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 adding two-digit numbers when the sum crosses the tens boundary.

So we've been thinking a bit about two-digit numbers recently and about addition.

So I wonder what strategies we'll be using today to help us.

Let's have a look at what's in our lesson.

We've got one keyword.

This is the word that was in my mind when I was thinking about adding two-digit numbers, it's partition.

I wonder if you can say it back after I've said it.

So my turn, partition, your turn.

Lovely.

I'm sure it's a word you know.

Partitioning numbers helps us to break them down, to make them easier to handle when we're adding.

So let's have a look at how partitioning is gonna help us today.

Well, in part one of our lesson, we're going to be partitioning both addends to add two-digit numbers.

And in the second part, we're going to be partitioning one addend to add two-digit numbers.

So let's make a start on part one.

And in this lesson, we've got Andeep and Izzy helping us out with our learning.

Izzy buys an astronaut and a football, and Andeep buys a rocket and a car.

How much does each child spend? Izzy says, "I spent 26 pounds," 20 pounds plus six pounds.

And Andeep says, "I spent 58 pounds," 50 pounds plus eight pounds.

I wonder how much the children spent altogether.

Izzy says, "I will write an equation to find out." Thank you, Izzy.

26 plus 58 is equal to something, and that something is the sum, is how much they spent altogether.

Izzy partitions the addends to help them add.

So she had 26 plus 58, so she's partitioned into 20 and 6 and 50 and 8, and she's put her tens together.

Then, she uses base 10 blocks to represent the tens and the ones.

So we've got 20 from the 26 and 50 from the 58, plus the six and the eight.

She says, "We can add the tens, and then the ones." So let's look at the tens.

We know that 20 plus 50 is equal to 70, and we know that six plus eight is equal to 14.

Oh, and 14 is a 10 and 4 ones.

"The ones crossed the tens boundary, so how can we recombine these numbers?", she says.

Let's have a look.

We've got our 70 and then we've got our 14.

"14 is made of 1 ten and 4 ones, so we can add the tens and then the ones." So she's partitioned the 14 that she made.

So we can add our 10 from our 14 in with our tens.

70 plus 10 is equal to 80, and then we've got our extra four.

So 20 plus 50 plus 6 plus 8 or 26 plus 58 is equal to 84.

So the children spent 84 pounds altogether.

Can you write an equation to help you to find the cost of the items shown? Remember to partition the addends, and then, use base 10 blocks to help you to solve the equation.

So you've got a T-shirt for 25 pounds and some boots for 57 pounds.

Pause the video, have a go, and when you're ready for the answer and some feedback, press Play.

How did you get on? So we've got 25 plus 57 to calculate to find the total cost of the two items. So now we're going to partition to help us.

25 is 20 and 5, 57 is 50 and 7.

So we're going to represent them with base 10 blocks to help us, 2 tens and 5 ones and 5 tens and 7 ones.

So let's start with our tens.

20 plus 50 is equal to 70, and then our ones we've got five plus seven.

Well, five plus seven is going to bridge through 10.

So we've taken three from the five to make our seven up to 10.

So five plus seven is equal to 12.

We've bridged through 10 again, haven't we? Now we need to recombine our 70 and our 12, so we can partition again.

Partition the 12 into 10 and 2.

70 plus 10 is equal to 80 plus another two is equal to 82.

So 25 plus 57 is equal to 82.

So the items cost 82 pounds altogether.

Andeep thinks he can solve this equation without using base 10 blocks.

Hm, I wonder what he's going to do.

He says, "I will partition the numbers, and then add them." So 36 plus 27, 30 and 6 plus 20 and 7.

He's going to add the tens first.

30 plus 20 is equal to 50, and six plus seven is equal to 13.

He might have used a near double there.

Six plus six is equal to 12 plus another one is equal to 13, so he's got 50 and 13.

Now he needs to recombine this.

He's going to partition the 13 into 10 and three.

50 plus 10 is equal to 60, add another three is equal to 63.

So 36 plus 27 is equal to 63.

Izzy solves the equation here without using base 10 blocks.

Is she correct? She's got 38 plus 25.

Well, that looks all right.

She's partitioned 30 and 8 plus 20 and 5.

She's added her tens, 30 plus 20 is equal to 50.

That looks all right.

And eight plus five is equal to 13.

Well, that looks right as well.

50 plus 13, she says, is 53.

Oh, I'm not sure about that.

Can you see what she's done? Oh, she says, "I forgot to partition the 13." Well spotted, Izzy.

13 is equal to 10 and 3.

So now she needs to add her 10 from her 13 in with her tens.

50 plus 10 is equal to 60 plus another 3 is 63.

