Lesson video

Hi, everyone, my name is Miss Ku and I'm really happy that you're joining me today.

Today we're going to be looking at bounds, upper bounds, lower bounds, degrees of accuracy, error intervals and more importantly, the real life application of these.

I hope you enjoy the lesson, I know I will.

Let's make a start.

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

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

Firstly, the upper bound.

Now, the upper bound of a rounded number is the smallest value that would round up to the next rounded value.

For example, when six has been rounded to the nearest integer, that means 6.

5 is the upper bound.

When 6.

3 has been rounded to one decimal place, this means 6.

35 is the upper bound.

The lower bound.

Well, the lower bound for a rounded number is the smallest value that the number could have taken prior to being rounded.

We'll also be looking at the word error interval and the error interval for a number x shows the range of possible values of x and it's written as an inequality, where a is less than or equal to x which is less than b.

Today's lesson will be broken into two parts.

We'll be looking at using rounding and estimation and bounds first and then we'll be considering bounds second.

So let's make a start using rounding, estimation and bounds.

When calculating, it's so important to reflect on the reasonableness of an answer.

For example, Laura and Sam are given this calculation, 401.

75 divided by 4.

8 squared, add 9.

1.

Sam says the answer is 125 and Laura says the answer is 12.

5.

How can you quickly work out who is most likely correct without a calculator? So you can give it a go.

Press pause if you need more time.

Well, we do it using estimation.

So let's have a look at our calculation.

Firstly, all I'm going to do is round my 401.

75 to one significant figure, giving me 400.

And I'm going to round my 4.

8 squared to one significant figure, giving me five squared.

And then I'm going to round my 9.

1 to the nearest 10, giving me 10.

From here, applying my priority of operations, I work out five squared to be 25 and then I have my fraction, 400 over 35, which I'm going to simplify to 80 over 7.

Converting this into a mix number, I have the estimate answer to my calculation to be 11 and three sevenths.

So that means I expect the answer to be around about 11 and three sevenths.

So that means Laura is most likely to be correct.

Now what I want you to do is a check.

Without working out the exact answer, which of the following is the correct answer to this calculation? See if you can give it go.

Press pause if you need more time.

Well done.

Well, hopefully you can spot it's C, but let's have a look at my working out.

Working this out, you might notice I've rounded each number to one significant figure.

This gives me an approximate answer of 10, so I expect my answer to be close or in the tens.

So it must be 23.

5888795, given the fact that all the other numbers are either too big or too small.

Choosing the correct operation for the situation is still important when estimating.

For example, I'm going to give you three scenarios and I want you to match the correct scenario with the correct calculation.

We have to estimate how many 95 millilitre cups of a jug holding 821 millilitres can fill.

We need to estimate the area of a rectangular face of wood which is measured as 821 millimetres by 95 millimetres.

And we have to estimate the total mileage when a car travelled 821 miles and then a further 95 miles.

So which calculation matches each situation? See if you can give it a go.

Press pause if you need.

Well done.

Well hopefully you've spotted we have these answers.

Look for the keywords in the context.

Because we're trying to find out how many times one number fits into another number, that's how we knew it was division.

We're finding the area of a rectangular face, so the formula to work out area is lengths and widths.

Lastly, the word total mileage indicate we are summing.

Well done if you got these.

Now let's move on to another check.

Here we have a wall in the shape of a trapezium and the Oak teacher wants to cover the wall in 20 centimetre square tiles.

Now, the tiles come in packs of 12, each pack costing 24 pounds.

I want you to estimate how much it will cost the teacher to tile the wall.

Ensure you show all your working out.

Press pause as you need more time.

Great work, let's have a look how you did.

Well, given the fact that we have different units of measure, metres and centimetres, it's best to always convert to the same units and because we are estimating, let's round as well.

So rounding and converting to the same units, I have 200 centimetres, 300 centimetres and 400 centimetres.

Now let's work out the area of each of our tile.

Well, the area of our tile would be 20 centimetres by 20 centimetres, giving the area to be 400 centimetres squared.

Now I can work out the approximate area for the trapezium.

