Proportion (Algebraic) · Study Notes · Edexcel IGCSE 4MA1 Higher

Algebraic direct and inverse proportion. Higher Tier expects you to derive the equation, find the constant, then use it.

At a glance

Direct: $y \propto x^{n} \iff y = kx^{n}$ .
Inverse: $y \propto 1/x^{n} \iff y = k/x^{n}$ .
Method: state, find $k$ , use.
Always check direction: as $x$ ↑, does $y$ ↑ or ↓?
Common $n$ : $1, 2, 1/2, 3$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.4 — Set up and use direct and inverse proportion equations.
2.4 — Find the constant of variation from given data.
2.4 — Solve problems involving multiple variables.

Direct proportion

▼

$y$ proportional to $x^{n}$ means $y = kx^{n}$ . Find $k$ , then use.

Forms of direct proportion:

$y \propto x$ → $y = kx$ (linear).
$y \propto x^{2}$ → $y = kx^{2}$ (quadratic).
$y \propto x^{3}$ → $y = kx^{3}$ (cubic).
$y \propto \sqrt{x}$ → $y = k\sqrt{x}$ .

Three-step method:

State the equation: $y = kx^{n}$ .
Use given values to find $k$ .
Substitute back to answer.

Every direct-proportion graph passes through the origin; the power of x sets its shape.

Example. $y$ directly proportional to $x^{3}$ . When $x = 2$ , $y = 24$ . Find $y$ when $x = 5$ .

$y = kx^{3}$ .
$24 = k \times 8$ , so $k = 3$ .
$y = 3x^{3}$ . When $x = 5$ : $y = 3 \times 125 = 375$ .

Worked qualitative. Why is the surface area of a sphere proportional to $r^{2}$ ?

Surface area $= 4\pi r^{2}$ .
$4\pi$ is the constant $k$ .
$A \propto r^{2}$ .

Edexcel tip. Always show the THREE steps — state, find $k$ , use. Mark schemes credit each step.

$y = kx^{n}$ .
Find $k$ from given data.
Use the equation to answer.
Common: $n = 1, 2, 1/2$ .

Inverse proportion

▼

$y$ proportional to $1/x^{n}$ means $y = k/x^{n}$ . Same method.

Forms of inverse proportion:

$y \propto 1/x$ → $y = k/x$ .
$y \propto 1/x^{2}$ → $y = k/x^{2}$ .
$y \propto 1/\sqrt{x}$ → $y = k/\sqrt{x}$ .

In inverse proportion y falls as x rises; the curve approaches but never reaches the axes.

Same three-step method.

Example. $P$ inversely proportional to $\sqrt{Q}$ . When $Q = 16$ , $P = 5$ . Find $P$ when $Q = 100$ .

$P = k/\sqrt{Q}$ .
$5 = k/4$ , so $k = 20$ .
$P = 20/\sqrt{Q}$ . When $Q = 100$ : $P = 20/10 = 2$ .

Worked qualitative. Force of gravity is inversely proportional to distance squared. If you double the distance, what happens to the force?

$F \propto 1/d^{2}$ .
Doubling $d$ : new force $= 1/(2d)^{2} = 1/(4d^{2}) =$ quarter of original.
Force becomes $1/4$ .

Edexcel tip. Doubling the input doesn't always halve the output — depends on the power.

$y = k/x^{n}$ .
Same method as direct.
Doubling $x$ : divide $y$ by $2^{n}$ .
Common: $n = 1, 2, 1/2$ .

Real-world applications

▼

Spheres, gases, gravity, electricity — all proportion.

Common direct proportion examples:

Distance ∝ time at constant speed.
Cost ∝ quantity at fixed price.
Area of square ∝ side².
Volume of cube ∝ side³.

Common inverse proportion examples:

Time inversely proportional to speed (fixed distance).
Pressure inversely proportional to volume (constant temperature gas).
Brightness inversely proportional to distance².
Loudness inversely proportional to distance.

