Part whole division – why?

I’ve been thinking a lot about division recently. I wrote this here a short while ago about dividing by fractions, then I was sent a document by Andrew Harris (2001) entitled “Multiplication and Division”, which I was asked to read as part of a series if CPD sessions from the local maths hub, then a number of different people have asked me questions about division recently too. I think probably for most this is due to helping their own kids with maths and meeting methods and structures that they aren’t familiar with, as they weren’t taught when they were at school themselves.

The main thing from friends that keeps popping up is using part whole models for division. And funnily enough it is one of the structures I was considering after reading the Harris document and looking at the distributive law and what higher level topics this underpins in later maths.

So what is it?

Using the part whole method for division is where you split a number into 2 or more parts before dividing then add your answers back at the end. For instance, if you want to divide 486 by 6 you can split it into 480 and 6. The benefit of choosing these numbers is that 48 is in the six times table. So you can see that 48÷6=8 so 480÷6=80, then you have 6÷6=1, add them together and you get 486÷6=81.

This structure, or method, is a very common mental strategy used by lots of people when dividing numbers in their head. Lots of those people will never have heard the term “part whole model” and will not have seen it laid out in a pictorial manner as students today will, but they will use that structure nonetheless. I myself was using it as a mental strategy a long time before I’d heard anyone refer to a part whole model or seen the visual representations.

What we are doing when we do this is using the distributive law of multiplication and division to break our problem into chunks that are easier to manage.

One of the questions I was asked was “is there a rule to how you split it up?” The person who asked me was wondering if you always split it up into hundreds, tens, ones etc or if you could do any. I explained that it didn’t matter, and that actually the divisor would normally be important in deciding. For instance if you were dividing 423 by 3 it wouldn’t make much sense to use 400 as this isnt divisible by 3. It would be more sensible to choose 300 (÷3=100), 120 (÷3=40) and 3(÷3=1).

But why not use short or long division?

This is a question I’ve seen a lot of times from a lot of people. They see the part whole method as a long and clumsy way to solve problems that they can solve easily using one of the two standard algorithms. I can see the point in asking, the algorithms are far more efficient as written methods. But that’s not why this model is taught. No one expects students to get to their GCSE and start drawing part whole models to solve division problems. The visual representations are their to help build an understanding of what is going on, an understanding of the relationship between numbers and mathematical operations. In this case it’s to build an understanding of how the distributive law works and to give a good mental strategy for division. It even helps understand how the long and short division algorithms work, as they are both based on splitting the dividend up into parts. There must come a point when these structures and representations are removed and students move to the abstract, but that doesn’t devalue their importance to that learning journey.

What else is the model used for?

The idea of a part whole diagram is introduced way earlier than this. Students get used to partitioning numbers into part whole models while working on addition and subtraction. It helps then see at that level that they have a relationship, that they are the inverse of each other. So when students come to meet this model for division it’s a small step on what they were already doing.

These are similar to some of the earliest part whole models my daughter did when she started school. They were being used to show place value, and also to show how addition and subtraction work and interact. For both these tasks this model is an excellent visual representation to help students understand the concepts.

Part whole models can also be represented as bar models. Here the one on the left can again be used to show either place value or addition/subtraction. The one on the right is an early algebraic model, and if we are told that x+2=9 we can use this representation to show why x must equal 7. This representation is more effective if students are familiar with it from their earlier mathematics.

Building on this we can show the distributive law when it comes to multiplication:

And show how that links to division:

As we go further into maths this idea of part whole division comes up again and again. One place that springs to mind is when calculus is first introduced at A level. One of the first things that we teach is how do differentiate and integrate polynomials with different powers of x. And a favourite style of question from examiners is this:

Or its derivative equivalent.

The easiest way to do this, when it comes to integration or differentiation, is to rewrite the fraction as separate terms:

What we have done here is used the part whole model to divide the expression on the numerator by x^2. We could draw that in our part whole model:

I wouldn’t advise that, its unnecessary, but having a secure knowledge of that model and how it works due to the distributive law is key to understanding how and why we can simplify this fraction in that way.

