📚 Year 11 CAIE Mathematics: Case Study Practical Exercises | 案例分析实战演练
A strong performance in CAIE IGCSE Mathematics often hinges on the ability to apply learned techniques to unfamiliar, real‑world problems. Case studies challenge you to move beyond routine textbook exercises and reason mathematically in context. This article guides you through four carefully chosen practical examples, highlighting the step‑by‑step analytical process and the core skills you need to demonstrate in your exams.
在CAIE IGCSE数学考试中获得高分，往往取决于你是否能将所学知识灵活运用到陌生的现实问题中。案例分析题要求你突破常规练习，在具体情境中进行数学推理。本文将通过四个精心挑选的实战例子，带你一步步拆解分析过程，并梳理考试中必须具备的核心技能。
1. Understanding the Case Study Approach | 理解案例分析的方法
A case study in mathematics is not a separate topic – it is an extended problem that draws together several branches of the syllabus. You might need to combine algebra with geometry, or statistics with ratio and proportion. The key is to read the problem carefully, identify what you are being asked to find, and then translate the written scenario into mathematical language.
数学中的案例研究并不是一个独立的课题，而是一道综合性的题目，往往融合了课程大纲中的多个知识板块。你可能需要把代数与几何结合起来，或者将统计与比和比例配合使用。关键是要仔细读题，明确需要求解的未知量，然后把文字描述的场景转译成数学语言。
2. Key Mathematical Skills Required | 所需的关键数学技能
Before diving into the examples, remind yourself of the fundamental skills assessed in case‑study questions: forming and solving equations, interpreting graphs, handling formulas with multiple variables, and justifying conclusions in everyday language. Accuracy with units and sensible rounding are equally important, as they reflect the real‑world context.
在进入实例之前，请回顾案例分析题考查的基本功：建立并求解方程、解读图像、处理含多个变量的公式，以及用日常语言合理解释结论。对单位的精确把握和合理取整同样重要，因为它们反映了现实背景中的实际意义。
3. Case 1: Optimising a Rectangular Garden | 案例一：优化矩形花园面积
You have 24 metres of fencing to enclose a rectangular garden against a wall. The wall forms one side of the rectangle, so fencing is needed only for the three remaining sides. Let the side parallel to the wall be y metres and the two equal perpendicular sides be x . The total fencing used is 2x + y = 24. The area of the garden, A, is given by A = x × y.
你有24米长的围栏，要靠着墙围出一个矩形花园。墙面构成矩形的一条边，所以只需在其余三边使用围栏。设平行于墙的边长为y米，两条相等的垂直边长为x米。使用的围栏总长为2x + y = 24。花园的面积A为 A = x × y。
4. Solving the Quadratic Model | 求解二次模型
Express y in terms of x from the fencing constraint: y = 24 – 2x. Substitute into the area formula to obtain A = x(24 – 2x) = 24x – id shape, x must be greater than 0 and less than 12. The quadratic expression is a concave‑ e square or using symmetry, the optimum x is 6, giving y = 12 and a maximum area of 72 m².
由围栏约束条件用x表示y：y = 24 – 2x。代入面积公式得 A = x(24 – 2x) = 24x – 2x²。由于围栏要构成有效的图形，x必须大于0且小于12。这个二次式对应开口向下的抛物线，最大值出现在顶点处。通过配方法或利用对称性，可求得最佳 x 为 6，相应的 y 为 12，最大面积为 72 平方米。
5. Interpreting the Results | 解释结果
case, the garden that maximises area with 24 m of fencing against a wall should have two 6‑m sides perpendicular to the wall and a 12‑ both dimensions are positive and the fencing length is 2×6 + 12 = 24 m, exactly as given.
只有把答案放回原情境中，解题才算完整。就这个例子而言，用24米围栏靠着墙围出最大面积时，应让垂直于墙的两条边各长6米，平行于墙的边长12米。可获得的最大面积是72平方米。你还应验证解是否符合取值范围：当x=6时，所有边长均为正，围栏总长2×6 + 12 = 24米，与条件完全一致。
6. Case 2: Compound Interest Over Time | 案例二：复利随时间增长
A person invests $2000 in an account that pays 4% d annually. The task is to find how many whole years it takes for the investment to exceed $ A = P(1 + r/100)ⁿ, where P is the principal, r is the rate, and n is the number of years.
某人将2000美元投入一个年利率为4%、每年复利一次的投资账户。问题是，需要经过多少整年投资额才能超过3000美元，并利用公式 A = P(1 + r/100)ⁿ 预测一定年限后的金额，其中P为本金，r为利率，n为年数。
7. Setting Up and Solving the Inequality | 建立并求解不等式
4) > log(1.5). Thus n > log(1.5) ÷ log(1.04). Using a calculator, log(1.5) ≈ 0.1761 and log(1.04) ≈ 0.0170, so n > 10.36. The investment will exceed $3000 after 11 whole years.
