Lesson video

Hi, my name is Mr. Peters.

Thanks for joining me for this lesson today.

In this lesson, we're gonna be building on all of our learning, around numbers of the three decimal places and we're gonna begin to apply this to solving a range of problems. When you're ready, let's get started.

By the end of this lesson today, you should be confident enough to say that I can solve problems with decimal numbers up to three decimal places.

Throughout this lesson, we're going to be referring to three key words.

I'll have a go at saying them and then you can repeat them after me.

The first one is represent.

Your turn.

The second one is operation.

Your turn.

And the last one is efficient.

Your turn.

let's her think about what these mean in a bit more detail.

A representation is a way of showing the mathematics that we're talking about.

We could use concrete objects, we could use pictures or we could use symbols.

And all of these would represent the mathematics that we're trying to share.

We use the term operation to describe mathematical processes that we use.

For example, when we are using addition, subtraction, multiplication or division, all of these can be described as mathematical operations.

And finally, the word efficient.

When we think about efficiency, we're thinking about what's the best strategy to use.

And when we think about what we mean by best, we're actually thinking about what's the most accurate strategy as well as the quickest strategy to get us to the answer that we're looking for.

This lesson today is broken down into two parts.

In the first cycle, we'll be thinking about representing problems with three decimal places, and in the second cycle we'll be solving problems, using efficient strategies.

Let's get started.

Throughout this lesson today, you'll meet Sophia and Jun who will share their thinking and their great questions that they have as they go through the lesson.

Wimbledon is a very prestigious tournament that takes place every year in England in the months of June and July.

Is a two week competition and it's played on grass courts and over the two weeks up to 675 matches are played.

Each match can last between 30 minutes or up to 11 hours, depending on how close it is between the players.

As you can imagine, before they begin the tournament, they have to spend a lot of time preparing the grounds to make sure it's fair for all of the players.

In order to make sure it's fair at the grounds themselves, they have 18 tennis courts that the matches are played on.

So to ensure that it's fair, they need to make sure that the grass is at the same height on every single tennis court.

This height has to be eight millimetres in the weeks leading up to the tournament.

The grounds person on court three recognises that the grass is only five millimetres tall.

How much taller does the grass need to grow in order for it to be playable? Sophia says we can represent this as a bar model.

Here, the purple bar represents how tall the grass is at the moment, and here, the green bar represents how tall the grass needs to be.

And then finally, the arrow represents the difference between the actual grass size and the expected grass height.

June's pointing out that he feels it's helpful to use a bar model as it allows him to see what operation we're going to need here in order to solve this problem.

He's added that we're looking to find the difference, aren't we, between the expected grass height and the actual grass height.

So when we're thinking about difference, it can be useful to use subtraction as an operation.

X says therefore we're going to need to take the expected grass height and subtract the actual grass height from this to find out the difference.

So the expected grass height was eight millimetres and the actual grass height was five millimetres.

So eight millimetres minus five millimetres would be equal to the difference, wouldn't it? I'll let you have a moment to have a think what that might be.

That's right, it'd be three millimetres, wouldn't it? We know that eight minus five is equal to three.

So eight millimetres minus five millimetres would be equal to three millimetres.

As the grounds people continue their preparation on court two.

One of the grounds people is painting on the court lines.

One of the lines he has to paint is called the baseline.

This is the line at the back of the court.

As he's painting it, he runs outta paint after 7.

95 metres.

He needs to paint an additional 78 millimetres.

So the question here is how long should the baseline be altogether? Let's represent this as a bar model again.

7.

95 is the length of the baseline that he has painted so far.

78 millimetres is the extra bit that he needs to paint on.

All of that together with combine to make the total length of the baseline.

Therefore, the light green bar here would represent the total length of the baseline that would be painted.

Jun is now saying that the bar model has allowed 'em to see that the operation we'd need to use here would be addition.

We need to add both of the parts together to find the total length of the baseline.

So in order to find the total length of the baseline, we'd need 7.

95 metres and we need to add on 78 millimetres.

We don't need to worry about working out this one just yet 'cause we're really interested here in how we can use the bar model to help us represent the problem.

Here's a different problem on court five.

They're putting the nets up to make sure they're the right height.

The centre of the net should be 0.

914 metres tall.

At the moment, the height of the net is 0.

923 metres tall.

Hmm, it feels like the net's a bit too high, isn't it? So how much does the net need to be lowered by? Again, let's represent this as a bar model.

Here, the green bar is representing how tall the net should be, 0.

914 metres.

And here the purple bar is representing how tall the net is at the moment, which is 0.

923 metres.

Again, you can see here we're gonna be looking for the difference between the purple bar and the green bar, aren't we? Which is represented by this arrow.

So again, for this kind of problem, it looks like we're going to need to find the difference, aren't we? And this time, because the numbers are so close together, we could count on to find the difference between these two amounts using addition, or again, we could use subtraction.

