What is variation in the context of Cambridge IGCSE Mathematics?

In Cambridge IGCSE Mathematics, variation refers to the relationship between two quantities where a change in one quantity results in a change in the other. This relationship can be direct, inverse, or joint, and is often expressed using algebraic equations.

How do you differentiate between direct and inverse variation?

In direct variation, two quantities increase or decrease together at a constant rate, represented by the equation y = kx, where k is the constant of proportionality. In inverse variation, one quantity increases while the other decreases, represented by the equation y = k/x. Understanding these differences is crucial for solving related problems in IGCSE Mathematics.

Can you explain joint variation with an example?

Joint variation occurs when a quantity varies directly with the product of two or more other quantities. For example, if z varies jointly with x and y, it can be expressed as z = kxy, where k is the constant of proportionality. This concept is often tested in Cambridge IGCSE Algebra problems.

What are some common applications of variation in real-world scenarios?

Variation concepts are widely used in real-world scenarios such as physics for calculating force and pressure, economics for understanding supply and demand, and biology for studying population growth. These applications help students appreciate the practical importance of variation in Cambridge IGCSE Mathematics.

How can I solve a variation problem using algebraic methods?

To solve a variation problem, first identify the type of variation (direct, inverse, or joint). Then, write the appropriate algebraic equation using the given information. Substitute known values to find the constant of proportionality, and use this constant to solve for unknown quantities. Practice with past Cambridge IGCSE exam questions can improve your problem-solving skills in this area.

Why is "Setting up $y = kx$ for an inverse-proportion question" a common mistake in Variation?

Why it happens: Students miss the word "inversely" and default to direct. How to avoid it: Underline "directly" or "inversely" before writing any equation. Source: 0580 Extended examiner reports — recurring

Why is "Missing the "square / cube / square-root" qualifier" a common mistake in Variation?

Why it happens: Quick reading. "y is proportional to the square of x" gets read as "y is proportional to x". How to avoid it: Highlight the modifier in the question and put the corresponding power on x.

Why is "Not finding $k$ first" a common mistake in Variation?

Why it happens: Students try to substitute new values directly into y = kx without solving for k. How to avoid it: Step 1 = find k from the given pair. Step 2 = use k with the new value.

Why is "Forgetting units when the variation describes a real-world quantity" a common mistake in Variation?

Why it happens: Focusing on the algebra, students drop units like cm, kg, m/s. How to avoid it: If the question gave units, your answer must include them.

Cambridge IGCSE Mathematics (0580)

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Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Variation (Direct, Inverse, Joint) — Cambridge IGCSE 0580 Maths Extended (2026)

How one quantity changes with another. Direct (proportional), inverse (reciprocal), joint (multiple variables), and powered variants. Same three-step method every time: write the formula, find the constant, then use it.

At a glance

$y \propto x$ → $y = kx$ . Direct: both grow.
$y \propto \dfrac{1}{x}$ → $y = \dfrac{k}{x}$ . Inverse: one grows, the other shrinks.
$y \propto x^{2}$ → $y = kx^{2}$ . Direct to a power.
$y \propto \dfrac{1}{x^{2}}$ → $y = \dfrac{k}{x^{2}}$ . Inverse-square law (gravity, light).
$y \propto xz$ (joint variation): $y = kxz$ .
Three steps every time: write the formula, find $k$ , use it.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.13 — Express direct and inverse variation in algebraic terms; use this form to find unknown quantities.

Direct variation

$y \propto x$ . Two quantities grow together at the same rate.

Direct variation: $y$ varies directly with $x$ when $y = kx$ for some constant $k$ .

The constant $k$ is the constant of proportionality.

Three-step method.

Write the equation: $y = kx$ .
Use given values to find $k$ .
Use $y = kx$ for any other case.

Worked. " $y$ varies directly with $x$ . When $x = 6$ , $y = 15$ . Find $y$ when $x = 10$ ."

$y = kx$ .
$15 = k \times 6 \Rightarrow k = 2.5$ .
$y = 2.5 \times 10 = 25$ .

Find k from the known pair, then use y = kx for the new value.

Direct variation graphs as a straight line through (0, 0); the gradient is the constant k.

Variants.

$y \propto x^{2}$ : $y = kx^{2}$ (e.g. area scales with square of length).
$y \propto x^{3}$ : $y = kx^{3}$ (e.g. volume scales with cube of length).
$y \propto \sqrt{x}$ : $y = k\sqrt{x}$ .

