Direct and inverse proportion

What you'll learn

Direct and inverse proportion questions appear regularly in CIE IGCSE Mathematics papers, testing your ability to recognise and solve problems involving variables that change in relation to one another. This topic builds on your understanding of linear relationships and ratio, extending into more complex algebraic manipulation and real-world problem-solving contexts including speed, density, currency conversion, and compound measures.

Key terms and definitions

Direct proportion — two variables are directly proportional when they increase or decrease at the same rate; if one doubles, the other doubles. Written as y ∝ x or y = kx where k is the constant of proportionality.

Inverse proportion — two variables are inversely proportional when one increases as the other decreases at a related rate; if one doubles, the other halves. Written as y ∝ 1/x or y = k/x where k is the constant of proportionality.

Constant of proportionality — the fixed value k that relates two proportional quantities; found by substituting known values into the proportional relationship.

Proportionality symbol ∝ — mathematical notation meaning "is proportional to"; indicates a relationship exists but does not yet specify the constant.

Square proportion — a relationship where one variable is proportional to the square of another, written as y ∝ x² or y = kx².

Compound proportion — situations involving multiple proportional relationships simultaneously, such as y ∝ x/z or y ∝ x²/z.

Core concepts

Recognising direct proportion

Two quantities are in direct proportion when their ratio remains constant. The graph of directly proportional variables always passes through the origin and forms a straight line.

Key characteristics to identify direct proportion:

• When x = 0, y = 0 (the relationship must pass through the origin)
• The ratio y/x remains constant for all values
• Doubling x results in doubling y
• The equation takes the form y = kx (never y = kx + c, as the +c means it doesn't pass through the origin)

Common exam contexts for direct proportion:

• Currency conversion (£ to $ at a fixed exchange rate)
• Cost of items at a fixed unit price
• Distance travelled at constant speed over time
• Mass and volume of the same substance (constant density)

Setting up and solving direct proportion problems

The standard method follows four clear steps:

1. Write the proportional statement: y ∝ x
2. Convert to equation form: y = kx
3. Find k using given values: substitute known pair of values to calculate k
4. Use the equation: substitute the new value to find the unknown

This systematic approach works for all direct proportion questions and matches the mark scheme expectations on CIE papers.

Recognising inverse proportion

Two quantities are in inverse proportion when their product remains constant. As one variable increases, the other decreases proportionally.

Key characteristics to identify inverse proportion:

• The product xy remains constant for all values
• Doubling x results in halving y
• Tripling x results in dividing y by 3
• The equation takes the form y = k/x or xy = k
• The graph forms a rectangular hyperbola, never touching either axis

Common exam contexts for inverse proportion:

• Speed and time for a fixed distance (distance = speed × time, so time = distance/speed)
• Number of workers and time to complete a job
• Pressure and volume of a gas at constant temperature
• Gear ratios and rotational speeds

Setting up and solving inverse proportion problems

The standard method mirrors direct proportion with one crucial difference:

1. Write the proportional statement: y ∝ 1/x
2. Convert to equation form: y = k/x (or more simply, xy = k)
3. Find k using given values: calculate k = xy using the known pair
4. Use the equation: substitute to find the unknown, remembering k/x or k/y as appropriate

Many students find the xy = k form simpler because it avoids fraction manipulation in step 4.

Square and higher-power proportion

Variables may be proportional to powers other than 1. The most common in CIE IGCSE Mathematics papers is square proportion.

For y ∝ x²:

• The equation becomes y = kx²
• If x doubles, y increases by a factor of 4 (since 2² = 4)
• If x triples, y increases by a factor of 9 (since 3² = 9)

Common exam contexts:

• Area relationships (area of a circle ∝ radius²)
• Kinetic energy ∝ velocity²
• Gravitational force ∝ 1/distance²

The solution method remains identical: state the relationship, form the equation, find k, then solve.

Compound proportion problems

Some CIE IGCSE questions combine multiple proportional relationships. For example, y might be proportional to x and inversely proportional to z simultaneously.

Written as: y ∝ x/z, which becomes y = kx/z

Approach systematically:

1. Identify each separate relationship (y increases with x, y decreases with z)
2. Combine into a single proportional statement
3. Form the equation with k
4. Substitute known values to find k
5. Use the complete equation to answer the question

Using proportion to solve practical problems

CIE examiners favour realistic contexts. When tackling word problems:

• Read carefully to identify which variables change and which remain constant
• Determine whether the relationship is direct (both increase together) or inverse (one increases, the other decreases)
• Extract the numerical information systematically
• Write your proportional statement before calculating
• Include units in your final answer where appropriate

Extended questions may require you to:

• Set up proportion from a written description
• Complete a table of values
• Plot a graph to verify proportionality
• Use your model to make predictions or solve problems

Worked examples

Example 1: Direct proportion (currency conversion)

