Lesson video

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

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

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

Today's lesson is from our unit calculating with decimal fractions.

The lesson is called multiply 1-digit numbers by decimals.

As we move through the learning today, we will develop our understanding of how we can multiply whole numbers by decimals.

We will think about using unitizing and our known facts, so our times tables, and we will also consider how we can use the distributive law to help us.

Once we've learned these, we will then think about how we can apply them to problem-solving.

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

Let's get started then, shall we? How can we multiply 1-digit numbers by decimals? These are the keywords that we will use in our learning today.

We have unitizing and distributive law.

It's always good to practise saying new words, so let's have a go.

My turn.

Unitizing.

Your turn.

Nice.

My turn.

Distributive law.

Your turn.

Fantastic.

Unitizing is when we treat groups that contain the same number of things as ones or units.

It's particularly important when we handle money and in understanding place value and it supports us to think multiplicatively.

You can see here we've got one 50 pence coins, but actually, that's unitizing because it's equal to 51 pennies.

The 50 p is a unit.

And the distributive law, well, that says that multiplying a number by a group of numbers added together is the same as doing each multiplication separately.

So, for example, five 3s added to three 3s is the same as eight 3s.

Let's start our learning today by thinking about how we can use unitizing and known facts to help us multiply 1-digit numbers by decimals.

In this lesson, we have got Lucas and Sofia to help us.

Let's have a look at this.

Lucas eats four slices of cheese.

Each slice has a mass of 25 grammes.

What mass of cheese does he eat? Good idea, Lucas.

When we have a word problem, it's always a very good idea to represent it in a bar model.

As our bar model, we don't know the whole.

That's what we are trying to find out.

And we can see that we have got four slices of cheese represented in the four parts, each has a value of 25 grammes.

And we can use this bar model to form an equation.

We've got four 25s, and that's what we need to determine.

But we know our multiples of 25 so we can use this as a known fact.

We could count up in 25's, 25, 50, 75, 100.

So we know 4 multiplied by 25 is 100.

Oh, good idea, Sofia.

We didn't have to count up in 25's, did we? We could use a repeated doubling strategy because we are multiplying by four, and we know when we multiply by four it is the same as doubling and doubling again.

So four 25s, well, that's the same as two 50s, which is 100.

Either way, we can say that Lucas has eaten 100 grammes of cheese.

Let's look at this.

Sofia eats four raspberries.

Each raspberry has a mass of 2.

5 grammes.

What mass of raspberries does she eat? Good idea, Sofia, let's represent this in a bar model.

We can see the whole is unknown, and each raspberry is worth 2.

5 grammes and there are 4 raspberries, so we have four parts.

And yes, Sofia, thank you.

We can use our bar model to form an equation.

4 multiplied by 2.

Ah, Sofia has spotted something.

I wonder if you've spotted something.

There is a relationship between this equation, 4 times 2.

5 and the equation Lucas formed, 4 times 25.

Can you see that? What can you see? What's the same and what is different? That's right.

Thank you, Sofia.

5 is one-tenth times the size of 25.

So in both equations, we have a 4 and in one of them, we have 25 and the other we have 2.

5, which is one-tenth times the size.

When we find one-tenth, we can multiply by 0.

1, which is the same as dividing by 10.

So then we can use that to help us determine the answer to 4 times 2.

5, because if 2.

5 is one-tenth times the size of 25, then 4 lots of 2.

5 must be one-tenth times the size of four 25s.

And when we multiply by 0.

1 or divide by 10, the digits move one place to the right, so we get 10.

Sofia has eaten 10 grammes of raspberries.

We can say that if one factor is made one-tenth times the size, the product will be one-tenth times the size.

So 25 was made one-tenth times the size, so the product 100 was made one-tenth times the size, but the other factor stayed the same, didn't it? It stayed at four.

5 is one-tenth times the size of 25 so 4 multiplied by 2.

5 is one-tenth times the size of 4 multiplied by 25.

Let's check your understanding on this.

Could you use this unitizing strategy that we've just done and this stem sentence to calculate 3 times 2.

