📚 IGCSE CCEA Further Mathematics: Case Study Practice Drills | IGCSE CCEA 进阶数学：案例分析实战演练
Welcome to this targeted revision resource on case study questions in CCEA IGCSE Further Mathematics. Unlike standard procedural tasks, these problems weave together multiple topics and require you to interpret real-world contexts, build mathematical models, and justify your conclusions. This article walks you through several worked examples and shares essential strategies to help you perform confidently in your examination.
欢迎使用这份针对 CCEA IGCSE 进阶数学案例分析题目的专项复习资料。与常规计算题不同，案例分析融合了多个知识点，要求你解读真实情境、建立数学模型并论证结论。本文将通过多个详尽的例题，分享关键策略，帮助你在考试中自信应对。
1. Understanding the Nature of Case Study Questions | 理解案例分析题的本质
Case study questions in CCEA IGCSE Further Mathematics are open-ended scenarios drawn from finance, engineering, physics, or everyday decision-making. They test your ability to extract quantitative information, define variables, and select appropriate techniques from algebra, calculus, sequences, trigonometry, or probability. Crucially, the final part often asks for an interpretation, such as commenting on the validity of a model or suggesting improvements.
CCEA IGCSE 进阶数学的案例分析题是来自金融、工程、物理或日常决策的开放式情境。它们考查你提取量化信息、定义变量以及从代数、微积分、数列、三角学或概率中选择合适方法的能力。关键的是，最后部分通常要求给出解释，例如评价模型的有效性或提出改进建议。
Each case study is broken into several sub-questions that guide you from simple calculations to more complex analysis. You must always link your mathematical working back to the original context. A strong response includes clearly labelled diagrams, precise notation, and a final statement that addresses exactly what the question asks.
每个案例分析都分为若干子问题，引导你从简单计算走向复杂分析。你必须始终将数学推理与原始情境联系起来。一份优秀的答案应包含清晰的标注图、精确的符号，以及准确切题的最终陈述。
2. Case Study 1: Optimising Resource Allocation | 案例一：资源分配优化
A craft workshop produces two types of decorative vases, Standard (S) and Deluxe (D). Each Standard vase generates a profit of £25, and each Deluxe vase £40. Production is limited by two resources: moulding time and glazing time. Each Standard vase requires 2 hours of moulding and 1 hour of glazing; each Deluxe vase requires 4 hours of moulding and 1 hour of glazing. In a week, the workshop has 80 hours of moulding time and 30 hours of glazing time available. The manager wants to maximise weekly profit.
一家手工艺作坊生产两种装饰花瓶：标准款（S）和豪华款（D）。每个标准款利润为 £25，豪华款为 £40。生产受两项资源限制：模具时间和上釉时间。每个标准款需要 2 小时模具和 1 小时上釉；每个豪华款需要 4 小时模具和 1 小时上釉。每周作坊有 80 小时模具时间和 30 小时上釉时间。经理想最大化周利润。
e hours: 2x + 4y ≤ 80 (moulding), x + y ≤ 30 (glazing), with x ≥ 0 and y ≥ 0. Graphically, the feasible region is a quadrilateral with vertices at (0,0), (30,0), (0,20), and the intersection of the two boundary lines.
设 x 为标准款数量，y 为豪华款数量。目标是最大化 P = 25x + 40y。约束来自可用工时：2x + 4y ≤ 80（模具），x + y ≤ 30（上釉），且 x ≥ 0，y ≥ 0。可行域是一个四边形，顶点为 (0,0)、(30,0)、(0,20) 以及两条边界线的交点。
s to 2x + 120 - 4x = 80, so -2x = -40, yielding x = 20 and y = 10. Evaluating the profit at each vertex:
解方程组 2x + 4y = 80 和 x + y = 30 求交点。将 y = 30 - x 代入第一个方程得 2x + 4(30 - x) = 80，简化得 2x + 120 - 4x = 80，即 -2x = -40，解得 x = 20, y = 10。计算各顶点利润：
| Vertex (x, y) | Profit P = 25x + 40y (£) |
|---|---|
| (0, 0) | 0 |
| (30, 0) | 750 |
| (0, 20) | 800 |
| (20, 10) | 25×20 + 40×10 = 500 + 400 = 900 |
The maximum weekly profit of £ + 10 = 30), so both resources are fully utilised. The solution demonstrates how linear programming identifies the optimal product mix under multiple constraints.
最大周利润 £900 通过生产 20 个标准款和 10 个豪华款实现。这恰好用尽 80 小时模具时间（2×20 + 4×10 = 80）和 30 小时上釉时间（20 + 10 = 30），两种资源均得到充分利用。该解法展示了线性规划如何在多重约束下确定最佳产品组合。
3. Case Study 2: Kinematics and Calculus | 案例二：运动学与微积分
ocity v(t) is the first derivative, and acceleration a(t) is the second derivative. Analyse the motion to determine when the particle is at rest, the total distance travelled, and the times at which it changes direction.
