📚 Year 8 AQA Further Mathematics: Case Study Practical Exercises | Year 8 AQA 进阶数学：案例分析实战演练
Welcome to a collection of hands-on case studies designed specifically for Year 8 AQA Further Mathematics. Instead of isolated textbook problems, you will step into real-life scenarios—designing a school garden, running a charity cinema night, surveying classmates, and building a birdhouse. Each case study brings together algebra, geometry, ratio, statistics, and financial skills to show how mathematics helps us make decisions, compare options, and solve authentic problems. By working through these tasks, you will strengthen your reasoning, practise clear working-out, and gain confidence using concepts such as linear equations, area and perimeter, data averages, and surface area. Let’s dive in and put your further maths knowledge into action.
欢迎来到专为 Year 8 AQA 进阶数学设计的一系列动手案例分析。这里不再只是孤立的课本习题，你将进入真实生活场景——设计学校花园、举办慈善电影之夜、调查同学偏好，以及搭建一个鸟屋。每个案例都融合了代数、几何、比率、统计和理财技能，展示数学如何帮助我们做决策、比较备选方案并解决实际问题。通过逐步完成这些任务，你将加强逻辑推理，练习清晰展示解题过程，并自信地运用线性方程、面积与周长、数据平均数以及表面积等概念。让我们一起把进阶数学知识付诸实践吧。
1. Setting the Scene: Understanding the Task | 场景设置：理解任务
In each case study you will be given a short scenario and a set of questions. The key is to identify the mathematical topic hidden in the words. Does the problem involve finding lengths and areas? That calls for geometry. Does it ask “how many” items or “how much” money? Ratio, proportion, or solving equations. Is there a survey? Statistics. Read the scenario twice: once for the story, once for the numbers. Underline key data and write down what you know and what you need to find. This approach, often called “RUCSAC” (Read, Understand, Choose, Solve, Answer, Check), helps you stay organised.
在每一个案例中，你会收到一个简短的情景和一组问题。关键是找出隐藏在文字背后的数学主题。问题是不是要求长度和面积？那就需要几何知识。是否问“多少个”物品或“多少钱”？比率、比例或解方程。有调查吗？那就是统计。先把情景读两遍：第一遍看故事，第二遍看数字。划出关键数据，写下已知量和待求量。这种常被称为“RUCSAC”（阅读、理解、选择、求解、回答、检查）的方法能帮你保持条理清晰。
2. Case Study 1: The School Garden Project – Area and Perimeter | 案例1：学校花园项目 – 面积与周长
Your school wants to create a rectangular vegetable garden. The space available is a rectangle measuring 8 m long and 5 m wide. Before planting, the gardening club needs to put a fence around the garden and lay a weed-proof membrane across the whole area. You are asked to calculate how much fencing and membrane to buy. First, find the perimeter using the formula: Perimeter = 2 × (length + width). Next, find the area: Area = length × width.
你的学校打算开辟一个长方形菜园。可用空间是一个长 8 m、宽 5 m 的矩形。在种植之前，园艺俱乐部需要在花园周围围上围栏，并在整个区域铺设防草膜。你的任务是计算需要购买的围栏和防草膜数量。首先，用公式求周长：周长 = 2 × (长 + 宽)。然后求面积：面积 = 长 × 宽。
Perimeter = 2 × (8 m + 5 m) = 2 × 13 m = 26 m
周长 = 2 × (8 m + 5 m) = 2 × 13 m = 26 m
Area = 8 m × 5 m = 40 m²
面积 = 8 m × 5 m = 40 m²
The fence is sold in 2 m panels; how many panels are needed? Membrane comes in rolls that cover 10 m² each. Work out both quantities, remembering to round up when necessary.
围栏以 2 m 为一片出售；需要多少片？防草膜每卷可覆盖 10 m²。请计算出两者数量，必要时记得向上取整。
- Fence panels: 26 ÷ 2 = 13 panels
- 围栏片数： 26 ÷ 2 = 13 片
- Membrane rolls: 40 ÷ 10 = 4 rolls
- 防草膜卷数： 40 ÷ 10 = 4 卷
Takeaway: Always check units and convert if needed. A simple diagram labelled with numbers is your best friend.
要点：务必检查单位，必要时进行转换。一张用数字标注的简单草图会帮大忙。
3. Case Study 1: Budgeting for Materials – Ratio and Proportion | 案例1：材料预算 – 比率与比例
The gardening club decides to mix their own potting compost using soil, sand, and peat in the ratio 5 : 2 : 3 by volume. They need 60 litres of compost in total. How many litres of each ingredient are needed? Add the parts: 5 + 2 + 3 = 10 parts. One part is 60 L ÷ 10 = 6 L. Therefore soil = 5 × 6 L = 30 L, sand = 2 × 6 L = 12 L, peat = 3 × 6 L = 18 L.