Combined proportion (less common at IGCSE):

If $y \propto x$ and $y \propto 1/z$ , then $y = kx/z$ .

Worked qualitative. A bridge cable's load capacity is proportional to its diameter squared. If the diameter doubles, how does load capacity change?

$L \propto d^{2}$ .
Doubling $d$ : $L$ multiplies by $4$ .
Load quadruples — explaining why thicker cables are dramatically stronger.

Edexcel tip. When given a real-world context, identify whether the relationship is direct or inverse FIRST, then write the equation.

Direct: same direction.
Inverse: opposite direction.
Powers matter: $x^{2}$ doubles when $x$ does $\sqrt{2}$ ≈ 1.41.
Use proportion type to decide method.

Quick recap

Direct: $y = kx^{n}$ .
Inverse: $y = k/x^{n}$ .
Method: state, find $k$ , use.
Powers matter: doubling $x$ in $kx^{n}$ multiplies $y$ by $2^{n}$ .
Common contexts: gravity, gas, area, volume.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Direct: $y = kx^{n}$ .
Inverse: $y = k/x^{n}$ .
Method: state $\to$ find $k$ $\to$ substitute back.
Doubling $x$ changes $y$ by $2^{n}$ (direct) or $1/2^{n}$ (inverse).

How it’s examined

Algebraic proportion appears Paper 2H (4-6 marks). Typical: state proportion, find $k$ , use to find unknown. Examiner reports flag (1) skipping the 'find $k$ ' step, (2) confusing direct with inverse.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

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

1Algebraic direct proportion
Higher• Adapted from 4MA1/2H Jan 2024 Q11• direct proportion
▼
Question
$y$ is directly proportional to $x^{3}$ . When $x = 2$ , $y = 24$ . (a) Find $y$ in terms of $x$ . (b) Find $y$ when $x = 5$ .
Step-by-step solution
1. Step 1
  (a) $y \propto x^{3} \implies y = kx^{3}$ .
2. Step 2
  Substitute given values: $24 = k \times 2^{3} = 8k$ , so $k = 3$ .
3. Step 3
  Equation: $y = 3x^{3}$ .
4. Step 4
  (b) When $x = 5$ : $y = 3 \times 125 = 375$ .
Answer
(a) $y = 3x^{3}$ (b) $y = 375$
2Algebraic inverse proportion
Higher• inverse proportion
▼
Question
$P$ is inversely proportional to $\sqrt{Q}$ . When $Q = 16$ , $P = 5$ . Find $P$ when $Q = 100$ .
Step-by-step solution
1. Step 1
  $P = \dfrac{k}{\sqrt{Q}}$ .
2. Step 2
  Find $k$ : $5 = \dfrac{k}{\sqrt{16}} = \dfrac{k}{4}$ , so $k = 20$ .
3. Step 3
  Equation: $P = \dfrac{20}{\sqrt{Q}}$ .
4. Step 4
  When $Q = 100$ : $P = \dfrac{20}{\sqrt{100}} = \dfrac{20}{10} = 2$ .
Answer
$P = 2$
3Reverse-engineer the proportion equation
Higher• proportion, reverse
▼
Question
$y$ is directly proportional to $(x + 1)^{2}$ . When $x = 2$ , $y = 18$ . Find the value of $y$ when $x = 5$ .
Step-by-step solution
1. Step 1
  $y = k(x + 1)^{2}$ .
2. Step 2
  Substitute: $18 = k(2 + 1)^{2} = 9k$ , so $k = 2$ .
3. Step 3
  Equation: $y = 2(x + 1)^{2}$ .
4. Step 4
  When $x = 5$ : $y = 2(5 + 1)^{2} = 2 \times 36 = 72$ .
Answer
$y = 72$
4Recognising inverse proportion from a table [1 mark]
Getting started• Paper 1H/2H style — 1 mark• inverse proportion, tables
▼
Question
Here are four tables of values.

