Lesson video

Hi, my name is Mr. Peters.

Thanks for joining me for today's lesson.

In this lesson, we're gonna be thinking about how we can apply the commutative and the associative law to help simplify a range of problems in different contexts.

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

So by the end of this lesson today, you should be able to say that I can apply the commutative and the associative law to simplify problems in a range of different contexts.

Throughout this lesson, we've got two key words we're gonna be referring to.

The first one is associative, and the second one is commutative.

Can you have a go at saying them? After me.

Associative.

Commutative.

Let's think about what these mean then.

The associative law states it doesn't matter how you group or pair values in a calculation.

For example, it doesn't matter which pair you calculate first, the result is still the same.

This applies to both addition and multiplication.

However, the commutative law states how you can order the values any way you like within a calculation, and the calculation will still remain the same, and therefore the result will still remain the same.

This applies to both addition and multiplication as well.

Our lesson today is broken down into two cycles.

The first cycle will solve problems with three factors, and the second cycle will solve problems with unknown factors.

If you're ready, let's dive into the first cycle.

Throughout this lesson today, you're gonna meet Izzy and Jacob.

They're gonna be sharing their thinking and their learning as they go throughout the lesson as well.

So let's start here.

A gardener is planting some seeds to grow leeks.

She's got a number of different trays that she can use to do this.

And within each tray cell, she plants four seeds.

How many leeks do you think she could grow altogether? Jacob suggests we start with this tray here.

Let's have a think about how this could work then.

Well, we can see in this tray here, there are two rows and there are three columns.

So we can start our calculation by writing two multiplied by three.

The two represents the rows and the three represents the columns.

This would give us the total number of cells in the tray.

We now can place four seeds into each one of the cells.

So to our calculation, this now adapts it so it becomes two multiplied by three, multiplied by four.

Have a look at the calculation then.

How would you go about solving it? Jacob's saying he'd use the associative law.

He'd group the three and the four together and multiply those first.

So three multiplied by four would be equal to 12.

Then he can multiply that by the remaining factor.

So two multiplied by 12 is equal to 24 seeds.

So in the smaller tray, there are 24 seeds all together.

Izzy's saying she lived on it slightly differently by multiplying the two and the three together first.

So she's associated the two and the three, that would give her six, and then she can multiply that by the four, which using her timetable knowledge, again would be 24 seeds.

So the smaller tray altogether has 24 seeds in it, which means 24 leeks could possibly grow from it.

So we know how many leeks are potentially gonna be able to grow in the smaller tray.

Let's have a look at the next tray, shall we? Following a similar process, we can see that we have three rows and we have four columns.

So we can represent this as three multiplied by four.

And in each tray cell, then we need to put the four seeds in.

So we can multiply the three and the four by another four, can't we? So how would we go about tackling this one then? Jacob says he would associate both fours together.

So he could represent this as three, multiplied by four multiplied by four, with the fours in brackets.

Four multiplied by four is 16.

So the calculation becomes three multiplied by 16.

Hmm, they're not the easiest numbers to work with.

It's doable, but they're not the easiest.

And Izzy thinks it might have been easier to associate the four and the three together.

Let's have a look.

If we associate the three and the four, that would give us 12, wouldn't it? And then we've got 12 multiplied by four.

And that's a times table fact we know, isn't it? We know 12 fours are 48.

So the number of seeds in this tray here would be 48 seeds.

So there's 48 possible leeks able to grow in this tray.

And then finally, the last tray.

How do you think we're gonna tackle this one? Well, let's have a look.

So in this tray there are four rows and there are five columns, aren't there? So we can represent that as four multiplied by five.

And as usual, we need to multiply each one of those cells by four because that's how many seeds we put into each cell.

So we're gonna multiply the four by five by the four as well.

Now we can think about which numbers we're gonna associate together.

Izzy is saying four multiplied by five is quite nice, isn't it? 'Cause that will give us 20.

And 20 is a nice number to multiply.

So let's associate the four with the five then.

So you've got two options here, haven't you? You could multiply the first four with the middle five, or you could multiply the middle five with the last four.

We are gonna do the first four and the middle five.

So four multiply by five is equal to 20, and then 20 multiplied by four, that will give us the total amount of seeds, which is 80 seeds altogether.

So the larger tray has got 80 seeds in it altogether.

