thought.

Then I showed him the bar method. I drew your diagram for him - and then he saw that the calculation was straightforward.

So thank you! I'll be getting him to practice bar diagrams some more over the summer. And I'm planning to email the maths lead at his school to suggest that they use it too.

For the ticket one - once I'd taken the tickets off the two end boxes, it became obvious that since John has six times as much money, the first two boxes represented 5x as much money. Wasn't really sure how to draw that - but figured it didn't really matter, since I knew what calculation to do.

@cornflakegirl, indeed. There is no point in giving a problem which I suspect the teacher himself doesn't understand. There is SO MUCH to be learnt, that it is a shame to let it go to waste.
@IStillDrinkCava, can you find out what your DC meant? I can't work it out.

@Arkadia re DS's 2s and 7s. For example, if the number of boys ended with a 7, the number of girls would end in a 2 (because 6 x 7 = 42), plus number of boys would give a number ending in a 9. Add that to the number of teachers (7-6=1) and you get a total ending in 0. Do that for all final digits 0 to 9, and 2 and 7 are the only final digits that give you a 0 in the units column of the total. Therefore you can reduce the solution space down to only numbers of boys ending in 2 or 7. This sounds long winded but it really wasn't.

Oh, I see... That was indeed different ;) it would have never crossed my mind. :D

Well, if you thought that problem was hard, try this one I have just come across. It has taken me a good few hours to solve it.

Mrs Anderson had twice as many chickens as ducks. She sold 272 chickens and 16 ducks. She then had half as many chickens as ducks. How many chickens did she have in the beginning?

Again, to be solved with the tools available to an Y3/4/5.

It's a very well known question isn't it? And indeed aimed at grade 4 children. If they know the bar model method, surely they can solve it.
I would use algebra myself though.

Irvine, i didn't know that problem at all until the other day (it is in a book we are working on) and I can assure you that even though I am very familiar with the bar modelling I couldn't see the solution (and without the bar modelling you couldn't even tackle it - unless you used algebra).

Think I've done it - is it 352 chickens? Fairly sure my yr4 would be totally stumped though. Does that count as a challenging question or an everyone in the class should be able to do it question? I'd have found algebra much easier!

Very good, @bkwrm ;)
No, I don't think anyone in the school would be able to do it (and that includes the staff ;) ). Even I, despite being accustomed to this kind of problems, couldn't do it, but had to think bout it at length.
I have even sent it to MindYourDecision (a YouTube channel where they do maths problems/puzzles) and they couldn't do it ;)

Once you get the right insight it is fairly easy, but getting the insight... that IS the problem ;)

22 boys, 132 girls and 16 teachers.

No idea on my workings though. Defo wouldn't get extra marks for showing workings!

www.singaporemathplus.com/2010/07/singapore-grade-four-question-without.html

We've got to think like a 7 year old child and use methods that might be more likely to be being taught to them...

So I think they'd be using "chunking" rather than algebra.

I imagine a teacher helping the child to make conclusions thusly:

There are 6 girls to 1 boy. That makes 7. Now we need to find out how many 7's go into 170.

We know (a chunk of) 10 x 7 is 70, so how about 20 x 7? That's 140.

Leaving us with 30 (but that's less than another chunk of 10 x 7) so we need smaller chunks.

So let's work with single numbers instead of 10's. If we try 1 x 7 we get 147 in total. But that's not it, because that's 21 boys and 23 teachers.

So how about another single chunk?

That's (10x7) + (10x7) + (1x7) + (1x7) which is... 154. That leaves us with 16 teachers, and that's 6 teachers less than the 22 boys we've counted!

Honestly when I first heard of chunking I thought it was stupid but it sure beats algebraic equations for the younger student, and I realised I'm using it all the time in real life (like when you count money made up of lots of different counselling for instance).

Nice link, Irvine. We used method 2-1, although I think they are pretty much similar. At least it wasn't just me who found it difficult...
One thing, though: all problems are constructed backwards, otherwise we would hardly ever get nice round figures ;)

It’s not unlike some of the extra challenge problems from White Rose maths’ barvember challenges.

Those are the ones I can never remember the method for answering though and have to play about with a model —before looking up the answer—.

Trial and error got me the chickens one pretty quickly - was sure it had to end in 0 (before the 272 added back on) and it had to be in the high single digit range, so tried it on the high x0's and hit it quickly.

I'm not sure I can articulate why it had to end in 0 - so it might've just been that I got lucky with an incorrect assumption.