Lesson video

Hello, how are you today? My name is Dr.

Shorrock, and I'm really excited to be learning with you today.

You have made a great choice to learn maths with me, and I'm here to guide you through the learning.

Welcome to today's lesson.

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

This lesson is called: Solve comparison and change problems in a range of contexts.

As we move through the learning today, we will look at strategies that we can use to solve a variety of problems that are related to different measures, such as area of an object or the length of an object.

Sometimes new learning can be a little bit tricky, but I know that you are going to work really hard today, and by working really hard, we can be successful.

And remember, I'm here to guide you through the learning.

So let's get started, shall we? Let's have a think about how can we solve comparison and change problems in a range of contexts.

These are the keywords that we will be using in our learning today.

We've got bar model, multiplicative, and times the mm.

You might be familiar with some of those words, but it's always useful to practise saying them aloud.

So let's have a go at practising them together.

My turn, bar model.

Your turn.

Lovely.

My turn, multiplicative.

Your turn.

Brilliant.

Well done.

Tricky one that one, isn't it? And then my turn, times the mm.

Your term.

Fantastic.

So when we talk about a bar model, we mean a pictorial representation of a problem, and bars are used to represent known and unknown quantities.

Usually we use bar models to solve number problems that are related to the additive or multiplicative relationships.

A multiplicative relationship is a comparison between numbers that relates to multiplication, and therefore it also relates to division.

In a multiplicative relationship, the parts are equal.

In this example, we can see here, we can compare the multiplicative relationship between 10 and five.

We can say that 10 is two times the size of five, or 10 is twice five, because two equal parts of five are the same as 10.

We could also say that five is one half times the size of 10.

And then times the mm, well that's a phrase we're going to be using throughout this lesson and we use it to compare measures multiplicatively.

We've got the word times in there, so that suggests we are comparing multiplicatively, so look out for those words.

And for example, one tree might be three times the height of another tree, it's three times as tall.

So today we are going to deepen our understanding of solving comparison and change problems in a range of contexts.

Let's start our learning by looking at how we can represent problems pictorially.

These are the children who will help us in our learning today.

We've got Lucas, Sam, and Izzy.

When we are given a word problem to solve, it is important to be able to visualise it, to show we can see the math in the real world, and visualise it means just see it in your head.

For example, if we had a problem about the difference in mass between a whole cake and a slice of cake, could you visualise it? What do you see in your head? Can you see a whole cake and can you see a slice of cake? So that's an important skill, to be able to visualise the words from the problem.

This is what I see, there's my whole cake and my slice of cake.

And then, once we've visualised it, we need to represent the word problem mathematically.

Because this then supports us to form an equation which we can then solve.

So I'm going to represent the word problem in a bar model.

So let's practise that by looking at this problem.

Some year five children collected data on the number of fans who attended football matches that weekend.

Can you visualise that? I wonder if any of you have ever been to a football match in a stadium.

What does it look like? What do you see? Ah, there's a picture of a football stadium.

It's an empty football stadium, isn't it? How many people do you think might fit in that stadium? It's a good practise at estimating there.

The year five collected the data on the number of fans that attended football matches, and here are their results.

They've presented their results in a bar graph.

What can you see from the graph? What can you tell? That's right, the year five children, well, by looking at this bar graph, they could see the difference in football match attendance.

Can you see which football match had the greatest attendance? How can you tell? That's right, it's got the longest bar, hasn't it? What about the football club that had the lowest attendance? That's right, you could tell that as well, couldn't you? Luton Town, because it's got the shortest bar.

Let's compare the attendance at the Manchester United game to the Crystal Palace game.

And really, we need to start by identifying the key information.

Have you spotted what the key information is? That's right, we need to know which clubs we are comparing.

So Manchester United, and we can find Manchester United on our graph, it's the first bar, and then we can use the bar to determine how many fans attended the match.

Hmm, but it's more than the scaled number, isn't it? It's more than the highest value of 70,000.

But what do you notice? That's right, the intervals are increasing by 10,000 each time, and each interval is divided into two equal parts.

So each part must be worth 5,000.

So the number of fans who attended the Manchester United game must be 75,000.

So 75,000 people attended the Manchester United game.

Then we were also asked to compare the attendance at Manchester United to the Crystal Palace game.

So we now need to look at the Crystal Palace game.

We can see from the bar from Crystal Palace, the attendance is halfway between 20,000 and 30,000, and we know each part is worth 5,000, so 25,000 people attend the Crystal Palace game.

So once we've used that key information, once we've found the key information, we can then represent it in a bar model.

I think Lucas is going to help us.

So Lucas is telling us, "Manchester United have the greater attendance." So they are our whole, and they are a known whole.