While using our formula for the area of the trapezium, you sum the 200 and the 400, divide by two, multiply by the perpendicular height, giving me the approximate area of our trapezium to 90,000 centimetres squared.

Given we know each tile is 400 centimetres squared, dividing 90,000 by 400 gives me 225 tiles are needed.

Now remember, they're sold in packs of 12.

So if we divide 225 by 12, it gives me an approximate answer of 20.

So that means I know that Oak teacher needs to buy 20 packs.

Remember, each pack cost 24 pounds, so you multiply 20 by 24 to give us an estimate cost of 480 pounds.

Really well done if you got this.

So using formulae can appear with bounds, but it can also appear with estimation as well and it's so important to know the correct formula to use as well as which bound to substitute in according to the operation.

So all I'm going to do is summarise this in a table.

If we have a multiplied by b, how would we find the lower bound? Well, it'll have to be the lower bound of a multiplied by the lower bound of b.

How would I find the upper bound? Well, it'd be the upper bound of a multiplied by the upper bound of b.

What I'd like you do, is copy this table down and have a think.

How do we find the lower bound and the upper bound given these different operations? See if you can give it a go.

Press pause if you need more time.

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

Well, if you're adding a to b and we're asked to find the lower bound, it must be the lower bound of a add our lower bound of b.

To work out the upper bound, it would be the upper bound of a, add our upper bound of b.

Now, what is we were subtracting? So a subtract b, this means to find the lower bound a subtract b, we must use the lowest possible value of a, subtract the highest possible value of b.

And to work out the upper bound, it'd be the upper bound of a, subtract the lower bound of b.

If we have a calculation where a is being divided by b, then we do the lower bound of a divided by the upper bound of b and to find the upper bound, it would be the upper bound of a divided by the lower bound of b.

Really well done if you got this and if you don't have this table down, copy it down, press pause if you need.

So now let's have a look at an example.

Here's a right angled triangle.

We know the angle x is 46 degrees, correct to the nearest degree and we know y is 14.

3 centimetres, correct to one decimal place.

Looking at this question, what do we need to identify so to work out the maximum value of z? Well, we need to know the error intervals of x and y and given the fact that we have lengths and angles in a right angled triangle, we also know we're using trigonometry to calculate the value of z.

So let's work out the bounds and the correct trigonometric ration.

First of all, let's identify our bounds in x and y.

Because you know x has been rounded to the nearest degree, here's our error interval for x.

And because we know y has been rounded to one decimal place, here is our error interval for y.

Now let's have a look at what trigonometric ratio we're using.

Well, using x as the focus angle, how would you label the triangle? Well, we'd label y as the hypotenuse, z as the opposite and have our unlabeled length as the adjacent.

Now which trigonometric function would be the best to use for this question? Well, given the information and we're asked to calculate the value of z, using the sine function would be the most efficient.

And to work out the opposite side length using sine, it would be the opposite side is equal to the hypotenuse, multiplied by sine theta.

So let's work this out.

We're going to use the sine rule and remember, h is our hypotenuse, let's substitute in.

Well, we know the opposite side length is z, the hypotenuse is y and the angle is labelled as x.

So from here, let's substitute so we're able to find the upper bound of z.

To do this, we're going to work out the upper bound of z by the upper bound of y, multiplied by sine, the upper bound of x.

Substituting this in, we have 14.

35 multiplied by sine 46.

5, giving me 10.

4 centimetres to three significant figures.

I just want to say, in the previous example, we used the upper bound of the angle to find the upper bound of the sine of the angle.

In other words, the upper bound of z equal the upper bound of y, multiplied by sine, the upper bound of x.

This did, in this case, find the upper bound, however, increasing the size of the angle does not always increase the value of the sine, cosine or tangent of the angle.

For example, if you wanted to find the upper bound of the cosine of an angle of 43 degrees to the nearest degree, the error interval of x would be 42.

5 less than or equal to x, which is less than 43.

5.

What do you notice about the cosine of the lower and upper bounds? While using a calculator, cos 42.

5 is greater than cos 43.

5.

So the upper bound of the cosine of the angle would use the lower bound of the angle here.

Therefor, careful attention must be paid when angles and the trigonometric ratio sine, cosine and tangent are used.

