📚 Year 11 CIE Maths: Case Study Practice | Year 11 CIE 数学案例分析实战演练
Case studies bring mathematics to life by connecting abstract theory to real-world scenarios. For Year 11 CIE IGCSE candidates, practising case studies sharpens problem-solving skills, builds confidence and deepens understanding of how different topics intertwine. This article presents eight carefully designed case studies covering algebra, geometry, trigonometry, statistics and more, each followed by step-by-step reasoning and exam-style commentary.
案例分析通过将抽象理论与实际场景相联系，让数学变得鲜活。对于 Year 11 CIE IGCSE 考生来说，练习案例分析可以锻炼解题技巧、增强信心并加深对各个知识点如何相互交织的理解。本文呈现了八个精心设计的案例，涵盖代数、几何、三角学、统计等内容，每个案例都配有逐步推理和考试风格的点评。
1. Function Application Case Study | 函数应用案例分析
A theme park models the path of a roller-coaster drop using the quadratic function h(t) = – nager needs to know the maximum height reached and the time when the coaster is exactly 10 m above the ground.
某主题公园用二次函数 h(t) = –4.9t² + 15t + 20 模拟过山车的一段俯冲路径，其中 h 为高度（米），t 为俯冲开始后的时间（秒）。管理者需要知道最大高度以及过山车恰好位于地面以上 10 m 的时刻。
To find the maximum height, use the vertex formula: t = –b/(2a) = –15/(2 × –4.9) = –15/–9.8 = 1.53 s (to 3 sf). Then h(1.53) = –4.9(1.53)² + 15(1.53) + 20 ≈ 31.5 m. For h = 10, solve –4.9t² + 15t + 20 = 10 ⇒ –4.9t² + 15t + 10 = 0. Multiply by –10 to get 49t² – 150t – 100 = 0. Using the quadratic formula: t = [150 ± √(150² – 4×49×(–100))] / (2×49) = [150 ± √22500 + 19600] / 98 = [150 ± √42100] / 98. √42100 ≈ 205.2, giving t ≈ (150 + 205.2)/98 ≈ 3.62 s (positive root only).
要找出最大高度，使用顶点公式：t = –b/(2a) = –15/(2 × –4.9) = 1.53 秒（3 位有效数字）。然后 h(1.53) ≈ 31.5 米。对于 h = 10，解方程 –4.9t² + 15t + 20 = 10 得到二次式 49t² – 150t – 100 = 0，代入求根公式得 t ≈ 3.62 秒（只取正值）。
2. Modelling with Equations and Inequalities | 方程与不等式建模案例
A small business produces custom mugs. Fixed daily costs are £80, and each mug costs £2.50 to produce. The selling price is £6 per mug. The owner wants to know how many mugs must be sold to make a profit of at least £200, and express the break-even condition as an inequality.
一家小企业生产定制马克杯。每日固定成本为 80 英镑，每个杯子的生产成本为 2.50 英镑，售价 6 英镑。老板想知道至少需要卖出多少个杯子才能盈利不低于 200 英镑，并将盈亏平衡条件用不等式表示。
Let n be the number of mugs sold. Profit P = 6n – (80 + 2.5n) = 3.5n – 80. For a profit of at least £200: 3.5n – 80 ≥ 200 ⇒ 3.5n ≥ 280 ⇒ n ≥ 80. Break-even occurs when profit is zero or positive, so 3.5n – 80 ≥ 0 ⇒ n ≥ 22.86, thus at least 23 mugs must be sold to avoid loss.
设 n 为卖出的杯子数量。利润 P = 6n – (80 + 2.5n) = 3.5n – 80。要盈利至少 £200：3.5n – 80 ≥ 200 ⇒ n ≥ 80。盈亏平衡意味着利润大于等于零：3.5n – 80 ≥ 0 ⇒ n ≥ 22.86，因此至少卖出 23 个杯子才能不亏损。
If VAT of 20% is added to the selling price, the new selling price becomes £7.20. The new profit per mug is 7.20 – 2.50 = £4.70. The inequality for at least £200 profit becomes 4.70n – 80 ≥ 200 ⇒ n ≥ 59.57, so 60 mugs.
