Lesson video

Hello, everyone, my name is Ms. Ku, and I'm really happy that you're joining me today because it's one of my favourites again, compound measures.

Compound measures is one of my favourites because it's the real-life application of mathematics.

I really hope you enjoy the lesson.

Let's make a start.

Welcome to this lesson on advanced problem solving with compound measures, and it's 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.

Today's lesson will be broken into two parts.

First of all, we'll be looking at rate of flow, and then we'll be looking at further problems with compound measures.

So let's make a start looking 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 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 1 metre cubed per hour.

This means a flow of 2 litres per second.

And this means a flow of 3 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, multiply by width, multiply 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.

And 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 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're 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 2 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 2 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 2 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 2 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? See if 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 the 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 multiply 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 1 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 2 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 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, whereas 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 trapezius.

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 give 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'll 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 2.

"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 question 3.

Question 3, "There's a trough and it's filled with water, but water is leaking out at a rate of 0.

2 litres per second." 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.

Great work.

So let's move on to question 4.

Question 4 states, "A water container is full of water and is in the shape of a cylinder and cone.

In two hours, the cylinder is empty.

How long, in hours and minutes, to the nearest minutes, does it take for the whole container to empty?" I've been very kind here and I've given you the formula for the volume of a cone.

It's one-third times by pi, times by radius squared, multiply by the height.

See if you can give it a go.

Press pause for more time.

Well done.

Let's see how you got on.

Well, question 1, you should have had these answers.

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

Well done.

For question 2, 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 3, 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've got this.

For question 4, let's have a look at this working out.

Lots of working out here.

Please press pause if you need more time.

Basically, for the time to empty the container, it will take two hours and 33 minutes.

Really well done if you've got this.

Great work, everybody.

So let's have a look at further problems with 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 7 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, gives 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 ask 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 multiplied by two-thirds, 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've 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 with the 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 increase 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 that 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 per centimetre squared is not 0.

203 newtons per centimetre squared.

So therefore, Aisha is incorrect.

So let's represent this algebraically.

Well, we know before it's N newtons, and the area is pi times, I'm gonna call the radius x, so therefore pi times x squared.

To work out the pressure, remember it's force per unit area, so I'm going to divide by pi times x squared.

Algebraically, I'm going to leave it as N over pi x squared.

Now let's have a look at the force applied after the increase.

Well, we know the force would have to be 1.

1 newtons.

Remember before it was simply 1 newton, simply N, and then increasing by 10% gives me 1.

1 newtons.

With the radius was x centimetres, so now what I'm going to do is increase that by 10%, so it's 1.

1X as the radius.

Using the formula for the area of the circle, it's pi times our radius squared, so it's pi times 1.

1x all squared, giving me 1.

21 pi x squared.

Just like before, remember pressure is force per unit area, so I'm going to divide by 1.

21 pi x squared.

Algebraically, 1.

1 N over 1.

21 pi x squared.

I can simplify removing a common factor of 1.

1 giving me N over 1.

1 pi x squared.

Well, let's find out if the pressure did decrease by 10%.

Well, 90% of our N over pi x squared is N over 9 pi x squared, and that is not the same as N over 1.

1 pi x squared.

So it's exactly the same processes of what we applied in the question where we're given numerical values, in this case, we're just given algebraic values, and using our knowledge on simplifying terms. Now let's have a look at a check.

Here are two different size cubes with different masses.

Write the ratio of densities of cube A to cube B in the form of 1 to n.

Notice how cube A has a length of y and mass of x.

Notice how cube B has a length of 4y and a mass of 4x.

See if you can give it a go.

Press pause one more time.

Great work.

Let's see how you got on.

Well, putting it into a ratio table, we know the mass of A is x, and the volume of A is given as y multiply by y, multiply by y, which is y cubed.

Remember, density is mass per unit volume, so I'm going to divide by y cubed.

Algebraically, this means the density of cube A is given as x over y cubed per centimetres cubed.

Looking at cube B, we know the mass is 4x and the volume can be found by multiplying 4y by 4y by 4y, which simplifies to 64y cubed.

