📚 Case Study Combat Training for Year 11 CIE Mathematics | Year 11 CIE 数学：案例分析实战演练
Case studies are a hallmark of the CIE IGCSE Mathematics (0580) exam. They push you beyond routine number crunching into the realm of real‑world modelling, interpretation and justification. This article walks you through a hands‑on training programme that builds the skills needed to tackle any case study with confidence.
案例分析是剑桥 IGCSE 数学（0580）考试的标志。它把你从机械计算推向真实世界的建模、解读与论证。本文将带你完成一套实战训练，逐步培养应对任何案例分析题的信心和能力。
1. Understanding the CIE Case Study Format | 理解CIE案例分析题型
A typical CIE case study opens with a contextual scenario – a business launch, a sports tournament, an architectural design. It then poses several sub‑questions that gradually increase in demand. You must extract numerical and descriptive information, select appropriate mathematics, perform calculations, and often explain your reasoning in everyday language.
典型的 CIE 案例分析以一个真实情境开场——比如新产品上市、体育赛事或建筑设计。随后它会提出几个难度递增的子问题。你需要提取数字和文字信息、选择合适的数学方法、完成计算，并且经常要用日常语言解释你的推理过程。
Key success factors include focused reading, clear labelling of variables, and a habit of writing down every step. Diagrams, tables and bullet points are encouraged even if not explicitly asked for; they demonstrate structured thinking.
成功的关键在于专注的阅读、清晰地标记变量以及写下每一步的习惯。即使题目没有明确要求，图表、表格和要点式列示也是值得提倡的；它们能展现你的结构化思维。
2. Data Interpretation: Quadratic Modelling of Sales Revenue | 数据解读：二次函数建模销售收入
A start‑up company’s monthly revenue R, in thousands of dollars, can be modelled by R = 20t – 0.5t², where t is the number of months since launch. The case study asks you to interpret the model, find the month of maximum revenue, and evaluate its reliability.
一家初创公司的月收入 R（千美元）可以用模型 R = 20t – 0.5t² 表示，其中 t 为成立后的月份数。案例分析要求你解释模型、找出收入达到最高的月份并评估其可靠性。
Start by identifying the type of function: it is a quadratic with a negative coefficient of t², so the parabola opens downward. The maximum occurs at the vertex. Using t = –b / (2a) with a = – . That means maximum revenue is achieved at 20 months. Substituting t = 20 gives R = 20×20 – 0.5×20² = 400 – 200 = 200, i.e. $200 000.
首先识别函数类型：这是一个二次函数，且 t² 的系数为负，因此抛物线开口向下。最大值出现在顶点处。利用 t = –b / (2a)，其中 a = –0.5, b = 20，可得 t = 20。即第 20 个月收入达到最高。代入 t = 20 得 R = 20×20 – 0.5×20² = 400 – 200 = 200，即 20 万美元。
The model predicts revenue rising and then falling – possibly due to market saturation. When asked about reliability, mention that real data might fluctuate because of competitors or seasonal effects, so the model is only an approximation.
模型预测收入先升后降——可能是市场饱和所致。被问及可靠性时，可以指出真实数据会因竞争对手或季节效应而波动，因此模型只是一个近似。
3. Geometry in Action: Designing a Storage Box | 几何应用：设计储物箱
A factory produces open‑topped boxes by cutting squares of side x dboard sheet and folding up the flaps. The case study asks for an expression for the volume, the maximum possible volume, and the limitations of the model.
一家工厂从一块 40 cm × 30 cm 的硬纸板的每个角剪去边长为 x cm 的正方形，再折叠成无盖盒子。案例分析要求你写出容积表达式、求出最大容积并讨论模型的局限。
The resulting box will have length (40 – 2x), width (30 – 2x) and height x. Volume V = x(40 – 2x)(30 – 2x). Expand to get a cubic function: V = 4x³ – board dimensions are fixed, x must be between 0 and 15 (otherwise width becomes negative).