Same three-step method — just substitute into the appropriate formula shape.

$y = kx$ .
Find $k$ from one (x, y) pair.
Up together: as $x$ grows, $y$ grows.
Variants: $kx^{2}$ , $kx^{3}$ , $k\sqrt{x}$ — same method.

Inverse variation

$y \propto \dfrac{1}{x}$ . Product is constant: as $x$ grows, $y$ shrinks.

Inverse variation: $y \propto \dfrac{1}{x}$ → $y = \dfrac{k}{x}$ , equivalently $xy = k$ .

Worked. " $y$ varies inversely with $x$ . When $x = 4$ , $y = 9$ . Find $y$ when $x = 12$ ."

$xy = k$ .
$4 \times 9 = 36 = k$ .
$12y = 36 \Rightarrow y = 3$ .

As x triples (4 → 12), y is divided by 3 (9 → 3), so the product k = 36 is unchanged.

Inverse variation graphs as a curve that approaches both axes without touching them.

Variants.

$y \propto \dfrac{1}{x^{2}}$ : $y = \dfrac{k}{x^{2}}$ (inverse-square law — gravity, light intensity).
$y \propto \dfrac{1}{\sqrt{x}}$ : $y = \dfrac{k}{\sqrt{x}}$ .

Real contexts. Speed and time (constant distance), pressure and volume (constant temperature, Boyle's law), workers and time.

$y = \dfrac{k}{x}$ , or $xy = k$ .
Up one, down the other.
Test: is the product constant across pairs?
Inverse-square: $y = k/x^{2}$ (Newton, light).

Joint variation

$y$ depends on two or more variables at once.

Joint variation combines two relationships: $y$ varies with multiple quantities simultaneously.

Examples:

$y \propto xz$ : $y = kxz$ (direct in both).
$y \propto \dfrac{x}{z}$ : $y = \dfrac{kx}{z}$ (direct with $x$ , inverse with $z$ ).
$y \propto x^{2}z$ : $y = kx^{2}z$ (square of $x$ , direct in $z$ ).

Method. Write the formula, plug in one set of given values to find $k$ , then use the formula for new values.

Worked. " $y$ varies jointly with $x$ and the square root of $z$ . $y = 36$ when $x = 4$ and $z = 9$ . Find $y$ when $x = 5$ and $z = 16$ ."

$y = kx\sqrt{z}$ .
$36 = k \times 4 \times \sqrt{9} = 12k \Rightarrow k = 3$ .
$y = 3 \times 5 \times \sqrt{16} = 3 \times 5 \times 4 = 60$ .

Combine every relationship into one formula with a single k, find k, then use it.

Real contexts. Volume of cylinder ( $V = \pi r^{2} h$ — joint in $r^{2}$ and $h$ ), centripetal force ( $F \propto \dfrac{mv^{2}}{r}$ ).

Joint = two or more variables.
Write all relationships into one formula, with one constant $k$ .
Find $k$ from one complete data set.
Then plug in new values.

Quick recap

$y \propto x$ : $y = kx$ (direct).
$y \propto 1/x$ : $y = k/x$ or $xy = k$ (inverse).
Joint: $y = kxz$ (or with powers).
Always find $k$ FIRST from the given data.
Then substitute new values.
Variants: $x^{2}$ , $\sqrt{x}$ , $1/x^{2}$ — same three-step method.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Direct variation — $y = kx$ for some constant $k$ ; both quantities grow together.
Inverse variation — $xy = k$ for some constant $k$ ; one grows, the other shrinks.
Joint variation — $y$ varies with two or more quantities, e.g. $y = kxz$ .
Constant of proportionality — the constant $k$ linking the variables.

How it’s examined

Variation appears most years on Paper 4 as a 4-5 mark question — often inverse-square or joint variation. Paper 2 has the simpler direct/inverse cases as 2-3 mark items. Examiner reports flag the missed power (e.g. forgetting $x^{2}$ when $y \propto x^{2}$ ) and forgetting to find $k$ before substituting new values.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.13); 0580/42 Oct/Nov 2024 — Q9 (inverse-square variation); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Variation

Step-by-step solutions to past-paper-style questions on variation, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Variation

Almost every variation exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

$y$ is proportional to … with one pair of values given and a second value to find — direct, inverse, joint or power variation.