5? And the stem sentence is shown below.

5 is one-tenth times the size of 25, so 3 multiplied by 2.

5 must be one-tenth times the size of 3 multiplied by 25.

Pause the video while you think about what would go in the missing blank.

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

How did you get on? Did you say it must be 7.

5? It must be 7.

5 because 25 has been made one-tenth times the size so we need to make 75 one-tenth times the size.

75 divided by 10 is 7.

Well done.

Let's look at these equations.

4 multiplied by 25 is equal to 100.

So what does that mean about 4 multiplied by 0.

25 do you think? Can you see something that is the same or something that is different? That's right.

There is a relationship between these equations.

25 is 100th times the size of 25.

Did you spot that? When we find 100th, we can multiply by 0.

01 or divide by 100.

If 0.

25 is 100th times the size of 25, then 4 multiplied by 0.

25 is 100th times the size of 4 multiplied by 25.

And when we divide by 100, it is the same as moving the digits two places to the right.

100th made 100 times smaller is 1.

And we can summarise this.

If one factor is made 100th times the size, the product will be 100th times the size.

25 is 100th times the size of 25, so 4 multiplied by 0.

25 must be 100th times the size of 4 multiplied by 25.

So we can use our known facts, 4 multiplied by 25, to work out the answer to 4 multiplied by 0.

25.

Let's summarise our learning.

You can see we knew four 25s were equal to 100, and then we looked at 4 multiplied by 2.

Well, that's the same as 25-tenths, which would be 100-tenths, and we know 10 tenths is equal to 1, so 100 tenths is equal to 10.

And then 4 multiplied by 0.

25, well, that's 25 hundredths.

So 4 multiplied by 25 hundredths would be 100 hundredths, and 100 hundredths is equal to 1.

Let's look at this equation and use our learning to help us solve it.

We've got 5 multiplied by 2.

Hmm, how could we solve that? What could we do? Ah yes, thank you, Lucas.

To multiply by 2.

5, we could convert that number to a whole number by multiplying by 10, 2.

5 multiplied by 10 is 25.

We can then calculate using our known facts.

We know five 25s are 125, but how do we then solve the original equation? That's right, we need to divide the product by 10 to solve the original equation.

125 divided by 10 is 12.

We can say 5 times 25 ones is equal to 125 ones, so 5 times 25 tenths, which is 2.

5, is equal to 125 tenths or 12.

Let's look at this equation.

We've got 6 multiplied by 0.

25.

We know to multiply by a decimal number, we first need to transform it to a whole number, so we can multiply 0.

25 by 100 because there are 25 hundredths.

So if we multiply by 100, that will give us 25.

We can then use our known facts to calculate.

We would get 150, but that's not the answer to the original equation, is it? So what do we need to do? Well, that's right.

We transformed 0.

25 into 25 by multiplying by 100.

So we need to divide the product by 100 to solve the original equation.

150 divided by 100 is 1.

We can say 6 times 25 ones is equal to 150 ones so 6 times 25 hundredths is equal to 150 hundredths or 1.

Let's check your understanding with this.

True or false? 0.

5 multiplied by 0.

9 is equal to 45.

Maybe find someone to chat to about this.

Pause the video and when you are ready for the answers, press play.

How did you get on? Did you say it must be false, but why is it false? Is it because A, we would need to make both decimal fractions 10 times larger to make them both whole numbers.

We would then need to make the product 100 times smaller.

It should be 0.

45.

Or is it B, five-tenths times by nine-tenths is equal to 45-tenths.

It should be 4.

Pause the video while you have a think about that.

When you are ready for the answer, press play.

How did you get on? Did you say it must be option A, but why? Well, we would need to make both decimal fractions 10 times larger to make them both whole numbers.

Then we would have to adjust the product 100 times and make it 100 times smaller, not just 10 times smaller.

We need to adjust for both decimal fractions being made 10 times larger.

So the answer should be 0.

45.

Well done if you spotted that.

It's your turn to practise now.

