📚 Case Study Practice: Real-World Applications in Cambridge IGCSE Mathematics | 案例分析实战演练：剑桥 IGCSE 数学的实际应用
Case study questions in Cambridge IGCSE Mathematics are designed to bridge the gap between textbook theory and real-life problem solving. They often present a scenario with numerical information, diagrams, and a series of linked tasks. To succeed in these extended questions, you need to read carefully, identify relevant mathematical techniques, and communicate your reasoning clearly. This article walks you through a variety of typical case studies, showing step-by-step methods and highlighting common pitfalls so you can approach your exams with confidence.
剑桥 IGCSE 数学中的案例分析题旨在连接课本理论与现实问题解决。这类题目通常会给出一个情境，结合数据信息、图表和一系列相关联的子问题。要想在这种大题中拿高分，你需要仔细阅读、识别合适的数学方法，并清晰地呈现推理过程。本文将带你遍历多个典型案例，逐步展示解题方法并揭示常见陷阱，帮助你自信地应对考试。
1. Understanding Case Study Questions | 理解案例分析题型
Case study questions appear in both Core and Extended papers, usually carrying a high mark weighting. They often cover multiple topics within a single scenario. The key is to break the problem down into smaller, manageable parts. Start by highlighting numerical data, units, and any constraints given. Identify the subject areas involved—such as algebra, geometry, statistics, or ratio. Sketching a diagram or creating a table can help organise information before you begin calculations.
案例分析题出现在核心和扩展试卷中，通常分值很高。这类题目常常在一个情境中融合多个知识点。关键是将其拆解为小而可控的部分。第一步是标出所有数值、单位和约束条件。确定涉及的学科领域——比如代数、几何、统计或比例。在开始计算之前，绘制示意图或创建表格有助于整理信息。
Always show your working step by step. Cambridge examiners award method marks even if the final answer is incorrect. Use correct mathematical notation and state any formulas before substituting numbers. If the question asks for a conclusion or interpretation, make sure you refer back to the context—for example, explaining whether a budget is sufficient or a measurement is realistic.
始终要一步步展示解题过程。即使最终答案错误，剑桥考官也会给出方法分。使用正确的数学符号，在代入数值前先写出所用公式。如果题目要求得出结论或进行解释，务必回扣情境——例如，说明预算是否足够或尺寸是否现实。
2. Case Study 1: Personal Finance and Percentages | 案例研究一：个人理财与百分比
Imagine a scenario: Emma earns £2400 per month. She pays 22% in tax and National Insurance, and then spends 35% of her remaining income on rent. Of the money left, she saves ⅖ and spends the rest on living expenses. The question might ask: How much does she save each month? What percentage of her gross income is spent on rent? If she receives a 5% pay rise, how much extra will she save per month?
设想一个场景：Emma 每月赚取 2400 英镑。她缴纳 22% 的税和国民保险，然后将剩余收入的 35% 花在房租上。在剩下的钱中，她储蓄 2/5，其余用于生活开销。题目可能会问：她每月储蓄多少？房租支出占她总收入的百分比是多少？如果她加薪 5%，每月将多存多少钱？
First, calculate the tax: 22% of £2400 = 0.22 × 2400 = £528. So her net income is £2400 – £528 = £1872. Rent is 35% of net income: 0.35 × 1872 = £655.20. After rent, remaining = £1872 – 655.20 = £1216.80. Savings are ⅖ of this: (2/5) × 1216.80 = £486.72. The rent as a percentage of gross income is (655.20 / 2400) × 100% = 27.3%.
先计算税收：2400 英镑的 22% = 0.22 × 2400 = 528 英镑。因此净收入为 2400 – 528 = 1872 英镑。房租占净收入的 35%：0.35 × 1872 = 655.20 英镑。付完房租后剩余 1872 – 655.20 = 1216.80 英镑。储蓄为其中的 2/5：(2/5) × 1216.80 = 486.72 英镑。房租占原始总收入百分比为 (655.20 / 2400) × 100% = 27.3%。
With a 5% rise, new gross income = £2400 × 1.05 = £2520. Repeating the steps gives new savings of £511.06 approximately. The increase in savings is £511.06 – £486.72 = £ This type of multi-step percentage calculation is extremely common in IGCSE case studies.
