A-Level Maths · Exam guide · OCR A (H240)

OCR A-Level Maths (H240): the exam, the marks, the method

OCR A-Level Mathematics A is examined by three equally weighted papers spanning pure mathematics, statistics and mechanics. This guide explains exactly how the qualification is assessed, where OCR examiners say marks are lost, and works through four original exam-style questions with full solutions and a note on what the examiner is looking for in each.

Dr Nicky Grant · Cambridge PhDA-Level Maths exam guideOCR A · H240

OCR A-Level Mathematics A (H240) is assessed by three written papers, each 100 marks, each 2 hours and each worth one third of the A Level: Paper 01 Pure Mathematics, Paper 02 Pure Mathematics and Statistics, and Paper 03 Pure Mathematics and Mechanics. Marks are split across three assessment objectives, with AO1 (standard techniques) the largest at roughly 60%, and AO2 (reasoning and communication) and AO3 (problem solving and modelling) each around 20%. The most common reasons for lost marks are careless algebra, incomplete reasoning on "show that" and "prove" questions, and answers that do not match the command word used.

Who this guide is for

This is a guide for students sitting OCR A-Level Mathematics A, specification code H240, and for parents trying to understand how the exam works. It assumes you are partway through the course and want to convert the content you already know into marks. The structural details below reflect the current OCR A (H240) specification; the worked questions are written in OCR's style but are entirely original, not reproduced from any past paper. For one-to-one help see A-Level maths tuition.

1. How H240 is assessed

There is no coursework in OCR A-Level Maths. The whole grade comes from three written papers sat at the end of the course. They are equally weighted, so no single paper can be treated as the soft option.

Paper 01 — Pure Mathematics

100 marks, 2 hours, one third of the A Level. A single section of pure mathematics with a deliberate gradient of difficulty: short, accessible questions early on building to longer multi-step problems. Covers proof, algebra and functions, coordinate geometry, sequences and series, trigonometry, exponentials and logarithms, differentiation, integration, numerical methods and vectors.

Paper 02 — Pure Mathematics and Statistics

100 marks, 2 hours, one third of the A Level. Two sections of roughly 50 marks each: pure mathematics, then statistics. The statistics section covers sampling, data presentation and interpretation, probability, statistical distributions and hypothesis testing, and some questions are set on the OCR pre-release large data set.

Paper 03 — Pure Mathematics and Mechanics

100 marks, 2 hours, one third of the A Level. Two sections of roughly 50 marks each: pure mathematics, then mechanics. The mechanics section covers quantities and units, kinematics, forces and Newton's laws, and moments.

A practical consequence: pure mathematics is examined on all three papers and makes up the clear majority of the qualification, while statistics and mechanics each appear on only one. Strong, fluent pure technique is the single highest-leverage thing to drill, because it pays off three times over.

2. The assessment objectives, and why they matter

Every mark in H240 is tagged to one of three assessment objectives (AOs). Understanding them changes how you read a question, because the AO tells you what kind of response earns the marks.

AO1 — Use and apply standard techniques (≈ 60%)

The largest objective. Selecting and carrying out routine procedures accurately: differentiating, integrating, solving equations, manipulating algebra. These are the marks you secure by being fast and accurate at the mechanics.

AO2 — Reason, interpret and communicate (≈ 20%)

Constructing arguments, proofs and chains of reasoning, and communicating them with correct notation and language. "Show that", "prove" and "explain" questions live here. Sloppy notation costs AO2 marks even when the maths is right.

AO3 — Solve problems and model (≈ 20%)

Translating an unstructured or real-world situation into mathematics, solving it, and interpreting the result. OCR reports AO3 in two strands, AO3(PS) for problem solving and AO3(M) for modelling. These are the questions where you must decide the method yourself.

The exact percentages move a little between the three papers within OCR's published ranges, but the headline is stable: roughly three fifths of the qualification is AO1, with the remaining two fifths split between reasoning (AO2) and problem solving and modelling (AO3). Students who are excellent at technique but treat reasoning as optional typically cap out in the low grade bands, because they are conceding two marks in five.

3. Where OCR candidates lose marks

OCR examiner reports, published after each series, are strikingly consistent year on year about what separates strong scripts from weak ones. The recurring themes are these.

None of these is about not knowing the maths. They are about exam discipline. That is exactly why they are recoverable with deliberate practice.

4. Four worked exam-style questions

The following four questions are written in OCR's H240 style and format, one from each of the main flavours you will meet: a pure "show that" with calculus, a binomial hypothesis test from the statistics section, a mechanics problem with connected particles, and a problem-solving question on sequences. They are original, not past-paper questions. Work each one before reading the solution, then read the examiner note to see where the marks actually sit.

Question 1 — Differentiation and a "show that" (Pure)

Paper 01 · Pure Mathematics · ~7 marks

A curve has equation \(y = x^3 - 6x^2 + 9x + 2\).