For question one, could you solve these equations by using the relationships that you spot, and then, once you finish those, could you make up a set of your own equations using this pattern? For question two, could you solve these equations? First, make the decimal fraction a whole number, so by multiplying by 10 or 100, then calculate using your known facts, so your times tables, and then adjust the product accordingly.

So divide by 10 or 100 depending on what you first multiplied by.

Pause the video while you have a go at both questions.

When you are ready for the answers, press play.

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

So for question one, you are asked to solve equations using the relationships that you spot.

You were given six 15s are 90, and then did you spot that that 6 remained the same but the factor 15, well, that was made 10 times smaller to 1.

5, so we needed to make the product 10 times smaller.

90 made 10 times smaller is 9, and then for the third one, the 15 was made 100 times smaller to 0.

15, so we needed to make the 90, 100 times smaller, 0.

There was a similar relationship in the second set of questions.

We know nine 11s are 99, the factor 9 was remained the same but the factor 11 was made 10 times smaller, then 100 times smaller, which meant the product was made 10 times smaller, 9.

9 and then 100 times smaller, 0.

99.

And it was the same relationship for the third set of questions.

We had 3 multiplied by 12 is 36, but then the 12 was made 10 times smaller and the product was made 10 times smaller but the 3 then must have remained the same.

And the same with the last question, the 3 must have remained the same.

You then might have made up your own set of equations like this where you kept one factor the same, but the second factor was made 10 times smaller and then 100 times smaller.

So the product had to be 10 times smaller, then 100 times smaller.

For question two, you were asked to solve the equations, first by making the decimal fraction a whole number, then calculating using your known facts and then remembering to adjust the product accordingly.

So 8 times 2.

5 was equal to 20.

Then for part A, we had 4 multiplied by 1.

5, well, that would equal 6 and you would have converted the decimal fraction to 15, so four 15s would be equal to 60.

Then you would've had to make that 10 times smaller to get 6.

4 multiplied by 0.

06, you've got six hundredths there so you would've multiplied by 100 to get four 6s or 24 and then make that product 100 times smaller, 0.

24.

For part C, you had nine hundredths.

We'd need to make that 100 times larger to get nine 4s, which are 36, and then make the 36, 100 times smaller, 0.

36.

For part D, you had 2 multiplied by 0.

75.

We'd need to multiply it by 100 to get 75.

Two 75, well, 75 doubled is 150, make that 100 times smaller, 1.

For part E, we had five hundredths, so multiplied by 100 to get 5, five 8s are 40, but then devising the 40 by 100 to get 0.

For part F, 3 multiplied by 0.

9, we've got nine-tenths.

So we're going to multiply it by 10 to get 9, three 9s are 27, but then we need to divide the product by 10 to get 2.

For part G, 2 multiplied by 0.

15, we've got 15 hundredths, so we'll multiply by 100 to get 15.

Double 15 is 30, but then we would have to make the product 100 times smaller, 0.

For part H, 8 multiplied by 0.

7, we've got seven-tenths, so we're going to multiply by 10 to get eight 7s or 56, but then we would need to make 56, 10 times smaller, 5.

And our last one, 0.

07, seven hundredths, we'd need to multiply it by 100, which would give us 7 multiplied by 5 is 35 and then we would need to make that 100 times smaller, 0.

35.

Well done on your learning so far.

You are working really hard.

We're going to move on now and think about how we can use the distributive law to help us multiply 1-digit numbers by decimals.

Let's look at this calculation.

7 multiplied by 0.

13.

"To multiply by 0.

13," Lucas is saying, "we first need to convert this to a whole number by multiplying by 100." 0.

13 multiplied by 100.

Well, we know when we multiply by 100, we move the digits of the decimal fraction two places to the left to get a whole number.

So the calculation can be transformed into 7 multiplied by 13.

And then we can multiply this but it's not a known fact, is it? We only knew half times table facts up to 7 times 12.

So this time we're going to calculate using the distributive law.

I wonder if you remember the distributive law.

It's all about decomposing our numbers.

So we've got 13 is composed of 10 and 3, so what we're going to do is we can multiply 7 by 10 and then 7 by 3 and then sum the products.

