📚 Case Study Practical Exercises in Cambridge Year 7 Advanced Mathematics | 剑桥七年级进阶数学：案例分析实战演练
Mathematics is not just about numbers on a page; it is a powerful toolkit for solving real-world problems. In this article, we bring together key topics from the Year 7 Cambridge Advanced Mathematics curriculum – including arithmetic, algebra, geometry, data handling and probability – and apply them through a series of practical case studies. You will plan a school fete, design a garden, conduct a survey and even test the fairness of a dice game. Each case study is broken down into manageable steps, showing you how different mathematical skills connect and reinforce each other. By working through these exercises, you will develop a deeper understanding of concepts and see why mathematics matters in everyday life.
数学不仅仅是纸上的数字；它还是一个解决现实问题的强大工具箱。本文汇集了剑桥七年级进阶数学课程中的关键主题——包括算术、代数、几何、数据处理和概率，并通过一系列实际案例研究加以应用。你将策划一场学校嘉年华、设计一个花园、进行一次调查，甚至检验一个骰子游戏的公平性。每个案例被分解成可以操作的步骤，展示不同的数学技能如何相互关联与强化。通过这些练习，你将对概念有更深的理解，并明白数学在日常生活中为何如此重要。
1. Real-World Case Studies: Why They Matter | 现实世界案例研究：为何如此重要
Case studies bridge the gap between classroom theory and everyday situations. In Year 7 Advanced Mathematics, you have learned about fractions, decimals, algebraic expressions, area, perimeter, bar charts and basic probability. When these topics appear in realistic scenarios, they become more memorable and meaningful. Working through a budget, for instance, reinforces operations with money and decimals, while designing a garden deepens your sense of measurement and scale. The aim is not just to find an answer, but to decide which mathematics to use and how to communicate your reasoning clearly.
案例研究弥合了课堂理论与日常情境之间的鸿沟。在七年级进阶数学中，你已经学习了分数、小数、代数式、面积、周长、条形图和基础概率。当这些主题出现在真实场景中时，它们会变得更加难忘和有意义。例如，处理预算会强化货币和小数运算，而设计花园则会加深你对度量与比例的理解。目的不仅仅是找到一个答案，而是决定使用哪种数学知识以及如何清晰地展示推理过程。
2. Case 1 – School Fete Budget: Listing Expenses | 案例一：学校嘉年华预算——列出各项开支
Imagine your class is organising a school fete to raise money for new sports equipment. The first task is to estimate all the costs. You need to hire a bouncy castle for £85, buy decorations costing £34.60, and purchase prizes for games worth £57.85. There is also a fixed permit fee of £25. To find the total expense, add these amounts carefully: £85 + £34.60 + £57.85 + £25. When working with decimals, align the decimal points and carry over as needed. The sum is £202.45. This total gives you a clear target for the income you must generate just to break even.
假设你们班正在组织一场学校嘉年华，为新的体育器材筹集资金。第一步是估算所有成本。你们需要花85英镑租一个充气城堡，花34.60英镑购买装饰品，再花57.85英镑购买游戏奖品。此外还有25英镑的固定许可费。要计算总支出，请仔细相加：£85 + £34.60 + £57.85 + £25。处理小数时，对齐小数点并按需进位。总和为202.45英镑。这个总额给出了你们必须达到的收入目标才能不赚不赔。
3. Case 1 – Setting Ticket Prices with Algebra | 案例一：用代数设定门票价格
Now that the costs are known, you want to raise an extra £300 for the sports equipment fund. Let the ticket price be p pounds. If you expect to sell 200 tickets, the income equation is 200p = total expenses + £300. Substitute the total expense: 200p = £202.45 + £300 → 200p = £502.45. To solve for p, divide both sides by 200: p = 502.45 ÷ 200. Using division, you get p = 2.51225. Rounding to the nearest penny gives a ticket price of £2.51. Setting p = 2.51 means you will actually collect 200 × £2.51 = £502, just slightly below target – alistic figure you can adjust by selling a few more tickets or slightly raising the price.
