Up until now, we’ve only been talking about multiplication with counting-numbers; now we’re going to be dealing with measuring-numbers as well. Try answering these questions.

If I can walk $100$ metres in $1$ minute, how far can I walk in $7$ minutes? Show answer

You can think of this as $7$ groups of $100$, and $100\times7 = 700$, so the distance will be $700$ metres.

One minute is $60$ seconds; how many minutes is $720$ seconds? Show answer

This is like asking how many groups of $60$ makes $720$, and $720\div 60 = 12$, so $720$ seconds is $12$ minutes.

Even though these questions involve situations we haven’t dealt with yet, you can answer them with the skills you’ve already learnt, but what if I changed the numbers? I could have asked how far I can walk in $6.2$ minutes, and might only be able to walk $85.1$ metres in a minute. You can’t think of that as $6.2$ groups of $85.1$; that doesn’t make any sense. I could have asked how many inches $50.2\,\mathrm{cm}$ is, knowing that one inch is $2.4\,\mathrm{cm}$.

We’re going to need to think about multiplication in a new way to deal with these situations.

Before we work out how to perform calculations like these, I want to explore situations like Question 4.2.1(a) in more detail. If you just want to know the procedure for multiplying measuring-numbers, you can skip ahead to Section 4.2.4: A Previously Solved Problem. A calculator can tell you what $1.21\times0.07$ is, for example, but it can’t tell you which calculation to do in the first place. You will need to understand proportion to know that.

When something travels at a constant speed (without slowing down or speeding up), there’s a special relationship between the distance it’s travelled and the time taken. It’s called ‘proportion’, and you could say ‘the distance is in proportion to the time’ or ‘the distance is proportional to the time’. That means that if I pick a unit of time, for example $1$ minute as in Question 4.2.1(a), then the distance traveled in that unit of time is always the same ($100\,\mathrm{m}$ in this case), no matter what part of the journey it’s in. The distance increases by $100\,\mathrm{m}$ for each extra minute.

If I changed my pace, walking more slowly for a while, then the distance traveled would not be in proportion to the time taken, because I might travel $100\,\mathrm{m}$ in the first minute, but then only $90\,\mathrm{m}$ in another minute.

There are many different situations that involve proportion. Here are some more examples:

If you have a map where every inch on the map represents $5$ miles in real life, and the distance between two towns is $5$ inches on the map, how far apart are the towns? The distance on the map is proportional to the distance in real life. (On a world map, where the surface of the globe has to be distorted to fit on a flat piece of paper, the distances on the map are not in proportion to the distances in real life.) Show answer

In this case, the distance between the towns is $5\times5 = 25$ miles.

A recipe requires $20$ grams of sugar for every cup of water. If you want to use $80$ grams of sugar, how much water should you use? In this case, the mass of the sugar is proportional to the volume of the water (and the volume is proportional to the mass – proportion goes both ways). Show answer

$80/20 = 4$, so you need $4$ cups of water.

If a pump can fill a $120$ litre tank in $15$ minutes, how much water can it pump in $1$ minute? The volume of water is proportional to the time. Show answer

This is a lot like dividing $120$ things into $15$ groups. $120/15 = 8$ (see Question 4.1.18 for a fast way to calculate $120/15$) so the pump pumps $8\,\mathrm{L}$ in $1$ minute.

Now that you have a feel for what proportion is, let’s see what happens when we’re not restricted to counting-numbers.

Imagine you have a machine that turns seawater into clean drinking water. The machine works day and night, and a constant stream of water comes out – a total of $4$ tonnes every day (every $24$ hours). The water goes into a tank. At the start the tank is empty, or in other words, after $0$ days there is $0$ tonnes of water in the tank. After $1$ day there are $4$ tonnes, after $2$ days there are $8$ tonnes, after $3$ days there are $12$ tonnes, and so on. We can write down these facts as equations: \begin{align*} 4\times0 &= 0\\ 4\times1 &= 4\\ 4\times2 &= 8\\ 4\times3 &= 12 \end{align*}

How much water comes out of the machine in half a day ($12$ hours)? How should we write this down as an equation? A ‘half’ is actually a number, and is usually written as $\frac{1}{2}$. Show hint

If you wait for half a day, twice, you’ll have waited for a whole day, and there’d be $4$ tonnes of water in the tank.However much water comes out in the first half of the day must also come out in the second half of the day, and make a total of $4$ tonnes. So we’re looking for a number which when added to itself makes $4$. It must be $2$. If we wrote down $4\times3 = 12$ to say that $12$ tonnes comes out in $3$ days, then maybe we should write down $4\times\frac{1}{2} = 2$ to say that $2$ tonnes comes out in $\frac{1}{2}$ a day. Section 4.2.5: Fractions will cover this sort of problem in more detail.

How much water comes out in $2$ and a half days? How should we right this down as an equation? Two and a half is also a number, and is written as $2\frac{1}{2}$. Show answer

We know that $2$ tonnes comes out in $\frac{1}{2}$ a day, and $8$ tonnes comes out in $2$ days, so $10$ tonnes must come out in $2\frac{1}{2}$ days: $4\times2\frac{1}{2} = 10$.

