# Lesson video

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Hello and welcome to today's lesson.

My name is Miss Thomas.

I'll be getting through the lesson with you today.

I'm really looking forward to it because it's so wet and miserable outside.

It's so nice to come down, sit here and prepare a lesson for you.

So let's not waste any time and let's begin.

In today's lesson agenda.

First, we'll be comparing multiplication strategies.

Then we'll move on to our Talk Task.

After that, we'll be deciding when to use certain multiplication strategies.

Finally, we'll finish with our end of lesson quiz.

For today's lesson, you're going to need a pencil paper and a ruler.

Pause the video, if you need to gather your equipment.

Here we have a word problem.

All we did and you can follow along with me.

Suri travelled 212 kilometres to visit Niagara Falls.

Niagara Falls is a waterfall in North America.

Josh travels six times as far.

How far does Josh travel? Here are some questions about the problem.

You should pause the video to explain.

The first question says, "What is known, what is unknown? Explain your answer.

Great, welcome back.

We'll go through the answers in a moment.

The second question says, "What calculation is needed? Explain your thinking out loud.

Pause the video now.

Welcome back.

We know that Suri travels 212 kilometres and let Josh travels six times as far.

What we don't know is the distance that Josh travels yet.

Our calculation will be, 212 multiplied by six.

We know that the factors are 212 and six, and we need to find the product.

Are you ready for the next parts? Wait.

The next part you need to represent the problem using a bar model, then explain out loud what strategy you would use to solve the equation? Pause the video to complete the next two steps.

Welcome back.

Here it's a bar model to represent the problem.

We're going to find out what six groups of 212 is.

It's great that you have already thought about the strategies that you would use, we're going to explore a few different strategies together.

So now you've thought a bit about how you might find the product 212 and six.

Let's look at a new strategy.

We could use the distributive law.

We haven't styled phrase here.

My turn distributive law, your turn? Distributive law is when we solve parts separately.

Here, 212 could be partitioned into separate parts of 200 and 12.

We can calculate 200 multiplied by six.

I need six groups of 200, I've got my first group of 200 on top of my area model, so I need another group of 200, three groups of 200, four groups of 200, five groups of 200, six groups of 200.

The calculation would be 200 multiplied by six.

If I know that six times two is 12, then I know that six times 200 will be a hundred times greater.

It'll be 1,200.

Next, 12 needs to be multiplied by six.

So I've got one group of 12 at the top.

I've got one 10 and two one, so I need five more to make six groups.

so another group of 12, three groups of 12, four groups of 12, five groups of 12, six groups of 12.

I know my 12 times table and my multiplication facts so, 12 multiplied by six is equal to 72.

The final step is to add the products, 1,200 plus 72.

That's equal to 1,272.

Was this method suitable for this equation? Is there a most suitable method? Pause the video now and explain your thinking.

Welcome back, great job.

I really like you have to explain out loud to your screen.

You may have explained this is a useful method because there is no regrouping when we add the products.

It's also helpful when we know our six times table well.

This is something we could do mentally.

You may have other reasons too, and that's great.

Next let's look at how we can partition our factors.

We can partition six into two groups of three.

First, we would multiply 212 by three.

So I've got one group of 212, two groups of 212, three groups of 212.

This is equal to 636.

Now we weren't meant to multiply 212 by three.

We were meant to multiply 212 by six.

So I need to times the product by two, because six is two times greater than three.

So I've got another group of three of 363.

So 363 multiplied by two is 1,272.

So I know that 212 multiplied by six it's 1,272.

Here we have two different strategies represented pictorially.

These strategies we've just looked at.

Explain out loud what's the same and what's different? Pause the video to explain.

Welcome back.

Welcome back.

You may have noticed the way the factors are petitioned was different.

In the area model, it was part of the number was partitioned into 212, into 200 and 12.

In the array, the values were partitioned in the six, into two groups of three.

You may have noticed that the products were both the same.

Both showed that 212 multiplied by six is 1,272.

This tells us that there are different methods to solve multiplication problems. It's up to us to find the most efficient for the calculation.

Next, let's take a look at how we can use short multiplication to solve the equation.

Hello, so we're going to use the counters to solve our equation using short multiplication.

And at the same time, we're going to practise writing, algorithm, for short multiplication too.

So the question is 212 multiply by six.

So I've got my columns, I've got my thousands, my hundreds, my tens and my ones ready.

