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 our understanding of the commutative and associative laws to solving problems that are based around volume.

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

So by end of this lesson today, you should be able to say confidently that I can apply the commutative and associative laws to simplify problems that are involving volume.

Throughout our lesson today, we've got three key words that we're gonna be referring to.

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

Can you have a go at saying them after me? Ready? Represent, associative, commutative.

Well done.

Let's think about what these mean in a bit more detail then.

To represent something is to show it in a different way.

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

For example, it doesn't matter which pair you do first, the result will still remain the same, and this is applicable to both addition and multiplication.

The commutative law states that you can reorder the values in a calculation, and yet the calculation itself will remain the same.

Therefore, the result will remain the same.

And once again, this can also be applied to both addition and multiplication.

Our lesson today is broken down into two cycles.

The first cycle, think about solving volume problems with three factors, and the second cycle, we'll think about solving volume problems with unknown factors.

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

Throughout our lesson today, we've got Lucas and Sam who will be joining us, sharing their thinking and developing their understanding as well as we go throughout this lesson.

Okay, so let's start here first of all then.

A swimming pool is 50 metres long, five metres wide and two metres deep.

What is the volume of the swimming pool? Lucas is saying, "It would help me if I could represent this using an image or a drawing." He says, "I could draw a cuboid to represent the swimming pool." Sam's saying that he's not particularly good at drawing cuboids and was wondering if Lucas could give him a hand in doing so.

Lucas suggests that we start with the front face.

Once you've drawn a rectangle from the front face, you can then draw exactly the same rectangle, but you can draw that above and across given the impression that it is behind the first rectangle.

Now we can start to join up the corners.

And there you go.

We have a cuboid that's been drawn and looks like a 3D cuboid on the page, which could help represent this problem for us.

Okay, quick check for understanding.

Can you finish drawing this cuboid for me? Well done, let's see how you got on.

Sam's saying that he could have drawn it in two different ways.

He could have drawn the back face here like that and then joined up or he could have drawn a slightly longer one.

Exactly the same process.

There's no right or wrong to how you drew your cuboid, as long as it looks like a 3D cuboid on the page.

So let's return to our representation so far.

Lucas has drawn a cuboid and he is now saying we can apply the measurements, can't we? So the swimming pool is 50 metres long, it has a width of five metres, and it has a depth of two metres.

Sam's saying, "Here's my representation of the swimming pool." What do you think about Sam's representation of the swimming pool in comparison to Lucas's so far? Hmm, Lucas is saying that the width and the depth of the swimming pool almost look the same, don't they? Although the width should represent five metres and the depth should represent two metres.

So they should look quite different as well, shouldn't they? So Lucas is saying, "Maybe we should make the height a bit shorter, Sam." So Sam says, "There we go, like this?" That's better, isn't it? Now we can start thinking about the length a bit more.

Bearing in mind the width is five metres, the length should be 50 metres.

So it should look again quite significantly longer than the five metres.

It looks like the 50 metres is barely over double the size of the five metres width, isn't it? So in order to do that, we're going to need to make it a bit longer.

Now, it doesn't need to be perfectly accurate, does it? But it does need to be representative of what the swimming pool might roughly look like.

So therefore, it is important that our representation is a rough idea of what the measurements we want it to be will look like.

And this example is a better example than what Sam had from before.

Okay, so now we've drawn our representation and we've added all the measurements, we can now start thinking about how we can solve this.

So we know that to find the volume of a cuboid, we need to multiply the length by the width by the height.

So the length is 50, the width is five, and the height is two.

So we can use the associative law to associate the five and the two together here, and we've placed brackets around them to help recognise that we're gonna multiply those two together first.

That means five multiplied by two would be equal to 10, and therefore, we then need to multiply that by 50 to find the total volume.

So the total volume of the swimming pool would be 500 cubic metres.

Well done if you got that.

Sam's saying we could have also applied the commutative law here, couldn't we? Currently, the factors are in a position of 50 times by five times by two.

We could have reordered these factors so that it was 50 multiplied by two, multiplied by five.

Then we can associate the 50 and the two together if we wanted to.

That would give us 100.

And then we've got 100 multiplied by five, which would be equal to 500 again.

There we go, using the commutative and the associative law to help us solve problems with volume.

Okay, another check for understanding here.

Which representation best represents the problem? An inflatable paddleboard is packaged in a box which is 180 centimetres long, 50 centimetres wide and 50 centimetres deep.

