Mathematics · AQA 8300 §R2

Proportion

Spec: AQA 8300 §R2 ⭐⭐⭐ ⏱ 45 mins AQA · Edexcel · OCR Grade 9

Set up and use direct proportion equations of the form $y = kx$
Set up and use inverse proportion equations of the form $y = \dfrac{k}{x}$
Find the proportionality constant $k$ from a given pair of values
Recognise proportional relationships from graphs and tables of values
Solve problems where $y$ is proportional to powers, roots, or combinations of $x$

🔑 Core Concepts

Direct Proportion — $y \propto x$

Two quantities are in direct proportion when they increase and decrease at the same rate. If $x$ doubles, $y$ doubles; if $x$ is halved, $y$ is halved. The ratio $\dfrac{y}{x}$ is always constant.

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DEFINITION — Direct Proportion

$y$ is directly proportional to $x$, written $y \propto x$, if there exists a non-zero constant $k$ such that $y = kx$ for all values. The graph is a straight line through the origin with gradient $k$.

Direct Proportion

$$y = kx \qquad \Longleftrightarrow \qquad \frac{y}{x} = k$$

$y$ = dependent variable $x$ = independent variable $k$ = proportionality constant (gradient)

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EXAM TIP

Always state the equation explicitly: write $y = kx$, substitute the known pair, solve for $k$, then rewrite $y = (\text{value})x$. Examiners award method marks for showing each step.

✗

COMMON MISTAKE

Students often say "proportional" but mean "increases". Proportionality is a specific mathematical relationship — the ratio $y/x$ must be constant, not just increasing.

How to find $k$: Given that $y \propto x$ and $y = 18$ when $x = 6$: substitute into $y = kx$ to get $18 = 6k$, so $k = 3$. The equation is $y = 3x$.

Inverse Proportion — $y \propto \dfrac{1}{x}$

Two quantities are in inverse proportion when one increases at the same rate as the other decreases. If $x$ doubles, $y$ halves. The product $xy$ is always constant.

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DEFINITION — Inverse Proportion

$y$ is inversely proportional to $x$, written $y \propto \dfrac{1}{x}$, if there exists a non-zero constant $k$ such that $y = \dfrac{k}{x}$. The graph is a hyperbola in the first (and third) quadrant, never touching either axis.

Inverse Proportion

$$y = \frac{k}{x} \qquad \Longleftrightarrow \qquad xy = k$$

$k$ = proportionality constant (product $xy$) $x$ = independent variable ($x \neq 0$) $y$ = dependent variable ($y \neq 0$)

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EXAM TIP

For inverse proportion, the product $xy = k$ is constant. You can use this as a quick check: multiply any $x$–$y$ pair in the table; if the products are all equal, the relationship is inverse proportion.

✗

COMMON MISTAKE

Do not write $y \propto \dfrac{1}{x}$ as $y = \dfrac{1}{kx}$. The proportionality constant $k$ sits in the numerator: $y = \dfrac{k}{x}$.

Proportionality to Powers and Roots

The proportionality symbol $\propto$ can connect $y$ to any function of $x$. These are called non-linear proportional relationships, essential for Grade 9.

$y \propto x^2$ — Square Proportion

$$y = kx^2$$

Graph: parabola through the origin If $x$ doubles, $y$ multiplies by $4$

$y \propto x^3$ — Cube Proportion

$$y = kx^3$$

Graph: cubic curve through the origin If $x$ doubles, $y$ multiplies by $8$

$y \propto \sqrt{x}$ — Square Root Proportion

$$y = k\sqrt{x}$$

Graph: half-parabola through the origin ($x \geq 0$) If $x$ quadruples, $y$ doubles

$y \propto \dfrac{1}{x^2}$ — Inverse Square

$$y = \frac{k}{x^2}$$

Graph: steeper hyperbola than $1/x$ If $x$ doubles, $y$ divides by $4$

🎯

EXAM TIP — Scale Factor Rule

For $y = kx^n$: if $x$ is multiplied by a scale factor $s$, then $y$ is multiplied by $s^n$. For $y = k/x^n$: if $x$ is multiplied by $s$, then $y$ is divided by $s^n$.

