What you'll learn
Direct and inverse proportion questions appear regularly across both Foundation and Higher tier AQA GCSE Mathematics papers, typically worth 2-5 marks per question. This topic tests your ability to recognise proportional relationships, form equations, solve problems involving ratio and scale, and interpret real-world contexts. Understanding proportion is essential for topics including graphs, equations, and problem-solving throughout the course.
Key terms and definitions
Direct proportion — Two quantities are directly proportional when they increase or decrease at the same rate. If one quantity doubles, the other also doubles. Written as y ∝ x or y = kx.
Inverse proportion — Two quantities are inversely proportional when one increases at the same rate as the other decreases. If one quantity doubles, the other halves. Written as y ∝ 1/x or y = k/x.
Constant of proportionality (k) — The fixed value that links two proportional quantities in an equation. Finding k is the key step in solving proportion problems.
Proportional symbol (∝) — The symbol meaning "is proportional to", used before forming the full equation with the constant k.
Direct proportion squared/cubed — Advanced relationships where y ∝ x² or y ∝ x³, tested on Higher tier papers.
Graph of proportion — Direct proportion produces a straight line through the origin; inverse proportion produces a reciprocal curve (hyperbola).
Core concepts
Recognising direct proportion relationships
Two quantities are in direct proportion when their ratio stays constant. Real-world examples tested on AQA papers include:
- Cost and number of items purchased (at fixed unit price)
- Distance travelled and time (at constant speed)
- Currency conversion at fixed exchange rates
- Recipes scaled up or down proportionally
- Wages and hours worked (at fixed hourly rate)
The key test: if you divide y by x for any pair of values, you always get the same number (the constant k).
Critical skill for exams: The question may not use the word "proportion" explicitly. Phrases like "varies directly with", "is proportional to", or "at a constant rate" all indicate direct proportion.
Setting up and solving direct proportion problems
Follow this four-step process for every direct proportion question:
- Write the proportional relationship: y ∝ x
- Convert to an equation: y = kx (where k is the constant of proportionality)
- Find k using given values: Substitute known values of x and y, then solve for k
- Answer the question: Use your equation with k to find the unknown value
This method works for all direct proportion contexts. On AQA papers, you must show the proportionality statement (y ∝ x) to earn full marks — writing y = kx immediately without the ∝ step typically loses a method mark.
Recognising inverse proportion relationships
Two quantities are in inverse proportion when their product stays constant. Typical AQA contexts include:
- Time taken and speed to cover a fixed distance (faster speed = less time)
- Number of workers and time to complete a job
- Pressure and volume of gas (at constant temperature)
- Gear ratios in mechanics problems
The key test: if you multiply x and y for any pair of values, you always get the same number (the constant k).
Watch for the word "inversely" — this is your signal to use the inverse proportion method, not direct proportion.
Setting up and solving inverse proportion problems
Follow this parallel four-step process:
- Write the inverse relationship: y ∝ 1/x
- Convert to an equation: y = k/x or xy = k
- Find k using given values: Substitute known values and solve for k
- Answer the question: Use your equation to find the unknown
The form xy = k is often easier for calculation. Both forms are algebraically equivalent and equally acceptable on mark schemes.
Direct proportion with powers (Higher tier only)
Higher tier papers test relationships where:
- y ∝ x² means y = kx² (y is proportional to the square of x)
- y ∝ x³ means y = kx³ (y is proportional to the cube of x)
- y ∝ √x means y = k√x (y is proportional to the square root of x)
Examples from real papers include:
- Area of circles (A ∝ r²) or squares
- Volume of spheres or cubes (V ∝ r³)
- Kinetic energy (E ∝ v²)
- Time for falling objects (t ∝ √h)
The solution method remains identical: write the relationship, form the equation, find k, then solve. Just substitute x² or x³ rather than x alone.
Graphical representation of proportion
Direct proportion graphs:
- Always a straight line passing through the origin (0,0)
- Gradient of the line equals the constant k
- Equation is y = kx (compare to y = mx + c with c = 0)
Inverse proportion graphs:
- Produces a smooth curve called a hyperbola
- Never touches either axis (asymptotic behaviour)
- As x increases, y decreases towards zero
- Shape is the same in all four quadrants for y = k/x
AQA questions may ask you to identify which type of proportion from a graph, or to sketch the relationship. Remember: straight line through origin = direct; smooth curve avoiding axes = inverse.
Inverse proportion with powers (Higher tier only)
More complex inverse relationships include:
- y ∝ 1/x² means y = k/x²
- y ∝ 1/x³ means y = k/x³
Physics contexts appear frequently: gravitational force (F ∝ 1/r²), light intensity (I ∝ 1/d²), sound intensity decreasing with distance squared.