现在成本已知，你们想为体育器材基金额外筹集300英镑。设门票价格为 p 英镑。如果预计售出200张票，收入方程为 200p = 总支出 + £300。代入总支出：200p = £202.45 + £300 → 200p = £502.45。两边同除以200得 p = 502.45 ÷ 200。使用除法，得到 p = 2.51225。四舍五入到分，门票价格为2.51英镑。若设定为2.51英镑，实际收入为200 × £2.51 = £502，仅略低于目标——这是一个现实的数字，你可以通过多售几张票或略微提价来调整。
4. Case 2 – Designing a Rectangular Garden: Perimeter and Area | 案例二：设计矩形花园——周长与面积
A school garden is planned as a rectangle with length 9 m and width 6 angle is calculated as P = 2 × (length + width). Substitute: P = 2 × (9 + 6) = 2 × 15 = 30 m. The area is A = length × width, so A = 9 × 6 = 54 . Knowing both values allows you to order the correct amount of materials and avoid waste.
学校花园设计为长9米、宽6米的矩形。需要两项基本测量：用于围栏的周长和用于铺设草皮的面积。矩形周长计算公式为 P = 2 × (长 + 宽)。代入得：P = 2 × (9 + 6) = 2 × 15 = 30 米。面积公式为 A = 长 × 宽，所以 A = 9 × 6 = 54 平方米。知道了这两个数值，你就能订购准确数量的材料，避免浪费。
5. Case 2 – Costing Fencing and Turf | 案例二：围栏与草皮成本计算
Fencing material costs £11.80 per metre, and turf costs £3.90 per square metre. The total fencing cost is 30 m × £11.80 = £354.00. The turf cost is 54 m² × £3.90 = £210.60. Adding these gives a combined material cost of £ tion, reinforcing place value skills. Always double-check by estimating: fence cost ≈ 30 × £12 = £360, turf cost ≈ 50 × £4 = £200, giving a rough total of £560, which confirms the calculation is sensible.
围栏材料每米售价11.80英镑，草皮每平方米售价3.90英镑。围栏总成本为30米 × £11.80 = £354.00。草皮成本为54平方米 × £3.90 = £210.60。两者相加，材料总成本为£564.60。这个练习运用了小数乘法和货币表示法，强化了数位值技能。务必通过估算进行复查：围栏成本 ≈ 30 × £12 = £360，草皮成本 ≈ 50 × £4 = £200，粗略总计£560，这确认了计算是合理的。
6. Case 2 – Paving a Path Around the Garden | 案例二：围绕花园铺设小径
To make the garden accessible, a 1.2 he rectangle. The overall dimensions including the path become length = 9 + 2 × 1.2 = 11.4 m and width = 6 + 2 × 1.2 = 8.4 m. The area of the path is the larger rectangle minus the garden area: 11.4 × 8.4 = 95.76 m², minus 54 m² gives 41.76 m². If each bag of gravel covers 2.5 ou must buy 17 bags. This example blends geometry with rounding up in a practical context.
为了便于进入花园，将在矩形外围铺设一条1.2米宽的砾石小径。包含路径后的总尺寸为：长 = 9 + 2 × 1.2 = 11.4 米，宽 = 6 + 2 × 1.2 = 8.4 米。小径的面积等于大矩形面积减去花园面积：11.4 × 8.4 = 95.76 平方米，减去54平方米得41.76平方米。如果每袋砾石可铺2.5平方米，所需袋数为41.76 ÷ 2.5 = 16.704，因此必须购买17袋。这个例子将几何与实际情境中的向上取整结合起来。
7. Case 3 – Survey: Collecting Data on Favourite Fruits | 案例三：调查——收集最爱水果的数据
Your class decides to investigate the most popular fruit among Year 7 students. Each student chooses one from apples, bananas, oranges, grapes or strawberries. The raw data for 30 students is recorded as tally marks and then organised into a frequency table. For example: Apple 8, Banana 7, Orange 5, Grape 4, Strawberry 6. The total frequency must sum to 30, which acts as a check. This step practises data collection, tallying and organisation – essential skills for any statistical investigation.
你们班决定调查七年级学生中最受欢迎的水果。每位学生从苹果、香蕉、橙子、葡萄和草莓中选择一种。30名学生的原始数据用画记法记录，然后整理成频率表。例如：苹果8，香蕉7，橙子5，葡萄4，草莓6。总频数之和必须为30，这是一个检查点。这一步练习了数据收集、画记与整理——任何统计调查的基本技能。
8. Case 3 – Drawing a Bar Chart and Calculating the Mode | 案例三：绘制条形图并计算众数
Using the frequency table, you can construct a bar chart with the fruit types on the horizontal axis and frequencies on the vertical axis. The height of each bar represents how many students chose that fruit. The mode – the most frequently occurring category – is clearly 'apple' with a frequency of 8. A well-drawn bar chart also needs a title, labelled axes and evenly spaced bars. This visual representation makes it easy to compare categories at a glance, a key objective in data handling.