What is the volume of the box? So look at the images below, which one would best represent the problem that we have above? That's right, it's B, isn't it? We know the width and the depth of the box are both the same.

So we know in this example, the front face is gonna have a width and a height of 50 centimetres, which would make it a square.

And therefore, we would just need to add on the length and the length in this case is 180 centimetres, so it's gonna be a lot longer than it is taller or wider.

Well done if you managed to get that.

Okay, and onto your first task for today, what I'd like you to do is read through the problems and then draw an appropriate representation and mark on the measurements for each one.

Then I'd like to solve the problem by calculating, and it might be useful to use the associative or commutative law to help you with that as well.

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

Okay, welcome back, let's have a look.

So for question one, we've got a book which is two centimetres wide, 30 centimetres tall and 15 centimetres long.

So you can see I've drawn this representation on the right hand side, and the proportions are roughly about correct.

They're not exactly, but they're roughly about correct.

And then, I've got the calculation underneath.

So I've put 15 multiplied by two, multiplied by 30, and we're gonna associate the 15 and the two together 'cause that's gonna leave us with 30 and then 30 multiplied by 30.

Well, I know three times three is equal to nine.

So if 30 times three is equal to 90, then 30 times 30 would be equal to 900.

So the volume of the book would be 900 cubic centimetres.

For question two, a sports coach packs his first aid kit into the boot of his car.

The first aid kit is 40 centimetres long, has a height of 20 centimetres and a width of 15 centimetres.

How much space does it take up in the boot? Well, let's have a look at our drawing again.

We can see the length is 40, the width is 15 and the height is 20.

And to calculate that, we can write an equation which says 40 multiplied by 15, multiplied by 20.

This example, we've associated the 15 and the 20 together.

I know 15 times two is 30, so 15 times 20 would be equal to 300.

And then, we would need to do 300 times by 40.

Well, 300 times by four is 1200 or 1,200.

So 300 times by 40 would be 12,000.

12,000 cubic centimetres.

Well done if you've got both of those.

Okay, so onto our second phase of our lesson then, solving volume problems with unknown factors.

Okay, so let's have a look at this problem then, shall we? The volume of a cuboid is 220 cubic centimetres.

The width of the cuboid is five centimetres.

What could the length and the height be? Hmm, well, how are we gonna work this out? Lucas has got an idea of maybe if you draw it first of all to try and give us a rough flavour of what it might look like.

Well, if we draw a cuboid, we know we can represent it by placing on the width, can't we? Which we know is five centimetres, but the length is unknown and the height is unknown.

So we can represent this as an equation, can't we? We can say something multiplied by something.

Those two somethings represent the length and the height, and then we can multiply those by the width and all of that would be equal to 220 cubic centimetres.

However, if we decided to treat the product of both the height and the length as one number, we could rewrite our equation like this.

So we now have something multiplied by five is equal to 220 cubic centimetres.

And that something has come about from multiplying the length and the height together.

Now we have a missing factor, don't we? And we know if we have a missing factor problem, we can divide the product by the remaining factor to find out the size of the factor that's missing.

For example, I know that two multiplied by something is equal to 10.

Well, if I divide 10 by two, that will give me five, which would be the missing factor.

So let's have a look at this example here then.

We've got 220 as our product, and if we divide that by five, the factor that we have, then we can find out the size of the missing factor.

So 220 divided by five.

We know that there are 20 fives in 100, so there would be 40 fives in 200, and there's an extra four fives in 20.

So if we combine that together, the total number of fives we'd have would be 44 fives, wouldn't there? So there are 44 fives in 220.

So we could say that 44 multiplied by five is equal to 220.

Now remember, this isn't finished here because at the moment, that 44 is representing the product of both the length and the height when they're multiplied together.

So we need two numbers now that multiply together and make 44, don't we? Hmm, can you think of any numbers that might do that? So take a moment for yourself.

What are the factor pairs of 44? Remembering that a factor pair are two numbers that can multiply together to make a product.

That's right, one and 44 would be a factor pair, two and 22 would be a factor pair, four and 11 would also be a factor pair.

So what could the missing height and the missing length be? Well, we know the factors are 44 or one and 44, two and 22 and four and 11.

So it could be any one of these pairs, couldn't it? We've opted for the four and the 11, and it doesn't matter which order we place those into our calculation 'cause as we know, using the commutative law, any of the factors can be placed in any order we like.

So we've just placed the four to multiply by the 11 to multiply by the five.

