ESAT Maths Section 2027: Worked Solutions by Argument Shape

The ESAT maths section is really two separate modules: Mathematics 1, a 27-question, 40-minute module every ESAT candidate sits, and Mathematics 2, a further 27-question, 40-minute module required only by some courses. Neither permits a calculator, and every correct answer scores one mark with nothing deducted for a wrong one. What separates a 5.0 from a 7.0 is rarely a missing formula — it is recognising, within seconds, which kind of mathematical argument a question demands.

Mathematics 1 draws on GCSE-level number, algebra, geometry, statistics and probability. Mathematics 2 assumes all of that and adds calculus, radians, logarithms and the sine and cosine rules. Both modules stay multiple choice throughout, which is exactly why the five reasoning shapes below matter more than any topic list.

That is the organising idea of this page. Instead of grouping problems by topic — "algebra questions", "trigonometry questions" — the five sections below group them by the shape of the reasoning that solves them: substitution that collapses a hard-looking equation into a quadratic, elimination that beats solving when four answer options are already in front of you, a constraint you must spot before the algebra even starts, a demand for an exact value rather than a decimal, and working backwards from the form the answer must take. Every worked problem below is our own, written in the style of the published ESAT specification, not reproduced from a UAT-UK past paper — and every final answer has been checked twice, independently, before publication.

Which Module Tests Which Kind of Reasoning — Mathematics 1 or Mathematics 2?

Before working through the argument shapes, it matters which module you are actually sitting, because several of the techniques below only exist in one of the two. The UAT-UK content specification for assessment in October 2026 and January 2027 sets out five modules in total: Mathematics 1, Biology, Chemistry, Physics and Mathematics 2. "Most courses will require candidates to answer Mathematics 1 and two further modules," the specification states, and which two is set by the university and course, not by the candidate.

Two exclusions are worth knowing precisely because they trip candidates up in both directions. The specification states plainly, for Mathematics 1: "Candidates are not expected to recall or use the sine or cosine rules." Mathematics 2 then requires exactly those two formulae, including the "ambiguous case" where an angle-side-side triangle admits two valid solutions rather than one. And within Mathematics 2's logarithm content, the specification is equally direct the other way: "Questions requiring knowledge of the change of base formula will not be set." A candidate who spends revision time memorising change-of-base manipulations for the ESAT is preparing for a question type that, by the board's own specification, will not appear.

Which Substitution Makes the Question Collapse?

The first shape is the most common trap in a 40-minute module: an equation that looks unfamiliar — a mixed power equation, or a trigonometric equation with both sine and cosine in it — collapses into an ordinary quadratic the moment you name the right substitution. Spotting it early is worth far more than being fast at algebra once you have spotted it.

Worked example 1. Solve 2^2x+1 − 9·2^x + 4 = 0 for x.

Write 2^2x+1 as 2·(2^x)², then let u = 2^x. The equation becomes 2u² − 9u + 4 = 0, an ordinary quadratic. The discriminant is 9² − 4(2)(4) = 81 − 32 = 49, so u = (9 ± 7) / 4, giving u = 4 or u = 0.5. Converting back: 2^x = 4 gives x = 2; 2^x = 0.5 gives x = −1. Answer: x = −1 or x = 2. Check by substitution: at x = 2, 2⁵ − 9(4) + 4 = 32 − 36 + 4 = 0. At x = −1, 2⁻¹ − 9(0.5) + 4 = 0.5 − 4.5 + 4 = 0. Both roots check exactly.

Worked example 2. Solve 2cos²x − sin x − 1 = 0 for 0° ≤ x < 360°.

Two trig functions in one equation is the signal: use sin²x + cos²x = 1 to substitute cos²x = 1 − sin²x, turning the whole equation into one variable. That gives 2(1 − sin²x) − sin x − 1 = 0, which simplifies to −2sin²x − sin x + 1 = 0, or (multiplying by −1) 2sin²x + sin x − 1 = 0. This factorises as (2sin x − 1)(sin x + 1) = 0, so sin x = 0.5 or sin x = −1. sin x = 0.5 gives x = 30° or x = 150° in the given range; sin x = −1 gives x = 270°. Answer: x = 30°, 150°, 270°. Checking x = 270°: cos²(270°) = 0, sin(270°) = −1, so 2(0) − (−1) − 1 = 0, confirmed.

