Lesson video

Hi there.

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

If you are all ready, we can get started.

Here's the outcome of this lesson: I can choose the most efficient strategy to subtract from a three-digit number.

So, that's what we want you to be able to say by the end of this lesson.

Here are the key words.

I'm going to say them, and you're gonna repeat them back to me.

So, I'll say "my turn," say the word, and then "your turn," and you can repeat it back.

My turn: minuend.

Your turn.

My turn: subtrahend.

Your turn.

My turn: difference.

Your turn.

My turn: partition.

Your turn.

Well done.

Now let's have a look and see what each of these words means.

The minuend is the number being subtracted from.

A subtrahend is a number subtracted from another.

The difference is the result after subtracting one number from another.

you can see an equation towards the bottom of the screen.

Seven subtract three is equal to four.

In this equation, seven is the minuend, three is the subtrahend, and four is the difference.

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

The outline for today's lesson on choosing the most efficient strategy to subtract from a three-digit number is this.

In the first part, we're gonna be looking at choosing the most efficient strategy.

And in the second part, we're gonna look at solving problems efficiently.

Now, prior to this lesson, you might have come across a few strategies for mental subtraction.

Keep those in mind, because they're going to be useful, particularly in this first part.

Let's get going.

Here are two friends that we are going to meet during these slides: Alex and Jacob.

They're gonna help us to discuss maths problems. They're gonna give us some of the answers, and they might help to deepen our thinking.

Alex and Jacob look at a bar model; you can see it there.

The total is 620, and we have one part that's 270, and a missing part that needs to be calculated.

They consider how to write this as an equation.

Alex says the subtraction equation is: 620 subtract 270 is equal to question mark, an unknown.

Then they think about what strategy they might choose.

Alex says, "I'm going to use partitioning and bridging here." Jacob says, "I think I'm going to find the difference by counting on in tens." Who will be more efficient? Well, let's discover.

Alex partitions 270 into tens and hundreds to subtract.

He shows the steps on a number line.

You can see he says there, "I will partition 270 into 200 and 70 to subtract mentally." There it is.

He says, "I will subtract 200 from the minuend." He ends up on 420, having done that.

Alex says, "I now need to subtract 70, but that's not easy.

I know! I can partition again and bridge through 100.

I can partition 70 into 20 and 50 to help me bridge 400.

420 subtract 20 is 400, which is an easier number to use." And he shows his thinking on the number line.

You can see it there.

"400 subtract 50 is 350, so the difference is 350." He shows that thinking on the number line, too, and he says, "Sometimes partitioning twice makes calculating easier." Jacob also draws a number line to think about the difference between 270 and 620.

He says, "I will start on 620 and count back in tens to 270." Here we go.

He gets to 560 so far.

And he says, "Erm, this is gonna take a long time." But he perseveres.

Always important to be resilient when you're learning maths.

He says, "The minuend and subtrahend are too far apart." He keeps going.

Then he gives up.

He says, "It's too inefficient! I might need to partition some of the steps." And Alex says, "You mean like I did?" "Oh yeah! So, finding the difference is inefficient here." Alex says, "I agree.

It's best to use partitioning when the difference is large." Alex and Jacob look at a second bar model.

The total is 780.

One of the parts is 720, and then there's a missing part that we don't know the value of yet.

Again, they consider how to write it as an equation.

"The subtraction equation is 780 subtract 720." Then they think about what strategy they might choose.

Jacob says, "I'm gonna try counting on again." Alex says, "I'll use partitioning and bridging.

Then we can compare efficiency." Who will be more efficient? What do you think? Well, let's discover.

Jacob draws a number line to help.

He says, "I will start on 780 and count back in tens.

The numbers are closer this time!" He says, "I end up on 720, having counted back six lots of 10 in total.

Six lots of 10 makes 60, so I counted back 60.

The difference between 780 and 720 is 60." And he writes it in to complete the equation, but he's not finished there.

He goes on to say, "I think I can also count forwards to find the difference, and the result will be the same." It's always good to conjecture in maths, to have a thought, and see if you can prove it.

He says "I was right! I counted on six lots of 10, which is 60.

Then he says, "I don't even think I needed to count.

I know that the difference between 2 and 8 is 6, so I also know that the difference between 20 and 80 is 60 even quicker." So, he's disregarded the hundreds here because he knows that the difference is between 20 and 80.

The hundreds, in a sense, are irrelevant here.

Alex partitions 270 into tens and hundreds to subtract.

He shows the steps on a number line.

He says, "I will partition 720 into 700 and 20 to subtract mentally." You can see he's done it there.

He takes away 700, and then he takes away 20, and it gives him a difference of 60.

Alex and Jacob compare number lines.

There's Alex's, and there's Jacob's.

Which is more efficient? Have a look at those.

Which do you think is more efficient? You might like to point at it.

Alex says, "I think that yours is more efficient," and he's talking to Jacob now.

He says, "There's only one step." Jacob says, "I agree, as long as you use known facts like 8 subtract 2." Alex sticks with partitioning, and Jacob uses find the difference, and they look at a new bar model.

