Ratio, Proportion and Rates of Change
KS4MA-KS4-D003
Applying ratio, proportion and multiplicative reasoning to solve problems involving percentages, compound measures, scale, direct and inverse proportion, and rates of change.
National Curriculum context
Ratio, proportion and rates of change at KS4 formalises and extends the multiplicative reasoning developed at KS3, applying it in more complex and applied contexts. Pupils work with ratio to solve problems involving maps, recipes, mixing, division into parts, currency conversion and similar situations, and extend percentage work to include reverse percentages, compound interest and repeated proportional change. The curriculum requires pupils to understand gradient as a rate of change and to apply this in practical contexts such as speed-time and distance-time graphs, unit rates and compound measures such as density and pressure. Rates of change provide the conceptual bridge between arithmetic, algebraic thinking and graphical interpretation. Higher tier pupils additionally work with direct and inverse proportion expressed algebraically, including recognising and using y = kx^n relationships.
4
Concepts
2
Clusters
1
Prerequisites
4
With difficulty levels
Lesson Clusters
Use ratio and scale to solve problems and calculate percentage change
introduction CuratedRatio/scale (simplifying, dividing, scale drawings) and percentage change/financial mathematics are the applied proportional reasoning targets at GCSE foundation and higher.
Apply direct and inverse proportion and interpret compound measures
practice CuratedDirect/inverse proportion (algebraic, graphical, numerical) and compound measures/rates of change (speed, density, gradient) are the advanced proportion cluster at GCSE.
Prerequisites
Concepts from other domains that pupils should know before this domain.
Concepts (4)
Ratio and Scale
process AI FacilitatedMA-KS4-C013
Expressing, simplifying and using ratios to solve problems including division into parts, scale drawings, maps, and comparison of quantities.
Teaching guidance
Distinguish clearly between part:part and part:whole representations — many problems require conversion between the two. Use unitary ratio as a bridge between ratio and rate reasoning. Scale drawings provide a concrete application with built-in accuracy checking. Ratio notation (a:b) should be fluently connected to fraction (a/(a+b)) and percentage representations.
Common misconceptions
Pupils often divide in ratio a:b by computing a/(a+b) × total for the first part correctly, but then subtract to find the second part instead of computing b/(a+b) × total directly. Many pupils treat 2:3 as meaning ⅔ rather than 2 parts to 3 parts (so 2 out of 5 total). Ratio is frequently confused with fractions in questions involving 'the ratio of A to B is 3:5' meaning A/B = 3/5 (fraction of A to B), not A/total = 3/5.
Difficulty levels
Can write and simplify ratios and share a quantity in a given ratio.
Example task
Share £240 in the ratio 3:5.
Model response: Total parts = 8. Each part = £30. Shares: £90 and £150.
Solves ratio problems including finding one share when another is known, and interprets scale drawings and maps.
Example task
A map scale is 1:25,000. A road is 8 cm on the map. What is the real length in km?
Model response: Real length = 8 × 25,000 = 200,000 cm = 2,000 m = 2 km.
Solves complex ratio problems including combining ratios, using ratio to set up and solve equations, and connecting ratio to fraction/percentage representations.
Example task
In a school, the ratio of boys to girls is 4:5. There are 45 more girls than boys. How many students in total?
Model response: Difference in parts = 5 - 4 = 1 part = 45 students. So 1 part = 45. Total = 9 parts = 9 × 45 = 405 students.
Solves algebraic ratio problems and problems involving changing ratios.
Example task
Cement, sand and gravel are mixed in the ratio 1:2:3. 15 kg of sand is used. The total is then increased by 50%. Find the new total mass.
Model response: Sand = 2 parts = 15 kg. So 1 part = 7.5 kg. Original total = 6 × 7.5 = 45 kg. Increased by 50%: 45 × 1.5 = 67.5 kg.
Delivery rationale
Secondary maths process concept — problem-solving benefits from structured AI delivery with facilitator for extended reasoning.
Percentage Change and Financial Mathematics
process AI FacilitatedMA-KS4-C014
Calculating percentage increase/decrease, original value (reverse percentage), simple and compound interest, and repeated proportional change in financial contexts.
Teaching guidance
Use multiplier method as the standard approach for efficiency — 15% increase means ×1.15. Reverse percentages require dividing by the multiplier, not subtracting the percentage from 100 first. Compound interest is the most important application: explicitly model the difference between simple and compound interest with tables before revealing the formula A = P(1+r)ⁿ.
Common misconceptions
For reverse percentages, pupils frequently calculate the wrong value by subtracting x% from the given value rather than dividing by (1 + x/100). Many confuse compound interest (interest on interest) with simple interest (interest on principal only). Percentage change direction (increase vs decrease) and application to the original value are persistently confused.
Difficulty levels
Can calculate a percentage of an amount and find percentage increase or decrease.
Example task
A coat costs £80. There is a 15% discount. Find the sale price.
Model response: Discount = 15% of £80 = £12. Sale price = £80 - £12 = £68. Or: 80 × 0.85 = £68.
Uses multipliers for percentage change and solves reverse percentage problems (finding the original value).
Example task
After a 20% increase, a price is £96. Find the original price.
Model response: £96 = 120% of original. Original = 96/1.2 = £80.
Solves compound interest problems using the formula A = P(1 + r)ⁿ and compares simple and compound interest.
Example task
£5,000 is invested at 4% compound interest for 3 years. Find the total amount.