Table A $\quad x$ : $1, 2, 4$ $\quad y$ : $12, 6, 3$

Table B $\quad x$ : $1, 2, 4$ $\quad y$ : $12, 24, 48$

Table C $\quad x$ : $1, 2, 4$ $\quad y$ : $12, 3, 0.75$

Table D $\quad x$ : $1, 2, 4$ $\quad y$ : $12, 10, 6$

Write down the letter of the table in which $y$ is inversely proportional to $x$ . [1]
Step-by-step solution
1. Step 1
  If $y = \dfrac{k}{x}$ then $xy = k$ , so the product $xy$ must be the same for every pair.
  $\text{A: } 12, 12, 12 \qquad \text{C: } 12, 6, 3 \qquad \text{D: } 12, 20, 24$
2. Step 2
  B1 Only Table A has a constant product. (Table B is direct proportion, $y = 12x$ ; Table C is $y = \dfrac{12}{x^{2}}$ .)
Answer
Table A
Examiner tip
Examiner tip: Test for proportion with a constant: $\dfrac{y}{x}$ constant means direct, $xy$ constant means inverse, $x^{2}y$ constant means inverse square.

Common mistake: Choosing Table D because $y$ goes down as $x$ goes up. Decreasing is not enough; inverse proportion needs $xy$ to stay the same.
5Direct proportion with a decimal constant [2 marks]
Getting started• Paper 1H/2H style — 2 marks• direct proportion, constant of proportionality
▼
Question
$y$ is directly proportional to $x$ .

When $x = 2.5$ , $y = 9$

Find the value of $y$ when $x = 7$ [2]
Step-by-step solution
1. Step 1
  M1 Write $y = kx$ and find $k$ .
  $9 = 2.5k \quad\Rightarrow\quad k = 3.6$
2. Step 2
  A1 Use $y = 3.6x$ with $x = 7$
  $y = 3.6 \times 7 = 25.2$
Answer
$25.2$
Examiner tip
Examiner tip: Even for a simple direct proportion, write $y = kx$ first. On harder questions that line is worth a method mark on its own.

Common mistake: Finding $k$ the wrong way up, $k = \dfrac{2.5}{9}$ , which gives $y = 1.94\ldots$ . Check: $y$ should be bigger when $x$ is bigger.
6Finding x from a cube proportion [3 marks]
Getting started• Paper 1H/2H style — 3 marks• direct proportion, cube roots
▼
Question
$y$ is directly proportional to $x^{3}$ .

When $x = 0.5$ , $y = 1.5$

Find the value of $x$ when $y = 96$ [3]
Step-by-step solution
1. Step 1
  M1 Write $y = kx^{3}$ and substitute. Note $0.5^{3} = 0.125$
  $1.5 = k \times 0.125 \quad\Rightarrow\quad k = 12$
2. Step 2
  M1 Substitute $y = 96$ and make $x^{3}$ the subject.
  $96 = 12x^{3} \quad\Rightarrow\quad x^{3} = 8$
3. Step 3
  A1 Take the cube root.
  $x = \sqrt[3]{8} = 2$
Answer
$x = 2$
Examiner tip
Examiner tip: Dividing by a small number such as $0.125$ makes $k$ larger. Here $1.5 \div 0.125 = 12$ .

Common mistake: Using $0.5^{3} = 1.5$ (multiplying by $3$ instead of cubing), which gives $k = 1$ and $x = 4.58\ldots$
7Frequency of a vibrating string [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• inverse proportion, context
▼
Question
The frequency, $f$ hertz, of a vibrating string is inversely proportional to its length, $L$ metres.