The method stays the same, but take extra care with calculations involving x² or x³ in the denominator.
Worked examples
Example 1: Direct proportion (Foundation/Higher)
Question: The cost of hiring a car is directly proportional to the number of days hired. It costs £180 to hire the car for 6 days.
(a) Find a formula connecting cost (C) and number of days (d). (b) Work out the cost of hiring the car for 10 days.
Solution:
(a) C ∝ d [1 mark for proportionality statement] C = kd [1 mark for equation form]
When C = 180, d = 6: 180 = k × 6 k = 180 ÷ 6 = 30 [1 mark]
Formula: C = 30d [1 mark with correct k]
(b) When d = 10: C = 30 × 10 C = £300 [1 mark]
Total: 5 marks
Example 2: Inverse proportion (Foundation/Higher)
Question: The time taken to paint a fence is inversely proportional to the number of painters working. When 3 painters work together, it takes 4 hours to paint the fence.
How long would it take 6 painters to paint the same fence?
Solution:
Let t = time (hours) and n = number of painters
t ∝ 1/n [1 mark] t = k/n or tn = k [1 mark]
When n = 3, t = 4: 3 × 4 = k k = 12 [1 mark]
When n = 6: t × 6 = 12 t = 12 ÷ 6 t = 2 hours [1 mark]
Total: 4 marks
Example 3: Direct proportion with square (Higher tier only)
Question: y is directly proportional to the square of x. When x = 5, y = 200.
(a) Find a formula for y in terms of x. (b) Calculate the value of y when x = 8.
Solution:
(a) y ∝ x² [1 mark] y = kx² [1 mark]
When x = 5, y = 200: 200 = k × 5² 200 = k × 25 k = 200 ÷ 25 = 8 [1 mark]
Formula: y = 8x² [1 mark]
(b) When x = 8: y = 8 × 8² y = 8 × 64 y = 512 [1 mark]
Total: 5 marks
Common mistakes and how to avoid them
Confusing direct and inverse proportion — Read the question carefully. "Inversely proportional" or contexts where one quantity decreasing causes another to increase signal inverse proportion. When in doubt, test: if doubling one value doubles the other, it's direct; if doubling one halves the other, it's inverse.
Not showing the proportionality statement — Writing y = kx immediately without first writing y ∝ x loses a method mark on AQA papers. Always show the ∝ symbol in your working, even though it feels like an extra step.
Using the wrong form for inverse proportion — Writing y = kx when you mean y = k/x is a fundamental error that leads to completely wrong answers. Check your equation matches your proportionality statement: 1/x must appear in the equation if it appears in the relationship.
Forgetting to find k before solving — You cannot skip the step of calculating k. Each proportion problem has a unique value of k based on the given information. Questions are designed so you must find k from one pair of values before calculating others.
Mixing up x² and 2x in power relationships — When y ∝ x², the equation is y = kx² (x squared), not y = 2kx (2 times x). These are completely different relationships. Always square the entire value of x, not just multiply by 2.
Incorrect manipulation with fractions — When solving y = k/x for x, multiply both sides by x to get xy = k, then divide by y to get x = k/y. Common error: writing x = k - y or x = y/k instead. Use inverse operations correctly: multiplication undoes division.
Exam technique for Direct and inverse proportion
Command words matter: "Show that..." requires you to demonstrate every step leading to the given answer. "Find a formula..." needs the proportionality statement, equation form, calculation of k, and final formula clearly stated. "Calculate..." can be more concise if you've already found k.
Mark allocation guides your working: A 4-mark proportion question typically awards 1 mark for y ∝ x (or y ∝ 1/x), 1 mark for y = kx, 1 mark for finding k, and 1 mark for the final answer. Show four distinct steps to access all marks.
Context clues help identify proportion type: Time and speed for fixed distance = inverse. Cost per item at fixed price = direct. Number of workers and time = inverse. Distance and time at constant speed = direct. Currency exchange = direct. Train these recognition patterns.
Include units in final answers: If the question gives values with units (£, hours, metres, etc.), your answer must include the appropriate unit. This is explicitly mentioned on mark schemes as required for the final mark.
Quick revision summary
Direct proportion (y ∝ x, so y = kx) occurs when quantities increase together at the same rate; the graph is a straight line through the origin. Inverse proportion (y ∝ 1/x, so y = k/x) occurs when one quantity increases as the other decreases; the graph is a smooth curve. Always write the proportionality statement first, convert to an equation, find k from given values, then solve. Higher tier includes squared and cubed relationships. Check which type from context: same direction = direct, opposite direction = inverse.