Lesson video

Hi there, I'm Mr. Taziman and I'm really excited to be doing some learning with you today.

If you are all ready, we can get started.

Okay, let's get started with this lesson.

Today, the outcome is that I want you to be able to say that you can choose the most efficient strategy to add two three-digit numbers.

Here are some of the key words that you are gonna see during today's lesson.

I'm gonna say them and then you are gonna repeat them back to me, so I'll say, "My turn", say the word.

Then I'll say, "Your turn" and you can repeat it back.

Are you ready? Let's do it.

My turn, partition, your turn.

My turn, adjust, your turn.

My turn, redistribute, your turn.

My turn, redistribution, your turn.

My turn, efficient, your turn.

Okay, here's an explanation of these keywords.

Partitioning is the act of splitting an object or value down into smaller parts.

When you adjust, you make a small change to a number.

This can make a calculation easier to solve mentally.

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

And working efficiently means finding a way to solve a problem quickly whilst also maintaining accuracy.

In today's lesson on choosing the most efficient strategy to add two three-digit numbers, there are gonna be two parts.

Firstly, we're gonna remind ourselves of three different strategies that you may have come across before, and then in the second part we're gonna think about choosing which of those strategies would be the most efficient.

Okay, let's get started with the first part of the lesson.

We are gonna have two friends who are joining us in this lesson.

Sam and Jun.

Sam and Jun will help us in our discussions, our answers, and even sometimes give us clues when we are having a go ourselves at some maths problems. Let's begin, Sam and Jun are working as a pair in a football skills game.

They're kicking a ball at a target wall.

SAM scores 202 points and Jun scores 290 points.

There they are.

How many points do they score as a pair? We can write this as a sum to help.

202 points plus 290 points equals it's an unknown.

We're not sure yet.

And it can be calculated mentally using different strategies.

Let's look at some of those.

Jun says, "Partitioning is where we split the addends into their place value groups and recombine them to get the sum.

First, we partition the addends into their place value groups." And you can see there we've taken 202 and 290 and we've partitioned it into 200 plus 200 plus two plus 90.

Then he says, "We combine the hundreds, tens, and ones to make two groups.

Lastly, we combine the groups to get the sum." That's partitioning, it gives us the answer of 492 points.

Sam tells us about adjustment.

"Adjustment is where we adjust one of the addends so that it is easier to count on from.

Then adjust the sum at the end using the inverse.

This transforms the calculation to make it easier to find the sum." Let's have a look.

"First, we choose an addend that is closest to the nearest boundary, usually a hundred boundary when adding three-digit numbers." So we've selected 202 here.

"Next, we adjust the addend so that it's a multiple of a hundred.

You can see we've subtracted two to make 200.

Then we calculate the sum with the adjusted addend, 490.

Lastly, we adjusted the sum using the inverse." And on this occasion the inverse was addition, so we added two onto our adjusted sum to create a total sum, which is 492 points.

Jun's gonna tell us about redistribution.

"Redistribution is where we take some ones or tens from one addend and redistribute them to the other addend to transform the calculation and make it easier." Let's have a look.

"First, we identify the addend that is closest to a boundary, usually a hundred.

Then, we take some ones or tens from one addend and redistribute them to the other so that the calculation is easier." We subtracted two from 202, so we'll add two to 290 and there we get 200 plus 292.

"Lastly, we add together the redistributed addends to find the sum." We get 492 points.

Here's partitioning, here's adjustment, and here's redistribution.

What do you notice? Hmm.

Well, Jun says, "All three strategies give the same answer.

They are all accurate methods." And we remember that accuracy is a really important part of efficiency.

If you can be quick and accurate, then you're being efficient.

Okay, it is time for you to have a go to check your understanding.

Sam and Jun have another go at that football game.

Sam scores 303 points, Jun scores 190 points below are three different strategies they've used to calculate a total score.

Can you name the strategies? You've got to match these jottings with either partitioning, adjustment, or redistribution just as Sam's told us there.

Pause the video, have a go and I'll be back in a moment to give you the answers so you can check how you've got on.