So 38 plus 25 is equal to 63.

So can you partition these numbers in the equation and solve it? We've got 42 plus 39, so use partitioning and don't fall into Izzy's trap or learn from Izzy's mistake as she did.

And remember to partition if your ones digits are going to bridge through 10, which I think they are.

Pause the video, have a go, when you're ready for the answer and some feedback, press Play.

How did you get on? Did you partition first? So 42 partitions into 40 and 2, and 39 into 30 and 9.

Then we can add our tens.

40 plus 30 is equal to 70, and then our ones, 2 plus 9 is equal to 11.

Now what do we need to do? Remember Izzy's step? That's right, we need to partition our 11 into 10 and 1.

70 plus 10 is equal to 80, add another 1 is equal to 81.

So 42 plus 39 is equal to 81.

Andeep wants to find the missing whole.

He thinks he can record the steps more efficiently.

Ah, so he's got 48 plus 24 is equal to something, and he's represented that as a bar model.

The parts are 48 and 24, and the whole, the sum, is the bit we're trying to work out.

He says, "I'll partition the numbers." So 48 is 40 and 8, 24 is 20 and 4.

He says, "I'll add the tens." 40 plus 20 is equal to 60.

And then "I'll add the ones," he says, eight plus four is equal to 12, and now "I'll recombine," he says, 60 plus 12 is equal to 72.

So he's done that stage all in one, hasn't he? He's realised that 12 is a 10 and 2 ones, so he's added a 10 onto his 60 and the 2 ones, but he's done that all in one go.

Can you use Andeep strategy to find the missing whole in the bar model? Remember, first, partition the numbers.

Pause the video, have a go.

When you're ready for some feedback, press Play.

How did you get on? Did you remember to partition the numbers first? So we had 30 and 7 and 50 and 4.

Then, we're going to add the tens.

30 plus 50 is equal to 80.

Then add the ones, seven plus four is equal to 11.

And finally, recombine, 80 plus 11, well, that's 10 and one more.

So 80 plus 10 is 90 plus one more is 91.

So 37 plus 54 is equal to 91.

Our whole is 91.

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

So in A, B, and C, you've got two pairs of equations, and you're going to partition both addends to solve the equations, but look carefully at what you've got.

Is there anything you notice as you're working? Pause the video, have a go at A, B, and C.

And when you're ready for the answers and some feedback, press Play.

How did you get on? Did you spot anything? So let's have a look at A, 37 plus 53.

So we've got 30 and 7 plus 50 and 3.

Well, 30 plus 50 is equal to 80, and seven plus three is equal to 10.

So our sum is 90.

What did you notice about the next equation? We had 37 plus 53, now we've got 53 plus 37.

Ah, it's 90 again, both equations have the same number of tens and ones.

In fact, they have the same numbers just in a different order.

So they both have the same sum.

What about B? We've got 37 plus 54 this time.

Ah, do you notice that's going to be one more than we had before? So that's 91.

And then we've got 37 plus 55, and another one, so 92.

One more we increased one of our addends by one, so our sum must also increase by one.

And in each equation, there's an addend of 37.

The other addend increases by one, so the sum will increase by one.

And what about C? Oh, did you see we had 37 plus 55 in B? This time we've got 36 plus 55, so we've decreased one of our addends by one.

So our sum must also decrease by one, it's 91, and then we've decreased again, 36 plus 55 has become 35 plus 55.

So our sum is 90.

Both equations have an addend of 55.

So when the other addend decreases by one, the sum will decrease by one as well.

Well done if you spotted those connections between the equations, and you were able to use your knowledge of increasing and decreasing addends to help you to solve those.

And on into the second part of our lesson.

This time we're just going to look at partitioning one addend to add two-digit numbers.

Izzy wants to solve this equation by partitioning both the addends just like we've been doing, but Andeep thinks he can find a more efficient way to solve this equation.

Go on, Andeep, show us.

He says, "I think it will be more efficient to partition one number instead of both numbers." So he's going to leave the 36 and just partition the 28.

He says, "I will show this on the number line." Oh, well done, Andeep.

So there's 36 on our number line.

So he's partitioned the 28 that he's adding on into 20 and 8.

So 36 plus the 20 first.

36 plus 20 is equal to 56.

We've added two more tens.

Now we've got to add on eight.

Hmm, that's going to bridge through 10, isn't it? He says, "When I add 8 to 56, I will cross the tens boundary, so I will bridge through 10." So what does he need to add on to 56 to get to the next 10? It's four, isn't it? 56 plus 4 will take him to 60.

But remember we are adding on 8 ones, so we need another 4 ones.

