📚 Case Study Practice for Year 12 WJEC Maths | 案例分析实战演练
Case study questions in Year 12 WJEC Mathematics challenge you to apply pure, statistics, and mechanics concepts to real-world scenarios. These problems bridge the gap between textbook theory and practical reasoning, testing your ability to model, solve, and interpret within a single coherent narrative. This article walks through seven carefully selected cases, highlighting the modelling cycle, common pitfalls, and exam-style strategies.
WJEC 十二年级数学中的案例研究题要求你将纯数、统计和力学的概念应用到真实情境中。这些问题在课本理论与实际推理之间架起桥梁，考查你在一个连贯的叙述中进行建模、求解和解释的能力。本文将通过七个精选案例，突出建模循环、常见陷阱以及考试风格的应对策略。
1. Introduction to Case Studies | 案例研究简介
Unlike standard exercise problems, a case study presents a situation with data, assumptions, and often multiple parts that build on one another. Your job is to identify relevant mathematical structures—quadratics for optimisation, exponentials for decay, geometric series for finance—and translate the description into equations. The WJEC specification emphasises that marks are awarded not only for the correct answer but for clear communication of the modelling process.
与标准练习题不同，案例研究会呈现一个带有数据、假设的情景，且常常包含多个相互关联的部分。你的任务是识别出相关的数学结构——优化问题用二次函数、衰减用指数函数、金融用等比数列——并将文字描述转化为方程。WJEC 大纲强调，得分不仅依赖正确答案，还依赖于对建模过程的清晰阐述。
2. The Art of Mathematical Modelling | 数学建模的艺术
Every case study follows a modelling cycle: simplify by stating assumptions, define variables with units, formulate equations, solve analytically or numerically, and finally interpret the results in context. Checking whether the answer is sensible—e.g., lengths and probabilities must be positive—is an essential part of the cycle. In WJEC exams, failing to relate a numerical answer back to the real-world question can cost valuable marks.
每一个案例研究都遵循一个建模循环：通过陈述假设进行简化，定义带单位的变量，建立方程，解析或数值求解，最后在情境中解释结果。检查答案是否合理——例如长度和概率必须为正——是该循环的重要环节。在 WJEC 考试中，若未将数值答案联系回实际问题，可能会丢失宝贵分数。
3. Case 1: Optimising Fencing Area | 案例一：优化围栏面积
A farmer has 100 metres of fencing and wants to enclose a rectangular area against a long barn. The barn forms one side of the rectangle, so no fencing is needed there. Let the width perpendicular to the barn be x metres. Then the side parallel to the barn uses (100 − 2x) metres of fencing. The area A is given by
一位农民有 100 米长的围栏，想靠着一个长谷仓围出一个矩形区域。谷仓构成矩形的一条边，因此该边无需围栏。设垂直于谷仓的宽度为 x 米，则平行于谷仓的边所用围栏为 (100 − 2x) 米。面积 A 由下式给出
A = x(100 − 2x) = 100x − 2x²
This is a quadratic concave downward, so its maximum occurs at the stationary point. Differentiate and set to zero:
这是一个开口向下的二次函数，最大值出现在驻点处。求导并设为零：
dA/dx = 100 − 4x = 0 ⇒ x = 25
The second derivative d²A/dx² = −4 < 0 confirms a maximum. The maximum area is A = 25 × 50 = 1250 m². The context satisfies domain constraints: x must be between 0 and 50, and our solution lies inside.
二阶导数 d²A/dx² = −4 < 0 确认是最大值。最大面积为 A = 25 × 50 = 1250 平方米。情境满足定义域限制：x 必须在 0 到 50 之间，我们的解正好落在该区间内。
4. Case 2: Car Braking Distance | 案例二：汽车制动距离
A car travels at 20 m s⁻¹ when the driver spots a hazard. The reaction time is 0.7 s, after which brakes produce a constant deceleration of 6 m s⁻². Find the total stopping distance. During reaction, distance = speed × time = 20 × 0.7 = 14 m. For braking, use the kinematic equation v² = u² + 2as with v = 0, u = 20, a = −6:
一辆汽车以 20 m s⁻¹ 的速度行驶，司机发现危险。反应时间为 0.7 s，之后刹车产生 6 m s⁻² 的恒定减速度。求总制动距离。反应期间，距离 = 速度 × 时间 = 20 × 0.7 = 14 m。制动阶段使用运动学公式 v² = u² + 2as，其中 v = 0，u = 20，a = −6：
0 = (20)² + 2(−6)s ⇒ 0 = 400 − 12s ⇒ s = 33.3 m (to 3 s.f.)
