GCSE Maths revision · Higher: number and algebra

Proportion equations

Higher only

The idea

Direct proportion: $y = kx$ (to $x^2$ : $y = kx^2$ ; to $\sqrt{x}$ : $y = k\sqrt{x}$ ). Inverse proportion: $y = \frac{k}{x}$ (to $x^2$ : $y = \frac{k}{x^2}$ ). Find $k$ from one pair of values, then use the equation.

Write the equation with $k$ , using the right power.
Substitute the pair you are given and solve for $k$ .
Write the full equation, then use it to answer the question.

Watch out 'Inversely proportional to the square' means $y = \frac{k}{x^2}$ .

See it worked

Worked exampleNon-calculator3 marks

$y$ is directly proportional to $x^2$ . When $x = 3$ , $y = 18$ .

(a) Find a formula for $y$ in terms of $x$ .

(b) Work out $y$ when $x = 5$ .

(a) Write the proportion with a constant. $y = kx^2$
Substitute. $18 = k \times 9 \Rightarrow k = 2$
(b) $y = 2 \times 25 = 50$ .

Answer (a) $y = 2x^2$ ; (b) $50$

Your turn

Question 1Non-calculator3 marks

y is directly proportional to $x^{2}.$ When x = 5, y = 75. Work out y when x = 8.

Question 2Non-calculator5 marks

$y$ is inversely proportional to $\sqrt{x}$ . When $x = 16$ , $y = 6$ .

(a) Find $y$ when $x = 36$ .

(b) Find $x$ when $y = 12$ .

Question 3Calculator allowed5 marks

The brightness, $I$ , of a light is inversely proportional to the square of the distance, $d$ metres, from the light. When $d = 2$ , $I = 45$ .

(a) Work out $I$ when $d = 3$ .

(b) The distance from the light is increased by $50\%$ . Work out the percentage decrease in the brightness. Give your answer to 1 decimal place.

In past papers

48 official questions in 47 papers since 2017. The latest:

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