How to approach it

Write the relationship with $k$ ( $y=kx$ , $y=\frac{k}{x}$ , $y=kxz$ , $y=kx^n$ ), substitute the given pair to find $k$ , then substitute the new value.

Common trap

Missing the square / cube / square-root qualifier, or substituting the wrong variable for the new value. Examiner reports flag both — underline directly/inversely and the power before writing anything.

Multi-step problem

Recognise it by

A percentage-change twist — if $x$ is increased by 20%, find the percentage change in $y$ — where no numerical pair is given at all.

How to approach it

Write $y=kx^n$ , replace $x$ with the scaled value (e.g. $1.2x$ ), and compare $y_{\text{new}}$ to $y$ as a multiplying factor to read off the percentage change.

Common trap

Assuming the percentage change in $y$ equals that in $x$ . Examiner reports note a square law turns a 20% rise into a 44% rise — the exponent magnifies the change.

1Set up direct variation
CoreDirect calculation• direct variation
Question
$y$ is directly proportional to $x$ . When $x = 4$ , $y = 20$ . Find $y$ when $x = 7$ .
Step-by-step solution
1. Step 1
  Direct variation: $y = kx$ .
  $20 = k \times 4 \implies k = 5$
2. Step 2
  Substitute new $x$ .
  $y = 5 \times 7 = 35$
Answer
$y = 35$
2Variation with a power
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q9• direct variation, power
Question
$y$ is directly proportional to the square of $x$ . When $x = 3$ , $y = 27$ . Find $y$ when $x = 5$ .
Step-by-step solution
1. Step 1
  $y = kx^{2}$ .
  $27 = k(3)^{2} = 9k \implies k = 3$
2. Step 2
  Substitute new $x$ .
  $y = 3 \times 25 = 75$
Answer
$y = 75$
Examiner tip
Read carefully: "directly proportional to the square of $x$ " gives $y = kx^{2}$ , not $y = kx$ .
3Inverse variation
ExtendedDirect calculation• inverse variation
Question
$y$ is inversely proportional to $x$ . When $x = 6$ , $y = 10$ . Find $x$ when $y = 4$ .
Step-by-step solution
1. Step 1
  $y = \dfrac{k}{x}$ .
  $10 = \dfrac{k}{6} \implies k = 60$
2. Step 2
  Substitute new $y$ .
  $4 = \dfrac{60}{x} \implies x = 15$
Answer
$x = 15$
4Inverse-square variation
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q11• inverse-square
Question
$F$ is inversely proportional to the square of $r$ . When $r = 2$ , $F = 25$ . Find $F$ when $r = 5$ .
Step-by-step solution
1. Step 1
  $F = \dfrac{k}{r^{2}}$ .
  $25 = \dfrac{k}{4} \implies k = 100$
2. Step 2
  Substitute.
  $F = \dfrac{100}{25} = 4$
Answer
$F = 4$
5Find the constant of proportionality
CoreDirect calculation• direct variation, constant
Question
$d$ varies directly as $t$ , and $d = 45$ when $t = 9$ . Find the constant of proportionality and write the formula linking $d$ and $t$ .
Step-by-step solution
1. Step 1
  Write the proportional relationship.
  $d = kt$
2. Step 2
  Substitute the given pair.
  $45 = 9k \implies k = 5$
3. Step 3
  State the formula.
  $d = 5t$
Answer
$k = 5,\ d = 5t$
