Lesson video

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

It's from our unit; Adding and Subtracting Ones and Tens To and From Two Digit Numbers.

So we're going to be looking at two digit numbers and we're going to be thinking about adding and subtracting and hopefully using our known facts to help us.

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

And in this lesson we're going to be using bridging to solve addition and subtraction problems. You may have done some bridging recently.

We're going to be using it for adding and subtracting.

So thinking about multiples of 10 and our known facts to and from multiples of 10.

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

We've got one key word in our lesson today and that's bridge.

So I'll say it first and then it'll be your turn.

My turn, bridge.

Your turn.

Well done.

I hope you know what that means when we're thinking about it in maths.

About going through a multiple of 10 to help us to solve a calculation.

So look out for it as we go through this lesson.

There are two parts to our lesson today.

In the first part we're going to be solving word problems and in the second part we're going to be solving problems in other contexts.

So let's make a start with some word problems. And we've got Alex and Sam helping us in our lesson today.

So we can bridge 10 to help us solve problems more efficiently.

Can you see some building bricks here? Let's look at this problem and think about how to represent it.

Alex had 57 yellow bricks and 4 red bricks, how many bricks did he have altogether? Sam says, "Is this an addition or a subtraction problem?" Well, he had 57 yellow bricks and 4 red bricks, how many did he have all together? And she says, "When we combine the parts to make the whole "it is addition." So that's what we're doing this time, isn't it? Which equation would represent the problem? Well, we've got 57 yellow bricks and 4 red bricks.

So our equation here represents those 57 yellow bricks and another 4 red bricks and we're using an addition.

The ones digits add to more than 10.

We've got 7 ones and 4 ones.

"So I think I can bridge 10 to solve this," says Sam.

Sam says, "I will use a number line to solve this." 57 has 7 ones, I need 3 more ones to reach the next multiple of 10.

I will partition the 4 into 3 plus 1.

So she's partitioned her addend of 4.

then 57 add 3 is equal to 60.

So here's her number line.

57 add 3 is equal to 60.

She's bridged through the next multiple of 10.

And 60 add 1 is equal to 61.

So 57 plus 4 is equal to 61.

So Alex must have 61 bricks in total.

and she can check that we've added 4 by combining the 3 and the 1 on the number line.

Sam thinks she can use Alex's bricks to make up a subtraction story.

I wonder what it could be.

Can you think of a subtraction story? This time she says, "I will start with the whole".

Altogether there are 61 bricks.

61 is the whole.

Four bricks are red, so four is a part.

Whole subtract part is equal to part so she can work out what colour the other bricks are.

61 subtract 4 is equal to something.

She says, "I will write this subtraction problem." Alex had 61 bricks, some were red and some were yellow.

Four of them were red, how many bricks were yellow? "Now let's solve it," she says.

So 61 is our total number of bricks and four is the number of red bricks.

So if we subtract the number of red bricks from the total, we'll find out how many yellow bricks there are.

She says, "61 has 1 one, "I need to subtract one "to reach the previous multiple of 10.

"I will partition the 4 into 1 and 3." So there it is.

Then 61 subtract 1 is equal to 60.

So she's showing that on a number line.

61 subtract 1 is equal to 60, and then 60 subtract 3 is equal to 57.

We can imagine regrouping one of those tens into 10 ones, subtracting three of them and we know that 10 subtract 3 is equal to seven, so she must have 57 bricks.

So that means that 57 of the bricks are yellow.

And she can check that she's subtracted four, 3 and 1 is equal to 4, so she's subtracted four in total.

Time to check your understanding.

Can you write an equation and then draw a number line to solve each of these problems? So there were 27 children standing up and 8 children sitting down.

How many children were there altogether? That's our first problem.

And our second problem says, there were 35 children, some were sitting down and some were standing up, eight of them were sitting down, how many were standing up? So pause the video, have a go at writing the equation and drawing a number line to solve each problem.

And when you're ready we'll get together for some feedback.

How did you get on? Did you spot in the first one.

We were combining the children standing up and sitting down to find out how many altogether, so it was an addition.

27 plus eight is equal to something.

27 on our number line and we were adding eight, so it made sense to partition our eight into three and five.