When Does Eliminating Three Options Beat Solving Directly?

Because every ESAT question is multiple choice and nothing is awarded for method, 27 questions in 40 minutes leaves under 90 seconds per question on average. When four numerical options are already printed on the screen, substituting each one into the original equation is frequently faster than solving from first principles — and it is the technique the Factor Theorem exists to formalise, which the specification lists explicitly as assumed Mathematics 2 knowledge.

Worked example 3. Which of the following is a root of 3x² − 5x − 2 = 0: −2, −1/3, 1, or 5/3?

Rather than factorising the quadratic from scratch, substitute each option in turn. At x = −2: 3(4) − 5(−2) − 2 = 12 + 10 − 2 = 20 ≠ 0. At x = −1/3: 3(1/9) − 5(−1/3) − 2 = 1/3 + 5/3 − 2 = 2 − 2 = 0. That already answers the question in two lines of arithmetic — no need to test the remaining two options once one substitution hits zero. Answer: x = −1/3. (For completeness, the quadratic factorises as (3x + 1)(x − 2), confirming the two roots are exactly −1/3 and 2, and 2 was not among the four options offered.)

Worked example 4. Which of the following is not a root of 2x³ − 3x² − 11x + 6 = 0: −2, 1/2, 3, or 4?

Testing a cubic by full algebraic division against four candidates would burn most of a minute. The Factor Theorem says instead: if f(a) = 0, then (x − a) is a factor, so just evaluate f at each option. Let f(x) = 2x³ − 3x² − 11x + 6. f(−2) = 2(−8) − 3(4) − 11(−2) + 6 = −16 − 12 + 22 + 6 = 0 (a root). f(1/2) = 2(1/8) − 3(1/4) − 11(1/2) + 6 = 0.25 − 0.75 − 5.5 + 6 = 0 (a root). f(3) = 2(27) − 3(9) − 33 + 6 = 54 − 27 − 33 + 6 = 0 (a root). f(4) = 2(64) − 3(16) − 44 + 6 = 128 − 48 − 44 + 6 = 42, which is not zero. Answer: x = 4 is not a root. The cubic in fact factorises completely as (x + 2)(2x − 1)(x − 3), confirming the other three values are its only roots.

Which Constraint Must You Spot Before the Algebra Starts?

A different trap produces algebra that is entirely correct and an answer that is entirely wrong, because a constraint attached to the question — a square root cannot be negative, a discriminant must be positive for real roots to exist — rules out a solution the algebra itself never flags. Spotting the constraint has to happen before you accept the algebraic answer, not after.

Worked example 5. Solve √(x + 7) = x − 5.

Squaring both sides gives x + 7 = (x − 5)² = x² − 10x + 25, which rearranges to x² − 11x + 18 = 0, factorising as (x − 2)(x − 9) = 0, so x = 2 or x = 9. But squaring introduces a constraint that was not visible before squaring: a square root is never negative, so the right-hand side x − 5 must also be ≥ 0, meaning x ≥ 5. That constraint eliminates x = 2 immediately, before any further checking. Answer: x = 9 only. Verify: √(9 + 7) = √16 = 4, and 9 − 5 = 4 — both sides agree exactly. At the rejected x = 2, the left side would be √9 = 3, but the right side is 2 − 5 = −3, which a square root can never equal.

Worked example 6. For which values of k does x² + kx + 9 = 0 have two distinct real roots?

Before reaching for the quadratic formula, the constraint that decides the whole question is the discriminant: two distinct real roots require b² − 4ac > 0. Here a = 1, b = k, c = 9, so the condition is k² − 36 > 0, which factorises as (k − 6)(k + 6) > 0. Answer: k < −6 or k > 6. This is worth checking at a boundary: at k = 6, the discriminant is exactly 36 − 36 = 0, giving a single repeated root rather than two distinct ones, which is why 6 itself is excluded from the answer.

Where Does an Exact Answer Beat a Decimal Approximation?

Because no calculator is permitted and every question is multiple choice, the four options on an ESAT maths question are written to the exact value the algebra produces — a surd, a fraction, or a multiple of π. The specification is direct about this: candidates must "calculate exactly with fractions, surds and multiples of π." A decimal produced by rough mental estimation will not match any option on the screen.