Great work, everybody.

So now it's time for your task.

Aisha and Lucas use a calculator to work out this answer.

Aisha says it is 142.

0827761 and Lucas say it's 14.

20827761.

One of the answers is correct.

Use estimation to find out who is correct.

Press pause if you need more time.

Well done, let's move on to question two.

Which of the following calculations will give the upper bound of a squared? See if you can give it go.

Press pause for more time.

Well done.

For question three, here's a right angled triangle.

A is 5.

6 centimetres, correct to one decimal place and c is 8.

3 centimetres, correct to one decimal place.

You're asked to work out the minimum length of b and give your answer to three significant figures.

See if you can give it a go.

Press pause if you need more time.

Great work, let's move on to question four.

Here's a right angled triangle.

A is 12 centimetres correct to the nearest centimetre, b is five centimetres correct to the nearest millimetre.

You're asked to work out the maximum angle of theta, giving your answer to three significant figures.

See if you can give it a go.

Press pause if you need more time.

Well done, let's go through question five.

The Oak teacher travelled from Stoke to London.

She travelled 240 kilometres to the nearest five kilometres and it took her 200 minutes to the nearest five minutes.

Calculate the lower bound for the average speed of the journey and give the answer in kilometres per hour, correct to three significant figures.

Notice how I've given you the formula for the average speed, which is the total distance over the total time.

See if you can give it a go.

Press pause if you need more time.

Well done, let's go through these answers.

Well, here's the working out for question one and Aisha is correct, because our answer should be approximately 125 and Aisha's answer is closest to that.

Well done.

For question two, well done, you should've got B.

For question three, here's our working out.

Always work out those error intervals first as they always help.

Press pause if you need more time to look at these answers.

Well done.

Let's move on question four.

Always work out those error intervals as they're always helpful.

From here, press pause if you need more time to mark your work.

Now let's have a look at question five.

Let's have a look at these error intervals for distance and time first.

Our distance is given in kilometres, but our time is given in minutes.

So converting our time to minutes allows us to write our answer in the correct units of measure stated in the question.

So from here, you should have this error interval for distance and this error interval for time given in hours.

From here, we can work out the lower bound of the speed to be 70.

4 kilometres per hour.

Well done if you got this one right.

Great work, everybody.

So now it's time for the second part of our lesson where we are considering bounds.

Upper and lower bounds are crucial when identifying the accuracy of the answer and the smaller the difference between the limits of accuracy, the more accurate the answer.

As a results, it's important to state a number to a degree of accuracy which has considered both the upper and lower bounds.

For example, here we have the upper and lower bounds of a number x and they've been calculated.

The lower bound of x is equal to 48.

32384412 and the upper bound is 48.

32448392.

What we're going to do is, we're going to consider these bounds and identify the value of x to a suitable degree of accuracy.

Now, Alex has considered these bounds and said x is 48.

3 to three significant figures.

Now what I want you to do, is have a think.

Do both lower and upper bounds round to 48.

3 to three significant figures? Have a look.

Well, yes they do.

Both the lower bound and the upper bound round to three significant figures, but can we improve this degree of accuracy? Well, yes, we can.

So let's have a look at what Sophia says.

Sophia has considered the bounds and said, "Well, x is 48.

32 to two decimal places." So do both the lower and upper bounds round to 48.

32 to two decimal places? Well, yes, they do.

But can we improve the degree of accuracy? Let's have a look at what Sam says.

Sam's considered the bounds and says, "Well, x is 48.

324 to three decimal places." I want you to have a look at the upper and lower bounds and think, do the upper and lower bounds round to 48.

324, to three decimal places? Yes, they do, but can we improve the degree of accuracy? Well, Lucas says he's considered bounds and he thinks x is 48.

3245 to four decimal places.

Now, do both the upper and lower bounds round to 48.

3245, to four decimal places? Well, no, so therefore we know x is equal to 48.

324 to three decimal places.

We've considered both the upper bound and the lower bound and both of these bounds round to 48.

324 to three decimal places.

So that means it's the most accurate value for x.

