📚 Mastering CIE KS3 Advanced Mathematics through Case Studies | CIE KS3 进阶数学案例分析实战演练
Mathematics comes alive when we use it to solve real problems. This article presents a series of case studies designed for CIE KS3 Advanced Mathematics, guiding you through practical applications of algebra, geometry, statistics, and more. Each case will break down a realistic scenario, ask key questions, and demonstrate step‑by‑step reasoning.
当我们用数学解决真实问题时，它才变得生动。本文为 CIE KS3 进阶数学设计了一系列案例分析，带领你体验代数、几何、统计等知识的实际应用。每个案例会拆解真实情境，提出关键问题，并展现一步步的推理过程。
1. Introduction to Problem‑Solving | 问题解决入门
Before diving into cases, it’s helpful to recall structured problem‑solving: understand the problem, devise a plan, carry out the plan, and check your answer. In these case studies we apply this framework to topics you meet in the CIE Advanced Mathematics syllabus, building both confidence and fluency.
在深入案例之前，回顾一下结构化解题方法会很有帮助：理解问题、制定计划、执行计划、检查答案。在这些案例研究中，我们将这一框架应用于 CIE 进阶数学大纲中的课题，同时建立信心和流畅度。
2. The Fencing Problem – Optimising Area | 围栏问题 – 面积最优化
A farmer has 60 metres of fencing and wants to enclose a rectangular field against a straight river, so only three sides need fencing. What is the maximum possible area the field can have?
一位农民有 60 米围栏，想靠一条笔直的河围一个矩形田，因此只需围三面。这块田可能的最大面积是多少？
Let the two equal sides perpendicular to the river be x metres each, and the side parallel to the river be (60 − 2x) metres. The area A = x(60 − 2x) = 60x − 2x². This is a quadratic function opening downwards. To find its maximum, we can complete the square: A = −2(x² − 30x) = −2[(x − 15)² − 225] = 450 − 2(x − 15)². The maximum A is 450 m² when x = 15.
设垂直于河的两条等边各为 x 米，平行于河的一边为 (60 − 2x) 米。面积 A = x(60 − 2x) = 60x − 2x²。这是一个开口向下的二次函数。为了求最大值，我们可以配方：A = −2(x² − 30x) = −2[(x − 15)² − 225] = 450 − 2(x − 15)²。当 x = 15 时，最大面积 A = 450 m²。
This classic optimisation illustrates how a quadratic expression can model a real constraint and give a precise answer. Think about why the symmetric shape – a width twice the depth – yields the largest area under a fixed perimeter condition.
这个经典优化展示了二次表达式如何对实际约束建模并给出精确答案。想一想为什么对称的形状——宽度是深度的两倍——在固定周长条件下产生最大面积。
3. Number Pattern Challenge – Sequences and Rules | 数字模式挑战 – 数列与规则
A pattern of dots forms the first three shapes of a sequence: Shape 1 has 4 dots, Shape 2 has 7 dots, Shape 3 has 10 dots. Each shape adds a row of dots with a fixed increment. Find the number of dots in Shape n and the shape that uses exactly 100 dots.
一个点阵模式给出了数列的前三个图形：图形 1 有 4 个点，图形 2 有 7 个点，图形 3 有 10 个点。每个图形增加一行点，增量固定。求第 n 个图形的点数以及恰好使用 100 个点的图形。
The sequence 4, 7, 10, … is arithmetic with first term a = 4 and common difference d = 3. The nth term formula is T(n) = a + (n − 1)d = 4 + 3(n − 1) = 3n + 1. To find the shape with 100 dots, set 3n + 1 = 100 → 3n = 99 → n = 33. Shape 33 has 100 dots.
序列 4, 7, 10, … 是等差数列，首项 a = 4，公差 d = 3。第 n 项公式为 T(n) = a + (n − 1)d = 4 + 3(n − 1) = 3n + 1。求恰好 100 个点的图形：令 3n + 1 = 100 → 3n = 99 → n = 33。第 33 个图形有 100 个点。
Spotting arithmetic and geometric patterns equips you to make predictions far beyond the initial data. This is the foundation of algebraic modelling in many strands of mathematics.
