Lesson video

Hello, my name's Mrs. Hopper, and I'm really looking forward to working with you in this lesson.

This lesson comes from the unit on addition and subtraction of two-digit numbers.

Is this new for you? Have you done anything like this before? I'm sure you've got lots of skills about numbers, and addition and subtraction, that you are going to be able to bring to this lesson.

So if you're ready, let's make a start.

In this lesson, we're going to be solving problems involving addition of two-digit numbers.

Have you been thinking about two-digit numbers recently, and have you been thinking about addition? Well, let's bring all that together and solve some problems, adding two-digit numbers.

We've got two keywords in our lesson today.

We've got partition and efficient.

So I'll take my turn to say them and then it'll be your turn.

I'll say my turn.

Partition.

Your turn.

My turn.

Efficient.

Your turn.

Well done.

I expect you are quite familiar with those words and you've done quite a lot of partitioning, and you've been working hard to find efficient ways to solve problems and to do calculations.

So let's see how those words are going to be used in our lesson today.

There are two parts to our lesson.

The first part, we're going to be looking at problems with a missing whole.

And in the second part, problems with a missing part.

So let's make a start on part one, where we're looking at a missing whole.

And we've got Andeep and Izzy helping us in our lesson today.

Izzy is watching the minibeasts in the garden.

Have you done some work watching minibeasts perhaps? I wonder what she's seen.

She counts 35 ants and 23 woodlice.

How many minibeasts does she count altogether? Hmm.

How are we going to solve this one? She says, "We can use a bar model to understand this problem." So she's drawn a bar model.

She says, "There were 35 ants.

This is one part." And there were 23 woodlice.

This is another part.

She says, "I must combine the parts to find the missing whole." Part plus a part is equal to the whole.

So in this case, our equation is 35 plus 23.

She says, "First, I'm going to partition." So she's partitioned both of her parts.

35, she's partitioned into 30 and five, and 23 into 20 and three.

Then she says, "I'm going to add the 10s." So 20 plus 30 is equal to 50.

And then, "Add the ones," she says.

Five plus three is equal to eight.

And now she can recombine her totals.

The total of her 10s was 50 and the total of her ones was eight.

50 plus eight is 58.

She says, "I saw 58 minibeasts altogether." 58 was the missing whole.

Izzy decides to give the minibeasts some water.

Good idea, Izzy.

They need water, don't they? She pours 35 millilitres into a jug, and then another 36 millilitres.

How much water is now in the jug? Hmm.

I wonder how we can work this one out.

She's drawn another bar model.

So the first part that she poured in was 35 millilitres, and then she poured in the other part which was 36 millilitres, and we need to work out how much water there is altogether.

So we need to do 35 add 36.

Izzy notices something different about this problem.

Have you noticed something? She says, "The ones digits in 35 and 36 will bridge 10 when we add them." So she says, "It's more efficient to partition one addend." Ah, so she's just going to partition one addend and count on from the first addend.

Let's have a look at what she does.

So she's not going to partition the 35, but she is going to partition the 36.

Now she's going to record the stages on the number line.

So she's going to add on the 30 to the 35.

35 add 30.

Well, we're adding on three more 10s.

We're not changing the ones, so just the 10s will change.

35 add 30 is equal to 65.

And now she can add the ones, but she's going to bridge through 10.

So she's got six ones to add.

She knows that she can partition the ones into five and one.

So six ones is equal to five plus one.

We add the five.

We get to the next multiple of 10, which is 70, and then we add on the other one.

So we've got to 71.

She's added on 30 and then six, but she's partitioned the six into five and one so she can bridge through 10, and her total is 71.

She says, "There's now 71 millilitres of water in the jug." I think they're still outside, but they've moved on to slightly bigger animals.

They're looking at bird seed in a bird feeder.

So the children pour some bird seed into the bird feeder.

Andeep pours in 55 grammes, then Izzy pours in another 43 grammes.

So how much is there altogether? Well, one part was Andeep's part, which was 55 grammes, and the other part is Izzy's part that she poured in, which was 43 grammes.

So our parts of 55 and 43.

But we don't know what the whole is.

So how much seed has been poured into the bird feeder? Andeep says, "To decide which strategy to use, I must think about whether the ones digits bridge 10." Do you think they're going to this time? He says, "Five plus three." Those are our ones digits in our two parts.

Five plus three sums to less than 10.

So he says, "I will partition both addends this time." I've not got to worry about bridging 10.