We know how many people attended the Manchester United game, because we've read that from the graph, 75,000 people.

And then Crystal Palace, well they have the smaller attendance, so that is our known part.

We have now represented this problem pictorially in the bar model.

So we could then use this to form an equation to solve.

We could form an equation to compare the values additively, and find how many more people attended Manchester United's game.

Or we could compare the values multiplicatively, and find out how many times greater the attendance was at Manchester United's game.

So there's two equations that we could form there depending on what we want to find out.

Let's check your understanding.

Could you look at this graph, and then choose a bar model which represents a comparison in attendance between Aston Villa and the Luton Town matches? Pause the video, maybe compare what you think with a friend, and justify why, and when you are ready, press play.

How did you get on? Did you realise it must be bar C? It can't be bar A, because Aston Villa was 40,000 and Luton town is 10,000.

So the 25,000, well that is Crystal Palace, so we've used the wrong club there.

B, it can't be bar B because Crystal Palace again, was 25,000, not Luton Town, so it must be C.

We've got Aston Villa with 40,000 and we have got Luton Town with 10,000.

Well done if you chose bar C.

Let's look at a different problem now.

For their geography learning, some children estimated the distance all the way around Earth.

So all the way around the equator.

Can you visualise that? Can you visualise Earth? Can you see the equator go all the way around? What do you see in your head? This is what I see.

There's my image of Earth, and the equator going all the way around, and the children are estimating that distance all the way around Earth.

And here are their results.

So you can see, the children have presented their data in a table.

So Izzy, she estimated 8,000, and it's kilometres, we can see from the title of the heading.

Lucas, 16,000, Jacob 24,000 and Sam, 48,000.

What could the year five children do with this data? What could they compare? That's right, they could compare their different estimates to see how different they are to one another.

They could do it additively, or they could do it multiplicatively.

And I'm going to tell you actually, because I know I would want to know this, the actual distance all the way around the equator is about 40,000 kilometres.

It's still an estimate, but it's about 40,000 kilometres.

What could we do with that piece of information? That's right, the year five children could also compare their estimate to the real distance, to the real estimated distance.

So let's have a go.

Let's compare the actual estimate of 40,000 kilometres to Izzy's estimate of 8,000 kilometres.

Yes, good thinking, Izzy, we really should.

Once we visualise it, we should always represent the information in a bar model.

The actual estimate is the greater value, and this is our whole, and it is a known whole, we know it is 40,000 kilometres.

And we know Izzy's estimate is a smaller value, so it is our known part.

We have now represented this problem pictorially in our bar model.

We could then use this bar model to form an equation which we could then go ahead and solve.

We could compare the values additively, and find how many more kilometres the actual estimate is to Izzy's estimate.

Or we could compare the values multiplicatively, and find out how many times greater the actual estimate is to Izzy's estimate.

Let's check your understanding.

Could you look at this table? Could you then draw a bar model to represent a comparison between Izzy and Jacob's estimates? Pause the video while you draw a bar model, and when you are ready to see the bar model that I drew, press play.

How did you get on? This is the bar model that I drew.

Jacob has the larger value of 24,000, so that is our whole, and Izzy had the smaller value of 8,000, so that is a part of the whole.

Your turn to practise now.

For question one, could you represent this information in a bar model? So we've got a table which shows the approximate areas of the countries in the United Kingdom.

Could you draw a bar model to compare the area of Wales and England.

For part B, Lucas's wardrobe is 58 centimetres in width.

He has made a model wardrobe that is 58 millimetres wide.

Could you draw a bar model to compare the width of the wardrobe and the model? For question two, could you represent this problem in a bar model? 25 grammes of strawberries are in a small carton.

Can you visualise that? A small carton is just like a little plastic box.

And you've got 0.

5 kilogrammes of strawberries in a large box.

Can you visualise that? Could you then draw a bar model to compare the mass of the carton and the box? Have a go at questions one and two, pause the video, and when you are ready to go through your answers, press play.

How did you get on? So for question one, you were asked to represent information in a bar model.

And for part A, you had to draw a bar model to compare the area of Wales and England, and you had to find those values reading that table.

We can see the value for England was 140,000 kilometres squared, and for Wales, 20,000 kilometres squared.

140,000 is larger than 20,000, so that is our known whole.

And the value for Wales, 20,000, is a part of the whole.

For question B, the question about the wardrobe and the width of the wardrobe.

Well 58 centimetres was our whole, because it is the larger value, and then the part of the whole is the width of the model, which is 58 millimetres wide.

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

We know that 0.

5 kilogrammes, well that's the same as 500 grammes, so it's larger than 25 grammes, so it must be our whole, and then the 25 grammes is a part of the whole.

