Lesson video

Hello, how are you today? I trust you are feeling well and ready for your math lesson.

My name is Dr.

Shorrock, and I am here to guide you through the learning today.

This lesson is from our unit "Rounding and Solving Problems with Numbers with up to Seven Digits".

This lesson is called "Explore Written and Mental Strategies to Solve Problems".

As we move through the learning today, we are going to look at what mental strategies can be used to solve problems, such as the number line.

We will also look at a concept known as the constant difference, and this is a strategy that makes the formal column subtraction algorithm more efficient.

Now, sometimes new learning can be a little bit tricky, but if we work really hard together, then I know we can be successful, and I am here to guide you through the learning.

Let's get started then, shall we? How can we explore written and mental strategies to solve problems? Our keywords for the learning today are difference and efficient.

It's always good practise to say new words aloud.

Let's have a go together.

My turn, difference, your turn.

Nice.

My turn, efficient, your turn.

Brilliant.

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

It's how much one number differs from another, and it's represented by how far apart they are on a number line.

So, the difference between eight and three is five.

Being efficient means that we're going to find a way to solve a problem quickly, but we are also going to maintain that accuracy.

Let's get started with our learning today then.

We're gonna start by looking at how we can represent and solve calculations mentally.

To help us in our learning today, we have Aisha and Lucas.

Aisha is dreaming that she won £40,000 on the lottery and she is researching the cost of her dream car.

I wonder what your dream car is.

Aisha has found that her dream car costs £29,999.

So, if Aisha bought this car, how much money would she have left? How can we solve that problem? Good idea, let's start by representing this as a bar model.

The whole is known, and it is the amount that Aisha won, £40,000.

One part is known, and that's the cost of the car, £29,999, and the unknown part, that's the difference between the amount she won and the total cost of the car, and that's the amount of money she would have left.

Let's use the bar model to form an equation.

We know that 29,999, added to the missing part or the missing addend, must equal the sum of £40,000, our whole amount.

We can rearrange the equation, and if we subtract the known part from the whole, that would leave us with the difference.

And how would you then solve these calculations? What strategy would you use? Lucas is saying, "We should solve this using a written method because 29,999 is a precise amount and it will be far too tricky to calculate mentally with." Do you agree with Lucas? Aisha doesn't, she's respectfully challenging him.

She's suggesting we can use a mental strategy.

Can you think of a mental strategy that we could use? Let's start by representing this on a number line, good idea.

So, we get to draw our number line, and we're starting with 29,999.

And have you noticed something about that number? That's right, it's really close to the next multiple of 10,000, isn't it, which would be 30,000.

If we add one, that takes us to 30,000, and then, well, we need to try and get to our whole amount of 40,000, so what can we add? That's right, we're just going to add £10,000.

So, by using a number line, we have managed to work out that the missing part, or that difference, is £10,001.

So, actually, that really was quite straightforward using that number line, wasn't it? So, good call, Aisha, to get us to use a mental strategy.

So, if Aisha bought her dream car, she would have £10,001 left.

Let's check your understanding with that.

Could you solve this problem using a number line? Lucas dreams that he wins £60,000 and buys his dream car for £39,997.

How much money does he have left? I've started you off by drawing the start of a number line, and we're going to start at 39,997, and I want you to have a think about what would you add to that to take you to the next multiple of 10,000, and then what would you add to that to take you to our whole amount of 60,000? Pause the video while you have a go, and when you're ready to go through the answer, press play.

How did you get on? Did you realise we were only three away from the next multiple of 10,000, which is 40,000? And then, we can use our known facts.

We know four add two is equal to six, so 40,000, add 20,000 must be equal to 60,000.

Then we can work out our difference.

So, if Lucas buys his dream car, he would have £20,003 left.

So, don't forget the number line.

It's a really useful mental strategy to help us solve problems. It's your turn to practise now.

For question one, could you solve these problems? Part A: Lucas and Aisha play a game.

Lucas wins 49,995 points and Aisha wins 90,002 points.

How many more points does Aisha have? Part B: A great white shark has a massive 1,993 kilogrammes.

A blue whale has a mass of 150,000 kilogrammes.

How much heavier is the blue whale than the great white shark? In her dream, Aisha has £3,100,002, whereas Lucas only has £99,999.

