Lesson video

Hello everyone, my name is Ms. Ku and I hope you enjoy today's lesson as I'm really happy you've chosen to be learning with me today.

In today's lesson, it's gonna be easy and perhaps hard in some parts, but don't worry, I'm here to help as well as some of our other pupils too.

Now, you might come across some keywords that you may or may not know, but we'll go through those in the lesson, so don't worry.

I really hope you enjoy today's lesson so let's make a start.

Hi everyone and welcome to this lesson on problem solving with estimation and rounding, and it's under the unit estimation and rounding.

And by the end of the lesson you'll be able to use your knowledge of estimation and rounding to solve problems. Let's have a look at a quick keyword.

We'll be looking at the keyword estimate and a quick estimate for a calculation is obtained from using approximate values, often rounded to one significant figure.

And when we show estimations, we use this approximation sign, two squiggly lines to identify that calculations are approximately the same but not equal.

For example, 892 multiplied by 176.

9 is approximately 900 multiplied by 200.

Today's lesson will be broken into three parts.

The first we'll be looking at problem solving with estimation.

Then we'll look at problem solving with error intervals and then we're going to be using Fermi estimation.

So let's make a start.

Remember, a quick estimate for a calculation is obtained from using approximate values, often rounded to one significant figure.

When we estimate it's not required to be exact.

So why do we estimate? Well, we estimate to make it easier and quicker to do calculations where the answer is very close to the actual answer.

Identifying the strategy first and then completing the calculation breaks the problem solving question into chunks.

So let's have a look at an example here We're given the height of the Statue of Liberty is 62 times the height of an average secondary school pupil.

So how would you go about estimating the height of the Statue of Liberty? Well first of all, we need to have an estimate for the height of a secondary school pupil and then multiply by 60, because I've rounded 62 to 60.

What's really important is recognising the question wants us to estimate.

So that means when we round, so the average height of a pupil is around about 1.

5 metres.

The Statue of Liberty then is around about 1.

5 x 60, thus giving me the Statue of Liberty is approximately 90 metres.

This gives us an approximate answer of the Statue of Liberty.

But let's have a look at the actual answer.

Well, the actual height of the Statue of Liberty is 93 metres high, so we were quite close there.

So let's have a look at a check question.

Here's a rectangle with a 3.

78 centimetre length square removed and the area of the shaded is 198.

7 centimetre squared.

The question wants us to work out an estimate for the missing length.

See if you can give it a go and press pause if you need more time.

Well done.

So remember, because the question is asking us to estimate, we round.

So let's round those numbers.

I'm going to use 10 centimetres to represent my 9.

86 centimetres and I'm going to use four centimetres to represent the length of 3.

78 centimetres.

From here, I know that the area of the rectangle subtract the area of the square, gives me approximately 200 centimetre squared because I've rounded 198.

7 centimetre squared up to 200.

Thus I can work out the area of the square or the area of square is approximately 16 centimetres squared.

Therefore 10 multiplied by our length subtract 16 must give us the approximate 200 centimetres squared.

I can then add 16.

So I know 10 multiplied by our length gives me approximately 216 centimetres squared.

So therefore dividing by 10, I know my length has to be around about 21.

6 centimetres.

This was a great question and just remember if a question asks you to estimate always round.

Now it's time for your task.

Here is a scale drawing of a ground floor house.

Given the dimensions, can you work out an estimate for the missing dimensions, ensure to show you're working out so you can give it a go and press pause if you need more time.

Well done, let's see how you got on.

Well there are lots of different ways you could have worked this out.

So let's start with rounding appropriately.

All I'm going to do here is I'm going to round each number to the nearest five.

From here, given the diagram is to scale the 15 metres can be approximately divided in the ratio of 1:2.

So that means I can split my 15 metres into five metres and 10 metres.

Now given the fact that this area makes 30 metres squared immediately I know, this has to be a length of roundabout three metres.

Now given that this length should be slightly greater than three because we have 35 metres squared, a good estimate would be four metres.

That means this area is then expected to be around about 40 metres squared.