When $L = 0.6$ , $f = 440$

Work out the frequency when $L = 0.8$ [3]
Step-by-step solution
1. Step 1
  M1 Write the inverse proportion with a constant.
  $f = \frac{k}{L}$
2. Step 2
  M1 Substitute $L = 0.6$ , $f = 440$
  $440 = \frac{k}{0.6} \quad\Rightarrow\quad k = 264$
3. Step 3
  A1 Use $f = \dfrac{264}{L}$ with $L = 0.8$
  $f = \frac{264}{0.8} = 330$
Answer
$330$ hertz
Examiner tip
Examiner tip: A longer string gives a lower note, so the frequency must go down. $330 < 440$ confirms the inverse relationship.

Common mistake: Dividing $440$ by $0.6$ to get $k$ , which treats the relationship as direct and gives $f = 587$ (3 s.f.).
8Percentage decrease for a square-root proportion [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• direct proportion, percentage change
▼
Question
$y$ is directly proportional to $\sqrt{x}$ .

$x$ is decreased by $36\%$ .

Work out the percentage decrease in $y$ . [3]
Step-by-step solution
1. Step 1
  M1 A $36\%$ decrease means $x$ is multiplied by $0.64$
  $y = k\sqrt{x}, \qquad y_{\text{new}} = k\sqrt{0.64x}$
2. Step 2
  M1 Take the square root of the multiplier.
  $k\sqrt{0.64x} = \sqrt{0.64} \times k\sqrt{x} = 0.8 \times k\sqrt{x}$
3. Step 3
  A1 A multiplier of $0.8$ is a decrease of $20\%$
  $(1 - 0.8) \times 100 = 20\%$
Answer
$20\%$
Examiner tip
Examiner tip: Change the percentage into a multiplier ( $0.64$ ), apply the power to the multiplier, then change back.

Common mistake: Rooting the percentage: $\sqrt{36} = 6\%$ . The root must act on $0.64$ , not on $36$ .
9Formula, then both values of x [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• direct proportion, square roots
▼
Question
$y$ is directly proportional to $x^{2}$ .

When $x = 2$ , $y = 20$

(a) Find a formula for $y$ in terms of $x$ . [2]

(b) Find the values of $x$ when $y = 45$ [2]
Step-by-step solution
1. Step 1
  M1 (a) Write $y = kx^{2}$ and substitute.
  $20 = k \times 2^{2} = 4k$
2. Step 2
  A1 So $k = 5$
  $y = 5x^{2}$
3. Step 3
  M1 (b) Substitute $y = 45$
  $45 = 5x^{2} \quad\Rightarrow\quad x^{2} = 9$
4. Step 4
  A1 Both square roots are needed.
  $x = 3 \quad\text{or}\quad x = -3$
Answer
(a) $y = 5x^{2}$ (b) $x = 3$ or $x = -3$
Examiner tip
Examiner tip: "The values of $x$ " (plural) is a clue. With no context and no condition such as $x > 0$ , give both roots.

Common mistake: Giving only $x = 3$ . Taking only the positive root costs the final A1.
10Chained proportion: show that p is inversely proportional to r [5 marks]
Stretch• Paper 1H/2H style — 5 marks• chained proportion, inverse proportion
▼
Question
$p$ is inversely proportional to $q^{2}$ . When $q = 4$ , $p = 5$

$q$ is directly proportional to $\sqrt{r}$ . When $r = 9$ , $q = 6$

(a) Show that $p = \dfrac{20}{r}$ [4]

(b) Find the value of $p$ when $r = 8$ [1]
Step-by-step solution
1. Step 1
  M1 (a) First proportion: $p = \dfrac{k}{q^{2}}$
  $5 = \frac{k}{16} \quad\Rightarrow\quad k = 80, \quad p = \frac{80}{q^{2}}$
2. Step 2
  M1 Second proportion, with a different constant: $q = m\sqrt{r}$
  $6 = m\sqrt{9} = 3m \quad\Rightarrow\quad m = 2, \quad q = 2\sqrt{r}$
3. Step 3
  M1 Substitute $q = 2\sqrt{r}$ into the formula for $p$ , keeping the brackets.
  $p = \frac{80}{(2\sqrt{r})^{2}}$
4. Step 4
  A1 Square both parts of the bracket: $(2\sqrt{r})^{2} = 4r$
  $p = \frac{80}{4r} = \frac{20}{r}$
5. Step 5
  B1 (b) Use the printed formula.
  $p = \frac{20}{8} = 2.5$
Answer
(a) $p = \dfrac{80}{(2\sqrt{r})^{2}} = \dfrac{20}{r}$ , as required (b) $2.5$
Examiner tip
Examiner tip: Part (b) can be done even if (a) goes wrong — the formula is printed. Never leave a follow-on part blank.

