Lesson video

Hello, my name is Mr. Tazzyman and I'm gonna be teaching you today's lesson from the unit that's all about equivalence, compensation and addition.

I hope that with some of the learning that you experience today, you'll have a better understanding of some of the underlying structures for addition problems that you face in maths.

Here's the outcome for the lesson.

By the end, we want you to be able to say, "I can use redistribution with addition of decimal fractions." Here are the keywords that you might hear during the lesson.

I'm gonna 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 repeat it.

Ready, my turn.

Redistribution.

Your turn.

My turn, decimal fraction.

Your turn.

My turn, whole.

Your turn.

My turn, part.

Your turn.

Here's what those words mean then.

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

It is sometimes knows as the same sum rule.

A decimal fraction is a fraction written as a decimal.

A decimal point separates whole numbers from decimal fractions.

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

A part is some of the whole.

And the bar model below shows the relationship between whole and part.

Okay, here's the outline of the lesson.

To begin with, we're gonna look at redistribution with ones and tenths.

Then we're gonna move on to looking at redistribution with hundredths.

Let's get started.

Here's two friends we're gonna meet during this lesson, Alex and Lucas.

They're gonna help us by discussing some of the maths that we think about, giving us some prompts, giving us some hints and sometimes revealing some answers.

Hi, Alex, hi, Lucas.

All right, let's get going.

Alex has an addition involving decimal fractions to solve.

Something is equal to 2.

95 plus 1.

71.

How would you solve this? So have a look at that expression.

What would you do to find what it's equal to? Alex says, "This look like one for column addition.

Decimal fractions make it a bit awkward and I know how to do columns really well." Lucas says, "Do you always get columns correct though? Does it always work quickly? I think there's a quicker, mental method.

Let's do some exploring to find out if we can use redistribution and equivalence." Lucas explores with cubes and bucket scales.

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

Then he adds 15 red cubes.

Lucas put 30 blue cubes into the other side of the bucket balance.

Then he adds 14 red cubes.

What do you notice? Look at that bucket balance and look at the quantities in each bucket.

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 is the first expression.

30 plus 14 is the second expression.

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

They are equivalent although paired parts differ.

So there's the equals sign.

Which of these two expressions would be easiest to solve mentally? What do you think, have a look? 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.

Alex makes a suggestion.

What if we took the value of each cube to be one tenth? What would change and what would stay the same? Have a think about that.

Imagine that the 29 in that first expression wasn't 29 ones, it was 29 tenths and do the same with the other parts as well.

What would stay the same and what would change? "The place value of each part would have to change," says Lucas.

29 tenths plus 15 tenths is equal to 30 tenths plus 14 tenths.

The numbers would become one tenth the size.

So they would turn into decimal fractions with a decimal point.

29 tenths would be two and nine tenths.

That's 2.

9 as a decimal fraction.

Let's change the expressions, and there they are.

2.

9 plus 1.

5 is equal to 3.

0 plus 1.

4.

The two expressions are still equivalent.

We can use redistribution to transform addition calculations making them easier to solve mentally.

Base 10 can be used to represent redistribution, even with decimal fractions as long as each cube represents a decimal fraction.

Did you see the cube disappear? It's redistributed to the other side making 3.

0.

Altogether, that's equal to 4.

4.

Remember, each cube is no longer thought of as representing one, instead it now represents one tenth.

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.

This is true for both decimal fractions and integers.

The difference is in place value not quantity.

Jottings can be useful to help with redistribution.

Can you see that we've taken one cube away from the expression on the right and the jotting is up there on the top left of the screen have been written in to represent this.

We've got an arrow with a subtract 0.

1, subtract one tenth.

The cube goes over the other side and you can see in the jottings, we've put plus one tenth.

That gives us 3.

0.

We combine those cubes together, giving us a result of 4.

4.

These jottings help me to calculate mentally.

I might not need them all of the time.

If I write it down in this way, I don't have to hold lots of information in my head.

It's really quick too.

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

Okay, let's check your understanding.

Use counters or base 10 to model the redistribution below.

Treat each ones block as one tenth.

And if you're using counters, you can treat each counter as one tenth.

Pause the video here and have a go at that and I'll be back in a little while.

Welcome back.

Here is the redistribution modelled using base 10.

We start with 19 tenths and 15 tenths.

We redistribute from the second part, giving us 20 tenths and then 14 tenths and that gives us 3.

4 when we have combined all the cubes together.

Did you do something like that? I hope so.

Okay, let's move on.

Lucas and Alex compare three different examples of redistribution.

There's the first.

There's the second and there's the third.

What's the same and what's different? Look closely at them.

Compare them and what do you notice? Remember, we've already got the answers, this is about a comparison of the process rather than just trying to find the answer.

Lucas says, "The first part is different in each addition.

It is decreasing by one tenth each time.

Consequently, the whole is also decreasing by one tenth each time." The redistribution is increasing by one tenth each time.

