Lesson video

Hi, my name is Mr. Tazzyman, and I'm looking forward to learning with you today.

This lesson is from the unit all about equivalence and compensation with addition.

It'll give you some extra hints and tips on how to face addition problems and solve them mentally, rather than jumping straight to a written method.

Let's get started then.

Here's the outcome for today's lesson.

By the end, we want you to be able to say, "I can use redistribution with addition of integers." Here are the key words that you're going to hear during this lesson.

Let's practise them.

I'll say them, and I want you to repeat them back to me.

So I'll say my turn, say the word, and then I'll say your turn, and you can say it back, ready? My turn, redistribution, your turn.

My turn, sum, your turn.

My turn, whole, your turn.

My turn, part, your turn.

Let's look at what each of these keywords actually means.

Redistribution is where part of one addend is moved to another addend.

It is sometimes known as the same sum rule.

The sum is the total when numbers are added together.

The whole is all the parts or everything, the total amount.

A part is some of the whole.

And you can see that there is a bar model at the bottom there just showing the relationship between whole and part.

Here's the outline then for today's lesson on using redistribution with addition of integers.

To begin with, we're gonna look at redistribution with two digit numbers.

Then we're gonna look at redistribution with larger numbers.

Let's start with the first part.

Here are two maths friends, Sofia and Andeep.

They're gonna help us throughout this lesson by discussing some of the maths prompts, giving us some thoughts and ideas, and even giving us some hints and tips to get to some of the answers.

Hi, Sofia, hi, Andeep.

Let's go.

Sofia has an addition with larger numbers to solve.

299,999 plus 455,672 is equal to.

It's an unknown represented by a question mark.

How would you solve it? I'm not asking you to solve it now, but just have a think.

How would you go about it? What would you do? What method would you use? Sofia says, "I think I need to use column edition "because they are six digit numbers "and would need lots of regrouping." Andeep replies, "I don't think you need to.

"There is a quicker way, "if you look closely at the parts you are adding together." "But I know columns really well, "and it always works." "Does it," says Andeep.

"Well, no, but I know the steps off by heart." "Let's try it mentally using jottings instead, "using redistribution.

"Why not try it?" "Okay, I'll give it a go." Andeep shows Sofia a bucket balance.

Andeep puts 29 blue cubes into one side of a bucket balance.

You can see the bucket balance changing because the blue cubes have weighed one side down.

Then he adds 15 red cubes.

It changes slightly more.

Andeep puts 30 blue cubes into the other side of the bucket balance.

There they go.

Then he adds 14 red cubes.

What do you notice? Have a look at both of those sides.

What can you see? What's the same, what's different? Sofia says, "Both sides have an array of cubes "with one missing, "but the missing cube is a different colour in each bucket.

"The bucket scale is balanced, "so the number of cubes on each side must be equal." We can write this as an equation.

29 plus 15.

There's the expression for one side of the bucket balance.

30 plus 14.

There's the expression for the other side of the bucket balance.

The bucket scale is balanced, so we know that the totals are of equal value.

Which of these two expressions would be the easiest to solve mentally? What do you think? If someone were to ask you a question, would you prefer to answer it if it was 29 plus 15 or 30 plus 14? Sofia says, "I am confident with both, "but the second one is easier "because one of the parts is a multiple of 10.

"That makes it easier to add onto "because you don't have to bridge twice." We can use redistribution to transform addition calculations, making them easier to solve mentally.

Base 10 blocks can be used to represent redistribution.

Here are the two parts to start with.

So Andeep is taking one of the expressions that we had from the bucket scales.

Watch what happens.

What do you notice about the total number of cubes? The total number of cubes, the whole, remains the same, even though the parts have changed through redistribution.

If one part is increased by an amount and the other part is decreased by the same amount, the whole remains the same.

Jottings can be useful to help with redistribution.

We've got our representation on the right hand side here, and then we're going to put the jottings down on the left hand side.

There goes the cube that's being redistributed, and you can see in the jottings that one has been subtracted from the second addend to give 14 instead of 15.

Then we need to redistribute that one over to the first addend, so we take 29 and add one to give us 30.

Altogether, that's equal to 44.

And there are 44 cubes there.

These jottings help me to calculate mentally.

I might not need them all of the time.