Combined Proportionality (Grade 9)

At the highest level, $y$ may depend on more than one variable simultaneously.

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DEFINITION — Combined Proportionality

If $y$ is directly proportional to $x$ and inversely proportional to $z^2$, we write $y \propto \dfrac{x}{z^2}$, giving the equation $y = \dfrac{kx}{z^2}$. Find $k$ by substituting a known set of values $\{x, y, z\}$.

🧠

MEMORY TRICK — Proportion Steps

WEBS: Write the proportionality statement → Equation (replace $\propto$ with $= k \cdot$) → Bind $k$ (substitute known values) → Solve for any unknown.

Graphs of Proportional Relationships

⚠️

IMPORTANT — Graph Identification

Being able to identify the type of proportionality from a graph is a Grade 8–9 skill. Key features:

$y = kx$: straight line through origin, positive gradient
$y = kx^2$: parabola through origin, symmetric about $y$-axis if $x$ allows both signs
$y = k/x$: hyperbola, never touches axes, in 1st and 3rd quadrants
$y = k/x^2$: steeper hyperbola, always positive, in 1st quadrant only (for $k>0$)
$y = k\sqrt{x}$: starts steep near origin, flattens out, $x \geq 0$ only

Determining Proportionality from Data Tables

To decide what type of proportionality holds, test the data systematically.

Compute $\dfrac{y}{x}$ for each row

→

Constant? → Direct $y \propto x$

→

Compute $xy$ for each row

→

Constant? → Inverse $y \propto 1/x$

→

Try $y/x^2$, $y/\sqrt{x}$, $yx^2$ …

→

Constant? → Identify type

🗺️ Visual Notes

Proportion

Direct $y \propto x$

Equation: $y = kx$
Graph: straight line through origin
$y/x = k$ (constant)
$x$ doubles → $y$ doubles

Inverse $y \propto 1/x$

Equation: $y = k/x$
Graph: hyperbola in quadrants 1 & 3
$xy = k$ (constant)
$x$ doubles → $y$ halves

Powers & Roots

$y = kx^2$ — parabola through origin
$y = kx^3$ — cubic through origin
$y = k\sqrt{x}$ — root curve ($x \geq 0$)
$y = k/x^2$ — steep hyperbola

Finding $k$

Write equation with $k$
Substitute known pair
Solve for $k$
Rewrite full equation

Combined

$y \propto x/z^2$: $y = kx/z^2$
Need 3 known values
Substitute to find $k$
Use equation for predictions

Real-World Uses

Speed, distance, time
Hooke's Law ($F \propto x$)
Newton's gravity ($F \propto 1/r^2$)
Compound interest models

Comparison of Proportionality Types

Type	Symbol	Equation	Graph Shape	Constant check	$x \times 3$ gives $y$…
Direct	$y \propto x$	$y = kx$	Straight line through origin	$y/x = k$	$\times 3$
Inverse	$y \propto 1/x$	$y = k/x$	Hyperbola (Q1 & Q3)	$xy = k$	$\div 3$
Square	$y \propto x^2$	$y = kx^2$	Parabola through origin	$y/x^2 = k$	$\times 9$
Cube	$y \propto x^3$	$y = kx^3$	Cubic through origin	$y/x^3 = k$	$\times 27$
Square root	$y \propto \sqrt{x}$	$y = k\sqrt{x}$	Root curve, $x \geq 0$	$y/\sqrt{x} = k$	$\times \sqrt{3}$
Inverse square	$y \propto 1/x^2$	$y = k/x^2$	Steep hyperbola (Q1)	$yx^2 = k$	$\div 9$

Decision Tree — Identifying the Relationship

Look at data table or graph

→

Does $y/x$ = constant? YES → $y = kx$ (direct)

→

NO: Does $xy$ = constant? YES → $y = k/x$ (inverse)