Welcome back.

Let's see how you got on.

I'm gonna reveal the answers.

The first set was adjustment.

The set on the top right was partitioning and the set towards the bottom was redistribution.

How did you do? Did you spot them? Did you recognise them? Okay, let's move on.

Here is the same equation again.

202 points plus 290 points equals 492 points.

Which strategy do you think the base 10 represents? Remember, it's either partitioning, adjustment, or redistribution.

Let's have a look.

We've got our equation in base 10.

Then each of the parts of base 10 seem to group together.

Then they're combined to give us 492.

Sam says, "This was partitioning.

The addends were partitioned into place value groups." Okay, we are gonna do the same thing again, I wonder which strategy is gonna be represented this time.

We start with our equation again, the two ones are taken away from the first addend and put into the second, and then they're combined to give us 492.

Sam says, "This was redistribution.

Two ones were redistributed to the other addend to transform the calculation." Great.

Okay, let's have a last go.

Here's the equation again.

You can see two has been removed from the first addend.

The groups have been combined and then that two has been added back on.

Sam says, "This was adjustment.

The first addend was adjusted by two to make it a multiple of hundred and make the calculation easier.

Then the sum was adjusted using the inverse operation.

Two was added to give the final sum." Alright, it's your turn to check that you've understood what we've just been learning about.

With a partner and some base 10, model finding the sum below using partitioning, adjustment, or redistribution.

Afterwards, your partner will need to say which they think you used.

So our equation is 105 plus 190 equals 295.

Pause the video here and give it a go.

I'll return in a moment to see how you've got on.

Okay, here's an example.

If I took five from the first addend and moved it to the second addend and then combined those groups, it would've been redistribution.

As Jun said, "That was definitely redistribution.

Five ones were redistributed to transform the calculation into 100 plus 195, which gives 295." How did you all get on? Did you manage to work out what your partner had shown you, which strategy they'd used? Alright, it's time for you to have some practise.

We've got one question here with parts A and B.

Sam and Jun play the football target game again.

They each have two goes.

Work out the total three times using each of the methods, partitioning, adjusting and redistribution.

In A, they scored 202 points and 197 points.

And for B they scored 360 points and 230 points.

So for both A and B, I'd like you to use all three strategies and see if you come up with the same answer for each.

Have a go.

Pause the video here and I'll be back with some feedback in a little while.

Welcome back.

Let's do some feedback.

Here's one A.

If you partitioned it, it may have looked like this and you would've got 399.

With adjustment, you might have adjusted the first addend by taking away two and adjusted the second addend by adding three and then at the end you might have added two and taken away three to give you 399.

It says here, "You might have only adjusted one addend, but we chose to do two because it's a near double." And redistribution might have looked like this.

202 take away two gives you 200, so you redistribute that two to give you 199 and that scores 399 again.

All three answers were 399 points.

Okay, let's look at one B, similar kind of thing.

Here's partitioning and it would've given you 590.

Here's adjustment, we chose to adjust by taking 30 away from the second addend and adding it back onto the sum at the end.

And then redistribution and we decided to add 40 and then subtract 40 from the second addend for our redistribution.

All three scored 590 points.

"We thought that partitioning was the most efficient because the addends were multiples of 10." Okay, that's a recap of some strategies that you may have seen before.

We are ready to move on to the second part of the lesson now, which is where we start to consider which is the most efficient strategy.

Are you ready? Sam and Jun are thinking about efficiency.

They want to know when to use each strategy.

They use digit cards to create a sum to solve, 353 plus 206.

Sam says, "Let's try all three strategies and see what happens.

And Jun replies, "Okay, good idea.

Then we can see what strategy is most efficient for different addends." Which method was the most efficient? Here they all are.

Which do you think would be the most efficient for these addends? Jun said, "I think redistribution is best for these addends.

One of the addends is just over a hundred's boundary, so it makes it easy to redistribute." And Sam agrees.

She says, "I don't think partitioning is very efficient because you end up with five groups to combine." They try some new addends, which is most efficient here? 198 plus 201.