How much water comes out in $0.1$ days? It might help you to consider how big the answer will be – is it greater or less than $1$? How should we write this down as an equation? Show hint

If you waited $0.1$ days ten times, you would have waited $1$ whole day.However much water comes out in $0.1$ days, ten times that amount must be $4$ tonnes. If $1$ tonne came out in $0.1$ days, then in a whole day, $10$ tonnes would come out. So $1$ tonne is too big. We’re looking for a number smaller than $1$.

We essentially want to split $4$ tonnes into $10$ parts. If we split $1$ tonne into $10$ parts, we get a tenth of a tonne, or $0.1$ tonnes. $4$ tonnes divided by $10$ must be $4$ times this much – $4$ tenths of a tonne, or $0.4$ tonnes. It is less than $1$ tonne, so that’s a good sign, but if you’re still not confident about the answer, you could find $10$ times $0.4$ by adding $0.4+0.4+0.4+0.4+0.4+0.4+0.4+0.4+0.4+0.4$. You’d find that it does add up to $4$.

So $0.4$ tonnes comes out in $0.1$ days. We can write this as $4\times0.1 = 0.4$. You might think that’s obvious, $4\times0.1$ must be $0.1+0.1+0.1+0.1$ which is $0.4$. But remember that we just chose to write the equation like that, and it may not behave the same way with measuring-numbers as it does with counting-numbers.

Suppose that at midnight on Monday, you’ve forgotten how much water there is in the tank, so you mark the current level, and decide to talk about the amount of water in the tank as the number of tonnes more than the Monday midnight level. So $3$ days later the amount of water is $+12$ tonnes, which we could write as $4\times3 = 12$, or to be clearer, $4\times(+3) = +12$. How much water was there $2$ days before, at midnight on Saturday? How should we write this as an equation? Show hint

You may want to review Section 3.3: Negative Numbers.There were $8$ tonnes less. If we wrote this as $4\times2 = 8$ it would mean that there are $8$ more tonnes $2$ days later. We can represent $2$ days before as $-2$ and $8$ tonnes less as $-8$, so $4\times(-2) = -8$. Section 4.2.7: Multiplying Negative Numbers will cover this sort of problem in more detail.

With the old kind of multiplication, we could only ask how much water would come out in $1$ day, $2$, $3$, $4$ days, and so on. Now we could ask how much comes out in an hour, a minute, a second, and even smaller units of time if we wanted. Of course, with any real machine, there would be small changes in the flow of water so that the mass of the water isn’t actually in proportion to the time when we use small units like seconds. In fact, all of the examples we’ve looked at aren’t really proportion if you look closely enough. For example, the grains of sugar prevent you from having exactly $20$ grams for every cup of water even if you had a scale that could measure that accurately, but it’s still very useful to treat these situations as if they were true mathematical proportion. In Section 4.2.1: Average Rate we’ll talk more about treating situations that aren’t proportional as if they were. For the rest of this section, we’re going to do some proportion problems requiring a bit more ingenuity.

Instead of saying ‘for every’, people often say ‘per’. For example, ‘the car was travelling at $20$ kilometres per hour’ means that it went $20$ kilometres for every hour.

If a car travels at $8$ miles per hour, how far does it go in $15$ minutes? Show hint

An hour is $4$ times $15$ minutes.

You can think of $15$ minutes as the unit of time. It travels $8$ miles in $4$ units of time, so it travels $8\div 4 = 2$ miles in $1$ unit of time.

A wheel turns around $500$ times every $4$ minutes, and travels $80$ centimetres per turn.

What is the ‘rpm’ of the wheel? That stands for ‘revolutions per minute’ – the number of turns the wheel makes in $1$ minute. Show answer

In $1$ minute, the wheel turns $500\div4 = 125$ times. In other words, the wheel turns at $125$ rpm.

How far does it travel in $1$ minute? Show answer

We already worked out that the wheel turns at $125$ rpm, so it travels $80 \times 125 = 10,000$ centimetres in $1$ minute.

It’s also possible to work out the answer like this: in $4$ minutes, the wheel travels $500\times80 = 40,000$ centimetres, so in $1$ minute it travels $40,000\div4 = 10,000$ centimetres. With the first method, we’re calculating $(500\div4)\times 80$, and with the second, $(500\times80)\div4$, which are of course both the same because the order doesn’t matter with multiplication and division.

There is a mark somewhere on the wheel. If the mark is at the top of the wheel when it starts rolling, and it stops as soon as the mark gets to the bottom, how far has the wheel travelled? Show hint

If the wheel stopped as soon as the mark got back to the top, that would be $1$ whole turn, and the wheel would have travelled $80$ centimetres.You can think of a whole turn as two half turns. When the wheel makes a full turn, the mark starts at the top and goes to the bottom, then goes from the bottom to the top. The distance travelled should be the same in each of these half turns, and they have to add up to $80$ centimetres, so the distance travelled must be $80\div2 = 40$ centimetres.