27 add 3 is equal to 30.

30 add 5 is equal to 35.

So we partitioned our addend of eight into three and five so that we could bridge through the next multiple of 10.

So there were 35 children altogether, so there were 35 children altogether and we can check to see that we have added eight, 3 plus 5 is equal to 8.

What about the next one? Well, this time there were 35 children and some of them were sitting down and some of them were standing up.

So 35 was our whole.

We knew that eight of them were sitting down, so we had to subtract those eight to find the other part, which is the children that were standing up.

So 35 subtract 8.

So there's 35 on our number line.

Well, it makes sense if we're going to partition through the previous multiple of 10.

So subtract the five first.

So subtract five will get us to 30, but we need to subtract eight in total and 8 is equal to 5 and 3.

So we've got to subtract another three.

30 subtract 3 is equal to 27.

Because we know that 10 subtract 3 is equal to 7.

And we can check that we've subtracted eight, 5 and 3 is equal to 8.

We have subtracted eight, so 27 children, were standing up.

We've got another problem here and we're trying to spot what's different about this problem.

First, Alex had 36 pencils, then he found some more pencils, now he has 43 pencils.

Alex says, "I'll draw a bar model "to help me understand the problem." First there were 36 pencils, so there are the 36 pencils, then some were added.

So 36 is a part and another part was added.

Now there are 43 pencils, so our whole must be 43.

So we are missing a part.

36 add something is equal to 43.

"I wonder how we can solve this," says Alex.

Let's have a think.

He says, "I'm going to use a number line".

First I have 36, but when I add some I have 43.

So we've got to find out what the difference is.

How many has he added? He says, "I know there is a multiple of 10 "between 36 and 43, "so I can use the bridge 10 strategy." "I wonder which multiple of 10 I'll use to bridge," he says.

Do you know which one it is? He says, "I know that the 40s come after the 30s, "so 40 is the multiple of 10 after 36." So we can fill in the 40 on the number line.

That's the multiple of 10 we're going to bridge through.

Now we can bridge to help us find the difference.

36 has 6 ones, so I must add four to reach the next multiple of 10.

So we're going to add four.

And I need three more to reach 43.

4 plus 3 is equal to 7.

So seven is the difference.

Seven pencils were added.

"This problem was different," he says, "because I had to find a missing addend".

And he did it by finding the difference.

Sam takes away some of Alex's 43 pencils and she puts them away.

Now he has 36 pencils, how many did Sam put away? Oh, Alex says, "I'm going to draw a bar model "to understand the problem again." First there were 43 pencils, then some were subtracted.

So 43 is our whole this time.

Some were taken away we dunno how many.

Now there are 36 pencils.

36 is the other part.

43 subtract something is equal to 36.

"I wonder how I can solve this," he says.

He says, "I'll use a number line".

First Alex had 43 pencils and now he has 36 pencils.

How many pencils were subtracted? That's what we need to find out.

We're finding the difference again.

Sam says, "I know there is a multiple of 10 "between 43 and 36.

"So I can use the bridge 10 strategy." Which multiple of 10 will I bridge? Well she says, "43 is in the 40s decade "and the multiple of 10 "at the start of the 40s decade is 40, "so it must be 40." Alex says, "43 has 3 ones.

"So I must subtract three "to reach the previous multiple of 10." So we can fill in subtract three.

I need to subtract four more to reach 36.

He knows that a multiple of 10 subtract four will get us a number with the 6 in the ones.

So subtract four.

"Oh, 4 and 3 is 7 "so seven pencils were subtracted," he says.

"This problem was different "because I had to find the part that was subtracted".

Over to you to check your understanding.

Fill in the multiple of 10 that you would use to bridge on each number line.

And what do you notice about each multiple of 10? Pause the video, have a go.

And when you're ready we'll get together for some feedback.

How did you get on? Well, we know that the 80s come after the 70s.

So 80 is the next multiple of 10 after 78.

What about the other number line? Well we know that 83 is in the 80s and so the multiple of 10 at the start of the 80s is 80.

So the previous multiple of 10 before 83 is 80.

What do you notice? Well, each multiple of 10 is the same because 80 is the multiple of 10 between 78 and 83.

