Lesson video

Hello, everyone.

My name is Ms. Coe and today we're looking at compound measures, a fantastic topic looking at pressure, speed, density, volume, lots of real-life applications.

I hope you really enjoy the lesson.

Let's make a start.

Hi, everyone and welcome to this lesson on problem solving with compound measures under the unit compound measures.

And by the end of the lesson, you'll be able to use your knowledge of compound measures to solve problems. So let's have a look at some keywords, starting with rate of flow.

Now, rate of flow measures the volume of fluid that passes through a particular pipe or channel per unit of time.

We'll be looking at this a lot in our lesson.

Today's lesson will be broken into two parts.

Firstly, we'll be looking at the rate of flow and secondly, we'll move on to real-life problems involving compound measures.

So let's have a look at rate of flow.

Now, the rate at which a fluid or substance flows into or out of an object is called the rate of flow.

So can you think of some examples where you've experienced rate of flow today? Have a little think.

Well, some examples would be running a tap in order to have a bath or perhaps have a shower or perhaps make a cup of tea or coffee or even pouring some fluid in order to have a drink.

All of these are examples of rate of flow and rate of flow is everywhere.

Now, rate of flow measures the volume of fluid that passes through a particular pipe or channel per unit of time and there are different units to measure rate of flow.

So what do you think these units represent? See if you can give it a go.

Press pause for more time.

Well done.

Let's have a look.

Well, this means a flow of one metre cubed per hour.

This means a flow of two litres per second, and this means a flow of three centimetres cubed per second.

There are lots of units used to measure rate of flow, so conversions between units is important to know.

So let's have a look at a question.

We can work out rate of flow using our knowledge of volume and time.

Here we have a cuboid and this cuboid has a leak.

It was full of water and after three hours, it's completely empty.

So what is the rate of flow of the water leaving the container? And the question wants us to give our answer in centimetres cubed per minute.

So step one, let's work out that volume first.

To work out the volume of our cuboid, we do length multiplied by width multiplied by our height, which is 50 multiplied by our 20 multiplied by our 36, giving me 36,000 centimetres cubed.

Now, this means 36,000 centimetres cubed flowed out in three hours.

Let's insert this into a ratio table.

36,000 centimetres cubed in three hours.

Now, remember, the question wants it per minute.

So I'm gonna work out per hour first by simply dividing by three.

This means it was 12,000 centimetres cubed per hour.

Given we know there are 60 minutes in an hour, we simply divide by 60.

So that means 200 centimetres cubed per minute and that is the rate of flow of the leak.

It's 200 centimetres cubed per minute.

Now I want you to do a check.

A hose is filling a tank in the shape of a cuboid and the rate of flow of the hose is constant and it takes the hose 4.

5 hours to fill the tank.

Now, what I want you to do is work out the rate of flow of the hose, giving your answer in centimetres cubed per minute.

See if you can give it a go.

Press pause one more time.

Great work.

Let's see how you got on.

Well, first of all, let's work out the volume of our cuboid, which is 32,400 centimetres cubed.

So that means we know a volume of 32,400 centimetres cubed was filled in 4.

5 hours.

But given the fact that we want our answer in centimetres cubed per minute, let's divide by 4.

5 to give me the 7,200 centimetres cubed per hour.

But we want per minute, so I'm going to divide by 60, giving me 120 centimetres cubed per one minute, which is the rate of flow.

The rate of flow of the hose is 120 centimetres cubed per minute.

Really well done if you got this.

But sometimes the question will give the rate of flow and it's important to use this to calculate time given the context of the question.

Once again, there are lots of units used to measure the rate of flow, so conversions between units are important to know.

Let's have a look at a farmhouse sink, which is generally in the shape of a cuboid.

Now, water flows into the sink at a constant rate of 0.

4 litres per minute and we're asked to calculate the number of minutes it will take to completely fill the sink.

Well, to do this, just like we did before, let's work out the volume of our sink.

To work out the volume, length times width times height, gives me 41,600 centimetres cubed.

From here, notice how we are given the rate of flow, which is in litres per minute.

So that means we need to convert our volume of centimetres cubed into litres.

The conversion is one litre is always 1,000 centimetres cubed.

So that means the volume of our farmhouse sink in litres is 41.

6 litres.

From here, we can now insert our information into a ratio table and work it out in minutes.

We know the rate of flow is 0.

4 litres per minute.

We also know our farmhouse sink is 41.

6 litres in capacity.

So what do we multiply by? Well, we multiply by 104.