发现等差和等比规律能让你做出远超初始数据的预测，这是许多数学分支中代数建模的基础。
4. Designing a Water Tank – Volume and Surface Area | 设计水箱 – 体积与表面积
A factory needs an open‑top square‑based tank that holds exactly 32 m³ of water. The base side length is s, and the height is h. Find the dimensions that minimise the material used (surface area of the four walls and the base).
一家工厂需要一个无盖正方底水箱，恰好容纳 32 m³ 水。底面边长 s，高度为 h。求使材料用量（四个侧面和底面的表面积）最小的尺寸。
Volume constraint: s²h = 32 → h = 32/s². Surface area M = base + four sides = s² + 4sh = s² + 4s(32/s²) = s² + 128/s. To minimise M, we can test values or use calculus (or graph the function). A table of values shows M decreases then increases. At s = 4, M = 16 + 32 = 48; at s = 5, M = 25 + 25.6 = 50.6; at s = 3, M = 9 + 42.67 = 51.67. The minimum occurs near s = 4. Check s = 4: h = 32/16 = 2 m. So the optimal tank is 4 m by 4 m by 2 m, using 48 m² of material.
体积约束：s²h = 32 → h = 32/s²。表面积 M = 底面 + 四个侧面 = s² + 4sh = s² + 4s(32/s²) = s² + 128/s。为求 M 最小，我们可以试值或使用微积分（或画函数图）。取值表显示 M 先减后增。s = 4 时 M = 48；s = 5 时 M = 50.6；s = 3 时 M = 51.67。最小值在 s = 4 附近。验证 s = 4：h = 32/16 = 2 m。最优水箱尺寸为 4 m × 4 m × 2 m，用料 48 m²。
This problem combines volume and surface area in an optimisation context. It shows how changing shape affects material cost – an important industrial design principle.
这个问题在优化背景下融合了体积和表面积，它展示了形状变化如何影响材料成本——一个重要的工业设计原理。
5. Probability in Games – Expected Outcomes | 游戏中的概率 – 期望结果
A game involves rolling a fair six‑sided die. If you roll a 6, you win £5; if you roll a 1, you lose £2; otherwise you win nothing. What is the expected gain per roll? Is it worth paying £0.50 to play each round?
一个游戏投掷一枚均匀的六面骰子。掷出 6 点赢 5 英镑；掷出 1 点输 2 英镑；其他情况无输赢。每次投掷的期望收益是多少？每轮支付 0.50 英镑玩是否划算？
Probability of 6 = 1/6, gain = +5; probability of 1 = 1/6, loss = −2; probability of 2,3,4,5 = 4/6 = 2/3, gain = 0. Expected value = (1/6)×5 + (1/6)×(−2) + (2/3)×0 = 5/6 − 2/6 = 3/6 = £0.50. So the expected gain per roll is £0.50. If it costs £0.50 to play, the net expected outcome is £0, making it a fair game in the long run.
概率：6 点的概率 = 1/6，收益 +5；1 点的概率 = 1/6，损失 −2；2,3,4,5 点的概率 = 4/6 = 2/3，收益 0。期望值 = (1/6)×5 + (1/6)×(−2) + (2/3)×0 = 5/6 − 2/6 = 3/6 = £0.50。每轮期望收益为 0.50 英镑。若玩一次花费 0.50 英镑，净期望结果为 0，长期来看这是一个公平游戏。
Expected value is a powerful tool for comparing risks and rewards. Always convert all outcomes to their numeric values and multiply by probabilities – that reveals the hidden “average” result.
期望值是比较风险与回报的有力工具。始终将所有结果转化为数值并乘以概率——这会揭示隐藏的“平均”结果。
6. Interpreting Statistical Data – Which Average to Use? | 解读统计数据 – 使用哪种平均数？
A small company pays its five employees the following annual salaries: £18 000, £20 000, £22 000, £25 000, and £95 000 (the managing director). Calculate the mean, median, and mode. Which average best represents the typical salary, and why?