So it can be used when finding the difference between them, whether we're counting on or counting back.

Alex notices something about these two number lines.

He says, "In the first example I added four and then three.

"And in the second example "I subtracted three and then four." Do you remember he was finding out how many pencils had been put away or how many pencils he'd found.

He says, "In both I added and subtracted seven "because the difference between 36 and 43 is always seven." It doesn't matter if we're counting on to find that difference or counting back, the difference will always be seven.

Time to check your understanding.

Use the number line to help you find the missing numbers.

And what do you notice? Pause the video.

Have a go and when you're ready we'll get together for some feedback.

How did you get on? Well, 58 has 8 ones.

So 8 plus 2 is equal to 10.

So I must add two to reach the next multiple of 10, which is 60.

And then 60 plus 4 is equal to 64.

And altogether I added 6, 58 plus 6 is equal to 64.

What about the other number line? Well, 64 has four ones, so I must subtract four to reach the previous multiple of 10.

64 subtract 4 is equal to 60, and then 60 subtract 2 is equal to 58, altogether I subtracted six.

And I noticed that in each case the total number added or subtracted was six because the difference between 58 and 64 is always six, whether we're counting on or counting back.

Time for you to do some problems. Match the equations to the problems and then use the bridge 10 strategy to solve them.

So we've given you the equations that match the problems and we want you to match them and then solve them using a number line to bridge 10.

And you've got six to have a go at.

Pause the video, have a go.

And when you're ready we'll get together for some feedback.

How did you get on? Did you spot the.

Question one, match the equation 48 plus 7.

Sam collected 48 pebbles and Alex collected 7, how many pebbles did they collect altogether? And if you used a number line to help you, you'd have worked out that the answer was 55.

Question two matched with the first equation.

55 subtract 7.

Sam and Alex collected 55 pebbles.

Alex collected seven, how many did Sam collect? We knew one part and the whole, so we had to subtract that known part from the whole, and if we did the subtraction the answer was 48.

So question three was this last equation.

Sam and Alex collected 55 pebbles, but Sam dropped some, now they have 48 pebbles, how many did Sam drop? So 55 subtract something is equal to 48.

And if we work that out on the number line, we knew that the missing number was 7.

Oh, did you spot the same numbers in all of those equations? What about the next ones? So the first equation, 69 plus something is equal to 73 matched with question five.

There were 69 children in the hall, some came in and now there are 73 in the hall.

How many children came in? So we had to find the difference.

And if we used our number line to bridge through 10, we'd have found out that the difference was four.

69 plus four is equal to 73.

The second equation matched question six.

There were 69 children in the hall and 4 more came in.

How many children are in the hall now? We knew the two parts, so we had to add them to find the whole.

69 plus 4 is equal to 73.

So question four must have matched with our final equation.

73 subtract 4 is equal to something.

There were 73 children in the hall, four were sitting down, how many were not sitting down? So we knew the whole and one part, so we could subtract the part from the whole.

And if we did that subtraction using a number line and bridging through 10, we'd find that the answer was 69.

I hope you were successful with those.

And on into the second part of our lesson.

Alex is sorting some equations according to whether they can be solved by bridging a multiple of 10.

So he's got four equations there.

How can we tell if an equation is going to bridge a multiple of 10? These are all additions.

So let's look at this first one.

Sam's helping us out with this.

She says, I know that five plus eight is greater than 10, so I can bridge the next multiple of 10 to cross the tens boundary.

So that one will need to bridge through 10.

What about the next one? Oh, she's looking at the ones again.

I know that 5 plus 3 is less than 10, so I do not bridge because I do not cross the tens boundary.

35 plus 3 will not bridge through the next multiple of 10, so we don't need to.

So it goes there on the table.

What about the next one? 28 plus 3.

What do you notice about the ones digits? Yes, she says I know that 8 plus 3 is greater than 10, so I can bridge the next multiple of 10 to cross the tens boundary.

And what about the last one? 57 plus 2.

That's right, 7 plus 2 is less than 10.

So we will not have to bridge because we don't cross a tens boundary.

So well done.

Sam's helped Alex to sort those equations into the right places in the table.

Time to check your understanding.

In which of the following equations will you bridge 10? Can you have a look? Pause the video, have a go, and when you're ready we'll get together for some feedback.

