Lesson video

Hello, my name's Mrs. Cornwell, and I'm going to be helping you with your learning today.

We're going to be finding out all about money.

So we're going to find out how we can use what we already know to help us with our learning, and we're also going to find out how we can work in the most efficient way.

Okay, so I know you're going to work really hard in today's lesson and I'm really looking forward to it.

So let's get started.

So our lesson today is called, Solve Problems including those involving giving change.

And it comes from the unit Recognise Coins and use the Pounds and Pence Symbol.

Okay, so in our lesson today, we're going to solve money problems, including those which involve giving change.

So our keyword for today is change.

My turn, change, your turn.

Well done.

And the first part of our lesson is called addition and subtraction problems. So we're going to be learning how to solve those problems using money.

And in our lesson today, we will meet Jun and also Sofia.

They're going to be helping us with our learning today.

The children set up a role play shop in their classroom.

Sofia buys the dragon and the knight, how much does she spend? So we can see that the dragon costs 25 p and the knight cost 62 p.

So 25 p plus 62 p.

Hmm, how will she work that out? Well, she thinks about it before she tries to solve it and she says the ones sum to less than 10 so it won't bridge 10.

So that's helping her think about the best strategy to use, isn't it? I will partition both parts then add.

20 p plus 60 p is equal to 80 p, and five p plus two p is equal to seven p.

80 p plus seven p is equal to 87 p.

I spent 87 p on the dragon and the knight.

So now let's check your understanding.

Jun buys the football and the rocket, how much does he spend? Write the equation then solve this.

Pause video now while you do that.

Okay, and let's see how you got on.

So the ones sum to less than 10 because it's 13 p plus 84 p, isn't it? So they won't bridge 10.

So Jun's saying I will partition the parts, then add.

If you partition the tens, you have 10 p plus 80 p is equal to 90 p.

And then the ones, three p plus four p is equal to seven p.

And then when you recombine 90 p plus seven p is equal to 97 p.

So Jun spends 97 p, doesn't he? Well done if you work that out, the children visit their role play shop.

Jun has 92 p and he buys the rocket.

How much money does he have left? So we can see that the rocket costs 84 p, doesn't it? So we need to subtract 92 p minus 84 p is equal to.

So he says if I subtract four from 92, it will bridge 10.

So again, he's thinking about that before he actually tries to solve it.

I will partition one part and then subtract the tens, then the ones.

And we can see that he's partitioned the 84 p, hasn't he? 92 p minus 80 p is 12 p, then 12 p minus four p is equal to eight p.

So he will have eight p left, won't he? So now let's check your understanding of that.

Sofia has 71 p and she buys the robot.

How much money will she have left? Write the equation, then solve this.

Pause the video now while you try that.

So let's see how we got on.

We had 71 p minus 56 p.

So we need to partition one part, then subtract the tens, then the ones, don't we? So 71 p minus 50 p is 21 p, then 21 p minus six p is 15 p.

So Sofia has 15 pence left, doesn't she? Well done if you did that.

So Jun spends exactly 76 pence on two toys.

Oh, I wonder which two it will be.

"I wonder which two toys he bought." says Sofia.

We need two toys that combine to make 76 pence.

Both toys must cost less than 76 pence.

So that means it's not the rocket.

We can choose one toy and see if we can make 76 p when we add the cost of another toy.

Sofia chooses 56 p because it has a ones digit of six.

So she spotted something that can help her there, hasn't she? So we know 76 p.

If one part is 56 p, then she says, "I know when I add a multiple of 10, the ones digit does not change, so I will add 20 p." And then that will combine to make 76 pencil together, won't it? 56 p plus 20 p equals 76 p.

It must be the robot and the aeroplane.

We could have also used to subtraction to solve this.

76 p is the whole and the cost of the other toys are the parts.

So we can subtract one part to find the other part.

So if 76 p is the whole, Jun's saying 76 p minus 20 p is equal to 56 p.

So I know I am right.

So he checked Sofia's calculation there, didn't he? By doing a subtraction.

Sofia spends 72 p on two toys.

Let's find out which toys she bought.

So 72 p is the whole.

And when added, the ones digit must be two.

"I can't see any toys that have a ones digit below two." say Sofia, "So I must have to bridge 10." That means the ones must sum to 12 when added.

Hmm, are there any ones digits that when combined will sum to 12? Seven plus five equals 12.