Here are the three sets of jottings again, which do you think would be the most efficient? Well, Jun says, "I think redistribution is best for these addends because they are both close to a hundred's boundary." Sam says, "I'd rather use adjustment because these addends are a near double so I can transform it into an easy sum." Jun says, "Again, I think partitioning isn't as efficient because there are five groups to combine." Another set of addends, which is the most efficient here? 308 plus 260.

There are the three sets of jottings.

Jun says, "I think I found partitioning the most efficient here." Sam says, "I agree.

There was a placeholder in each addend, which meant you only had to combine four groups." You can see them there.

Jun says, "Also, you didn't need to bridge when you combined the groups because there was a group of ones and a group of tens." And they're highlighted there.

Sam replies, "Yes, I suppose there's no right or wrong answer though.

All the strategies are accurate, this is about speed." Sam and Jun create a new sum.

Once again, they consider the most efficient of the three strategies, 349 plus 574.

And Sam says, "I can't see a quick way to transform this calculation to something easier using either adjustment or redistribution.

Jun says, "I could partition, but there's still loads of regrouping and then bridging.

That's hard to do mentally." Okay, it's time for us to check your understanding of what we've just gone through.

For each of these sums, would you use adjusting, redistribution or partitioning and why? We've got 140 plus 302, 511 plus 398 and 297 plus 301.

Pause the video here, look at each of those pairs of addends and discuss which strategy you'd prefer to use.

I'll be back in a moment to see how you've got on.

Welcome back, let's have a look and see what Sam and Jun thought.

Well, Jun thought he'd use partitioning for the first one because both addends have placeholders.

Sam said, "Redistribution for the second because 398 is close to a hundred's boundary.

And Jun said, "I'd use adjustment for the third one because it's a near double so I can create an easier calculation.

It's time for you to have a practise now.

Number one asks you to sort these sums into groups by how you would solve them.

Adjusting, redistribution, or partitioning.

A is 109 plus 280.

B is 399 plus 420.

C is 121 plus 296 and D is 303 plus 298.

Remember, we're not necessarily looking for you to get the answers.

We're looking for you to choose which strategy you'd use to get those answers.

And Sam gives a little tip here.

She says, "If you're not sure, you could try the methods to see which one works best." Number two, we have Sam and Jun playing the football target a last time.

They each have two goes.

Choose the most efficient method to work out the total for each.

351 points and 204 points is A and 209 points and 460 points is B.

Pause the video here.

Have a really good go at these practise questions and I'll be back in a little while for some feedback, good luck.

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

One A was probably best suited to partitioning.

Sam says, "A is a great sum to partition as there are placeholders in both addends.

This means that there's no bridging to do and only four groups to combine." "B would've gone into redistribution.

It featured an addend near to a boundary, so redistribution works well.

This might also work with adjustment for the same reason." C would've gone into both groups as well.

296 is only four away from 300 and "D is a near double, so adjustment will work well.

Makes a super easy calculation." Thanks, Jun.

"These aren't definite though", says Sam.

"Different people prefer different methods." Okay, let's look at two.

A, 351 points plus 204 points.

There's adjustment which gave 555.

There's redistribution which also gave 555.

"Adjustment or redistribution are most efficient here." "Partitioning creates too much regrouping." Okay, let's look at two B there.

The workings below.

They've used partitioning.

They've partitioned 209 and 460 into their place value groups.

Then they've grouped them back again and they've got a final sum of 669.

"We thought that partitioning was the most efficient because both addends contained placeholders." Well done, Sam.

Jun says, "Also, neither addend was near a hundred's boundary." Okay, thanks for participating in today's lesson.

Here's a summary of all the things that we've learned and thought about.

Partitioning, adjusting and redistribution are all effective methods for adding three-digit numbers mentally.

When addends are varied, these different methods can be more or less efficient.

If both addends contain placeholders then partitioning is an efficient method.

If one or both the addends can be transformed to make an easier sum, then redistribution or adjustment are more efficient.

Thanks very much for joining in this lesson.

My name's Mr. Zaziman and I hope to see you again soon, thank you.