如果售价加上 20% 增值税，新售价为 7.20 英镑，每个杯子利润变为 4.70 英镑。盈利 ≥ £200 的不等式变为 4.70n – 80 ≥ 200 ⇒ n ≥ 59.57，因此需要 60 个杯子。
3. Geometry Problem Solving | 几何问题实战
An architect designs a rectangular courtyard with a diagonal pathway. The courtyard measures 24 m by 10 m. Two circular flower beds, each of radius 1.5 m, are placed at opposite corners along the diagonal. Find the length of the diagonal and the remaining area of the courtyard taken up by paving, giving your answer in terms of π.
一位建筑师设计了一个带有对角线通道的长方形庭院。庭院尺寸为 24 米 × 10 米。沿对角线的两个对角各有一个半径为 1.5 米的圆形花坛。求对角线长度以及除去花坛后铺装区域的面积，答案用 π 表示。
Using Pythagoras: diagonal = √24² + 10² = √576 + 100 = √676 = 26 m. Area of courtyard = 24 × 10 = 240 m². Area of one circle = π × 1.5² = 2.25π. Two circles cover 4.5π m². Thus paved area = 240 – 4.5π m².
运用勾股定理：对角线 = √24² + 10² = √676 = 26 米。庭院总面积 = 240 平方米。每个圆的面积 = π × 1.5² = 2.25π。两个圆总面积为 4.5π 平方米。铺装面积 = 240 – 4.5π 平方米。
The client also asks for a 0.5 m wide border of gravel along all four sides of the courtyard. The inner rectangle then measures 23 m by 9 m. Its area is 207 m². The paved area (excluding flower beds) inside the border becomes 207 – 4.5π m².
客户还要求在庭院四周铺设 0.5 米宽的砾石边。内部矩形尺寸变为 23 米 × 9 米，面积为 207 平方米。边界内的铺装区域（除去花坛）变为 207 – 4.5π 平方米。
4. Mensuration and Trigonometry | 测量与三角学实战
A surveyor marks three points A, B and C such that AB = 150 m, BC = 200 ngle ABC = 110°. She needs the distance AC and the area of triangle ABC for land registry.
一位测量员标出三点 A、B 和 C，其中 AB = 150 米，BC = 200 米，∠ABC = 110°。她需要计算距离 AC 及三角形 ABC 的面积以登记土地。
By the cosine rule: AC² = AB² + BC² – 2×AB×BC×cos(110°) = 150² + 200² – 2×150×200×cos110°. cos110° ≈ –0.3420, so AC² = 22500 + 40000 – 60000×(–0.3420) = 62500 + 20520 = 83020. AC = √83020 ≈ 288.1 m.
用余弦定理：AC² = 150² + 200² – 2×150×200×cos110° = 22500 + 40000 – 60000×(–0.3420) = 83020，AC ≈ 288.1 米。
50 × 200 × sin110°. sin110° ≈ 0.9397, so area = 75 × 200 × 0.9397 = 15000 × 0.9397 ≈ 14096 m², or 1.41 ha.
面积 = ½ × 150 × 200 × sin110° ≈ 0.5 × 150 × 200 × 0.9397 ≈ 14096 平方米，合 1.41 公顷。
coordinates can be set: let B = (0,0), A = (150,0). Then C = (200cos110°, 200sin110°) ≈ (–68.40, 187.94). Midpoint of AC = ((150 – 68.40)/2, (0 + 187.94)/2) = (40.80, 93.97), distance from B ≈ √40.80² + 93.97² ≈ 102.5 m.