Worked example 7. Rationalise the denominator of (2 + √3) / (√3 − 1).

Multiply numerator and denominator by the conjugate, √3 + 1. The denominator becomes (√3 − 1)(√3 + 1) = 3 − 1 = 2. The numerator becomes (2 + √3)(√3 + 1) = 2√3 + 2 + 3 + √3 = 5 + 3√3. Answer: (5 + 3√3) / 2, exactly — not the decimal 5.098, which is what a candidate reaching for a non-existent calculator would be tempted to write down and which would not appear as an option.

Worked example 8. Find the exact value of sin 60° cos 30° + cos 60° sin 30°, without a calculator.

This only needs the exact values the specification lists for 0°, 30°, 45°, 60° and 90°: sin 60° = √3/2, cos 30° = √3/2, cos 60° = 1/2, sin 30° = 1/2. Substituting: (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4. Answer: 1, exactly — no rounding, no decimal, and no addition formula required, just the exact-value table applied twice and added.

Worked example 9. A sector of a circle has radius 6 cm and angle π/3 radians. Find its exact area.

Sector area is ½r²θ, which the Mathematics 2 content on radian measure requires you to apply directly in radians rather than converting to degrees first. Here ½ × 6² × π/3 = ½ × 36 × π/3 = 18 × π/3. Answer: 6π cm², exactly — not the decimal 18.85 cm², which is what the same working rounded early would produce.

How Do You Work Backwards From the Shape of the Answer?

The last shape only exists because the answer options are visible before you start working: sometimes it is faster to expand a candidate answer and check it matches, or to apply a known transformation rule directly to a given point, than to derive the result forwards from nothing.

Worked example 10. Which of the following is the correct factorisation of 6x² + x − 12: (2x + 3)(3x − 4), (2x − 3)(3x + 4), (3x + 2)(2x − 6), or (6x − 4)(x + 3)?

With a leading coefficient of 6 and a constant of −12, there are several possible factor pairs to try forward, which is slow. Working backwards is faster: expand each candidate and check which reproduces 6x² + x − 12. Testing (2x + 3)(3x − 4): 2x × 3x = 6x²; 2x × (−4) = −8x; 3 × 3x = 9x; 3 × (−4) = −12. Combining the middle terms: −8x + 9x = x. Total: 6x² + x − 12 — an exact match. Answer: (2x + 3)(3x − 4). None of the other three options need to be expanded once a match is confirmed, which is the entire point of working backwards from a fixed set of candidates rather than factorising forwards from the original expression.

Worked example 11. The graph of y = f(x) has a minimum point at (2, −3). State the coordinates of the minimum point of y = f(x − 1) + 4.

Rather than trying to reconstruct f(x) itself, apply the two transformations directly to the known point. f(x − 1) is a horizontal shift of +1 (right), so the x-coordinate becomes 2 + 1 = 3. Adding 4 outside the function is a vertical shift of +4 (up), so the y-coordinate becomes −3 + 4 = 1. Answer: (3, 1). This is the same logic as example 10 in a different setting: the four transformation rules in the specification tell you exactly what happens to a known point, so applying the rule directly to the point is faster and less error-prone than deriving and re-analysing an entirely new function.

Frequently Asked Questions About the ESAT Maths Section

Every worked problem above is our own, constructed in the style of the published content specification to demonstrate the argument shape rather than reproduced from a UAT-UK past paper — reproducing past-paper questions wholesale would breach UAT-UK's own guidance that "the sharing of test content either during or after the admissions cycle is strictly prohibited." For structured, syllabus-wide preparation, our ESAT preparation programme works through the full Mathematics 1 and Mathematics 2 specification with a tutor, and our ESAT preparation guide for international students covers registration, test centres and the wider application timeline this page does not. Candidates weighing the ESAT against the separate Test of Mathematics for University Admission should see our TMUA preparation programme — the two assessments are run by the same body but test different content, and are not interchangeable.

Read against the specification directly rather than a summary of it: the official ESAT Content Specification from UAT-UK lists every topic labelled M (Mathematics 1) and MM (Mathematics 2) referenced throughout this page, and the official ESAT test page confirms the format, scoring scale and 2026-2027 sitting dates cited above.