→

NO: Try $y/x^2$, $y/x^3$, $y/\sqrt{x}$, $yx^2$

→

Whichever ratio is constant → that is the type

Scale Factor Summary

If $x$ is multiplied by $s$…	$y = kx$	$y = kx^2$	$y = kx^3$	$y = k\sqrt{x}$	$y = k/x$	$y = k/x^2$
…$y$ is multiplied by:	$s$	$s^2$	$s^3$	$\sqrt{s}$	$1/s$	$1/s^2$
Example ($s = 2$)	$\times 2$	$\times 4$	$\times 8$	$\times \sqrt{2}$	$\div 2$	$\div 4$
Example ($s = 3$)	$\times 3$	$\times 9$	$\times 27$	$\times \sqrt{3}$	$\div 3$	$\div 9$

✏️ Worked Examples

Grade 4–5 · Direct Proportion

$y$ is directly proportional to $x$. When $x = 5$, $y = 35$.
(a) Find the equation connecting $y$ and $x$.
(b) Find $y$ when $x = 9$.
(c) Find $x$ when $y = 63$.

Write the proportionality statement

Since $y \propto x$, the equation is $y = kx$ for some constant $k$.

Substitute the known pair

Substituting $x = 5$, $y = 35$: $\quad 35 = k \times 5$

Solve for $k$

$$k = \frac{35}{5} = 7$$ So the equation is $y = 7x$.

Answer parts (b) and (c)

(b) When $x = 9$: $y = 7 \times 9 = 63$
(c) When $y = 63$: $63 = 7x \Rightarrow x = 9$

$$y = 7x \quad;\quad \text{(b) } y = 63 \quad;\quad \text{(c) } x = 9$$

Grade 6–7 · Non-Linear Proportion

$P$ is proportional to the square of $Q$. When $Q = 4$, $P = 48$.
(a) Express $P$ in terms of $Q$.
(b) Find $P$ when $Q = 7$.
(c) Find $Q$ when $P = 108$ (give a positive answer).

Write the equation

$P \propto Q^2 \Rightarrow P = kQ^2$

Find $k$

$48 = k \times 4^2 = 16k$
$$k = \frac{48}{16} = 3$$

Write the full equation

$$P = 3Q^2$$

Answer part (b)

$P = 3 \times 7^2 = 3 \times 49 = 147$

Answer part (c)

$108 = 3Q^2 \Rightarrow Q^2 = 36 \Rightarrow Q = 6$ (taking positive root)

$$P = 3Q^2 \quad;\quad \text{(b) } P = 147 \quad;\quad \text{(c) } Q = 6$$

Grade 9 · Combined Proportionality

The force $F$ between two objects is directly proportional to the product of their masses $m_1$ and $m_2$, and inversely proportional to the square of the distance $d$ between them. When $m_1 = 2$, $m_2 = 3$, and $d = 6$, the force $F = 5$.

(a) Find $F$ when $m_1 = 4$, $m_2 = 5$, and $d = 10$.
(b) The distance $d$ is halved whilst $m_1$ and $m_2$ remain the same. Find the factor by which $F$ increases.

Write the combined proportionality equation

$$F \propto \frac{m_1 m_2}{d^2} \quad \Rightarrow \quad F = \frac{k \, m_1 m_2}{d^2}$$

Substitute the known values to find $k$

$$5 = \frac{k \times 2 \times 3}{6^2} = \frac{6k}{36} = \frac{k}{6}$$ $$k = 30$$

Write the full equation

$$F = \frac{30 \, m_1 m_2}{d^2}$$

Calculate $F$ for part (a)

$$F = \frac{30 \times 4 \times 5}{10^2} = \frac{600}{100} = 6$$

Analyse the scale factor for part (b)

If $d \to d/2$, then $d^2 \to d^2/4$. Dividing by $d^2/4$ is the same as multiplying by $4/d^2$, so $F$ is multiplied by $4$. That is, $F$ increases by a factor of $\mathbf{4}$.