How much more money does Aisha have? Now, try using a mental strategy, such as a number line.

You will find it will be much more efficient.

Pause the video while you solve those problems, and when you are ready to go through the answers, press play.

How did you get on? Let's have a look.

You were asked to solve these problems. For part A, Lucas and Aisha were playing a game.

We can represent the words as a bar model.

Our whole amount is 90,002, it's the larger amount, and Lucas wins £49,995, which is a part, and we need to find that missing part.

So, we can form an equation, 49,995, add something, well, that equals our whole of 90,002.

We can rearrange that equation and subtract the known part from the whole, and then we can use a number line to help us.

So, 49,995, well, that's close to the next multiple of 10,000, 50,000, isn't it? It's five away, so we can add five, and that will take us to the next multiple of 10,000.

Then we know we can add 40,000 to get to 90,000, but we've still got to remember to add that remaining two.

So, Aisha has 40,007 more points than Lucas.

Part B, when we were asked to compare the mass of a great white shark and the blue whale, again, we can represent this as a bar model.

The whole amount, 150,000, and the mass of the shark is apart, 1,993.

We can form equations from this.

We've got a missing part, so 1,993, add something is equal to our sum of 150,000.

We can rearrange that equation and subtract the known part from the whole, and then we can use a number line to determine the difference.

We are starting at 1,993.

We need to add seven to take us to the next multiple of 1,000, which is 2,000.

We then know we can add 148,000, which takes us to 150,000.

So, the mass of the whale is 148,007 kilogrammes more than the mass of the shark.

Part C, we can represent this again as a bar model.

The whole amount is 3,100,002, and one of the parts is 99,999.

Again, we can form equations from this showing that we are looking for a missing part or a missing addend.

So, we can rearrange the equation and subtract the known part from the whole, and again, we can use a number line to help us because the number we are subtracting is close to the next multiple of 100,000.

We can add one, and then we know we can add 3,000,000 to take us to 3,100,000, but we've still got to add the remaining two to take us to 3,100,002.

So, in total, Aisha has £3,000,003 more than Lucas.

How did you get on with those questions? Brilliant! Fantastic learning so far.

I can see how hard you are trying.

We're going to move on now and look at this concept called the constant difference.

Let's look at this problem.

A charity aims to raise £200,000, and so far, it has raised 158,436, and we want to work out how much more money does the charity need to raise to reach its target amount? Good idea, let's represent this as a bar model.

Our whole amount is £200,000, and one of the parts is 158,436, and we need to find a missing part.

We can then form equations from the bar model.

We know we've got one of the parts, 158,436, and if we add it to the missing part, we would get the sum, or the whole, of 200,000.

We can then rearrange the equation and subtract the known part from the whole.

And then, yes, let's use a number line, like we have done in the previous part of the learning, to calculate this.

I'm going to set up my number line.

I'm going to start at 158,436, and we know we need to get to 200,000.

It's a bit trickier this time, isn't it? Do you think that a number line is the most efficient method to use here? What's tricky about it? A number line is a tricky strategy to use on this occasion because it's not obvious what to add or subtract.

Our part of 158,436 is not close to the nearest multiple of 1,000 or 10,000, is it? So, it's a little bit trickier.

What could we do instead? Good idea, Lucas, we could use a formal written method this time, so I'm going to set up my column subtraction.

Let's have a go at working through this, but I've spotted something already, have you? Zero ones subtract six ones, well, there's insufficient ones, so I'm going to have to regroup.

Ooh, but there's a zero, so I can't regroup, so I need to look to the left of that zero.

Ooh, that's a zero, look to the left of that zero.

Ooh, that's a zero.

It's going to be a little bit tricky, isn't it? So, what can we do instead? There's going to be a lot of regrouping.

I would have to finally regroup from that two in the hundred thousands place, wouldn't I? So, regroup from the 100,000s place, and then I would have to regroup from the 10,000s place, and then I would have to regroup from the 1,000s place, and then I would have to regroup from the 100s place, and then I would have to regroup from the 10s place to finally then be able to calculate.

10 ones subtract six ones are four ones.

Nine tens subtract three tens are six tens.

Nine 100s subtract four 100s are five 100s.