Okay, so when I look at that equation, I know that that means I need to have six groups of 212.

So I've got two ones, one, two, I've got one 10, and I've got two hundreds.

I'm going to write that in my short multiplication now as well.

So I've got two ones, one 10 and two hundreds.

And when multiplying by six.

I've got my first group, but need to have six groups.

So what my first group of two ones, second group of two ones, third group of two ones, fourth group of two ones.

fifth group of two ones and six two ones.

Now I know that six multiplied by two is equal to 12.

If I've got 12 in my ones column, do I need to regroup? Explain your thinking out loud to your screen.

Great.

You might have said, "Yes, we do need to regroup," "because 12 has a 10 in it." And that one 10 in 12 belongs in the tens column.

So if I was to take away 10 from 12, I'd be left with two.

So I can take way 10 two, four, six, eight, 10, taken away my 10 ones.

And I need to regroup one group of 10 and then put it up here.

Now I'm not going to multiply this group of 10 because it's been already been multiplied in the ones column.

So let's just show what that looks like.

So we've got, I'll draw my lines make it clear.

So we have 12 to ones remained in the ones column and we regrouped the one 10 into the tens column.

So we're not going to multiply back one because it's already been multiplied so I'm just need to have another six groups in my one 10.

So that's the game.

But first group of ten, second group of 10, third group of ten, fourth group of 10, fifth group of 10, sixth, grade 10.

Now all I know that 10 times six is equal to 60.

So I've got 60, 60 in here plus the extra group of 10 I regrouped I've got 70.

I've got seven tens or 70, do I need to regroup to my hundreds column? Call out your answer.

No, we don't, you might have said we don't need to because 70 doesn't have a hundred in it.

So we can leave them.

Our seven tens in our tens column.

Next I've got two hundreds, I've got my first group, I need to have six groups so, my second group of 200, third group of 200 fourth group of 200 fifth group 200, sixth group of 200.

So I've got here six groups of 200.

How can I use my derived my known facts of six times two to tell you what six times 200 is? Call out your answer.

You might have said that if we know that six times two is 12, we know that six times two, we know that six times 200 is going to be a hundred times greater.

So it's going to be 1,200.

I've got 1,200 in my hundreds column.

Do I need to regroup to my thousands? Shout out your answer.

Yes we so because to 1,200 has got one group of a thousand in it.

So let's take away our one group of a thousand, so I've left my two hundreds.

I've taken my one group of a thousand and I'm going to put in regroup my thousands counter, okay.

So now I've got, we don't, we just add what we regroup so we've got, let's go to our written method.

So now we've got, so let's write it up in our written method.

So let's go to where we were.

We did two times two which gave, two times six which gave us 12.

So we've put two ones in, we grouped our one 10.

Then we had one 10 multiplied by six, which gave us six tens plus the one 10 we regrouped, which gave us seven tens.

Then we moved on to our hundreds where we did two hundreds times six, which we found was 1,200, so we had, we left our two hundreds here and we regrouped our 1000 and now we're going to add what we regroup in multiplication so they bring it straight down so we've got 1,272 let's check, 1,272 Great job guys, well done.

Let's see what we've got to do next.

Here we have a word problem.

Marcus runs tours on the ice hockey stadium.

Over the course of a week, Marcus shows 146 groups around the stadium.

How many people took the tour all together? I'm going to show you some questions now you need to pause the video and answer out loud to your screen.

First question says, "What is known, what is unknown?" Pause the video and explain you're thinking out loud.

Welcome back, fantastic.

We know that there are eight people in a group and there has been 146 groups.

What we don't know is how many people took the tour all together? The next question says, "What calculation is needed? Pause the video now and decide.

Welcome back.

You might have spotted that the equation is eight multiplied by 146 to find how many people took the tool all together.

Now you need to pause the video and draw a bar model to represent this calculation, eight multiplied by 146.

Pause the video now.

Let's take a look.

Here is a bar model, that represents the problem.

We're going to find out what eight groups of 146 is? You might've found this the fastest way.

It would take a long, long, long time to draw a bar model where they were apart at eight, where the parts had a value of eight and there were 146 different equal parts that would take long time to draw.

We need 146 different bars.

Here we have Antoni.

He says, "In short multiplication the only way to solve the problem" "is using the short multiplication method." Is this the only way to find the product? How else could the problem be solved? See if Antoni is right or wrong? Pause the video and try some of your own strategies to solve the problem.