Formally: new $F = \dfrac{30\,m_1 m_2}{(d/2)^2} = \dfrac{30\,m_1 m_2 \times 4}{d^2} = 4 \times$ old $F$.

$$\text{(a) } F = 6 \qquad \text{(b) } F \text{ increases by a factor of } 4$$

❓ Exam Questions

Q1 1 mark

$y$ is directly proportional to $x$. When $x = 3$, $y = 12$. Write down the value of $y$ when $x = 8$.

Mark Scheme:
$y = kx \Rightarrow k = 12/3 = 4 \Rightarrow y = 4x$
When $x = 8$: $y = 4 \times 8 = \mathbf{32}$ ✓ [1 mark — accept $y = 32$ or $k = 4$ shown]

Q2 2 marks

$y$ is inversely proportional to $x$. When $x = 5$, $y = 8$. Find $y$ when $x = 20$.

Mark Scheme:
[M1] $y = k/x \Rightarrow 8 = k/5 \Rightarrow k = 40$, equation $y = 40/x$
[A1] When $x = 20$: $y = 40/20 = \mathbf{2}$

Q3 3 marks

The table below shows values of $x$ and $y$.

$x$	$y$
2	20
5	125
8	320

Show that $y$ is proportional to $x^2$ and find the equation connecting $y$ and $x$.

Mark Scheme:
[M1] Test $y/x^2$: $20/4 = 5$; $125/25 = 5$; $320/64 = 5$ — all equal to 5
[A1] Since $y/x^2 = 5$ is constant, $y \propto x^2$
[A1] $k = 5$, so $y = 5x^2$

Q4 4 marks

$T$ is proportional to the square root of $L$. When $L = 9$, $T = 6$.
(a) Find $T$ when $L = 25$. (b) Find $L$ when $T = 10$.

Mark Scheme:
[M1] $T = k\sqrt{L}$; substituting: $6 = k\sqrt{9} = 3k$, so $k = 2$
[A1] Equation: $T = 2\sqrt{L}$
[A1] (a) $T = 2\sqrt{25} = 2 \times 5 = \mathbf{10}$
[A1] (b) $10 = 2\sqrt{L} \Rightarrow \sqrt{L} = 5 \Rightarrow L = \mathbf{25}$

Q5 4 marks

$y$ is inversely proportional to $x^2$. When $x = 3$, $y = 4$.
(a) Find $y$ when $x = 6$. (b) A student says: "When $x$ increases by 50%, $y$ decreases by 50%." Is the student correct? Show working to justify your answer.

Mark Scheme:
[M1] $y = k/x^2$; $4 = k/9 \Rightarrow k = 36$, equation $y = 36/x^2$
[A1] (a) $y = 36/36 = \mathbf{1}$
[M1] (b) Increasing $x$ by 50%: new $x = 1.5x$, so new $y = 36/(1.5x)^2 = 36/(2.25x^2)$
[A1] New $y = (4/9) \times$ old $y$. This is a decrease of $5/9 \approx 55.6\%$, not 50%. The student is incorrect.

Q6 6 marks

The pressure $P$ exerted by a gas is directly proportional to its temperature $T$ (in Kelvin) and inversely proportional to its volume $V$.
When $T = 300$, $V = 5$, $P = 120$.
(a) Find $P$ when $T = 400$ and $V = 8$.
(b) The temperature is doubled and the pressure is kept constant. Find the factor by which the volume changes.

Mark Scheme:
[M1] $P \propto T/V \Rightarrow P = kT/V$
[M1] Substituting: $120 = k \times 300/5 = 60k \Rightarrow k = 2$
[A1] Equation: $P = 2T/V$
[M1] (a) $P = 2 \times 400/8 = 800/8 = \mathbf{100}$
[M1] (b) $P = 2T/V$. Constant $P$ means $T/V = $ constant. If $T \to 2T$, then $V \to 2V$.
[A1] Volume doubles (increases by factor of 2).