How did you get on with drawing your bar models? Well done.

Fantastic learning so far.

You are really deepening your understanding of how we can represent problems pictorially, and the real importance of visualising first, and then representing in a bar model.

Because as you'll see now, we can use that bar model to form and solve equations.

So let's revisit our first problem, and use the bar model that we made to form some equations and to compare the attendance at those football games.

If you can remember Manchester United, there were 75,000 people who attended their match, and Crystal Palace was 25,000 people.

So we could look at this additively, and determine how many more people attended Manchester's game, and we can do that by finding the unknown part, and there is one missing part.

To find a missing part, we subtract the known part from the whole.

75,000, subtract 25,000, that is 50,000.

So 50,000 more people attended the Manchester United game.

We can also compare the attendances multiplicatively to determine how many times the attendance is at one match than the other.

So we can use our bar model to form a multiplication equation.

25,000 is my smaller part, and I want to know what do I times it by to get 75,000.

There is one way of doing this.

We could just count up in 25,000.

If we count up in 25,000 once, that's 25,000.

25,000 times two is 50,000 25,000 times three is 75,000.

So we can say that 75,000 is three times 25,000.

Oh, thank you Izzy, for prompting us to think deeply.

Could we have formed a different equation, do you think? That's correct, we could use the bar model, and this time we could form a division equation.

How many 25,000s are in 75,000? That's what we want to find out.

So 75,000 divided by 25,000.

Well to determine this, we need to work out how many 25,000s are in 75,000.

We can count up in 25s to calculate this.

So 25, 50, 75, there are three 25s in 75, which means there must be 3 25,000s in 75,000.

If we look at both those equations that we formed from the bar model, 75,000 divided by 25,000, well that was three, and also, we worked out that we had to multiply 25,000 by three to get 75,000.

So if we look at both equations, we can see that the attendance at Manchester's game was three times the attendance at Crystal Palace.

And we worked out that there are three 25,000s in 75,000.

So 25,000 must be one third of 75,000.

So we could also say that the attendance at Crystal Palace was one third times the attendance at Manchester.

Let's check your understanding with that.

Which of these equations can be formed from the bar model? You've got a bar model showing the attendance at Aston Villa and at Luton Town.

Pause the video while you have a look through all options, and when you think you know which ones are correct, press play.

How did you get on? A must be correct, because we can multiplicatively compare 40,000 and 10,000.

So what do we multiply 40,000 with to get 10,000? B can't be correct, because that would lead to a larger number.

We are multiplying 40,000 by 10,000, that would be much larger.

C, well that's correct.

How many 10,000s are in 40,000.

So 10,000 times what is 40,000? Well D, we could also do that.

That describes the additive relationship.

If we subtract 10,000 from 40,000, we would find the missing part.

We would find how much more 40,000 is than 10,000.

E, that's correct, 10,000 add what would give us 40,000? Ah, F is correct as well.

It's also describing the multiplicative relationship.

40,000 divided by 10,000, so how many 10,000s are in 40,000? Well done if you've got all five of those.

Let's revisit our other problem, where children estimated the distance around the equator.

We can use the bar model that we formed to form some equations to compare the actual estimate and Izzy's estimate.

So let's have a look at the bar model and how we can use it to help us.

We could compare the estimates additively, by calculating the difference between the values.

There is a missing part.

To find a missing part, we subtract the known part from the whole.

40,000 subtract 8,000.

Well, I know 40 subtract eight is 32.

So 40,000 subtract 8,000 must be 32,000.

So the actual estimate is 32,000 kilometres more than Izzy's estimate.

We could also compare these estimates multiplicatively.

And Izzy has a go.

So we need to form a multiplication equation this time.

8,000 multiplied by what would be equal to 40,000.

Well we could calculate this by counting up in 8,000s until we reach 40,000.

So 8,000 times one is 8,000.

8,000 twice 16,000, 8,000 three times 24,000, 8,000 four times is 32,000, and 8,000 five times is 40,000.

So we could say that 40,000 is five times larger than 8,000.

But is there a different way we could have calculated this? Did we need to count up in 8,000s? Ah, thank you, Sam.

That's right.

We could have used our known facts.

Which known facts do we know? Ah, we know eight fives are 40, so 8,000 fives must be 40,000.

So here again we can see that 40,000 is five times greater than 8,000.

But could we have formed a different equation from our bar model? A different equation, so not a multiplication equation.

That's right, we could have formed a division equation.

So we've got 40,000 divided by 8,000.

So how many 8,000 are in 40,000? And we can use known facts to calculate this.

We know 40 divided by eight is five.

So 40,000 divided by 8,000 must also be five.