This is because the first part is getting further from the next whole number.

Okay, it's time to check your understanding then.

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

We've got 1.

7 plus 5.

7 is equal to, then we've got 2.

6 plus 5.

5 is equal to.

And both of those have unknown answers as yet.

So which of those needs the greatest amount redistributed? Pause the video, have a think and have a go.

Welcome back.

Here's the first one.

0.

3 was redistributed, so three tenths from the second part to the first part, giving a transformed calculation of 2.

0 plus 5.

4 which is equal to 7.

4.

Now let's look at the second one.

We had 0.

4 or four tenths being redistributed, giving a transformed calculation of 3.

0 plus 5.

1 which is equal to 8.

1.

And Alex says, "The second has a greater redistribution." Is that what you found? Okay, let's move on.

Lucas and Alex both use redistribution for an addition.

There they are.

What do you notice and whose is correct? So have a good look.

They're both solving 2.

7 plus 6.

2, but they've done it using different amounts redistributed.

I redistributed a different amount.

I chose to redistribute three tenths from the second part.

I chose to redistribute two tenths from the second part.

Both ways transformed a part so that it was an integer, making the addition easier.

Lucas and Alex both complete another.

There they are.

Again, what do you notice? Have a look at both of those.

What can you see? Lucas says, "I redistributed differently.

I chose to redistribute from the second part to the first part." I chose to redistribute from the first part to the second.

I suppose it depends on which you find quickest.

That will be your most efficient way.

They both achieved the same whole or sum.

They are both valid methods.

Time to check your understanding again.

Here's a true or false.

Just like integers, you can redistribute tenths from one part to another and the whole remains the same.

Is that true or is that false? Pause the video, have a think and I'll be back in a moment.

Welcome back, that was true.

But how can we justify that? Here's two justifications and your job is going to be to select the one that you think backs up that answer of true.

So for A, we've got tenths are a unit of place value the same as ones, tens and hundreds, et cetera.

They can be used in redistribution.

Or B, tenths are integers, because you can count in them.

What do you think? Pause the video, maybe have a chat about it and I'll be back to reveal the answer shortly.

Welcome back.

The first one was the best justification there.

Tenths are a unit of place value the same as ones, tens and hundreds, et cetera.

They can be used in redistribution.

For B, it's true that you can count in tenths, but they're not integers because they are decimal fractions.

They are a unit that is less than one.

Okay, let's do your first practise task.

Your job for one A and B is gonna be to fill in the missing numbers.

Now notice, this isn't just about finding the whole at the end, it's also about putting in the missing numbers that are part of the jottings.

For number two, you're gonna use redistribution to calculate the sums. For number three, you've got to solve this addition mentally with jottings using redistribution but in three different ways.

So you might choose to redistribute three different amounts.

Then Lucas asks, "Which do you prefer and why?" 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 I'll be back in a little while with some feedback.

All right, let's mark the first ones then.

One A, here are the missing numbers.

We had 0.

3 being redistributed and that gave us a transformed calculation of 3.

0 plus 6.

2 is equal to 9.

2.

For B then, the second part was 7.

5, which you could've worked out by using the inverse on the redistribution at the second part from the transformed calculation.

0.

2 is being redistributed and the transformed calculation read 4.

0 plus 7.

7 is equal to 11.

7.

Let's look at two then, two A.

We chose to redistribute one tenth, giving 5.

0 plus 2.

7 is equal to 7.

7.

For B, we chose to redistribute two tenths, giving 4.

0 plus 3.

6 is equal to 7.

6.

Okay, here's number three.

Here were three possible ways.

In the first one, we redistributed three tenths from the second part to the first part, giving 5.

0 plus 1.

9 is equal to 6.

9.

For the second one, we redistributed two tenths from the second part to the first part, giving 4.

9 plus 2.

0 is equal to 6.

9.

And lastly, we redistributed seven tenths from the first part to the second part, giving 4.

0 plus 2.

9 is equal to 6.

9.

All of those redistributions gave the same result, but they were slightly different amounts.

I prefer the middle one, because there's no bridging and simpler redistribution.

Which did you prefer? Okay, we're gonna move on now to number four.

In the examples below, redistribution has been used and you were asked to explain what happens to the whole if an amount is redistributed between the parts.

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

Okay, well done, we finished the first part of the lesson, let's move on.

Redistribution with hundredths.

Alex looks again at his tricky question.

So do you think you could use red to solve this problem? Yes, but I'm still desperate to use columns.

I think redistribution might end up being more efficient.

This problem has hundredths in it, so could we explore those a bit using redistribution? Yes, great plan.

They revisit the cubes in the balance scales.

Last time we changed from ones to one tenths.

What is each cube was one hundredth this time? What would change and what would stay the same? What do you think? The cubes would now be one hundredth the size.

29 hundredth is 0.

29 as a decimal fraction.

