What you'll learn
This topic forms a substantial component of Edexcel GCSE Mathematics, accounting for approximately 25% of examination content across both Foundation and Higher tiers. Questions test your ability to work with ratios, solve direct and inverse proportion problems, handle percentage change, compound measures, rates of change, and apply these to real-world contexts. Mastery of this topic requires fluency in multiple calculation methods and the ability to recognise which approach suits each problem.
Key terms and definitions
Ratio — a comparison between two or more quantities showing how many times one value contains another, written in the form a:b or a:b:c.
Proportion — the equality of two ratios, indicating that two ratios are equivalent (e.g., 2:3 = 4:6).
Direct proportion — when two quantities increase or decrease at the same rate; if one doubles, the other doubles (y ∝ x).
Inverse proportion — when one quantity increases at the same rate as another decreases; if one doubles, the other halves (y ∝ 1/x).
Compound measure — a measure formed from two or more other measures, such as speed (distance/time) or density (mass/volume).
Rate of change — how quickly one quantity changes in relation to another, often represented as a gradient on a graph.
Percentage multiplier — a decimal used to calculate percentage increases or decreases in a single step (e.g., 1.15 for 15% increase).
Scale factor — the multiplier used to enlarge or reduce quantities proportionally.
Core concepts
Simplifying and sharing in ratios
Ratios must be simplified to their lowest terms by dividing all parts by their highest common factor. For 24:36, divide both by 12 to give 2:3.
When sharing amounts in a given ratio:
- Add all ratio parts together to find total parts
- Divide the total amount by the number of parts to find one part
- Multiply each ratio part by this value
Example: Share £240 in the ratio 3:5.
- Total parts = 3 + 5 = 8
- One part = £240 ÷ 8 = £30
- First share = 3 × £30 = £90
- Second share = 5 × £30 = £150
For three-part ratios such as 2:3:5, the same process applies with total parts = 10.
Edexcel papers frequently test the reverse process: given one share and the ratio, find the total amount or other shares.
Direct and inverse proportion
Direct proportion problems require identifying that two quantities change at the same rate. The unitary method provides a reliable approach:
- Find the value of one unit
- Multiply to find the required amount
Alternatively, use the multiplier method by finding what one quantity is multiplied by, then applying the same multiplier to the other.
For algebraic direct proportion where y ∝ x:
- Write y = kx (where k is the constant of proportionality)
- Substitute known values to find k
- Use the equation to find unknown values
Inverse proportion occurs when the product of the two quantities remains constant. If y ∝ 1/x:
- Write y = k/x or xy = k
- Find k using given values
- Solve for unknowns
Example context: If 8 workers complete a job in 12 days, the product is 8 × 12 = 96 worker-days. For 6 workers: 6 × days = 96, so days = 16.
Percentage calculations
Edexcel GCSE Mathematics examinations test percentages extensively within this topic.
Finding a percentage of an amount:
- Convert percentage to decimal (divide by 100)
- Multiply by the amount
- Alternatively, find 1% then multiply
Percentage change:
Percentage increase/decrease = (change ÷ original) × 100
The change is always compared to the original value, not the new value.
Percentage multipliers provide the most efficient method for increase/decrease calculations:
- 15% increase: multiply by 1.15
- 15% decrease: multiply by 0.85
- General increase of x%: multiply by (1 + x/100)
- General decrease of x%: multiply by (1 - x/100)
Reverse percentage problems require working backwards when given the result after a percentage change:
- Identify what percentage the final amount represents (e.g., after 20% increase, final = 120%)
- Divide the final amount by this percentage (as a decimal) to find 100%
Example: After a 35% increase, a price is £189. Original price = £189 ÷ 1.35 = £140.
Compound interest and depreciation
Compound interest occurs when interest is calculated on both the original amount and previously earned interest.
Formula: Final amount = P × (multiplier)^n
Where P = principal amount, n = number of time periods.
For 4% compound interest over 3 years: Final amount = P × (1.04)³
Depreciation uses the same formula with a multiplier less than 1. For 12% annual depreciation: Multiplier = 0.88
Edexcel questions may require finding the original amount, the rate, or the number of years through equation manipulation or trial and improvement.
Compound measures and conversions
Speed = distance ÷ time (common units: m/s, km/h, mph)
Density = mass ÷ volume (common units: g/cm³, kg/m³)
Pressure = force ÷ area (common units: N/m², pascals)
The triangle method helps rearrange these formulas. For speed:
- Cover the quantity you want to find
- The remaining arrangement shows the calculation
Unit conversions require careful attention:
- 1 km/h = 1000 m ÷ 3600 s = 5/18 m/s
- To convert km/h to m/s: multiply by 5/18 or divide by 3.6
- To convert m/s to km/h: multiply by 3.6
For area and volume conversions:
- 1 m² = 10,000 cm² (length conversion squared)
- 1 m³ = 1,000,000 cm³ (length conversion cubed)
Growth and decay graphs
Exponential growth appears when quantities increase by a constant percentage. The graph curves upward, getting steeper.
Exponential decay occurs with constant percentage decrease. The graph curves downward, approaching but never reaching zero.
