What you'll learn
This topic forms a substantial component of CIE IGCSE Mathematics papers, testing your ability to work with ratios, solve proportion problems, and apply rates of change in real-world contexts. You'll encounter these concepts in both Paper 2 (Extended) and Paper 4 (Extended), often combined with algebraic manipulation and problem-solving. Mastery of this content is essential for tackling currency conversion, scale drawing, density calculations, and percentage change questions.
Key terms and definitions
Ratio — A comparison of two or more quantities of the same type, written in the form a:b, showing how many times one value contains another.
Proportion — A statement that two ratios are equal, often written as a/b = c/d or a:b = c:d, fundamental to solving scaling problems.
Direct proportion — When two quantities increase or decrease at the same rate, such that y = kx where k is the constant of proportionality.
Inverse proportion — When one quantity increases as another decreases, following the relationship y = k/x or xy = k.
Rate — A ratio that compares quantities with different units, such as speed (distance per time) or density (mass per volume).
Percentage change — The relative increase or decrease in a quantity expressed as a percentage of the original value.
Scale factor — The multiplier used to enlarge or reduce quantities proportionally, crucial in map work and similar figures.
Unit rate — A rate expressed with a denominator of 1, such as price per kilogram or kilometres per hour.
Core concepts
Simplifying and working with ratios
Ratios express relationships between quantities and must always compare like units. To simplify ratios, divide all parts by their highest common factor (HCF):
- Simplifying ratios: 24:36 = 24÷12:36÷12 = 2:3
- Three-part ratios: 15:20:25 = 3:4:5 (dividing by 5)
- Converting units first: 2 kg:500 g becomes 2000 g:500 g = 4:1
To divide a quantity in a given ratio:
- Add the ratio parts together to find total shares
- Divide the quantity by total shares to find one share
- Multiply each ratio part by the value of one share
For example, dividing £240 in the ratio 3:5:
- Total shares = 3 + 5 = 8
- One share = £240 ÷ 8 = £30
- First part = 3 × £30 = £90
- Second part = 5 × £30 = £150
Direct proportion problems
When quantities are in direct proportion, their ratio remains constant. If x doubles, y doubles; if x is multiplied by 3, y is multiplied by 3.
The unitary method works efficiently for these problems:
- Find the value for one unit
- Multiply to find the required amount
CIE IGCSE Mathematics papers frequently test direct proportion through:
- Recipe scaling (ingredients for different numbers of servings)
- Currency conversion (exchanging money at fixed rates)
- Wages and time worked
- Bulk buying and unit costs
The algebraic form y = kx means the graph passes through the origin with constant gradient k. To find k, substitute known values of x and y.
Inverse proportion calculations
In inverse proportion, as one quantity increases, the other decreases so that their product remains constant. The relationship xy = k (constant) characterises these problems.
Common exam contexts include:
- Workers completing a task (more workers, less time)
- Speed and journey time for fixed distances
- Pressure and volume relationships
Method for solving inverse proportion:
- Identify that the product of the two quantities is constant
- Calculate k using initial values: k = x₁y₁
- Use k = x₂y₂ to find the unknown value
For instance, if 6 workers take 15 days to complete a job, how long for 10 workers?
- k = 6 × 15 = 90
- 10 × time = 90
- time = 9 days
Percentage calculations in context
CIE IGCSE Mathematics assesses various percentage applications:
Percentage increase/decrease:
- New value = original × (1 + percentage/100) for increase
- New value = original × (1 - percentage/100) for decrease
- A 15% increase: multiply by 1.15
- A 20% decrease: multiply by 0.80
Finding percentage change: Percentage change = (change ÷ original value) × 100
Reverse percentage problems require working backwards when given the final amount:
- Original value = final value ÷ multiplier
- If £84 is the price after a 20% discount: original = 84 ÷ 0.80 = £105
Compound interest and repeated percentage change: Final amount = P(1 + r/100)ⁿ where P is principal, r is rate, n is number of time periods
Speed, distance and time calculations
The fundamental relationship speed = distance/time generates three formulas:
- Speed = Distance ÷ Time
- Distance = Speed × Time
- Time = Distance ÷ Speed
These are commonly tested using:
- Average speed over journeys with multiple stages
- Conversion between units (km/h to m/s: divide by 3.6)
- Distance-time graphs interpretation
- Meeting problems (two objects travelling towards each other)
For multi-stage journeys, calculate average speed using: Average speed = total distance ÷ total time
Never calculate average speed by averaging the speeds.
Rates involving compound measures
Density = mass/volume creates problems requiring formula rearrangement:
- Mass = Density × Volume
- Volume = Mass ÷ Density
Population density = population/area follows identical logic.
