How important is the Ratio, proportion and rates of change area?

This area is heavily examined and very applied, appearing on every paper through real-life contexts such as money, recipes, maps, speed and growth. It builds directly on the number work with fractions and percentages and connects to graphs, so it rewards confident proportional reasoning.

Which topics in this area are Higher tier only?

Inverse proportion with squares, cubes and roots, setting up proportion equations with a constant of proportionality, and estimating the gradient of a curve using a tangent are weighted to Higher tier. Foundation tier covers ratio and scale, simple direct and inverse proportion, percentage change and interest, and reading distance-time and velocity-time graphs.

What is the multiplier method and why does it matter?

The multiplier method turns a percentage change into a single multiplication: a $12\%$ increase uses $\times 1.12$ and a $12\%$ decrease uses $\times 0.88$. It makes percentage change, reverse percentages and compound interest quick and reliable, and it is the expected method for compound growth and depreciation.

What is the most common mistake in this area?

Confusing direct and inverse proportion, and handling reverse percentages by multiplying again instead of dividing by the multiplier, are the most common errors. Mixing up simple and compound interest, and reading speed instead of acceleration on a velocity-time graph, also lose marks regularly.

How should I revise this area?

Master ratio and scale first, then direct and inverse proportion, then percentage change and interest with the multiplier method, then rates of change from graphs. Practise forming the proportion equation, finding the constant, and reading gradients and areas from graphs. Use real-life worded problems to rehearse choosing the right method.

EnglandMaths

AQA GCSE Mathematics Ratio, proportion and rates of change: a complete overview of scale, proportion, interest and rates

A deep-dive AQA GCSE Mathematics guide to the Ratio, proportion and rates of change area. Covers ratio and scale, direct and inverse proportion, percentage change and interest, and rates of change from graphs, with the multiplier and proportion methods and exam patterns AQA repeats.

Generated by Claude Opus 4.816 min read8300Updated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What this area demands
Ratio and scale
Direct and inverse proportion
Percentage change and interest
Rates of change from graphs
How this area is examined
Check your knowledge

What this area demands

Ratio, proportion and rates of change is the most applied area of the course. It takes the number skills of fractions and percentages and uses them to solve real problems about money, maps, recipes, speed and growth. AQA tests proportional reasoning: recognising the type of relationship, choosing the right method, and reading rates of change from graphs.

This guide walks through the four topics in specification order, then sets out the exam patterns AQA repeats. Each topic has a matching dot-point page with practice questions; this overview ties them together.

Ratio and scale

The area opens with ratio and scale: simplifying ratios, dividing a quantity in a given ratio, using scale factors and map scales, and combining ratios that share a part. This links directly to scale drawings in geometry and to the proportion topics that follow.

Direct and inverse proportion

Direct and inverse proportion covers recognising each type, setting up an equation with a constant of proportionality, using it to find unknown values, and interpreting the graphs. Direct proportion gives a straight line through the origin; inverse proportion gives a reciprocal curve.

Percentage change and interest

Percentage change and interest is one of the most useful topics on the course. It covers percentage increase and decrease using multipliers, reverse percentages to find an original amount, and simple and compound interest including depreciation. The multiplier method is the key technique.

Rates of change from graphs

Rates of change and graphs covers distance-time and velocity-time graphs, the gradient as a rate of change, and the area under a graph as a total. At Higher tier you estimate the gradient of a curve using a tangent to find an instantaneous rate.

How this area is examined

A typical AQA profile for ratio, proportion and rates of change:

Applied calculations. Sharing in a ratio, scaling recipes, map distances, and percentage change in real contexts.
Proportion. Forming and using direct and inverse proportion equations.
Money problems. Compound interest, depreciation and reverse percentages.
Graph interpretation. Reading speed and acceleration from distance-time and velocity-time graphs, and gradients of curves at Higher tier.

Check your knowledge

A mix of recall and calculation questions covering this area. Attempt them under timed conditions, then check against the solutions.

Share $£180$ in the ratio $4:5$ . (3 marks)
On a $1:25000$ map two points are $8$ cm apart. Find the real distance in kilometres. (2 marks)
$y$ is inversely proportional to $x$ , and $y = 6$ when $x = 2$ . Find $y$ when $x = 4$ . (3 marks)
Increase $£90$ by $20\%$ using a multiplier. (2 marks)
A price is $£84$ after a $20\%$ increase. Find the original price. (2 marks)
$£1500$ is invested at $4\%$ compound interest for $2$ years. Find the value. (3 marks)
A distance-time graph rises $80$ m over $16$ s. Find the speed. (2 marks)
State what the area under a velocity-time graph represents. (1 mark)

Solutions

Step 1: Share a quantity in a given ratio

Find the value of one share by dividing the total by the sum of the ratio parts. Multiply that one share by each ratio number to find the two amounts.

Total shares $= 4 + 5 = 9$ . One share $= \frac{180}{9} = 20$ .

4 \times 20 = \pounds 80, \quad 5 \times 20 = \pounds 100

Final answer: $\pounds 80$ and $\pounds 100$ .

Step 2: Convert a map distance to a real distance

Multiply the map distance by the map scale to find the real distance in the map's units, then convert to the required unit. The scale gives the ratio of map length to real length.

8 \times 25\,000 = 200\,000 \text{ cm} = 2\,000 \text{ m} = 2 \text{ km}

Final answer: 2 km.

Step 3: Use inverse proportion to find an unknown value

For inverse proportion, the product of the two variables is constant: $k = xy$ . Find $k$ using the given pair of values, then use $k$ to find the unknown value.

k = 2 \times 6 = 12, \quad y = \frac{12}{4} = 3

Final answer: $y = 3$ .

Step 4: Increase an amount by a percentage using a multiplier

A percentage increase of $r\%$ is achieved by multiplying by the decimal multiplier $(1 + r/100)$ . A $20\%$ increase has multiplier $1.2$ .

90 \times 1.2 = \pounds 108

Final answer: $\pounds 108$ .

Step 5: Find an original amount from a percentage increase (reverse percentage)

After a $20\%$ increase the amount has been multiplied by $1.2$ , so dividing by $1.2$ reverses the increase and recovers the original. Never subtract $20\%$ from the given price, as that gives the wrong answer.

\frac{84}{1.2} = \pounds 70

Final answer: $\pounds 70$ .

Step 6: Calculate compound interest

Compound interest applies the same percentage growth each year to the accumulated total, not just the original principal. Multiply by the yearly multiplier raised to the power of the number of years.

1500 \times 1.04^2 = 1500 \times 1.0816 = \pounds 1622.40

Final answer: $\pounds 1622.40$ .

Step 7: Find speed from a distance-time graph

The gradient of a distance-time graph equals the speed, because speed is distance divided by time. Read the rise and run from the graph and divide.

\text{Speed} = \frac{80}{16} = 5 \text{ m/s}

Final answer: $5$ m/s.

Step 8: Interpret the area under a velocity-time graph

On a velocity-time graph, the area under the graph represents the total distance travelled, because multiplying velocity by time gives distance. This is the integral interpretation of area under a curve.

The area under a velocity-time graph represents the distance travelled.

Final answer: The distance travelled.

Sources & how we know this

AQA GCSE Mathematics (8300) specification — AQA (2015)