So he partitions both the addends.

So that's his first step, to partition.

Then he can add the 10s.

So we've got 50 and five, and 40 and three.

So our 10s are 50 plus 40, which is equal to 90.

Then he can add the ones.

Five plus three, which is equal to eight.

And now he can recombine.

90 plus eight is equal to 98.

So 98 grammes of bird seed was poured into the feeder.

Izzy wonders whether she could partition one addend to solve the same problem.

So she's going to use the strategy that we've used when the ones digits do bridge 10, but this time they don't.

But will the strategy work? Have a go, Izzy.

So she hasn't partitioned 55, but she has partitioned the 43 into 40 and three.

She says, "The ones digits don't bridge 10, but I think it will still be efficient to partition one addend." Let's see, Izzy.

So first, partition.

So she just partitioned the 43 into 40 and three.

Then, add the 10s.

This time she's going to add 40 onto 55.

55 plus 40 is 95.

Why is it equal to 95? Well, all we've done is added four more 10s, the ones haven't changed.

So five 10s add four 10s is equal to nine 10s, and we've still got our five, so 95.

Then add the ones.

So instead of doing five and three, she's doing 95 plus three.

But again, we're not bridging 10, so it's only the ones that are changing.

Five plus three is equal to eight.

So 95 plus three is equal to 98.

She says, "When the ones digits don't bridge 10, we can use known facts, so it is efficient to use either strategy." It's whichever one you prefer and feel most confident with.

Time to check your understanding.

Can you draw a bar model and write an equation to represent the problem below? Then, practise using each strategy to solve it.

Which one do you prefer? So practise by doing a partitioning of both addends and then just partitioning one of the addends.

And this is the problem.

There were 72 woodlice under the log.

13 more crawled under.

How many woodlice are under the log now? So pause the video, have a go at drawing a bar model and solving that using both strategies.

And when you've decided on which one you prefer, press play and we'll have some feedback.

How did you get on? So this is what our bar model looks like.

There were 72 woodlice under the log and 13 more crawled under.

How many woodlice are under the log now? So we're doing 72 plus 13.

Well, our ones digits are two and three.

They don't bridge 10.

So we could partition both addends.

So there's step one.

We've partitioned 70 and two, 10 and three.

Then we can add the 10s.

70 add 10 is equal to 80.

Then add the ones.

Two add three is equal to five.

And then recombine.

We've got 80 when we added our 10s and five when we added our ones.

80 plus five is equal to 85.

So there are 85 woodlice under the log now.

What about the other strategy? So you could just have partitioned one addend.

So we've kept 72 altogether and we've just partitioned the 13.

Now we can add the 10s, but this time we're adding 10 onto our 72.

72 add 10 is equal to 82, just the 10s have changed.

And then we add the ones, but we're adding 82 and three.

So we're adding three onto our 82.

82 add three is 85 because two add three is equal to five.

So we've still got 85 woodlice under the log now.

Which strategy did you prefer? Which did you find the most efficient? Andeep counted 24 ants on the path, then another 28 ants crawled onto the path.

How many ants were on the path altogether? So he's drawn a bar model.

His parts or his ant parts were 24 and another 28, but he wants to know what the whole is the missing whole.

He says, "To decide which strategy to use, I must think about whether the ones digits bridge 10." Do you think they're going to this time? We've got 24 and 28.

Hmm.

Well, he says, "Eight add four is greater than 10, so I will partition one addend," he says.

So he's just going to partition the 28.

So first, he's partitioned, then he's going to add the 10s, and he's using a number line to help him keep track.

24 add 20.

So we're adding on two more 10s, but the four isn't changing.

So we've got 44.

24 add 20 is equal to 44.

Now we've got to add on eight, but we're going to bridge through 10.

So what should we partition our eight into? We've got 44.

We know that 44 add six will take us to the next multiple of 10, so we can partition our eight into six and two.

So there we go.

So in adding the ones, we're going to partition our eight into six and two.

44 add six is equal to 50, and then the other two, 50 add two is equal to 52.

So we've added on 20 and eight, and we've got to a total of 52.

There were 52 ants altogether on the path.

Time to check your understanding again.

Can you draw a bar model and write an equation to represent this problem, and then choose the most efficient strategy to solve it.

There were 38 woodlice under the log.

19 more crawled under.

How many woodlice are under the log now? Pause the video, have a go.

When you're ready for the answer and some feedback, press play.