加薪 5% 后，新的总收入为 2400 × 1.05 = 2520 英镑。重复上述步骤可得新月储蓄约为 511.06 英镑。储蓄增加额为 511.06 – 486.72 = 24.34 英镑/月。这种多步百分比计算在 IGCSE 案例题中极为常见。
3. Case Study 2: Designing a Garden with Geometry | 案例研究二：几何设计花园
A landscape architect designs a rectangular garden measuring 15 m by 8 m. It includes a circular pond with radius 1.8 m and a square patio of side 3.5 m. Grass covers the remaining area. The case study asks for the area of grass, the length of fencing needed to enclose the garden, and the number of paving slabs (each 0.5 m by 0.5 m) required for the patio, including 10% extra for wastage.
一位景观设计师设计了一个长 15 米、宽 8 米的矩形花园。花园包含一个半径 1.8 米的圆形池塘和一个边长 3.5 米的正方形平台。剩余区域为草坪。本案例要求计算草坪面积、围栏总长，以及平台所需铺路石板的数量（每块 0.5 米 × 0.5 米），需额外增加 10% 以考虑损耗。
3.142 × 3.24 = 10.18 m² (using π ≈ 3.142). Patio area = 3.5² = 12.25 m². Grass area = 120 – ab = 0.5 × 0.5 = 0.25 m². Number needed without wastage = 12.25 / 0.25 = 49 slabs. With 10% extra: 49 × 1.1 = 53.9, so order 54 slabs.
花园总面积 = 15 × 8 = 120 平方米。池塘面积 = π × (1.8)² ≈ 3.142 × 3.24 = 10.18 平方米（π 取 3.142）。平台面积 = 3.5² = 12.25 平方米。草坪面积 = 120 – (10.18 + 12.25) = 97.57 平方米。围栏为周长：2 × (15 + 8) = 46 米。一块石板面积 = 0.5 × 0.5 = 0.25 平方米。不计损耗所需数量 = 12.25 / 0.25 = 49 块。增加 10%：49 × 1.1 = 53.9，故订购 54 块。
Notice how geometry case studies blend area, perimeter, and arithmetic with practical rounding decisions. Always check whether rounding up or down is appropriate, and make your reasoning explicit.
请注意几何案例如何将面积、周长和算术与实际的取舍相结合。务必判断该向上取整还是向下取整，并明确写出理由。
4. Case Study 3: Survey Data and Statistics | 案例研究三：调查数据与统计
A school surveys 120 students about the number of hours they spend on homework per week. The results are grouped. The case study may ask you to estimate the mean, complete a cumulative frequency table, draw a cumulative frequency curve, and find quartiles. Finally, you compare two groups, such as Year 10 and Year 11, using box plots.
一所学校对 120 名学生每周做作业的小时数进行了调查。数据已分组。案例题可能要求你估算平均值、完成累积频数表、绘制累积频数曲线并找出四分位数。最后用箱线图比较两组，如十年级和十一年级。
e upper boundary of each interval against the running total. The median is the value at 50% of total frequency, the lower quartile at 25%, and the upper quartile at 75%. The interquartile range (IQR) = UQ – five key values visually. When comparing, comment on medians and IQRs in context, such as 'ar 11 students typically spend more time on homework, and their hours are more consistent.'
对于分组数据，均值估计 = Σ（组中值 × 频数）/ 总频数。绘制累积频数图时，将每个区间的上界与累计频数描点。中位数在总频数的 50% 处，下四分位数在 25%，上四分位数在 75%。四分位距 IQR = UQ – LQ。箱线图直观地展示了这五个关键值。比较时要结合背景，例如“十一年级学生通常花更多时间做作业，且用时更稳定”。
Be precise when reading values from graphs; draw faint construction lines onto the graph and label them clearly. State any assumptions made, such as data being evenly spread within intervals, which is necessary for estimating the mean.
