Direct and inverse proportion

The unitary method is used for simple calculations with direct and inverse proportion. Usually we know three related values and have to work out a fourth. In these problems we must carefully tell direct and inverse proportion apart, because they are calculated differently.

Why we need it

You go to the shop to stock up on lemonade. For £2 you bought 5 bottles of lemonade. How many would you buy if you had £4? This is a typical problem that can be solved with this method.

At this point we can do some simple thinking – in the first case we had £2 to spend, in the second £4. So we can expect that if we have twice as much money, we can buy twice as many bottles. For £4 we would buy 2 · 5 = 10 bottles of lemonade.

The reasoning “the more… the more” is not entirely obvious, because we can have the following problem: 10 bricklayers build a house in four months. How long would it take 20 bricklayers to build the house? If we use the previous reasoning – there are twice as many bricklayers, so there will be twice as many months – we get that 20 bricklayers would build the house in 2 · 4 = 8 months.

This is of course not correct, because the more bricklayers there are, the fewer months it takes them to build the house. We must go the other way round: twice as many bricklayers build the house in half the time. That gives the correct result: 20 bricklayers build the house in 4/2 = 2 months.

The two different methods above have names: direct and inverse proportion.

Direct proportion

If “the more… the more” holds, it is direct proportion. Examples:

• The more money we have, the more lemonades, cucumbers or scooters we can buy.
• The more articles a journalist writes, the more money they earn.
• The more diggers there are digging, the more they dig up.
• The longer we leave the pump running, the more water we pump out.

Example: A car uses 6 litres of petrol per 100 kilometres. How many litres of petrol does it use after driving 250 kilometres? This is a direct proportion problem, because the further the car travels, the more petrol it uses.

Method: First we write down what we know. We use the following table, which makes it clear what we know and what we want to work out:

$$\begin{array}{rcl}100\text{ km}& \dots & 6\text{ litres}\\ 250\text{ km}& \dots & x\text{ litres}\end{array}$$

In the first row we have the values we know completely. In the second row, on the right we have the value we don’t know, and on the left the value for which we want to work out the unknown. In the columns there must always be the same kind of quantity; in this case each column must contain either always kilometres or always litres.

There is a memory aid that will help you solve this problem. First draw an arrow on the right, pointing from bottom to top, like this:

$$\begin{array}{rcl}100\text{ km}& \dots & 6\text{ litres}\\ 250\text{ km}& \dots & x\text{ litres}\uparrow \end{array}$$

If it is direct proportion, then on the left we draw the same arrow, from bottom to top:

$$\begin{array}{rcl}100\text{ km}& \dots & 6\text{ litres}\\ \uparrow 250\text{ km}& \dots & x\text{ litres}\uparrow \end{array}$$

In the next step we build fractions from the columns, following the arrows. The numerator is the number where the arrow starts and the denominator is the number where the arrow ends. We set these fractions equal. From the first column (kilometres) we get the fraction $\frac{250}{100}$ and from the second (litres) the fraction $\frac{x}{6}$. Written as an equation:

$$\frac{250}{100}=\frac{x}{6}$$

This is an ordinary linear equation, which we can solve by equivalent steps. We multiply the whole equation by six, which gets rid of the fraction on the right-hand side:

$$6\cdot \frac{250}{100}=x$$

Now we just work out x:

$$x=6\cdot \frac{250}{100}=6\cdot \frac{5}{2}=\frac{30}{2}=15$$

The result is that after driving 250 kilometres the car uses 15 litres of petrol.

Inverse proportion

If “the more… the less” holds, it is inverse proportion. Examples:

• The more people share a pizza, the smaller the slice each one gets.
• The faster our internet is, the sooner we download a film.
• The faster we drive, the sooner we get to our destination.
• The more workers there are working on a house, the sooner the house will be built.

Example: On your computer you have internet with a speed of 2 MB per second, and you downloaded a video of a concert in 450 seconds. How long would it take to download the same video if you had internet with a speed of 6 MB per second?

The method will be the same as for direct proportion – we write the data in a clear table:

$$\begin{array}{rcl}2\text{ MB/s}& \dots & 450\text{ seconds}\\ 6\text{ MB/s}& \dots & x\text{ seconds}\end{array}$$

Now the arrows. The right-hand arrow is again from bottom to top, nothing changes there. But because this is inverse proportion, the left-hand arrow goes the opposite way, from top to bottom.

$$\begin{array}{rcl}2\text{ MB/s}& \dots & 450\text{ seconds}\\ \downarrow 6\text{ MB/s}& \dots & x\text{ seconds}\uparrow \end{array}$$

The rest of the method is the same – we make fractions from this table, again following the arrows. We get the fractions $\frac{2}{6}$ and $\frac{x}{450}$, which we set equal:

$$\frac{2}{6}=\frac{x}{450}$$

We multiply the equation by 450:

$$450\cdot \frac{2}{6}=x$$

And work out x:

$$x=450\cdot \frac{2}{6}=450\cdot \frac{1}{3}=\frac{450}{3}=150$$

With the faster connection we download the video in 150 seconds.