We could subtract the expected height of the net from the actual height of the net.

So we've got two operations here.

We could use 0.

914 plus something is equal to 0.

923.

Or we could say that 0.

923 minus 0.

914 would be equal to the difference that we're looking for.

Okay, time for a check for understanding them.

Which of these bar models represents the problem? Each line on the core needs to be 50 millimetres wide, except for the baseline, which needs to be a hundred millimetres wide.

How much wider is the baseline? Take a moment to have a think That's right.

It could have been either A or B, couldn't it? We could say that the green bar represents the width of each line and the purple bar represents the width of the baseline and therefore the arrow represents a difference between these two.

A, it's just a vertical representation of the bio model, whereas actually B is exactly the same.

It's just represented horizontally, isn't it? Why is it not C? Well in C, it shows that the baseline is a hundred millimetres wide and then each other line is 50 millimetres wide and they've been joined together to make a whole, which that means we'd have to add them together, which isn't the right operation to solve this problem, is it? Well done if you got that.

Okay, here's another check then.

Which equation would this bar model represent? I'll give you a moment to have a think.

That's right.

It's B isn't it? The purple bar is 175 millimetres and the green bar is 28 millimetres.

And they're joined together, aren't they? So which means we're going to need to add these two together to find our total amount, which would be the hole underneath that represented by the white bar.

So 0.

028 metres plus 0.

175 metres would be equal to a whole, even though the add ends are the other way around, it still represents the bar model, doesn't it? Okay, onto our first task for today, then, what I'd like you to do here is match each of these problems to the bar model that would represent each of these problems. And then for task two, what I'd like you to do is draw a bar model to represent the problem here for the carpenter who's looking to create the correct length shelf.

And then for task three, I've given you the bar model.

And in this case, what I'd like you to do is write your own problem, which is bar model could represent.

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

Okay, welcome back.

Let's go through these problems then.

So the first bar model would represent an addition problem and that here is the problem at the bottom where we're working out how long the snail travelled altogether over two hours and firstly in the first hour and then in the second hour added together.

The middle bar model represents the middle problem.

In this problem, we know how far the snail travelled in the first hour and we know how far the snail travelled altogether, but we don't know how far the snail travelled in the second hour and that's what we're trying to work out and therefore we're going to use the difference to help us solve that.

And finally, the third bar model is representative by this top problem here.

Here we know how far the snail travelled in the first hour and we know the difference this time, don't we? We know how much less it travelled in the second hour, so hopefully the green bar would be representing the distance it travelled in the second hour this time.

Well done if you've got those.

Okay, for task two then the purple bar represents the current length of the shelf and the green bar represents how long the shelf needs to be.

So again, the arrow represents a difference between these two, how much we need to shave off of the shelf.

Hopefully your bar model looks similar to that.

And then finally, Jun has come up with a problem for this bar model here at the bottom.

Jun says, if I had half a metre of ribbon and I used 0.

15 metres of the ribbon, how much ribbon would I have left? And again, this is representative of our bar model, isn't it? The purple bar represents the total amount of ribbon.

The green bar represents how much ribbon he used and the arrow represents how much ribbon would be left.

Well done if you managed to get that or draw something very similar to that.

Okay, onto cycle two of our lesson now then.

Every year at Wimbledon in the summer because the competition is played on grass courts, there has to be very wary of the rain that might fall.

And hopefully, because it's the summer and the sun's shining, the grounds people are more able to keep the ground safe.

However, sometimes that's not always the case and there can be some rain in the summer, can't there? And that can affect the play at the competition.

Here is a bar chart which represents the rain that falls during one week of the competition.

The green bars represent the expected amount of rainfall, how much rain was forecasted for each day, and the purple bar represents the actual amount of rain, how much rain actually fell on each day during that first week of the competition.

The amount of rainfall can be measured in metres or millimetres, and that's represented on the Y axis on our bar chart.

And each weekday is represented on the x axis of our bar chart.

So let's have a look at our first problem then.

What was the expected amount of rain on Monday and Tuesday on the first week? Sophia's saying that on the Monday the expected amount of rain was 0.

008 metres or eight millimetres, and on the Tuesday the expected amount of rain was 0.

002 metres or two millimetres.

Let's represent this as a bar model.

The purple bar represents the amount of rain expected on the Monday, and the green bar represent the amount of rain expected on the Tuesday.

So again, we can see here that we're going to need to use addition and we can represent this using this equation.

0.

008 metres plus 0.

002 metres is equal to the total amount we are looking for.

Jun says we can use our known facts to help us solve this.

I know that eight plus two is equal to 10, so eight thousandths of a metre plus two thousandths of a metre would be equal to 10 thousandths of a metre.