How did you get on? So there's the bar model, showing that there were 38 and then 19 more crawled under.

So those are our two parts, and we've got to combine them to find that missing whole.

What's going to be the most efficient strategy, do you think? We've got 38 add 19.

What about those ones? Ah yes, eight plus nine is greater than 10.

The ones digits bridge 10.

So I'm going to partition just one addend.

So I'm going to keep my 38 whole and I'm just going to partition my 19.

38 add 10, just add one more 10.

So I've got three 10s and eight, I'm going to have four 10s and 8, 48.

And I can show that on the number line.

Now I've got to add on my ones.

I've got to do 48 add nine.

Well, I can see I'm going to bridge through 10 here, so I can partition my nine into two and seven.

48 add two is equal to 50, and 50 add seven is equal to 57.

So I can see that 38 add 19 is equal to 57.

So that meant there were 57 woodlice under the log altogether.

And it's time for you to do some practise.

So there are three questions here.

You're going to draw a bar model to represent each problem and then solve it in the most efficient way.

So for A, there are 47 children having school dinners and 28 bringing a packed lunch.

How many children are having lunch altogether? In B, there were 49 children out in the playground, then 36 more went out.

How many were in the playground in total? And in C, the children had two packets of bird seed.

One packet had a mass of 36 grammes and the other had a mass of 52 grammes.

How much did the packets weigh altogether? Pause the video.

Have a go at drawing a bar model and solving those problems. And when you're ready for the answers and some feedback, press play.

How did you get on? So A, there were 47 children having school dinners and 28 bringing in a packed lunch, and we had to work out how many there were altogether.

So our bar model might look like this.

One part is 47 and the other part is 28, and the whole is missing.

So our equation is 47 plus 28.

What did you notice about the ones? That's right.

Seven plus eight is greater than 10.

The ones digits bridge 10.

So we're just going to partition one addend.

So we're going to partition the 28 into 20 and eight.

Now we're going to add the 10s.

47 add 20 is equal to 67.

We can see that with a jump on the number line, just the 10s have changed.

We've only added two 10s, no extra ones.

Now we've got to add the ones and we're going to bridge through 10, aren't we? And we're going to bridge through the next multiple of 10, which is 70.

So we can partition our eight into three and five.

Three add five is equal to eight.

67 add three will take us to 70, add the other five will take us to 75.

So we know that 67 add eight is equal to 75.

So there are 75 children having lunch.

For B, there were 49 children out in the playground and then 36 more went out.

So how many children were out there altogether? So our parts this time were 49 and 36, and we had to add those together to work out the missing whole.

Again, six plus nine is greater than 10.

So the ones digits will bridge 10.

So we're going to partition just one addend.

So we've partitioned the 36.

We're going to add the 10s.

49 add 30 is equal to 79.

Then we're going to add on the six.

And we can partition our six into one to get us to 80, and then add the other five on.

So 79 add six is equal to 85.

So the total number of children in the playground now is 85.

And finally, we were looking at bird seed.

There were two packets.

One had a mass of 36 grammes and the other of 52 grammes.

So there's our bar model.

36 add 52.

What do you notice about the ones digits this time? Yes, they're going to be less than ten.

Six add two is less than 10.

They don't bridge 10.

So we could partition both addends or just one addend.

It depends which you prefer.

We've partitioned both of them.

30 and six add 50 and two.

Add the 10s.

30 add 50 is equal to 80.

Add the ones.

Six plus two is equal to eight.

And recombine.

80 plus eight is equal to 88.

So the packets have a mass of 88 grammes together, or you may have partitioned just the one addend.

So you may have kept the 36 as one whole number and then partitioned the 52.

So again, we can add the 10s.

36 add 50.

Well, that's three 10s.

Add five 10s and another six, that's 86.

And then 86 add two is equal to 88.

Maybe that was slightly more efficient, I wonder.

Which did you choose to do? And on into the second part of our lesson, we've got problems with a missing part here.

Izzy fed the rabbits at the weekend.

They ate 98 grammes of food.

They ate 42 grammes of food on Saturday.

How much food did they eat on Sunday? Hmm.

What do we know about this time? Well, this time we know the whole, which was 98 grammes of food, and we know that they ate one part of it on Saturday, and that was 42 grammes.

So we're missing the part they ate on Sunday.

Let's draw a bar model.

The whole has nine 10s and eight ones.

It's 98.

The part we know about is 42.

Four 10s and two ones.

So 42 add something is equal to 98.