I’ve thought a lot about division recently, and I’m sure I will continue to do so, so if you agree,disagree or have anything else to add please get in touch either in the comments or via social media as I’d love to hear your views.

Circle theorems and rectangles

Yesterday I was support an colleague during a live lesson. The topic was angles in parallel lines and followed on from other angles rules including triangles. During the lesson my colleague put up the following question and asked students to have a go:

It was a past GCSE question and one I’ve seen many like before. The students took one of 2 approaches, both using the fact that the diagonals bisect each other to make isoceles triangles. Some considered the triangle on the left and calculated the base angles to be 65. Subtracted that from 90 to get 25 then used the fact the top triangle was also isoceles to get x=25. Some uses angles on a straight like to get 130 as the top angle, then used the isoceles triangle to get x = 25. Both perfectly good, logical and well reasoned solutions.

However, when I saw it initially I saw it slightly differently. I know that all rectangles are cyclic quadrilaterals and as such can be circumscribed by circles. I also know that as the diagonals bisect they are diameters so the point they meet it the centre of the circumcircle:

This leads to x=25 as angles at the circumference are half angles at the centre.

This is not something I’ve thought about before when looking at questions of this type. I’m not sure why it jumped out at me today but never before. I do like it though, it’s a lovely neat solution. I assume it would carry the same amount of marks as the other solutions, it also made me wonder if this was a solution that others would use? Over the years I’ve had some student’s who often chose solutions I wouldn’t have thought were as obvious and this is the sort of thing I imagine they would do with this question.

I’d love to hear if this would be your go to answer for this question, and if so why? I’d.also love to hear any anecdotes of you, or your students giving out of the box answers. Please let me know in the comments or on social media.

Late tiering decisions

Last week year 11 sat their mocks. Some did really well, others did really poorly. It’s the latter group that has me purplexed. Students sitting the higher tier paper but only scoring single digits per paper, or even earlt teens per paper. What to do with them?

Some of them asked if they could move to foundation, I think its best for them. 1 student got 32 marks over 3 higher papers, did the 3 foundation and was well over 100. 1 student got 40 marks over 3 higher papers spent 30 mins in a foundation paper and got 60 marks. The grade 5s they want seem more achievable on foundation.

My issue lies with a few students desperate to do higher and try for 6s. Scoring around 50 marks over 3 higher papers it seems a risk. But having taught them both i feel that it’s within their capabilities. But from November to march they have made only tiny gains in marks. On the ine hand, foundation means they cant get a 6 and for at least one of them means rethinking post 16 choices, but on the other hand sitting higher means they might end up with only a 4 or less and thst would mean rethinking post 16 again. It’s tricky, any thoughts are welcomed.

Proof by markscheme

While marking my Y11 mocks this week I came across this nice algebraic proof question:

The first student had not attempted it. While looking at it I ran through it quickly in my head. Here is the method i used jotted down:

I thought, “what a nice simple proof”. Then I looked at the markscheme:

There seemed no provision made in the markscheme for what I had done. (Edit: It is there, my brain obviously just skipped past it) How did you approach this question? Please let me know via the comments or social media.

Anyway, some of my students gave some great answers. None of them took my approach, but some used the same as the markscheme:

And one daredevil even attempted a geometric proof…….

Cereal Percentages

This week my Y11s are sitting mock exams. One of the questions that came up on paper 1 stumped a lot them.

They came out if the exam on monday, and said the paper was very difficult. One of them asked me one of the questions:

“Sir, if you have a box of cereal and increase it by 25% but keep the price the same, what percentage would you need to decrease the price of the original box by to get the same value?”

I immediately said “20%”, an answer which flummoxed the student and the others stood around. They couldn’t work out how I had got that answer, never mind so quickly.

I tried to explain it to them, but in that moment, on the corridor, I didn’t do a very good job. For me, it was intuitive. A 25% increase and a 20% decrease would yield the same value as in one you are changing the top of a fraction and the other the bottom of a fraction so you need to use the reciprocal, 4/5 is the reciprocal of 5/4 and 4/5 is 80% hence it needs to be a 20% decrease. Cue blank looks and pained expressions. I was seeing the students again later in an intervention session so I promised to go through it in more detail then.