⭐ Grade 9 Model Answers

Model Answer — Question 6 (Combined Proportionality, 6 marks)

✏️

FULL MODEL ANSWER

Part (a):

[Setting up the equation — 1 mark]
Since $P$ is directly proportional to $T$ and inversely proportional to $V$:

$$P = \frac{kT}{V}$$

[Substituting to find $k$ — 1 mark]
Using $T = 300$, $V = 5$, $P = 120$:

$$120 = \frac{k \times 300}{5} = 60k \quad \Rightarrow \quad k = 2$$

[Writing the equation — 1 mark]

$$P = \frac{2T}{V}$$

[Computing $P$ — 1 mark]
When $T = 400$, $V = 8$:

$$P = \frac{2 \times 400}{8} = \frac{800}{8} = \boxed{100}$$

Part (b):

[Identifying the constraint — 1 mark]
Constant pressure means $P = $ constant, so $\dfrac{kT}{V} = $ constant, which implies $\dfrac{T}{V} = $ constant.

[Concluding the scale factor — 1 mark]
If $T$ is doubled (multiplied by 2), then $V$ must also double (be multiplied by 2) to keep the ratio $T/V$ constant.

The volume increases by a factor of $\boxed{2}$.

🎯

WHY EACH PART EARNS MARKS

Mark 1 — Correctly translating the verbal description into a combined proportionality equation. Many students forget one of the variables or invert the fraction.
Mark 2 — Substituting all three known values correctly and solving for $k$. Common error: substituting incorrectly into $P = kTV$ (missing the denominator).
Mark 3 — Writing the complete equation $P = 2T/V$ explicitly. Even if arithmetic is wrong, method marks can be earned.
Mark 4 — Correct numerical answer for part (a). Follow-through marks may apply if $k$ is wrong but method is correct.
Mark 5 — Correctly identifying that constant $P$ implies $T/V$ is constant. This requires algebraic reasoning, not just number work.
Mark 6 — Stating the conclusion clearly: volume doubles. A bare answer of "2" is acceptable but "volume doubles" is the ideal Grade 9 response.

Key Examiner Observations

⚠️

IMPORTANT — Grade 9 Communication

Grade 9 answers are distinguished not just by the correct answer, but by clear, logical presentation. Always:

State the proportionality relationship first (e.g. $P \propto T/V$)
Write the equation with $k$ before substituting
Solve for $k$ in a clearly labelled step
Restate the final equation before using it
For scale-factor questions, show the algebraic reasoning, not just an example with numbers

📋 Revision Sheet

Key Definitions

Term	Meaning
Direct proportion	$y/x = k$ constant; $y = kx$
Inverse proportion	$xy = k$ constant; $y = k/x$
Proportionality constant	$k$, found by substituting a known pair
$y \propto x^2$	$y$ grows as the square of $x$
$y \propto \sqrt{x}$	$y$ grows as the square root of $x$
$y \propto 1/x^2$	Inverse square law

Essential Formulae

$$y \propto x \Rightarrow y = kx$$

$$y \propto \frac{1}{x} \Rightarrow y = \frac{k}{x}$$

$$y \propto x^2 \Rightarrow y = kx^2$$

$$y \propto x^3 \Rightarrow y = kx^3$$

$$y \propto \sqrt{x} \Rightarrow y = k\sqrt{x}$$

$$y \propto \frac{1}{x^2} \Rightarrow y = \frac{k}{x^2}$$

$$y \propto \frac{x}{z^2} \Rightarrow y = \frac{kx}{z^2}$$

Memory Hooks

WEBS: Write → Equation → Bind $k$ → Solve
Direct: same direction — both up or both down
Inverse: opposite direction — one up, one down
Direct graph passes through origin; inverse graph is a hyperbola
If $x \to s \cdot x$: for $y = kx^n$, multiply $y$ by $s^n$
For inverse $y = k/x^n$: divide $y$ by $s^n$
Test $y/x^n$ or $yx^n$ — whichever is constant tells you the type

Exam Tips

Always write the equation before substituting
Show the calculation of $k$ explicitly — it earns method marks
State the final equation clearly: $y = kx^n$ with $k$ filled in
For "find $x$" questions, square root (or cube root) at the end
Read whether the question wants a positive root
In scale factor questions, use algebra, not specific numbers
Check: does your answer make sense? (bigger $x$ should give bigger/smaller $y$ depending on type)

🔄 Flashcards

Click a card to reveal the answer. Work through all 15 before moving on.

