Ratio, Proportion and Rates of Change

What you'll learn

This topic assesses your ability to work with ratios, solve proportion problems, and handle rates of change including percentages and compound measures. You'll need to simplify and compare ratios, share quantities in given ratios, work with direct and inverse proportion, calculate percentage changes, and understand compound measures like speed and density. These skills appear across both Foundation and Higher tier papers.

Key terms and definitions

Ratio — a comparison of two or more quantities showing the relative size of one quantity to another, written in the form a:b

Proportion — a statement showing that two ratios are equal; when quantities are in proportion, they maintain a constant multiplicative relationship

Direct proportion — when two quantities increase or decrease at the same rate; if y is directly proportional to x, then y = kx where k is the constant of proportionality

Inverse proportion — when one quantity increases as another decreases at a related rate; if y is inversely proportional to x, then y = k/x

Compound measure — a measure involving two or more units combined, such as speed (distance/time), density (mass/volume), or pressure (force/area)

Percentage change — the increase or decrease in a quantity expressed as a percentage of the original value

Scale factor — the multiplier used to enlarge or reduce a quantity in direct proportion

Unit rate — a ratio comparing a quantity to one unit of another quantity, such as price per kilogram or miles per hour

Core concepts

Simplifying and writing ratios

Ratios must be expressed in their simplest form by dividing all parts by their highest common factor (HCF). Always ensure ratios use the same units before simplifying.

Method for simplifying ratios:

1. Convert all quantities to the same units
2. Find the HCF of all parts
3. Divide each part by the HCF
4. Write in the form a:b or a:b:c

Example: Simplify 250 g : 1.5 kg

• Convert to same units: 250 g : 1500 g
• HCF is 250
• Divide both parts: 1 : 6

Comparing ratios requires converting them to the same total or finding equivalent ratios. To compare 2:3 with 3:5, express both with a common second part (e.g., 2:3 = 10:15 and 3:5 = 9:15) or use fractions (2/5 compared to 3/8).

Sharing in a ratio

To divide a quantity in a given ratio, calculate the total number of parts, then find the value of one part.

Sharing method:

1. Add the ratio parts to find the total parts
2. Divide the quantity by total parts to find one part
3. Multiply each ratio number by the value of one part

For three-part ratios, the principle remains identical. To share £420 in the ratio 3:5:6, total parts = 14, so one part = £420 ÷ 14 = £30. The shares are £90, £150, and £180.

Finding the whole from a part: If one share is known, work backwards. If 3 parts equal £90, then 1 part = £30, so the total (14 parts) = £420.

Direct and inverse proportion

Direct proportion exists when y/x remains constant. Doubling x doubles y. The graph of directly proportional quantities is a straight line through the origin.

Solving direct proportion problems:

1. Identify the constant of proportionality: k = y/x
2. Write the relationship: y = kx
3. Use this to find unknown values

Example: If 5 kg of apples cost £7.50, find the cost of 8 kg.

• Set up proportion: cost/weight is constant
• £7.50/5 kg = cost/8 kg
• cost = (£7.50 × 8)/5 = £12

Inverse proportion exists when xy remains constant. Doubling x halves y. Written as y ∝ 1/x or y = k/x.

Solving inverse proportion problems:

1. Find the constant: k = xy
2. Write the relationship: y = k/x
3. Calculate unknown values

Example: 6 workers complete a job in 8 days. How long for 4 workers?

• Constant = 6 × 8 = 48 worker-days
• Time = 48/4 = 12 days

Percentages and percentage change

Finding a percentage of an amount: Convert the percentage to a decimal and multiply.

• 17% of £240 = 0.17 × 240 = £40.80

Percentage change formula: Percentage change = (change/original) × 100%

Increase and decrease: Use multipliers for efficiency.

• Increase by 15%: multiply by 1.15
• Decrease by 23%: multiply by 0.77

Reverse percentages find the original amount when the final value after a percentage change is known.

Method:

1. Identify what percentage the final value represents
2. Divide by this decimal to find 100%

Example: After a 12% increase, a price is £89.60. Find the original.

• £89.60 represents 112% (100% + 12%)
• Original = £89.60 ÷ 1.12 = £80

Compound interest involves repeated percentage increases: Final amount = P × (1 + r/100)^n

where P is principal, r is rate (%), and n is number of time periods.