Both follow the general form y = ka^x where:
- k affects the starting value (y-intercept when x = 0)
- a determines growth rate (a > 1) or decay rate (0 < a < 1)
Edexcel Higher tier papers may require interpreting these graphs, reading values, or identifying the type of relationship shown.
Rates of change on graphs
The gradient of a distance-time graph represents speed.
The gradient of a speed-time graph represents acceleration.
For non-linear graphs, the gradient at a point is found by drawing a tangent and calculating its gradient using:
Gradient = change in y ÷ change in x
The area under a speed-time graph represents distance travelled. For irregular shapes, count squares or divide into triangles and rectangles.
Worked examples
Example 1: Ratio and best buy (Foundation/Higher)
Question: A 400g jar of coffee costs £5.60. A 750g jar costs £9.90. Which jar offers better value for money?
Solution:
400g jar: £5.60 ÷ 400g = £0.014 per gram = 1.4p per gram
750g jar: £9.90 ÷ 750g = £0.0132 per gram = 1.32p per gram
The 750g jar offers better value (lower price per gram).
Alternative method:
Find cost per 100g:
- 400g jar: (£5.60 ÷ 400) × 100 = £1.40 per 100g
- 750g jar: (£9.90 ÷ 750) × 100 = £1.32 per 100g
750g jar is better value. [3 marks]
Example 2: Reverse percentage (Higher)
Question: After a 12% increase, the population of a town is 39,200. What was the population before the increase?
Solution:
12% increase means the new amount represents 112% of the original.
112% = 39,200
1% = 39,200 ÷ 112 = 350
100% = 350 × 100 = 35,000
Alternative method using multipliers:
Original × 1.12 = 39,200
Original = 39,200 ÷ 1.12 = 35,000
The population before the increase was 35,000. [2 marks]
Example 3: Inverse proportion (Higher)
Question: The time taken to complete a journey is inversely proportional to the average speed. When the average speed is 60 mph, the journey takes 2.5 hours. How long would the journey take at an average speed of 75 mph?
Solution:
For inverse proportion: speed × time = constant (k)
Find k: 60 × 2.5 = 150
For 75 mph: 75 × time = 150
Time = 150 ÷ 75 = 2 hours
Algebraic method:
T ∝ 1/S so T = k/S
When S = 60, T = 2.5:
2.5 = k/60, so k = 150
T = 150/S
When S = 75: T = 150/75 = 2 hours [3 marks]
Common mistakes and how to avoid them
Mistake: Adding ratios incorrectly when sharing amounts. Students calculate 3 + 5 = 8 but then divide £240 by 3 or 5 instead of 8.
Correction: Always add all parts of the ratio first to find total parts, then divide the amount by this total.
Mistake: Confusing direct and inverse proportion. Using the wrong relationship leads to answers that are too large or too small.
Correction: Check whether quantities increase together (direct) or one increases while the other decreases (inverse). Test with simple numbers: if workers double, does time double (direct) or halve (inverse)?
Mistake: Calculating percentage change from the wrong value, particularly using the new amount instead of the original.
Correction: Percentage change always uses the original value as the denominator: (change ÷ original) × 100.
Mistake: In reverse percentage problems, subtracting the percentage from the final amount instead of using division.
Correction: Identify what percentage the final amount represents (e.g., 115% after 15% increase), then divide the final amount by this percentage (as a decimal).
Mistake: Unit conversion errors, especially with compound measures. Converting 5 km/h to m/s by only changing km to m.
Correction: Convert both units: 5 km/h = 5000m/3600s = 1.39 m/s (approximately). Remember the shortcut: ÷ 3.6 for km/h to m/s.
Mistake: In compound interest, adding the percentage each year instead of using the multiplier method.
Correction: Use the formula: Final = P × (multiplier)^years. For 3% over 5 years: P × (1.03)⁵, not P × (1 + 0.03 × 5).
Exam technique for Ratio Proportion and Rates of Change
Command words matter: "Calculate" requires a numerical answer with working. "Show that" means you must demonstrate through clear steps how the given answer is reached, with sufficient detail for full marks. "Hence" tells you to use a previous answer. "Determine" requires finding a value through calculation or reasoning.
Show your method clearly: Even if you use a calculator for the final answer, write the calculation you're entering. For a 3-mark question on sharing in a ratio, write the total parts, value of one part, and both shares. Edexcel mark schemes award method marks before accuracy marks.
Check answers make sense: If sharing £200 in ratio 1:3 gives you £150 and £200, something has gone wrong (total exceeds the original). If a 20% increase gives a smaller number, you've used 0.8 instead of 1.2. Quick sense-checks prevent careless errors.
Read units carefully: Edexcel questions frequently require unit conversions. Circle or underline different units in the question. When finding speed, ensure distance and time units match or convert your final answer to the units requested in the question.
Quick revision summary
For ratios, always find total parts before dividing amounts. Use percentage multipliers for efficient calculations: 1.15 for 15% increase, 0.85 for 15% decrease. Reverse percentages require division by the multiplier. Direct proportion means quantities change together; use y = kx. Inverse proportion means one increases as the other decreases; use y = k/x or xy = k. Compound interest uses Final = P × (multiplier)ⁿ. For compound measures, remember speed = distance/time, density = mass/volume. Convert km/h to m/s by dividing by 3.6. Always show full working for method marks.