Lesson video

My name is Ms. Lambel.

I'm really glad that you decided to join me today to do some maths.

I hope you'll enjoy it and of course you'll enjoy it.

Let's get going.

Welcome to today's lesson.

The title of today's lesson is Problem Solving with Standard Form Calculations, and that's in the unit Standard Form Calculations.

By the end of this lesson, you will be able to use your knowledge of standard form to solve a variety of problems. Quick recap as to what standard form is.

Standard form is when a number is written in the form A multiplied by 10th power of n, where A is greater than or equal to one, but less than 10, and n is an integer.

Exponential form we will be referring to in today's lesson also.

This is when a number is multiplied by itself multiple times and it can be written more simply in its exponential form.

We will also refer to the commutative and associative laws.

The commutative law states that you can write the values of a calculation in a different order without changing the calculation.

The result is still the same.

It applies for addition and multiplication.

The associative law states that it doesn't matter how you group or pair values, ie, which we calculate first, the result is still the same.

It also applies for addition and multiplication.

In the first of today's learning cycles, we are going to be looking at data and standard form.

And in the second one we are going to be looking at standard form in geometry problems. Let's get going with that first one.

We're going to be looking at some data here.

Here we have a table showing us the approximate radius of the planets of our solar system.

What is the mean radius of the eight planets? That's the question that we need to be answering.

Aisha says, "My mum told me a great phrase to help me remember which the mean average is." There are different averages, different types of averages and we will be using each of them today.

But Aisha's mum has told her a good way to remember which is the mean, and that is, "Don't be mean, share it out." I wonder how that applies to what we are going to be doing to see whether it's really as useful as Aisha thinks.

Can you remember how we calculate the mean? The mean is the sum of all of the values divided by the number of values.

Ah, right, I can see now why Aisha's mom has said, "Don't be mean, share it out." We work at how much there is altogether and then we share it out.

We divide it by the number of values.

So here it's not very fair that Jupiter is so big and Mercury is so small.

What we would do is we would work out the total of all of their radiuses and then divide it by how many planets there are to make them all have the same radius.

What is the sum of the eight radii? And here you can use your calculator.

Pause the video, use your calculator to work out the sum of the eight radii.

What did you get? It's 1.

985 multiplied by 10th power of five.

Did you get that? Super.

The sum of the radii is 1.

985 multiplied by 10th power of five.

The mean is the sum of the values, so the sum of the radii in this case, divided by eight because we added together the radii of eight planets.

Our answer is 24812.

5.

Don't forget this value has a unit and the unit is kilometres.

I'd like you please now, and you may use your calculator to calculate the mean of these numbers.

Pause the video when you've got your answer, come back.

What did you get? The correct answer was B.

Hopefully you got B.

A was incorrect value and it was also not in standard form.

C was the correct value but not in standard form.

And D was incorrectly entered into the calculator.

The brackets were omitted from the sum of the values.

It is really important when you are summing the values that you put them in brackets, or alternatively, you can press equals at the end of your sum before doing your division.

What is the median radius of the eight planets? Give your answer in standard form.

Aisha says, "Oh, I know how to do this." Let's see what Aisha's doing.

So she's crossing off one from each end, which leaves her with two in the middle.

She says, "Now I find the mean of Mars and Jupiter." Unfortunately, Aisha's made a mistake.

What mistake has Aisha made? She's forgotten to put the radii in numerical order.

Remember, when we're finding the median, we must make sure that our data is in numerical order.

Let's write the radii in numerical order.

2.

5 multiplied by 10 cubed, three multiplied by 10 cubed, six multiplied by 10 cubed, another six multiplied by 10 cubed.

It is important we write both of those.

We can't leave out Earth or Mars.

2.

5 multiplied by 10th power of four, 2.

5 multiplied by 10th power of four, six multiplied by 10th power of four, and 7.

1 multiplied by 10th power of four.

Aisha's method was correct by crossing one off from each end so that we find the middle, it was just she'd forgotten to order them.

So let's use Aisha's method of crossing off the first and the last and then repeating that until we end up, and in this case, we end up with two in the middle.

We find the mean of these two middle values.

To find the mean of those, we add them together and divide by two by giving us an answer of 1.

55 multiplied by 10 to power of four.

And again, don't forget we need the kilometres.

We are measuring the radii of these planets in kilometres.

Find the median of those numbers.