And we know this is the same as saying that eight millimetres plus two millimetres is equal to 10 millimetres.

So the amount of rain that was expected to fall on Monday and Tuesday was 10 millimetres.

Here's another problem.

What was the difference in the amount of rain that was expected to fall on Sunday compared to the amount of rain that actually fell on Sunday? Take a moment to have a look at the bar chart yourself.

Jun says there was 0.

011 metres of rainfall expected on Sunday compared to 0.

005 metres of rainfall that actually fell on Sunday.

Again, we can represent this as a bar model, can't we? The green bar represents the amount of rain that was expected to fall on Sunday, and the purple bar represents the amount of rain that actually fell on the Sunday, and therefore we are looking for the difference between these two, which is represented by our arrow.

Therefore, we need to subtract, don't we? We need to subtract the actual amount of rain from the expected amount of rain.

If we were to do that, we could use potentially a bridging strategy to solve this.

So let's represent this bridging strategy on our number line.

Here, we've got 11 millimetres or 0.

011, and we need to subtract five millimetres, don't we? So 0.

005 millimetres.

What we can do here is partition this five millimetres into one millimetre and four millimetres, 0.

001 and 0.

004.

Then we can subtract the one millimetre, the 0.

001 to get back to 10 millimetres, and then we can bridge through the hundredth boundary by subtracting the additional four millimetres, the 0.

004, which would lead us to 0.

006 metres or six millimetres.

So we can say that the difference in the expected and actual rainfall on Sunday was 0.

006 metres or six millimetres.

Jun says that is a strategy we could use.

However he feels that maybe using a known fact would be a lot easier.

Let's have a look at what he says.

He says that 11 minus five is equal to six.

So again, the total difference between the expected and the actual rainfall on the Sunday would be 0.

006 metres or six millimetres.

And that is exactly what Jun is pointing out.

He's saying that 11 millimetres minus five millimetres would be equal to six millimetres.

Sophia's agreed with him.

She thinks that's a lot more efficient than potentially drawing a number line or even having to partition number and subtract both of those parts.

Great thinking Jun.

Okay, and one more problem for us to think about here.

The second week of the competition recorded 18 millimetres of rain.

So we're now asking what was the total amount of rain that fell over the whole fortnight? Hmm, we can represent this with a bar model.

The purple bar represents the amount of rain that fell on the first week and the green bar will represent the amount of rain that fell on the second week.

And obviously the white bar would represent the total amount, wouldn't it? So firstly, we're going to need to work out how much rain fell throughout the first week, aren't we? Here is the amount of rain that fell throughout the first week of the competition.

Have a look at the numbers.

How could we go about adding these? I think it's important that we think about how we can add these efficiently to do that.

Sophia thinks we could do this mentally and it might help us if we converted each of these units from metres into millimetres.

So let's do that now.

Now that we've done that, if it was me, I'd be looking for numbers that could combine together to make lots of 10 millimetres.

So let's have a look.

Can you see any? There we go.

Tuesday and Saturday we'd combine together to make 10 millimetres.

Wednesday and Sunday we'd also combine together to make another 10 millimetres.

So, so far we've got 20 millimetres of rain and then we've just got three more to add, haven't we? So if we add the 12, that would make it 32 millimetres of rain and then add the extra one.

So the total amount of rain that fell throughout the first week of the competition was 33 millimetres of rain.

Gosh, that's a lot.

I wonder if they've got any play in at all.

So now we can adjust our bar model while putting the numbers in place, can't we? So the amount of rain that fell on the first week, which was represented by the purple bar is now 33 millimetres or 0.

033 metres.

And the amount of rain that fell in the second week was 18 millimetres or 0.

018 metres.

Again, we can see we're going to need to add these together, aren't we, to solve this? And Sophia has decided she wants to use column addition to solve this.

Let's have a look if this was a useful strategy for Sophia to use, she's lined up her columns correctly, and now she says that 3000th of a metre plus 8000th of a metre would be 11000th of a metre.

So therefore we need to regroup 10000th of a metre for 100th of a metre and we leave 1000th in the thousandths column.

Then we add 300th plus 100th would be equal to four hundredths of a metre plus an additional one, which is underneath, which would be five hundredths of a metre, and then zero temps.

And zero temps would be zero temps of a metre and zero ones and zero ones would be zero hole metres.

So the total amount of rain that fell over the whole fortnight would be 0.

051 metres or 51 millimetres of rain.

Jun's now saying again that actually writing out in column could take quite a while and maybe we could have done this slightly more efficiently by thinking about it mentally by adjusting the numbers slightly.

Jun's saying the 1800 of a metre is nearly 20000th of a metre.

So we could take our starting number, which was 33 millimetres or 33000th of a metre and then add on the 20000th of a metre or 20 millimetres, which would take us to 53 millimetres or 53000th of a metre.