(a) Find \(\dfrac{dy}{dx}\) and hence find the coordinates of the two stationary points. [4]

(b) Show that the curve has a point of inflection at \(x = 2\), and determine whether the stationary point at the larger value of \(x\) is a minimum or a maximum. [3]

Solution, part (a). Differentiate term by term:

$$ \frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3). $$

Stationary points occur where \(\frac{dy}{dx} = 0\), so \(x = 1\) or \(x = 3\). Substituting back into the original equation:

$$ y(1) = 1 - 6 + 9 + 2 = 6, \qquad y(3) = 27 - 54 + 27 + 2 = 2. $$

The stationary points are \((1, 6)\) and \((3, 2)\).

Solution, part (b). Differentiate again:

$$ \frac{d^2y}{dx^2} = 6x - 12. $$

At \(x = 2\), \(\frac{d^2y}{dx^2} = 12 - 12 = 0\). To confirm a point of inflection we check that the second derivative changes sign: for \(x < 2\) we have \(6x - 12 < 0\) and for \(x > 2\) we have \(6x - 12 > 0\), so the concavity changes at \(x = 2\). Hence there is a point of inflection at \(x = 2\), as required. At the stationary point with the larger \(x\), namely \(x = 3\),

$$ \left.\frac{d^2y}{dx^2}\right|_{x=3} = 6(3) - 12 = 6 > 0, $$

so \((3, 2)\) is a minimum.

What the examiner is looking for

Part (a) is pure AO1: one method mark for a correct derivative, one for setting it to zero and solving, and accuracy marks for both correct coordinate pairs. The most common slip is an arithmetic error substituting back, so the marks for \(x=1\) and \(x=3\) are awarded independently. Part (b) is AO2 reasoning. For the "show that" you must do more than state \(\frac{d^2y}{dx^2}=0\) at \(x=2\): a second-derivative of zero alone does not prove inflection, so the sign-change argument (or equivalent) is the load-bearing step examiners want to see. Asserting the printed result without it typically loses the final mark. The minimum/maximum decision is one accuracy mark, and stating "minimum" without the supporting \(\frac{d^2y}{dx^2}>0\) value is penalised.

Question 2 — Binomial hypothesis test (Statistics)

Paper 02 · Statistics · ~6 marks

A manufacturer claims that at most 15% of the components it produces are faulty. A quality inspector takes a random sample of 20 components and finds that 6 are faulty. Test, at the 5% significance level, whether there is evidence that the proportion of faulty components exceeds 15%.

Solution. Let \(p\) be the proportion of faulty components and let \(X\) be the number of faulty components in the sample, so under the null hypothesis \(X \sim B(20, 0.15)\). Set up the hypotheses:

$$ H_0: p = 0.15, \qquad H_1: p > 0.15. $$

This is a one-tailed test at the 5% level. We find the probability of observing a result at least as extreme as \(6\) faulty, assuming \(H_0\) is true:

$$ P(X \ge 6) = 1 - P(X \le 5). $$

Using the binomial cumulative distribution (calculator) with \(n = 20\), \(p = 0.15\):

$$ P(X \le 5) = 0.9327\ (\text{4 d.p.}), \qquad \text{so } P(X \ge 6) = 1 - 0.9327 = 0.0673. $$

Compare with the significance level: \(0.0673 > 0.05\). The result is not in the critical region, so we do not reject \(H_0\).

$$ 0.0673 > 0.05 \;\Rightarrow\; \text{insufficient evidence to reject } H_0. $$

There is insufficient evidence at the 5% level to conclude that the proportion of faulty components exceeds 15%. The manufacturer's claim is not contradicted by this sample.

What the examiner is looking for

This question spans AO1 (computing the probability), AO2 (communicating the conclusion) and AO3(M) (setting up the model). The marks break down roughly: one for both hypotheses stated in terms of \(p\) — writing them in words or in terms of \(X\) is a frequent way to lose it; one for identifying \(X\sim B(20,0.15)\); method and accuracy marks for the correct tail probability \(P(X\ge 6)\); and crucially a final mark for a conclusion in context. OCR reports flag candidates who compute \(0.0673\) correctly, then write only "accept \(H_0\)". You must say what it means for the faulty-component claim. A second classic error is comparing the wrong tail or testing \(P(X\ge 5)\); the observed value is \(6\), so the inclusive tail is \(X\ge 6\).

Question 3 — Connected particles (Mechanics)

Paper 03 · Mechanics · ~7 marks

Two particles \(A\) and \(B\), of masses \(3\,\text{kg}\) and \(2\,\text{kg}\) respectively, are connected by a light inextensible string that passes over a smooth fixed pulley. The system is released from rest with the string taut and both particles hanging vertically. Take \(g = 9.8\,\text{m s}^{-2}\).