So that means in order for the water to fill the sink with 41.

6 litres, it will take 104 minutes.

Now it's time for a check.

A water tank has a height of 1.

5 metres and a radius of 45 centimetres, and water is leaking out the tank at a rate of two litres per minute.

How many minutes, rounded to the nearest minute, will it take to empty the tank? 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, first of all, let's work out the volume of the cylinder.

We know the formula is pi times radius squared times our height.

Substituting what we know, we now have the volume of our cylinder to be 303,750 pi centimetres cubed.

Notice how I've kept it in terms of pi just for that level of accuracy.

From here, the rate of flow is given in litres.

So let's convert the volume of our cylinder into litres by simply dividing by 1,000.

So that means our 303,750 pi centimetres cubed is 303.

75 pi litres.

From here, we can insert our information into a ratio table.

The rate of flow is two litres per minute, which is represented here, but we know the volume is 303.

75 pi litres.

So identifying the multiplier of 151.

875 pi allows me to convert my two litres into that 303.

75 pi litres, thus giving me 477.

1 minutes.

So that means it will take 477 minutes for the cylinder to be completely empty.

Massive well done if you got this one.

Sometimes questions can involve compound shapes and other units to be converted.

For example, this diagram shows a swimming pool in the shape of a prism and the swimming pool is filled using a hose.

Now, the hose pumps water at a constant rate of two litres per second.

Will the pool be filled in less than a day? Well, let's have a look.

Let's work out the volume of our pool given we know it's a prism and to work it out, we have to look at this cross-sectional area as we have our compound shape.

How do you think we could work out this cross-sectional area? Well, there are lots of different ways, but this is just one example by forming a rectangle and a trapezium.

Now, once we form the rectangle and trapezium, can you identify these missing lengths so we can work out that cross-sectional area? So you can give it a go.

Now, hopefully you spotted we should have these lengths given in metres.

Very well done.

Now we have all the information that we need to work out the cross-sectional area of our prism.

The area of this section, which is a rectangle, is simply 0.

3 metres squared.

And the area of this section, which is a trapezium, gives us 16.

38 metres squared.

So that means I know the total cross-sectional area is 16.

68 metres squared.

Now we know the cross-sectional area, I can easily work out the volume of our swimming pool.

The volume of the swimming pool is the cross-sectional area, multiplied by that 10, giving me 166.

8 metres cubed.

Once again, the rate of flow is given in litres per second.

So that means we need to convert our metres cubed into litres.

It's important to recognise one metre cubed is 1,000 litres.

So that means we know the volume of our swimming pool in litres is 166,800 litres.

So let's insert what we know into our ratio table.

We know the rate of flow was two litres per second and we know the volume of our swimming pool is 166,800 litres.

So what's our multiplier? Well, it has to be 83,400.

So multiplying by 83,400 gives us 83,400 seconds.

But remember, the question asked us will the pool be filled in less than a day? So we need to convert our 83,400 seconds into hours.

Well, to convert it into minutes, we divide by 60 to give me 1,390 minutes, and to convert it into hours, we divide by 60 again to give me 23.

2 hours to one decimal place.

So will the pool be filled in less than one day? Yes, it will be filled in less than one day because remember, one day is 24 hours and the pool will be filled in less than 24 hours.

Well done.

So let's have a look at a check.

A hose is filling a trough with water at a constant rate of 2.

5 litres per minute.

We're asked to work out in hours how long it will take for the trough to be completely filled.

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, let's identify the cross-sectional area.

It's made up of two trapezia.

So to work out the cross-sectional area, we substitute the values that we know.

So we have the cross-sectional area to be 3,500 centimetres cubed.

So therefore, to work out the volume, we multiply it by 150 to give me 525,000 centimetres cubed.

Remember, the question gives me the rate in litres per minute.

So I need to convert to litres.

In other words, our trough is 525 litres.

I simply divided our 525,000 by 1,000.

So now let's insert our information into our ratio table.

We know the rate was 2.

5 litres per minute.

We know the volume of our trough was 525 litres, so the multiplier would be 210, so therefore, it'd take 210 minutes.

But remember, the question wanted it in hours.

210 minutes is 3.

5 hours.

So it'll take 3.

5 hours to fill the trough.

Well done.

Great work, everybody.

So now it's time for your task.

I want you to work out the rate of flow for the following.

Press pause as you'll need more time.

Well done.

Let's move on to question two.

A garden has a water butt to store water for a garden and it's in the shape of a cylinder with a radius of 30 centimetres and a height of 1.

2 metres.