Now we can see the total amount of seeds for tray A, for tray B and tray C.

And I've represented this in a bar model now.

We can see each part is tray A, tray B, and tray C, and we wanna find out the total number of seeds that have implanted altogether, 'cause this will tell us the total number of leeks that could possibly grow.

So we can add 24 and 48 and 80 altogether, and we know that 24 and 48 is equal to 72, and then we need to add on the additional 80.

So 72 plus 80 is 152.

So the total number of seeds or the total number of leeks that could grow would be 152.

Okay, check for understanding here then.

Can you use the associative law to help you solve this problem? The gardener plants two batches of cucumbers over the year.

In each packet there are 50 cucumber seeds and she uses four packets for each batch.

How many cucumbers could she hope to grow over the year? Take a moment to have a think.

Well, let's work through this then.

We know the equation would be two multiplied by 50 multiplied by four.

The two represents the number of batches.

The 50 represents the number of seeds in each packet, and the four represents the number of packets that she uses.

Now, using the associative law, we could associate the two with the 50, couldn't we? So two multiplied by 50 is equal to 100, and then we could multiply that by four, and that would give us 400 altogether.

So the number of seeds that were planted were 400, and therefore number of cucumbers that she could hope to grow would be 400 as well over the year.

You may have associated the 50 and the four instead.

So 50 multiply by four is equal to 200 and 200 multiplied by two is also equal to 400.

Well done if you chose either one of those strategies.

Let's have a look at a slightly different problem this time.

The gardener sets up a sprinkler system to water the flowers in her garden.

The sprinkler system is set to run for 25 minutes each day.

And she does this for 28 days in a row.

How many minutes did the water run for altogether? Hmm, well, let's have a think about this.

How could we write an equation to represent this? Well, the water runs for 25 minutes every day, and that runs for 28 days.

So we could represent that as 25 multiplied by 28.

Hmm, as Izzy is pointing out, that's a two digit by two digit calculation, which she's not that confident with at the moment, actually.

So is there a way we could tackle it without that I wonder? Ah, there we go, Jacob's got an idea.

He said we could use our factors to help us here.

Let's see what he's thinking.

Jacob's noticed that 28 is a multiple of four.

So four is a factor of 28.

If we were to find a factor pair of 28, we could use four along with seven, can't we? We know that four multiplied by seven is equal to 28.

We can now express our calculation slightly differently.

We could say that 25 multiplied by four multiplied by seven is the same as 25 multiplied by 28.

Now knowing this as well, we can use the commutative law to order the calculation how we want to.

So now that we know that 28 can be broken down into two factors of four and seven, we can write an equation using the commutative law to make it easier for us to calculate this.

So we've got our calculation now of four multiplied by 25 multiplied by seven.

Hmm, well, why did Jacob decide to try and find four as a factor within one of the numbers? That's right, we're multiplying by 25, aren't we? And four multiplied by 25 gives us 100.

And that's a really easy number to multiply with again, isn't it? So we can associate here the four and the 25, can't we? Let's do that, we've put the brackets around them.

Four multiplied by 25 is equal to 100, and then 100 multiplied by seven is equal to 700.

Gosh, that made that calculation so much easier, didn't it? And we no longer have to think about how are we gonna solve 25 times by 28 just because we were able to manipulate one of the factors and change that from a two factor calculation into a three factor calculation.

Great thinking, Jacob.

As Izzy's pointed out, finding factors of numbers like 10 and 100 are really useful for helping us solve problems efficiently.

Okay, another check for understanding now.

Use both the commutative and the associative law to help solve the problem.

Jacob's dad goes out running each day for 25 minutes.

He does this for 16 days in the month of January.

How many minutes does Jacob's dad run for? Take a moment to have a think.

Okay, well we can represent this as 25 times by 16.

And we know that 16 can be broken down into four multiplied by four.

These are both factors of 16.

So now we can associate the 25 and the four together, that would give us the 100, and then 100 multiplied by four would equal 400.

So we can say Jacob's dad ran for 400 minutes over the course of January.

Well done Jacob's dad, that'll keep you really strong and fit.

Okay, so your first task for today, then.

Cupcakes are baked in the following trays.

If each cupcake is sold for £2.

50, how much money is made per tray? Once you've tackled that one, have a go at this.

Izzy is swimming lengths of the pool.