But don't use the short multiplication method.

Let's explore some different multiplication strategies for the problem.

It's important that we consider different strategies in maths to make sure we're using the most efficient way for that particular problem.

Here are two examples, but you may have found others of your own and that is great too.

Defined eight most applied by 146.

I could partition that 146 into 100, 40 and six.

As you can see on my area model, I've partitioned the counters.

I'm going to use the counters in my area model to show my thinking.

I've already got one group of 146 on top.

So I need seven more, to have eight groups.

So I need to have eight groups of 100, three groups of a hundred, four groups of a hundred, five groups of a hundred, six groups of a hundred, seven groups of a hundred, eight groups of a hundred.

Next I need seven more groups of my four tens.

I've already got my first group of four tens, third group, fourth group of four tens, five group of four tens, six group of four tens, seven groups of four tens and eight groups of four tens.

Finally, we need to have six, my six ones I need to have eight groups of them.

So I've got my first group, I need my second group of six ones, third group of six ones, fourth group, fifth group, sixth group, seventh group, eight groups.

We're going to show you another example now.

In the next example, I noticed that eight was two doubled and then doubled again.

So I use this in my strategy.

I started with one group of 146, and then I doubled it because doubling means times by two and then as doubled it again, I multiplied it by two again.

So all together after I multiplied it by two, and then by two again, I had eight groups.

When I added the products, I got 1,168.

We reached the Talk Task, great work so far.

Choose an expression and decide on an efficient strategy to solve it.

So that might be with an array or with a bar model or an area model.

Use a star words to justify your strategy choice.

Justify means that you can explain improve why your strategy is helpful in solving the problem quickly.

Once you have solved and represented the expression, you should just fight out loud using the star words.

Fantastic work.

You should have represented, solved and justified an expression.

If you didn't don't worry, pause the video and take some time to do that.

Let's take a look at the answers.

There are many, many different strategies you could have used to find the answer.

You might have used the distribution law to partition the factors, you may have used column multiplication, you might have used continuous addition.

Hopefully, you noticed that certain methods are more useful for particular problems than others and it's important in maths we consider different strategies.

Okay, here we have a word problem.

VIP tickets for an ice hockey match cost 87 pounds.

Marcus sells nine tickets.

How much money does he make? The fast question is, "What is known, what is unknown?" Pause the video and explain.

Welcome back.

We know that tickets cost 87 pounds.

And Marcus sold nine tickets.

What we don't know is how much money he made all together from selling his tickets? So the second question is, "What will the equation be? Great, the equation will be 87 multiplied by nine, and I've got my bar model that to represent it.

I've got nine groups of 87.

Here, we have full possible strategies for solving nine multiplied by 87.

You'll need to take some time now to follow the strategies, to solve the equations.

You should be able to explain which strategy you would use to solve the problem and why? Think, which would be the most efficient for this particular equation, 87 multiplied by nine.

Pause the video and try each strategy.

Well done for your perseverance in solving the problem, using four different strategies.

The answer to the problem was 783.

So they made 783 pounds selling tickets.

Which strategy did you prefer? Call out the colour of the strategy.

Great.

Why did you prefer it? Try and use the star words, when you explain yourself.

Pause the video and explain why you preferred it for this equation? Fantastic, you might have noticed that 87 could be partitioned into 80 and seven using the distributive law.

You might have noticed that nine is close to 10, so you could have times by 10 and then taken away one group of 87.

Whichever way you found to be the most efficient well done.

It's fantastic that you're able to justify your choice using key vocabulary.

Select a word problem and choose an efficient strategy to solve it.

You should record your workings out.

You may wish to draw a bar model or an area model alongside the problem to support your understanding if this will help you.

Great job, here are the answers.

I will not talk about the different strategies you could have used with each one because we won't have enough time.

Hopefully though, you've used a range of strategies.

In most applications, some problems suit strategies more than others.

If a factor is close to 10, you could times it by 10 and then take away one group of, if needed.

One of the number of that you need to take away.

If there is little regrouping needed, partitioning, I'm using the distributive law, can be done mentally.

Sometimes when there's a lot of regrouping, short multiplication method can be efficient, Whichever way you find suitable for you it's fabulous.

Fantastic work, this lesson.

You can now go and complete your end of lesson quiz.