I talked about the idea of value, how you could consider mass/price and get grams per penny – how many grams for each penny you spend – or you could consider price/mass and get penny per grams – how much you pay per gram. I said either of these would give an idea of value and you can use either in a best value problem.

I showed them the idea of the fraction, said you could call the price x and the size y.

The starting scenario is:

y/x

The posed scenario is:

1.25y/x

but we know 1.25 is 5/4 so that becomes:

(5/4)y / x

which in turn is:

5y/4x

I then showed that the second scenario meant getting to the same value but altering x. To do this you would need to mutiply x by 4/5:

y/(x(4/5))

(y/x)÷(4/5)

(y/x) × (5/4)

5y/4x

This managed to show some of them what was going on, but others still massively struggled. I tried showing them with numbers. 100 grams for £1. This again had an effect for some but still left others blank.

I’m now racking my brains for another way to explain it. If you have a better explanation, please let me know in the comments of via social media!

Proving Products

Just now one of the great maths based pages I follow shared this:

So naturally I figured I would have a go. I thoughts just get stuck in with the algebra and see what happens, normally a good approach to these things.

My first thought was that if I use 2n – 3, 2n -1, 2n + 1 and 2n +3 then tgere would be less to simplify later. I know that (2n + 1)(2n – 1) = 4n^2 – 1 and (2n – 3)(2n + 3) = 4n^2 – 9 so I multiples these together.

(4n^2 – 1)(4n^2 -9) = 16n^4 – 40n^2 + 9

I thought the best next move would be to complete the square:

(4n^2 – 5)^2 – 16

This shows me that the product of 4 consecutive odd numbers is always 16 less than a perfect square and as such that the product of 4 consecutive odd numbers plus 16 is always a square.
(4n^2 – 5)^2 – 16 + 16 = (4n^2 – 5)^2 
A nice little proof to try next time you teach it to your year 11s.

Consultation time again

Is it cynical of me to question the DoE’s repeated tactic of releasing consultations either just before the summer, when most teachers are in the midst of high stakes exam testing, or over the summer when a lot of teachers are either away or spending time catching up with their families who they haven’t seen through the heavy term time?

Anyway, this year they have released another one. It focusses around the new GCSEs, and more specifically the awarding of grades. The consultation states that for the first award there will be a heavier reliance on statistical methods to set the grade boundaries, allowing the same proportion of grade 4s as we currently have of grade Cs, likewise similar proportions of 1s to Gs and of 7s to As. The rest will be split arithmetically ie the boundaries in between will be equally spread. From Year 2 onwards it will revert back to examiner judgement, but use the statistical analysis as a guide as well as the national reference tests.

This immediately raises questions – how do we know that the first year to sit it should have a similar proportion of 4s as Cs? It seems that this has been decided without much thought about the prior attainment; the consultation certainly doesn’t mention it for the first year. It does going forward, but that doesn’t really explain how this prior attainment will be measured. I have been under the impression that the KS2 SATs are moving from level based assessments to assessments where the students’ scores will be reported as percentiles – surely then comparisons of prior assessment will always be the same? “This year, bizarrely, we saw exactly 10% score above the 90th percentile, what’s more bizarre is that is exactly the same proportion as last year!”

It seems strange to me to put such a heavy reliance on these prior attainment targets anyhow. We live (for now) in a society that has a fairly fluid immigration system, so the students who get to year 11 haven’t always been through year 6 in this country. There is also a question of the validity of the assumption that every year group will progress over the 5 years of secondary at the same rate.

The obvious elephant in the room is floor targets. By setting the boundaries so the same proportion of students get above a grade 4 as get above a C, but switching the threshold to a grade 5 you immediately drop the results of a whole host of schools down, what happens then remains to be seen, but I can imagine lot of departments will become under pressure and scrutiny for something that is statistically inevitable given the new grading formula.