She swims 24 lengths of the pool.

Each length is 25 metres long.

How far does she swim altogether? Izzy's saying, "Can you solve this in two different ways?" It'd be really interesting to see what you come up with.

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

Okay, let's go through this then.

Well, for the first tray, there are three rows and two columns, so that's three multiplied by two, and we know we need to multiply that by £2.

50 as well.

If we associate the two and the 2.

50 together, that gives us £5, and then three lots of £5 would be £15.

So £15 is the cost from the first tray.

In the second tray, there are four rows and two columns.

That's four multiplied by two, multiplied by £2.

50.

Again, we can associate the two with the £2.

50 this time, that would give us £5, and then four multiplied by five would be £20.

In the third tray, there are three rows and four columns.

So three multiplied by four, multiplied by £2.

50.

Hmm, if we use the commutative law to rearrange our calculation and then we can associate the four and the £2.

50 together, four lots of £2.

50 is equal to £10.

So £10 multiplied by three would equal to £30.

And then finally for the last one, there are four rows and five columns.

Four multiplied by five multiplied by £2.

50.

Once again, if we use the commutative law to rearrange our factors, then we can have four multiplied by £2.

50 and associate those two together, so that would be £10, and then we multiply that by five, becomes £50.

Well done if you've got all of those.

Okay, and then for Izzy, well we know that Izzy swims 24 lengths and each length is 25 metres long.

So we can represent that as 25 times by 24 or 24 times by 25.

Now we could find factors of 24 as four and six.

Because we've got a 25, again, it makes sense to try and find the four, doesn't it? So multiplying 25 by four would give us the 100.

So our new calculation would be 25 multiplied by four multiplied by six.

If we associate the 25 and the four together, that gives us 100, and then we multiply that by six, that gives us 600 altogether.

So that would be 600 metres that she swam.

A different strategy, might have been to find factors of 24 again, but maybe to use eight and three as an idea.

So this time our calculation becomes 25 multiplied by eight multiplied by three.

If I know that 25 multiplied by four is 100, and we could associate the 25 multiplied by eight, couldn't we? 'Cause that would just be double that.

So 25 multiplied by eight would be 200, and then we multiply that by three, which would be 600 altogether.

So as we said, Izzy swims 600 metres altogether.

Good swimming Izzy.

Okay, that's the end of our first cycle.

Let's have a look at our second cycle now.

Solving problems with unknown factors.

So we're gonna revisit our cupcake trays to help us with this problem here.

A tray of cupcakes makes £22.

50 at the cake sale.

If each cupcake was sold for £2.

50 how many cupcakes were on the tray together? I wonder if we could draw a picture for the size of the tray that was used.

But first of all, we need to figure out how we're gonna do this, don't we? Let's represent this, first of all, as a multiplication equation.

We know the cost of each cupcake was £2.

50, but we don't know the number of rows and columns in the tray, do we? So we're gonna lead those as two unknown factors.

We know the total cost we made was £22.

50.

So we can adapt our calculation now, 'cause we can express these three factors as two factors, can't we? By multiplying both of the unknown factors together to give us one product.

So now we've represented this as a multiplication equation.

Jacob's suggesting that we think about converting it from pounds into pence to make our lives a little bit easier as well.

£2.

50 is the equivalent of 250 pence.

Therefore £22.

50 is the equivalent of 2,250 pence.

At this point we can apply the associative law to make our calculation easier.

If we put the brackets around these two unknown factors, that means we're gonna find the product of these two unknown factors.

So we can rewrite our equation underneath that as the product of those two unknown factors multiplied by 250 is equal to 2,250.

Now, when we have a missing factor, we know that we can find the missing factor by dividing the product by the factor that we do know.

So 2,250 divided by 250.

Hmm, how many 250s are there in 2,250? Well, we know that 10 lots of 250 is 2,500.

So if we take off one group of 250, that would leave us with nine groups of 250 and that would be 2,250.

So we can say that 2,250 divided by 250 is equal to nine.

Right, now we know that the product of those two missing factors in the original equation is nine.

We just need to find the two factors, don't we? So if nine was the product, let's think of some factors of nine.

So it could be one and nine, could there? Or it could have been three and three.

So now you've gotta have a think.

Look at these two trays underneath.

One of them is one row and nine columns, and one of them is three rows and three columns.