There it is.

The total is 110 this time, with 80 as one of the parts, and a missing part that we need to calculate.

Here's Alex's workings.

He decides to partition, and he counts back.

So, he takes 10 away first to get to 100, and then he takes 20 away because he knows that 100 take away 20 is 80.

All together, he's taken away 30, so he knows the difference is 30.

Whereas Jacob counts back in 10 three times to get from 110 to 80.

Three lots of 10 make 30, so he also knows that 30 is the difference.

But which is more efficient? Hmm.

Well, Alex says, "I think both of these are quick and accurate," and Jacob agrees.

He says, "They're both efficient even though there's bridging." And if you think about it, although there are more steps in Jacob's, you can do them relatively quickly.

Again, Alex sticks with partitioning, and Jacob uses find the difference.

They look at another bar model.

408 and 393 is one of the parts, so we need to calculate that missing part again.

Here's Alex's workings.

He starts on 408, and he partitions, so he subtracts 8 first, then he subtract 7 to get to 393.

Again, he's working to the nearest hundred because it's easier to calculate any jumps from that.

In total, he's taken away 15, so the difference is 15.

Let's look at Jacob's.

You can see it's very different.

Jacob has decided to count on this time, but because neither of the numbers are multiples of 10, he's been forced to count on in ones.

He's had to count 15 times.

Which is more efficient? Alex says, "I think both of these are quick and accurate." Jacob says, "I counted on in ones, but I think partitioning is far more efficient here." Okay, it's time for you to have a go and put into practise what we've just learned.

Let's check you've understood.

Below, you can see we've got some equations on the left, and on the right there are some strategies.

What you need to do is connect the equations to the most efficient strategy, and I'd like you to give some reasons, as well, for your choices.

Pause the video here, have a go, and I'll be back in a moment to give you the answers.

Welcome back.

Let's see what we thought.

So, Jacob says that the middle one, 230 subtract 190, is best for finding the difference, and his reasoning is that the minuend and subtrahend are close together, and they're multiples of 10.

The other two equations need partitioning, and Alex says that's because "The minuend and subtrahend are quite far apart, and some bridging will be needed." Did you agree? Well, it might be that you would've done it differently.

Remember, efficiency is quite an individual thing.

However, in general, I think these would probably be the most efficient strategies in these equations.

Time for your practise.

Here's what I'd like you to do.

You can see you've got some bar models with missing parts, just like the ones that Alex and Jacob have been working with.

Your job is to find the missing part, firstly, using partitioning, and then finding the difference.

You can count on or back; it's up to you.

For each of these bar models, once you've calculated them, I want you to say which was the most efficient strategy, and why? Pause the video here, have a good go at these, and I'll be back in a little while for some feedback.

Welcome back.

Let's have a look at A, to begin with.

Partitioning: you might have drawn a number line out to help your thinking.

You might have subtracted 20 first, then 30.

Altogether, that's a subtraction of 50, which means the missing part was 50.

Finding the difference: you might have counted back in tens five times, so you would've subtracted 50 in total.

Alex says, "I think partitioning was more efficient here; fewer steps." Jacob says, "I agree, although the counting back in tens can be very fast." Here's B.

We'll do the same thing again and look at partitioning first.

We've got a number line; 512 and 491.

We can start by counting back 12, and then count back 9.

In total, we've counted back 21, which gives us a difference then of 21.

Let's see how we might have done that if we were finding the difference.

Same number line again, but you can see we've had to count back in ones, and that's because neither of the numbers are multiples of 10.

Alex says, "Partitioning is more efficient here; quick and simple." Jacob says "Yes, counting back has to be done in ones, which takes time, and you can easily lose track." Look at that number line.

I know what he means.

Let's do the same again for C.

We've got 670 on our number line.

We're gonna subtract 600, and then take away 10 to finish on 60, which means that the difference is 60.

For finding the difference, well, you might have just done one jump of 60, provided you can use your known facts.

Alex says, "Finding the difference is more efficient here." Jacob agrees, but he says, "Yes, it is, as long as you can recognise that 10 plus 60 equals 70." We finished the first part of the lesson there.

I hope you enjoyed it, but there's more to come.

Now we're gonna move on to looking at solving problems efficiently.

There are 360 children in the school.

250 go on a school trip to the zoo.

How many children do not go on the trip? How can this be efficiently solved? Hmm.

I wonder what strategy would be best to use.

But let's start with how can the problem be represented, move on to looking at the most efficient strategy, and then think about whether or not the answer we've managed to get is reasonable.

Does it solve the problem? It's these three stages that can help us to clarify our thinking when it comes to solving problems efficiently.

So, we'll start with how can the problem be represented? Alex says, "I'll use a bar model to show the unknown." Good thinking, Alex.

So, he draws one out.

We've got a total of 360.

250 is one of the parts, and then we have our missing part, which we've yet to calculate.

Which strategy is most efficient? Alex says, "I think partitioning is most efficient here because it involves bridging." So, he's looked at the numbers, and he can see that he's going to have to bridge across 300.