Notice that the method agrees with intuition. In the last step we divide 450 by three, which is what we would expect. The speed has increased from 2 MB/s to 6 MB/s, which is three times as fast. So we would expect the time it takes to download the video to be a third as long → and that is exactly what the fraction $\frac{450}{3}$ expresses.

Examples

First example: A room in a luxury hotel costs three hundred and fifty pounds for seven nights. How much would the same room cost if we wanted to stay for twelve nights?

First we write the data in a table:

$$\begin{array}{rcl}7\text{ nights}& \dots & 350\text{ pounds}\\ 12\text{ nights}& \dots & x\text{ pounds}\end{array}$$

Now we must decide whether this is direct or inverse proportion. Is it true that the longer we stay there, the less we pay? No, the opposite is true – the longer we stay, the more we pay. So this is direct proportion. We add the arrows:

$$\begin{array}{rcl}7\text{ nights}& \dots & 350\text{ pounds}\\ \uparrow 12\text{ nights}& \dots & x\text{ pounds}\uparrow \end{array}$$

Using the arrows we build the fractions:

$$\frac{12}{7}=\frac{x}{350}$$

We solve the equation:

$$\begin{array}{rcl}\frac{12}{7}& =& \frac{x}{350}\\ 350\cdot \frac{12}{7}& =& x\\ x& =& 350\cdot \frac{12}{7}\\ x& =& \frac{350\cdot 12}{7}\\ x& =& 600\end{array}$$

The room would cost us £600 for 12 nights.

Second example: 10 pickers are picking strawberries and pick 50 kilograms of strawberries a day. How many kilograms of strawberries will 7 pickers pick in a day?

We write the data in a table:

$$\begin{array}{rcl}10\text{ pickers}& \dots & 50\text{ kg}\\ 7\text{ pickers}& \dots & x\text{ kg}\end{array}$$

Is this direct or inverse proportion? Here there is a bit of a catch, because unlike in the previous examples, in the row with the unknown we have a smaller number of pickers than in the first row. Don’t let that confuse you; it is still direct proportion, because the more pickers there are, the more strawberries are picked. Or the other way round – the fewer pickers, the fewer strawberries. That is also direct proportion. So both arrows will go from bottom to top.

$$\begin{array}{rcl}10\text{ pickers}& \dots & 50\text{ kg}\\ \uparrow 7\text{ pickers}& \dots & x\text{ kg}\uparrow \end{array}$$

We make the fractions:

$$\frac{7}{10}=\frac{x}{50}$$

and work out x:

$$\begin{array}{rcl}\frac{7}{10}& =& \frac{x}{50}\\ 50\cdot \frac{7}{10}& =& x\\ x& =& 50\cdot \frac{7}{10}\\ x& =& 5\cdot 7\\ x& =& 35\end{array}$$

In a day, seven pickers will pick 35 kilograms of strawberries.

Third example: Jack’s average step length is 80 cm. On the way home from school he counted that he took 1,300 steps. How many steps would he take if each one were exactly one metre long?

First the table. Be careful: in the question we have 80 cm and one metre. In the table, though, we need the data in the same unit, so instead of one metre we use 100 cm – we could also have gone the other way and written 1 metre and 0.8 metres.

$$\begin{array}{rcl}80\text{ cm}& \dots & 1300\text{ steps}\\ 100\text{ cm}& \dots & x\text{ steps}\end{array}$$

And which kind of proportion is it? The longer our step, the fewer steps we need to cover a given distance, so this is inverse proportion. The left-hand arrow goes from top to bottom, the right-hand arrow from bottom to top.

$$\begin{array}{rcl}80\text{ cm}& \dots & 1300\text{ steps}\\ \downarrow 100\text{ cm}& \dots & x\text{ steps}\uparrow \end{array}$$

The fractions will look like this:

$$\frac{80}{100}=\frac{x}{1300}$$

We rearrange and work out x:

$$\begin{array}{rcl}\frac{80}{100}& =& \frac{x}{1300}\\ 1300\cdot \frac{80}{100}& =& x\\ x& =& 1300\cdot \frac{4}{5}\\ x& =& \frac{1300\cdot 4}{5}\\ x& =& 1040\end{array}$$

Jack would need only 1,040 steps to cover the same distance.

Formulae

As you can see from the calculations above, the methods are always the same and lead to simple formulae. First we write the table symbolically:

$$\begin{array}{rcl}a& \dots & b\\ c& \dots & x\end{array}$$

Then these formulae hold. For direct proportion:

$$x=b\cdot \frac{c}{a}$$

And for inverse proportion:

$$x=b\cdot \frac{a}{c}$$