Nine 1,000s subtract eight 1,000s is one 1,000.

Nine 10,000s subtract five 10,000s is four 10,000s, and then one 100,000s subtract 100,000, well, that's just zero, no hundred thousands.

Gosh, what a lot of regrouping! Do you think there's an easier way? Yes, Lucas does, "There must be a more efficient way," and luckily for us, there is.

We can use our number-sense superpower and this concept called the constant difference.

If the minuend and subtrahend, so that's if the whole and the part, both decrease, or increase, by a given amount, the difference will remain constant.

So, let's have a look how this will help us.

Our equation that we're trying to solve is 200,000 subtract 158,436, and that would give us the difference, but if we adjust the minuend and subtrahend by one, we would get 199,999, subtract 158,435, but that difference would be the same.

And then, we can use a formal written method, and we will not have to regroup.

Much easier, let's have a look.

So, I've set up my column subtraction.

We know the difference is going to be the same as if we had used the 200,000 as our whole.

Nine ones subtract five ones is four ones nine.

Nine 10s subtract three 10s is six tens.

Nine 100s subtract four 100s is five 100s.

Nine 1,000s subtract eight 1,000s is one 1,000.

Nine 10,000s subtract five 10,000s is four 10,000s, and then one 100,000 subtract one 100,000 is zero.

So, the charity needs to raise another £41,564 to reach its target amount.

Now, we got the same difference whichever calculation we did, whichever column subtraction, but actually, if you use the constant difference concept, there's no regrouping, so it makes the calculation much easier.

You've just got to remember to decrease the minuend and subtrahend, or the whole and part, by the same amount, and usually we can just adjust by one.

Let's check your understanding with that.

True or false, is it more efficient to solve 3,000,000 subtract 2,195,077 using a number line? Do you think that's true or false? Pause the video while you have a think about it.

When you're ready to hear the answer, press play.

How did you get on, did you say that was false? It's false because that 2,195,077, that's not close to a multiple of the power of 10, so it's difficult to know what to add or subtract.

But why is it false? Is it because a formal written method with 3,000,000 and 2,195,077 is more efficient, even with all the regrouping, or is it part B, a formal written method, where the minuend and subtrahend both decrease by one, would be more efficient and would involve no exchanging.

So, this is using that constant difference concept.

And we would alter the minuend and subtrahend by one, so we would get 2,999,999, subtract 2,195,076, and there would be no regrouping there.

Pause the video while you have a discussion about how you would justify your answer, option A or option B.

When you are ready to hear the answer, press play.

How did you get on? Yes, option B is the most efficient method.

We need to use that constant difference, and then there would be no regrouping.

We're less likely to make mistakes then as well.

It's your turn to practise now.

For question one, could you use that constant difference concept to calculate the answer to these questions? For question two, could you solve these problems? "Lucas' school has a budget of 350,000.

So far this year, the school has spent £36,287.

How much money does the school have left?" Part B, "Last weekend, the prize fund on a lottery was £2,000,000.

The winner gave £139,601 to charity.

How much money does the winner have remaining?" And part C, "Lucas wins 4,500,000 points in a game he is playing.

He spends some of the points in the game shop and has 350,915 points left.

How many points did he spend?" Pause the video while you have a go at those questions, and when you are ready to go through the answers, press play.

How did you get on? Let's have a look.

The first question, we needed to use the constant difference, so we adjust the whole and the part, or the minuend and the subtrahend, by one, then we can set up our column subtraction algorithm and no regrouping we'll be needing here.

Nine ones subtract one one is eight ones.

Again, nine 10s subtract one 10 is eight 10s.

Nine 100s subtract nine 100s is zero 100s.

Nine 1,000s subtract five 1,000s is four 1,000s.

Nine 10,000s subtract six 10,000s is three 10,000s, and then nine 100,000s subtract three 100,000s is six 100,000s.

The difference is 634,088.

For part B, again, we can use the constant difference concept.

I am adjusting the minuend and subtrahend by one, then I can set up my column subtraction algorithm.

Nine ones subtract one one is eight ones.

Nine 10s subtract zero 10s is nine 10s.

Nine 100s subtract one 100 is eight 100s.

Nine 1,000s subtract eight 1,000s is one 1,000.