So Sofia is saying that the jottings that have been written down there help her, but as she gets better, she might not even need to make those.

She might be able to do it all mentally.

"If I write it down this way, "I don't have to hold lots of information in my head." I feel like that sometimes, Sofia.

Writing and jotting down is a really good thing to do because it helps you to remember something without having to use your brain to do it.

It's really quick too.

This will help me do that tricky question at the start, I think.

Okay, it's your turn.

Let's check your understanding so far.

Use redistribution to transform the parts in this calculation and find the whole mentally, jottings if they help.

39 plus 27 is equal to an unknown, we don't know yet.

Pause the video here and have a go at that using redistribution, good luck.

Welcome back.

Let's look at the jottings that you may or may not have used to use redistribution to solve this expression.

Did you add one to give you 40? So subtract one to give you 26, making an easier calculation.

The answer was 66.

Okay, let's move on.

Sofia and Andeep compare three different examples of redistribution.

We've got 99 plus 53, that's been redistributed by taking one away from the second addend and putting it onto the first.

That gives us 100 plus 52, which is equal to 152.

Then we've got 97 plus 53, and we've redistributed three from the second addend to the first addend, giving us 100 plus 50 equals 150.

And lastly, we've got 98 plus 53, and we've redistributed two from the second addend to the first addend, giving us 100 plus 51 equals 151.

What's the same? What's different? Have a look at all three of those sets of jottings.

What similarities are there, what differences are there? Sofia says the first part is different in each edition.

It is decreasing by one each time.

Consequently, the whole is also decreasing by one each time.

Andeep says the amount redistributed is increasing by one each time.

This is because the first part is getting further from the next multiple of 100.

Okay, let's check your understanding again.

If you redistribute from the second to the first part, which of the following needs the greatest amount redistributed? Solve them both mentally with jottings using redistribution.

We've got 27 plus 32 and 46 plus 55.

Pause the video here and have a go at that.

Welcome back, let's see how we got on.

So in the first one, you might have redistributed three from the second addend to the first.

That would give you 30 plus 29, which is 59.

And in the second one, you might have redistributed four away from the second addend to the first, giving you 50 plus 51, which is equal to 101.

The second has a greater redistribution.

We redistributed four in the second one and only three in the first.

Okay, ready to move on? Let's go for it.

Sofia and Andeep both use redistribution for an addition.

There's the two different jottings.

What do you notice? Have a look at them, compare them.

"I redistributed a different amount," says Sofia.

"I chose to redistribute three "from the second part to the first part." "I chose to redistribute two "from the second part to the first." Both ways transformed apart to a multiple of 10, making the addition easier.

You can see that in Sofia's, her final addition that she had to make was 30 plus 59, which is equal to 89.

And Andeep's was 29 plus 60, which is also equal to 89.

They both got the same result, but they both had different multiples of 10 in their transformed calculation.

Sofia and Andeep both complete another.

What do you notice? Have a look at them again, compare them.

What's the same and what's different? "I redistributed a different amount.

"I chose to redistribute from the second part "to the first part." So Sofia took the second part, 41, and redistributed two from that over to the first part of 38.

38 plus two is 40.

41 subtract two was 39.

So her transformed calculation was 40 plus 39, which is equal to 79.

Whereas Andeep decided that what he would do was redistribute from the first part.

He subtracted eight from 38 to give him 30, redistributed that eight across to the second addend, 41 plus eight is equal to 49, so his transformed calculation was 30 plus 49, which is also equal to 79.

They both got the same result.

Andeep says, "I chose to redistribute "from the first part to the second." They both achieve the same whole or sum.

They are both valid methods.

"I suppose it depends which you find quickest.

"That will be your most efficient way." Okay, let's check your understanding again.

True or false, if a part is decreased by a certain amount and redistributed to another part, then the whole changes.

Have a think about that statement and decide whether you believe it to be true or false.

Pause the video here.

Welcome back.

That statement was false, but let's think about why.

Here are two justifications, and your next job is gonna be to choose which one of these you think gives us the best reason as to why that statement was false.

Pause the video here.

Welcome back.

A was the correct justification here.

If amounts are redistributed between parts, then the whole remains the same.

Okay, ready to move on? Let's go for it.

Here's the first practise task.