Did you spot that it was A.

When 8 and 7 are added, the total is more than 10, so we can bridge 10 to cross the tens boundary.

And another check, is it true or false? The bridge 10 strategy can be used to solve each of these equations.

So we've got A and B, decide whether it's true or false and when you're ready we'll get back together for some feedback.

What did you think? In example, A, this is true, we know that 5 plus 8 is greater than 10.

So 15 plus 8 will be greater than 20.

It will cross the tens boundary so we can bridge 10.

And there we can see 15 plus 5 is equal to 20, plus another 3 is equal to 23.

What about in B? Well it's also true in B.

When there are 2 ones we will cross the tens boundary if we subtract 3 ones.

So we will need to bridge 10 to subtract.

42 subtract 2 is equal to 40, subtract another one is equal to 39.

Alex wants to solve the equation by bridging the next multiple of 10.

He's got 8 plus 54.

He says, "I know addition is commutative, "so I will change the order of the addends.

8 plus 54, he can write as 54 plus 8, they're the same.

I will partition the second addend so I can bridge the next multiple of 10.

It doesn't matter which way around the addends are, but this way it's perhaps easier to see our partitioning.

So he is going to partition 8 into 6 and 2.

54 plus 6 is equal to 60 and 60 plus 2 is equal to 62.

And he can check that he's added eight altogether.

So 54 plus 8 is equal to 62.

And Sam says she can check Alex is correct by using a known number fact.

Oh, that's an interesting one.

Which number fact should she use? Sam says 8 ones are added to 4 ones, 8 plus 4 is equal to 12, so I can use this fact.

I know there must be 2 ones in the sum and there are 2 ones in 62 another way to check.

Sam writes a pair of equations, 59 plus 4 is equal to 63, and 64 plus 9 is equal to 73.

Let's think about what's the same and what's different about each equation.

Sam says the tens digits are different.

We've got 59 plus 4 and 64 plus 9.

In each equation the tens digit in the sum is one more than the tens digit in the first addend.

59 plus 4 is equal to 63, and 64 plus 9 is equal to 73.

The ones digits in the sum are the same in each equation.

Oh yes, so we've got 63 with 3 ones and 73 with 3 ones.

The ones digits in the addends are the same but in a different order.

Oh, let's look at that.

We've got 59 plus 4, and then we've got 64 plus 9.

So we've got 9 ones and 4 ones in each equation.

But in the first one, the 9 ones are with the tens.

And in the second one, the 4 ones are with the tens.

But Sam says that means both can be solved using the known fact 9 plus 4 is equal to 13 or 4 plus 9 is equal to 13.

Good thinking Sam.

Perhaps you could write some similar examples.

Alex says if you subtract seven from a number ending in a three, it will always leave a number with 6 ones.

Woo, right, let's think about that.

So he's got 43 subtract 7 and he thinks it will have a difference with 6 ones.

He says to reach the previous multiple of 10, I must subtract 3 to reach 40 and then subtract 4.

So I'll partition 7 into 3 and 4.

And he's drawn a number line.

43 subtract 3 is equal to 40, and subtract another 4 is equal to 36, so he does have a 6 in the ones.

Sam says, "Did you still subtract 7 from 43?" He says, "I subtracted 3 "and then I subtracted 4, "so I subtracted 7 in total." He says, "If I subtract 7 "from any number with 3 ones, "I will always partition it into 3 and 4 "to reach the previous multiple of 10 "and then subtract 4.

So Sam says, "So this means "you will always reach a number ending in a 6." So good thinking there Alex, you were right.

And we've changed our number line to show that it doesn't matter what our tens digit is.

If we subtract three, we'll get to the next multiple of 10.

And if we subtract 4, we'll end up with a number with a 6 in the ones.

Time to check your understanding, which of the following is true and can you prove it.

So A says if we add a number with a ones digit of 9 to a number with a ones digit of 3, it will always have a ones digit of 6.

B says, if we subtract 5 from a number with a ones digit of 3, the number we reach will always have a ones digit of 8.

And C says, if we add a number with a ones digit of 4 to a number with a one digit of 5, it will always have a one digit of 1.