(a) Find the acceleration of the system and the tension in the string. [5]

(b) State one modelling assumption you have used and explain its effect. [2]

Solution, part (a). The heavier particle \(A\) accelerates downwards and \(B\) upwards with the same magnitude of acceleration \(a\), and the tension \(T\) is the same throughout the string because it is light and the pulley smooth. Apply Newton's second law to each particle, taking the direction of motion as positive.

For \(A\) (moving down):

$$ 3g - T = 3a. $$

For \(B\) (moving up):

$$ T - 2g = 2a. $$

Adding the two equations eliminates \(T\):

$$ 3g - 2g = 3a + 2a \;\Rightarrow\; g = 5a \;\Rightarrow\; a = \frac{g}{5} = \frac{9.8}{5} = 1.96\ \text{m s}^{-2}. $$

Substitute back into the equation for \(B\):

$$ T = 2a + 2g = 2(1.96) + 2(9.8) = 3.92 + 19.6 = 23.52\ \text{N}. $$

The acceleration is \(1.96\,\text{m s}^{-2}\) and the tension is \(23.52\,\text{N}\) (or \(23.5\,\text{N}\) to 3 s.f.).

Solution, part (b). The string is modelled as light (massless), so the tension is the same at both ends and we do not have to account for the weight of the string. If the string had mass, the tension would differ along its length and both the acceleration and the two tensions would change.

What the examiner is looking for

Part (a) is AO1 and AO3(M). The marks reward two correct equations of motion (one each), eliminating \(T\) to find \(a\), and substituting back for \(T\). The most common error is a sign slip — writing \(T - 3g = 3a\) for the descending particle — which examiners treat as a method error that loses the accuracy marks downstream. A clear force diagram, even though not always demanded, protects against this. Part (b) is AO2: a single isolated word such as "light" earns little; the mark is for naming an assumption and explaining its consequence ("so tension is equal throughout"). The "smooth pulley" assumption is an equally valid answer. Leaving \(g\) symbolic until the end and only then substituting \(9.8\) reduces rounding errors and keeps the working transparent.

Question 4 — Sequences and problem solving (Pure)

Paper 01 · Pure Mathematics · ~6 marks

An arithmetic sequence has first term \(a\) and common difference \(d\). The sum of the first \(10\) terms is \(155\), and the sum of the first \(20\) terms is \(610\).

Determine the values of \(a\) and \(d\), and hence find the first term of the sequence that exceeds \(100\). [6]

Solution. The sum of the first \(n\) terms of an arithmetic sequence is \(S_n = \tfrac{n}{2}\big(2a + (n-1)d\big)\). Apply this to both pieces of information.

$$ S_{10} = \frac{10}{2}\big(2a + 9d\big) = 5(2a + 9d) = 155 \;\Rightarrow\; 2a + 9d = 31. $$
$$ S_{20} = \frac{20}{2}\big(2a + 19d\big) = 10(2a + 19d) = 610 \;\Rightarrow\; 2a + 19d = 61. $$

Subtract the first equation from the second to eliminate \(a\):

$$ (2a + 19d) - (2a + 9d) = 61 - 31 \;\Rightarrow\; 10d = 30 \;\Rightarrow\; d = 3. $$

Substitute \(d = 3\) into \(2a + 9d = 31\):

$$ 2a + 27 = 31 \;\Rightarrow\; 2a = 4 \;\Rightarrow\; a = 2. $$

The \(n\)th term is \(u_n = a + (n-1)d = 2 + 3(n-1) = 3n - 1\). We need the first term exceeding \(100\):

$$ 3n - 1 > 100 \;\Rightarrow\; 3n > 101 \;\Rightarrow\; n > 33.67. $$

So the smallest integer is \(n = 34\), giving \(u_{34} = 3(34) - 1 = 101\). The first term to exceed \(100\) is \(101\), the 34th term.

What the examiner is looking for

This is AO1 with a strong AO3(PS) element, since you must decide to form and solve simultaneous equations. Marks: one for a correct sum formula applied to give a usable equation, one for the second equation, method and accuracy for solving to get \(d\) and then \(a\), and the final two marks for the problem-solving step. A common trap is the last part: solving \(3n-1>100\) gives \(n>33.67\), and candidates either round down to \(33\) (which gives \(98\), not exceeding \(100\)) or quote \(n=34\) but forget to state the actual term value \(101\). Read the command: "determine the values" expects both \(a\) and \(d\) shown clearly, and "hence" signals that you must use your values, not start a fresh method. Stating \(n=34\) without the term, or the term without identifying which it is, leaves a mark on the table.

5. OCR-specific exam technique

Beyond knowing the content, a handful of habits are worth more on OCR papers specifically.

A-Level Maths tuition

Most marks lost on OCR A-Level Maths are lost to exam technique, not unknown content. For focused one-to-one help with H240 — pure technique, statistics, mechanics, or exam discipline on "show that" and problem-solving questions — see A-Level maths tuition, explore A-Level and admissions tutoring, or book a free consultation.

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