Compound measures

Speed = distance/time (units: m/s, km/h, mph)

Density = mass/volume (units: g/cm³, kg/m³)

Pressure = force/area (units: N/m², pascals)

Use the triangle method to rearrange formulas:

• For speed: D at top, S and T at bottom (D = S × T)
• For density: M at top, D and V at bottom (M = D × V)

Unit conversions are essential:

• 1 km/h = 1000 m/3600 s (simplifies to 5/18 m/s)
• To convert km/h to m/s: multiply by 5/18
• To convert m/s to km/h: multiply by 18/5

Average speed = total distance/total time (not the mean of individual speeds unless times are equal)

Rates and best value

Unit pricing helps compare value by calculating cost per unit (e.g., pence per 100 g, £ per litre).

Method:

1. Calculate price per standard unit for each option
2. Compare values
3. Select the lowest unit price for best value

Currency conversion uses direct proportion: £1 = $1.27 means £50 = $63.50

Exchange rates may involve a commission or service charge, requiring two calculations: apply the rate, then subtract fees.

Working with rates of change: Gradients on distance-time graphs show speed; steeper gradients indicate higher speeds. Horizontal sections show zero speed (stationary).

Worked examples

Example 1: Ratio and proportion problem (Foundation/Higher, 4 marks)

Question: Emma, James and Sarah share £540 in the ratio 4:5:6. James spends 3/5 of his share. How much does James have left?

Solution:

• Total parts = 4 + 5 + 6 = 15 [1 mark for method]
• One part = £540 ÷ 15 = £36 [1 mark]
• James's share = 5 × £36 = £180 [1 mark]
• Amount left = 2/5 × £180 = £72 [1 mark]

Answer: £72

Example 2: Inverse proportion (Higher, 3 marks)

Question: The time taken to complete a journey is inversely proportional to the average speed. At 60 km/h, the journey takes 2.5 hours. How long would the journey take at 75 km/h?

Solution:

• Since time ∝ 1/speed, then time × speed = constant
• Constant (k) = 2.5 × 60 = 150 [1 mark for method]
• At 75 km/h: time = 150 ÷ 75 [1 mark]
• time = 2 hours [1 mark]

Answer: 2 hours

Example 3: Compound measures with conversion (Higher, 5 marks)

Question: A rectangular metal sheet measures 80 cm by 50 cm by 0.5 cm. The density of the metal is 8.4 g/cm³. Calculate the mass of the sheet in kg.

Solution:

• Volume = 80 × 50 × 0.5 = 2000 cm³ [1 mark]
• Using density = mass/volume, rearrange: mass = density × volume [1 mark]
• Mass = 8.4 × 2000 = 16 800 g [1 mark]
• Convert to kg: 16 800 ÷ 1000 = 16.8 kg [1 mark for conversion, 1 mark for answer]

Answer: 16.8 kg

Common mistakes and how to avoid them

• Not converting to the same units before simplifying ratios — Always check units first. Converting 2 hours : 45 minutes requires both in minutes (120:45 = 8:3)

• Finding one part incorrectly when sharing — Ensure you add all parts of the ratio. For 2:3:7, total parts = 12, not 10

• Confusing direct and inverse proportion — Ask: do both increase together (direct) or does one increase as the other decreases (inverse)? Test with simple values

• Using the wrong percentage as the denominator in reverse percentage problems — When a value after a 20% increase is £72, divide by 1.2 (not 0.8) because £72 represents 120%

• Calculating mean speed incorrectly — Average speed requires total distance ÷ total time, not the mean of two speeds

• Forgetting to use brackets with compound interest — The formula requires (multiplier)^n, not multiplying the multiplier by n

Exam technique for "Ratio, Proportion and Rates of Change"

• Show clear working for ratio problems — Examiners award method marks for calculating total parts and finding one part, even if the final answer is incorrect. Write "Total parts = ..." explicitly

• Identify proportion type early — Questions often contain key phrases: "directly proportional" or "inversely proportional." Underline these and write the correct formula (y = kx or y = k/x) before starting

• Check units in compound measure questions — Multi-mark questions on speed, density or pressure typically award 1 mark for correct unit conversion. Show this step separately

• Use multipliers for percentage questions — Writing "multiply by 1.15" is clearer than "find 15% then add" and reduces calculation errors. Higher tier mark schemes expect efficient methods

Quick revision summary

Simplify ratios by converting to the same units and dividing by the HCF. Share quantities by finding total parts, calculating one part, then multiplying. Direct proportion maintains constant y/x; inverse proportion maintains constant xy. Calculate percentage change as (change/original) × 100%. For reverse percentages, divide by the decimal representing the final percentage. Use compound interest formula for repeated growth. Compound measures include speed (distance/time) and density (mass/volume). Always check units match before calculating, and show clear method steps for maximum marks.