## What you need to know

### Direct and Inverse Proportion

In maths, we say that two quantities are proportional if as one of them changes, the other one also changes in a specific way. There are two types of proportionality that you need to be familiar with, direct and inverse proportion. Having good knowledge of the following algebra topics will help with this:

**Direct Proportionality**

If two quantities, A and B, are directly proportional, then as one increases the other also increases at the same rate, e.g. as one doubles, the other one also doubles. To express this, we use the proportionality symbol, \propto, and write

A \propto B,

which reads like “A is directly proportional to B”. However, there isn’t much we can do with A \propto B as it is, so we turn it into the equation

A = kB,

where k is the number which tells us how A and B are related. It is called the constant of proportionality. When you know the value of this constant, you can put it to good use. Let’s see an example.

**Note:**You should be able to recognise graphs that represent direct proportionality. The equation y = 3x for example, is a straight-line graph with gradient 3 and y-intercept 0. This last point is true of all direct proportion graphs: they all pass through the origin (the point (0,0)).

If you see a graph that doesn’t, then the two quantities on the axes must not be directly proportional.

**Inverse Proportionality**

If two quantities, C and D, are inversely proportional, then as one increases the other decreases at the same rate, e.g. as one get three times bigger, the other one gets three times smaller. To express this, we write

C \propto \dfrac{1}{D},

which reads like “C is inversely proportional to D”, which is the same as saying “C is directly proportional to \frac{1}{D}”. As before, there isn’t much we can do with C \propto \frac{1}{D}, so we turn it into the equation

C = \dfrac{k}{D},

where k is the constant of proportionality which we want to find and then use.

**Note:** You should be able to recognise graphs that represent inverse proportionality. The equation y = \frac{6}{x} for example, is a reciprocal graph (see: right) with asymptotes along both axes. Indeed, all inverse proportion graphs will have this same shape with asymptotes in the same places.

If you see a graph that doesn’t, then the two quantities on the axes must not be inversely proportional.

### Example 1: Direct Proportion

y is directly proportional to x. When y = 24, x = 8. Work out the value of y when x = 2.

We have that y is directly proportional to x, i.e. y \propto x. Then, expressing this as an equation we get

y = kx

In the question we’re given that when y = 24, x = 8. Substituting these into the equation above, we get 24 = 8 \times k, and so k = 24 \div 8 = 3. Thus, the proportionality equation becomes

y = 3x

We can now use this to work out that when x = 2, y = 3 \times 2 = 6. If you are a foundation student, you won’t explicitly be asked to form an equation like y = 3x. However, you will be expected to know how to work with constants of proportionality and use such an equation, so all of this content is still important to you.

### Example 2: Inverse Proportion

y is inversely proportional to x. When y = 2, x = 3. Work out the value of y when x = 18.

We have that y is inversely proportional to x, so we write y \propto \frac{1}{x}. Then, expressing this as an equation we get

y = \dfrac{k}{x}

In the question we’re given that when y = 2, x = 3. Substituting these into the equation above, we get 2 = \frac{k}{3}, and so k = 2 \times 3 = 6. Thus, the proportionality equation becomes

y = \dfrac{6}{x}

Therefore, when x = 18, y = \frac{6}{18} = \frac{1}{3}.

### Example Questions

1) F is inversely proportional to the square of r. When F = 50, r = 3. Calculate the value of F when r = 12.

F is inversely proportional to the square of r (i.e. r^2), so

F \propto \dfrac{1}{r^2}

We can rewrite this equation using k to represent the constant of proportionality:

F = \dfrac{k}{r^2}

We know that when F = 50, r = 3, so we can work out k by substituting these known values into the equation:

50 = \dfrac{k}{3^2}

so

50 = \dfrac{k}{9}

Since we want to work out the value of k, we need to rearrange the formula to make k the subject which we can do by multiplying each side by 9:

k = 50 \times 9

Therefore, k = 450.

We can now rewrite the original equation with a value for the constant k:

F = \dfrac{450}{r^2}.

Since

F = \dfrac{450}{r^2}

then

F = \dfrac{450}{12^2}

so F = 3.125

2) Given that y is inversely proportional to x, complete the following table.

