Lesson video

Hi, my name's 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 key words 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 lesson 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 the session 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 5 multiplied by 2? Take a moment to have a think.

Well, Aisha's saying she's gonna start by multiplying the 5 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 5 multiplied by 2 multiplied by 28.

If we were to work left to right in this calculation now, we could do 5 multiplied by 2, 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 taken 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 wasn't our strategy, Andeep, but I think we can be even more efficient with that." Let's see how Jun tackles it.

Jun's 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's 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, 5 multiplied by the 2 to 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.

I should realise these laws are quite useful, aren't they? To help in making our calculations more efficient.

Let's see how else we can apply them, shall we? Here's another example.

9 multiplied by 4 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 9 multiplied by 4, 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, having a look at the numbers, what numbers would multiply together quite nicely and make our life easier? That's right, the 25 and the 4.

So if we rearranged our calculation to 25 multiplied by 4, we know that's gonna give us 100, isn't it? So now we've got 100 multiplied by the 9 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 the 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 4 and the 25 to 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 9, which gives us 900 altogether.

Yeah, and quite that, Jun.

That was working efficiently at its finest.

Okay, time for you to check your understanding now.

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 2 multiplied by 5 multiplied by 47.

We know that having the 2 and the 5 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 of 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 5 here.

We know that 2 times 5 is 10, so 20 multiplied by 5 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 66 hundreds 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 5 multiplied by 2 multiplied by 125, which gives us 1,250.

The second one could've been 2 multiplied by 5 multiplied by 125, which gives us 1,250 again.

This time, we could rearrange the factors so that 20 multiplied by 5 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've done the 50 multiplied by the 2 together this time, 'cause 50 times by 2 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've done 0.

5 multiplied by 2, couldn't we? 0.

5 multiplied by 2, that's the same as saying one hole, isn't it? So we could then have 1 multiplied by 72, which is equal to 72.

Would did you notice this time? That's right.

The decimal now is not from 0.

5 is 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 5 to give us the one whole and then multiply that by 72, which again would give us the 72.

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

So we can do 2/100 times by the 5 to give us that 1/10 and then multiply that by 72.

72/10 can be written as 7.

2, can't it? Well done if you managed to get one 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 two together, that would give us the 10.

And then 4 multiplied by 10 would be equal to 40.

And the one underneath that, we've got 6 multiplied by 20 multiplied by 5.

This time that needs to be equal to something multiplied by five, so if we group the 6 and the 20 together this time, that would give us 120.

And then 120 multiplied by 5 would be equal to 600.

Again, at the next column at the top, we can have 6 multiplied by 20 multiplied by 5 is equal to 6 times something.

So this time we're going to need to put brackets around the 20 multiplied by the 5.

That gives us 100, 6 multiplied by 100 is 600.

And then 2 multiplied by 50 multiplied by 8 is equal to something multiplied by 8.

So 2 times 50 is 100, 100 times by 8 is equal to 800.

And then finally at the bottom, 2 multiplied by 50 multiplied by 8 is equal to something times by something.

Hmm.

How would you have gone around that one? Well, the one above was multiplying the 2 by the 50.

What if we multiplied the 50 by the 8 this time? Well, we know this would be equal to 2 multiplied by 400 because 5 times 8 is 40, so 50 times 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 2 multiplied by 4 multiplied by 5, 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 is suggesting, to rearrange the factors to calculate how many melons there were.

So although 2 times 4 is a really simple calculation that we should know, 2 times 5 is also very simple, isn't it? 2 times 5 gives us 10.

So by rearranging the factors, we could put 2 multiplied by 5 multiplied by 4.

This gives us 10 multiplied by 4, and again, that's a really simple calculation.

We know our times tables, 10 multiplied by 4 is equal to 40.

There are 40 melons altogether.

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

2 multiplied by 4 multiplied by 5, well, we could of grouped the 4 and the 5 together.

4 times 5 is a nice comfortable calculation, because that gives us 20, doesn't it? So then our calculation becomes 2 multiplied by 20, and then that would be equal to 40 as well.

So again, there will 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? Write a calculation to solve this.

That's right, it would be 5 multiplied by 12 multiplied by 2, wouldn't it? You then might've applied the commutative law to make it 5 multiplied by 2 multiplied by 12.

We know that 5 multiplied by 2 is 10, so 10 times 12 with be equal to 120.

So that'd be 120 pips.

Okay, onto our final tasks for today then.

What I'd like 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 the second task, what I'd like 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 5 multiplied by 30 multiplied by 2.

And you could then either apply the commutative law to rearrange the factors, so it's 5 multiplied by 2 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 2 together, which gave us 60, and then we know that 60 multiplied by 5 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 2 multiplied by 9.

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 2 multiplied by 9 multiplied by 5.

Again, you could apply the commutative law so that you could put the 2 and the 5 together at the beginning of the calculation.

So 2 times 5 would give us the 10 and then multiply that by the 9 to give us the 90 or you may have decided to use the associative law, for example, multiply the 9 and 5 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 have 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 costs 50 p.

How much money did Alex's Mom 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 mom has got three boys.

She buys them each two packets, so we can represent that as 3 multiplied by 2.

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

And then 50 p is the price of each packet, so we can represent all of this as 3 multiplied by 2 multiplied by 50.

Hmm, how would you go about doing that? There's a number of ways we could've done this one.

You could apply the 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 2 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 mom spent 300 pence on stickers for the boys after school, lucky lads.

Okay, and then thinking about a problem to represent this equation here.

4 multiplied by 25 multiplied by 5 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 in 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 some to take away, and again, you can apply this to your everyday maths yourselves.

Take care, and I'll see you again soon.