But of course, we didn't want to add 20,000 of a metre of a metres is we only wanted to add 18,000 of a metre.

So we need to subtract another 2000 of a metre from our total, which would take us to the same amount, 51000th of a metre or 51 millimetres.

Again, Sophia's had a think about that and she agrees with Jun.

Actually that was a lot quicker than writing out all those numbers into those columns, wasn't it? So here maybe using an adjusting strategy would've been more efficient than writing a column strategy.

Okay, time to check your understanding now.

Jun is adding 0.

007 and 0.

005 together.

Tick the most efficient strategy.

Take a moment to have a think.

That's right.

It would be B, wouldn't it? Using the known fact would be the most efficient here.

If I know seven plus five is equal to 12, then 7,000 plus 5000th would equal to 12000th.

That's a lot quicker and easier, isn't it? Okay, a second check here, look at the bar model.

Which strategy would be most efficient for solving this? That's right, it's C, isn't it? Having a look at these numbers, they look quite complex, don't they? They're both four digit numbers and it looks like there might be several regrouping in this.

So by looking at this bar model, we can see that we're gonna be finding the difference.

So it looks like maybe a column subtraction strategy might be the best approach in order to find the total amount missing.

Okay, onto our final task for today then.

What I'd like to do here is draw a bar model to represent each of these problems and then solve each of those problems as well.

And then for task two, we have a carpenter who's received several different planks of wood.

So again, here's a number of problems here that I'd like you to have a go at solving using efficient strategies.

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

Okay, welcome back, let's go through these together then.

Parcel A weigh 0.

856 kilogrammes and parcel B weighed 1.

265 kilogrammes.

What was their combined weight? So the bar model here, the purple bar represents parcel A and the green bar represents parcel B and they're joined together to find the total amount and the total amount was 2.

121 kilogrammes.

For that one, I probably would've used a column strategy to solve that.

Question B, what was the difference between their weights? So again, here the green bar represents parcel B and the purple bar represents parcel A, and then the arrow represents the difference between these two.

The difference here is 0.

409 kilogrammes.

Once again here, this is looking very much like maybe a column subtraction to solve this one.

Due to the complexity of the numbers and some of the regrouping required.

For question C, parcel A has there some extra stuff put inside of it, so the purple bar represents the weight of parcel A, the green bar represents the additional item added into that parcel, and therefore we're looking for the total amount again here.

So this looks like another addition problem here.

To do this, I think we could have done this mentally.

Parcel A shows eight tenths of a kilogramme and we are adding three tenths of a kilogramme, so that would be 11 tenths of a kilogramme.

So that would be one whole kilogramme and additional one 10th of a kilogramme.

So the total amount would be 1.

156 kilogrammes.

And then for question D, we've got parcel B here, haven't we? Which has had something else put inside it.

Although we don't know the weight of that, we now know the total weight of the parcel.

So the total weight is represented by the white bar.

And then parcel A's original weight was represented by the green bar.

Again here there's two options, I think.

We could have probably done this mentally, but using accounting on strategy or we could have used the column subtraction as well.

The total amount was 0.

255 kilogrammes of the parcel that was put inside of parcel B.

Well done if you've got all of those.

Okay, and onto our final one for today, then.

Which piece of wood was the longest? Well, that would've been plank C.

Which pieces of wood would a join together to make 1.

1 metres of length of wood? That would've been plank D and plank E.

If he sanded plank B down by 0.

006 metres or six millimetres, how long would plank B be now? That would be 0.

844 metres.

And again, we could have done that as a mental strategy.

Plank B has 850 thousandth of a metre and we're subtracting six thousandths of a metre.

So 850 minus six would be 844, so that would be 0.

844 metres or 844 millimetres.

And finally wasting as little wood as possible.

What would be the best way to make a length of 1.

5 metres? Hmm, this made me think quite a bit actually.

I had to have a go at combining a range of woods to see what we came up with, and I finally came up with glueing planks, B and E together and then cutting off 0.

075 metres of wood.

That was the best way I found to make a 1.

5 metre length of wood and waste as little wood as possible.

Well done if you managed to get all of those.

So that's the end of our learning today.

Hopefully you enjoyed thinking about the different context, which numbers with three decimal places can be used within and the different problems that we can solve as a result of that.

To summarise our learning for today, it can be helpful to represent problems using bar model, which enables us to see which operations we can use to help solve the problem.

It's a way into the mass and for additive problems, when we're adding and subtracting, we can use a range of different strategies, can't we? We could use column methods, we could use adjusting methods, we could use bridging methods.

However, some are more efficient than others and develop our mathematical thinking.

It's always worth our while taking time to think about what would be the most efficient strategy rather than jumping for a method straight away.

Thanks for learning with me today.

I really enjoyed that lesson and hopefully you did too.

Take care, I'll see you again soon.