Let's change the expressions then.

They've changed them, we've got 0.

29 plus 0.

15 equals 0.

30 plus 0.

14.

The two expressions are still equivalent.

What's the same, what's different between these jottings? Have a look.

Lucas says, "Each redistribution has the same digits but with different place value.

In the first, it's 29 ones added to 15 ones.

The second is 29 tenths added to 15 tenths.

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

In the first, it's one, the second it's 0.

1 and in the third, it's 0.

01.

Let's do the same thing again then.

What's the same and what's different between these? The number of digits is different.

The numbers in the second are larger.

But the amount being redistributed is the same.

It's 0.

01 each time.

The second is simpler than it looks at first.

Yeah, by looking closely at the parts being added, you can see that it is ready for redistribution.

Alex revisits the tricky question from the start.

Okay, I'm ready.

I'll have a go at the tricky questions with decimal fractions.

He writes out the expression.

And then he says, "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 five hundredths from three, so that would work.

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

I'll do it mentally with jottings." He's redistributed five hundredths, giving a transformed calculation of 3.

00 plus 1.

66 is equal to 4.

66.

Wow, by looking closely, I didn't need to use columns.

It's a very useful method and much more efficient.

It doesn't always work though.

Now it's your turn to have a go.

Solve the following mentally with jottings by redistributing.

1.

37 plus 4.

97, what does that equal? Lucas gives us a tip here, "Remember to look closely at the parts first to see which is best for redistribution." All right, pause the video here and have a go.

Welcome back.

Here's how we solved this using redistribution.

We took three hundredths from the first part and redistributed them to the second part.

That gave a new calculation of 1.

34 plus 5.

00 which is equal to 6.

34.

Did you do something similar? I hope so.

Okay, let's move on.

Lucas and Alex look at two addition equations with decimal fractions.

They categorise each by thinking about whether they were likely solved by redistribution or not.

We got 2.

86 plus 4.

99 is equal to 7.

85.

Then we've got 2.

43 plus 1.

38 is equal to 3.

81.

What do you think? Which of those two do you think were solved using redistribution? Lucas says, "Let's look at the parts for each closely." Alex says, "What are we looking for?" If either of the parts are near to an integer that will make the addition easier.

I think the first equation was solved using redistribution.

Do you agree with Alex and why? Look at that first addition.

What is it about that addition that looks like it might have been solved using redistribution.

"I agree," says Lucas, "The second part is only one hundredth away from five, so easy to redistribute to." The second equation doesn't have any parts that are close to an integer.

I think I would have used a written method for that one.

Alex and Lucas look at a missing inequality problem.

We've got 4.

02 plus 3.

56 and then we've got 3.

62 plus 3.

97 and we're comparing them.

Which inequality symbol would go between? What do you think? Have a look.

Lucas says, "Let's look at the parts to see if we can use redistribution." Both expressions have parts close to an integer.

Okay, I'll do this side and you do the other.

So Lucas solves the first expression.

And he redistributed two hundredths and he gets seven 7.

58.

Alex solves the second expression by redistributing three hundredths and he gets 7.

59.

7.

58 is less than 7.

59 so the symbol is less than.

Okay, it's time for your task.

For number one, you've got to fill in the missing numbers, very similar to what you did in the first part.

And then for number two, you've gotta use redistribution to calculate these additions.

For number three, you've got to decide which inequality symbol goes between the expressions.

Pause the video here, have a go at those.

Think carefully, enjoy and I'll be back in a little while with some feedback.

Welcome back.

Time to mark.

For A, the missing redistribution was 0.

02 and the missing part in the transformed calculation was 5.

27 and the whole was 9.

27.

Let's look at the second one.

The redistribution was 0.

03 and the transformed calculation read 1.

31 plus 8.

00 is equal to 9.

31.

Okay, let's look at number two then.

Two A, we chose to redistribute four hundredths, giving 5.

28 plus 4.

00 is equal to 9.

28.

And then for B, we chose to redistribute one tenth, giving 5.

0 plus 3.

57 is equal to 8.

57.

Here's number three then.

We were looking to find out which inequality symbol went between these expressions.

Remember here to mark the inequality symbol of course, but make sure that you've got the correct answer for the right reasons, that you have calculated these expressions.

So on the first expression, we redistributed three hundredths and we got 8.

79 as the whole.

In the second expression, we redistributed one tenth and we got 8.

66 as the whole.

So the symbol that went in between them was greater than.

B, we got 8.

51 for our first expression after redistributing five hundredths and then we got 8.

52 for the second expression after redistributing four hundredths.

So the symbol that went between them was less than.

That brings us to the end of today's lesson then, but here's a quick summary of the things that we've learned and thought about.

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

This works for decimal fractions as well.

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 is Mr. Tazzyman and I've enjoyed learning with you today.

I hope that we'll meet again on another maths lesson.

Thank you very much.

Bye-bye.