Lesson video

Hello, welcome to today's maths lesson.

My name is Dr.

Shorrock.

I am really excited to be learning with you today.

We are going to have great fun as we move through the learning together.

Welcome to the lesson today.

Today's lesson is from our units, compare and describe measurements using knowledge of multiplication and division.

This lesson is called, solve comparison and change problems using division.

As we move through the learning today, we will be looking at problems where we compare two lengths or describe the change in one length.

Sometimes new learning can be a little bit tricky, but it's okay because I'm here to guide you, and I know if we work really hard together, then we can be successful.

So, shall we find out how do we solve comparison and change problems using division? These are the keywords that we will be using in our learning today.

We've got, comparison and change.

Let's practise those words together.

My turn, comparison.

Your turn.

Nice.

My turn, change.

Your turn.

Brilliant.

Look out for those words as we move through the learning today.

So when we make a comparison, we are determining how different two objects are.

In the case of this lesson, we're thinking about how many times longer, taller or deeper one object is than another.

But we can also make a comparison between an object before and after a change.

Examples of such a change include the change in height of a flower due to growth, or change in the depth of a puddle due to rainfall.

Today, we are going to deepen our understanding of solving comparison and change problems using division.

We are going to start by looking at how we represent comparison and change problems. We have Lucas and Sam to help us with our learning today.

Let's look at this change problem.

So, a pencil was 20 centimetres long when it was new.

Lucas has been writing with the pencil now and it is now one-quarter times its original size.

Can you visualise that? What do you see in your head? That's what I see, I've got my new pencil, which is longer, and I've got my shorter pencil now.

What might the question be do you think here? That's right, how long is the pencil now? So we know the pencil was 20 centimetres long when it's new but we don't know the length of the pencil now.

That's what we are going to find out.

So, we know that the pencil was 20 centimetres long when it was new, so that's where we start.

And Lucas has been writing with the pencil and it's now one-quarter times it's original length.

So, I'm going to represent that in my table with an arrow and multiply by one-quarter.

And then we want to find out how long is the pencil now, I'm representing that with a question mark.

We can also represent change problems as a bar model.

So if we remember, our pencil was 20 centimetres long when it was new, and Lucas has been writing with it, so it's now one-quarter times it's original size.

And we want to find out how long the pencil is now.

So, the pencil was 20 centimetres long when it was new.

So this must be our whole because it is the original length.

So, I've got my whole in my bar model, and I've labelled that with 20 centimetres.

And Lucas has been writing with the pencil, and it's now one-quarter of its original size.

So there must be four equal parts.

And how long is the pencil now? This will be the unknown part, because it's smaller in length and it will be one of those four equal parts.

Let's check your understanding.

A 42-centimeter tower of bricks is knocked down to one-half times its original size.

Oh dear.

What is the height of the tower now? Can you represent this change problem in a table and as a bar model? Pause the video while you draw those, and when you are ready to look at the answers, press play.

How did you get on? 42 centimetres was how tall the tower of block was to start with, and it was knocked down to one-half times its original site.

So that's represented by an arrow and multiplying by one-half.

And the question mark represents the height of the tower now.

We can also represent this in a bar model.

The whole tower was 42 centimetres tall and the height of the tower now is one-half times the original height.

So, I've split my whole bar into two equal parts and a question mark in one of those parts.

Let's look at a comparison problem.

A door in Sam's house is two metres tall.

Sam is going to make a door for her toy house.

The height of the toy house door needs to be one-10th times the height of the actual door.

Can you visualise that? What do you see in your head? That's what I see, I've got a door in my house and I've got a toy house door.

What might the question be? That's right, how tall does the model door need to be? What are we going to do? That's right, thank you, Lucas.

"To solve a comparison problem, it is useful to represent it as a bar model." So, if we have a look at our problem, we've got the actual door in Sam's house is two metres tall, and the height of the toy house door needs to be one-10th times the height.

And we want to work out how tall that toy house door needs to be.

So, I've got my bar model, and the actual door must be the whole because it's the original length and it's the greater length.

So I've labelled my bar model with two metres.

And then, I know the height of the toy house door is one-10th, so there must be 10 equal parts.

And you can see my bar model has been split up into 10 equal parts.

And then the height of the toy house door is the unknown part because it's smaller in height and it is equivalent to one of those equal parts.

Let's check your understanding.

Can you represent this comparison problem as a bar model? The height of a bush is one-third times the height of a tree.