So add another four takes us to 64.

So 36 plus 28 is equal to 64.

He's added on those 8 ones in total but partitioned them into four and four, so we could bridge through the multiple of 10.

So 36 plus 28 is equal to 64.

Let's compare the two strategies.

So Izzy solved hers by partitioning.

What's the same and what's different? So Izzy partitioned both the numbers, 30 plus 20 is equal to 50, 6 plus 8 is equal to 14, and 50 plus 14 is equal to 64.

So she can partition her 14 into a 10 and a 4.

Add on the 10 and then add on the 4.

Let's look at Andeep's strategy.

We can picture the number line that he used as well.

36 plus 20 plus 8.

36 plus 20 is equal to 56, and 56 plus 8 is equal to 64.

We bridged through 10, didn't we? So we can use what we already know to help us in each strategy, and we can get the correct answer in both of them.

There are fewer steps when we use Andeep's strategy because we only partition one number, but we need to make sure that we are very confident in adding on a multiple of 10 to a two-digit number and thinking about our partitioning to bridge through 10 as well.

He says, "My method is more efficient." Andeep, your method is more efficient if you are as confident as you are with adding multiples of 10 and bridging through the next multiple of 10.

Izzy's strategy allows us a little bit more time to think about those slightly trickier moments in our equation.

When we're ready to move on to Andeep's, Izzy's strategy though is more efficient, I do agree.

So let's use Andeep's strategy to solve this equation.

Remember, we're only going to partition one of our addends.

We're going to partition the 28.

Pause the video, have a go, and when you're ready for some feedback, press Play.

How did you get on? So Andeep was just partitioning the second addend.

He was going to keep 48 as 48.

So he's got 48 on his number line, and he's going to add on 20 and then 8.

So 48, add 20, that's another 2 tens.

We've got 4 tens, 2 more tens, 6 tens, so 68.

Now we've got to add on eight.

Well, that's going to bridge through 10, isn't it? So we can partition the eight into two to get to 70 and six to get to 76.

So we've still added on eight.

So 48 add 20 is equal to 68 and 68 add eight is equal to 76.

So 48 add 28 is equal to 76 using Andeep's strategy.

So they're working on another equation here.

Each child partitions one addend to solve this equation.

Which way do you prefer? 69 plus 27 and 27 plus 69.

Let's watch their strategies.

So Izzy is starting with 69, and she's going to partition the 27 into 20 and 7.

She's going to add on the 20 to 69, which will get her to 89, and then she's going to add on the seven and we need to bridge through a 10.

So she's going to partition the seven into one to get to 90 and six to get to 96.

So she's still added on the seven.

So she's added on 20 and she's added on seven and she's landed on 96.

So 69 plus 27 is equal to 96.

Well done, Izzy.

Let's look at Andeep.

Andeep has put his addends in a different order.

He's starting with the 27, and he's going to add on the 69.

So there's 27 on his number line, and he's going to count on 60.

27 add 60 is 87.

Now he's got to add on the nine, so he can partition that into three to get to 90 and six to get to 96, but he's still added on a total of nine.

So he's added on the 60 and he's added on the nine.

And 27 plus 69 is, of course, the same sum, it's 96.

Which way do you prefer, do you think? Well, both methods are, of course, correct and they both get the same answer.

But you might have found Izzy's methods slightly easier because there was a slightly smaller number to add on.

So in Izzy's, we started with the larger number and we added on the 27, the smaller value.

With Andeep, we started with the smaller value and added on the larger value.

We get to the same answer, of course, but Izzy's might be slightly easier and maybe slightly more efficient.

Oh, now Izzy solves this equation and then hides one addend.

Let's find the missing addend, 57 plus something is equal to 83.

And we've got a two-digit number hiding there.

Andeep says, "I will add a multiple of 10 until I can't add any more tens." Ah, so he's going to see how close he can get to 83 by adding tens.

So he is got 57, what can he add? Well, he can add 10 to get to 67.

He can add another 10 to get to 77.

Another 10 would take him to 87.

Oh, that's too far, isn't it? He says, "If I added 30, I will reach 87, which is too much, so I will add 20." So 57 add 20 is equal to 77.

So we know we're adding on 20, we're getting close, but we need some ones, don't we? Now he says, "I will bridge 10 to find out how many ones are needed to reach 83." So we've got 77, he's going to add on three, which we'll get him to 80 and another three to get him to 83.

So how many ones has he added altogether? He's added six.

So altogether he's added 26 onto 57.

So 57 add 26 is equal to 83.

So he was able to use his number line and bridging through 10 to help him work out the missing addend.

Well done, Andeep.

I really like that strategy.