一家小公司支付其五名员工的年薪如下：18 000 英镑、20 000 英镑、22 000 英镑、25 000 英镑和 95 000 英镑（主管）。计算平均数、中位数和众数。哪种平均数最能代表典型工资，为什么？
| Measure | Calculation | Value |
|---|---|---|
| Mean | (18+20+22+25+95)/5 | £36 000 |
| Median | Middle value of ordered list: 18,20,22,25,95 → 22 | £22 000 |
| Mode | No value repeats | None |
The mean is pulled up to £36 000 by the single very high salary, while four out of five employees earn £25 000 or less. The median (£22 000) better reflects a typical employee’s earnings. This example highlights why choosing the correct measure of central tendency matters.
平均数被一个特高工资拉高到 36 000 英镑，而五分之四的员工收入为 25 000 英镑或以下。中位数（22 000 英镑）更好地反映了典型员工的收入。这个例子凸显了选择正确的集中趋势度量为何重要。
7. Speed, Distance, Time – Reading Graphs | 速度、距离与时间 – 阅读图像
A cyclist travels from town A to town B. The journey has three stages: a fast 10 km ride in 0.4 hours, a 15‑minute rest, then a slower 5 km ride in 0.5 hours. Draw and interpret the distance‑time graph, and find the average speed for the whole journey.
一名骑行者从 A 镇前往 B 镇。行程分三阶段：用 0.4 小时快速骑行 10 km，休息 15 分钟，然后用 0.5 小时慢骑 5 km。画出并解读距离‑时间图，并求全程平均速度。
The total distance is 10 + 5 = 15 km. Total time includes riding 0.4 h + rest 0.25 h + riding 0.5 h = 1.15 hours. Average speed = total distance / total time = 15 km / 1.15 h ≈ 13.04 km/h. On the graph, the first segment is steep (fast), the rest shows a horizontal plateau, and the final segment is less steep (slower). The gradient of each segment equals the speed at that stage.
总距离为 10 + 5 = 15 km。总时间包括骑行 0.4 h + 休息 0.25 h + 骑行 0.5 h = 1.15 小时。平均速度 = 总距离 / 总时间 = 15 km / 1.15 h ≈ 13.04 km/h。在图上，第一段陡峭（快），休息段呈水平平台，最后一段坡度较缓（慢）。每段的斜率等于该阶段的速度。
This graphical approach turns a multi‑stage motion problem into a visual story, making it easier to spot constant speed, stops, and acceleration.
这种图像方法将多阶段运动问题转化为视觉故事，更容易发现匀速、停顿和加速。
8. Real‑Life Algebra – Age and Coin Puzzles | 生活化代数 – 年龄与硬币谜题
Three years ago, Jamal was four times as old as his sister Zara. In five years from now, Jamal will be twice as old as Zara. Find their current ages.
三年前，贾马尔是他妹妹扎拉年龄的四倍。从现在起五年后，贾马尔将是扎拉年龄的两倍。求他们现在的年龄。
Let Jamal’s present age be j and Zara’s be z. Three years ago: j − 3 = 4(z − 3) → j − 3 = 4z − 12 → j = 4z − 9. In five years: j + 5 = 2(z + 5). Substitute j: (4z − 9) + 5 = 2z + 10 → 4z − 4 = 2z + 10 → 2z = 14 → z = 7. Then j = 4×7 − 9 = 19. So Jamal is 19 and Zara is 7. Check: three years ago Jamal 16, Zara 4 (4×); in five years Jamal 24, Zara 12 (2×). Correct.
设贾马尔现在的年龄为 j，扎拉为 z。三年前：j − 3 = 4(z − 3) → j − 3 = 4z − 12 → j = 4z − 9。五年后：j + 5 = 2(z + 5)。代入 j：(4z − 9) + 5 = 2z + 10 → 4z − 4 = 2z + 10 → 2z = 14 → z = 7。于是 j = 4×7 − 9 = 19。贾马尔 19 岁，扎拉 7 岁。验证：三年前贾马尔 16，扎拉 4（4 倍）；五年后贾马尔 24，扎拉 12（2 倍）。正确。
Age problems train you to translate words into equations. Set up variables carefully and look for two distinct time points – the system of equations will do the rest.