Now what I want you to do, is have a look at these upper and lower bounds of x and I want you to consider these bounds and identify the value of x to a suitable degree of accuracy.

See if you can give it a go.

Press pause if you need more time.

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

While considering bounds, x can be rounded to 0.

4 to one significant figure, but we can't be more accurate.

Now looking to two significant figures, you might notice, well, they don't round to the same number, so we definitely can't use two significant figures , but considering bounds, x can be 0.

385 to three significant figures, and this is the most accurate value for x.

So therefore, x is equal to 0.

385 to three significant figures or you may have written x to be 0.

385 to three decimal places.

Continuing with this check question, I want you to explain why Laura is wrong.

Laura says x is 0.

3850 to four significant figures, and I also want you to think to make Laura's comment correct, what bound would need changing and to what value? See if you can give this a go.

Press pause if you need more time.

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

Well, Laura's wrong, because if you round both the upper bound and the lower bound to four significant figures, they give two different numbers, 0.

3850 and 0.

3581.

So therefore, x is not 0.

3850 to four significant figures.

But to make Laura's comment right, one example would be if the upper bound had the fourth significant figure as zero.

This would then make Laura's comment correct.

Well done if you got this.

Let's continue with another check question.

Andeep is given this formula, m is equal to the square root of p over q.

We have p, which is 3.

495, correct to three decimal places and q is 1.

13, correct to three significant figures.

I want you to work out the error intervals for p and q, work out the upper and lower bound for m and then Andeep says he's considered bounds and he says m is equal to 1.

8 to two significant figures.

Is Andeep correct, and I'd like you to show your working out.

See if you can give it a go.

Press pause if you need more time.

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

Well, first of all, we should have these error intervals for p and q.

Working out the upper bound for m, remember, we're using division so it should've been the upper bound of p divided by the lower bound of q.

Then to work out the lower bound of m, it should've been the lower of p divided by the upper bound of q.

Considering these bounds, let's have a look if Andeep is correct.

Andeep says m is 1.

8 to two significant figures.

Yes, both bounds agree to this number and the degree of accuracy.

Well done if you got this one right.

Great work, everybody.

So now it's time for your task.

Considering bounds of the following, work out the value of x and give your answer to a suitable degree of accuracy using significant figures.

I've done the first one for you, see if you can continue the rest.

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

Question two gives you c is equal to the cube root of a, all divided by b, where is 3.

84, correct to two decimal places and b is 4.

493, correct to three decimal places.

By considering bounds, work out the value of c to a suitable degree of accuracy and you must show all your working out and give a reason for your answer.

See if you can give it a go.

Press pause if you need more time.

Well done, let's move on to question three.

A sphere has a radius of 2.

43 metres, correct to the nearest centimetre.

It has a mass of 2150 kilogrammes, correct to the nearest gramme.

Given that the volume for the volume of a sphere is four third PI r cubed and the formula for density is mass over volume.

By considering bounds, work out the value of the density in kilogrammes per metres cubed of the sphere to a suitable degree of accuracy.

You must also show your working out.

Great question, see if you can give it a go.

Press pause if you need more time.

Well done, let's go through these answers.

For question one, you should've had these values of x to the appropriate degree of accuracy.

Well done if you got this.

For question two, working out those error intervals always helps and then identifying c is 0.

35, correct to two significant figures or maybe you put correct to two decimal places.

Both are fine.

Well done.

Moving on to question three, this was a great question.

Remember, same again, identify those error intervals for our radius and mass.

From here, we can calculate our bounds for our volume.

Press pause if you need more time to look at this.

From our bounds of our volume that we can substitute into the formula for density, giving the density as 36 kilogrammes per metres cubed to two significant figures.

All to the nearest integer.

Massive well done if you got this one right.

Great work, everybody.

So in summary, calculating answers is as equally important as reflecting on the reasonableness of the answer and estimating allows us to reflect on the appropriateness of the answer obtained.

Upper and lower bounds are crucial when identifying the accuracy of the answer and the smaller the difference between the limits of accuracy, the more accurate the answer.

Given this, we can work out a single number to a degree of accuracy which has considered both the upper and lower bounds.

Really well done, everybody.

It was great learning with you.