How to Solve Inversely Proportional Questions
Two quantities are inversely proportional when one increases as the other decreases, so that their product stays constant. This Edexcel IGCSE Maths page shows how to recognise an inverse relationship, find the constant of proportionality k, and use the formula y = k/x (and the square and root laws) to calculate missing values, with worked examples in full exam layout and auto-marked practice below.
How to Solve Inverse Proportion Questions
Two quantities are inversely proportional when one gets bigger as the other gets smaller, so that their product stays constant: double one and the other halves. The method mirrors direct proportion, but the constant divides instead of multiplies.
- Write the relationship with a constant \(k\): \(y \propto \dfrac{1}{x}\) becomes \(y = \dfrac{k}{x}\) (or \(y = \dfrac{k}{x^2}\), \(y = \dfrac{k}{\sqrt{x}}\) for a square or root law).
- Find \(k\) by substituting one known pair of values: here \(k = y \times x\) (the product), or \(y \times x^2\) for a square law.
- 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 \(y\) and rearrange to find \(x\).
Core ideas
Inverse-proportion graphs never touch the axes
A reciprocal curve dives toward both axes as a limit but never reaches them. The product \(xy\) stays constant, so as \(x\) doubles, \(y\) for \(y = \tfrac{k}{x}\) halves.
Worked examples
💡 Example 1: inverse (find then predict)
For two meshed gears, the turning speed \(N\) (rpm) of a gear is inversely proportional to its number of teeth \(T\). A gear with 24 teeth turns at 150 rpm. Find the speed of a meshed gear with 40 teeth.
\(N \propto \dfrac{1}{T}\), so \(N = \dfrac{k}{T}\)
Find \(k\):
\[ \begin{array}{rcl} k &=& N \times T \\ &=& 150 \times 24 \\ &=& 3600 \end{array} \]Formula: \(N = \dfrac{3600}{T}\)
At \(T = 40\):
\[ \begin{array}{rcl} N &=& \dfrac{3600}{40} \\ &=& \mathbf{90}\ \text{rpm} \end{array} \]What's happening?
The constant is the product \(N \times T\). More teeth means a slower turn, so \(N\) falls as \(T\) rises.
💡 Example 2: inverse (work backwards)
The width \(w\) of a rectangle of fixed area is inversely proportional to its length \(L\). When \(L = 8\) cm, \(w = 15\) cm. Find the length when the width is 10 cm.
\(w = \dfrac{k}{L}\)
Find \(k\):
\[ \begin{array}{rcl} k &=& w \times L \\ &=& 15 \times 8 \\ &=& 120 \end{array} \]Formula: \(w = \dfrac{120}{L}\)
Set \(w = 10\) and rearrange:
\[ \begin{array}{rcl} 10 &=& \dfrac{120}{L} \\ L &=& \dfrac{120}{10} \\ &=& \mathbf{12}\ \text{cm} \end{array} \]What's happening?
\(k = 120\) is the fixed area. Given the width, rearrange \(w = \tfrac{k}{L}\) to get \(L = \tfrac{k}{w}\).
💡 Example 3: inverse square \(\left(y \propto \tfrac{1}{x^2}\right)\)
The force \(F\) (newtons) between two magnets is inversely proportional to the square of the distance \(d\) (cm) between them. \(F = 45\) when \(d = 2\). Find \(F\) when \(d = 3\).
\(F = \dfrac{k}{d^2}\)
Find \(k\):
\[ \begin{array}{rcl} k &=& F \times d^2 \\ &=& 45 \times 2^2 \\ &=& 180 \end{array} \]Formula: \(F = \dfrac{180}{d^2}\)
At \(d = 3\):
\[ \begin{array}{rcl} F &=& \dfrac{180}{9} \\ &=& \mathbf{20}\ \text{N} \end{array} \]What's happening?
Square the distance first: \(2^2 = 4\). With an inverse-square law, tripling \(d\) divides \(F\) by \(9\).
💡 Example 4: inverse root \(\left(y \propto \tfrac{1}{\sqrt{x}}\right)\)
A machine sorts \(n\) items per second, inversely proportional to the square root of each item's weight \(w\) (grams). At \(w = 9\) it sorts 8 per second. Find the rate when \(w = 36\).
\(n = \dfrac{k}{\sqrt{w}}\)
Find \(k\):
\[ \begin{array}{rcl} k &=& n \times \sqrt{w} \\ &=& 8 \times \sqrt{9} \\ &=& 24 \end{array} \]Formula: \(n = \dfrac{24}{\sqrt{w}}\)
At \(w = 36\):
\[ \begin{array}{rcl} n &=& \dfrac{24}{\sqrt{36}} \\ &=& \dfrac{24}{6} \\ &=& \mathbf{4}\ \text{per second} \end{array} \]What's happening?
Root the weight first: \(\sqrt{9} = 3\). Choose weights that are perfect squares so the root is whole.
🔑 Key points
- \(y \propto \dfrac{1}{x}\) means \(y = \dfrac{k}{x}\); for powers, \(y \propto \dfrac{1}{x^n}\) means \(y = \dfrac{k}{x^n}\).
- Find \(k\) from a known pair: it is the product \(k = x^n y\) (for \(y = \tfrac{k}{x}\), simply \(k = xy\)).
- Then the formula works both ways: find \(y\) from \(x\), or rearrange for \(x\).
- Deal with the power or root of \(x\) first before dividing by it.
- Every inverse-proportion graph is a reciprocal curve that approaches the axes but never crosses them.
⚠️ Common pitfalls
- Multiplying instead of dividing: inverse proportion is \(y = \tfrac{k}{x}\), not \(y = kx\).
- Finding \(k\) as \(\tfrac{y}{x}\) (that is direct proportion): for inverse, \(k\) is the product \(xy\).
- Forgetting the power: for \(y = \tfrac{k}{x^2}\), use \(x^2\), not \(x\).
- Assuming "twice \(x\) gives half \(y\)" for a square law: that only holds for \(y = \tfrac{k}{x}\).