Seven 10s and seven 3s.

Well, seven 10s we know is 70 and seven 3s are 21, and if we sum those products, we get 91.

But we need to remember that wasn't the original equation, was it? We need to divide the product by 100 to solve the original equation.

So we need to move the digits two places to the right.

91 divided by 100 is 0.

91.

Let's look at a different calculation.

We've got 2.

3 multiplied by 6.

We know that if we have a decimal fraction, we need to convert it to a whole number, we got 23 tenths.

So we're going to multiply by 10 and when we multiply by 10, the digits of the decimal fraction move one place to the left and that gives us a whole number, 23.

And then we can multiply 23 by 6, but that's not a known fact, is it? It's not a times table fact.

So we're going to calculate using the distributive law.

We know 23 is composed of 20 and 3 so we can multiply 6 by 20 and 6 by 3 and then sum the products.

20 sixes are 120, 3 sixes are 18, if we sum those, we get 138.

But is that the answer to our original equation, 2.

3 times 6? It's not, is it? We need to divide the product by 10 to solve the original equation.

138 divided by 10.

Well, we know we need to move the digits one place to the right.

138 divided by 10 is 13.

Let's check your understanding and practise one together.

I'm going to look at 5 multiplied by 4.

So I've got a decimal fraction so I know I need to convert it to a whole number.

I've got 43 tenths, I'm going to multiply by 10.

That gives me my whole number, 5 times 43.

That's not a times table fact.

So I'm going to use the distributive law.

43 is composed of 40 and 3, so I can do 5 multiplied by 40 and 5 multiplied by 3.

Five 40s are 200 and five 3s are 15 and then I can sum them to get 215.

But then I know I must remember to divide by 10 to answer the original equation.

5 multiplied by 4.

3 is equal to 21.

I'm going to give you some time now just to practise on your own using that structure.

So pause the video and when you've had a go and you want to go through the answers, press play.

How did you get on? Did you say that you needed to multiply the 2.

6 by 10 to get 26 and then we can use the distributive law.

Four 20s added to four 6s.

Well, four 20s are 80, four 6s are 24, that gives us 104.

However, that's not the answer to the original equation.

We needed to divide by 10.

104 divided by 10 is 10.

4 times 2.

6 is equal to 10.

How did you get on with that? It's your turn to practise now.

For this question, could you solve these equations? First, make the decimal fraction a whole number, so by multiplying by 10 or 100, then calculate using the distributive law and then remember to adjust the product accordingly.

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

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

So for the first equation, we needed to multiply by 10 to make 4.

5 into a whole number, 45 and then we can use the distributive law.

Three 40s added to three 5s is equal to 135, but then we needed to make that 135 10 times smaller, which is 13.

For the second question, 0.

05 times 18.

I've got 5 hundredths so I need to multiply by 100 to make the 0.

05 into a whole number, 5.

And five 18s, well, I know 18 is composed of 10 and 8, so I can do 5 multiplied by 10 and 5 multiplied by 8, 50 and 40 is 90, but then I need you to remember to divide that product 90 by 10, 0.

For the next question, we have 13 multiplied by 0.

I need to convert the 0.

3 into a whole number by multiplying by 10, then I can use the distributive law to help me.

13 is composed of 10 and 3 so I can do 10 threes, 30, and 3 three is 9 and sum those to get 39.

But then I needed to remember to adjust the product by dividing by 10.

39 divided by 10 is 3.

And for the next question, 6 multiplied by 1.

I need you to transform the 1.

4 to a whole number, 14, by multiplying by 10.

We could then use the distributive law.

14 is composed of 10 and 4.

six 10s, 60, and six 4s, 24, if we sum those, we get 84.

We then needed to adjust the product by dividing by 10.

84 divided by 10 is 8.

03 multiplied by 24, we've got three hundredths, so we need to multiply by 100 and then we get 3 multiplied by 24.

24 is composed of 20 and 4, so I can do three 20s, 60, and three 4s, 12, sum those together to get 72 and then adjust the product accordingly.

We multiply it by 100 so I need to divide by 100, 0.