If we look at both multiplicative equations that we have created, 40,000 divided by 8,000 was equal to five, and 8,000 times five was equal to 40,000, well, from both equations we can see that the actual estimate of 40,000 is five times Izzy's estimate.

We know there are five 8,000s in 40,000.

So 8,000 must be one fifth of 40,000.

So we could multiply 40,000 by one fifth, and we would get 8,000.

So we can also say that Izzy's estimate, it was 1/5 times the actual estimate.

So from that one bar model, there are lots of equations that we could form.

We could form an additive relationship equation to find how many more.

We could then have created two multiplicative equations, one that was a multiplication and one that was a division.

Let's check your understanding with this.

Could you look at the bar model, which equations could be used to compare these amounts multiplicatively.

Pause the video, have a look through all the six options.

When you are ready and you think you know, press play.

How did you get on? Equation A is correct, because that is one way we could describe this amount multiplicatively.

What do we multiply 24,000 by to give us 8,000? It's a smaller amount, so we know it would be a fraction.

What about part B? No, that can't be correct, because then it would give us a greater whole.

Part C there.

That is correct.

What do we multiply 8,000 by to get 24,000? Part D, well that can't be correct because that's not a multiplicative equation, that is for determining an additive relationship, and the same with E.

But 24,000, that is correct, we could form a division equation.

How did you get on with those? Well done.

So it's your turn to practise now.

We're going to revisit these questions from task A, but this time, I want you to use your bar models from task A, form equations and then solve the problem.

So part A is about comparing how many times the size Wales is than England? And for part B, Lucas' wardrobe is 58 centimetres in width, and he has made a model that is 58 millimetres wide.

Can you tell me and work out how many times the width of the wardrobe is in comparison to the model? Be careful.

What do you notice with the units in part B? For question two, could you use the bar model that you drew in Task A, to form an equation for this problem and solve it? 25 grammes of strawberries are in a small carton, 0.

5 kilogrammes of strawberries are in a large box.

How many times the small cartons can make one large box? Pause the video while you have a go at the questions, and when you are ready for the answers, press play.

How did you get on? You were asked to use your bar model from task A to form equations.

We are trying to find how many times the size of Wales is England.

And times the size, so that tells us we need a multiplication equation 20,000 times something is 140,000, and we need to find that missing factor.

We can use our known facts, two sevens are 14.

So we now know that 20,000 sevens would be 140,000.

So we have got seven parts of 20,000 that would be equivalent to 140,000.

We might have represented this as a division equation.

140,000 divided by 20,000.

Again, we can use our known facts.

14 divided by two is seven.

So 140,000 divided by 20,000 would also be seven.

So we can say that England is seven times the size of Wales.

For part B, again, we were asked to form an equation and solve it using our bar model from Task A.

We've got 58 centimetres, and we need to multiply it by something and it will give us 58 millimetres.

So again, we are finding a missing factor.

Well, our units are different, aren't they? So this time we needed to convert.

We know one centimetre is the same as 10 millimetres, so 58 centimetres must be 580 millimetres.

So we are trying to find out what we multiply 580 millimetres by to give us 58 millimetres.

And we could also do a division equation.

We've got 580 millimetres and we're going to divide by 58, and that gives us 10.

So we know divided by 10 is same as finding 1/10.

So 580 millimetres multiplied by 1/10 would also be equal to 58 millimetres.

So the width of the model is 1/10 times the width of the wardrobe.

For question two, you could use your bar model to form an equation.

Again, how many times? So it tells us we are using and looking at the multiplicative relationship.

So I could form a multiplication equation.

25 grammes times what, a missing factor, is 0.

5 kilogrammes? Again, our units are different, so it's really useful to convert.

We know one kilogramme is 1000 grammes, so half a kilogramme or 0.

5 of a kilogramme must be 500 grammes.

So we can form our equation, 25 grammes multiplied by something is 500 grammes.

We know 25 fours are 100.

So we can use that to help us find out that 25 times 20 would then be 500.

We could also have formed a division equation, 500 grammes divided by 25 grammes, and that would be 20.

So we can say that 20 times the small cartons would make one large box.

How did you get on with both of those questions? Very well done.

Fantastic learning today everybody.

You have really made excellent progress at your ability to solve comparison and change problems in a range of contexts.

You now know that to help solve a problem, the context should first be visualised, and then represented pictorially as a bar model.

We know that equations which help us solve the problem can then be formed from the bar model, and we know that careful attention needs to be paid to the language used, so we know if we are looking for an additive or a multiplicative relationship.

I have had a lot of fun learning with you today.

You should be very proud of yourselves, and I look forward to learning with you again soon.