So the sum of 47 plus 25 will have a ones digit of two.

47 p plus 20 p equals 67 p.

Then 67 p plus five p equals 72 p.

Sofia must have bought the teddy and the dragon.

We can check this by subtracting.

72 p minus 47 p.

72 p minus 40 p is equal to 32 p.

Then 32 p minus seven p is equal to 25 p.

So there's Jun saying to Sofia, "You were right." So now let's check our understanding again.

I have 89 p and I buy two toys.

Which two toys did I buy? Okay, so pause the video while you think about that.

Okay, and let's see how you got on.

<v ->The ones digit must combine to make nine</v> unless you bridge 10.

Six p plus three p equals nine p.

So let's try 56 p plus 33 p.

So 56 p plus 30 p is equal to 86 p.

Then 86 p plus three p is equal to 89 p.

So the car and the robot must have been in the two items that were bought.

So here's the task for the first part of your lesson.

Work out the cost of the following items, so they've all got a price.

And we can see you've got to work out the cost of the knight in the ball, the teddy in the car, and the knight in the dragon.

So let's look at the second part of your task.

Sofia has 86 p, she can buy one toy from the shelf.

Work out how much Sofia will have left if she buys the following items. So you need to think about how much she would have left from 86 pence if she bought the teddy, then how much she would have left from 86 pence if she bought the robot, then how much she would have left from 86 pence if she bought the plane, and then how much she would have left from 86 pence if she bought the dragon, okay? So pause the video now while you think about that.

Okay, so let's see how you got on.

Did you find the cost of these items? So first of all, the knight the ball.

So we know that we have to combine the items and add them, don't we? And you can see there that because 62 is the greater amount, we've put that first, haven't we? You get the same answer if you put the 13 first, but it's usually easier to start with a greater amount.

So 62 p plus 13 p, the ones digits do not bridge 10.

So we can partition both parts.

So 60 plus 10 is equal to 70, then two plus three is equal to five.

So the total cost is 75 p.

Now let's look at B.

47 p plus 33 p.

So 40 p plus 30 p is equal to 70 p.

Then seven p plus three p is equal to 10 p.

So the total cost is 80 p this time.

Now let's look at the knight and the dragon.

So 62 p plus 25 p.

If we look at the ones, we can see that we will not bridge 10.

So we can partition both parts.

So 60 p plus 20 p is equal to 80 p.

Then two p plus five p is equal to seven p.

So 87 p.

So well done if you did that.

So now let's look at the second part.

So did you work out how much Sofia would have left if she bought these items? So first of all, if she bought the teddy.

So she started off with 86 p and she spent 47 p.

So in subtraction we partition one part.

So we would partition the 47 into 40 and seven, wouldn't we? So 86 p minus 40 p minus seven P 86 p minus 40 p is equal to 46 p.

Then 46 p minus seven p is equal to 39 p.

So she would have 39 p left, wouldn't she? Okay, then let's look at the robot.

86 p minus 56 p.

In subtraction, we partition one part again, so we can partition the 56.

86 p minus 50 p is equal to 36 p.

And then 36 p minus six p is equal to 30 p.

So Sofia would have 30 p left that time.

Let's look at the plane, so 86 p minus 20 p.

So in subtraction we partition one part, don't we? But we can see that 20 is a multiple of 10.

So actually we don't need to partition that, do we? So 86 p minus 20 p is equal to 66 p.

Sofia would have 66 p left.

And then finally, when she buys a dragon, 86 p minus 25 p.

In subtraction, we partition one part again.

So we partition the 25 into 20 and five.

86 minus 20 is equal to 66 p.

Then 66 p minus five is equal to 61 p.

So she would have 61 p left if she bought the dragon, wouldn't she? So well done if you did that.

So now the second part of our lesson is called solve problems involving giving change.

After school, Sofia takes a five pound note to her local shop.

She wants to buy the lolly.

And we can see the lolly costs two pounds, doesn't it? "I don't have exactly the right money, so can I still buy the lolly?" She asks.

"That lolly costs less than five pounds.

So you must have enough to buy it." says Jun.

"But if I give my five pound, I will have given too much money." Say Sofia.

"Don't worry, the shopkeeper will give you the extra money you paid back has change." Says Jun.

Let's look at this on a bar model.

So here's our bar model and the whole amount paid is five pounds.