72.

And finally, we had 16 multiplied by 0.

9, nine-tenths.

So I'm going to multiply by 10 to make the 0.

9 into a whole number, 9.

16 is composed of 10 and 6.

So 10 nines, 90, and 6 nines, 54, would give us 144, but then we needed to adjust the product and make it 10 times smaller, 14.

How did you get on with all those questions? Well done.

Brilliant learning, everyone.

I can see how hard you are trying and how much your learning has developed.

We are now going to use what we have learned and apply it to some problem-solving.

Let's have a look at this problem.

Lucas has 1.

6 kilogrammes of flour.

Sofia has four times as much.

How much flour does Sofia have? Good idea, Lucas, let's represent this as a bar model.

So we don't know how much Sofia has, but we know it's four times as much as Lucas and Lucas has 1.

6 kilogrammes.

And we can use this bar model to help form an equation.

We know we need four times that 1.

And to multiply by 1.

6, can you remember what we do? That's right, we have to convert it to a whole number first.

We've got 16-tenths, so we're going to multiply by 10 to give us a whole number, 16 and then we have to make a decision.

How are we going to multiply 4 and 16? What would you do? Well, we can't use our time-table facts because we only know up to 12 times 4.

What else could we do then? Well, we could use the distributive law or we could use repeated doubling, can we? Because one of the factors is four.

What should we do? And it doesn't matter.

You're right, Lucas.

We will get the same product at the end.

Let's use repeated doubling, okay.

We want to work out 4 multiplied by 16.

So I'm going to double the 16 to get 32, but we still need to multiply that by 2.

We get 64.

Then we must remember this, don't we? We need to divide the product by 10 to solve the original equation.

64 divided by 10 is 6.

So Sofia has 6.

4 kilogrammes of flour.

Let's look at a different problem.

Strawberries cost 4.

23 pounds per kilogramme.

Sofia buys five kilogrammes of strawberries to make some jam.

How much does she spend? Good idea, Sofia, let's represent this as a bar model.

Our whole is unknown but it is equivalent to 5 lots of 4.

23 pounds.

So we can use this bar model to form an equation.

5 multiplied by 4.

23.

But how are we going to do that? How are we going to multiply by a decimal fraction? That's right.

We need to convert it to a whole number first and we can do this by multiplying by 100.

23 multiplied by 100 is 423.

Then we need to think how are we going to multiply 5 and 423.

It's not in our times-table, is it? We only know up to 12 times 5.

What else could we do? That's right.

Thank you, Sofia.

We can use the distributive law.

We know 423 is composed of 400, 20, and 3, so we can multiply each of those separately and then sum them together.

Five 400s are 2,000, five 20s are 100, and five 3s are 15.

2,115 is the total, but that's right, thank you for reminding me, Sofia.

We need to divide the product by 100 to solve the original equation.

2,115 divided by 100, well, the digits move two places to the right, we have 21.

15.

Sofia spends 21 pounds and 15 pence.

Let's check your understanding with this.

I wonder if you can tell me which strategy would be most efficient to solve this calculation.

6 multiplied by 1.

23 is equal to 7.

38.

Would it be A, would you adjust 1.

23 to 123 and use a distributive law? Is it B, would you adjust 1.

23 to 123 then use repeated doubling? Or would you just use short multiplication algorithm? Pause the video while you have a think about this.

Maybe chat to somebody else about it, see if you agree.

When you are ready for the answer, press play.

How did you get on? Did you say that actually the most efficient way would be to adjust the decimal fraction to a whole number and then use the distributive law before then remembering to readjust the product.

Your turn to practise now.

For question one, could you solve these problems? It might be useful to represent them in a bar model to help you form an equation to solve.

And then think about the most efficient way to calculate the answer.

Part A, Sofia buys nine raffle tickets.

They cost 0.

31 pounds each.

How much do they cost altogether? Part B, Sofia makes a cupcake that has a mass of 0.

2 kilogrammes.

Lucas makes a cake that is eight times this.

How much heavy is Lucas' cake? And for part C, Sofia has a jug with a capacity of 0.

