IGCSE · Practice questions · Direct and inverse proportion

Find k first.

Extended. Six original Cambridge-style questions on direct and inverse proportion: forming the equation, finding the constant and using the formula in context.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Equation, constant, formula, answer.

Equation: replace \(\propto\) by \(= k \times\) (direct) or \(= \dfrac{k}{\ldots}\) (inverse).
Power: use the exact one in the question: square, cube, square root, cube root.
Constant: substitute the given pair of values and solve for \(k\).
Formula: rewrite the equation with the number for \(k\).
Answer: substitute; if finding \(x\), undo the power at the end.

Recall

[2 marks]

Write each statement as an equation using a constant \(k\). (a) \(p\) is proportional to the cube of \(q\). (b) \(m\) is inversely proportional to the square root of \(n\).

(a) \(p = kq^3\) ✓
(b) \(m = \dfrac{k}{\sqrt{n}}\) ✓

Calculation

[3 marks]

\(y\) is directly proportional to \(x\). When \(x = 4\), \(y = 18\). (a) Find \(y\) when \(x = 10\). (b) Find \(x\) when \(y = 63\).

\(18 = 4k\), so \(k = 4.5\) and \(y = 4.5x\). ✓
(a) \(y = 4.5 \times 10 = 45\). ✓
(b) \(63 = 4.5x\), so \(x = 14\). ✓

Calculation

[3 marks]

\(y \propto x^2\). When \(x = 5\), \(y = 75\). (a) Find \(y\) when \(x = 8\). (b) Find the positive value of \(x\) when \(y = 27\).

\(75 = 25k\), so \(k = 3\) and \(y = 3x^2\). ✓
(a) \(y = 3 \times 64 = 192\). ✓
(b) \(27 = 3x^2\), so \(x^2 = 9\) and \(x = 3\). ✓

Calculation

[2 marks]

\(y\) is inversely proportional to \(x\). When \(x = 6\), \(y = 8\). Find \(y\) when \(x = 16\).

\(8 = \dfrac{k}{6}\), so \(k = 48\). ✓
\(y = \dfrac{48}{16} = 3\). ✓

Analysis

[4 marks]

The frequency \(f\) hertz of a pendulum is inversely proportional to the square root of its length \(l\) metres. When \(l = 1\), \(f = 0.5\).

(a) Find \(f\) when \(l = 0.25\). (b) Find the length of a pendulum with frequency 0.2 hertz.

\(f = \dfrac{k}{\sqrt{l}}\): \(0.5 = \dfrac{k}{1}\), so \(k = 0.5\). ✓
(a) \(f = \dfrac{0.5}{\sqrt{0.25}} = \dfrac{0.5}{0.5} = 1\) hertz. ✓
(b) \(0.2 = \dfrac{0.5}{\sqrt{l}}\), so \(\sqrt{l} = 2.5\). ✓
\(l = 2.5^2 = 6.25\) m. ✓

Analysis

[5 marks]

The mass \(m\) grams of a steel ball is proportional to the cube of its radius \(r\) cm. A ball of radius 1.5 cm has mass 54 g.

(a) Find a formula for \(m\) in terms of \(r\). (b) Find the mass of a ball of radius 2 cm. (c) Find the radius of a ball of mass 250 g. (d) One ball has twice the radius of another. How many times heavier is it?

(a) \(m = kr^3\): \(54 = k \times 1.5^3 = 3.375k\), so \(k = 16\) and \(m = 16r^3\). ✓
(b) \(m = 16 \times 2^3 = 128\) g. ✓
(c) \(250 = 16r^3\), so \(r^3 = 15.625\) and \(r = \sqrt[3]{15.625} = 2.5\) cm. ✓
(d) \(16(2r)^3 = 8 \times 16r^3\), so it is 8 times heavier. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.