Cambridge IGCSE Mathematics 0580: Proportion – Study Guide

This study guide teaches subtopic E2.8 Proportion from Topic 2, Algebra and graphs, of the Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027. E2.8 is Extended only: the Core column lists C2.8 as “Extended content only”, so it is examined on Paper 2 (non-calculator) and Paper 4 (calculator), not on Papers 1 and 3. If you are entered for Core, you can skip this page; the numerical ratio work in C1.11 is a different subtopic.

For the whole of Topic 2, start with the Algebra and graphs study guide, the Algebra and graphs revision notes and the practice sets (Algebra and graphs practice, Algebra (Extended) practice). Those pages give the basic y = kx and y = k/x method in a line or two. This page goes further: every type of proportion the syllabus names, finding either unknown, the effect of a change, and spotting the type from a table.

What this subtopic covers

Syllabus	What you must be able to do	Tier
E2.8	Express direct proportion in algebraic terms (y ∝ x, y ∝ x², y ∝ √x, y ∝ x³, y ∝ ∛x)	Extended only
E2.8	Express inverse proportion in algebraic terms (y ∝ 1/x, 1/x², 1/√x, 1/x³, 1/∛x)	Extended only
E2.8	Use the algebraic form to find unknown quantities	Extended only
E2.8	Know and use the proportional symbol ∝	Extended only

The syllabus note says E2.8 “includes linear, square, square root, cube and cube root proportion”, and that “knowledge of proportional symbol (∝) is required”. So you need ten relationships in all: five direct and five inverse.

The symbol ∝ and the constant k

“y ∝ x” is read “y is proportional to x”. The symbol ∝ is not an equals sign, so you cannot calculate with it. You turn it into an equation by replacing ∝ with “= k ×”, where k is a constant called the constant of proportionality:

y ∝ x²        becomes   y = kx²
y ∝ 1/√x      becomes   y = k/√x

The ten forms

Words in the question	Using ∝	Equation
y is (directly) proportional to x	y ∝ x	y = kx
y is proportional to the square of x	y ∝ x²	y = kx²
y is proportional to the square root of x	y ∝ √x	y = k√x
y is proportional to the cube of x	y ∝ x³	y = kx³
y is proportional to the cube root of x	y ∝ ∛x	y = k∛x
y is inversely proportional to x	y ∝ 1/x	y = k/x
y is inversely proportional to the square of x	y ∝ 1/x²	y = k/x²
y is inversely proportional to the square root of x	y ∝ 1/√x	y = k/√x
y is inversely proportional to the cube of x	y ∝ 1/x³	y = k/x³
y is inversely proportional to the cube root of x	y ∝ 1/∛x	y = k/∛x

“Varies directly as” means the same as “is directly proportional to”, and “varies inversely as” means “is inversely proportional to”.

The method, every time

1. Write the relationship with ∝, then as an equation with k.
2. Substitute the one pair of values you are given.
3. Solve for k. Write the full formula with the number in place of k.
4. Use the formula to find the unknown. If the unknown is the variable inside a power or root, rearrange and undo the power last.

If the question says “find y in terms of x” or “find a formula”, the answer is the full equation (for example y = 3√x), not just the value of k.

Direct proportion

Worked example 1: square root (non-calculator)

P is directly proportional to the square root of Q. When Q = 16, P = 12. (a) Find P when Q = 49. (b) Find Q when P = 27.

P ∝ √Q   so   P = k√Q
12 = k × √16 = 4k
k = 3          formula: P = 3√Q

(a) P = 3 × √49 = 3 × 7 = 21
(b) 27 = 3√Q
    √Q = 9
    Q = 9² = 81

In (b) you finish by squaring, because squaring undoes the square root.

Worked example 2: cube, in context (non-calculator)

The mass, m grams, of a solid metal ball is proportional to the cube of its radius, r cm. A ball of radius 3 cm has mass 54 g. (a) Find the mass of a ball of radius 5 cm. (b) Find the radius of a ball of mass 128 g.

m = kr³
54 = k × 3³ = 27k
k = 2          formula: m = 2r³

(a) m = 2 × 5³ = 2 × 125 = 250 g
(b) 128 = 2r³
    r³ = 64
    r = ∛64 = 4 cm

Worked example 3: calculator, with rounding

The time, T seconds, for a process is proportional to the cube root of the volume, V cm³. When V = 27, T = 4.2. (a) Find T when V = 50. (b) Find V when T = 7.

T = k∛V
4.2 = k × ∛27 = 3k
k = 1.4        formula: T = 1.4∛V

(a) T = 1.4 × ∛50 = 1.4 × 3.6840...
      = 5.1576...  = 5.16 seconds (3 s.f.)
(b) 7 = 1.4∛V
    ∛V = 5
    V = 5³ = 125 cm³

The syllabus asks for non-exact answers to 3 significant figures unless the question says otherwise, so 5.16 is right. Keep the full calculator value until the last line.

Worked example 4: proportional to a bracket (calculator or non-calculator)

The quantity can be an expression, not just a single letter. y is directly proportional to (x + 2)², and y = 36 when x = 1. Given that x is positive, find (a) y when x = 3 and (b) x when y = 64.

y = k(x + 2)²
36 = k(1 + 2)² = 9k
k = 4          formula: y = 4(x + 2)²

(a) y = 4(3 + 2)² = 4 × 25 = 100
(b) 64 = 4(x + 2)²
    (x + 2)² = 16
    x + 2 = 4   (or −4)
    x = 2       (x = −6 is rejected because x is positive)

Square the whole bracket. A common slip is to work out x² + 2 instead of (x + 2)².