若要在 AC 中点放置标记，可设坐标 B(0,0), A(150,0)，则 C(200cos110°, 200sin110°) ≈ (–68.40, 187.94)。AC 中点约 (40.80, 93.97)，距 B 约 102.5 米。
5. Statistics and Probability | 统计与概率案例
A school records the number of late arrivals over 30 school days: 0,1,2,0,0,1,3,2,1,1,0,2,4,1,0,2,1,3,1,0,2,2,1,0,4,1,2,0,3,1. The leadership team wants a five-number summary and a box plot to discuss punctuality.
一所学校记录了 30 个上课日的迟到人数：0,1,2,0,0,1,3,2,1,1,0,2,4,1,0,2,1,3,1,0,2,2,1,0,4,1,2,0,3,1。校领导希望获得五数概括和箱线图以讨论守时情况。
= (30+1)/2 = 15.5, so median = average of 15th and 16th values. Cumulative counts: up to 0:9; 1:19; so 15th and 16th are both 1, median = 1. Lower quartile Q1: position (30+1)/4 = 7.75, average of 7th and 8th: both 0 (since 7th is within first 9 zeros), Q1 = 0. Upper quartile Q3: 3/4 × 31 = 23.25, average of 23rd and 24th: 23rd is 2, 24th is 2, Q3 = 2. Five-number summary: 0, 0, 1, 2, 4. Interquartile range = 2 – 0 = 2.
排序后计数：0 出现 9 次，1 出现 10 次，2 出现 7 次，3 出现 3 次，4 出现 2 次。最小值 0，最大值 4。中位数位置 15.5，取第 15、16 个值的均值，两者均为 1，中位数为 1。下四分位数 Q1：第 7 和第 8 个值的均值，均为 0，故 Q1 = 0。上四分位数 Q3：第 23 和 24 个均值，均为 2，故 Q3 = 2。五数概括：0, 0, 1, 2, 4。四分位距 = 2。
and 1)/30 = (9+10)/30 = 19/30 ≈ 0.633. This suggests room for improvement in punctuality.
随机选择一天迟到人数少于 2 的概率为 (9+10)/30 = 19/30 ≈ 0.633，表明守时方面仍有提升空间。
6. Graph Interpretation | 图表解读实战
A water tank is filled at a constant rate of 10 litres per minute for 5 minutes, then the rate increases to 25 litres per minute for the next 3 minutes. After a pause of 2 minutes, the tank is emptied at 30 litres per minute until empty. The tank initially contains 50 litres. Draw the volume-time graph and find the total time until the tank is empty.
一个水箱最初有 50 升水。以 10 升/分钟的恒定速率注水 5 分钟，然后速率提高到 25 升/分钟继续注水 3 分钟。暂停 2 分钟后，以 30 升/分钟的速率排水，直到排空。绘制体积–时间图并计算水箱排空的总时长。
First 5 min: gain 10×5 = 50 L, volume becomes 100 L. Next 3 min: gain 25×3 = 75 L, volume = 175 L. Pause 2 min: volume stays 175 L. Then drain at 30 L/min: time taken = 175 / 30 ≈ 5.833 min. Total time = 5 + 3 + 2 + 5.833 = 15.833 min (15 min 50 s).
前 5 分钟：增加 50 升，总体积变为 100 升。接下来 3 分钟：增加 75 升，体积达到 175 升。暂停 2 分钟，体积保持 175 升。排水阶段：175 ÷ 30 ≈ 5.833 分钟。总时间 ≈ 15.833 分钟（15 分 50 秒）。
The graph consists of: (0,50) to (5,100) – line with slope 10; (5,100) to (8,175) – slope 25; (8,175) to (10,175) – horizontal; (10,175) to (15.833,0) – slope –30. This piecewise linear graph helps visualise filling and draining processes, a common CIE exam task.
