Applied Mathematics Problem-Solving Strategies

Paper 3 of AQA 7367 is the applications paper. You answer the two options your school has chosen from Mechanics (7367/3M), Statistics (7367/3S) and Discrete (7367/3D) — never all three. Crucially, Paper 3 carries a different assessment-objective balance from the Pure papers: AO1 40% / AO2 25% / AO3 35%. Compared with the Pure papers' 20% AO3, problem-solving here nearly doubles. That single fact should reshape how you revise applications: the marks are weighted towards setting up a model, stating assumptions, choosing a method, and interpreting the answer in context — not merely executing a technique you were handed.

This lesson gives strategy and worked walkthroughs across all three option routes — Mechanics, Statistics and Discrete — so that, whichever pair you sit, you have a reliable, well-rehearsed attack plan for the applied paper.

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The modelling cycle (the AO3 spine of Paper 3)

The defining feature of Paper 3 is that questions arrive wrapped in a real situation — a particle on a string, a factory schedule, a road network, a sample of measurements — rather than as a bare instruction to differentiate or solve. That wrapping is the point: the elevated AO3 weighting means a substantial share of marks is for turning the situation into mathematics, choosing the right tool, and turning the answer back into a statement about the situation. Almost every applied question, whatever the option, rewards the same underlying process. Make it explicit on the page rather than doing it silently in your head:

1. Read the context and define variables/notation — let $v$v be the speed, $X$X the random variable, the network the road map.
2. State the modelling assumptions you are using — "model the particle as a point mass", "the string is light and inextensible", "events occur independently at a constant rate", "the sample is random". These are marked in many questions.
3. Set up the governing equation(s) — Newton's second law, an energy equation, a probability distribution, an algorithm.
4. Solve using the appropriate technique, showing method.
5. Interpret in context and, where asked, comment on the model ("the assumption of no air resistance means the true range is shorter").

Exam Tip: "State your assumptions" and "comment on your model" look like throwaway parts but are pure AO3/AO2 marks. On Paper 3 they appear often precisely because the paper is problem-solving-weighted. Never leave them blank.

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Mechanics (7367/3M)

Circular motion

The most heavily examined Mechanics topic. Core relations:

Quantity	Formula
Centripetal acceleration	$a = \dfrac{v^2}{r} = r\omega^2$a=rv2​=rω2
Centripetal (resultant) force toward centre	$F = \dfrac{mv^2}{r} = mr\omega^2$F=rmv2​=mrω2
Angular velocity	$\omega = \dfrac{v}{r} = \dfrac{2\pi}{T}$ω=rv​=T2π​

Strategy for motion in a vertical circle:

1. Draw a force diagram at the point of interest (top, bottom, or general angle).
2. Resolve towards the centre; the resultant equals $\dfrac{mv^2}{r}$rmv2​.
3. Use conservation of energy between two positions to relate the speeds.
4. The "critical" condition (e.g. completing the circle) is usually where the tension or normal reaction reaches zero.

Exam Tip: At the top of a vertical circle the weight points towards the centre; at the bottom it points away. Resolving weight the wrong way is the most common circular-motion error and corrupts the whole equation.

Elastic strings and Hooke's law

$T = \frac{\lambda x}{l}, \qquad \text{EPE} = \frac{\lambda x^2}{2l},$T=lλx​,EPE=2lλx2​,

where $T$T is tension, $\lambda$λ the modulus of elasticity, $x$x the extension and $l$l the natural length. Where the motion involves a varying force, prefer energy conservation ($\text{KE} + \text{GPE} + \text{EPE} = \text{constant}$KE+GPE+EPE=constant) over $F=ma$F=ma: energy methods sidestep the variable acceleration entirely.

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Worked example 1 — Mechanics (vertical circle, with mark scheme)

Specimen-style. A particle of mass $m$m is attached to one end of a light inextensible string of length $r$r; the other end is fixed at $O$O. The particle is set moving in a complete vertical circle. Show that the speed $v_{\text{top}}$vtop​ at the highest point satisfies $v_{\text{top}}^2 \ge gr$vtop2​≥gr for the string to remain taut. (4 marks)

Solution. At the top, both the weight $mg$mg and the tension $T$T act towards the centre $O$O. Newton's second law towards the centre:

$T + mg = \frac{m v_{\text{top}}^2}{r}.$T+mg=rmvtop2​​.