What mistakes have been made when finding the median of these numbers? Pause the video, and then when you've decided what the mistakes are, you can come back and we'll check we agree.

What did you decide? The numbers are not in order.

Why do you think that mistake has been made? Have a look at those numbers and why do you think that mistake was made? The reason for it was the magnitude of the number has not been considered by looking at the powers of 10.

This person here, they've just looked at the number part, 2, 4, 5, and six.

They've not looked at the powers of 10 to see that actually two multiply by 10 to the power five is clearly greater than six multiplied by 10 cubed.

The second mistake is that the middle value has not been found.

Can you think of a way that this could be avoided? You could count the number of values either side of the median and it must be the same.

Really good tip: when you've crossed off your numbers, always check you've crossed the same number off to the left and the same number to the right and then you can be pretty confident then that you're gonna get your answer right.

So here we can see that two have been crossed off to the left and only one to the right.

That's how that mistake could have been avoided.

What is the modal radius of the eight planets? Give your answer in standard form.

Let's see what Aisha's got to say about this one.

"I know that modal means the mode and that is the most frequently occurring radius." Do you agree with Aisha? Aisha is correct, we need to find the radius which is most frequent.

Which is the most frequently occurring radius? This data is bimodal.

That means it has two modes, both six multiplied by 10 cubed kilometres and 2.

5 multiplied by 10 to the four kilometres occur most frequently.

There are two of each of them, so therefore the data is bimodal and both of these are considered the mode.

True or false, the mode of those numbers is zero? I'd like you to decide true or false and then also decide for me what your justification is.

Pause the video and then come back when you're ready to check your answer with me.

It was false.

A really common mistake though.

What was the justification? It was B.

If there is no value that appears more times than any other, the data has no mode.

Really important we write no mode.

'Cause think about it, if we've got a list of numbers or a list of data, some of that data could be zero and therefore, it's really important we write no mode.

What's the range of the radii of the eight planets? Aisha says, "The range is the difference between the largest and the smallest radius." Which planet has the largest radius? And that's Jupiter.

If we look, we're looking for the highest exponent, which is four, and then we're looking for the highest number part, which is 7.

1.

Which planet has the smallest radius? And that's Mercury.

Now, you may have known both of those.

You may have known Mercury was the smallest planet and Jupiter is the largest.

Mercury, if we look, as an exponent of three, and the number part of all of the ones with exponents of three was the smallest, 2.

5.

What's the range then of the radii? It is the radius of Jupiter subtract the radius of Mercury giving us 6.

85 multiplied by 10 to the power of four.

Which of the following is the correct calculation to find the range of those numbers? Pause the video, make your decision, and then you can come back.

And the correct answer was C.

The largest number in that list.

We look at the highest exponent.

We had two multiplied by 10th power of five and five multiplied by 10th power of five.

Five is obviously larger than two, so that was the largest.

And then again the smallest, we were looking for the smallest power of 10, and then looking at the smallest of the two numbers, which was the four multiplied by 10 cubed.

The mean radius of four planets closest to the sun is 4,375 kilometres.

The mean radius of the four planets farthest from the sun is 1.

8 multiplied by 10 to the power of five kilometres.

Aisha says, "The planets farthest from the sun have a higher mean.

Therefore, on average, they are larger than the planets closest to the sun." And I agree with Aisha.

The range of radii of the four planets closest to the sun is 3.

8 multiplied by 10 cubed kilometres.

The range of the radii of the four planets farthest from the sun is 4.

6 multiplied by 10 to the power of four kilometres.

Aisha says this time, "The range of radii of the planets farthest from the sun is greater than the planets closest to the sun.

Therefore, the radii of the planets farthest from the sun are more varied." Remember, the larger the range, the more varied the data is.

This table shows the widths of viruses that do not take on a circular form.

So I had a look at some pictures of some viruses.

Some of them look like they're circular and others didn't.

And these are the ones that didn't look circular, in my opinion anyway.

What is the mean width of the viruses? So we've got our little help, our little hint there.

Aisha's mum helps us.

What is the mean of the viruses? What's the sum of the five widths? That is 1.

542 multiplied by 10th power of negative six.

The sum of the widths, the mean is the sum of the widths divided by the number of values, and we added together the widths of five different viruses.

This means that the mean width of these viruses is 3.

084 multiplied by 10th power of negative seven.

And that's metres.

What about median? I'm gonna write them in order from the smallest to the largest and then we identify the middle.