图像由以下线段组成：(0,50) 至 (5,100) 斜率 10；(5,100) 至 (8,175) 斜率 25；(8,175) 至 (10,175) 水平线；(10,175) 至 (15.833,0) 斜率 –30。这种分段线性图帮助直观理解注水与排水过程，是 CIE 考试常见题型。
7. Sequences Case Study | 数列案例分析
A staircase is built using concrete blocks. The first step uses 3 blocks, the second step uses 7 blocks, the third uses 11 blocks, and so on, forming an arithmetic sequence. The staircase has 15 steps. How many blocks are needed for the 15th step, and what is the total number of blocks for the whole staircase?
某楼梯用混凝土砖块砌成。第一级台阶用 3 块砖，第二级用 7 块，第三级用 11 块，以此类推，形成等差数列。楼梯共有 15 级台阶。第 15 级台阶需要多少块砖？整段楼梯总共需要多少块砖？
Sequence: 3, 7, 11, … common difference d = 4. nth term formula: aₙ = a₁ + (n– . Sum of first 15 terms: S₁₅ = n/2 × (a₁ + a₁₅) = 15/2 × (3 + 59) = 7.5 × 62 = 465 blocks.
数列：3, 7, 11,… 公差 d = 4。通项公式 aₙ = 3 + (n–1)×4。第 15 项 a₁₅ = 3 + 14×4 = 59 块。前 15 项和 S₁₅ = 15/2 × (3+59) = 7.5 × 62 = 465 块。
If the design changes to a geometric pattern for a decorative pillar: 2, 6, 18, 54, … m of first 8 terms. Here a₁ = 2, r = 3. 8th term = 2 × 3⁷ = 2 × 2187 = 4374. Sum S₈ = 2(3⁸ – 1)/(3 – 1) = 2(6561 – 1)/2 = 6560.
如果设计改为装饰柱的几何图案：2, 6, 18, 54,…，求第 8 项及前 8 项和。首项 2，公比 3。第 8 项 = 2 × 3⁷ = 4374。和 S₈ = 2(3⁸ – 1)/(3 – 1) = 6560。
8. Vectors and Transformations | 向量与变换实战
A drone delivery service plans routes using position vectors. Depot O is at (0,0). A parcel is taken from O to P with vector (8, 5) and then from P to Q with vector (–2, 7). All units are kilometres. The drone then returns directly from Q to the depot. Find the coordinates of Q, the direct return vector and the total distance flown, correct to 3 significant figures.
一家无人机送货公司用位置向量规划路线。仓库 O 位于 (0,0)。包裹从 O 运到 P，向量为 (8, 5)，再从 P 到 Q，向量为 (–2, 7)。单位为公里。然后无人机从 Q 直接返回仓库。求 Q 的坐标、直接返回的向量以及飞行总距离，保留 3 位有效数字。
Coordinates of P = (8,5). Q = P + (–2,7) = (8–2, 5+7) = (6, 12). Return vector QO = O – Q = (0–6, 0–12) = (–6, –12) or simply the opposite of (6,12). Magnitudes: |OP| = √8²+5² = √89 ≈ 9.434 km. |PQ| = √((–2)²+7²) = √53 ≈ 7.280 km. |QO| = √6²+12² = √180 ≈ 13.416 km. Total distance ≈ 9.434 + 7.280 + 13.416 = 30.13 km ≈ 30.1 km.
P 点坐标为 (8,5)。Q = (8,5) + (–2,7) = (6, 12)。返回向量 QO = (–6, –12)。各段距离：|OP| = √89 ≈ 9.434 km，|PQ| = √53 ≈ 7.280 km，|QO| = √180 ≈ 13.416 km。总距离 ≈ 30.1 km（3 位有效数字）。
ransformed by a rotation of 90° anticlockwise about O after reaching Q, the new position Q' is found by multiplying by the rotation matrix: (0 –1; 1 0) × (6,12) = (–12, 6). This demonstrates linking vectors with transformations.
如果无人机在到达 Q 后绕原点逆时针旋转 90°，新位置 Q' 由旋转矩阵计算：(0 –1; 1 0) × (6,12) = (–12, 6)。这展示了向量与变换的结合。
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