Lesson video

Hi, my name is Mr. Peters.

Thanks for joining me today for this lesson.

In this lesson, we're gonna be thinking about how we can apply our understanding of the commutative and the associative law to make our calculations even easier to work with.

This is a really exciting lesson, as it opens up so many more opportunities to think more flexibly of our calculations.

If you're ready, let's get started.

So by end of this lesson today, hopefully you should be able to say with confidence that you can apply the commutative and associative laws to simplify calculations with three or more numbers.

Okay, we've got some keywords today that we're gonna be referring to throughout.

I'll have a go at saying them, and then you can repeat them after me.

The first one, efficient.

Your turn.

The second one, commutative.

Your turn.

And the last one, associative.

Your turn.

So let's think about what these mean then.

Working efficiently means finding a way to solve a problem quickly, but also doing it accurately as well.

The commutative law states that you can write the values of a calculation in any order and the result will still remain the same.

This applies for both addition and subtraction.

And the associative law states it doesn't matter how you group or pair the values in a calculation at all.

The result, once again, would remain the same.

This also applies to addition and multiplication.

So in this session today, we've got two cycles.

The first one is calculating efficiently, and the second one is calculating problems with three factors efficiently.

Let's get started with the first cycle.

Throughout this lesson today, you're gonna meet three of our students, Aisha, Andeep, and Jun, and throughout the lesson, they'll be sharing their thinking and their learning as we go.

So we're gonna start off with this calculation here.

How would you have a go at calculating 28 multiplied by five multiplied by two? Take a moment to have a think.

Well, Aisha's saying she's gonna start by multiplying the five with 28 to start off with.

So she thinks she might need a pen and paper to jot some of her thinking down.

Andeep's saying wait, you don't need a pen and paper.

We can do this mentally.

Andeep says we can use the commutative law, as multiplication is commutative.

This means we can swap the position of the factors and the product would remain the same.

So we can see now that Andeep has reordered the calculation, so it says five multiplied by two multiplied by 28.

If we were to work left to right in this calculation now, we could do five multiplied by two, which would be equal to 10, and then we'd have to multiply that by the 28, so that would give us 280.

Aisha's realised, wow, that was really efficient, wasn't it? That's taking no time at all and it was really accurate.

So actually, that's a really good strategy to use going forward, isn't it? Jun's saying that was a nice strategy Andeep, but I think we can be even more efficient with that.

Let's see how Jun tackles it.

Jun is saying we could use the associative law this time instead.

This means it doesn't matter how we pair the factors in the calculation, we'll still get the same product as well.

So Aisha is asking does that mean we don't need to swap the factors around? No, we don't need to, do we? Here, we can see that the five and the two are next to each other, so we could draw brackets around the five and the two to show that we're gonna multiply these two first.

Once again, the five multiplied by the two would give us the 10, and then that makes our calculation really easy because it's 28 multiplied by 10 this time, which again would be 280.

Aisha realised these laws are quite useful, aren't they? To helping making our calculations more efficient.

Let's see how else we can apply them, shall we? Here's another example, nine multiplied by four multiplied by 25.

How would you have a go at doing that one? Take a moment to have a think.

Well, Aisha's saying, typically, I'd start by doing nine multiplied by four, which would give me 36, and then I'd multiply that by the 25.

Again, that could take a little while, couldn't it? So Andeep's thinking, should we apply the commutative law then? Let's see how we could do that.

Well, have a look at the numbers, what numbers we've multiplied together quite nicely and make our life easier? That's right, the 25 and the four.

So if we rearranged our calculation to 25 multiplied by four, we know that's gonna give us 100, isn't it? So now we've got 100 multiplied by the nine, and that would leave us with 900 altogether.

Gosh, that was easy, wasn't it? As Andeep said, that was a lot more efficient.

So we've used commutative law in the last example.

Could we apply the associative law this time and not have to change the order of the factors? Look carefully.

Which pair would you multiply together first? That's right, Aisha, we could multiply the four and the 25 together first, couldn't we? 'Cause they're next to each other, so that would give us the 100 again, and then we can multiply that by the nine, which gives us 900 altogether.

Yeah, and quite right, Jun, that was working efficiently at its finest.

Okay, time for you to check your understanding.

Can you use the commutative law to make this calculation easier to manage? Then you need to find the product.

Take a moment to have a think.

Well done, you may have rearranged this equation, so it is two multiplied by five multiplied by 47.