Can you put that into practise and use a number line to find the missing addend in this equation? 35 plus something is equal to 72.

Remember when you're adding on the tens, don't go too far.

So we're looking for a two-digit number to add to 35 that equals 72.

Draw a number line, have a go, pause the video.

When you're ready for the feedback, press Play.

How did you get on? So first, we're going to add a multiple of 10 until you can't add any more tens, but remember we've got to stop before we get to 72.

So 35 add what? How many tens can we add before we get too far? Well, we can add 30, can't we? If we reach 75, if we'd added another 10, we'd have added too many tens, but 35 plus 30 is equal to 65, 3 tens and 3 more tens is 6 tens.

And then we've got our five.

So we're getting close, but we now need to think about the ones.

Now we're going to bridge through 10 to find out how many ones are needed to reach 72.

So 65 add five gets us to 70, add another two gets us to 72.

So how many ones have we added altogether? Well, five plus two is equal to seven, we've added 7 ones.

35 plus 37 is equal to 72.

Well done if you were able to use Andeep's number line strategy to work out the missing addend, it's 37.

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

So for this question, you're going to partition one addend to solve the equations and draw a number line to solve it.

And again, can you find any patterns? So in A and B, we've got a missing sum to find each time.

And then in C and D, we've got some missing addends to find.

So pause the video, make good use of your number lines, and when you're ready for the answers and some feedback, press Play.

How did you get on? So did you do something like this? So for A, we had 45 plus 26 is equal to something, and we were just going to partition the 26.

So 45 and then we're going to add on 20 and 6, 45 add 20 is equal to 65, and then we've got six to add on and we can partition that into a five and a one, and we can bridge through 10.

So we've added six in total.

So 45 plus 26 is equal to 71.

Now what about the second one? 45 plus 26 is equal to 71.

Now we've got 55 plus 26, we've got 10 more, haven't we? So we could move that whole equation 10 more up the number line or we could work it out for ourselves that we know that the sum is going to be 81 this time, 10 more because we've added 10 more.

So in set A, one addend was the same, and the first addend in each equation increased by 10.

So the sum increased by 10.

Let's have a look at B.

We've got 45 plus 27 this time, it's one more.

We started A with 45 plus 26.

Now we've got one more, so our sum must be 72, and then we've got 45 plus 28.

Another one more, so our sum must be 73.

In set B, one addend was the same, and the second in each equation increased by one.

So the sum increased by one as well.

So we really only had to calculate the first equation, 45 add 26, all the others we could work out by knowing that the sum of that was 71.

And then changing things as one of the addends changed.

Well done if you spotted all of that.

So let's look at C.

So in C, we had a missing addend, 71 is equal to 45 plus something.

So let's use the number line to help us.

We're going to add tens until we get close, but we don't go beyond our 71.

So we can add 20 to 45, and we get to 65, which is just less than 70 and just less than 71.

Now we need to think about how many ones to add.

Well, we can add five to get to 70 and another one to get to 71.

So how many have we added altogether? We've added 6 ones, 26.

Does that remind you of some equations you've looked up before? So 71 is equal to 45 plus 26.

Now then let's have a look at the second one in C.

72, oh, we've got a sum that's one greater.

45, one part is the same.

If we want our sum to be one more, then our other part has to be one more, so it must be 27.

So in set C, one addend was the same, the 45, and the sum increased by one.

So the missing addend also increased by one.

So in set D, we've still got an addend of 45, but this time our sum is 62.

So we know that 45 plus 27 is equal to 72.

We want 62, which is 10 less.

So our other addend must be 10 less as well.

45 plus 17 is equal to 62.

Now let's look at the second one in set D, we've got something plus 46 is equal to 62.

Oh, so our sum is the same, but our 45 is increased by one to 46.

So if we picture our base 10 blocks, our 17 has got to decrease by one to be 16.

So 16 plus 46 is equal to 62.

In set D, each equation had the same sum, but one addend increased by one.

So the missing addend decreased by one.

I really hope you were able to use your reasoning as well as your partitioning and number line skills to help you to solve those.

Well done and great mathematical thinking.

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

We've been adding two-digit numbers, crossing the tens boundary.

What have we thought about? Well, when adding two-digit numbers that cross the tens boundary, we can partition both addends if we want to.

But when adding those two-digit numbers that cross the tens boundary, we can also just partition one addend.

This is a more efficient strategy, and we used a number line to help us, didn't we? And that number line can be used to help to find a missing addend as well when the sum crosses the tens boundary.

So great partitioning, great number line work, and great reasoning today.

Thank you so much for all your hard work, and I hope I get to see you again soon.

Bye-bye.