Model response: A = 5000 × 1.04³ = 5000 × 1.124864 = £5,624.32.
Solves problems involving repeated proportional change, depreciation and growth models, and evaluates financial products critically.
Example task
A car depreciates by 15% each year. After how many complete years will it be worth less than half its original value?
Model response: After n years: value = P × 0.85ⁿ. Need 0.85ⁿ < 0.5. Taking logs: n > log(0.5)/log(0.85) = -0.301/-0.0706 = 4.27. So after 5 complete years. Check: 0.85⁴ = 0.522 (>0.5), 0.85⁵ = 0.444 (<0.5) ✓.
Delivery rationale
Secondary maths process concept — problem-solving benefits from structured AI delivery with facilitator for extended reasoning.
Direct and Inverse Proportion
knowledge AI DirectMA-KS4-C015
Recognising and applying direct proportion (y = kx) and inverse proportion (y = k/x), including algebraic, graphical, and numerical representations; extending to y = kxⁿ relationships.
Teaching guidance
Always begin with the unitary method (find the value for one unit, then multiply) before formalising as an equation. Direct proportion produces a straight-line graph through the origin; inverse proportion produces a hyperbola — visual representations reinforce conceptual understanding. For Higher, pupils should be able to identify which type of proportion applies from a given table of values by testing ratios and products.
Common misconceptions
Pupils frequently confuse direct and inverse proportion — both involve one quantity changing as another changes, but in opposite directions. Many pupils identify y = kx as direct proportion but do not recognise y = kx² as also a proportional relationship (just not linear). The phrase 'y is inversely proportional to x' is sometimes misread as 'y = 1/x' rather than 'y = k/x'.
Difficulty levels
Recognises direct proportion (when one quantity doubles, the other doubles) and can solve simple proportion problems using the unitary method.
Example task
5 pens cost £3.50. How much do 8 pens cost?
Model response: 1 pen = £3.50/5 = £0.70. 8 pens = 8 × £0.70 = £5.60.
Writes direct proportion as y = kx and inverse proportion as y = k/x, finding the constant k from given information.
Example task
y is directly proportional to x. When x = 4, y = 20. Find y when x = 7.
Model response: y = kx. 20 = 4k, so k = 5. When x = 7: y = 35.
Works with direct and inverse proportion in algebraic, graphical and tabular forms, identifying the type from given data.
Example task
When x is doubled, y is halved. What type of proportion is this? Write the equation if y = 12 when x = 3.
Model response: Inverse proportion: y = k/x. 12 = k/3, k = 36. y = 36/x. Check: when x = 6 (doubled), y = 36/6 = 6 (halved) ✓.
Works with non-linear proportion (y ∝ x², y ∝ 1/x², y ∝ √x) and solves complex problems in applied contexts. (Higher tier)
Example task
The force F between two magnets is inversely proportional to the square of the distance d between them. When d = 2 cm, F = 20 N. Find F when d = 5 cm.
Model response: F = k/d². 20 = k/4, so k = 80. When d = 5: F = 80/25 = 3.2 N. The force drops dramatically with distance — from 20 N to 3.2 N when distance increases from 2 to 5 cm.
Delivery rationale
Secondary maths concept — abstract, procedural, and objectively assessable.
Compound Measures and Rates of Change
process AI FacilitatedMA-KS4-C016
Using and interpreting compound units such as speed, density and pressure; interpreting gradient of graphs as a rate of change; solving problems involving compound measure conversions.
Teaching guidance
Use the formula triangle approach (Speed = Distance/Time, Density = Mass/Volume) as scaffolding initially, but develop algebraic rearrangement fluency so pupils can work from any form. Connect gradient of a distance-time graph explicitly to speed — this is a key transferable concept for science. Ensure pupils can perform unit conversions within compound measures (e.g. km/h to m/s).
Common misconceptions
Pupils commonly invert the formula — computing time/distance for speed rather than distance/time. Unit conversions for compound measures are particularly error-prone (e.g. converting m/s to km/h requires multiplying by 3.6 not by 1000/60). The distinction between density as mass/volume versus volume/mass is regularly confused.
Difficulty levels
Can use the formula speed = distance/time in straightforward calculations and knows the standard units.
Example task
A car travels 180 km in 3 hours. What is its average speed?
Model response: Speed = 180/3 = 60 km/h.
Uses compound units (speed, density, pressure) and can rearrange the formulae to find any variable.
Example task
A block has density 8 g/cm³ and volume 15 cm³. Find its mass.
Model response: Mass = density × volume = 8 × 15 = 120 g.
Interprets graphs involving compound measures, including distance-time and speed-time graphs, understanding gradient as rate of change.
Example task
On a distance-time graph, the gradient between t = 2 and t = 5 is 40 km/h. What does this represent?
Model response: The gradient represents speed. Between t = 2 and t = 5, the object was travelling at 40 km/h.
Interprets area under speed-time graphs as distance, works with non-constant rates of change, and solves complex problems involving multiple compound measures.
Example task
A speed-time graph shows a car accelerating uniformly from 0 to 20 m/s in 10 seconds, then travelling at 20 m/s for 30 seconds. Find the total distance.
Model response: Phase 1 (triangle): distance = ½ × 10 × 20 = 100 m. Phase 2 (rectangle): distance = 30 × 20 = 600 m. Total = 700 m. The area under the speed-time graph equals the distance travelled.
Delivery rationale
Secondary maths process concept — problem-solving benefits from structured AI delivery with facilitator for extended reasoning.