Examiner tip
The mark scheme awards a method mark for explicitly stating $d = kt$ before substituting. Skipping that line is a recurring loss in the examiner reports.
6Joint variation
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q11• joint variation
Question
$z$ is jointly proportional to $x$ and $y$ . When $x = 4$ and $y = 3$ , $z = 60$ . Find $z$ when $x = 5$ and $y = 7$ .
Step-by-step solution
1. Step 1
  Joint variation gives $z = kxy$ .
  $60 = k(4)(3) = 12k \implies k = 5$
2. Step 2
  Substitute new values.
  $z = 5(5)(7) = 175$
Answer
$z = 175$
7Variation with a cube root
ExtendedDirect calculation• variation, cube root
Question
$y$ is proportional to the cube root of $x$ . When $x = 8$ , $y = 6$ . Find $y$ when $x = 125$ .
Step-by-step solution
1. Step 1
  $y = k\sqrt[3]{x}$ .
  $6 = k\sqrt[3]{8} = 2k \implies k = 3$
2. Step 2
  Substitute.
  $y = 3\sqrt[3]{125} = 3(5) = 15$
Answer
$y = 15$
8Inverse variation — find $x$ from $y$
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2024 Q11• inverse variation
Question
$y$ is inversely proportional to $(x + 1)$ . When $x = 3$ , $y = 8$ . Find the value of $x$ when $y = 2$ .
Step-by-step solution
1. Step 1
  $y = \dfrac{k}{x + 1}$ .
  $8 = \dfrac{k}{4} \implies k = 32$
2. Step 2
  Substitute $y = 2$ .
  $2 = \dfrac{32}{x + 1} \implies x + 1 = 16$
3. Step 3
  Solve.
  $x = 15$
Answer
$x = 15$
Examiner tip
The examiner report flags candidates often substitute $x = 3$ for $y = 2$ by reflex. Read which variable is given each time.
9Combined direct and inverse variation
ChallengeDirect calculation• Adapted from 0580/42 Feb/Mar 2024 Q14• combined variation
Question
$P$ varies directly as $x$ and inversely as the square of $y$ . When $x = 12$ and $y = 2$ , $P = 9$ . Find $P$ when $x = 20$ and $y = 5$ .
Step-by-step solution
1. Step 1
  Combine the variations.
  $P = \dfrac{kx}{y^{2}}$
2. Step 2
  Find $k$ .
  $9 = \dfrac{12k}{4} = 3k \implies k = 3$
3. Step 3
  Substitute new values.
  $P = \dfrac{3(20)}{5^{2}} = \dfrac{60}{25} = 2.4$
Answer
$P = 2.4$
Examiner tip
The mark scheme awards a method mark for setting up the combined formula $P = \dfrac{kx}{y^{2}}$ . The examiner report flags candidates often mishandle the inverse-square term by writing $\dfrac{1}{y}$ instead.
10Effect of changes on a varying quantity
ChallengeMulti-step problem• variation, percentage change
Question
$y$ varies directly as $x^{2}$ . If $x$ is increased by $20\%$ , find the percentage increase in $y$ .
Step-by-step solution
1. Step 1
  $y = kx^{2}$ . Let the new $x$ be $1.2x$ .
2. Step 2
  New $y$ .
  $y_{\text{new}} = k(1.2x)^{2} = 1.44 kx^{2} = 1.44 y$
3. Step 3
  Increase factor $= 1.44$ , so the percentage increase is $44\%$ .
Answer
$44\%$ increase
Examiner tip
The examiner report flags candidates often give $20\%$ or $40\%$ as the answer. The square law means the percentage change in $y$ is NOT the same as in $x$ .