Thus I can work up the missing area to be approximately five metres squared.

So there are lots of different ways you could have worked this out.

Anything between six metres and 10 metres at that length of the left is absolutely fine.

And anything between four metres squared and eight metres squared for that unknown area, is also acceptable.

Great work if you've got this one right.

Well done everybody, so let's move on to question two, looking at problem solving with error intervals.

Now we use error intervals to help us calculate the largest and smallest values of a calculation and the error interval has a upper limit and a lower limit.

Here you can see A represents the lower limit of our error interval and B represents the upper limit of our error interval.

But it is important to know which limits from our error intervals to use in order to calculate the largest and smallest values.

So let's have a look at what I mean.

In an example we're going to look at addition first, imagine during a summer day the temperature was 12.

4 degrees celsius measured to one decimal place and it's increased by 6.

8 degrees celsius to one decimal place.

What do you think the error interval for the new temperature would be? Let's start by working out the error intervals of what we have.

This means we have the error interval for the original temperature is 12.

35 is less than the equal to the original temperature, less than 12.

45 and the increase has an error interval of 6.

75 less than or equal to the increase less than 6.

85.

But how do we work out the error interval using addition? Well to work out the error interval using addition, it's really straightforward.

We simply sum the lower limits and we sum the upper limits.

So that means the error interval for our new temperature, is 19.

1 is less than or equal to our new temperature, which is less than 19.

3, well done if you've recognised this.

Now let's move on to another example looking at multiplication.

A square has length 8.

2 metres measured to one decimal place.

We need to work out the error interval for the area of the square.

Well let's work out the error intervals of what we have.

We know the error interval for the length of our square as 8.

15, less than equal to L, less than 8.

25.

Thus, how do you think we can work out the error interval using multiplication? Well to work out the error interval of the area of the square, we simply square those lower limits and upper limits, thus giving me the error interval of the area of our square to be 66.

4225 less than or equal to A, less than 68.

0625.

Well done if you spotted this.

So in summary, using multiplication, the lower limit is found by the product of the lower limits and the upper limit is found by using the product of the upper limits.

Great work if you spotted this.

Now let's have a look at subtraction.

A piece of string was 12 centimetres measured to the nearest centimetre and a length of 9.

7 centimetres to one decimal place was cut off.

What is the error interval for what's remaining? Well first let's work out the error intervals of what we have.

Here we have the error interval of our length and here we have the error interval of what was cut.

How do you think we can work out the error interval using subtraction? Well, to work out the error interval for the remaining length, we have to work out the lowest possible length that we have.

Thus it would be the lower limit of the original length of string, subtract the upper limit of what was cut and to work out the upper limit of the remaining length of string, it's simply the upper limit of the length of string subtract the lower limit of the length which was cut.

Therefore giving us the error interval to be 1.

75 less than what's remaining less than 2.

85.

What is so important to remember is, notice how our error interval does not include the inequality less than or equal to.

And this is because we use the upper limit of L or C in our calculations and that they cannot be included as a measurement.

So let's practise a little bit more.

Understanding how to use the limits or error intervals allows multi-step problem solving to be easier.

So here you can see a triangle and we have the length 6.

5 centimetres to one decimal place, 8.

2 centimetres to one decimal place.

And we know the perimeter is 16.

4 centimetres to one decimal place.

The question wants us to find the error interval of the unknown length.

What do you think the first step is before applying any calculations? Well done, the first step would be to find the error intervals of each length or measurement given.

So the error interval of our 6.

5 I've labelled as A, the error interval of our 8.

2 I've labelled as B and the error interval of our perimeter I've labelled as P.

But how do you think we can work out the error interval of the missing length? Well it would require us to identify the upper limit of the missing length.

And to do it, it's simply the biggest possible value for the perimeter subtract the lowest possible value of A, subtract the lowest possible value of C.

Substituting this in gives me the upper limit of the unknown length to be 1.

85.

To work out the lower limit of the length, it would have to be the lowest possible length of our perimeter, subtract the upper possible length of A, subtract the upper possible length of C.