He draws out a number line to help.

He takes a jump of 200 backwards to get to 160, and then he subtracts 50 to get to 110.

So, the difference is 110.

That means that 110 children did not go on the trip.

Does the answer solve the problem? Alex says, "Yes, I can see that it is a reasonable answer because 25 plus 11 equals 36, so 250 plus 110 equals 360." He's used something called unitizing to help him out.

Well done, Alex.

Good check.

Here's another problem.

Lucas has read 162 pages of his book.

Alex has read 45 pages.

How many more pages has Lucas read? How can the problem be represented? Alex says, "I'll use a part-part-whole model to show the unknown," and he's done that there.

Which strategy is most efficient, then? "Again, I think partitioning.

These aren't multiples of 10, so I think counting on or back would be time-consuming." That means it would take a lot of time to do it.

That wouldn't be very efficient.

He draws out his number line to clarify his thinking.

He subtracts 40 to begin with, then he subtracts 2, and then he subtracts 3.

You can see with those last two jumps, in total, they make 5.

He's had to partition into 2 and 3 in order to bridge through a 10.

Does the answer solve the problem? Alex says, "Yes, I can see that it is a reasonable answer because I know it'll be more than 100 and less than 120." Well done, Alex.

Good thinking.

Alex pours out 560 millilitres of squash drink.

He drinks 510 millilitres.

How much of his drink remains? Again, let's start with how this problem can be represented.

Alex says, "I'll use a bar model again to show the unknown." There it is.

560 is his total, one part is 510.

What's the missing part? Well, which strategy will be most efficient? He says, "I can count on here by using what I know.

1 plus 5 equals 6, so 51 plus 5 equals 56, which helps me calculate the missing part." There's his number line, and he jumps on 50 to get to 560, which means the difference is 50.

But does the answer solve the problem? He says, "Definitely.

I know that it's going to be a multiple of 10, and it will be less than 100." Well done, Alex.

I want you to have a go now.

We're gonna check your understanding.

Have a go at this problem, using the three stages.

Alex and Jacob play each other at a game of Scrabble.

Jacob scores 310 points, and Alex scores 260 points.

By how many points has Jacob won? Pause the video, have a go, and I'll be back in a moment to share with you what our maths friends think.

Welcome back.

Let's look at this problem.

First of all, how might we have represented it? Well, here's a bar model.

You might have used something different to represent the problem.

Which strategy is most efficient? Well, here we've got partitioning.

We've counted back in 200, then in 10, and then in 50, to give us a difference of 50.

Does the answer solve the problem? Alex says, "It is a reasonable answer.

I can see that because the difference is going to be less than 100." How did you get on with it? It might be that you had some slightly different answers.

Maybe you used a different strategy, maybe you used a different representation.

You might have even had a different justification for whether or not the answer was reasonable.

Ready to move on? Let's go.

Now it's time for you to have a practise.

In each of these questions, choose the most efficient strategy and solve it.

A: Alex pours 370 millilitres of squash into a glass.

He drinks 310 millilitres of it.

How much squash remains? B: Alex scores 470 points, and Jacob scores 180 points against each other on a computer game.

How many points does Alex win by? And C: Alex and Jacob each collect football cards.

Alex has 65, and Jacob has 186 cards.

How many more cards does Jacob have? Okay, pause the video here, have a go at all three of those using our three different steps, and I'll be back in a little while to give you some feedback.

Good luck.

Welcome back.

Let's have a look at each of these individually.

Here's our first problem.

You might have represented it using a bar model like this.

Strategy that was most efficient? Well, you may have counted on and found the difference using a known fact.

60 was the answer.

Does it solve the problem? Alex says, "It's a reasonable answer.

I've used the known addition fact of 1 plus 6 equals 7 to help." Here's B.

Here's the representation using a bar model.

The strategy that we chose to use this time was partitioning.

So, 100 was taken away first, then 70, and then 10.

And you can see those last two jumps were a second round of partitioning because we had to bridge through 300.

The answer was 180, because that's what was subtracted in total.

Does it solve the problem? "Yes, it looks reasonable," says Alex.

"I know that the missing part was more than Jacob's score." And finally, here's C.

Here it is represented as a bar model.

Strategy: well, partitioning might have been best here, subtracting 60 at first, and then subtracting 5 to give 121.

Does the answer solve the problem? Alex says, "It is a reasonable answer.

I can see that because the difference is going to be more than 100." Thank you for participating in today's lesson.

Here's a summary of all the different things that you might have learned.

We can subtract three-digit numbers by partitioning the subtrahend and subtracting this from the minuend.

We can also subtract three-digit numbers by finding the difference, which is more efficient when the subtrahend and minuend are close together and both are multiples of 10.

Sometimes we can use known addition facts to make finding the difference even more efficient.

And when solving subtraction problems, it's best to choose the most efficient strategy first.

My name's Mr. Tazzyman, and I've really enjoyed learning today With you.

I hope to see you again soon in another lesson.

Bye-bye.