Key Formulae — Variation

The formulae you need to memorise for variation on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Direct proportion
$y \propto x \iff y = kx$
$k$
constant of proportionality
When to use
When doubling $x$ doubles $y$ .
Inverse proportion
$y \propto \dfrac{1}{x} \iff y = \dfrac{k}{x}$
$k$
constant
When to use
When the product $xy$ is constant.
Variation with a power
$y \propto x^{n} \iff y = kx^{n};\quad y \propto \dfrac{1}{x^{n}} \iff y = \dfrac{k}{x^{n}}$
$n$
the given power (e.g. $\frac{1}{2}$ for square root)
When to use
Whenever the question mentions square / cube / square-root proportion.

Key Definitions and Keywords — Variation

Definitions to memorise and the exact keywords mark schemes credit for variation answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Direct proportion
Examiner keyword
Two quantities such that their ratio is constant — as one increases, the other increases at the same rate.
Inverse proportion
Examiner keyword
Two quantities whose product is constant — as one increases, the other decreases proportionally.
Constant of proportionality ( $k$ )
Examiner keyword
The fixed multiplier linking proportional quantities. Found from a given pair of values.
Joint variation
When one variable is proportional to the product of two others, e.g. $y = kxz$ .

Common Mistakes and Misconceptions — Variation

The traps other students keep falling into on variation questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Setting up $y = kx$ for an inverse-proportion question
0580 Extended examiner reports — recurring
Why it happens
Students miss the word "inversely" and default to direct.
How to avoid it
Underline "directly" or "inversely" before writing any equation.
✕Missing the "square / cube / square-root" qualifier
Why it happens
Quick reading. " $y$ is proportional to the square of $x$ " gets read as " $y$ is proportional to $x$ ".
How to avoid it
Highlight the modifier in the question and put the corresponding power on $x$ .
✕Not finding $k$ first
Why it happens
Students try to substitute new values directly into $y = kx$ without solving for $k$ .
How to avoid it
Step 1 = find $k$ from the given pair. Step 2 = use $k$ with the new value.
✕Forgetting units when the variation describes a real-world quantity
Why it happens
Focusing on the algebra, students drop units like cm, kg, m/s.
How to avoid it
If the question gave units, your answer must include them.

Variation — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Mathematics (0580)

Variation

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Short Notes - Variation

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Variation (Direct, Inverse, Joint) — Cambridge IGCSE 0580 Maths Extended (2026)

At a glance

$y \propto x$ → $y = kx$ . Direct: both grow.
$y \propto \dfrac{1}{x}$ → $y = \dfrac{k}{x}$ . Inverse: one grows, the other shrinks.
$y \propto x^{2}$ → $y = kx^{2}$ . Direct to a power.
$y \propto \dfrac{1}{x^{2}}$ → $y = \dfrac{k}{x^{2}}$ . Inverse-square law (gravity, light).
$y \propto xz$ (joint variation): $y = kxz$ .
Three steps every time: write the formula, find $k$ , use it.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.13 — Express direct and inverse variation in algebraic terms; use this form to find unknown quantities.

Direct variation

$y \propto x$ . Two quantities grow together at the same rate.

Direct variation: $y$ varies directly with $x$ when $y = kx$ for some constant $k$ .

The constant $k$ is the constant of proportionality.

Three-step method.

Write the equation: $y = kx$ .
Use given values to find $k$ .
Use $y = kx$ for any other case.

Worked. " $y$ varies directly with $x$ . When $x = 6$ , $y = 15$ . Find $y$ when $x = 10$ ."

$y = kx$ .
$15 = k \times 6 \Rightarrow k = 2.5$ .
$y = 2.5 \times 10 = 25$ .

Find k from the known pair, then use y = kx for the new value.

Direct variation graphs as a straight line through (0, 0); the gradient is the constant k.

Variants.

$y \propto x^{2}$ : $y = kx^{2}$ (e.g. area scales with square of length).
$y \propto x^{3}$ : $y = kx^{3}$ (e.g. volume scales with cube of length).
$y \propto \sqrt{x}$ : $y = k\sqrt{x}$ .

Same three-step method — just substitute into the appropriate formula shape.

$y = kx$ .
Find $k$ from one (x, y) pair.
Up together: as $x$ grows, $y$ grows.
Variants: $kx^{2}$ , $kx^{3}$ , $k\sqrt{x}$ — same method.

Inverse variation

$y \propto \dfrac{1}{x}$ . Product is constant: as $x$ grows, $y$ shrinks.

Inverse variation: $y \propto \dfrac{1}{x}$ → $y = \dfrac{k}{x}$ , equivalently $xy = k$ .

Worked. " $y$ varies inversely with $x$ . When $x = 4$ , $y = 9$ . Find $y$ when $x = 12$ ."

$xy = k$ .
$4 \times 9 = 36 = k$ .
$12y = 36 \Rightarrow y = 3$ .

As x triples (4 → 12), y is divided by 3 (9 → 3), so the product k = 36 is unchanged.

Inverse variation graphs as a curve that approaches both axes without touching them.

Variants.

$y \propto \dfrac{1}{x^{2}}$ : $y = \dfrac{k}{x^{2}}$ (inverse-square law — gravity, light intensity).
$y \propto \dfrac{1}{\sqrt{x}}$ : $y = \dfrac{k}{\sqrt{x}}$ .

Real contexts. Speed and time (constant distance), pressure and volume (constant temperature, Boyle's law), workers and time.

$y = \dfrac{k}{x}$ , or $xy = k$ .
Up one, down the other.
Test: is the product constant across pairs?
Inverse-square: $y = k/x^{2}$ (Newton, light).

Joint variation

$y$ depends on two or more variables at once.

Joint variation combines two relationships: $y$ varies with multiple quantities simultaneously.

Examples:

$y \propto xz$ : $y = kxz$ (direct in both).
$y \propto \dfrac{x}{z}$ : $y = \dfrac{kx}{z}$ (direct with $x$ , inverse with $z$ ).
$y \propto x^{2}z$ : $y = kx^{2}z$ (square of $x$ , direct in $z$ ).