Common mistake: Writing $(2\sqrt{r})^{2} = 2r$ , squaring the root but not the $2$ . It is the same slip as substituting $5x$ into $x^{2}$ without brackets, and it loses the A1 in a "Show that".

Question 1
Paper 1H/2H style — 1 mark1 mark
$y$ is directly proportional to $\sqrt{x}$ .

The value of $x$ is multiplied by $25$ .

Write down the number that $y$ is multiplied by. [1]
Model answer
$y = k\sqrt{x}$ , so $k\sqrt{25x} = \sqrt{25} \times k\sqrt{x} = 5k\sqrt{x}$

Answer: $5$
Why this scores
Mark scheme:
- B1 for $5$ cao
Why this scores full marks: The square root acts on the multiplier: $\sqrt{25} = 5$ . The usual wrong answers are $25$ (ignoring the root) and $625$ (squaring instead of rooting).
Question 2
Paper 1H/2H style — 2 marks2 marks
$y$ is inversely proportional to $x^{2}$ .

When $x = 4$ , $y = 2.5$

Find a formula for $y$ in terms of $x$ . [2]
Model answer
$y = \dfrac{k}{x^{2}}$

$2.5 = \dfrac{k}{16}$ , so $k = 40$

Answer: $y = \dfrac{40}{x^{2}}$
Why this scores
Mark scheme:
- M1 for $2.5 = \dfrac{k}{4^{2}}$ oe
- A1 for $y = \dfrac{40}{x^{2}}$ oe
Why this scores full marks: The script squares $x$ before using it ( $4^{2} = 16$ ) and writes the full formula on the answer line, not just $k = 40$ .
Question 3
Paper 1H/2H style — 3 marks3 marks
The gravitational force, $F$ newtons, between two objects is inversely proportional to the square of the distance, $d$ metres, between them.

When $d = 2$ , $F = 720$

Work out the value of $F$ when $d = 6$ [3]
Model answer
$F = \dfrac{k}{d^{2}}$

$720 = \dfrac{k}{4}$ , so $k = 2880$

$F = \dfrac{2880}{6^{2}} = \dfrac{2880}{36} = 80$

Answer: $80$ newtons
Why this scores
Mark scheme:
- M1 for $F = \dfrac{k}{d^{2}}$ (not $F = kd^{2}$ )
- M1 for $k = 2880$ or $F = \dfrac{2880}{d^{2}}$
- A1 for $80$ cao
Why this scores full marks: Each stage is written separately: the proportion equation, the constant, then the substitution. Check: the distance is trebled, so the force is divided by $3^{2} = 9$ , and $720 \div 9 = 80$ .
Question 4
Paper 1H/2H style — 3 marks3 marks
$y$ is inversely proportional to $x$ .

Here is a table of values.

$x$ : $\quad 2 \quad 5 \quad b$

$y$ : $\quad 30 \quad a \quad 7.5$

Find the value of $a$ and the value of $b$ . [3]
Model answer
$y = \dfrac{k}{x}$ , so $30 = \dfrac{k}{2}$ and $k = 60$

$a = \dfrac{60}{5} = 12$

$7.5 = \dfrac{60}{b}$ , so $b = \dfrac{60}{7.5} = 8$

Answer: $a = 12$ , $b = 8$
Why this scores
Mark scheme:
- M1 for $k = 60$ or $xy = 60$ oe
- A1 for $a = 12$
- A1 for $b = 8$
Why this scores full marks: The constant is found once and used for both gaps. For $b$ the script rearranges $7.5 = \dfrac{60}{b}$ correctly, multiplying by $b$ before dividing by $7.5$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
The speed, $v$ m/s, of a wave in shallow water is directly proportional to the square root of the depth, $h$ metres, of the water.