Izzy fills the water butt with a hose and water flows out of the hose at a constant rate of 0.

4 litres per minute.

How long, to the nearest minute, will it take to fill the water butt? See if you can give it a go.

Press pause if you need more time.

Well done.

Let's move on to wuestion three.

Question three is a trough and it's filled with water, but water is leaking out at a rate of 0.

2 litres per second.

Now, Jun says it'll take less than one hour for the trough to be empty from full.

Is Jun correct? I'd like you to show your working out.

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, question one, you should have had these answers.

Press pause if you need more time to copy those answers.

Well done.

For question two, we should have had this working out, but essentially, it should have taken 848 minutes to fill the water butt.

Well done if you got this.

For question three, we should have had this working out so you know that the volume of our trough is 510 litres, therefore the time to empty would be 0.

708 hours.

So therefore, we know it will take less than one hour.

Great work if you got this.

Fantastic work, everybody.

So let's have a look at some real-life problems involving compound measures.

Now, we've seen many real-life applications of compound measures.

When algebra is incorporated in the question, it's important to note that the process does not change.

So let's have a look at two questions.

One question will use numerical values and the other question will use algebra.

And you'll notice how the process does not change.

For example, a car travels at 36 miles per hour for 20 minutes, then for 42 miles per hour for 10 minutes and we're asked to find the average speed of the whole journey.

So we know 36 miles per hour in 20 minutes can be represented in this ratio table, 36 miles per hour, 60 minutes.

And then to find the distance in 20 minutes, we simply divide by three.

So we know 12 miles was travelled in 20 minutes.

Next, we know the speed of 42 miles per hour is the same as 42 miles per 60 minutes.

We need to know what was the distance in 10 minutes.

So we simply divide by six, thus giving me seven miles in 10 minutes.

Then just like we've done before, to work out the average speed, we total the distances and we total the time.

The total distance is 19 miles and the total time is 30 minutes.

Therefore, because we know speed is measured per one unit of time, converting it into hours, I'm gonna simply multiply by two, which gives me 60 minutes and the distance would be 38 miles.

So therefore, the average speed is 38 miles per hour.

Now let's have a look at exactly the same question but written using algebra.

A car travels at 15x miles per hour for 20 minutes, then for 12x miles per hour for 10 minutes and we're asked to work out the average speed for the whole journey.

Just like before, 15x miles per hour means 15x miles per 60 minutes, but we want 20 minutes.

So we simply divide by three, thus giving me 5x miles in 20 minutes.

The next speed was 12x miles per hour.

Well, we know 12x miles per hour is the same as 12x miles per 60 minutes.

We also know the question says it travelled this for 10 minutes, so we're dividing by six, thus giving me 2x miles in 10 minutes.

Just like before, we need to total the distance, 5x miles add 2x miles give 7x miles and the total time, well, 20 minutes add 10 minutes is 30 minutes.

Just like before, speed is measured per unit of time.

So we need to make it an hour.

In other words, 60 minutes.

So we multiply by two giving me 14x miles in 60 minutes, which gives an average speed of 14x miles per hour.

Exactly the same processes, just one simply uses algebra.

So let's have a look at a check.

A car travels from A to B in 10 minutes at a speed of 24y kilometres per hour.

Then the car travels from B to C in 40 minutes at a speed of 9y kilometres per hour.

And you're asked to work out the average speed of the car in kilometres per hour from A to C.

See if you can give it a go.

Press pause if you need more time.

Well done.

So let's see how you got on.

Well, the distance from A to B is 24y kilometres per hour, which is 24y kilometres per 60 minutes.

And we know the time is 10 minutes, so we simply divide by six.

Thus telling me 4y kilometres was travelled in 10 minutes.

Now, from B to C, we know the speed was 9y kilometres per hour, which is the same as 9y kilometres per 60 minutes.

But we know the time was 40 minutes.

So we simply multiply by 2/3, thus giving me a distance of 6y kilometres in the 40 minutes.

So to work out the average speed, we sum the total distance.

4y add 6y is 10y kilometres.

We sum the total time.

10 add 40 is 50 minutes.

And remember, speed is measured per unit of time, so I want it per hour, in other words per 60 minutes.

So I'm multiplying by six over five, thus giving me 12y kilometres per 60 minutes, which then tells you that the average speed from A to C is 12y kilometres per hour.

Really well done if you got this one right.

So alongside knowledge of compound shapes, proportion is often incorporated in the problem solving.

Once again, ratio tables are excellent optional approaches.