年龄问题训练你将文字转化为方程。仔细设定变量，寻找两个不同的时间点——方程组会完成剩下的工作。
9. Angles and Bearings – Navigation | 角度与方位角 – 导航
A ship sails 8 km on a bearing of 060° from port P to buoy Q, then turns to a bearing of 150° and sails 6 km to reach point R. Find the direct distance and bearing from P to R.
一艘船从港口 P 沿方位角 060° 航行 8 km 到达浮标 Q，然后转向方位角 150° 航行 6 km 到达点 R。求 P 到 R 的直线距离和方位角。
Draw triangle PQR. The change in bearing from 060° to 150° is a 90° turn (since 150° − 060° = 90°), so angle PQR = 90°. Thus triangle PQR is right‑angled at Q, with PQ = 8 km, QR = 6 km. By Pythagoras, PR = √(8² + 6²) = 10 km. To find the bearing of R from P, calculate angle QPR: tan(θ) = 6/8 → θ = tan⁻¹(0.75) ≈ 36.87°. The bearing from P to R = initial bearing 060° + θ ≈ 096.87° (or approximately 097°). So the direct distance is 10 km on a bearing of about 097°.
画出三角形 PQR。方位角从 060° 变为 150° 是 90° 的转向（因为 150° − 060° = 90°），因此角 PQR = 90°。于是三角形 PQR 在 Q 处为直角，PQ = 8 km，QR = 6 km。根据勾股定理，PR = √(8² + 6²) = 10 km。求 R 相对于 P 的方位角：计算角 QPR：tan(θ) = 6/8 → θ = tan⁻¹(0.75) ≈ 36.87°。从 P 到 R 的方位角 = 起始方位角 060° + θ ≈ 096.87°（约 097°）。因此直线距离为 10 km，方位角约为 097°。
Bearings are always measured clockwise from north. Recognising the right angle here simplifies the problem – a reminder to look for perpendicular relationships in navigation diagrams.
方位角总是从正北顺时针测量。认出这里的直角简化了问题——这提醒我们在导航图中寻找垂直关系。
10. Linear Programming – Maximising Profit | 线性规划 – 利润最大化
A small factory produces chairs and tables. Each chair requires 1 hour of assembly and 2 hours of finishing, giving £30 profit. Each table requires 3 hours of assembly and 1 hour of finishing, giving £50 profit. The factory has 15 assembly hours and 12 finishing hours available per day. Find the production plan that maximises daily profit.
一个小工厂生产椅子和桌子。每把椅子需要 1 小时组装和 2 小时打磨，利润 30 英镑。每张桌子需要 3 小时组装和 1 小时打磨，利润 50 英镑。工厂每天有 15 组装小时和 12 打磨小时可用。求使日利润最大化的生产计划。
Let c = number of chairs, t = number of tables. Constraints: Assembly: c + 3t ≤ 15; Finishing: 2c + t ≤ 12; c ≥ 0, t ≥ 0. Profit P = 30c + 50t. Plotting the feasible region, the vertices are (0,0), (0,5), (6,0), and the intersection of c+3t=15 and 2c+t=12. Solve simultaneously: from first, c = 15 − 3t; substitute: 2(15 − 3t) + t = 12 → 30 − 6t + t = 12 → 30 − 5t = 12 → 5t = 18 → t = 3.6, c = 15 − 10.8 = 4.2. Since we need whole products, check integer points near (4.2, 3.6): (4,3) and (4,4). (4,4) fails assembly: 4+12=16>15. (4,3) works: assembly 4+9=13≤15, finishing 8+3=11≤12. Profit: 30×4 + 50×3 = 120+150=270. Check other vertices: (0,5) profit 250; (6,0) profit 180. Maximum profit is £270 with 4 chairs and 3 tables.