Once again, remember density is measured per unit of volume, so I'm gonna divide by 64y cubed, thus giving me 4x over 64y cubed as the mass of B.

So, simplifying this gives me x over 16y cubed per 1 centimetres cubed.

In other words, we have the density of B to be x over 16y cubed per centimetres cubed.

So, writing this in the form of 1 to n, we know x over y cubed is the density of cube A, two, x over 16y cubed is the density of cube B.

It can be simplified down to 1 to 1 over 16.

Really well done if you got this one right.

Problem solving with compound measures can incorporate forms of proportion too.

And ratio tables and summary tables continue to be excellent representations to support the understanding of these types of ratio questions.

Now, we're gonna have a look at a question where an alloy is made by mixing the masses of copper, nickel, and zinc in the ratio of three to one to one.

Now, given copper has a density of 8.

96 grammes per centimetres cubed, and nickel has a density of 8.

9 grammes per centimetres cubed, and zinc has a density of 7.

14 grammes per centimetres cubed, the question wants us to work out the density of the metal alloy to two decimal places.

Now, we have no masses here, we only have proportions, so this means regardless of the masses, as long as the ratio between the metals is correct, the density of the alloy will always be the same.

For example, this is just one example where I've said the mass of copper is 30 grammes, the mass of nickel is 10 grammes, and the mass of zinc is 10 grammes.

Notice how the masses are still in proportion, three to one to one.

From here, we can work out the volume, then we can work out the total mass and the volume, and then we can work out the density of the alloy, which is 8.

51 grammes per centimetres cubed.

Another example will show exactly the same answer.

Notice how I've chose different numerical values for mass, 600, 200, and 200, still in that ratio of three to one to one, and I get exactly the same density for our alloy.

So now let's show this algebraically.

We're going to write the ratio in terms of x.

Copper is 3x, nickel is x, and zinc is x, still in the ratio of three to one to one.

We can then work out the volume in terms of x.

So the volume for copper will be 3x over 8.

96.

The volume for nickel would be x over 8.

9.

And the volume for zinc would be x over 7.

14.

Working out the total mass and the total volume, summing up all those masses, we know it's 5x, summing up all those volumes, really awkward fractions here, gives you 85,295x over 145,248.

You can represent it as a decimal if you want.

The density would still be 8.

51 grammes per centimetres cubed.

Lovely example just to show you that you can use numerical values for the mass as long as the proportions are the same, or algebraic values still giving you that same density.

Great work, everybody.

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

Question 1 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 2 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 drones 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 3.

"An object with an area of 20 centimetres squared acts on a desk with an 80 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 incorrect? And you must show your working out." See if you can give it a go.

Press pause if you need more time.

Great work.

Let's move on to question 4.

"The mass of cube A to the mass of cube B is in the ratio of one to two.

The ratio of the length of the cubes is in the ratio of one to two.

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

Is Sophia correct?" And you have to show you're working out.

See if you can give it a go.

Press pause if you need more time.

Great work.

Let's move on to the last question, which is really tough.

"A chocolate making factory is making a new type of sweet.

Milk chocolate with a density of 1.

4 grammes per centimetres cubed in the shape of a cone is bonded with a hemisphere of pink chocolate with a density of 1.

2 grammes per centimetres cubed.

Now the height of the cone to the radius of the hemisphere is given in the ratio of three to one, and you're asked to work out the average density of the new sweet to three significant figures." I've given you the volume of the cone and the volume of the sphere.

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've got these.

Press pause if you need more time.

Well done.

Let's move on to question 3.

Here's my working out and the answer.

We know 70% of 4 newtons per centimetre squared is not 2.

8 newtons per centimetre squared, so therefore, Jun is not correct.

Well done if you've got this.

Press pause if you need.

Great work, everybody.

Here are the answers to question 4.

Press pause if you need.

Well done.

And for question 5, here are our answers.

Press pause if you need.

And overall, in summary, the average density of our sweet was 1.

39 grammes per centimetres cubed.

Well done if you got this.

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.