折叠后盒子的长为 (40 – 2x)、宽为 (30 – 2x)、高为 x。容积 V = x(40 – 2x)(30 – 2x)。展开得到三次函数：V = 4x³ – 140x² + 1200x。由于纸板尺寸固定，x 必须在 0 到 15 之间（否则宽度变为负值）。
es works well). The maximum volume occurs around x ≈ 5.66 cm, giving approximately 3032 cm³. Always check end points: at x = 0 and x = 15, volume is zero.
要求最大值，可以使用图像或微积分（尽管 IGCSE 不总要求微积分，但用图像或通过代入数值表格尝试效果也很好）。最大容积大约发生在 x ≈ 5.66 cm 处，约为 3032 cm³。务必检查端点：x = 0 和 x = 15 时容积为零。
Limitations include the model ignoring thickness of cardboard, crease width, and the fact that an actual box may not be perfectly rigid. In an exam, highlighting these practical constraints shows deeper understanding.
模型的局限性包括忽略了纸板厚度、折痕宽度，以及实际盒子可能不够刚硬等。在考试中，指出这些实际限制能体现更深层次的理解。
4. Sequences and Patterns: Tiling a Courtyard | 数列与规律：庭院铺砖
A designer arranges square tiles in patterns that form black and white rings. Pattern 1 has 1 black tile; Pattern 2 has 8 black tiles; Pattern 3 has 16 black tiles. The case study asks you to extend the pattern, find a formula for the number of black tiles, and justify your reasoning.
一位设计师用正方形瓷砖排成黑白相间的环状图案。图案 1 有 1 块黑砖；图案 2 有 8 块；图案 3 有 16 块。案例分析要求你扩展图案，找出黑砖数量的公式并进行论证。
ely look like a common linear or quadratic sequence. But if you analyse the geometry, each ‘ tiles forms a border around a smaller square. For Pattern n, the central white area is (2n–3)² when n≥2, so black tiles = (2n–1)² – (2n–3)². Simplify: (4n² – 4n + 1) – (4n² – 12n + 9) = 8n – 8. For n=2 gives 8, n=3 gives 16; for n=1 the formula gives 0, but the special case is 1.
数列 1, 8, 16, … 初看不像常见的等差或等比数列。但若分析几何结构，每一圈黑砖都围绕着更小的白色正方形。对图案 n，中心白色区域当 n≥2 时为 (2n–3)²，故黑砖数 = (2n–1)² – (2n–3)²。化简得：8n – 8。n=2 得 8，n=3 得 16；n=1 代入得 0，实际上特例为 1。
Thus the general term for n≥2 is 8n – 8, with Pattern 1 as an initial case. You can then predict Pattern 10 will have 8×10 – study may also ask for the total number of tiles, including white ones. Pattern n has (2n–1)² total tiles.
因此当 n≥2 时通项为 8n – 8，图案 1 作为初始特例。你可以预测图案 10 有 72 块黑砖。案例分析还可能询问包含白砖的总瓷砖数。图案 n 共有 (2n–1)² 块。
Forming your own sequence and testing a rule teaches you to move from numerical observation to algebraic proof – a skill that examiners love.
自己构建数列并检验规律，训练你从数值观察迈向代数证明——这是考官青睐的能力。
5. Statistical Analysis: Customer Satisfaction Survey | 统计分析：客户满意度调查
A café conducts a survey asking 200 customers to rate their experience on a scale from 1 (poor) to 5 (excellent). The results are summarised in a frequency table. The case study asks for the mean rating, a cumulative frequency graph, and the interquartile range to assess consistency.
一家咖啡馆调查了 200 位顾客，请他们从 1（差）到 5（优）评分。结果汇总成频数表。案例分析要求计算平均评分，绘制累积频数图并求四分位距以评估评分的一致性。
Suppose the frequencies are: score 1 – 20, 2 – 35, 3 – 60, 4 – 55, 5 – 80 + 220 + 150 = 640. Mean = 640 / 200 = 3.2. So the average rating is just above neutral.
假设频数为：评分 1 – 20, 2 – 35, 3 – 60, 4 – 55, 5 – 30。均值 = Σ(fx) / Σf。Σfx = 1×20 + … + 5×30 = 640，均值 = 3.2。可见平均评分略高于中等。
dary of the score. Draw a smooth curve and find the median (≈3), lower quartile (≈2.5) and upper quartile (≈4). IQR = 4 – 2.5 = 1.5, indicating moderate spread. A small IQR suggests consistent satisfaction.