Method. Write the formula, plug in one set of given values to find $k$ , then use the formula for new values.

Worked. " $y$ varies jointly with $x$ and the square root of $z$ . $y = 36$ when $x = 4$ and $z = 9$ . Find $y$ when $x = 5$ and $z = 16$ ."

$y = kx\sqrt{z}$ .
$36 = k \times 4 \times \sqrt{9} = 12k \Rightarrow k = 3$ .
$y = 3 \times 5 \times \sqrt{16} = 3 \times 5 \times 4 = 60$ .

Combine every relationship into one formula with a single k, find k, then use it.

Real contexts. Volume of cylinder ( $V = \pi r^{2} h$ — joint in $r^{2}$ and $h$ ), centripetal force ( $F \propto \dfrac{mv^{2}}{r}$ ).

Joint = two or more variables.
Write all relationships into one formula, with one constant $k$ .
Find $k$ from one complete data set.
Then plug in new values.

Quick recap

$y \propto x$ : $y = kx$ (direct).
$y \propto 1/x$ : $y = k/x$ or $xy = k$ (inverse).
Joint: $y = kxz$ (or with powers).
Always find $k$ FIRST from the given data.
Then substitute new values.
Variants: $x^{2}$ , $\sqrt{x}$ , $1/x^{2}$ — same three-step method.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Direct variation — $y = kx$ for some constant $k$ ; both quantities grow together.
Inverse variation — $xy = k$ for some constant $k$ ; one grows, the other shrinks.
Joint variation — $y$ varies with two or more quantities, e.g. $y = kxz$ .
Constant of proportionality — the constant $k$ linking the variables.

How it’s examined

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.13); 0580/42 Oct/Nov 2024 — Q9 (inverse-square variation); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Variation

Step-by-step solutions to past-paper-style questions on variation, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Variation

Almost every variation exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

$y$ is proportional to … with one pair of values given and a second value to find — direct, inverse, joint or power variation.

How to approach it

Write the relationship with $k$ ( $y=kx$ , $y=\frac{k}{x}$ , $y=kxz$ , $y=kx^n$ ), substitute the given pair to find $k$ , then substitute the new value.

Common trap

Multi-step problem

Recognise it by

A percentage-change twist — if $x$ is increased by 20%, find the percentage change in $y$ — where no numerical pair is given at all.

How to approach it

Write $y=kx^n$ , replace $x$ with the scaled value (e.g. $1.2x$ ), and compare $y_{\text{new}}$ to $y$ as a multiplying factor to read off the percentage change.

Common trap

Assuming the percentage change in $y$ equals that in $x$ . Examiner reports note a square law turns a 20% rise into a 44% rise — the exponent magnifies the change.

