How to Solve Direct Proportion Questions
Direct proportion means two quantities grow together at a constant rate, written y = kx. This page shows how to solve direct proportion questions for IGCSE Maths: write the relationship, find the constant of proportionality k from a known pair of values, and use it to predict unknown values, including power relationships such as y = kx² and y = kx³. The worked examples follow the exam layout so you can secure full method marks, and the auto-marked practice questions below let you build fluency at your own pace.
How to Solve Direct Proportion Questions
Two quantities are in direct proportion when one is always a fixed multiple of the other: as one grows, so does the other, at a constant rate. The method is the same every time.
- Write the relationship with a constant \(k\): \(y \propto x\) becomes \(y = kx\) (or \(y = kx^2\), \(y = kx^3\), \(y = k\sqrt{x}\) for a power or root law).
- Find \(k\) by substituting one known pair of values and solving.
- Write the full formula with the value of \(k\) filled in.
- Use the formula to predict an unknown: put in the given \(x\) to find \(y\), or the given \(y\) to find \(x\).
Core ideas
Every direct-proportion graph passes through the origin
Linear, square and root laws all start at \((0,0)\): when \(x = 0\), \(y = 0\). What changes is the shape, set by the power of \(x\).
Worked examples
💡 Example 1: linear (find then predict)
The cost \(C\) of petrol is directly proportional to the volume \(V\) (litres). It costs £8.40 for 6 litres. Find the cost of 10 litres.
\(C \propto V\), so \(C = kV\)
Find \(k\): \(\ 8.40 = 6k \Rightarrow k = 1.40\)
Formula: \(\ C = 1.40V\)
At \(V = 10\): \(\ C = 1.40 \times 10 = \mathbf{14.00}\) (£14.00)
What's happening?
Proportional means \(C = kV\). Use the known pair \((6,\ 8.40)\) to find the rate \(k\), then put any volume into the formula.
💡 Example 2: linear (work backwards)
A worker's pay \(P\) is directly proportional to the hours \(h\) worked. The pay is £90 for 6 hours. How many hours give a pay of £120?
\(P = kh\)
Find \(k\): \(\ 90 = 6k \Rightarrow k = 15\)
Formula: \(\ P = 15h\)
Set \(P = 120\): \(\ 120 = 15h \Rightarrow h = \mathbf{8}\) hours
What's happening?
Once you have \(P = 15h\) you can go backwards too: given the pay, divide by \(k\) to find the hours.
💡 Example 3: square law \((y \propto x^2)\)
The kinetic energy \(E\) (joules) of an object is directly proportional to the square of its speed \(v\) (m/s). \(E = 180\) when \(v = 6\). Find \(E\) when \(v = 10\).
\(E = kv^2\)
Find \(k\): \(\ 180 = k(6)^2 = 36k \Rightarrow k = 5\)
Formula: \(\ E = 5v^2\)
At \(v = 10\): \(\ E = 5 \times 100 = \mathbf{500}\) J
What's happening?
Square the speed first: \(6^2 = 36\). With a square law, doubling \(v\) multiplies \(E\) by \(4\).
💡 Example 4: cube law \((y \propto x^3)\)
The mass \(M\) (kg) of a metal sphere is directly proportional to the cube of its radius \(r\) (cm). \(M = 24\) when \(r = 2\). Find \(M\) when \(r = 5\).
\(M = kr^3\)
Find \(k\): \(\ 24 = k(2)^3 = 8k \Rightarrow k = 3\)
Formula: \(\ M = 3r^3\)
At \(r = 5\): \(\ M = 3 \times 125 = \mathbf{375}\) kg
What's happening?
Cube the radius first: \(2^3 = 8\). A cube law grows fast, so a modest rise in \(r\) gives a large rise in \(M\).
🔑 Key points
- \(y \propto x\) means \(y = kx\); for powers, \(y \propto x^n\) means \(y = kx^n\).
- Always find \(k\) first from a known pair: \(k = \dfrac{y}{x^n}\).
- Then the formula works both ways: find \(y\) from \(x\), or \(x\) from \(y\).
- Deal with the power or root of \(x\) first (square it, cube it, root it) before using \(k\).
- Every direct-proportion graph is a curve (or line) through the origin.
⚠ Common pitfalls
- Skipping \(k\): you cannot predict values without the constant of proportionality.
- Forgetting the power: for \(y = kx^2\), substitute \(x^2\), not \(x\).
- Assuming "twice \(x\) gives twice \(y\)" for a square or cube law: that only holds for \(y = kx\).
- Confusing direct with inverse proportion (where \(xy\) is constant, not \(y/x\)).