We know that having the two and the five multiplied together gives us the 10, which is a really useful number to have, 'cause then we can multiply 10 by 47, which gives us 470.

And that's exactly what Aisha's saying.

She noticed that having a two and a five in a calculation, by multiplying those two together makes life really easy, doesn't it? 'Cause that allows us to get to the 10, which is a comfortable number to multiply with, isn't it? Choose a pair numbers to put some brackets around to make the calculation easier, and then find the product.

There we go, you might have put the brackets around the 20 and the five here.

We know that two times five is 10, so 20 multiplied by five is gonna be equal to 100, and again, 100's a nice, comfortable number to work with, isn't it? So 100 multiplied by 66 gives us 6,600 or 6,600.

Well done if you managed to get that one too.

Okay, brilliant, time for you to have a go at practising these for yourself now.

So what I'd like you to do is rewrite each one of these equations using the commutative law to make the calculations easier, and then also, find the product each time.

And then what I'd like you to do for task two is to put a pair of brackets around the two factors that you think would be easiest to multiply first, and then fill in the missing numbers by working through the next two steps.

The first one has been done for you.

Good luck with those and I'll see you back here shortly.

Let's go through the first set then.

The first one could be five multiplied by two multiplied by 125, which gives us 1,250.

The second one could have been two multiplied by five multiplied by 125, which gives us 1,250 again.

This, time we could rearrange the factors so that 20 multiplied by five came first, 'cause we know that's gonna give us 100, and then multiply that by the 36 to give us 3,600.

Then we could have done the 50 multiplied by the two together this time, 'cause 50 timesed by two gives us 100 isn't it? Multiply that by 36, that also gives us 3,600.

Hmm, this one's an interesting one.

We could have done 0.

5 multiplied by two, couldn't we? 0.

5 multiplied by two, that's the same as saying one whole, isn't it? So we could then have one multiplied by 72, which is equal to 72.

What did you notice this time? That's right, the decimal now is not from 0.

5, it's 0.

2, and we know that five lots of 0.

2 would make one whole again, so we could represent this as 0.

2 multiplied by five to give us the one whole, and then multiply that by 72, which again would give us the 72.

And what you notice for the last one? We've got 2/100ths this time, haven't we? And we know that 2/100ths timesed by five would give us 1/10th.

So we can do 2/100ths timesed by the five to give us that 1/10th and then multiply that by 72.

72/10ths can be written as 7.

2, can't it? Well done if you managed to get all of those.

Okay, and for these ones here, well, the second one required us to do four multiplied by five multiplied by two, and that was equal to four multiplied by something.

So if we grouped the five and the twos together, that would give us the 10, and then four multiplied by 10 would be equal to 40.

And the one underneath that, we've got six multiplied by 20 multiplied by five.

This time, that needs to be equal to something multiplied by five.

So if we group the six and the 20 together this time, that would give us 120, and then 120 multiplied by five would be equal to 600.

Again, at the next column at the top, we can have six multiplied by 20 multiplied by five is equal to six times something.

So this time, we're going to need to put brackets around the 20 multiplied by the five.

That gives us 100, six multiplied by 100 is 600.

And then two multiplied by 50 multiplied by eight is equal to something multiplied by eight.

So two times 50 is 100, 100 timesed by eight is equal to 800.

And then finally, at the bottom, two multiplied by 50 multiplied by eight is equal to something timesed by something.

Hmm.

How would you have gone around that one? Well, the one above was multiplying the two by the 50.

What if we multiplied the 50 by the eight this time? Well, we know this would be equal to two multiplied by 400 because five times eight is 40, so 50 timesed by eight is 400, and then two lots of 400 is 800.

Well done if you managed to get all of those.

Okay, onto our second cycle now, calculating problems with three factors efficiently.

So let's have a look here.

There are two rows of four melons in each box, and there are five boxes altogether.

How many melons are there altogether? Well, to calculate this, we could represent this as two multiplied by four multiplied by five, couldn't we? The two represents the number of rows in each box, the four represents the number of columns in each box, and the five represents the number of boxes altogether.

We could then apply the commutative law, couldn't we? As Aisha's suggesting, to rearrange the factors to calculate how many melons there were.

So although two times four is a really simple calculation that we should know, two times five is also very simple, isn't it? Two times five gives us 10, so by rearranging the factors, we could put two multiplied by five multiplied by four.

This gives us 10 multiplied by four, and again, that's a really simple calculation.

We know our times tables, 10 multiplied by four is equal to 40.

There are 40 melons altogether.