Thus giving me 1.

55.

Therefore the error interval for the missing length has to be 1.

55 less than L, less than 1.

85.

Now it's time for your check here you can see the area of the circle is given as 12 centimetres squared to two significant figures and the length of the square is 2.

8 centimetre squared to one decimal place.

The question wants you to work out the error interval of what is shaded.

So you can give it a go and press pause if you need more time.

Well done, let's see how you got on.

Well, let's work out those error intervals first.

The error interval for the area of the circle is given here.

So it's 11.

5 less than equal to A, less than 12.

5.

The error interval for the area of the square.

Well we have to work this out by squaring our 2.

75 and then squaring our 2.

85, thus giving us the error interval for our square to be 7.

5625 less than or equal to the area of our square, less than 8.

1225.

Now we need to work out the area of what's shaded.

To do this, let's work out our lower limit and to do it, we simply have the lowest possible area of our circle subtract the biggest possible area of our square.

This gives us our lower limit.

For the upper limit, it's the biggest possible of our circle subtract the lowest possible limit of our square, thus we have this error interval for the area of the shaded shape.

Amazing work.

If you've got any of this right, especially with the notation that we have.

Great work everybody so now let's move on to your task.

Here we have a rectangle and a square has been removed from the centre.

You need to work out the error interval of the shaded area.

See if you can give it a go and press pause if we need more time.

Well done so let's go through these answers.

First things first.

Always identify those error intervals.

Here's the error interval the length of our square.

So that means this is the error interval for the area of our square.

This is the error interval for the length of our rectangle and this is the error interval for the width of our rectangle.

Thus, to work out the error interval for the area of our rectangle, we are multiplying those lower limits of the length and width and we're multiplying the upper limits of our length and width, giving us the error interval for the area of the rectangle to be 4.

725, less than a equal to the error of our rectangle less than 6.

525.

So to work out the shaded area area interval, we simply do the smallest possible area of our rectangle, subtract the largest possible area of our square.

That will be the lower interval of the area of the shaded, and to give us the upper interval for the area of the shaded.

It will be the upper limit of the area of the rectangle.

Subtract the lowest limit of the area of the square.

So that means we have the area interval for the area of the shaded is 4.

0025, less shaded area, which is less than 5.

9625.

Fantastic work if you got this one right.

Fantastic work everybody so let's have a look at the last section of our problem solving lesson.

We're using Fermi estimation.

Have you ever been to a fair or event and there is a jar full of sweets and you win a prize if your guess is the closest to the actual number of sweets in the jar, if you have, how did you make your guess? Was it random or did you have a strategy? Aisha says, "I love these types of things, I just guess." But Sam says, "I love them too I look at the bottom layer of sweets, count how many there are, and then multiply by approximately how many layers there are." Sam continues to say, "I count around about five sweets in the bottom layer.

I think there are around about nine layers.

So I think there are around about 45 sweets." Now the method Sam is using is an approach used with Fermi problems. And Fermi problems, much like Sam's approach needed some initial data of information first in this case it was the first layer of sweets.

Then apply the relevant operation or structured working out.

And Fermi problems embrace a variety of strategies to estimate an answer.

So perhaps whenever you are making guess of the number of suites in a jar, you may have used a different approach, but still a structured approach to work out an estimate answer.

Sam brings up a really good point why are they called Fermi problems? Well, Enrico Fermi was an engineer and physicist born at the start of the 20th century.

He was born in Italy and later moved to America and his renowned for being the creator of the world's first nuclear reactor.

A Fermi estimation problem uses approximations to make calculations short and quick.

And these problems are named after Fermi as he was known for his ability to make good approximate calculations.

So let's see if we can look at a Fermi problem in a check.

The question says, how many times does the average person's heart beat in a lifetime? And you cannot use a calculator.

Now to tackle this question, I'm going to give you a hint.

How can you find some data on how many heartbeats there are per minute? Well perhaps, if you don't have this data to hand hand, why not count your own? See if you can work out this Fermi problem and remember, it's an estimation and you're not allowed to use the calculator.