When $h = 4$ , $v = 6$

Work out the speed of a wave when the depth is $10$ metres. Give your answer correct to $3$ significant figures. [3]
Model answer
$v = k\sqrt{h}$

$6 = k\sqrt{4} = 2k$ , so $k = 3$

$v = 3\sqrt{10} = 9.4868\ldots$

Answer: $9.49$ m/s
Why this scores
Mark scheme:
- M1 for $v = k\sqrt{h}$ with $h = 4$ , $v = 6$ substituted
- M1 for $3\sqrt{10}$ oe (ft their $k$ )
- A1 for awrt $9.49$
Why this scores full marks: The script keeps the exact value $3\sqrt{10}$ until the last step and rounds once, to 3 significant figures as asked.
Question 6
Paper 1H/2H style — 3 marks3 marks
$y$ is directly proportional to $x^{2}$ .

$x$ is increased by $30\%$ .

Work out the percentage increase in $y$ . [3]
Model answer
$y = kx^{2}$

New $y = k(1.3x)^{2} = 1.69kx^{2}$

$1.69 - 1 = 0.69$

Answer: $69\%$
Why this scores
Mark scheme:
- M1 for $(1.3x)^{2}$ oe, or a numerical example with $x$ increased by $30\%$
- M1 for $1.69$ seen
- A1 for $69$ cao
Why this scores full marks: The multiplier $1.3$ is squared, not the percentage. The common wrong answers are $60\%$ ( $2 \times 30$ ) and $9\%$ (squaring $0.3$ to get $0.09$ instead of squaring the multiplier $1.3$ ).
Question 7
Paper 1H/2H style — 4 marks4 marks
$y$ is inversely proportional to $x^{2}$ , where $x > 0$

When $x = 5$ , $y = 7.2$

(a) Find the value of $y$ when $x = 4$ [2]

(b) Find the value of $x$ when $y = 20$ [2]
Model answer
(a) $y = \dfrac{k}{x^{2}}$ , $7.2 = \dfrac{k}{25}$ , so $k = 180$

$y = \dfrac{180}{16} = 11.25$

Answer: $11.25$

(b) $20 = \dfrac{180}{x^{2}}$ , so $x^{2} = 9$

$x = 3$ (since $x > 0$ )

Answer: $3$
Why this scores
Mark scheme:
- M1 for $7.2 = \dfrac{k}{5^{2}}$ oe
- A1 for $11.25$ oe
- M1 for $x^{2} = \dfrac{180}{20}$ oe (ft their $k$ )
- A1 for $3$ (accept $3$ with $-3$ rejected)
Why this scores full marks: The script reads the condition $x > 0$ and gives only the positive root, saying why. Without that condition, $x = -3$ would also be required.
Question 8
Paper 1H/2H style — 5 marks5 marks
The time, $T$ hours, that $n$ identical machines take to complete a job is inversely proportional to $n$ .

$12$ machines take $5$ hours to complete the job.

(a) Find a formula for $T$ in terms of $n$ . [2]

The job must now be completed in less than $4$ hours.

(b) Work out the least number of extra machines needed. [3]
Model answer
(a) $T = \dfrac{k}{n}$ , $5 = \dfrac{k}{12}$ , so $k = 60$

Answer: $T = \dfrac{60}{n}$

(b) $\dfrac{60}{n} < 4$ , so $n > \dfrac{60}{4} = 15$

$15$ machines take exactly $4$ hours, so at least $16$ machines are needed.