Since y is inversely proportional to x, we can write this as an equation, as follows:

y \propto \dfrac{1}{x}

We can rewrite this equation using k to represent the constant of proportionality:

y = \dfrac{k}{x}

From the table, we know that when x = 4, y = 7.5. If we substitute these values into our equation, we can work out the value of the constant k:

Since

y = \dfrac{k}{x}

then

7.5 = \dfrac{k}{4}

Since we want to work out the value of k, we need to rearrange the formula to make k the subject which we can do by multiplying each side by 4:

k = 7.5 \times 4

k = 30

We can now rewrite the original equation with a value for the constant k:

y = \dfrac{30}{x}

Now that we have calculated the value of the constant k and rewritten the original proportionality equation, we can work out the value of y when x = 60:

When x = 60, y= \dfrac{30}{60}

So y = 0.5

We can also work out the value of x when y = 12:

When y = 12,

12 = \dfrac{30}{x}

If we multiply both sides by x and divide both sides by 12, we have rearranged the equation making x is the subject:

x = \dfrac{30}{12} .

Therefore when y = 12, x = 2.5

So, the completed table should look like this:

3) The amount of money earned by Sasha, M, is directly proportional to the number of hours she works, h. If she works for 9.5 hours she earns £155.80.

a) Express M in terms of h.

b) Using the equation formed in part a), or otherwise, find out how many hours it would take her to earn £688.80.

a) Since M is directly proportional to h, we can write this as an equation, as follows:

M \propto h

We can rewrite this equation using k to represent the constant of proportionality:

M = kh

We know that when M = \pounds155.80, h = 9.5 \text{ hours}, so we can work out k by substituting these known values into the equation:

155.80 = k \times 9.5

Since we want to work out the value of k, we need to rearrange the formula to make k the subject which we can do by dividing each side by 9.5:

k = \dfrac{155.80}{9.5}

Therefore, k = 16.4.

We can now rewrite the original equation with a value for the constant k:

M = 16.4h

b) Now that we have calculated the value of the constant k and rewritten the original proportionality equation, we can work out the value of h when M = \pounds688.80.

Since

M = 16.4h

then

\pounds688.80 = 16.4h

If

16.4h = \pounds688.80

then

h = \pounds688.80 \div16.4

so

h = 42

So it will take Sasha 42 hours to earn £688.80.

4) x is directly proportional to y. When x = 2, y = 8.

a) Write an equation connecting x and y

b) Calculate the value of x when y is 32

c) Calculate the value of y when x is 50

a) x is directly proportional to y, so

x \propto k

We can therefore write this as an equation where k is a constant:

x = ky

We now need to work out the value of the constant k which we can do by substituting in the known values for x and y:

Since:

x = ky

then

2 = k \times 8

If we make k the subject of the formula by dividing both sides by 8, we can work out the value of k:

\dfrac{2}{8} = k

Therefore k has a value of \frac{2}{8}, which can be simplified to \frac{1}{4}.

Now that we know the value of the constant k, we can write the equation connecting x and y.

Since x = ky

then

x = \frac{1}{4}y

or

x = \frac{y}{4}

b) We now have an equation connecting x and y, so we can work out the value of x when y is 32.

Since

x = \frac{y}{4}

then

x = \frac{32}{4}

32 \div4 = 8

So x = 8

c) We now have an equation connecting x and y, so we can work out the value of y when x is 50.

Since

x = \frac{y}{4}

then

y = x \times 4

50 \times 4 = 200

So y = 200

5) The time taken (t) for customers to receive their orders at a fast-food restaurant is inversely proportional to the square of the number of staff (s) on duty. It takes 20 minutes for customer orders to be taken when there are 4 staff members on duty.

a) Write an equation for t in terms of s

b) If the number of staff is doubled, how many times quicker will the customers receive their orders?

a) Since the time taken (t) is inversely proportional to the square of the number of staff on duty (s), we can write a basic equation as follows:

t \propto \dfrac{1}{s^2}

This formula can be rewritten using k as a constant that connects t and s:

t = \dfrac{k}{s^2}

We can now work out the value of k if we substitute in the known values for t and s:

Since

t = \dfrac{k}{s^2}

then

20 = \dfrac{k}{4^2}

so

20 = \dfrac{k}{16}

To work out the value of k, we need to rearrange the formula, making k the subject. We can do by multiplying both sides by 16:

20 \times 16 = k

So k = 320

Since we now know the exact value of k, we can rewrite the proportionality formula:

Since

t = \dfrac{k}{s^2}

then

t = \dfrac{320}{s^2}

b) It would be too easy to assume that if you double the staff, then the time taken would be halved! Sadly, this is not the case.

If the number of staff is doubled, then there would be 8 staff members on duty instead of 4. We know from part a) that k = 320 and we know that s = 8. If we substitute these values into our formula, it should look like this:

Since

t = \dfrac{k}{s^2}

then

t = \dfrac{320}{8^2}

so

t = \dfrac{320}{64}

320 \div 64 = 5 \text{ minutes}

Therefore, if you double the number of staff the orders take 5 minutes to be received instead of 20 minutes, so the orders are received 4 times faster.

### Worksheets and Exam Questions

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