Inverse proportion

When y is inversely proportional to something, the “something” goes in the denominator. As x increases, y decreases.

Worked example 5: inverse square (non-calculator)

The force, F newtons, between two objects is inversely proportional to the square of the distance, d metres, between them. When d = 3, F = 40. (a) Find F when d = 6. (b) Find d when F = 250.

F = k/d²
40 = k/3² = k/9
k = 360        formula: F = 360/d²

(a) F = 360/6² = 360/36 = 10 N
(b) 250 = 360/d²
    d² = 360/250 = 1.44
    d = √1.44 = 1.2 m

Notice that doubling d (from 3 to 6) divided F by 4, not by 2. That is the square at work.

Worked example 6: inverse cube root (non-calculator)

y is inversely proportional to the cube root of x. When x = 8, y = 6. (a) Find y in terms of x. (b) Find y when x = 27. (c) Find x when y = 3.

y = k/∛x
6 = k/∛8 = k/2
k = 12

(a) y = 12/∛x
(b) y = 12/∛27 = 12/3 = 4
(c) 3 = 12/∛x
    ∛x = 12/3 = 4
    x = 4³ = 64

To rearrange an inverse formula, multiply both sides by the denominator first, then divide: 3∛x = 12, so ∛x = 4.

Worked example 7: inverse linear, in context (non-calculator)

The time, t hours, to empty a tank is inversely proportional to the number of pumps, n, in use. With 4 pumps it takes 18 hours. How long does it take with 6 pumps, and how many pumps are needed to empty it in 8 hours?

t = k/n
18 = k/4,  k = 72     formula: t = 72/n

With 6 pumps:   t = 72/6 = 12 hours
For t = 8:      8 = 72/n,  n = 72/8 = 9 pumps

A quick check for any inverse linear relationship: the product t × n stays the same (4 × 18 = 6 × 12 = 9 × 8 = 72).

The effect of a change

Some questions give no numbers for x or y. They ask what happens to y when x changes. You can answer straight from the power in the formula.

Relationship	Change in x	Effect on y
y = kx²	x multiplied by 3	y multiplied by 3² = 9
y = k/x³	x doubled	y multiplied by 1/2³, so divided by 8
y = k/√x	x multiplied by 4	y multiplied by 1/√4, so halved
y = kx³	x multiplied by 0.8	y multiplied by 0.8³ = 0.512

Worked example 8: percentage change (calculator)

(a) y is proportional to x². x increases by 10%. Find the percentage increase in y. (b) y is inversely proportional to x². x increases by 25%. Find the percentage change in y.

(a) new x = 1.1x
    new y = k(1.1x)² = 1.21 × kx² = 1.21y
    y increases by 21%

(b) new x = 1.25x
    new y = k/(1.25x)² = (1/1.5625) × k/x² = 0.64y
    y decreases by 36%

The answer to (a) is not 20%. You square the multiplier 1.1, not the percentage.

Recognising the type from a table

Sometimes you are given values and must decide the relationship. Test each form: if y = kxⁿ, then y ÷ xⁿ is the same for every pair; if y = k/xⁿ, then y × xⁿ is the same for every pair.

x	1	2	4
y	5	20	80

y ÷ x gives 5, 10, 20 (not constant). y ÷ x² gives 5, 5, 5. So y = 5x².

x	1	4	9
y	6	3	2

y falls as x rises, so test inverse forms. y × x gives 6, 12, 18 (not constant). y × √x gives 6 × 1 = 6, 3 × 2 = 6, 2 × 3 = 6. So y = 6/√x.

Paper 2 or Paper 4?

E2.8 can appear on either Extended paper. On Paper 2 (non-calculator) you must work out squares, roots and cubes by hand, so learn the perfect squares to 15² and cubes to 10³. On Paper 4 (calculator) k or the final answer may be a decimal that needs rounding to 3 significant figures. Either way, show the line that finds k, so your method can be credited even if a later step goes wrong.

Common errors

• Writing y = x² + k or y = k + 1/x. Proportion always multiplies by k (or divides k by the quantity). It never adds k.
• Using y = k/x when the question says “inversely proportional to the square of x”. Read the whole phrase and underline the power.
• Stopping at k. If the question asks for y, substitute again. If it asks for a formula, write the equation with the value of k in it.
• Finding Q correctly as √Q = 9, then giving 9 as the answer. Undo the root: Q = 81.
• In inverse questions, rearranging 250 = 360/d² as d² = 250/360. Multiply by d² first, then divide by 250, so d² = 360/250.
• Taking the percentage change in x straight across to y (10% in x does not give 20% in y for square proportion). Work with multipliers.
• Rounding k to 2 or 3 significant figures before using it again on Paper 4, which can push the final answer off by 1 in the third figure.

Where to go next

• Recap the whole topic in the Algebra and graphs revision notes.
• Practise mixed Extended algebra, including inverse proportion, in the Algebra (Extended) practice questions.
• Find your weak areas with the free Extended diagnostic (or the Core diagnostic if you are entered for Core).
• See every topic for the course on the Cambridge IGCSE Mathematics hub, and tick off each learning outcome on the printable checklist. Other resources for syllabus 0580 are listed there.

Official syllabus

Cambridge International, Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027, Topic 2 Algebra and graphs, subtopic E2.8 Proportion (Extended subject content).