从图中读取数值时要精确；在图上画出辅助线并清晰标注。说明所做的任何假设，例如假设数据在区间内均匀分布，这是估算均值所必需的。
5. Case Study 4: Travel Planning – Speed, Distance, and Currency | 案例研究四：旅行规划——速度、距离与汇率
ave enough time if she drives at an average speed of 60 km/h, considering a 15-minute stop? dy ? The case study might also involve currency exchange: Maria then flies from London to Paris and changes £400 into euros at a rate of €1.15 = £1, with a 2% commission. How many euros does she receive?
Maria 从伦敦自驾前往布莱顿。距离 85 公里，她需在上午 10:30 前到达。如果她上午 9:00 出发，以平均时速 60 km/h 行驶并中途休息 15 分钟，时间是否足够？案例可能还涉及货币兑换：之后 Maria 从伦敦飞往巴黎，将 400 英镑兑换成欧元，汇率为 €1.15 = £1，收取 2% 的手续费。她能得到多少欧元？
am + 1 h 40 min = 10:40 am, which is later than 10:30 am. So she will not arrive on time; she needs 10 more minutes. For currency: without commission, euros = 400 × 1.15 = €460. Commission amount = 2% of €460 = €9.20, so she receives €460 – €9.20 = € lternatively, the commission may be taken in pounds first, so read the question carefully.
行驶时间 = 距离 / 速度 = 85 / 60 = 1.4167 小时 = 1 小时 25 分钟。总行程时间含休息 = 1 小时 25 分 + 15 分 = 1 小时 40 分。出发时间 9:00 + 1 小时 40 分 = 10:40，晚于 10:30。所以她无法准时到达；需要多 10 分钟。货币兑换：无手续费时，欧元 = 400 × 1.15 = €460。手续费 = €460 的 2% = €9.20，因此她得到 €460 – €9.20 = €450.80。也有可能是先从英镑中扣除手续费，所以要仔细审题。
stance / time, and proportional reasoning with exchange rates. Always check for mixed units (minutes and hours) and do not round prematurely.
这些现实任务考察时间单位的处理、速度公式以及汇率的比例推理。务必检查混合单位（分钟与小时），并且不要过早地进行四舍五入。
6. Case Study 5: Business Growth and Algebraic Modelling | 案例研究五：业务增长与代数建模
A company's profit, P (in thousands of pounds), is modelled by the formula P = 2n² + 5n – 3, where n is the number of years since 2020. The case study asks to calculate profit in the third year, find when profit reaches £50 000, and compare with a linear model P = 15n + 10. Which model predicts higher profit in 2025? Interpret the meaning of the constant term.
某公司的利润 P（千英镑）由公式 P = 2n² + 5n – 3 模拟，其中 n 为自 2020 年起的年数。案例要求计算第三年的利润，找出利润达到 50,000 英镑的时间，并与线性模型 P = 15n + 10 进行比较。哪个模型预测 2025 年的利润更高？解释常数项的含义。
For n = 3 (year 2023), P = 2(3)² + 5(3) – 3 = 18 + 15 – 3 = 30, so profit is £30 000. To find when P = 50, solve 2n² + 5n – 3 = 50 → 2n² + 5n – 53 = 0. Use the quadratic formula: n = [–5 ± √25 + 424] / 4 = [–5 ± √449] / 4. √449 ≈ 21.19, so positive n = (–5 + 21.19)/4 = 4.0475. Thus in the 5th year (during 2025) profit reaches £50 000. For 2025, n = 5. Quadratic model: P = 2(25) + 25 – 3 = 50 + 25 – 3 = 72 (£72 000). Linear model: P = 15(5) + 10 = 85 (£85 000). Linear predicts higher that year. Constant terms –3 and +10 give profit at n=0 (year 2020), though negative might indicate initial loss.