Total distance = 14 + 33.3 = 47.3 m. This simple model neglects road conditions, assuming constant deceleration. In an exam, always state such modelling assumptions explicitly.
总距离 = 14 + 33.3 = 47.3 m。这个简单模型忽略了路面状况，假设减速度恒定。在考试中，务必明确陈述此类建模假设。
5. Case 3: Quality Control with Normal Distribution | 案例三：正态分布质量控制
A factory produces bolts whose lengths are normally distributed with mean μ = 50 mm and standard deviation σ = 0.5 mm. The specification limits are 50 ± 1.2 mm. Bolts outside [48.8, 51.2] are rejected. To find the acceptance probability, calculate z-scores:
一家工厂生产的螺栓长度服从正态分布，均值 μ = 50 mm，标准差 σ = 0.5 mm。规格界限为 50 ± 1.2 mm，超出 [48.8, 51.2] 范围的螺栓将被拒收。为求接受概率，计算 z 分数：
z₁ = (48.8 − 50)/0.5 = −2.4, z₂ = (51.2 − 50)/0.5 = 2.4
Using standard normal tables, P(−2.4 < Z < 2.4) = 2 × P(Z < 2.4) − 1 = 2(0.9918) − 1 = 0.9836. Thus about 98.4% of bolts are acceptable. In an extended case, you might be asked to set limits to guarantee a given rejection rate—this involves solving in reverse.
使用标准正态分布表，P(−2.4 < Z < 2.4) = 2 × P(Z < 2.4) − 1 = 2(0.9918) − 1 = 0.9836。因此约 98.4% 的螺栓合格。在扩展案例中，可能会要求设定界限以保证给定的拒收率——这需要逆向求解。
6. Case 4: Equilibrium on an Inclined Plane | 案例四：斜面平衡
A block of mass m rests on a rough inclined plane with angle θ. The coefficient of static friction is μ. By resolving forces parallel and perpendicular to the slope, the limiting equilibrium condition is:
一质量为 m 的木块静置于倾角为 θ 的粗糙斜面上，静摩擦系数为 μ。通过分解沿斜面和平行于斜面的力，极限平衡条件为：
mg sinθ = μ mg cosθ ⇒ tanθ = μ
For the block to stay at rest without sliding, θ must satisfy θ ≤ arctan(μ). If μ = 0.4, then the maximum safe angle is arctan(0.4) ≈ 21.8°. This result arises frequently in WJEC Mechanics papers and illustrates the direct link between physical constraints and a trigonometric inequality.
要使木块保持静止而不滑落，θ 必须满足 θ ≤ arctan(μ)。若 μ = 0.4，则最大安全角度为 arctan(0.4) ≈ 21.8°。这一结果在 WJEC 力学试卷中经常出现，展示了物理约束与三角不等式之间的直接联系。
7. Case 5: Radioactive Decay | 案例五：放射性衰变
A radioactive substance decays according to m = m₀ e⁻ᵏᵗ, where m₀ is the initial mass and k is a positive constant. The half-life T½ is the time for mass to halve. Setting m = m₀/2 gives:
一种放射性物质按 m = m₀ e⁻ᵏᵗ 衰变，其中 m₀ 为初始质量，k 为正常数。半衰期 T½ 是质量减半所需的时间。令 m = m₀/2 可得：
m₀/2 = m₀ e⁻ᵏᵀ½ ⇒ ½ = e⁻ᵏᵀ½ ⇒ −ln 2 = −k T½ ⇒ k = (ln 2)/T½
If T½ = 8 days, then after 10 days the fraction remaining is m/m₀ = e⁻(¹⁰ˡⁿ²)/⁸ = (½)^(10/8) ≈ 0.42. The model can be adapted to find the time for mass to fall below a given threshold, a common exam request.