This is all interesting, but it’s not much different to previous announcements and consultations, what is different is the formula for awarding grades 8 and 9. The formula looks to be a fair way of doing it, but it seems strange to me to use this formula just for the first year. Why then revert to examiner judgement about the grade standard? The government seem to be happy to use statistical analysis and similar grade proportions in parts of their grading system, but not in all of it, and that seems odd to me.

Have you responded yet? If not you can here (but hurry, the consultation closes June 17th). I’d love to hear other people’s views either in the comments or via social media.

Forming and Solving Equations

While checking the work of a year 11 student on Friday I came across a question that could have been a great one for the higher GCSE students to practice their skills together and also their selection of which mathematics to use.

The question was to find the area of this triangle:

A great question. One that to you or I is straightforward but that would take GCSE level students and below a bit of thinking and let’s them hone their skills.

The way to tackle it is to use Pythagoras’s Theorem to form an equation, solve for x then find the area. I feel is beneficial as it combines Pythagoras’s Theorem with a decent amount of algebra then includes the find the area bit at the end.

In this case though, that wasn’t the question. There was more information on offer and the question was:

Which is still a fairly nice form and solve an equation problem.

3x + 1 + 3x + x – 1 = 56
7x = 56
x = 8
A = 0.5×7×24 = 84

There is a niceness to this question that goes beyond the question itself.  It shows us a great way of differentiating within lessons. Just be leaving out a tiny portion of the information, in this case the perimeter, we can make the question much harder. This idea is something I’ve been working on in various places. M1 questions can be made much easier by providing a diagram, for example.

Have you used questions in a similar way? If so I’d love to see them, please do get in touch.

Cross-posted to Betterqs here.

Is one solution more elegant?

Earlier this week I wrote this post on mathematical elegance and whether or not it should have marks awarded to it in A level examinations, then bizarrely the next day in my GCSE class I came across a question that could be answered many ways. In fact it was answered in a few ways by my own students.

Here’s the question – it’s from the November Edexcel Non-calculator higher paper:

I like this question, and am going to look at the two ways students attempted it and a third way I think I would have gone for. Before you read in I’d love it if you have a think about how you would go about it and let me know.

Method 1

Before I go into this method I should state that the students weren’t working through the paper, they were completing some booklets I’d made based on questions taken from towards the end of recent exam papers q’s I wanted them to get some practice working on the harder stuff but still be coming at the quite cold (ie not “here’s a booklet on sine and cosine rule,  here’s one on vectors,” etc). As these books were mixed the students had calculators and this student hadn’t noticed it was marked up as a non calculator question.

He handed me his worked and asked to check he’d got it right.  I looked, first he’d used the equation to find points A (3,0) and D (0,6) by subbing 0 in for y and x respectively. He then used right angled triangle trigonometry to work out the angle OAD, then worked out OAP from 90 – OAD and used trig again to work out OP to be 1.5, thus getting the correct answer of 7.5. I didn’t think about the question too much and I didn’t notice that it was marked as non-calculator either. I just followed his working, saw that it was all correct and all followed itself fine and told him he’d got the correct answer.

Method 2

Literally 2 minutes later another student handed me her working for the same question and asked if it was right, I looked and it was full of algebra. As I looked I had the trigonometry based solution in my head so starter to say “No” but then saw she had the right answer so said “Hang on, maybe”.

I read the question fully then looked at her working. She had recognised D as the y intercept of the equation so written (0,6) for that point then had found A by subbing y=0 in to get (3,0). Next she had used the fact that the product of two perpendicular gradients is -1 to work put the gradient of the line through P and A is 1/2.

She then used y = x/2 + c and point A (3,0) to calculate c to be -1/2, which she recognised as the Y intercept, hence finding 5he point P (0,-1.5) it then followed that the answer was 7.5.

A lovely neat solution I thought, and it got me thinking as to which way was more elegant, and if marks for style would be awarded differently. I also thought about which way I would do it.

Method 3

I’m fairly sure that if I was looking at this for the first time I would have initially thought “Trigonometry”, then realised that I can essential bypass the trigonometry bit using similar triangles. As the axes are perpendicular and PAD is a right angle we can deduce that ODA = OAP and OPA = OAD. This gives us two similar triangles.