当 n = 3（2023 年），P = 2(3)² + 5(3) – 3 = 18 + 15 – 3 = 30，即利润为 30,000 英镑。求 P = 50 时，解 2n² + 5n – 3 = 50 → 2n² + 5n – 53 = 0。用求根公式：n = [–5 ± √25 + 424] / 4 = [–5 ± √449] / 4。√449 ≈ 21.19，正根 n = (–5 + 21.19)/4 = 4.0475。故在第五年（2025 年间）利润达到 50,000 英镑。对 2025 年，n = 5。二次模型：P = 2(25) + 25 – 3 = 72（72,000 英镑）。线性模型：P = 15(5) + 10 = 85（85,000 英镑）。线性模型当年预测更高。常数项 –3 与 +10 表示 n=0（2020 年）时的利润，负值可能意味着初期亏损。
Modelling questions assess your ability to substitute into expressions, solve equations, and evaluate the reasonableness of results in a business context. Be careful with units (thousands of pounds) and state answers clearly.
建模题评估你对代数的代入、解方程以及在商业背景下评估结果合理性的能力。注意单位（千英镑）并清晰地陈述答案。
7. Case Study 6: Mixture and Proportionality | 案例研究六：混合物与比例推理
A chemist mixes two liquids, A and B, in the ratio 3:2 by volume. Liquid A costs £12 per litre, and liquid B costs £15 per litre. Find the cost per litre of the mixture. If she needs 35 litres of the mixture, what volume of each liquid is required? The case study extends: she evaporates some water from 5 litres of liquid A to increase its concentration by 20%. What is the new volume?
一位化学家将液体 A 和 B 按体积比 3:2 混合。A 每升 12 英镑，B 每升 15 英镑。求混合物每升的成本。若需 35 升混合物，各需多少体积的 A 和 B？案例延伸：她从 5 升液体 A 中蒸发掉一些水，使其浓度增加 20%。新体积是多少？
For the mixture, in one 'batch' of 3+2 = 5 litres, cost = 3×12 + 2×15 = 36 + 30 = £66. Cost per litre = 66/5 = £ increase: if original volume is 5 litres with some solute, increasing concentration by 20% concentration. Suppose original concentration is c. After evaporation, concentration becomes c+0.2c = 1.2c. Since solute constant, c×5 = 1.2c × Vnew, so Vnew = 5c/(1.2c) = 5/1.2 ≈ 4.17 litres. The question would supply numerical details.
对于混合物，在 3+2 = 5 升的一个“批次”中，成本 = 3×12 + 2×15 = 36 + 30 = 66 英镑。每升成本 = 66/5 = £13.20。需 35 升时，总份数 = 35/5 = 7 批。因此 A 体积 = 7×3 = 21 升，B 体积 = 7×2 = 14 升。浓度增加：若原体积 5 升并含有一定溶质，浓度增加 20% 意味着溶质量不变，但体积减少。没有原始浓度数据时，题目通常会给出目标浓度。假设原浓度为 c。蒸发后浓度为 1.2c。由于溶质恒定，c×5 = 1.2c × Vnew，故 Vnew = 5c/(1.2c) = 5/1.2 ≈ 4.17 升。题目会提供具体数字。
Mixture problems test ratio, proportion, and sometimes inverse proportionality. When water evaporates, the volume decreases but the amount of dissolved substance stays the same—this is a classic inverse proportion scenario.
混合问题考察比和比例，有时也涉及反比例。当水蒸发时，体积减少但溶质质量不变——这是一个典型的反比例场景。
8. Dissecting the Mark Scheme | 剖析评分方案
Cambridge mark schemes for case studies are structured into method (M), accuracy (A), and sometimes communication (B) marks. M marks are awarded for a correct approach, even if subsequent arithmetic is flawed. A marks require correct final answers. For example, in a compound interest calculation, using the formula correctly earns an M1, while the correct final amount earns an A1. If you make an early arithmetic slip, you can still gain M marks later by using your incorrect value consistently.