为了绘制累积频数图，计算累积值：≤1:20, ≤2:55, ≤3:115, ≤4:170, ≤5:200。选取评分上界作图并绘制平滑曲线，找出中位数（≈3）、下四分位数（≈2.5）和上四分位数（≈4）。四分位距 IQR = 1.5，表示评分离散程度适中。较小的 IQR 意味着满意度比较一致。
Comment on the shape of the curve: if it is steep in the middle, ratings cluster around the median. Use this to advise the café whether its service is consistent.
对曲线形状加以评论：如果中间段陡峭，说明评分集中在中位数附近。借此可向咖啡馆建议服务是否稳定。
6. Financial Mathematics: Loan Repayment Plan | 金融数学：贷款还款计划
A student borrows $5000 at an annual compound interest rate of 4%, compounded monthly. The case demands a calculation of the total owed after 3 years, and then proposes a monthly repayment plan with the question: how much must the monthly payment be to clear the debt in 3 years?
一名学生以年利率 4% 按月复利借入 5000 美元。案例要求计算 3 年后的总欠款，然后提出按月还款方案并询问：每月需还多少才能在 3 年内还清贷款？
Compound interest formula: A = P(1 + r/n)nt. Here P=5000, r=0.04, n=12, t=3. A = 5000 × (1 + 0.04/12)12×3 = 5000 × (1.003333...)36 ≈ 5000 × 1.12749 ≈ $5637.45. So if no repayments are made, the debt grows to about $5637.
复利公式：A = P(1 + r/n)nt。这里 P=5000, r=0.04, n=12, t=3。A = 5000 × (1 + 0.04/12)36 ≈ 5000 × 1.12749 ≈ 5637.45 美元。如果不还款，债务将增至约 5637 美元。
For the repayment plan, use the present value of an annuity formula: Loan = R × [1 – (1 + i)–N] / i, where i = 0.04/12, N=36. Solve for R: 5000 = R × [1 – 1.003333...^(–] / 0.003333... Using the known annuity factor (approx 33.8708) gives R ≈ 5000 / 33.8708 ≈ $ study expects you to set up the equation and perhaps use trial and error or a given table.
对于还款计划，使用年金现值公式：贷款 = R × [1 – (1 + i)–N] / i，其中 i = 0.04/12，N=36。解出 R ≈ 147.60 美元/月。题目可能简化计算，但案例分析期望你建立方程，或许使用试错法或给出的表格。
Interpret the result: paying $147.60 monthly clears the debt. Compare with paying only interest-only payments, showing the power of amortisation. This scenario links algebra, functions, and financial literacy.
解读结果：每月支付 147.60 美元可还清债务。与仅付利息的还款方式对比，显示摊还的力量。这个情境连接了代数、函数和金融素养。
7. Probability in Game Design: Lottery Expectation | 概率与游戏设计：抽奖期望值
A school funfair designs a game: a player draws two marbles without replacement from a bag containing 3 red, 5 blue, and 2 green. It costs $1 to play; a player wins $10 for two reds, $2 for one red and one blue, and nothing otherwise. The case study asks whether the game is fair, and what the organiser’s expected profit is over 500 plays.
学校游园会设计了一个游戏：玩家从装有 3 红、5 蓝、2 绿的袋中无放回地摸出两个弹珠。每次花费 1 美元；如果摸到两个红球赢 10 美元，一红一蓝赢 2 美元，其他无奖。案例分析要求判断游戏是否公平，并计算 500 次游戏后主办方的预期利润。
(15+15)/90 = 30/90 = 1/3. All other outcomes have zero prize. Expected winnings per game = $10 × (1/15) + $2 × (1/3) = $10/15 + $2/3 = $2/3 + $2/3 = $4/3 ≈ $1.33.