✗ Common Mistakes

✗

MISTAKE 1 — Writing $y = \dfrac{1}{kx}$ instead of $y = \dfrac{k}{x}$

What students do wrong: When setting up inverse proportion, they put $k$ in the denominator: $y = 1/(kx)$.
Why marks are lost: This gives a completely different value of $k$ and wrong subsequent answers.
How to avoid: The proportionality symbol $\propto$ is replaced by $= k \times$. So $y \propto 1/x$ becomes $y = k \times (1/x) = k/x$.

✗

MISTAKE 2 — Forgetting to square (or cube) when substituting

What students do wrong: For $y = kx^2$, when given $y = 48$ and $x = 4$, writing $48 = 4k$ instead of $48 = 16k$.
Why marks are lost: $k$ becomes 12 instead of 3, and every subsequent answer is wrong.
How to avoid: Always evaluate the power first: write $4^2 = 16$, then substitute $48 = 16k$.

✗

MISTAKE 3 — Not taking the square root when solving for $x$

What students do wrong: When finding $x$ from $y = kx^2$, getting $x^2 = $ (answer) and then forgetting to square root.
Why marks are lost: The final answer for $x$ is the value of $x^2$, which is wrong.
How to avoid: Box the step: "I have $x^2 = \ldots$, so $x = \pm\sqrt{\ldots}$". Take the positive root unless told otherwise.

✗

MISTAKE 4 — Confusing direct and inverse proportion in context

What students do wrong: Reading "more workers → less time" and writing $T = kW$ instead of $T = k/W$.
Why marks are lost: Incorrect equation type, so $k$ and all answers are wrong.
How to avoid: Ask: "as $x$ increases, does $y$ increase (direct) or decrease (inverse)?" More workers means less time → inverse.

✗

MISTAKE 5 — Using scale factors additively instead of multiplicatively

What students do wrong: For $y \propto x^2$, if $x$ increases by 3 (e.g. from 4 to 7), they compute $y$ increases by $3^2 = 9$ (adding) rather than using the ratio correctly.
Why marks are lost: Scale factor reasoning only works with multiplicative changes. The ratio method ($x \times 2 \Rightarrow y \times 4$) requires multiplying, not adding.
How to avoid: Always phrase as "multiplied by" not "increased by". Use $x_{\text{new}} / x_{\text{old}}$ as your scale factor.

✗

MISTAKE 6 — Not stating the proportionality relationship clearly

What students do wrong: Jumping straight to "let $k = \ldots$" or computing $k$ without ever writing the proportionality statement or equation.
Why marks are lost: Method marks often require writing $y \propto x$ and $y = kx$ explicitly. Skipping these loses typically 1–2 marks even if the numerical answer is correct.
How to avoid: Always start with the symbol: write $y \propto x^n$, then $y = kx^n$, then substitute. Never skip the first two steps.

✅ Final Checklist

Click each item when you are confident. Aim for 100% before your exam.

I can state the equation for direct proportion and explain what $k$ represents
I can state the equation for inverse proportion and explain what $k$ represents
I can find $k$ by substituting a known pair of values into the equation
I can identify a direct proportion relationship from a graph (straight line through origin)
I can identify an inverse proportion relationship from a graph (hyperbola)
I can write and use the equation $y = kx^2$ and find unknown values
I can write and use the equation $y = kx^3$ and find unknown values
I can write and use the equation $y = k\sqrt{x}$ and find unknown values
I can write and use the equation $y = k/x^2$ and find unknown values
I can determine the type of proportionality from a table of values by testing ratios
I can apply the scale factor rule: if $x \to sx$ then $y \to s^n y$ for $y = kx^n$
I can set up combined proportionality equations (e.g. $y \propto x/z^2$)
I can solve multi-step proportion problems and interpret answers in context
I always show my working: proportionality statement → equation → find $k$ → answer
I can identify Grade 9 proportion questions in context (e.g. physics, engineering models)

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