$16 - 12 = 4$

Answer: $4$ extra machines
Why this scores
Mark scheme:
- M1 for $5 = \dfrac{k}{12}$ oe
- A1 for $T = \dfrac{60}{n}$ oe
- M1 for $\dfrac{60}{4}$ or $n > 15$ oe (ft their $k$ )
- M1 for $16$ machines identified
- A1 for $4$ cao
Why this scores full marks: "Less than $4$ hours" rules out $15$ machines, which take exactly $4$ hours, so the script rounds up to $16$ . It then answers the question actually asked — the number of extra machines — rather than stopping at $16$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
$y$ is directly proportional to one of $x$ , $x^{2}$ or $x^{3}$ .

When $x = 2$ , $y = 5$

When $x = 4$ , $y = 40$

(a) Find a formula for $y$ in terms of $x$ . [3]

(b) Work out the value of $y$ when $x = 6$ [2]
Model answer
(a) $x$ is doubled ( $2 \to 4$ ) and $y$ is multiplied by $\dfrac{40}{5} = 8 = 2^{3}$

So $y \propto x^{3}$ and $y = kx^{3}$

$5 = k \times 2^{3}$ , so $k = \dfrac{5}{8} = 0.625$

Answer: $y = 0.625x^{3}$

(b) $y = 0.625 \times 6^{3} = 0.625 \times 216 = 135$

Answer: $135$
Why this scores
Mark scheme:
- B1 for identifying $y \propto x^{3}$ with a reason (e.g. $x \times 2$ gives $y \times 8 = 2^{3}$ )
- M1 for $5 = k \times 2^{3}$ oe
- A1 for $y = 0.625x^{3}$ oe
- M1 for substituting $x = 6$ into their formula
- A1 for $135$ cao
Why this scores full marks: The script compares the multipliers for $x$ and $y$ to find the power, and says so. It then checks the formula with the second pair: $0.625 \times 4^{3} = 40$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
$P$ is directly proportional to $\sqrt{Q}$ .

When $Q = a$ , $P = 6$

When $Q = a + 20$ , $P = 9$

Find the value of $a$ . Show clear algebraic working. [5]
Model answer
$6 = k\sqrt{a}$ and $9 = k\sqrt{a + 20}$

Divide: $\dfrac{9}{6} = \dfrac{\sqrt{a + 20}}{\sqrt{a}}$

Square: $2.25 = \dfrac{a + 20}{a}$

$2.25a = a + 20$

$1.25a = 20$

Answer: $a = 16$
Why this scores
Mark scheme:
- M1 for $6 = k\sqrt{a}$ and $9 = k\sqrt{a + 20}$
- M1 for eliminating $k$ , e.g. $\dfrac{9}{6} = \dfrac{\sqrt{a + 20}}{\sqrt{a}}$ or $6\sqrt{a + 20} = 9\sqrt{a}$
- M1 for squaring correctly to a linear equation, e.g. $2.25a = a + 20$ or $36(a + 20) = 81a$
- M1 for isolating $a$ , e.g. $1.25a = 20$ or $45a = 720$
- A1 for $16$ cao
Why this scores full marks: Dividing the two equations removes $k$ in one step. Squaring the whole of each side, not term by term, gives a linear equation. Trial of values scores 0 here, because algebraic working is demanded. Check: $k = \dfrac{6}{4} = 1.5$ and $1.5\sqrt{36} = 9$

4MA1 Proportion (Algebraic)

At a glance

What you’ll learn

Direct proportion

Inverse proportion

Real-world applications

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Algebraic direct proportion

2Algebraic inverse proportion

3Reverse-engineer the proportion equation

4Recognising inverse proportion from a table [1 mark]

5Direct proportion with a decimal constant [2 marks]

6Finding x from a cube proportion [3 marks]

7Frequency of a vibrating string [3 marks]

8Percentage decrease for a square-root proportion [3 marks]

9Formula, then both values of x [4 marks]

10Chained proportion: show that p is inversely proportional to r [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10