And I'm going to do that by crossing off the first and last and repeating that and then I'm left with one in the middle.

This means then the median width of these five viruses is nine multiplied by 10 to the power of negative eight metres.

The mode or width of the five viruses.

Which is the most frequently occurring width? There is no mode.

None of those numbers appear more frequently than any others, so there's no mode.

What about the range? Which virus had the largest width? And that was Ebola.

Which is the smallest? And that was Hepatitis B.

The range then is Ebola subtract Hepatitis B, giving us 9.

28 multiplied by 10th power of negative seven metres.

Now you can have a go at Task A.

This table shows the mass of planets in our solar system, correct to two significant figures.

I'd like to calculate the mean, median, mode and range.

And of course, you may use a calculator for this.

Pause the video and then come back when you're ready.

In question two, I've given you here some viruses which appear circular in shape.

So when I looked at them, they looked like they were fairly circular.

Can you please calculate the mean, median, mode, and range of these? Question three, calculate the mean, median, mode, and range of these viruses, which do not take on a circular form.

And then for question four, I'd like you to compare the circular and non-circular viruses using the mean and the range.

Pause the video now and then come back and you're ready.

Superb work.

There was quite a lot to do there, wasn't there? Let's check our answers.

Question 1A, the answer is 3.

3 multiplied by 10 to the power of 26.

B, 4.

65 multiplied by 10 to the power of 25.

C, there was no mode.

And D is 1.

89967 multiplied by 10 to the power of 27.

You may have there just 1.

9 multiplied by 10th power of 27, which is fine.

Onto question two where we were looking at viruses that appeared circular in shape.

So, A was 1.

1167 multiplied by 10 to the power of negative seven.

B, 1.

2 multiplied by 10 to the power of negative seven.

C, 1.

2 multiplied by 10 to the power of negative seven.

And D, 1.

5 multiplied by 10 to the power of negative seven.

And then if you notice, these were the questions that we actually went through in the lesson, so you may have gone back and you may have copied those answers and then answered question four or you may have worked them out for yourself.

I'm not gonna read them out.

If you need to pause the video you can, but for question four we needed to say something like, "On average, the non-circular viruses are wider as their mean is higher." Or you could say, "On average, the circular viruses are narrower as their mean is lower." And then a comment about the range.

"The widths of the circular viruses are less varied as their range is lower." Or on the flip side, you could say, "The widths of the non-circular viruses are more varied as their range is higher." Now, we'll move on to standard form in geometry.

What information would I need to know about this solid chocolate sphere and the toilet roll tube in order to work out how many chocolate spheres will fit into the toilet roll tube? I'd need to know the volume of each of them and the percentage of the tube that can be filled as there will be gaps.

If I'm putting spheres into a cylinder, I'm not going to be able to fill the entire volume of the cylinder.

The volume of the chocolate is 1.

4 centimetres cubed.

The percentage of space consumed in a cylinder if we put spheres into it, is approximately 64%.

The dimensions of the toilet roll tube are its diameter is four centimetres and its height is 10 centimetres.

This is the information we need then to be able to solve this problem.

What is the formula for finding the volume of a cylinder? Now, it may be a while since you've done this, finding the volume of a cylinder, but can you dig into the depths of your memory and remember what the formula is? The formula for finding the volume of a cylinder is the volume is equal to pi r squared h.

What do the r and the h represent? r represents the radius and h represents the height.

What is the radius of our tube? If the diameter is four centimetres, then the radius is half of this, which is two centimetres.

We can now substitute r and h into the formula and find the volume of the toilet roll tube.

B is equal to pi multiplied by r, which we know is two, and that's squared multiplied by h, and we can know the height of the toilet roll tube is 10 centimetres.

This means that the volume of the toilet roll tube is 125.

7 centimetres cubed.

And I've given that to one decimal place.

This is the information we've got so far.

We've got the volume of the tube, we've got the volume of the chocolate and the percentage of space consumed.

And the percentage of space consumed is approximately 64%.

Let's now work out the volume of the toilet roll tube that will be consumed by the chocolate spheres.

We need to find 64% of 125.

7.

64% as a multiplier is 0.

64 and we multiply that by 125.

7, which is 80.

448 centimetres cubed.

We'll be able to fill that volume of the cylinder with the chocolate spheres.

The number of chocolates therefore, is going to be the volume (indistinct) tubes divided by the volume of one chocolate sphere, which is 57.