You are going to start by filling in the missing numbers, and A and B are both sets of jottings that feature redistribution.

So you're gonna have to think not just about what the answer might be, but also about how much has been redistributed and what the new parts would be.

In number two, A and B, you're going to use redistribution to calculate those expressions that have been chosen with digit cards.

For number three, you're gonna solve this addition mentally with jottings using redistribution in three different ways.

So think about three different ways that you could redistribute to solve that sum.

Sofia also asks, "Which do you prefer and why?" So be ready to justify which method you think is best.

And lastly, for number four, in the examples below, redistribution has been used.

What happens to the whole if an amount is redistributed between the parts.

Explain your answer.

Okay, pause the video here and have a go at those practise tasks.

Welcome back, let's get some answers, so be ready to mark.

One, A, those with the missing numbers.

Three was redistributed from the second addend to the first, giving a transformed calculation of 40 plus 62, which is equal to 102.

For B, this time, two was redistributed from the first part to the second part, giving a transformed calculation of 20 plus 77, which is equal to 97.

Because you knew that 77 was the second addend in your transformed calculation, that meant that you knew the second part in the initial calculation was 75, because two had been added onto it.

Let's look at number two then.

Here are some jottings of how you might have chosen to do redistribution.

We've got one redistribution from the second part to the first part, giving a calculation of 36 plus 40, which is equal to 76.

You might have chosen to have redistributed five instead.

If you've got the answer 76, that's okay.

Two, B.

Two has been redistributed from the first part to the second part.

Again, you might have chosen to redistribute six instead, but if you've got an answer of 78, that's okay.

Here's number three, then.

Solve this addition mentally with jottings using redistribution in three different ways.

Here's one way where you've redistributed three and you've ended up with 30 plus 29, which is equal to 59.

Here's another way where you've redistributed two and you've ended up with 29 plus 30 is equal to 59.

Sounds very familiar, doesn't it? And then lastly, you might have chosen to redistribute seven, giving you 20 plus 39, which is equal to 59.

So three different redistributions.

The first was three, the second was two and the last was seven.

Sofia asked, "Which do you prefer and why?" Well, Andeep says that, "Here are three possible ways.

"I prefer the middle one because there's no bridging "and simpler redistribution." What did you think? You might have had a different answer to that.

Just make sure that you've written a good justification.

Here's number four, then.

We had some examples with redistribution.

The question was, what happens to the whole if an amount is redistributed between the parts? Explain your answer.

If an amount is redistributed between parts, the whole remains the same.

Did you get that? I hope so.

Okay, we're gonna move on.

It's time for the second part of the lesson, redistribution with larger numbers.

Sofia looks again at her tricky question.

"So do you think we could use redistribution "to solve this problem?" "I think so.

"Let's look at some examples with larger numbers now." "Yes, because I'm still desperate to use columns, "but I think redistribution "might end up being more efficient." What's the same and what's different between these three redistributions and their jottings.

Have a look, compare them.

What's similar, what's different? What do you notice? Sofia says, "The parts have the same digits, "but with different place values shown by the placeholders." "The first is 29 ones added to 15 ones.

"The second is 29 tens added to 15 tens.

"The third is 29 hundreds added to 15 hundreds." "Each redistribution is only one unit, "but the place value in each is different.

"In the first it's one, "the second it's 10 and in the third is 100." What's the same, what's different between these three? The numbers of digits is different.

The numbers get larger, but the amount being redistributed is the same.

It's one each time.

It's simpler than it looks at first glance.

Yes, by looking closely at the parts being added, you can see that it's ready for distribution.

Sofia returns to her tricky problem to have a go.

What do you think now? So have a look at that tricky problem.

We saw it earlier in the lesson.

What do you think of it now that we've learned a little bit more or remembered a little bit more about redistribution? "I'll look closely at each part "to see if I can redistribute "to make them easier to work with mentally.

"The first part is only one from 300,000, "so that would work.

"The second part would be awkward to redistribute to, "so I won't do it that way.

"I'll do it mentally with jottings." So she writes out the expression initially.

She redistributes one from the second part to the first part.

Now she has 300,000 plus 455,671.

That feels a bit easier to do.

That's equal to 755,671.