The string remains taut provided $T \ge 0$T≥0. Setting $T \ge 0$T≥0:

$\frac{m v_{\text{top}}^2}{r} - mg \ge 0 \implies v_{\text{top}}^2 \ge gr. \quad \blacksquare$rmvtop2​​−mg≥0⟹vtop2​≥gr.■

How the 4 marks are earned:

• M1 — resolving towards the centre at the top with both $T$T and $mg$mg directed inward.
• A1 — correct equation $T + mg = \dfrac{mv_{\text{top}}^2}{r}$T+mg=rmvtop2​​.
• M1 — applying the taut condition $T \ge 0$T≥0.
• A1 — reaching the printed inequality $v_{\text{top}}^2 \ge gr$vtop2​≥gr, with the conclusion stated.

Where candidates lose marks: writing $T - mg = \dfrac{mv^2}{r}$T−mg=rmv2​ (weight resolved outward — wrong at the top); using $T \le 0$T≤0; or not stating the condition that defines "taut". The mechanics is short, but the reasoning about the constraint is where the AO2/AO3 marks sit.

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Statistics (7367/3S)

Chi-squared goodness-of-fit tests

A high-frequency 7367/3S question gives observed frequencies and asks you to test against a model. A reliable structure:

1. State $H_0$H0​ and $H_1$H1​ in words (e.g. "$H_0$H0​: the data follow a Poisson distribution").
2. If you estimate a parameter from the data, say so — it costs a degree of freedom.
3. Compute expected frequencies $E_i$Ei​.
4. Combine adjacent classes where $E_i < 5$Ei​<5, and state what you merged.
5. Compute $X^2 = \sum \dfrac{(O_i - E_i)^2}{E_i}$X2=∑Ei​(Oi​−Ei​)2​ in a clear table.
6. State $\nu = (\text{number of classes after combining}) - 1 - (\text{parameters estimated})$ν=(number of classes after combining)−1−(parameters estimated).
7. Compare with the critical value and conclude in context.

Exam Tip: Incorrect degrees of freedom is the classic 7367/3S error. Write it explicitly: "after combining there are $k$k classes and $p$p parameters were estimated, so $\nu = k - 1 - p$ν=k−1−p."

Continuous random variables

A typical PDF question chains four standard tasks:

1. Find the constant by solving $\displaystyle\int f(x)\,dx = 1$∫f(x)dx=1 over the support.
2. Find $E(X) = \displaystyle\int x f(x)\,dx$E(X)=∫xf(x)dx and $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$Var(X)=E(X2)−[E(X)]2.
3. Find the CDF $F(x)$F(x) by integrating the PDF.
4. Find the median by solving $F(m) = 0.5$F(m)=0.5.

Lay each integral out with explicit limits; for piecewise PDFs, treat each piece and check continuity at the joins.

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Worked example 2 — Statistics (continuous random variable, M/A + grade-band)

Specimen-style. A continuous random variable $X$X has PDF $f(x) = kx^2$f(x)=kx2 for $0 \le x \le 2$0≤x≤2 and $f(x) = 0$f(x)=0 otherwise. (a) Show that $k = \tfrac{3}{8}$k=83​. (3) (b) Find $E(X)$E(X). (3)

Solution to (a). A PDF integrates to $1$1 over its support:

$\int_0^2 kx^2 \, dx = 1 \implies k\left[\frac{x^3}{3}\right]_0^2 = 1 \implies k \cdot \frac{8}{3} = 1 \implies k = \frac{3}{8}. \quad \blacksquare$∫02​kx2dx=1⟹k[3x3​]02​=1⟹k⋅38​=1⟹k=83​.■

• M1 — setting $\int_0^2 kx^2\,dx = 1$∫02​kx2dx=1. A1 — correct integration to $\tfrac{8k}{3}$38k​. A1 — solving to the given $k = \tfrac38$k=83​.