Nine 10,000s subtract four 10,000s is five 10,000s.

Four 100,000s subtract five 100,000s, ah, there's insufficient 100,000s, so I need to regroup from the digit to the left.

Fourteen 100,000s subtract five 100,000s is nine 100,000s, and then the two subtract the two is zero.

So, the difference here is 951,898.

Part C, again, I'm going to adjust the minuend and subtrahend by one, then I can set up my column algorithm.

Nine ones subtract zero ones is nine ones.

Nine 10s subtract three 10s is six 10s.

Nine 100s subtract six 100s is three 100s.

Nine 1,000s subtract five 1,000s is four 1,000s.

Nine 10,000s subtract six 10,000s is three 10,000s.

Eight 100,000s subtract two 100,000s is six 100,000s, and then five subtract zero, that just leaves me with the five.

The difference here is 5,634,369.

For part D, one of our parts was missing, so we need to rearrange that equation, and we're going to subtract the known part from the whole.

Then I'm going to use the constant difference concept to adjust my minuend and subtrahend and set up my column algorithm.

Nine ones subtract one one is eight ones.

Nine 10s subtract three 10s is six 10s.

Nine 100s subtract five 100s is four 100s.

Nine 1,000s subtract six 1,000s is three 1,000s.

Nine 10,000s subtract seven 10,000s is two 10,000s.

Nine 100,000s subtract one 100,000 is eight 100,000s, and then the three subtract the zero is three.

So, the difference here is 3,823,468.

And then we can reorder the equation as it was to begin with, and we can see that that missing part is 3,823,468.

For question two, you were asked to solve some problems. We can represent problems as a bar model.

Our whole is 350,000, and of the parts is 36,287, and we can form an equation from that.

We know we can subtract the known part from the whole to find the difference, then I'm going to use the constant difference concept to adjust my minuend and subtrahend, and then I can set up my column algorithm.

Nine ones subtract six ones is three ones.

Nine 10s subtract eight 10s is one 10.

Nine 100s subtract two 100s is seven 100s.

Nine 1,000s subtract six 1,000s is three 1,000s.

Four 10,000s subtract three 10,000s is one 10,000, and then the three 100,000s subtract zero is three.

So, the school has £313,713 left.

Part B, again, we can represent this in a bar model.

Our whole is 2,000,000, and one of our parts is 139,601.

We need to find the missing part, so we are going to subtract the known part from the whole.

Again, we're going to use the constant difference concept and adjust the minuend and subtrahend by one to make the column algorithm easier for us.

Nine ones subtract zero ones is nine ones.

Nine 10s subtract zero 10s is nine 10s.

Nine 100s subtract six 100s is three 100s.

Nine 1,000s subtract nine 1,000s is zero 1,000s.

Nine 10,000s subtract three 10,000s is six 10,000s.

Nine 100,000s subtract one 100,000 is eight 100,000s, and that 1,000,000 subtract zero is 1,000,000.

So, the winner has £1,860,399 remaining.

Part C, again, we can represent this in a bar model.

The whole is 4,500,000, and one of the parts is 350,915.

We can form a subtraction by subtracting the known part from the whole.

We can adjust the minuend and subtrahend by one to make the column subtraction easier, setting up our column subtraction.

Nine ones subtract four ones is five ones.

Nine 10s subtract one 10 is eight 10s.

Nine 100s subtract nine 100s is zero 100s.

Nine 1,000s subtract zero 1,000s is nine 1,000s.

Nine 10,000s subtract five 10,000s is four 10,000s.

Four 100,000s subtract three 100,000s is one 100,000, and the 4,000,000 subtract zero is 4,000,000.

So, Lucas spent 4,149,085 points.

How did you get on with those questions? Well done.

Fantastic learning today.

I can see how hard you have tried.

We've really deepened our understanding of how we can use mental or written strategies to solve problems, depending on the context.

To solve problems, we need to look at the numbers involved, and using our number-sense can support us to know whether to use a written or a mental strategy.

A number line is a useful mental strategy when it's obvious what to add or subtract.

When we use a formal written method, the concept of constant difference can be used to make the calculation easier.

Well done today, I have had great pleasure working with you, and I look forward to learning with you again soon.