Pause the video while you draw your bar model, and when you are ready to see the answers, press play.

How did you get on? This is what my bar model looks like.

I represented the height of the tree as my whole, and I divided the whole into three equal parts.

And the bar for the bush is equivalent to one of those parts because the height of the bush is one-third times the height of the tree.

Well done! Your turn to practise now.

For question one, could you compare the height of the house and the block of flats? Which sentences could describe the image? So maybe put a tick or a cross.

So, the house is three times the height of the tower block.

The tower block is three times the height of the house.

The house is one-third times the height of the tower block.

The tower block is one-third times the height of the house.

The height of the tower block multiplied by one-third is equal to the height of the house.

For question two, could you represent these change problems in a table and as a bar model? So for part A, the rainwater in a puddle was 10 centimetres deep.

Over the course of the day, the water evaporated until it was one-half times the depth.

What is the depth of the water in the puddle now? Part B, Lucas built his snowman that was one-meter 20 centimetres tall.

As the day went on, it melted to one-quarter times the height.

How tall is the snowman now? And for part C, a chocolate bar is 110 millimetres long.

Sam eats some of the chocolates so that it is one-10th times the length.

How long is the chocolate bar now? And for question three, could you represent this comparison problem as a bar model? Snail A travels down the garden path.

Snail B travels one-quarter of the distance of snail A.

And snail C travels three times the distance of snail B.

Which snail travels further? And Sam's given us a good clue there, "Try visualising this first." Or maybe drawing something to help you, and then have a go at the bar model.

Pause the video while you have a go at all three questions, and when you are ready to go through the answers, press play.

How did you get on? Let's have a look.

So, the first question you had to compare the height of the house and the block of flats.

So, the house is three times the height of the tower block.

Well, that can't be correct, can it? Because the house isn't as tall.

The tower block is three times the height of the house.

Well, that could be correct because that tower block is taller.

The house is one-third times the height of the tower block.

That could be correct.

The tower block is one-third times the height of the house.

Mm, no, because the tower block is taller so it can't be one-third times the height of the house or it would be shorter.

The height of the tower block multiplied by one-third would be the height of the house.

Well, that's right, because the house could be one-third the height of the tower block.

For question two, you were asked to represent these change problems in a table and as a bar model.

So part A, you had a question about rainwater in a puddle that was 10 centimetres deep.

And so, that's where we are starting.

And the water evaporates as one-half times the depth.

So, I've represented that with an arrow and multiplying by one-half.

And the question mark represents the question, what is the depth of the water in the puddle now? And we could also represent that as a bar model.

The whole is 10 centimetres, the whole has been divided into two equal parts and the depth of the puddle now is equal to one of those parts.

For part B, question about the snowman.

I could say that my snowman was one-meter 20 centimetre tall to start, and it melted to one-quarter times the height.

So, that is represented by an arrow multiplying by a quarter.

And the question mark represents the height of the snowman now, and we can also represent that in the bar model.

The whole is one-meter 20 and it has been divided into four equal parts, and the height of the snowman is equivalent to one of those parts.

For part C about the question about the chocolate bar.

The chocolate bar was 110 millimetres long.

And Sam ate some of it, so that it is one-10th times the length, so we are multiplying by one-10th.

And the question mark represents the length of the chocolate bar now.

We can represent that in the bar model.

The whole chocolate bar is 110 millimetres long, and it has been divided into 10 equal parts, and the chocolate bar is equivalent to one of those parts.

For question three, you were asked to represent this comparison problem as a bar model.

So, snail A travels down the garden path.

Snail B travels one-quarter of the distance of snail A.

So you can see, I've divided the whole into four equal parts and snail B travels the equivalent of one of those parts.

Snail C travels three times the distance of snail B.

Which snail travels further? And you can see from our representations quite clearly which snail travels further.

Snail A travels further.

Well done on your learning so far, you have made great progress with representing comparison and change problems. We're going to progress our learning onto looking at how we solve them now.

Let's revisit our change problem.

A pencil was 20 centimetres long when it was new.

Lucas was writing with that pencil, wasn't he? And it's now one-quarter times its original size.

And we want to find out how long the pencil is now.

So we can use our representations to help us.

Looking at the table, we know we need to multiply 20 centimetres by one-quarter.

And from the bar model, we can also see that we've divided the whole into four equal parts, so we need to divide by four.