Jun's saying instead of the commutative law, we could have applied the associative law, couldn't we? Let's have a look how we could have done that.

Two multiplied by four, multiplied by five.

Well, we could have grouped the four and the five together.

Four times five is a nice, comfortable calculation because that gives us 20, doesn't it? So then our calculation becomes two multiplied by 20, and then that would be equal to 40 as well.

So again, there'll be 40 melons altogether and we could apply either the commutative or the associative law to calculate that efficiently.

Okay, quick check for understanding.

There are five oranges in each bag.

In each orange, there are 12 segments, and in each segment, there are two pips.

How many pips would there be altogether? You could write a calculation to solve this.

That's right, it would be five multiplied by 12 multiplied by two, wouldn't it? You then might have applied the commutative law to make it five multiplied by two multiplied by 12.

We know that five multiply by two is 10, so 10 times 12 would be equal to 120, so that'd be 120 pips.

Okay, onto our final tasks for today then.

What I'd like you to do is write an equation for each of these problems here and use an efficient strategy to help you solve it, either using the commutative law or the associative law.

And then here, for second task, what I'd like you to do is write your own problem that could be represented using this calculation.

Good luck with those, and I'll see you again shortly.

Okay, welcome back.

Let's go through these then.

So in the first problem, there are five classes getting ready for the whole-school gymnastics competition.

There are 30 children in each class, and each child puts two socks into their PE bag.

How many socks are there altogether? Well, we could represent this as five multiplied by 30 multiplied by two, and you could then either apply the commutative law to rearrange the factors, so it's five multiplied by two multiplied by 30, or you could use the associative law to group together some of the factors.

So you might have grouped the 30 and the two together, which gave us 60, and then we know that 60 multiplied by five is also equal to 300.

So there are 300 socks left.

Okay, for question B, Jun cycles two kilometres to school.

Each kilometre takes him nine minutes to cycle, and if he does that for every school day in one week, how long will he have cycled for? Well, if it takes him nine minutes to cycle each kilometre, we can represent that as two multiplied by nine.

That's two kilometres, and each one takes nine minutes, and then he does that for each day of the school week, so there's five days in the school week, aren't there? So our calculation would become two multiplied by nine multiplied by five.

Again, you could apply the commutative law so that you could put the two and the five together at the beginning of the calculation.

So two times five would give us the 10, and then multiply that by the nine to give us the 90, or you may have decided to use the associative law, for example, multiply the nine and five together, which would give us 45 using our times table facts, and then multiply that by two, which would also give us 90.

So Jun altogether would've had to cycle for 90 minutes over the whole week.

Okay, and for question C, Alex's mum takes Alex and his two brothers to the shop after school for a treat.

She buys them each two packets of stickers, and one packet of stickers cost 50p.

How much money did Alex's mum spend on the boys altogether? Let's have a think about it.

First of all, it says Alex and his two brothers were taken to the shop, so that's three of them, isn't it? That means Alex's mum has got three boys.

She buys them each two packets, so we can represent that as three multiplied by two.

Three for the number of boys, two for the number of packets of stickers they get each.

And then 50p is the price of each packet, so we can represent all of this as three multiplied by two multiplied by 50.

Hmm.

How would you go about doing that? There's a number of ways we could have done this one.

You could apply this associative law and multiply the three and the two together first of all, which would give us six, and then multiply that by the 50, which would give us 300 pence, or you could multiply the two and the 50 together, which would give us 100, and then multiply that by the three, which would also give us 300 pence.

Either way, Alex's mum spent 300 pence on stickers for the boys after school.

Lucky lads.

Okay, and then thinking about a problem to represent this equation here, four multiplied by 25 multiplied by five is equal to 500.

Here's an example of Andeep's problem.

Andeep says he has a packet of raisins every day in his school lunch, and he does this for four weeks in a row, and each packet are 25 raisins.

So the question would be, how many raisins did Andeep eat altogether? And the answer to that would've been 500.

Well done if you managed to come up with your own story similar to that.

Okay, thanks for joining me today.

That's the end of our lesson.

Let's have a little think about how we can summarise our learning.

So we can apply the commutative or the associative law to our calculations to make our calculations more easy.

And often, by using the commutative or the associative law, it makes our calculations more efficient, therefore, we're making our calculations quicker and also more accurate.

Thanks for joining me today.

Hopefully, you've got something to take away.

And again, you can apply this to your everyday maths yourselves.

Take care and I'll see you again soon.