利用频率表，你可以构建一张条形图，横轴为水果种类，纵轴为频数。每个条形的高度代表选择该水果的学生人数。众数——出现频率最高的类别——显然是“苹果”，频数为8。一张绘制得当的条形图还需要标题、带标签的坐标轴和等间距的条形。这种可视化表示让人们一目了然地比较各类别，这是数据处理的一个关键目标。
9. Case 3 – Mean, Median and Simple Probability from Data | 案例三：平均数、中位数与从数据中获得的简单概率
arison. To find the median, list the frequencies in order: 4, 5, 6, 7, 8; the middle value is 6. From the frequency table, you can estimate the probability that a student chosen at random prefers strawberries: 6 out of 30, or ⅕ s data handling with probability, showing how observed frequencies can inform predictions.
你还可以计算每类水果的平均选择次数：用总回答数除以类别数，即30 ÷ 5 = 6，这对比较很有用。要找到中位数，将频数按顺序排列：4, 5, 6, 7, 8；中间值为6。根据频率表，你可以估计随机选择一名学生偏爱草莓的概率：30分之6，即⅕。这将数据处理与概率联系起来，展示了观察到的频率如何为预测提供依据。
10. Case 4 – Dice Game: Listing All Possible Outcomes | 案例四：骰子游戏——列出所有可能结果
re are 6 × 6 = 36 equally likely pairs. You can construct a sample space diagram: rows for Die 1 (1–6), columns for Die 2 (1– ay to get a sum of 2 (1+1), but six ways to get a sum of 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1).
抛掷两枚均匀的六面骰子，记录点数之和。要分析这个游戏，你必须首先系统地列出所有结果。共有6 × 6 = 36种等可能的点数对。你可以构建一个样本空间图：行表示骰子1（1–6），列表示骰子2（1–6），每个单元格显示和。和的取值范围从2到12。这种结构化的方法确保没有组合被遗漏。例如，只有一种方式得到和为2（1+1），但有六种方式得到和为7（1+6, 2+5, 3+4, 4+3, 5+2, 6+1）。
11. Case 4 – Calculating Probabilities and Expected Value | 案例四：计算概率与期望值
Consider the rule: you win £2 if the sum is greater than 7, otherwise you lose £ le outcomes for sums 8, 9, 10, 11, 12: 5 ways for 8, 4 for 9, 3 for 10, 2 for 11, 1 for 12 – 15/36 = 5/12. Probability of losing = 21/36 = 7/12. The expected value per game is (5/12 × £2) + (7/12 × –£1) = £10/12 – £7/12 = £3/12 = £ s the powerful concept of expectation, linking probability and arithmetic decision-making.
考虑规则：如果点数之和大于7，你赢2英镑，否则你输1英镑。首先，数出有利结果（和为8, 9, 10, 11, 12）：8有5种，9有4种，10有3种，11有2种，12有1种——总共15种有利结果。赢的概率 = 15/36 = 5/12。输的概率 = 21/36 = 7/12。每局游戏的期望值是 (5/12 × £2) + (7/12 × –£1) = £10/12 – £7/12 = £3/12 = £0.25。25便士的正期望值意味着从长远看这个游戏对玩家有利。这引入了期望值这一强大概念，将概率与算术决策联系起来。
12. Reflection: Skills Reinforced and Next Steps | 反思：强化的技能与下一步
Across these four case studies, you have exercised mental and written arithmetic with decimals, solved simple linear equations, calculated perimeter and area, interpreted and constructed frequency tables and bar charts, and evaluated probabilities using sample spaces. Each task required you to select the appropriate mathematical method and check your answers for sense. The ability to move flexibly between numerical, algebraic and graphical representations is a core skill in Cambridge Advanced Mathematics. To extend your learning, try modifying a case study: what if the garden had a different shape, or the fete charged separate entry and activity prices? Real-world mathematics is full of such open-ended questions, and you now have the tools to tackle them with confidence.
在这四个案例研究中，你锻炼了小数的口算和笔算能力，求解了简单的一次方程，计算了周长和面积，解读并构建了频率表和条形图，还利用样本空间评估了概率。每项任务都要求你选择恰当的数学方法并检验答案的合理性。在数值、代数和图形表征之间灵活转换的能力是剑桥进阶数学的核心技能。为拓展学习，你可以尝试修改一个案例：如果花园形状不同，或者嘉年华分别收取门票和活动费用呢？现实世界的数学充满了诸如此类的开放性问题，而你如今已拥有自信应对它们的工具。
Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com
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