一个粒子沿直线运动。t 秒后它到固定点的位移 s（米）由 s(t) = t³ - 6t² + 9t + 2 给出，0 ≤ t ≤ 5。速度 v(t) 是一阶导数，加速度 a(t) 是二阶导数。分析运动，确定粒子静止的时刻、总路程以及改变方向的时刻。
Differentiate s(t) to obtain velocity:
对 s(t) 求导得速度：
v(t) = ds/dt = 3t² - 12t + 9
Set v(t) = 0 to find when the particle is at rest: 3t² - 12t + 9 = 0 ⇒ t² - 4t + 3 = 0 ⇒ o t = 1 s and t = 3 s. These are the turning points where the particle reverses direction.
令 v(t) = 0 求粒子静止时刻：3t² - 12t + 9 = 0 ⇒ t² - 4t + 3 = 0 ⇒ (t - 1)(t - 3) = 0，所以 t = 1 s 和 t = 3 s。这些是粒子改变方向的转折点。
al distance travelled from t = 0 to t = 5, calculate the displacement at key times: s(0) = 2, s(1) = 1 - 6 + 9 + 2 = 6, s(3) = 27 - 54 + 27 + 2 = 2, s(5) = 125 - 150 + 45 + 2 = 22. The particle moves forward from 2 to 6 (distance 4), then backward from 6 to 2 (distance 4), then forward again from 2 to 22 (distance 20). Total distance = 4 + 4 + 20 = 28 m.
加速度 a(t) = dv/dt = 6t - 12。在 t = 1，a(1) = -6 m/s²（正向减速）；在 t = 3，a(3) = 6 m/s²（加速）。要求从 t = 0 到 t = 5 的总路程，计算关键时刻的位移：s(0) = 2，s(1) = 1 - 6 + 9 + 2 = 6，s(3) = 27 - 54 + 27 + 2 = 2，s(5) = 125 - 150 + 45 + 2 = 22。粒子从 2 向前运动到 6（距离 4），然后从 6 向后运动到 2（距离 4），再从 2 向前运动到 22（距离 20）。总路程 = 4 + 4 + 20 = 28 m。
This case study illustrates how calculus translates a real-world motion into algebraic expressions that reveal velocity, acceleration, and displacement patterns. Always interpret the signs of v and a in the physical context: negative velocity means motion in the opposite direction.
本案例展示了微积分如何将真实运动转化为代数表达式，揭示速度、加速度和位移模式。务必从物理情境解读 v 和 a 的正负：负速度表示朝相反方向运动。
4. Case Study 3: Financial Sequences and Annuities | 案例三：金融数列与年金
An investor deposits £800 at the start of every year into a savings account that pays compound interest at 4% per annum. How much will the account be worth immediately after the 10th deposit is made, before any further interest is added that year? This scenario involves the sum of a geometric progression where the first deposit earns interest for 10 years and the last earns none in that moment.
一位投资者在每年年初存入 £800 到一个按年利率 4% 计算复利的储蓄账户。第 10 次存款刚存入后（该年尚未加计利息），账户价值是多少？该情境涉及等比数列求和，首笔存款计息 10 年，最后一笔此刻未计息。
ount A is the sum: A = 800(1.04) + 800(1.04)² + ... + 800(1.04)¹⁰. This is a geometric series with first term a = 800 × 1.04 = 832, common ratio r = 1.04, and n = 10 terms.
第一笔存款增长为 800 × (1.04)¹⁰，第二笔为 800 × (1.04)⁹，依此类推，直到第 10 笔为 800 × (1.04)¹ = 800 × 1.04。总金额 A 为：A = 800(1.04) + 800(1.04)² + ... + 800(1.04)¹⁰。这是一个等比数列，首项 a = 800 × 1.04 = 832，公比 r = 1.04，项数 n = 10。
Use the sum formula Sₙ = a(rⁿ - 1)/(r - 1):
使用求和公式 Sₙ = a(rⁿ - 1)/(r - 1)：
A = 832 × (1.04¹⁰ - 1) / (1.04 - 1) = 832 × (1.480244... - 1) / 0.04
Compute: 1.04¹⁰ ≈ 1.480244, so A ≈ 832 × 0.480244 / 0.04 = 832 × 12.0061 ≈ £ es, check that the number of terms matches the deposit schedule, and use correct rounding.
计算：1.04¹⁰ ≈ 1.480244，因此 A ≈ 832 × 0.480244 / 0.04 = 832 × 12.0061 ≈ £9,989.08。该舍入值给出了实际预测。金融案例分析通常要求你处理数列、核对项数与存款时间表是否匹配，并使用正确的舍入方式。
5. Case Study 4: Geometry and Trigonometry in Measurement | 案例四：测量中的几何与三角学
A surveyor needs to determine the height of a tower without climbing it. From a point A on level ground, the angle of elevation to the top of the tower is 32°. After walking 60 metres closer to the base to point B, the angle of elevation becomes 55°. Find the height of the tower and the distance from B to the base.
一名测量员需要在不攀爬的情况下测定塔的高度。在水平地面上 A 点，测得塔顶仰角为 32°。朝塔底走近 60 米到达 B 点后，仰角变为 55°。求塔高以及 B 到塔底的距离。
from the base is (x + 60) metres, so tan 32° = h / (x + 60). Substitute h and solve for x.