23 litres.

Lucas has a jug that has a capacity that is four times this.

What is the capacity of Lucas' jug? For question two, could you solve this problem? Again, represent it in a bar model to support you to form an equation, then decide the most efficient way to calculate the answer.

A shop sells shampoo in different bottle sizes.

A large bottle contains 0.

35 litres of shampoo and a small bottle contains 0.

25 litres.

Sofia buys two large bottles and five small bottles.

How much shampoo does she have altogether? And if you can give your answer in litres.

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

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

For question one, you had some problems to solve.

Sofia buys nine raffle tickets.

So we can see my bar model.

I've got nine parts and each part is worth 0.

31 and we can use that to form an equation.

There are nine of them, so I need to do 9 multiply by 0.

31.

I've got a decimal fraction, so I need to make that into a whole number.

And I've got 31 hundredths, so I'm going to multiply by 100.

Then I've got 9 times 31.

What strategy did I use? Well, I thought about the distributive law.

I can't use repeated doubling and I can't use my known facts.

So 31 is composed of 30 and 1, nine 30s added to nine 1s.

Well, nine 30s are 270, nine 1s are 9, that gives me 279, but I need to remember to adjust the product.

So I'm going to divide by 100.

279 divided by 100 is 2.

79.

So altogether the tickets cost 2.

79 pounds.

For part B, when Sofia made a cupcake while Lucas made a cake that was eight times this.

So my bar model has eight parts, each with a value of 0.

2, and I can use that to form an equation, 8 multiplied by 0.

I've got a decimal fraction, and it's two-tenths.

So I can multiply it by 10 to make a whole number, 8 multiplied by 2 is 16.

I could just use my known facts there, couldn't I? I know two 8s or eight 2s, but then I need to remember to adjust my product.

16 divided by 10 is 1.

So Lucas' cake has a mass of 1.

6 kilogrammes.

But the question actually asked me how much heavier is Lucas' cake.

So I needed to find the difference in their masses.

So I had 1.

6 and I need to subtract the mass of Sofia's cupcake, 0.

2, and that is 1.

4 kilogrammes.

So Lucas' cake is 1.

4 kilogrammes heavier than Sofia's cupcake.

For part C, Sofia had a jug with a capacity of 0.

23 litres and Lucas' was four times this.

So my bar model has four parts, each with a value of 0.

23.

I can use that to form an equation, and then because I've got a decimal fraction, I need to make that decimal fraction into a whole number.

I'm going to multiply by 100.

Then I could use two strategies here, couldn't I? I could double and double again, repeated doubling or I could use the distributive law, but I decided to double and double again.

4 times 23.

Well, 23 doubled is 46, I need to double again, 92.

But then I need to remember to adjust the product.

We multiplied by 100, so now I need to divide by 100, which gives me 0.

92.

So the capacity of Lucas' jug is 0.

92 litres.

For question two, the shop selling shampoo in different-sized bottles.

You can see my bar model.

I have represented the two larger bottles with 0.

35 and the five smaller bottles with 0.

25, and we can use the bar model to form equations.

I had two bottles that were 0.

35 litres and I could transform the 0.

35 into 35.

I know my double 35 is 70, so then I can adjust that to get 0.

And then I had five bottles at 0.

25.

I can use my known facts, 5 multiplied by 25 is 125 and adjust the product, 1.

25.

Then I had to sum those values together.

7 added to 1.

25 is equal to 1.

95 litres.

So altogether, Sofia has 1.

95 litres of shampoo.

How did you get on with those questions? Well done.

Fantastic learning today, everybody.

I am really impressed with how hard you have worked.

You have really deepened your understanding of multiplying 1-digit numbers by decimals.

We know that when we multiply 1-digit numbers by decimals, we make the decimal into a whole number by multiplying by 10 or 100.

We know that we can then use unitizing, known facts, or the distributive law to multiply the whole numbers, but we then remember to adjust the product by division.

So really, well done.

I've had a lot of fun learning with you, and I look forward to learning with you again soon.

I've finished the video