1Set up direct variation
CoreDirect calculation• direct variation
Question
$y$ is directly proportional to $x$ . When $x = 4$ , $y = 20$ . Find $y$ when $x = 7$ .
Step-by-step solution
1. Step 1
  Direct variation: $y = kx$ .
  $20 = k \times 4 \implies k = 5$
2. Step 2
  Substitute new $x$ .
  $y = 5 \times 7 = 35$
Answer
$y = 35$
2Variation with a power
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q9• direct variation, power
Question
$y$ is directly proportional to the square of $x$ . When $x = 3$ , $y = 27$ . Find $y$ when $x = 5$ .
Step-by-step solution
1. Step 1
  $y = kx^{2}$ .
  $27 = k(3)^{2} = 9k \implies k = 3$
2. Step 2
  Substitute new $x$ .
  $y = 3 \times 25 = 75$
Answer
$y = 75$
Examiner tip
Read carefully: "directly proportional to the square of $x$ " gives $y = kx^{2}$ , not $y = kx$ .
3Inverse variation
ExtendedDirect calculation• inverse variation
Question
$y$ is inversely proportional to $x$ . When $x = 6$ , $y = 10$ . Find $x$ when $y = 4$ .
Step-by-step solution
1. Step 1
  $y = \dfrac{k}{x}$ .
  $10 = \dfrac{k}{6} \implies k = 60$
2. Step 2
  Substitute new $y$ .
  $4 = \dfrac{60}{x} \implies x = 15$
Answer
$x = 15$
4Inverse-square variation
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q11• inverse-square
Question
$F$ is inversely proportional to the square of $r$ . When $r = 2$ , $F = 25$ . Find $F$ when $r = 5$ .
Step-by-step solution
1. Step 1
  $F = \dfrac{k}{r^{2}}$ .
  $25 = \dfrac{k}{4} \implies k = 100$
2. Step 2
  Substitute.
  $F = \dfrac{100}{25} = 4$
Answer
$F = 4$
5Find the constant of proportionality
CoreDirect calculation• direct variation, constant
Question
$d$ varies directly as $t$ , and $d = 45$ when $t = 9$ . Find the constant of proportionality and write the formula linking $d$ and $t$ .
Step-by-step solution
1. Step 1
  Write the proportional relationship.
  $d = kt$
2. Step 2
  Substitute the given pair.
  $45 = 9k \implies k = 5$
3. Step 3
  State the formula.
  $d = 5t$
Answer
$k = 5,\ d = 5t$
Examiner tip
The mark scheme awards a method mark for explicitly stating $d = kt$ before substituting. Skipping that line is a recurring loss in the examiner reports.
6Joint variation
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2023 Q11• joint variation
Question
$z$ is jointly proportional to $x$ and $y$ . When $x = 4$ and $y = 3$ , $z = 60$ . Find $z$ when $x = 5$ and $y = 7$ .
Step-by-step solution
1. Step 1
  Joint variation gives $z = kxy$ .
  $60 = k(4)(3) = 12k \implies k = 5$
2. Step 2
  Substitute new values.
  $z = 5(5)(7) = 175$
Answer
$z = 175$
7Variation with a cube root
ExtendedDirect calculation• variation, cube root
Question
$y$ is proportional to the cube root of $x$ . When $x = 8$ , $y = 6$ . Find $y$ when $x = 125$ .
Step-by-step solution
1. Step 1
  $y = k\sqrt[3]{x}$ .
  $6 = k\sqrt[3]{8} = 2k \implies k = 3$
2. Step 2
  Substitute.
  $y = 3\sqrt[3]{125} = 3(5) = 15$
Answer
$y = 15$
8Inverse variation — find $x$ from $y$
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2024 Q11• inverse variation
Question
$y$ is inversely proportional to $(x + 1)$ . When $x = 3$ , $y = 8$ . Find the value of $x$ when $y = 2$ .
Step-by-step solution
1. Step 1
  $y = \dfrac{k}{x + 1}$ .
  $8 = \dfrac{k}{4} \implies k = 32$
2. Step 2
  Substitute $y = 2$ .
  $2 = \dfrac{32}{x + 1} \implies x + 1 = 16$
3. Step 3
  Solve.
  $x = 15$
Answer
$x = 15$
Examiner tip
The examiner report flags candidates often substitute $x = 3$ for $y = 2$ by reflex. Read which variable is given each time.
9Combined direct and inverse variation
ChallengeDirect calculation• Adapted from 0580/42 Feb/Mar 2024 Q14• combined variation
Question
$P$ varies directly as $x$ and inversely as the square of $y$ . When $x = 12$ and $y = 2$ , $P = 9$ . Find $P$ when $x = 20$ and $y = 5$ .
Step-by-step solution
1. Step 1
  Combine the variations.
  $P = \dfrac{kx}{y^{2}}$
2. Step 2
  Find $k$ .
  $9 = \dfrac{12k}{4} = 3k \implies k = 3$
3. Step 3
  Substitute new values.
  $P = \dfrac{3(20)}{5^{2}} = \dfrac{60}{25} = 2.4$
Answer
$P = 2.4$
Examiner tip
The mark scheme awards a method mark for setting up the combined formula $P = \dfrac{kx}{y^{2}}$ . The examiner report flags candidates often mishandle the inverse-square term by writing $\dfrac{1}{y}$ instead.
10Effect of changes on a varying quantity
ChallengeMulti-step problem• variation, percentage change
Question
$y$ varies directly as $x^{2}$ . If $x$ is increased by $20\%$ , find the percentage increase in $y$ .
Step-by-step solution
1. Step 1
  $y = kx^{2}$ . Let the new $x$ be $1.2x$ .
2. Step 2
  New $y$ .
  $y_{\text{new}} = k(1.2x)^{2} = 1.44 kx^{2} = 1.44 y$
3. Step 3
  Increase factor $= 1.44$ , so the percentage increase is $44\%$ .
Answer
$44\%$ increase
Examiner tip
The examiner report flags candidates often give $20\%$ or $40\%$ as the answer. The square law means the percentage change in $y$ is NOT the same as in $x$ .