And we know that multiplying by a unit fraction is the same as dividing by the denominator.

So, to calculate 20 centimetres multiplied by one-quarter, it is the same as dividing 20 centimetres by four.

And we can use our known facts to help us.

20 centimetres divided by four is five centimetres.

So, that means 20 centimetres multiplied by one-quarter is also five centimetres.

So the pencil is now five centimetres long.

Let's look at these equations in more detail.

What does the 20-centimeter represent? That's right, thank you, Sam.

"The 20 centimetres represents the original length of the pencil." What about the one-quarter or the divide by four, what do they represent? Ah, thank you, Sam.

We know that multiplying by one-quarter is the same as dividing by four.

And what about the five? What does the five represent? The five represents the length of the pencil now.

Let's check your understanding by revisiting this problem.

A 42-centimeter tall tower of blocks is knocked down to one-half times its original height, and we want to know what the height of the tower is now.

Use the representations to write an equation and then solve it.

Pause the video while you're doing that, and when you are ready for the answers, press play.

How did you get on? The equation is 42 centimetres multiplied by one-half, and we know that's the same as dividing by two, 42 centimetres divided by two is the same as halving, that's 21 centimetres.

And that also means that 42 centimetres multiplied by one-half is 21 centimetres.

So, the tower is now 21 centimetres tall.

Well done if you've got that.

Let's revisit our comparison problem.

If we remember, a door in Sam's house is two metres tall.

And Sam is going to make a door for her toy house.

The height of the toy house door needs to be one-10th times the height of the actual door.

And we want to work out how tall the model door needs to be.

So we can use our bar model representation to form an equation to help us solve the problem.

And looking at the bar model, we can see that we need to multiply two metres by one-10th because we have divided the bar, the whole bar, into 10 equal parts.

So, multiplying by aunit fraction is the same as dividing by the denominator.

So, we've got two metres and we're multiplying by one-10th.

But that is the same as dividing two metres by 10.

So, two metres divided by 10.

Well, I think I agree with Sam, it will definitely be easier to convert to centimetres.

Two metres is the same as 200 centimetres.

And I can do that a lot more easily.

200 centimetres divided by 10 is 20 centimetres.

So, two metres multiplied by one-10th is also 20 centimetres.

So the toy house door needs to be 20 centimetres tall, and 20 centimetres is one-10th times the size of two metres.

Let's check your understanding.

Let's revisit this problem.

The height of a bush is one-third times the height of a tree.

What is the height of the bush? So could you use the bar model representation to write an equation and solve it? Pause the video while you do that, and when you are ready for the answers, press play.

How did you get on? Did you write the equation? 360 centimetres, which is our whole, and we're multiplying by one-third because the height of the bush is one-third times the height of the tree.

And we know that when we multiply by a unit fraction, one-third in this case, it is the same as dividing by the denominator, which is three in this case.

So 360 centimetres multiplied by one-third is the same as 360 centimetres divided by three.

And I can use my known facts to help me.

36 divided by three is 12.

So 360 divided by three is 120.

And that 360 divided by three is the same as 360 multiplied by one-third, so that must also be 120.

So the height of the bush is 120 centimetres.

Well done.

Let's look at an equation without a context.

What can we say from that equation? Ah, thank you, Sam.

We can say that 21 multiplied by one-third is equal to seven.

Anything else? Ah, that's right, thank you, Lucas.

We can say that seven is one-third times the size of 21.

And we know that when we multiply by a unit fraction, it is the same as dividing by three.

So as Sam says, we also know that 21 multiplied by one-third is the same as dividing 21 by 3.

21 divided by three is seven.

Let's check your understanding on this.

Which of these statements correctly describes this equation? Is it A, seven is one-sixth times the size of 42? Is it B, 42 multiplied by one-sixth is equal to seven? Or is it C, 42 is one-sixth times the size of seven? Pause the video, maybe find someone to talk to about this.

And when you are ready for the answers, press play.

How did you get on? So statement A is correct, seven is one-sixth times the size of 42.

Because if we do 42 multiplied by one-sixth, it's the same as dividing 42 by six, which is seven.

And then also, 42 multiplied by one-sixth is equal to seven is correct as well.

So, well done if you've got those.

Your turn to practise now.

For question one, I'd like you to use your representations from task A and form an equation, and then solve these change problems. So the problems are the same as they were in task A.

In task A, you represented them, but in task B now here, I would like you to form an equation and solve them.