若 T½ = 8 天，则 10 天后剩余的比例为 m/m₀ = e⁻(¹⁰ˡⁿ²)/⁸ = (½)^(10/8) ≈ 0.42。该模型可用于求出质量降至给定阈值以下所需的时间，这是常见的考试要求。
8. Case 6: Savings and Geometric Series | 案例六：储蓄与等比数列
At the start of each year, £1000 is deposited in a bank offering 5% annual compound interest. The total after 10 years is the sum of ten geometrically growing amounts. The first deposit earns interest for 10 years: 1000 × 1.05¹⁰, the second for 9 years: 1000 × 1.05⁹, …, the tenth for 1 year: 1000 × 1.05. The series is:
每年年初，将 1000 英镑存入一家提供 5% 年复利的银行。10 年后的总额是十个按等比增长的本利和之和。第一笔存款获息 10 年：1000 × 1.05¹⁰，第二笔 9 年：1000 × 1.05⁹，……，第十笔 1 年：1000 × 1.05。该级数为：
S = 1000 × 1.05 + 1000 × 1.05² + … + 1000 × 1.05¹⁰
This geometric series has first term a = 1000 × 1.05 = 1050, common ratio r = 1.05, and n = 10 terms. Sum:
该等比数列首项 a = 1000 × 1.05 = 1050，公比 r = 1.05，项数 n = 10。求和：
S = a(rⁿ − 1)/(r − 1) = 1050 × (1.05¹⁰ − 1)/0.05 ≈ £13 206
The WJEC paper may also ask for the effect of depositing at the end of each year or for the number of years needed to reach a target. Always identify whether the first term is compounded.
WJEC 试卷也可能问到年末存款的影响，或达到目标金额所需的年数。要始终辨认首项是否已计息。
9. Case 7: Decision Making with Probability | 案例七：概率决策
An insurance company sells a policy for £200. A claim occurs with probability 0.02, and the average payout is £8000. The expected claim cost per policy is £8000 × 0.02 = £160. The expected profit per policy is therefore £200 − £160 = £40. Over 10 000 policies, the expected total profit is £400 000. The model uses the mean of a binomial-like distribution and can be refined by including administrative costs.
一家保险公司以 200 英镑出售一份保单。索赔概率为 0.02，平均赔付额为 8000 英镑。每份保单的期望索赔成本为 £8000 × 0.02 = £160。因此每份保单的期望利润为 £200 − £160 = £40。若销售 10 000 份，期望总利润为 £400 000。该模型使用了类似二项分布的均值，并可通过计入管理成本进行完善。
This illustrates how expected value guides business decisions. In a case study, you might then calculate the probability of making a loss over a given portfolio size using normal approximation to binomial.
这展示了期望值如何指导商业决策。在案例研究中，你可能需要进一步使用二项分布的正态近似，计算在给定保单规模下出现亏损的概率。
10. Common Pitfalls and How to Avoid Them | 常见陷阱与避免方法
Watch out for units: mixing metres and centimetres, or forgetting to convert minutes to seconds. In modelling, stating assumptions is crucial but often overlooked—write “assume constant speed” or “neglect air resistance” clearly. Also, treat domain restrictions seriously: in optimisation, check that your solution lies within the feasible interval. Finally, always reconnect the final number to the real-world question being asked.
注意单位：混淆米和厘米，或忘记将分钟转换为秒。在建模中，陈述假设至关重要但常被忽略——要清楚地写出“假设速度恒定”或“忽略空气阻力”。此外，认真对待定义域限制：在优化问题中，检查解是否落在可行区间内。最后，务必将最终数值重新联系到所问的实际问题。
11. WJEC Exam-Style Tips | WJEC 考试风格技巧
WJEC rewards structured reasoning: open with a brief introduction defining your chosen symbols, then present equations methodically. Use accurate notation—show differentiation steps, substitute correctly, and round final answers sensibly (often 3 significant figures). If a question says “hence or otherwise”, a clear scaling of a previous result often saves time. In statistics cases, sketch a normal curve and shade regions to avoid sign errors with z-values.
WJEC 青睐结构化的推理：以简要的介绍开头，定义所选符号，然后有条理地列出方程。使用准确的符号——展示求导步骤，正确代入，最后合理舍入（通常保留 3 位有效数字）。如果题目写有“hence or otherwise”，对前一结果进行清晰的调整常常能节省时间。在统计案例中，画出正态曲线并标出区域，可避免 z 值的符号错误。
12. Summary and Key Takeaways | 总结与关键要点
Case study mastery comes from repeatedly practicing the full modelling cycle. Across WJEC papers, you will encounter optimisation with quadratics and calculus, kinematics with constant acceleration, normal distribution quality checks, mechanics on slopes, exponential growth and decay, financial series, and expectation decisions. The core skills of translating words into mathematics, checking internal consistency, and interpreting outputs are what turn an A-level student into a confident mathematical modeller.
案例研究的精通来自于反复练习完整的建模循环。在 WJEC 试卷中，你将遇到二次函数与微积分的优化、匀加速运动学、正态分布质量检验、斜面力学、指数增长与衰减、金融数列以及期望决策。将文字转化为数学、检查内在一致性以及解释输出结果这些核心技能，正是将一名 A-level 学生转变为自信的数学建模者的关键。
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