Pressure, force and area calculations may appear in Extended papers: Pressure = Force ÷ Area
All rate problems require consistent units throughout. Convert before calculating, then convert the answer if needed.
Map scales and scale drawings
Scale represents the ratio of drawing length to actual length. Scales appear in three formats:
Ratio scales: 1:50000 means 1 cm represents 50000 cm (500 m) Statement scales: 1 cm represents 5 km Linear scales: A marked scale bar
To solve scale problems:
- Write the scale as a ratio in the same units
- Set up a proportion equation
- Solve for the unknown
For area calculations with scale drawings: Actual area = Drawing area × (scale factor)²
For volume: Actual volume = Drawing volume × (scale factor)³
Worked examples
Example 1: Dividing in a ratio with algebraic elements
The angles in a quadrilateral are in the ratio 2:3:4:6. Find the size of the largest angle.
Solution: Total angle sum in a quadrilateral = 360°
Ratio parts: 2 + 3 + 4 + 6 = 15 parts
One part = 360° ÷ 15 = 24°
Largest angle = 6 × 24° = 144°
[2 marks: 1 for method showing 15 parts or correct part value, 1 for correct answer]
Example 2: Inverse proportion with currency context
A group of 8 students plan a trip costing $480. If only 6 students go, how much will each student pay?
Solution: This is inverse proportion: fewer students means more cost per person.
Original: 8 students, $480 ÷ 8 = $60 each
Method 1 (using constant total): Total cost remains $480 6 students: $480 ÷ 6 = $80 each
Method 2 (inverse proportion): Number of students × cost per person = constant 8 × 60 = 480 6 × cost = 480 Cost = 480 ÷ 6 = $80
[3 marks: 1 for recognising fixed total or inverse proportion, 1 for correct method, 1 for answer]
Example 3: Compound percentage change
A car worth $15000 depreciates by 12% each year. Calculate its value after 3 years.
Solution: Depreciation means multiplying by (1 - 0.12) = 0.88 each year
After 3 years: Value = 15000 × (0.88)³
Value = 15000 × 0.681472
Value = $10222.08
Value = $10222 (to nearest dollar)
[3 marks: 1 for using 0.88 or equivalent, 1 for raising to power 3, 1 for correct answer]
Common mistakes and how to avoid them
Mixing units in ratios — Students write 2 kg:500 g as 2:500. Always convert to the same units before simplifying: 2000 g:500 g = 4:1.
Adding ratios incorrectly — When combining 2:3 with 3:5, students might write 5:8. These ratios compare different quantities and cannot be combined without context. If they share a middle term representing the same quantity, scale to make middle terms equal first.
Confusing direct and inverse proportion — More workers complete a job faster (inverse), but more workers need more wages (direct). Check whether quantities increase together or oppositely.
Calculating average speed incorrectly — Taking the mean of two speeds (30 km/h and 40 km/h gives 35 km/h) is wrong. Calculate total distance ÷ total time instead.
Reversing percentage calculations wrongly — After a 20% increase to £60, students calculate £60 - 20% = £48 for the original. The correct method: £60 ÷ 1.20 = £50, because the original was multiplied by 1.20.
Scale factor errors with area and volume — If the linear scale is 1:50, students use 1:50 for areas too. Area scales by (scale factor)² and volume by (scale factor)³. With scale 1:50, areas scale by 1:2500.
Exam technique for Ratio, proportion and rates of change
Command words matter: "Express in the form 1:n" requires you to manipulate the ratio so the first part equals 1 by dividing through. "Calculate the ratio" usually accepts simplified whole number form. "Hence" or "Using your answer" signals that previous working connects to the current part.
Show clear working for method marks: Even with incorrect arithmetic, a clear method showing "total parts = 8" or "constant = 240" earns marks. Write the formula explicitly before substituting values, especially for compound measures like density or speed.
Unit conversion awareness: Check units in the question and units required for the answer. Currency, metric-imperial, and time unit conversions appear frequently. Convert at the start or end, never mid-calculation.
Read multi-step problems carefully: Many proportion questions embed several steps. A typical 5-mark question might require: unit conversion, finding a constant of proportionality, applying it to new values, then converting back. Plan before calculating.
Quick revision summary
Simplify ratios by dividing by HCF after converting to matching units. Divide quantities by adding ratio parts to find one share. Direct proportion: y = kx, quantities increase together; inverse proportion: xy = k, one increases as the other decreases. Percentage multipliers: increase by 15% means ×1.15, decrease by 15% means ×0.85. Reverse percentages: divide by the multiplier. Speed = distance/time; density = mass/volume. Map scales with ratio 1:n mean 1 cm represents n cm in reality. Area scales by (scale factor)², volume by (scale factor)³. Always check units match before calculating.