But what do we know about our number? Well, four 10s plus five 10s is equal to nine 10s, so the missing part must have five 10s in it.

And two ones plus six ones is equal to eight ones, so the missing part must have six ones in it.

So because our ones digit of our whole is larger than the ones digit of our part, we can just think about counting on up.

So 40 plus 50 is equal to 90, two plus six is equal to eight, so our missing part is 56.

The rabbits ate 56 grammes of food on Sunday.

Time to check your understanding.

In two days, Izzy counted 68 ants in total on the path.

She saw 45 ants yesterday.

How many ants were on the path today? And we can represent that as a bar model.

68 ants is our whole and 45 ants is one of the parts.

Can you work out the missing part? Pause the video, have a go.

When you're ready for the answer and some feedback, press play.

How did you get on? So we were looking at 45 plus something is equal to 68, and I can use known facts to see how many 10s and ones I need.

I've got to add on some 10s and some ones, so I can just count on the 10s and count on my ones.

Four 10s plus two 10s is equal to six 10s.

So the missing part must have two 10s, so it must be 20 something.

Five ones plus three ones is equal to eight ones.

So the missing part must have three ones.

So the missing part must be 23.

There were 23 ants on the path today.

Well done if you work that one out.

Here's another problem.

Two rabbits drank 72 millilitres of water, one drank 36 millilitres.

How much water did the other rabbit drink? So in total, they drank 72 millilitres of water.

That's our whole.

And one of them, so one part is equal to 36.

So we've got to work out 36 add something is equal to 72.

What do you notice here about those ones digits? This time, we can't just add the 10s and add the ones because six adds something is not going to give us a two, is it, in the way we are thinking about it.

Hmm.

I wonder what we could do here.

Three 10s plus four 10s is equal to seven 10s, but if I add four 10s, 40, I will reach 76.

And that's too much.

So this time, we've got to think carefully about our 10s to make sure that we don't add on too many 10s and go past our whole.

So let's think carefully Andeep says, "I will add a multiple of 10 until I can't add any more 10s." So 36 add 10 would be 46.

That's all right.

Add 20 would be 56.

That would be all right.

Add 60 would be 66.

Oh, that would be all right, wouldn't it? If I add another 10 though, I'll get to 76.

So I can add on three 10s.

He says, "If I add 40, I will reach 76, which is too much, so I will add 30." So 36 add 30 will take us to 66.

So we know that our 10s digit missing is a three representing 30.

What about those ones then? Right.

We've got to bridge through 10, haven't we? Because we've got to get from 66 up to 72.

So let's bridge through 70.

What do we need to add on to 66 to reach 70? We've got to add four to get to our next multiple of 10.

So 66 add four is equal to 70, and then we need to add another two to get to 72.

So we've added on six altogether.

So we've added on six ones altogether.

So we've added on 30 and six.

So our missing part is 36.

The other rabbit drank 36 millilitres of water as well.

Did you see that? 36 must be half of 72.

Let's have a look at another one.

The children have been counting the minibeasts.

Yesterday, Andeep counted 23 ladybirds.

By the end of today, he had counted 52 in total.

How many did he count today? So our total is 52, and we know that one part is 23.

So 23 add something is equal to 52.

Can you spot again? The ones digit in our sum is smaller than the ones digit of our addend.

So we're gonna have to think carefully when we add on our 10s.

We can't just add on three 10s because we'll get to 53.

And as Andeep says, "Two 10s add three 10s is equal to five 10s, but if I add three 10s, I will reach 53." So we can't add on three 10s.

He says, "I will add a multiple of 10 until I can't add any more 10s." So he can add on 20.

23 plus 20 is equal to 43.

That's less than 52, so we are all right.

So the 10s part of our missing part must be 20 or two 10s.

Now we need to bridge through from 43 up to 52, so we can find our ones digit.

So 43 add seven is equal to 50, and then we need another two to get to 52.

So altogether, Andeep has added on nine ones.

So he added on two 10s and nine ones.

Two 10s and nine ones gives us a missing addend of 29.

So he must have counted 29 ladybirds today.

Time to check your understanding.

Can you choose the most efficient strategy to solve this problem? On Monday and Tuesday, Izzy's rabbit ate 42 grammes of food altogether.

On Monday, it ate 17 grammes of food.

How much did it eat on Tuesday? And we can represent this with a bar model.

So altogether, it ate 42 grammes of food.

On Monday, it ate 17 grammes.