Key Formulae — Variation

The formulae you need to memorise for variation on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Direct proportion
$y \propto x \iff y = kx$
$k$
constant of proportionality
When to use
When doubling $x$ doubles $y$ .
Inverse proportion
$y \propto \dfrac{1}{x} \iff y = \dfrac{k}{x}$
$k$
constant
When to use
When the product $xy$ is constant.
Variation with a power
$y \propto x^{n} \iff y = kx^{n};\quad y \propto \dfrac{1}{x^{n}} \iff y = \dfrac{k}{x^{n}}$
$n$
the given power (e.g. $\frac{1}{2}$ for square root)
When to use
Whenever the question mentions square / cube / square-root proportion.

Key Definitions and Keywords — Variation

Definitions to memorise and the exact keywords mark schemes credit for variation answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Direct proportion
Examiner keyword
Two quantities such that their ratio is constant — as one increases, the other increases at the same rate.
Inverse proportion
Examiner keyword
Two quantities whose product is constant — as one increases, the other decreases proportionally.
Constant of proportionality ( $k$ )
Examiner keyword
The fixed multiplier linking proportional quantities. Found from a given pair of values.
Joint variation
When one variable is proportional to the product of two others, e.g. $y = kxz$ .

Common Mistakes and Misconceptions — Variation

The traps other students keep falling into on variation questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Setting up $y = kx$ for an inverse-proportion question
0580 Extended examiner reports — recurring
Why it happens
Students miss the word "inversely" and default to direct.
How to avoid it
Underline "directly" or "inversely" before writing any equation.
✕Missing the "square / cube / square-root" qualifier
Why it happens
Quick reading. " $y$ is proportional to the square of $x$ " gets read as " $y$ is proportional to $x$ ".
How to avoid it
Highlight the modifier in the question and put the corresponding power on $x$ .
✕Not finding $k$ first
Why it happens
Students try to substitute new values directly into $y = kx$ without solving for $k$ .
How to avoid it
Step 1 = find $k$ from the given pair. Step 2 = use $k$ with the new value.
✕Forgetting units when the variation describes a real-world quantity
Why it happens
Focusing on the algebra, students drop units like cm, kg, m/s.
How to avoid it
If the question gave units, your answer must include them.

Variation — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Variation

Short Notes - Variation

Detailed Study Notes

At a glance

What you’ll learn

Direct variation

Inverse variation

Joint variation

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Variation

1Set up direct variation

2Variation with a power

3Inverse variation

4Inverse-square variation

5Find the constant of proportionality

6Joint variation

7Variation with a cube root

8Inverse variation — find xxx from yyy

9Combined direct and inverse variation

10Effect of changes on a varying quantity

Direct proportion

Inverse proportion

Variation with a power

Direct proportion

Inverse proportion

Constant of proportionality (kkk)

Joint variation

✕Setting up y=kxy = kxy=kx for an inverse-proportion question

✕Missing the "square / cube / square-root" qualifier

✕Not finding kkk first

✕Forgetting units when the variation describes a real-world quantity

Variation — frequently asked questions

Variation

Short Notes - Variation

Detailed Study Notes

At a glance

What you’ll learn

Direct variation

Inverse variation

Joint variation

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Variation

1Set up direct variation

2Variation with a power

3Inverse variation

4Inverse-square variation

5Find the constant of proportionality

6Joint variation

7Variation with a cube root

8Inverse variation — find xxx from yyy

9Combined direct and inverse variation

10Effect of changes on a varying quantity

Direct proportion

Inverse proportion

Variation with a power

Direct proportion

Inverse proportion

Constant of proportionality (kkk)

Joint variation

✕Setting up y=kxy = kxy=kx for an inverse-proportion question

✕Missing the "square / cube / square-root" qualifier

✕Not finding kkk first

✕Forgetting units when the variation describes a real-world quantity

Variation — frequently asked questions

8Inverse variation — find $x$ from $y$

Constant of proportionality ( $k$ )

✕Setting up $y = kx$ for an inverse-proportion question

✕Not finding $k$ first

8Inverse variation — find $x$ from $y$

Constant of proportionality ( $k$ )

✕Setting up $y = kx$ for an inverse-proportion question

✕Not finding $k$ first