Using the equation as in both methods above we get the lengths OD = 6 and OA = 3. The length OD in triangle OAD corresponds to the OA in OAP, and OD on OAD corresponds to OP, this means that OP must be half of OA (as OA is half of OD) and is as such 1.5. Thus the length PD is 7.5.

Method 4

This question had me intrigued, so i considered other avenues and came up with Pythagoras’s Theorem.

Obviously AD^2 = 6^2 + 3^2 = 45 (from the top triangle). Then AP^2 = 3^2 + x^2 (where x = OP). And PD = 6 + x so we get:

(6 + x)^2 = 45 + 9 + x^2

x^2 + 12x + 36 = 54 + x^2

12x = 18

x = 1.5

Leading to a final answer of 7.5 again.

Another nice solution. I don’t know which I like best, to be honest. When I looked at the rest of the class’s work it appears that Pythagoras’s Theorem was the method that was most popular, followed by trigonometry then similar triangles. No other student had used the perpendicular gradients method.

I thought it might be interesting to check the mark scheme:

All three methods were there (obviously the trig method was missed due to it being a non calculator paper). I wondered if the ordering of the mark scheme suggested the preference of the exam board, and which solution they find more elegant. I love all the solutions, and although I think similar triangles is the way I’d go at it if OD not seen it, I think I prefer the perpendicular gradients method.

Did you consider this? Which way would you do the question? Which way would your students? Do you tuink one is more elegant? Do you think that matters? I’d love to know, and you can tell me in the comments or via social media!

Cross-posted to Betterqs here.

Isosceles triangles and deeper understanding

When marking paper 3 of the Edexcel foundation Sample Assessment Materials recently I came across this question that I found interesting:

It’s a question my year tens struggled with, and I think it is a clear marker to show the difference between the current specification foundation teir and the new spec.

The current spec tends to test knowledge of isosceles triangles by giving a diagram showing one, giving an angle and asking students to calculate a missing angle. This question requires a bit of thinking.

To me, all three answers are obvious, but clearly not to my year 10s who do understand isosceles triangles. The majority of my class put 70, 70 and 40. Which shows they have understood what an isosceles is, even if they haven’t fully understood the question. They have clearly mentally constructed an isoceles triangle with 70 as one of the base angles and written all three angles out.

What they seem to have missed was that 70 could also be the single angle, which would, of course, lead to 55 being the other possible answer for B. One student did write 55 55 70, so showed a similar thought process to most but assumed a different position for the 70.

I already liked this question, and then I read part b:

Now students are asked to explain why there can only be one other angle when A = 120. Thus they need to understand that this must be the biggest angle as you can’t have 2 angles both equal to 120 in a triangle (as 240 > 180), thus the others must be equal as it’s an isoceles triangle.

The whole question requires a higher level of thinking and understanding than the questions we currently see at foundation level.

In order to prepare our students for these new examinations, we need to be thinking about how we can increase their ability to think about problems like this. I think building in more thinking time to lessons, and more time for students to discuss their approaches and ideas when presented with questions like this. The new specification is going to require a deeper, relational, understanding rather than just a procedural surface understanding and we need to be building that from a young age. This is something I’ve already been trying to do, but it is now of paramount importance.

There is a challenge too for the exam boards, they need to be able to keep on presenting questions that require the relational understanding and require candidates to think. If they just repeat this question but with different numbers than it becomes instead a question testing recall ability – testing who remembers how they were told to solve it, and thus we return to the status quo of came playing and teaching for instrumental understanding, rather than teaching mathematics.

What do you think of these questions? Have you thought about the effects on your teaching that the new specification may have? Have you any tried and tested methods, or new ideas, as to how we can build this deeper understanding? I’d love to hear in the comments or social media if you do.

Further Reading:

Teaching to understand – for there thoughts in relational vs instrumental understanding

More thoughts on the Sample assessment materials available here and here.

Cross-posted to Betterqs here.