剑桥的案例分析评分方案分为方法分(M)、准确度分(A)以及有时出现的表达分(B)。方法分授予正确的解题思路，即便后续运算有误。准确度分需要最终答案正确。例如，在复利计算中，正确使用公式可得 M1，而正确算出最终金额则得 A1。如果你早期计算失误，但只要后续始终使用你那个错误值正确推理，仍可获得后面的方法分。
Always write down formulas before substituting numbers. In statistics, drawing graphs neatly to scale and plotting points correctly earn M marks. Label axes and curves; this can secure B marks. When a question says 'show that' a result is true, you must demonstrate every step—these often have marks specifically for the demonstration. Check the number of marks available: a 3-mark question suggests at least two intermediate steps are needed.
务必先写下公式再代入数值。在统计题中，按比例整洁地绘制图表并正确描点可获方法分。标注坐标轴和曲线；这可以确保表达分。当题目要求“证明”某个结果时，你必须展示每一步——这类题通常有专门用于证明过程的分数。根据可得分值判断：一道 3 分的题目意味着至少需要两个中间步骤。
9. Common Pitfalls and How to Avoid Them | 常见陷阱及规避方法
One frequent mistake is ignoring units. If speed is given in km/h and time in minutes, convert minutes to hours before using the formula. Another is misreading a percentage increase: a common error is to add the percentage to 100% incorrectly or to apply a multiplier such as 1.05 instead of 1.5. In geometry, students often confuse radius with diameter, or forget to multiply by 2 when finding circumference and perimeter.
一个常见错误是忽略单位。如果速度单位是 km/h 而时间是分钟，使用公式前必须将分钟转换为小时。另一个错误是误读百分比的增加：常见的错误包括在 100% 基础上增加百分比时算错，或错误地使用乘数，如将 1.5 当作 1.05。几何题中，学生常常混淆半径与直径，或在计算周长时忘记乘以 2。
When interpreting graphs, always check the scale. A tiny error in reading a value from a cumulative frequency curve can change quartiles and the interquartile range. In algebra, carefully handle signs when expanding brackets or solving equations. Finally, answer the exact question asked; if the context asks 'Is this sufficient?' you must give a yes/no conclusion based on your calculation, not just the number.
解读图表时，务必检查刻度。从累积频数曲线上读取数值的微小误差会改变四分位数和四分位距。代数运算中，去括号或解方程时要小心处理正负号。最后，要回答题目所问；如果背景问“是否足够？”你必须基于计算结果给出是/否的结论，而不只是报出数字。
10. Practice Recommendations and Final Tips | 练习建议与终极技巧
Build fluency by working through past paper case studies under timed conditions. Start by doing them open-book if necessary, but soon progress to closed-book. For each case study, write a brief plan before you start calculating: list what you know, what you need to find, and possible methods. After completing, check your work with the mark scheme, noting where you lost marks and why.
通过限时完成历年真题的案例题来培养熟练度。必要时可以先开卷练习，但应尽快过渡到闭卷。每做一道案例题，在开始计算前先写一个简短的规划：列出已知条件、需要求解的目标以及可能的方法。完成后，对照评分方案检查，记下失分之处及其原因。
Create a 'case study toolkit' in your revision notes: common percentage multipliers, area and volume formulas, the speed triangle, and the quadratic formula. Practice mental arithmetic and estimation so you can quickly verify if an answer is plausible. On exam day, read the entire case study first, then start with the parts you find easiest—this builds confidence and saves time. Remember, method marks are generous; never leave a part blank.
在复习笔记中创建一个“案例工具箱”：常见百分比乘数、面积与体积公式、速度三角形以及二次公式。练习心算和估算，以便快速判断答案是否合理。考试当天，先通读整个案例，然后从最简单的部分入手——这能建立信心并节省时间。记住，方法分给得很宽松；永远不要让任何一个小问空着。
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