Well, on average it's around about 60 beats per minute.

So let's multiply 60 by 60 to gimme the number of beats per hour.

So that would be 3,600 beats per hour.

Now I'm going to round 24 hours, which is a day to 20 just to make the calculation easier.

3,600 times 20 is 72,000 beats per day.

Now I'm gonna work it out per year.

So there are 365 days in a year.

So I'm gonna round this to 400.

So then I'm gonna do 72,000 multiply by 400 gives me a huge number, 28,800,000 beats per year.

So now we need to work out the average person beats in a lifetime.

On average people live around about 80 years.

So I'm going to multiply our 28,800,000 by 80, I'm gonna round again.

So I'm gonna call it 30 million multiply by 80, which is 2,400,000,000 beats per lifetime or 2.

4 billion beats per lifetime.

This is an amazing Fermi question where you use some initial data and rounding to come up with an approximate answer.

Well done if you got anywhere close to this answer.

And remember, it's an estimation so that there are a range of solutions here.

So don't worry if you didn't get exactly what I got on the screen.

Great work everybody, so let's move on to your task.

Question one gives you a Fermi problem.

How much rubbish in tonnes does the UK produce every year? Work it out without a calculator.

I'm gonna give you a hint.

If you don't have any data to hand, think about how much your household throws away each week.

So if you can give it a go.

Well done, let's have a look at another question.

How many pennies would it take, to make a stack the height of the Statue of Liberty? Remember, you're not allowed to use a calculator.

And also remember the height of the Statue of Liberty that we used at the start of the lesson.

Now if you don't know the thickness of a penny, why not measure one? See if you can give it a go.

Press pause if you need more time.

Lastly, question three, I don't have the answer to this, but it's still a very nice Fermi question.

How many tennis balls can you fit in your bedroom? See if you can give it a go.

Press pause if you need more time.

Great work, so let's see how you got on.

Well, looking at question one, it says, how much rubbish in tonnes does the UK produce every year? So many different ways to estimate this.

An average household produces around about 20 kilogrammes of rubbish per week.

So we need to look at it per year.

So I'm going to multiply it by 52, but 52 is a bit of an awkward number.

So I'm going to round 52 to 50.

So 20 kilogrammes multiply by 50 gives us a thousand kilogrammes of rubbish produced per household per year.

Now there are approximately 27 million households in the U.

K.

I'm gonna around 27 to 30.

So 30 million multiplied by a thousand is 30 billion kilogrammes of rubbish per year.

Writing this in tonnes, is 30 million tonnes of rubbish per year the UK produce.

This is a lovely Fermi question, but it also brings into light how much rubbish and average household produces and how much in accumulation we produce as a nation.

Let's have a look at question two.

How many pennies would it make to stack to the height of the Statue of Liberty? Well, the thickness of a penny is around about 1.

52 millimetres.

I'm gonna round to 1.

5 millimetres.

The Statue of Liberty, remember we looked at that before, is 93 metres high, so I'm going to use 90.

So the calculation would be 90 metres divided by 1.

5 millimetres.

Converting 90 metres into millimetres we have 90,000 millimetres.

Thus 90,000 millimetres divided by 1.

5 millimetres gives us 60,000 pennies.

Well done if you've got anything near that, fantastic work.

Patsy, I don't know how many tennis balls can fit in your bedroom, but there are lots of different strategies to work this out, and it is very pupil dependent.

Measuring the dimensions of your bedroom and given a tennis ball has a standard diameter of 6.

85 centimetres.

You can work out an approximation of how many tennis balls are needed to fill your room.

Well done everybody.

So in summary, with any type of problem solving question, understanding the context of the problem allows a strategy to be formed.

This can be said with estimating and error intervals.

Knowing how to find the error interval, I'm gonna apply those limits correctly to calculations, allows maximum minimum values to be determined.

Fermi problems are fantastic problems as they allow you to make reasonable estimates about the situation, so to determine an approximate answer.

Fantastic work, well done.