There is a story about Archimedes that says a king asked him to work out if a crown was pure gold, or a mix of gold and silver. Archimedes found out by measuring the mass and volume of the crown. Gold weighs $20$ grams for every millilitre, and silver weighs $10$ grams for every millilitre. If the crown weighed $250$ grams, and its volume was $15$ millilitres, was it pure gold? Show answer

If the crown were pure gold, it would weigh $20\times15 = 300$ grams. It is lighter than that, at $250$ grams, so it can’t be pure gold.

Guess how many millilitres of gold and how many of silver went into the crown, and see if you are right. (This question is answered with algebra in Question 7.5.36.) Show hint

How heavy would it be if it were pure gold? What if it were pure silver? Does it seem to be mostly gold, or mostly silver?We already worked out that if the crown were pure gold, it would weigh $300$ grams. If it were pure silver, it would weigh $10\times15 = 150$ grams. $250$ is closer to $300$ than it is to $150$. In fact, $250$ is only $50$ away from $300$, but is $100$ away from $150$, and $100$ is twice $50$, so my guess would be that there is twice as much gold as there is silver, by volume. I am guessing that there were $10$ millilitres of gold, and $5$ millilitres of silver.

I can check if this could be correct by working out how much the crown would weigh if my guess were true. The $10$ millilitres of gold would weigh $20\times10 = 200$ grams and the $5$ millilitres of silver would weigh $10\times5 = 50$ grams, making $200 + 50 = 250$ grams together. So my guess may be right.

The road from Town A to Town B is $360$ kilometres long. A car travels along the road from town A towards town B at $40$ kilometres per hour. Another car travels from town B towards town A at $50$ kilometres per hour. If they both leave at the same time, how far from town A will they be when they meet? Show hint

What does ‘when they meet’ mean? It means the distance between them is zero, so it’ll probably be helpful to think in terms of the distance between the cars – how much does it change every hour?When you have no idea how to work out a problem, it often helps to define exactly what the question means. Sometimes you think of a definition that isn’t very useful, for example you could define ‘when they meet’ as ‘when the distance of each car from town A is the same’ which isn’t very helpful (unless you know algebra – see Question 7.5.8). So if you get stuck, it can help to think of lots of different ways it could be defined. In this case, if you think of ‘when they meet’ as ‘when the distance between the cars is zero’ then you can work it out, because the distance between the cars decreases by $90$ kilometres each hour ($40 + 50 = 90$). The distance was $360$ kilometres at the start, so it will be zero after $4$ hours ($360\div90 = 4$).

But the question didn’t ask for the time, it asked for the distance from town A. After $4$ hours, the first car will be $4\times40 = 160$ kilometres from town A.

It is also possible to work it out by calculating that the second car travelled $4\times50 = 200$ kilometres, so it is $360 - 200 = 160$ kilometres from town A. Working it out both ways helps us check our answer.

A hiker starts walking along a trail at $80$ metres per minute. Six minutes later, a runner starts running along the trail at $200$ metres per minute. When will the runner overtake the walker? Show hint

How far ahead is the walker when the runner starts? How much closer do they get every minute?When the runner starts, the walker has already walked $480$ metres ($6\times80 = 480$). In each minute, the runner travels $200$ metres and the walker travels $80$ metres, so the runner closes the distance by $120$ metres ($200-80 = 120$). Since they’re $480$ metres apart when the runner starts, the runner will overtake the walker after$ 4$ minutes of running ($480\div120 = 4$).

A tank is being filled with water through two pipes. $80$ litres of water per minute flow through the first pipe. $600$ litres of water flow into the tank in $5$ minutes. How much water flows through the second pipe every minute? Show answer

In $1$ minute, $120$ litres of water flow into the tank ($600\div5 = 120$). The second pipe must carry $40$ litres of water every minute ($120 - 80 = 40$).

If pump A takes $30$ minutes to fill a $4500\,\mathrm{L}$ tank, and pump B takes $45$ minutes, how long will they take to fill the tank if they’re both pumping together?

You want to fill an aquarium with salt water. Your fish need $110$ litres of water, with $35$ grams of salt for every litre (in other words, the ‘concentration’ of the salt is $35$ grams per litre). You already have $100$ litres of salt water, but with a salt concentration of only $30$ grams per litre, and you want to add some salt and water to it to get the salt water you need. How much salt do you need to add? Show hint

How much salt is in the salt water already? How much salt should there be at the end?

You may have trouble seeing how to get from the information you are given, to the information you need (how much salt to add). It helps to ask yourself, what could I work out from this information?

Knowing that you have $100$ litres of salt water at $30$ grams per litre, you can work out that you already have $30\times100 = 3000$ grams of salt. You want to have $110$ litres of salt water at $35$ grams per litre, so there needs to be $35\times110 = 3850$ grams of salt in the water at the end. If you have $3000$ grams of salt, and you need $3850$ grams of salt, then you need to add $3850 - 3000 = 850$ grams of salt.