5, and that's to one decimal place.

We would be able to fit approximately 57 chocolates into the tube.

I'd like you now to put the following steps in order to solve this problem.

The problem is how many chocolate spheres can you fit in a kitchen roll tube? And the radius of the kitchen roll tube is 2.

5 centimetres and the height is 20 centimetres.

You're going to order those steps.

You can pause the video now, and then when you've got them in the right order, you can check back in with me.

And the correct order was D, A, B and C.

The first thing we need to do is to find the volume of the tube, which is D.

We then find what volume we can consume with the spheres, which is A, we then work out how many chocolates fit in there, which is the calculation for B, and then we finish that off by giving our answer of how many chocolates we would fit in the kitchen roll tube.

Now you can have a go at Task B.

So in this task, you are going to work out how many of the chocolate spheres you can fit in different real life objects.

Throughout the task, I'd like you please to use the percentage of space consumed is approximately 64%.

Question number one, how many chocolate spheres would fit into the great Pyramid of Giza? I've given you here the volume of the Great Pyramid, and the volume of the chocolate.

I'd like you to give your answer to two significant figures in standard form.

Pause the video.

You may use your calculator.

What I'd like you to do, please, is make sure you write down all steps of your work in, so if you make a mistake, you'll be able to see where that is.

Good luck with this, and we'll find out how many chocolate spheres we could fit in the pyramid.

Good luck.

Pause the video now.

Great work.

Were you surprised how many it was? You were? So was I.

Question two, how many chocolate spheres would fit into the Leaning Tower of Pisa? For the purpose of this question, we're going to assume that the diameter of the tower remains constant and that it is not leaning.

So, we're going to assume basically that it is a cylinder.

I've given you here the dimensions of the Leaning Tower of Pisa.

Its diameter is seven multiplied by 10 squared centimetres and its height is 5.

6 multiplied by 10 to the power of four centimetres.

We're using the same chocolate spheres.

And remember, you're using the same value of 64% of the volume can be consumed.

Pause the video, have a go at this, and then come back when you've worked out how many chocolate spheres we could fit into the Leaning Tower of Pisa.

Well done.

And now finally, we're going to model the Large Hadron Collider on a cylinder.

Work out how many chocolate spheres would fit inside.

This question is a little more challenging because I've actually given you information about the circumference of the Large Hadron Collider rather than its radius or its diameter, but I know that you'll be able to figure this out.

The circumference of the Large Hadron Collider is 2.

76 multiplied by 10 to the power of six centimetres.

The length of the Large Hadron Collider is 27 kilometres.

Pause the video now, work through this step by step.

You have all of the skills to be successful at this question.

Just take your time, write down each step, and then when you're ready, come back, and we'll see whether you agree with how many chocolate spheres we would be able to fit into those items. Good luck.

Like I said, you've got everything you need to be successful at this.

Come on, you can do it.

We're nearly there.

Just this final one last question to do.

I know it's a challenging one, but you've got all of the skills to be successful at it.

And pause the video now and when you get back we'll be done.

We just need to mark our answers and summarise our learning.

Question number one, the answer was 1.

2 multiplied by 10 to the power of 12.

And if you've not got that answer, I've given you there the calculations to get that answer.

So pause the video, and just take a look and see if you can see where you've gone wrong and correct your answer.

Question two, the answer was one multiplied by 10 to the power of 10.

Again, all of the calculations there if you need to check those.

And question number three was 1.

1 multiplied by 10 to the power of 18.

If you need to, you can pause the video, check through those calculations, but well done, well done for sticking with me right to the end.

There's been some challenging problems here, hasn't there? In summary, then, in the first part of the lesson, we have calculated averages and measures of spread where the numbers were written in standard form and they represented very big or very small numbers.

For example, the mean width of the viruses were very small.

We then looked at geometrical problem solving questions involving volume and capacity.

For example, how many chocolates fit into the Leaning Tower of Pisa? Throughout the lesson, it was important to have a secure knowledge of how to multiply, divide, add, and subtract numbers in standard form, as well as being able to convert between an ordinary number and standard form and vice versa.

It's been quite a challenging one today, I think you all agree, but hopefully you enjoyed especially working out how many chocolate spheres you could fit into those real life objects.

Thank you again for joining me.

Hope to see you again really soon.

Make sure you take care of yourself and goodbye.