How much did it eat on Tuesday? That's our missing part, isn't it? Pause the video.

Have a go at choosing an efficient strategy.

And when you're ready for the answer and some feedback, press play.

How did you get on? So 17 plus something is equal to 42.

One 10 add three 10s is equal to four 10s, but if I add four 10s, it will reach 47, and that's more than our total.

So it can't be four 10s that we're missing.

So first, we're going to add a multiple of 10 until you can't add any more 10s.

So 17 add something, it's got to get us to less than 42 if we add our 10s.

So it must be 20 to get to 37.

So the 10s part of our missing addend must be 20 or two 10s.

Now we need to add ones and bridge through up to 42.

If you'd reach 47, you'd have added too many 10s.

So now we're going to bridge 10 to find out how many ones are needed to reach 42.

37 plus three is equal to 40 and another two takes us to 42.

So altogether, we've added on five ones.

So we added on two 10s and five ones, which is 25.

So the rabbit ate 25 grammes of food on Tuesday.

And it's time for you to do some practise.

You're going to draw a bar model to represent each problem and then write the equation, and use the most efficient strategy to solve it.

So in A, there are 68 cows and sheep on the farm.

There are 45 cows.

How many sheep are there? In B, a shop had 74 red and green sweets.

There are 46 red sweets.

How many green sweets are there? And in C, Izzy and Andeep have 86 centimetres of string altogether.

Izzy's piece is 45 centimetres long.

How long is Andeep's piece of string? Pause the video, have a go.

When you're ready for the answers and some feedback, press play.

How did you get on? So here's A, with the cows and the sheep.

So our whole was 68 animals, 45 are cows.

How many sheep are there? So we've got to find that missing part.

45 plus something is equal to 68.

Well, this time if you notice, our ones digit of our whole is bigger than the ones digit of our part.

So we can think about adding on 10s and adding on ones.

So we can use known facts to see how many 10s and ones I need.

Four 10s plus two 10s is equal to six 10s.

So the missing part must have two 10s.

20.

Five ones plus three ones is equal to eight ones.

So the missing part must have three ones.

So the missing part is 23.

There are 23 sheep.

In B, we were thinking about 74 sweets altogether.

46 are red.

How many are green? So 74 is our whole and 46 is our part.

So 46 add something is equal to 74.

Ooh, can you see? The ones digit of our sum is smaller than the ones digit of our part? So we have to think very carefully about how many 10s we can add.

Four 10s plus three 10s is equal to seven 10s, but if I add three 10s, I reach 76.

And that's too many.

So first, we're going to add a multiple of 10 until we can't add any more.

If you reach 76, you've added too many 10s.

So 46 add 20, add two 10s, is equal to 66.

So our missing part must have a 10s value of 20 or a 10s digit of two.

Now we need to bridge through our 70 to find out how many ones are needed to reach 74.

So 66 add four is equal to 70, add another four is equal to 74.

So altogether, we've added on eight ones.

So we've added on 20 and eight, which is 28.

So our missing part is 28, and there are 28 green sweets.

And finally, part C, this was about their string.

They've got 86 centimetres of string altogether and Izzy's piece is 45 centimetres long.

How long is Andeep's piece? So there's our bar model.

86 is the total length of the string.

45 is the part that's Izzy's.

We need to work out Andeep's part.

45 plus something is equal to 86.

Again, this time, our ones digit is larger in our sum.

So we can think about using are known facts.

Four 10s plus four 10s is equal to eight 10s.

So the missing part must have four 10s.

It must have a 10s digit of four worth 40.

Five ones plus one one is equal to six ones.

So the missing part must have a one digit of one to represent one one.

So four 10s and one one is 41.

So the missing part is 41, and Andeep's string is 41 centimetres Long.

Well done if you solved all those and use an efficient strategy to get there.

And we've come to the end of our lesson.

We've been solving problems involving addition of two-digit numbers.

What have we been thinking about? Well, we know that partitioning the addends can help us to solve problems involving two-digit numbers.

We know that once we've written the equation, we must think carefully about the most efficient strategy to use.

We need to spot whether the ones digits in the addends will bridge through the next multiple of 10.

And we know that it's really important to be able to calculate mentally before adding two two-digit numbers.

We've been using our number facts and our partitioning understanding, haven't we? Thank you for all your hard work in this lesson today.

I've really enjoyed it and I hope you have too, and I hope I get to work with you again soon.

Bye-bye.