总弹珠数 10。两红的概率 = (3/10)×(2/9)=6/90=1/15。一红一蓝的概率 = (3/10)×(5/9)+(5/10)×(3/9)=30/90=1/3。其他结果无奖。每次游戏的期望奖金 = $10×1/15 + $2×1/3 = $10/15+$2/3 = $4/3 ≈ $1.33。
Since it costs $1 to play, the expected gain for a player is $0.33. The organiser’s expected profit per game = $1 – $1.33 = –$0.33; actually it appears the organiser loses money! Better re‑ The game is not favourable for the organiser. For 500 plays, expected loss = 500 × 0.33 = $165. If the organiser wants profit, they should reduce prizes or increase ticket price.
由于每次花费 1 美元，玩家的期望收益为 0.33 美元，主办方期望利润 = 1 – 1.33 = –0.33 美元；也就是说平均每局亏损 0.33 美元。对于 500 次，预期亏损 165 美元。若主办方想盈利，应降低奖金或提高票价。
This case demonstrates how probability drives business decisions. Always present dollar amounts clearly and interpret the expectation in plain words.
这个案例展示了概率如何驱动商业决策。务必清晰呈现金额并用平实的语言解释期望值。
8. Graphs and Motion: Interpreting a Speed‑Time Graph | 图像与运动：解读速度—时间图
A cyclist’s journey is modelled by a speed‑time graph: it rises linearly from 0 to 8 m/s over 10 s, stays constant for 20 s, then slows down uniformly to stop over the next 15 s. The case study asks for the total distance travelled and the average speed.
一位自行车骑手的行程用速度—时间图表示：前 10 秒从 0 匀加速至 8 m/s，匀速 20 秒，随后 15 秒匀减速至停止。案例分析要求计算总行驶距离和平均速度。
Distance is the area under the speed‑ ht 8), area = 160 m. Triangle for deceleration (base 15, height 8), area = ½ × 15 × 8 = 60 m. Total distance = 40+160+60 = 260 m. The total time is 10+20+15 = 45 s, so average speed = 260/45 ≈ 5.78 m/s.
距离等于速度—时间图下的面积。分为三个形状：加速三角形底 10、高 8，面积 = ½×10×8 = 40 m；匀速矩形底 20、高 8，面积 = 160 m；减速三角形底 15、高 8，面积 = ½×15×8 = 60 m。总距离 = 260 m。总时间 45 s，平均速度 ≈ 5.78 m/s。
533 m/s²). Relate these to the gradient of the line. Also, discuss whether the graph is realistic: in real life, acceleration is not always perfectly constant.
案例可能进一步询问瞬时加速度（8 ÷ 10 = 0.8 m/s²）和减速度（−8/15 ≈ −0.533 m/s²）。将其与线段斜率联系起来。还要讨论图像的现实性：现实中加速度并不总是完全均匀的。
9. Trigonometry in Navigation: Distance Between Two Ships | 三角学导航：两艘船的距离
Ship A sails from port O on a bearing of 070° for 12 km, while Ship B sails on a bearing of 150° for 8 km. The case study asks for the distance between the two ships and the bearing of A from B. This combines bearings, the cosine rule, and sine rule.
A 船从港口 O 以方位角 070° 航行 12 km，B 船以方位角 150° 航行 8 km。案例分析要求计算两船的距离以及从 B 到 A 的方位角。这结合了方位角、余弦定理和正弦定理。
Begin by representing the paths as vectors from O. Angle AOB = 150° – le AOB with OA=12 km, OB=8 km, and included angle 80°. Use the cosine rule: AB² = 12² + 8² – 2×12×8×cos 80° = 144 + 64 – 192×0.173648 = 208 – 33.34 ≈ 174.66, so AB ≈ 13.22 km.
首先将路径表示为从 O 发出的向量。角 AOB = 150° – 070° = 80°。于是三角形 AOB 中 OA=12, OB=8, 夹角 80°。用余弦定理：AB² = 144+64−192×cos80° ≈ 208−33.34 = 174.66，AB ≈ 13.22 km。
.4°. The bearing of Boat B’s line to A is the bearing of B from O plus the turn: from B, the direction to A involves the line BO reversed. Bearings can be worked out: North line at B, measure clockwise. This part requires careful angle chase – a typical high‑demand item.