Question: The cost of £5 is equivalent to $6.15. Assuming the exchange rate remains constant, find: (a) the value in dollars of £8 (b) the value in pounds of $24.60

Solution:

(a) Let D = dollars and P = pounds

D ∝ P

D = kP

When P = 5, D = 6.15: 6.15 = k × 5 k = 6.15 ÷ 5 = 1.23

Equation: D = 1.23P

When P = 8: D = 1.23 × 8 = $9.84 [2 marks]

(b) Using D = 1.23P, when D = 24.60: 24.60 = 1.23 × P P = 24.60 ÷ 1.23 = £20 [2 marks]

Example 2: Inverse proportion (speed and time)

Question: A car travels between two cities. At an average speed of 60 km/h, the journey takes 2.5 hours. How long would the journey take at an average speed of 75 km/h?

Solution:

Let s = speed (km/h) and t = time (hours)

Since distance is constant, time is inversely proportional to speed:

t ∝ 1/s

t = k/s or st = k

When s = 60, t = 2.5: k = 60 × 2.5 = 150

Equation: st = 150

When s = 75: 75 × t = 150 t = 150 ÷ 75 = 2 hours [3 marks]

Example 3: Square proportion

Question: The energy E (in joules) stored in a spring is proportional to the square of its extension x (in metres). When the extension is 0.4 m, the energy stored is 3.2 J.

(a) Find the formula for E in terms of x. (b) Calculate the energy stored when the extension is 0.7 m. (c) Find the extension when the energy stored is 12.8 J.

Solution:

(a) E ∝ x²

E = kx²

When x = 0.4, E = 3.2: 3.2 = k × (0.4)² 3.2 = k × 0.16 k = 3.2 ÷ 0.16 = 20

E = 20x² [2 marks]

(b) When x = 0.7: E = 20 × (0.7)² E = 20 × 0.49 = 9.8 J [2 marks]

(c) When E = 12.8: 12.8 = 20x² x² = 12.8 ÷ 20 = 0.64 x = √0.64 = 0.8 m [2 marks]

Common mistakes and how to avoid them

• Assuming all linear relationships are direct proportion — Students see y = 3x + 2 and treat it as direct proportion. Correction: Direct proportion requires the relationship to pass through the origin (no constant term). The equation must be y = kx, not y = kx + c.

• Confusing when to use direct versus inverse proportion — In speed/time problems, students incorrectly set up time ∝ speed. Correction: Think about what happens in reality — if speed increases, time decreases, indicating inverse proportion (t ∝ 1/s).

• Forgetting to find k before solving — Students jump straight to setting up a ratio without finding the constant. Correction: Always complete all four steps: state the relationship, form the equation, find k, then use it. This structured approach earns method marks even if calculation errors occur.

• Incorrect manipulation of k in inverse proportion — When using y = k/x, students substitute incorrectly when solving for x. Correction: Rearrange to xy = k or x = k/y before substituting values. The product form xy = k often simplifies calculations.

• Not squaring or square-rooting correctly in square proportion — Students forget to square when finding k, or forget to square root when working backwards. Correction: Write down explicitly whether you're substituting into y = kx² or rearranging to x = √(y/k).

• Omitting units or using inconsistent units — Mixing km/h with m/s, or pounds with pence within the same calculation. Correction: Check units before starting; convert all measurements to the same unit system. Include the appropriate unit in your final answer.

Exam technique for Direct and inverse proportion

• Command word "show that" appears frequently in proportion questions. You must show every step of working, including writing the proportional statement (y ∝ x), forming the equation (y = kx), calculating k explicitly, and then demonstrating the final calculation reaches the given answer. Simply stating the answer earns no marks.

• Method marks are available even with numerical errors. CIE mark schemes award marks for correctly identifying the type of proportion (M1), correctly forming the equation (M1), and applying the correct method to find k or the final answer (M1), with the final accuracy mark (A1) dependent on correct calculation. Show clear working to maximise partial credit.

• Table-completion questions test proportion recognition. If a table shows x and y values, calculate y/x for several pairs (direct proportion) or xy for several pairs (inverse proportion). If these ratios/products are constant, identify the relationship and complete missing values using k.

• Multi-step problems require proportion as one component of a longer question. Identify where proportion applies, complete that calculation, then use the result in subsequent parts. Extended questions may combine proportion with percentages, ratios, or equation solving — tackle each element separately and methodically.

Quick revision summary

Direct proportion (y ∝ x): both variables increase together at a constant ratio; equation y = kx passes through origin. Inverse proportion (y ∝ 1/x): one variable increases as the other decreases; product xy = k remains constant. Always follow the four-step method: state relationship, form equation, find k using given values, solve for unknown. Square proportion uses y = kx². Common contexts include currency conversion, speed-time-distance, and scaling problems. Show all working to earn method marks even if calculation errors occur.