IGCSE · Practice questions · Changing the subject of a formula

Reverse the operations.

Six original Cambridge-style questions on changing the subject of a formula, from simple linear formulas to roots and repeated subjects.

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

Treat it like an equation.

Fractions: multiply both sides by the denominator first.
Reverse order: undo the last operation done to the subject first.
Whole sides: divide, square or root the whole of each side.
Subject twice (Extended): collect, factorise, divide by the bracket.
Check: put numbers into the old and new formula.

Recall

[2 marks]

Make the letter in brackets the subject. (a) \(P = 4s\) \([s]\) (b) \(c = 5d - 8\) \([d]\)

(a) \(s = \dfrac{P}{4}\) ✓
(b) \(c + 8 = 5d\), so \(d = \dfrac{c + 8}{5}\) ✓

Calculation

[2 marks]

Make \(w\) the subject of \(P = 2(l + w)\).

\(\dfrac{P}{2} = l + w\) ✓
\(w = \dfrac{P}{2} - l = \dfrac{P - 2l}{2}\) ✓

Calculation

[3 marks]

(Extended) (a) Make \(v\) the subject of \(E = \dfrac{1}{2}mv^{2}\).

(b) Find \(v\) when \(E = 144\) and \(m = 8\).

(a) \(2E = mv^{2}\), \(v^{2} = \dfrac{2E}{m}\), \(v = \sqrt{\dfrac{2E}{m}}\) ✓
(b) \(v = \sqrt{\dfrac{288}{8}} = \sqrt{36} = 6\) ✓

Calculation

[3 marks]

(Extended) Make \(r\) the subject of \(V = \dfrac{1}{3}\pi r^{2}h\).

\(3V = \pi r^{2}h\) ✓
\(r^{2} = \dfrac{3V}{\pi h}\) ✓
\(r = \sqrt{\dfrac{3V}{\pi h}}\) ✓

Calculation

[3 marks]

(Extended) Make \(x\) the subject of \(ax + 3 = bx - 7\).

\(3 + 7 = bx - ax\) ✓
\(10 = x(b - a)\) ✓
\(x = \dfrac{10}{b - a}\) ✓

Analysis

[5 marks]

The formula \(F = \dfrac{9C}{5} + 32\) converts a temperature \(C\)°C into \(F\)°F.

(a) Make \(C\) the subject of the formula.

(b) A normal body temperature is 98.6°F. Convert this to °C.

(a) \(F - 32 = \dfrac{9C}{5}\), so \(5(F - 32) = 9C\) and \(C = \dfrac{5(F - 32)}{9}\) ✓
(b) \(C = \dfrac{5 \times 66.6}{9} = 37\)°C ✓
(c) Put \(F = C\): \(C = \dfrac{9C}{5} + 32\), so \(5C = 9C + 160\) ✓
\(-4C = 160\), \(C = -40\). So \(-40\)°C \(= -40\)°F. ✓

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.