Which of the following is true and can you prove it? Pause the video, have a think and then we'll get back together to discuss our answers.

What did you think? Oh, it's B isn't it? If we subtract 5 from a number with a ones digit of 3, the number we reach will always have a ones digit of 8.

And you can see that on the number line.

We know that to reach the previous multiple of 10, we must subtract 3 from a number with a ones digit of 3.

So 5 would always be partitioned into 3 and 2.

And if we subtract that 2 from a multiple of 10, we'll always end up with a number with an 8 in the ones.

Well done of you spotted that and were able to explain it.

So is Alex right here? Alex says, I know that 6 plus 5 is equal to 11, so the sum will have a ones digit of 1.

This equation must be correct.

He says 36 plus 5 is equal to 51, is he right? Sam says, when you add a single digit to bridge the next 10, you move into the next decade.

And they've added a single digit 36 add 5, have they moved into the next decade there with their answer? She says the tens digit in the sum will be one more than the tens digit in the first addend, so the sum will be 40 something.

36 plus 5.

So we're going to add 4 to get to the next multiple of 10 and then add 1.

So 36 plus 4 plus 1, equals 40, and another one is 1.

So it's 41, not 51.

Alex's thinking was right, but he hadn't thought about the tens.

Oh, he says the sum will be 41.

Time for you to do some practise.

Can you tick the equation that can be solved by bridging the next multiple of 10 in A? In B, can you spot the mistakes and correct them? And in C, which of the following facts could you use to check this equation? And in question two, can you complete the equations? Pause the video, have a go at your practise tasks, and when you're ready we'll get back for some feedback.

How did you get on? Let's look at A.

So in A, we were thinking about which ones we could solve by bridging 10, and it was the final one, 54 plus 9.

The ones digit sum to more than 10 so I could bridge the next multiple of 10 to solve this one.

What about B? You had to spot the mistakes.

So we had 67 plus 6 is equal to 83.

Well, 7 plus 6 is equal to 13.

So the ones digit is correct, but when we bridge 10, we cross 70, so the sum should be 73 for our first one.

What about the next one? Well, 7 plus 6 is equal to 13, so the ones digit of the sum must be a 3.

We partition the 7 into 4 and 3, so we get to our next multiple of 10 and then we add on another 3, so the sum should be 93, well corrected.

And in C, which of the following known facts could you use to check this equation? 34 plus 9 is equal to something.

Ah, there it is, isn't it? 4 plus 9 is equal to 13 or we could have used 9 plus 4 is equal to 13 because we know that addition is commutative.

So that would help us to check.

We'd need to know that there was a three in the ones digit of our answer because 9 plus 4 is equal to 13, and we'd know we were going on into the next decade, so we'd have an answer of 43.

And what about the missing numbers here? So in A, we knew that 26 plus 5 was equal to 31, so 36 plus 5 must be equal to 41.

In B, 53 subtract 9.

Woo, so we're going to subtract 3 and then subtract another 6 so it must be 44.

And then 73 subtract 9.

We're going back into the previous decade, so it must be 64.

In C, 58 plus 3.

Well we can add 2 to make 60, add another one, so that's 61.

And then 53 plus 8.

Did you spot? We've just swapped the 3 and the 8, the ones digits around, haven't we? So the sum will be the same, 61.

In D, 76 subtract 7 is equal to 69.

So 86 subtract 7 must be equal to 79.

In E, 58 plus 6 is equal to 64.

So 68 plus something is equal to 74.

Well it must be 6 again, mustn't it? And for the third one there, something plus 6 is equal to 84, well it must be 78 if we're following that pattern.

And then for F, 52 subtract 6 is equal to 46.

So 62 subtract 6 must be equal to 56 and 72, subtract 6 must be equal to 66.

I hope you used your thinking to work out lots of those missing answers in those equations.

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

We've been using bridging to solve addition and subtraction problems. So what have we learned? We know that we can use known facts to help us find new facts efficiently, and we can use facts to bridge 10 when we're solving addition and subtraction problems. Thank you very much, you've worked really hard in this lesson and we've used lots of mathematical thinking, haven't we? I hope you've enjoyed it as much as I have and I hope I get to work with you again soon.

Bye-Bye.