我们需要找出满足不等式 2000 × (1.04)ⁿ > 3000 的最小整数n。两边同除以2000得 (1.04)ⁿ > 1.5。取对数（以任意基底均可）得 n × log(1.04) > log(1.5)。因此 n > log(1.5) ÷ log(1.04)。利用计算器，log(1.5) ≈ 0.1761，log(1.04) ≈ 0.0170，所以n > 10.36。经过11个整年后投资额将超过3000美元。
8. Predicting Future Value and Checking | 预测未来值并验算
k reasonableness check: 8 years is shorter than the doubling time (which is roughly 72/4 = 18 years), so $2000 growing by about 37% feels plausible. Always ask yourself whether the answer makes sense in the real‑world scenario.
要计算正好8年后的金额，将n=8代入公式：A = 2000 × (1.04)⁸。计算得 (1.04)⁸ ≈ 1.3686，故 A ≈ 2737.20美元。快速合理性检验：8年短于翻倍所需时间（大约72/4=18年），所以2000美元增长约37%是合理的。务必时常反问自己，答案在真实情境中是否说得通。
9. Case 3: Distance–Time Graph of a Journey | 案例三：旅程的距离–时间图
A distance–time graph shows a car’s journey. The car travels at a constant speed for 30 minutes to cover 25 km, stops for 15 minutes, then returns home at a steady speed, arriving 90 minutes after starting. The graph consists of two straight sloping segments and one horizontal segment. Questions often ask for the speed on each leg, the average speed for the whole journey, and the total distance travelled.
一幅距离–时间图展示了汽车的旅程。汽车先以恒定速度行驶30分钟走了25公里，停留15分钟，然后以稳定速度返回起点，从出发到返回共用时90分钟。该图由两条倾斜的直线段和一段水平线段组成。题目通常会要求计算各段的速率、整个旅程的平均速度以及行驶的总路程。
10. Analysing Motion and Average Speed | 分析运动与平均速度
min stop, the return leg takes 45 minutes (0.75 h). Return speed = 25 km ÷ 0.75 h ≈ 33.3 km/h. Total distance travelled = 25 km out + 25 km back = 50 km. Average speed = total distance ÷ total time = 50 km ÷ 1.5 h ≈ 33.3 km/h.
第一段：速率 = 25公里 ÷ 0.5小时 = 50公里/小时。回程必须在剩余时间走完25公里。总行程时间为1.5小时，因此除去30分钟行驶和15分钟停留，回程需要45分钟（0.75小时）。回程速率 = 25公里 ÷ 0.75小时 ≈ 33.3公里/小时。总行驶距离 = 去程25公里 + 回程25公里 = 50公里。平均速率 = 总距离 ÷ 总时间 = 50公里 ÷ 1.5小时 ≈ 33.3公里/小时。
11. Case 4: Probability with Independent Events | 案例四：相互独立事件的概率
e probability that at least one is green. Draw a tree diagram to map the possibilities: P(first red) = 5/10 = 1/2. After removing one red, P(second red) = 4/9. So P(both red) = (1/2) × (4/9) = 4/18 = 2/9. For at least one green, consider the complementary event: no green drawn at all.
一个袋子里有5个红色、3个蓝色和2个绿色筹码。随机取出两个且不放回。求两个都是红色的概率，以及至少有一个是绿色的概率。画出树状图理清各种可能：P(第一个红) = 5/10 = 1/2。拿走一个红后，P(第二个红) = 4/9。所以 P(两个红) = (1/2) × (4/9) = 4/18 = 2/9。对于至少一个绿，可考虑互补事件：完全没有抽到绿色。
12. Drawing Conclusions and Exam Tips | 得出结论与考试技巧
Complement: P(no green) means both counters are from the 8 non‑en ones. P(no green) = (8/10) × (7/9) = 56/90 = 28/45. Therefore P(at least one green) = 1 – able decimal. Time management is crucial: if a case study feels overwhelming, break it into smaller parts and tackle them one by one.
互补情况：无绿即两个筹码都来自8个非绿色筹码。P(无绿) = (8/10) × (7/9) = 56/90 = 28/45。因此 P(至少一绿) = 1 – 28/45 = 17/45。考试中务必清晰写出方法，必要时定义符号，最终答案以化简分数或合适的小数呈现。时间管理至关重要：如果案例分析题令你感到庞大，就将其拆解成若干小部分，逐一攻克。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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