Solution to (b).

$E(X) = \int_0^2 x \cdot \tfrac{3}{8}x^2 \, dx = \tfrac{3}{8}\int_0^2 x^3 \, dx = \tfrac{3}{8}\left[\frac{x^4}{4}\right]_0^2 = \tfrac{3}{8}\cdot\frac{16}{4} = \tfrac{3}{8}\cdot 4 = \frac{3}{2}.$E(X)=∫02​x⋅83​x2dx=83​∫02​x3dx=83​[4x4​]02​=83​⋅416​=83​⋅4=23​.

• M1 — using $E(X)=\int x f(x)\,dx$E(X)=∫xf(x)dx with the correct $k$k. A1 — correct integral $\tfrac{3}{8}\cdot 4$83​⋅4. A1 — $E(X)=\tfrac32$E(X)=23​.

Grade-band model answers for part (b)

Mid-band. Writes $E(X) = \int_0^2 x f(x)\,dx$E(X)=∫02​xf(x)dx and reaches $\tfrac32$23​, but omits the substitution of $k=\tfrac38$k=83​ as a separate line and does not show the limits being applied.

Examiner-style commentary. The final value is correct and would gain the accuracy mark, but condensed working risks the method mark if a slip occurs, and an examiner must be able to see the limits substituted. Showing $\big[\tfrac{x^4}{4}\big]_0^2$[4x4​]02​ explicitly secures the method.

Stronger. Full working: the $E(X)$E(X) formula, $k$k substituted, integral evaluated with limits shown, exact answer $\tfrac32$23​.

Examiner-style commentary. Full marks. Clean and complete; the exact fraction (not $1.5$1.5 as a "rounded" decimal) is correctly retained.

Top-band. As the stronger answer, and notes that $E(X)=\tfrac32$E(X)=23​ lies sensibly within the support $[0,2]$[0,2] and to the right of the midpoint because the density $kx^2$kx2 is increasing — a quick contextual sense-check.

Examiner-style commentary. Full marks with genuine statistical insight. Checking that an expectation lies in the support, and reasoning about where from the shape of the density, is exactly the interpretive maturity AO2/AO3 reward.

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Statistics: hypothesis tests and confidence intervals

For a hypothesis test, lay out a fixed skeleton: (i) hypotheses; (ii) significance level; (iii) test statistic formula and value; (iv) critical value (or $p$p-value); (v) decision; (vi) conclusion in context. Identify whether the population variance is known before choosing the distribution.

For a confidence interval always give the formula, the critical value, the computed interval, and an interpretation.

Exam Tip: The interpretation of a confidence interval is a perennial AO2 trap. The defensible wording is: "the method produces an interval that contains the true mean in 95% of samples." Avoid "there is a 95% probability that $\mu$μ lies in this interval" — once computed, the interval either contains $\mu$μ or it does not.

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Discrete (7367/3D)

Algorithms: show every step

The 7367/3D staples are Dijkstra (shortest path), Prim and Kruskal (minimum spanning tree), route inspection (the "Chinese postman" problem), the travelling-salesperson bounds, critical path analysis, linear programming and matchings. The golden rule across all of them: follow the algorithm exactly and show every step — the marks are for the process, not just the final answer.

Algorithm	What to show
Dijkstra	Working values and the order vertices are made permanent
Prim / Kruskal	The order edges are added and the rejection of cycle-forming edges
Route inspection	The odd vertices, every possible pairing, and the least repeat
TSP	Nearest-neighbour route (upper bound) and a delete-vertex lower bound
Critical path	Forward pass (earliest) and backward pass (latest) at every event
Matchings	The alternating path and the augmented matching

Critical path analysis

• Forward pass = earliest event times (take the maximum of incoming paths).
• Backward pass = latest event times (take the minimum of outgoing paths).
• A critical activity has zero float; the critical path is the chain of critical activities (the longest path through the network).