要求从 B 到 A 的方位，需用正弦定理求角 OBA。sin(OBA) ≈ 12 sin80°/13.22 ≈ 0.8937, 角OBA ≈ 63.4°。再通过角度计算得出方位角——这部分需要仔细的角度推算，是典型的拉分题。
Always sketch the diagram and label all known angles. Use three‑figure bearings and round to one decimal place as appropriate. The case rewards clear method and correct interpretation of the bearing definition.
始终绘制示意图并标注所有已知角度。使用三位数方位角，适当保留一位小数。清晰的解题步骤和正确的方位角定义理解为得分关键。
10. Optimisation: Maximum Area of a Rectangular Enclosure | 优化问题：最大矩形围栏面积
A farmer has 100 m of fencing and wants to build a rectangular enclosure against an existing straight wall (so only three sides need fencing). The case asks for the dimensions that maximise the area and the maximum area itself. It also invites a discussion on constraints.
一位农民有 100 米围栏，想靠着一面笔直的墙搭建一个矩形围场（因此只需围三边）。案例要求找出使面积最大的尺寸及最大面积，并讨论约束条件。
et the side perpendicular to the wall be x m; then the side parallel to the wall uses (100 – 2x) m of fencing. Area A = x(100 – 2x) = 100x – 2x². This is a quadratic opening downward. The vertex at x = –b/(2a) = –100/(2×–2) = 100/4 = 25 m. The parallel side = 100 – 2×25 = 50 m. Maximum area = 25 × 50 = 1250 m².
设垂直于墙的边长为 x 米，则平行于墙的边长为 (100 – 2x) 米。面积 A = x(100 – 2x) = 100x – 2x²，是开口向下的二次函数。顶点 x = –b/(2a) = –100/(–4) = 25 m。平行边 = 100 – 2×25 = 50 m。最大面积 = 1250 m²。
Constraints include x > 0 and 100 – 2x > 0 → x < 50, which is satisfied. Answer the “why” effect if the farmer requires a gate that uses extra fencing, which would alter the model.
约束条件有 x>0 且 100–2x>0 → x<50，均满足。解答“为什么”的部分：只有当墙是直的且足够长，且没有其他障碍时，二次模型才适用。案例还可能询问如果农民需要安装门而占用额外围栏，会对模型产生什么影响。
11. Exam‑Style Preparation and Common Pitfalls | 考试型准备与常见错误
Success in case studies depends on more than mathematical technique. Read the entire scenario before tackling sub‑questions. Highlight key numbers, units and the question command words. Many students lose marks by ignoring the context: forgetting to convert units, not rounding to an appropriate degree of accuracy, or failing to write a concluding sentence.
案例分析的成功不仅取决于数学技巧。解决子问题前先通读整个情境。标出关键数字、单位以及问题中的指令词。许多学生因忽略语境而丢分：忘记转换单位、未按要求精度四舍五入，或没有写出总结句。
Pitfall #1: Misinterpreting the modelling assumption. For example, assuming linear growth when data clearly curves. Pitfall #2: wrong use of probability – applying “with replacement” when “without” is stated. Pitfall #3: not checking that an answer is sensible in context (negative length, impossible angle). Always do a quick sanity check.
常见错误一：误解模型假设，比如当数据明显呈弯曲趋势时仍假定线性增长。错误二：概率应用不当——题目明确“不放回”却按“放回”计算。错误三：未检查答案在情境中是否合理（如长度为负、不可能的角度）。务必做快速合理性检查。
Practice writing short “explain” answers: “The model is less reliable after month 20 because other market factors become dominant.” Such statements turn a calculation into a real‑world evaluation.
练习写出简短的“解释”答案，例如：“20 个月后模型可靠性下降，因为其他市场因素开始占据主导。”这样的表述能将计算升华为现实评估。
12. Full Practice Mini‑Case: The Theme Park | 综合实战演练：主题公园案例
A theme park plans a new roller‑coaster ride. The height h (m) of the track above ground can be modelled by h = – 2 for the first section, where d is horizontal distance from the start. The case study sub‑questions: (a) What is the initial height? height and the distance at which it occurs. (c) For safety, the track must not exceed 30 m
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