设 c = 椅子数，t = 桌子数。约束条件：组装：c + 3t ≤ 15；打磨：2c + t ≤ 12；c ≥ 0, t ≥ 0。利润 P = 30c + 50t。绘制可行域，顶点为 (0,0)、(0,5)、(6,0) 以及 c+3t=15 和 2c+t=12 的交点。联立求解：由第一个式子，c = 15 − 3t；代入：2(15 − 3t) + t = 12 → 30 − 6t + t = 12 → 30 − 5t = 12 → 5t = 18 → t = 3.6，c = 4.2。因为产品必须为整数，检查 (4.2, 3.6) 附近的整点：(4,3) 和 (4,4)。(4,4) 组装不满足：4+12=16>15。(4,3) 可行：组装 4+9=13≤15，打磨 8+3=11≤12。利润：30×4 + 50×3 = 270。检查其他顶点：(0,5) 利润 250；(6,0) 利润 180。最大利润为 270 英镑，生产 4 把椅子和 3 张桌子。
Linear programming turns limited resources into optimal decisions. When solutions aren’t integers, testing nearby lattice points ensures you stay within constraints.
线性规划将有限资源转化为最优决策。当解不是整数时，测试附近的格点可确保不超出约束。
11. Probability Trees – Dependent Events | 概率树 – 相依事件
A bag contains 4 red and 6 blue marbles. Two marbles are drawn without replacement. What is the probability of getting two different colours? Draw a tree diagram to support your answer.
一个袋子里有 4 颗红色和 6 颗蓝色弹珠。依次取出两颗，不放回。求两颗颜色不同的概率。画出树状图辅助解答。
Tree: First draw: P(Red) = 4/10, P(Blue) = 6/10. Second draw probabilities change: if Red first, remaining 3 red, 6 blue (9 total): P(Red|Red) = 3/9, P(Blue|Red) = 6/9. If Blue first, remaining 4 red, 5 blue (9 total): P(Red|Blue) = 4/9, P(Blue|Blue) = 5/9. Two different colours are (Red then Blue) or (Blue then Red). P(different) = (4/10)×(6/9) + (6/10)×(4/9) = 24/90 + 24/90 = 48/90 = 8/15.
树状图：第一次抽：P(红) = 4/10，P(蓝) = 6/10。第二次概率变化：若第一次红，剩下 3 红 6 蓝（共 9）：P(红|红) = 3/9，P(蓝|红) = 6/9。若第一次蓝，剩下 4 红 5 蓝（共 9）：P(红|蓝) = 4/9，P(蓝|蓝) = 5/9。两颗不同色为 (红然后蓝) 或 (蓝然后红)。P(不同) = (4/10)×(6/9) + (6/10)×(4/9) = 24/90 + 24/90 = 48/90 = 8/15。
Without replacement, the fractions change at each branch. Multiplying along branches and adding for favourable outcomes is a reliable method for all conditional probability problems.
不放回时，每一步的分数都会变化。沿分支相乘再将有利结果相加，是解决所有条件概率问题的可靠方法。
12. Summary and Exam Tips | 总结与应试技巧
These case studies demonstrate that mathematics is not a collection of isolated techniques but a toolkit for understanding the world. When you face a challenging problem, always: identify what is given, what is unknown, choose a representation (diagram, table, equations), and reflect on whether your answer makes sense.
这些案例研究表明，数学不是孤立技巧的集合，而是理解世界的工具箱。面对挑战性问题时，始终：明确已知和未知，选择表示形式（图、表、方程），并反思答案是否合理。
In CIE Advanced Mathematics assessments, marks are awarded for clear method, not just the final answer. Show your steps, label your axes, write down formulas before substituting numbers, and check units. Practice transforming words into mathematics – the more case studies you work through, the sharper this skill becomes.
在 CIE 进阶数学考评中，分数不仅给最终答案，清晰的方法也会给分。展示你的步骤，标注坐标轴，在代入数字前写出公式，并检查单位。练习将文字转化为数学——做的案例分析越多，这项技能就越敏锐。
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