And again, that would give us our volume of 220 cubic centimetres.

Well done if you managed to get that.

Lucas is now saying, well, actually he needs to adjust his representation slightly here, doesn't he? 'Cause this isn't representative of what the cuboid would actually look like.

We know it has a length of 11 centimetres and a height of four centimetres.

So we're gonna have to change our cuboid to now look a little bit more like this.

There we go.

That's more representative of the cuboid from the problem.

Well done, Lucas.

Okay, check for understanding now.

What could the missing factors be? Something times something multiplied by five is equal to 50.

Take a moment to have a think.

That's right, it could be C or D, it could be two and five, or it could be one and 10.

Let's see how we work that out.

Well, we know we can convert those two factors into a product of those two factors.

So now our equation becomes something multiplied by five is equal to 50.

We know that 10 multiplied by five is equal to 50.

So if that product is 10, we need to find the factor pairs of 10.

That could have been one on 10 or it could have been two and five.

Therefore, these are the missing factors that could be put in place into the original equation.

Well done if you've got those.

Time for you to have a go.

Now here's some different contexts with some missing lengths, widths or heights potentially.

What I'd like you to do is work through these and try and find the multiple solutions that it could be for each one of these.

You might need to think about the context a little bit and think about whether the numbers you give would be suitable.

For example, if it was a swimming pool context and you were finding the missing length and you decide the length was one metre long, well actually, would a swimming pool ever be one metre long? It doesn't sound very good, does it? So you might need to think about the context a little bit before you put your answers in.

Once you tackle question one and two, I'd like you to have a look at number three.

And here, you'll need to sketch several possible solutions for what the room could look like.

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

Okay, so for task one, a pack of cards has a width of two centimetres.

The volume of the pack is 126 cubic centimetres.

What could the length and the height be? Well, the answers to this, the length would be seven and the height would be nine, and let's walk through how we can work this out.

We know we can represent this as an equation.

Something multiplied by something multiplied by two is equal to 126.

And again, if we combine those two unknown factors as a product, we now have one missing factor for the next calculation.

So we've got something multiplied by two is equal to 126.

We now need to divide the 126 by the two.

This will give us 63 because if we know the product and we know one factor, we can divide the product by the remaining factor to find the missing factor.

So if 63 is the size of the missing factor from the equation above it, then we need to find the factor pairs for 63, don't we? That could be one in 63, it could be three and 21, or it could be seven and nine.

Now we have to think about what would be the most appropriate for a pack of cards.

Would it be appropriate for a pack of cards to be 63 centimetres tall or would it be appropriate for a pack of cards to have a length of three centimetres? No.

What is important that we do know that a pack of cards is slightly taller than it is longer.

So the best numbers to use here would be the seven and the nine.

And therefore, we would give the length of the pack of cards seven centimetres and the height we would give the nine centimetres.

Well done if you managed to get that one.

Okay, and onto number two then.

A small cereal packet has a length of 20 centimetres.

The volume of the packet is 2,800 cubic centimetres.

So what could the width and the height be? Well, we've gone forward the width being five centimetres and the height being 28 centimetres.

It could also have been seven centimetres in width and 20 centimetres in height.

Or it could have been 10 centimetres in width and 14 centimetres in height.

Lots of different solutions there that you might have come up with.

You might like to extend yourself with drawing each one of those boxes with a different length, width and height and seeing which you think most looks like the best cereal packet to have.

Okay, and then for question three, an architect is designing rooms in a new house.

The width of the room has to be four metres.

The volume of the room is 60 cubic metres.

Sketch all the possible solutions for what the room could look like.

So here are some examples.

We knew that the width had to be four metres, didn't we? And therefore, we needed two other numbers that would multiply with the four, which would then allow us to have a volume of 60 cubic metres.

Those two numbers were three and five.

So the way we could organise this is the length of the room could be five metres and the height could have been three metres, or the length could have been three metres and the height could have been five metres.

Which room would you prefer to have and why? Hopefully you managed to get both of those and well done if you did.

Okay, that's end of our lesson for today.

Thanks for joining me, and hopefully you're feeling more confident about applying the commutative and associative law to help us solve volume problems. Let's summarise what we've just learned today.

You know that you can represent volume problems by drawing a 3D image of the object, and you can also apply both the commutative and associative laws to help simplify the calculation to help you find the volume or find a missing dimension if the volume is known.

Well done today, take care of yourself and we'll see you again soon.