So let's remind ourselves of part A.

The rainwater in a puddle was 10 centimetres deep.

Over the course of the day, the water evaporated until it was one-half times the depth.

And what's the depth of the water in the puddle now? For part B, Lucas built a snowman that was one-meter 20 centimetres tall.

As the day went on, it melted to one-quarter times the height.

How tall is the snowman now? And for Part C, a chocolate bar is 110 millimetres long.

Sam eats some of the chocolate bar so that it is one-10th the length.

How long is the chocolate bar now? For question two, using your representations from task A and the information given here, can you calculate how far snail B and C travelled? I'm telling you that snail A travels 240 centimetres down the garden path, and we know snail B travels one-quarter of the distance of snail A, and snail C travels three times the distance of snail B.

And for question three, could you complete these statements? For part A, nine is mm times the size of 63.

For part B, 192 multiplied by mm is equal to 48 or 192 divided by mm is equal to 48.

And for part C, 28 multiplied by mm equals four, and 28 divided by mm equals four.

And so, four is mm times the size of 28.

Pause the video while you have a go at all three questions and when you're ready for the answers, press play.

Shall we see how you got on? For question one, you were asked to use your representations to form an equation to solve these problems. The first question was about rainwater in a puddle.

So, we know it was 10 centimetres deep to start with, and the water evaporated until it was one-half times the depth.

So we know we are multiplying by one-half.

And we know that when we multiply by a unit fraction, it is the same as dividing by the denominator.

So, 10 centimetres multiplied by one-half is the same as 10 centimetres divided by two, which is five centimetres.

So the puddle is now five centimetres deep.

Part B, the problem about the snowman.

So, the snowman was one-meter 20 centimetres tall.

So, that is how we're going to start our equation.

And we know it melted to one-quarter times the height, so we need to multiply by one-quarter.

And we know that if we multiply by one-quarter, it is the same as dividing by four.

120 centimetres divided by four is 30 centimetres.

So the snowman is now 30 centimetres tall.

For Part C, the question about the chocolate bar, the chocolate bar is 110 millimetres long.

So, that is where I'm going to start my equation.

Some of the chocolate bar is eaten so that it is one-10th times the length.

So we know we are multiplying by one-10th, and that is the same as dividing by 10.

110 millimetres divided by 10 is 11 millimetres.

So, the chocolate bar is now 11 millimetres long, and that is equivalent to one centimetre, one millimetre.

For question two, you were asked to use your representations from task A and the information given here to calculate how far snail B and snail C travelled.

I'm telling you that snail A travels 240 centimetres down the garden path, and this is our whole.

We know that snail B travels one-quarter of the distance of snail A.

So, I have divided my whole into four equal parts and the part for snail B is equivalent to one of those parts.

We've got 240 centimetres, which is our whole.

Snail B travels one-quarter of the distance, so we are multiplying by one-quarter, which is the same as dividing by four.

240 divided by four is 60 centimetres.

We know this because we can use our known facts.

24 divided by four is six, so 240 divided by four must be 60.

So snail B troubled 60 centimetres.

If we think about snail C, snail C travels three times the distance of snail B.

So snail B was 60 centimetres, we know snail C, it travels three times that, so we've got 60 centimetres multiplied by three.

Well, six threes are 18, so 60 threes must be 180.

So, snail C travelled 180 centimetres, and that is equivalent to one-meter 80 centimetres.

For question three, you were asked to complete some statements.

So for part A, 63 multiplied by one-seventh is equal to nine.

And we know that if we are multiplying by a unit fraction, it is the same as dividing by the denominator.

So that is the same as 63 divided by seven is equal to nine.

And so, nine is one-seventh times the size of 63.

For part B, 48 is one-quarter times the size of 192.

So we could do 192 multiplied by a quarter is equal to 48, or we know we could divide by four and that would be equal to 48.

And for part C, 28 multiplied by one-seventh is equal to four.

And so, 28 divided by seven must also be equal to four.

And we can say four is one-seventh times the size of 28.

Fantastic progress in your learning today.

I am really impressed with how far you have come in your ability to solve comparison and change problems using division.

A change in length can be described using our knowledge of division, and we can say that mm is mm times the size of the mm, where multiplication by a unit fraction represents the division.

And comparison and change problems can be represented visually in tables and or as bar models to help solve the problem.

So really well done today, it's been a pleasure learning with you, and I look forward to learning with you again soon.