设塔高为 h 米，B 到塔底的距离为 x 米。则在 B 点，tan 55° = h / x，所以 h = x tan 55°。在 A 点，到塔底的距离为 (x + 60) 米，因此 tan 32° = h / (x + 60)。代入 h 并求解 x。
.4281x = 0.6249x + 37.494, so (1.4281 - 0.6249)x = 37.494, hence 0.8032x = 37.494, giving x ≈ 46.7 m. Then h = 46.7 × 1.4281 ≈ 66.7 m.
令两式相等：x tan 55° = (x + 60) tan 32°。使用近似值 tan 55° ≈ 1.4281，tan 32° ≈ 0.6249：1.4281x = 0.6249(x + 60)。展开：1.4281x = 0.6249x + 37.494，则 (1.4281 - 0.6249)x = 37.494，因此 0.8032x = 37.494，得 x ≈ 46.7 m。于是 h = 46.7 × 1.4281 ≈ 66.7 m。
This type of task is common in case studies where trigonometry is applied to indirect measurement. Always ensure your calculator is in degree mode and present your final answers with appropriate units. An alternative approach uses the sine rule in the triangle formed by the tower top, A, and B, illustrating the flexibility of trigonometric tools.
这类任务在案例分析中很常见，即用三角学间接测量。务必确保计算器处于角度模式，并以恰当的单位呈现最终答案。另一种方法是在塔顶、A 和 B 构成的三角形中使用正弦定理，展现了三角工具的灵活性。
6. Case Study 5: Probability and Decision-Making | 案例五：概率与决策
A company is considering launching a new product. If the market is favourable, the profit will be £300,000; if not, the loss will be £80,000. The probability of a favourable market is estimated at 0.55. Alternatively, the company can hire a market research firm for £15,000, which predicts the market correctly 90% of the time. Determine the optimal decision strategy using expected values.
某公司正考虑推出一款新产品。若市场有利，利润为 £300,000；若不利，则亏损 £80,000。市场有利的概率估计为 0.55。公司也可以花 £15,000 聘请市场研究公司，其预测准确率为 90%。使用期望值确定最优决策策略。
without research: EV(no research) = 0.55 × 300,000 + 0.45 × (-80,000) = 165,000 - 36,000 = £ le)×P(correct) + P(unfavourable)×P(incorrect) = 0.55×0.9 + 0.45×0.1 = 0.495 + 0.045 = 0.54. Similarly, P(predict unfavourable) = 0.46.
首先计算无研究时的期望值：EV（无研究）= 0.55 × 300,000 + 0.45 × (-80,000) = 165,000 - 36,000 = £129,000。在有研究的情况下，构建决策树。研究预测“有利”的概率为 P(预测有利) = P(有利)×P(正确) + P(不利)×P(错误) = 0.55×0.9 + 0.45×0.1 = 0.495 + 0.045 = 0.54。类似地，P(预测不利) = 0.46。
redicts favourable, the posterior probability of a truly favourable market is P(favourable | ing given a favourable prediction is 0.9167×300,000 + 0.0833×(-80,000) = 275,010 - 6,664 ≈ £268,346. Subtract the research cost: net gain = 268,346 - 15,000 = £ the research predicts unfavourable, the company would not launch, yielding a net loss of -£00. Thus, the overall EV with research = 0.54×253,346 + 0.46×(-15,000) ≈ 136,807 - 6,900 = £129,907. Since £129,907 > £129,000, hiring the research firm is marginally beneficial.
若预测有利，事后真实市场有利的概率 P(有利|预测有利) = (0.55×0.9) / 0.54 ≈ 0.9167。如果根据有利预测推出产品，期望利润为 0.9167×300,000 + 0.0833×(-80,000) = 275,010 - 6,664 ≈ £268,346。减去研究成本：净收益 = 268,346 - 15,000 = £253,346。若预测不利，公司不会推出产品，净亏损为 -£15,000。因此，有研究时的总体期望值 EV = 0.54×253,346 + 0.46×(-15,000) ≈ 136,807 - 6,900 = £129,907。由于 £129,907 > £129,000，聘请研究公司是有微利的。
Case studies involving probability demand careful handling of conditional probabilities and the clear presentation of decision trees. Even small differences in expected value must be interpreted with awareness of risk attitude and the accuracy of estimates.
涉及概率的案例分析要求仔细处理条件概率，并清晰地呈现决策树。即使是期望值的微小差异，也须结合风险态度和估计的准确性进行解读。
7. Common Pitfalls and How to Avoid Them | 常见陷阱与规避方法
Many students lose marks not because they cannot do the mathematics, but because they overlook contextual details. Typical mistakes include misidentifying independent and dependent variables, using the wrong sign for acceleration, forgetting to check the feasibility of optimisation solutions, or providing an interpretation that is disconnected from the scenario.
许多学生丢分并非因为不会做数学题，而是因为忽略了情境细节。典型错误包括误判自变量和因变量、加速度符号使用错误、忘记检查优化解的可行性，或给出与情境脱节的解释。
Always read the question more than once. Highlight the quantities given and precisely what the question asks you to find
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