For example, a force of 17 newtons acts on a table by a cylinder.

Then the force increases by 10% and the radius increases by 10%.

Aisha says the pressure will then decrease by 10%.

Is Aisha correct? Well, let's have a look.

First of all, we know before, there was a force of 70 newtons and the area of our circle was pi times our 10 squared, thus giving me an area of 100 pi.

Notice how I've kept it in terms of pi of that accuracy.

But for pressure, we need to work out per unit of area.

So I'm going to divide by 100 pi to give me 0.

223 newtons.

This is the pressure applied before the increase.

Now let's work out after the increase.

Well, we know the force increased by 10%, so therefore, the force is 77 newtons.

We also know the radius increased by 10%.

So that means the radius is 11 centimetres.

Working out the area, it would be pi times 11 squared, which gives me 121 pi.

Once again, notice how I've kept it in terms of pi just for the accuracy.

Then we know pressure is measured per unit of area.

So I'm going to divide by 121 pi, giving me a force of 0.

23 newtons.

So let's compare 90% of 0.

223 newtons centimetre squared is not 0.

203 newtons centimetre squared.

So therefore, Aisha is incorrect.

Great work.

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

The mass of cube A to the mass of cube B is given in the ratio of four to five and the ratio of the length of the cubes are given in the ratio of one to two.

And the question wants you to work out the ratio of the densities in the form A to B.

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, first things first, we know that the mass of cube B must be 25 gramme and we know the lengths of cube B must be four centimetres given the ratio from the question.

Now we can simply insert them into a ratio table.

We have a mass of 20 grammes for cube A and a volume of two multiplied by two multiplied by two, which is eight centimetres cubed.

Remember, density is measured per unit of volume.

So that means we need to divide by eight, which gives me 2.

5 grammes per centimetres cubed.

That would be the density for cube A.

Now looking at cube B, we know the mass of 25 grammes and the volume is found by four multiplied by four multiplied by four, which is 64 centimetres cubed.

Remember, density is mass per unit volume.

So we need to divide by 64 to find our mass, which is 25 over 64 grammes per one centimetres cubed.

So that means we can write the ratio of densities of A to B as 2.

5 grammes per centimetres cubed to 25 over 64 grammes per centimetres cubed.

Now, remember because the units are the same, we don't have to include our units.

So the ratio is simply 2.

5 to 25 over 64.

A massive well done if you've got this ratio or an equivalent ratio.

Well done.

Great work, everybody.

So now let's have a look at your task.

Question one states a car travels at a speed of x kilometres per hour for 20 minutes and then for 3x kilometres per hour for 10 minutes and you're asked to find the average speed of the car.

Question two states a drone starts by flying x metres per second for five minutes before increasing the speed to 5x metres per second for 15 minutes.

Work out the drone's average speed.

See if you can give these a go.

Press pause if you need more time.

Well done.

Let's have a look at question three.

An object with an area of 20 centimetres squared acts on a desk with an 18 newton force.

The force increases by 20% and the area increases by 15 centimetres squared.

Jun says the pressure has decreased by 30%.

Is Jun correct? And you must show you're working out.

So if you can give it a go.

Press pause if you need more time.

Well done.

Let's have a look at question four.

The mass of cube A to the mass of cube B is given in the ratio of one to two, and the ratio of the lengths of the cubes is in the form of one to two.

Cube A has a mass of 12 grammes and a length of four centimetres.

Sofia says the density of the ratios must be one to two.

Is Sofia correct? And you must show your working out.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's move on to these answers.

Here's my working out and the final answer.

Really well done if you got these.

Press pause if you need more time.

Well done.

Let's move on to question three.

Here's my working out and the answer.

We know 70% of four newtons per centimetre squared is not 2.

8 newtons per centimetre squared.

So therefore, Jun is not correct.

Well done if you got this.

Press pause if you need.

For question four, we have this working out.

Sofia's incorrect as the ratio of densities of cube A to cube B is actually 1 to 1/4.

Really well done if you got this.

Press pause if you need more help with this working out.

Great work, everybody.

So in summary, the rate of flow measures the volume of fluid that passes through a particular pipe or channel per unit of time.

And there are lots of different units to measure rate of flow.

So it's important to understand what the unit represents.

Alongside knowledge of compound shapes, proportion is often incorporated into the problem solving and ratio tables can help to organise your calculations.

Finally, there are many real-life applications of compound measures.

When using algebra, the processes we use with numerical measures is exactly the same as algebraic measures.

Great work, everybody.

It was wonderful learning with you.