"Wow, by looking closely at the parts, "I redistributed and I didn't need to use columns." Wa-hey, "It's a very useful method "and much more efficient.

"Although the parts "aren't always suitable for redistribution." Okay, have a go yourselves now at redistribution featuring larger numbers.

Solve the following mentally with jottings by redistributing 134,087 plus 699,998.

Sofia gives us a tip here.

"Remember to look closely at the parts first "to see which is best for redistribution." Pause the video here and have a go.

Welcome back.

Let's look at the jottings, then.

Two was redistributed from the first part to the second part, making the calculation much easier because the second part was now 700,000.

So the expression now read 134,085 plus 700,000.

Those two added together were equal to 834,085.

How did you get on? Did you get it? I hope so.

Okay, let's move on.

Sofia and Andeep look at some addition equations with larger numbers.

They categorise them into those they think were sold using redistribution.

You've got 246,672 plus 599,995 is equal to 846,667.

The second one reads 581,742 is equal to 346,891 plus 234,851.

So which of these do you think was sold using redistribution? We've already got the answers.

We're thinking about the journey to that answer.

What do you think? Well, Andeep says, "Let's look at the parts in both equations closely." "What are we looking for," says Sofia.

"If either of the parts are near to a boundary "that will make the addition easier." "I think the first equation was sold using redistribution." Do you agree and why? So have a look.

Why do you think Sofia has said she thinks that the first equation was solved using redistribution? "I agree," says Andeep.

"The second part is only five away from 600,000, "so easy to redistribute to." "The second equation doesn't have any parts "that are close to a boundary.

"I think I would have used a written method for that one." Okay, let's move on to looking at missing inequalities.

Sofia and Andeep look at a missing equality problem.

400,002 plus 376,764, and then we've got a question mark because we don't know what symbol's gonna go in there.

And then we've got 376,759 plus 399,997.

What do you think? "Let's look at the parts "to see if we can use redistribution." "Both expressions have parts close to a boundary." "Okay, I'll do this side, "and you do the other." So Sofia has used redistribution.

She's redistributed two, and she's ended up with 776,766.

She writes that in.

Andeep calculates the other side using redistribution.

He redistributes three, and he gets an end result of 776,756.

Now they can compare them more easily.

776,766 is greater than 776,756 by 10.

So the symbol is greater than.

Well done, Sofia and Andeep.

Okay, here is the second practise task.

Your job to begin with is gonna be to fill in the missing numbers, quite similar to what you did last time.

For number two, you're gonna use redistribution to calculate these additions.

For number three, you're gonna look at which inequality symbol goes between these expressions.

Okay, pause the video here and have a go at those, and I'll be back in a little while with some feedback.

Welcome back.

Let's start with one, A.

Two was redistributed from the second part to the first part.

When you had the different calculation that you'd redistributed, the second part read 427,592, and the whole was 927,592.

Let's look at one, B.

We knew we were redistributing three, and our transform calculation read 234,591 plus 300,000 is equal to 534,591.

Let's look at two then.

Two, A, four was redistributed from the first part to the second part, and the transform calculation read 127,590 added to 700,000 equals 827,590.

Let's look at B.

10 was redistributed in B.

That gave a new transform calculation of 200,000 plus 235,776 is equal to 435,776.

Let's look at three then.

We were looking to see which inequality symbols would go between these expressions.

So in the first one, after redistribution, we found a total of 769,831.

And after redistributing in the second one, we had 769,826.

The symbol should have been greater than.

For B, we'd redistributed five and ended up with a whole of 835,818 on the left hand expression.

And on the right hand expression, we'd redistributed 10 and ended up with 835,824.

So the symbol was less than.

Remember, you do need to check that you've got the calculations correct because if you just simply mark the fact that you got greater than and less than, you could get those with a 50-50 chance.

You need to know that you've definitely got the maths correct behind it as well.

Okay, then, let's summarise all of our learning today.

Redistribution can be used to transform a calculation to make it a more efficient mental addition.

This works for smaller and larger numbers.

It can be achieved by increasing or decreasing an amount from a part and redistributing that amount to another part.

The process of redistribution doesn't change the value of the whole.

The whole remains the same.

My name's Mr. Tazzyman.

I've really enjoyed the learning today, and I hope you have too.

I'll see you again on another lesson.

Bye, bye.