The cost of the lolly is two pounds, okay? So we can see that we've given five pounds, but the lolly only costs two pounds.

So we need some change, don't we? We need the extra money back.

How much change will the shopkeeper give Sofia? Five pound is the whole, and the cost of the lolly is one part.

The other part is the extra money paid or the change.

"I will subtract to find the change." Say Sofia.

five pound minus two pound is equal to three pound.

The shopkeeper will give me three pound change.

And there we go.

Let's try an example using some two digit numbers.

Jun takes this 50 p coin to the shop.

He wants to buy the sweet shown.

How much change will he get? So he's spending 24 p and he's giving 50 p.

So he's giving too much.

"I paid 50 p, but the suite only costs 24 p.

So I have paid too much." He says.

50 p as the whole and 24 is one part.

I will subtract 24 from 50 to find the other part, the extra part that was paid.

50 p minus 24 p.

50 minus 20 is equal to 30, then 30 minus four is equal to 26.

"You will get 26 pence change." Say Sofia.

And there it is.

So now it is time to check your understanding.

I take one pound of the shop, I buy the chocolate.

How much change will I receive? Complete the bar model to represent the problem, then solve it.

So pause a video now while you do that.

And let's see how you got on.

One pound is the same as 100 pence, isn't it? So we can put 100 pence into the top of the bar model.

That's the whole amount.

100 pence is the whole, the chocolate costs 65 pence.

So this is a part I will subtract to find the change needed.

So 100 p minus 65 p.

100 p minus 60 p is 40 p, 40 p minus five p is 35 p.

So you will get 35 p change.

Well done if you did that.

The next day the children return to play in their role play shop.

They each want to buy a knight, which costs 62 pence, doesn't it? Which child will receive change? Explain how you know.

So Jun says, "I have 80 p, which is more than 62 p, so I cannot pay exactly 62 p." Sofia says, "I have 64 p, which is also more than 62 p, but I am able to pay using three 20 pence coins and two one pence coins.

I can pay exactly the right money, so I don't need change." She doesn't need to give all of her money 'cause she can just give the right amount, can't she? But Jun says, "I have to pay more money than I need." 'Cause he can't split that 20 up, can he? That last 20, so he will need to receive some change 'cause he has to give extra.

Let's find out how much change Jun will receive.

So, "80 p is the whole and 62 p is the part I spent." He says.

80 p minus 62 p.

80 p minus 60 p is equal to 20 p.

Then 20 p minus two p is equal to 18 p.

"I will receive 18 p change." Says Jun.

So here's the task for the second part of your lesson.

Find out how much change will be given if each item shown was bought with the money shown.

Remember to look for any patterns to help you.

Okay, so we've got three examples there, haven't we? And then we've got another three examples there as well.

So pause the video now while you try that.

Okay, so let's see how you got on.

Did you do this? So first of all, we had 50 pence, didn't we? And we had to spend 47 pence.

So there would be 50 p minus 47 p is equal to three p.

So there would be three p change there, wouldn't there? And then here we had 60 p minus 47 p, so we would have 13 p change.

And then here 70 p minus 47 p, so we would have 23 p change.

Did you notice that in each case the whole amount was 10 p more, but the amount paid for the astronaut stayed the same, didn't it? So the change increased by 10 p each time, didn't it? Well done if you spotted that.

Okay, now let's have a look at these next examples.

60 p minus 25 p is equal to? We know 60 minus 20 is equal to 40 p.

And then 40 p minus five p is equal to 35 p.

So we get 35 p change there, wouldn't we? And then this time we have got 80 p minus 56 p.

So 80 p minus 50 p is equal to 30 p.

30 p minus six p is equal to 24 p.

24 p change there.

And then this time we've got 100 p, or a pound.

And we have to spend 68 p.

So 100 p minus 68 p is, 100 p minus 60 p is 40 p, 40 p minus eight p is 32 p.

32 pence change, so well done.

Hopefully you are feeling much more confident about working with money and giving change now.

So excellent, you've worked really hard.

So let's think about what we've learned in today's lesson.

When solving money problems, we can use strategies we already know to help us work efficiently.

We can also use addition and subtraction facts to help us solve money problems efficiently.

If we don't have the exact money to make a given value, we can give a larger value coin or note and receive the extra money back as change.

So well done, you've worked really hard in our lesson today, and hopefully you are feeling much more confident about working with money.

So excellent work.