National 5 Mathematics: expressions and formulae
Surds, indices and algebraic manipulation turn up on both National 5 papers, and on the non-calculator paper there is nothing to fall back on — so the marks turn on whether the manipulation is automatic. This set drills the ten things that come up over and over: simplifying a surd, the index laws, a fractional index, expanding a pair of brackets, factorising a trinomial and a difference of two squares, cancelling an algebraic fraction, changing the subject of a formula, rounding to significant figures, and substituting a negative value.
Every wrong option is a slip candidates actually make rather than a random number — a sign lost when the middle terms are collected, the root or the power dropped from a fractional index, the wrong number of significant figures kept, a negative squared and left negative. The explanations show the working line by line and, in most items, name which specific mistake produces the closest wrong answer, so a wrong click tells you what went wrong and not just that something did.
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- Simplify a surd by pulling out the largest square factor
- Apply the index laws for multiplying, dividing and raising a power to a power
- Evaluate a fractional index by taking the root first and then the power
- Expand a pair of brackets and collect the middle terms without losing a sign
- Factorise a trinomial and recognise a difference of two squares
- Simplify an algebraic fraction by factorising and cancelling
- Change the subject of a formula using inverse operations in the right order
- Round to significant figures, counting from the first non-zero digit
- Substitute negative values correctly, keeping a squared negative positive
Evaluate $8^{\frac{2}{3}}$ as a whole number. — The denominator 3 is the root and the numerator 2 is the power, so $8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4$.
Sample question
Given the formula for the volume of a cone $V = \frac{1}{3}\pi r^2 h$, make $h$ the subject.
- $h = \frac{3V}{\pi r^2}$
- $h = \frac{V}{3\pi r^2}$
- $h = \frac{3V}{(\pi r)^2}$
- $h = \frac{V - 3}{\pi r^2}$
See the answer
$h = \frac{3V}{\pi r^2}$
Multiply both sides by 3 to get $3V = \pi r^2 h$. Then divide by $\pi r^2$ to get $h = \frac{3V}{\pi r^2}$.
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↑ National 5 and Higher (Scotland)
- National 5 Maths: Trig Identities, the Sine Rule for an Angle and Bearings
- National 5 Maths: Trigonometric Graphs, Related Angles and Equations
- National 5 Maths: Circle and Shape Geometry, 3D Pythagoras and Volume
- National 5 Maths: Quadratic Formula, Completing the Square and Sketching
- National 5 Maths: Equations and Inequations in Context
- National 5 Mathematics: geometry and trigonometry
National 5 and Higher qualifications are awarded by Qualifications Scotland, which took over the awarding functions of the Scottish Qualifications Authority (SQA) in February 2026. Zestly is not affiliated with or endorsed by Qualifications Scotland or the SQA. These materials are our own and reproduce no past-paper content.