Which one of those would be the most suitable to use to bake cupcakes in an oven? That's right, it would be the three by three one.

Jacob saying he's never seen a one row by nine column tray before.

So using a three row by three column tray would be the way to go, wouldn't it? So we're gonna apply that back into our calculation.

So therefore the missing factors in this example were both three and three.

Okay, another check for understanding then.

What could the missing factors be here? I'll give you a moment to think about it.

Okay, let's work through it then.

The missing factors could have been B or D.

So it could have been five and four or 10 and two.

And let's see how we work that one out.

We know that those missing factors could be converted into a product.

So we now got the product of those two missing factors multiplied by 20 is equal to 400.

So that product is now acting as an unknown factor in this equation, isn't it? When we've got an unknown factor, we can divide the product by the factor that we know, can't we? So 400 divided by 20 would leave us with 20.

So now all we need to do is find the factors of 20.

That could be one and 20, two and 10, or four and five.

And as we've pointed out in our answers, the answers were B and D, weren't they? Well done if you got that.

Okay, and now for you to have a go at completing the following task to finish today, could you have a go at completing the following questions? Question one is about cans of fizzy drink and how they're stored in the warehouse at the back of a shop.

And then question two is about finding as many different solutions where one factor and the product is known.

And finally, for question three, could you have a go at creating your own three factor story with one factor known and the product also known? Izzy says she's gonna make this as challenging as she can so she can give it to Jacob and see if he can tackle it for her.

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

Okay, let's work through this first one then.

Which arrangements would be most sensible for the cans to be stacked in? Well, let's think about how we could tackle this.

We could write this as an equation, couldn't we? We don't know how many rows or columns they're gonna be stacked in, do we? But we do know how many cans there are in one box and how many cans there are altogether.

So we can represent this as something multiplied by something multiplied by 12 is equal to 240.

We could then use the associative law to associate the two unknown factors and multiply them together to make an unknown product.

And then we can use that product as a factor in the equation underneath.

So we've got an unknown factor multiplied by 12 is equal to 240.

We know that we have an unknown factor, we can use division.

So 240 divided by 12 is equal to 20, and then we just need to find the missing factors.

And then as that 20 was the product of those two unknown factors, we now just need to think about the different factor pairs that would make 20.

So it could be one and 20, two and 10 or four and five.

We then just need to think about what would be the most appropriate way of stacking the drinks.

And I think by using a four by five arrangement, that would be the best way to make sure we take up as least space on the floor and not go too high before they start falling over each other.

Well done if you managed to get that.

Okay, the product is 140 and if one other factors is two, what could the other two factors be? There are lots of different solutions that you could have had for this.

Following the same principles where we have two unknown factors and then we can multiply them by the two, which would leave us to 140.

We can then associate those two factors and multiply them together to make a product which we can then use as a factor in the equation underneath.

So now we've got an unknown factor multiplied by two is equal to 140.

When we've got an unknown factor, we can do division.

So 140 divided by two is equal to 70 and therefore we need the factor pairs of 70, don't we? So the factor pairs of 70 could be one and 70, two and 35, five and 14 or seven and 10.

Lots of different choices you could have chosen there.

And as Izzy's pointing out, if we didn't know the product and we were trying to work this out, then actually it'd probably be best to use a five and 14 or a seven and 10, as five and 10 are good numbers to work with, particularly if you have a two to multiply them with as well.

Well done if you managed to get any of those.

Okay, and for the last one as well today, creating your own three-factor story where we know one of the factors and we also have the known product.

Here's Izzy's example.

"The volume of the carton of juice is 700 cubic centimetres.

The height of the carton is 20 centimetres.

What is the width and the length of the carton?" That's a great question Izzy, and I wonder if Jacob's gonna be find a way to solve this.

Oh, and here he is now, "Did someone ask for a challenge?" Yep, I think that's you, Jacob, good luck with it.

Okay, and that's it for today's learning.

Thank you for joining me.

Hopefully you're feeling a lot more confident thinking about how you can apply the associative and the commutative laws in our everyday calculations of mathematics that we work with.

So just to summarise what we thought about today, examples where you multiply three factors together can be found in everyday life and we can apply the associative or commutative law to help us solve problems with three factors, can't we? Thank you for joining me again today.

Take care and I'll see you again soon.