园艺俱乐部决定按照体积比 5 : 2 : 3 将土壤、沙子和泥炭混合起来制作盆栽堆肥。他们总共需要 60 升堆肥。每种原料各需要多少升？先把总份数相加： 5 + 2 + 3 = 10 份。每份是 60 L ÷ 10 = 6 L。因此土壤 = 5 × 6 L = 30 L，沙子 = 2 × 6 L = 12 L，泥炭 = 3 × 6 L = 18 L。
Now, the club has a £50 budget. Soil costs £1.20 per litre, sand £0.80, and peat £2.50. Is it possible to buy all the ingredients? Calculate costs:
现在俱乐部有 £50 预算。土壤每升 £1.20，沙子每升 £0.80，泥炭每升 £2.50。能买齐所有原料吗？计算费用：
Soil cost = 30 × £1.20 = £36.00
土壤费用 = 30 × £1.20 = £36.00
Sand cost = 12 × £0.80 = £9.60
沙子费用 = 12 × £0.80 = £9.60
Peat cost = 18 × £2.50 = £45.00
泥炭费用 = 18 × £2.50 = £45.00
Total = £36.00 + £9.60 + £45.00 = £90.60. This exceeds the budget. The club needs to adjust the ratio or use cheaper alternatives. This shows how ratio and proportion directly connect to real spending decisions.
总计 = £36.00 + £9.60 + £45.00 = £90.60，超出了预算。俱乐部需要调整配方比例或选用更便宜的替代品。这体现了比率与比例如何直接影响真实消费决策。
4. Case Study 2: Planning a Charity Event – Linear Equations | 案例2：策划慈善活动 – 线性方程
The student council wants to put on a film night to raise money for charity. The hire of the school hall costs £40, and the film licence is £20. So fixed costs = £40 + £20 = £60. They sell tickets for £4 each. Let n be the number of tickets sold. The total income is £4n. Write an equation for the profit (or loss): Profit = Income − Fixed costs = 4n − 60. Find the number of tickets needed to break even (profit = 0).
学生会打算举办一场电影之夜为慈善机构筹款。租用学校礼堂的费用为 £40，电影放映许可证为 £20。因此固定成本 = £40 + £20 = £60。他们以每张 £4 的价格售票。设 n 为售出票数。总收入为 £4n。写出利润（或亏损）方程：利润 = 收入 − 固定成本 = 4n − 60。求使盈亏平衡（利润 = 0）的售票数量。
4n − 60 = 0 → 4n = 60 → n = 15
4n − 60 = 0 → 4n = 60 → n = 15
They need to sell at least 15 tickets just to cover costs. What if they want to donate £80 to charity? Then they need profit ≥ 80: 4n − 60 ≥ 80, solve to get n ≥ 35. So 35 tickets guarantee the donation.
他们需要售出至少 15 张票才能刚好覆盖成本。如果想为慈善机构捐出 £80 呢？那么需要利润 ≥ 80：4n − 60 ≥ 80，解出 n ≥ 35。因此售出 35 张票即可确保这笔捐款。
5. Case Study 2: Ticket Price Analysis – Inequalities | 案例2：票价分析 – 不等式
One committee member suggests lowering the ticket price to £3 to attract more people. Another suggests raising it to £5. Using the same fixed costs of £60, let’s examine the inequalities. For price £3: income = 3n, profit = 3n − 60. To have any profit, 3n − 60 > 0 → n > 20. For price £5: profit = 5n − 60, profit > 0 → n > 12. Notice that a higher ticket price requires fewer tickets to become profitable.
一位委员会成员建议把票价降到 £3 以吸引更多人。另一位则建议涨到 £5。基于相同的固定成本 £60，我们来分析不等式。票价 £3 时：收入 = 3n，利润 = 3n − 60。要有盈利，需满足 3n − 60 > 0 → n > 20。票价 £5 时：利润 = 5n − 60，利润 > 0 → n > 12。可以看出，票价越高，实现盈利所需售出的票数越少。
Now, if the hall can only seat 50 people, is it possible to raise £100 for charity with each pricing plan? Solve for profit ≥ 100. For £3: 3n − 60 ≥ 100 → 3n ≥ 160 → n ≥ 53.33, so at least 54 tickets, which exceeds the capacity. For £5: 5n − 60 ≥ 100 → 5n ≥ 160 → n ≥ 32, which is viable. This kind of analysis links inequalities directly to practical constraints.
现在假设礼堂最多容纳 50 人，那么各种票价方案下是否可能筹到 £100 善款？求解利润 ≥ 100。票价 £3 时：3n − 60 ≥ 100 → 3n ≥ 160 → n ≥ 53.33，因此需要至少 54 张票，超出容量限制。票价 £5 时：5n − 60 ≥ 100 → 5n ≥ 160 → n ≥ 32，这个数量可行。这类分析将不等式与实际约束条件直接联系起来。
6. Case Study 3: Surveying Year 8 Preferences – Collecting Data | 案例3：调查 Year 8 偏好 – 数据收集
To decide what film to show, the council surveys 40 Year 8 students, asking their favourite film genre. The results: Action (14), Comedy (10), Sci‑fi (8), Horror (5), Romance (3). You are asked to organise this data. First, check the total: 14+10+8+5+3 = 40. Good. A frequency table helps summarise:
为了决定放映哪部电影，学生会调查了 40 名 Year 8 学生，询问他们最喜爱的电影类型。结果为：动作片 (14)，喜剧 (10)，科幻片 (8)，恐怖片 (5)，爱情片 (3)。你需要整理这些数据。首先验证总数：14+10+8+5+3 = 40，正确。频率表有助于汇总：
| Genre / 类型 | Frequency / 频数 |
|---|---|
| Action / 动作 | 14 |
| Comedy / 喜剧 | 10 |
| Sci-fi / 科幻 | 8 |
| Horror / 恐怖 | 5 |
| Romance / 爱情 | 3 |
The modal genre is Action (highest frequency). The range of frequencies = 14 − 3 = 11, showing the spread. Next, we can calculate relative frequency as decimals to compare proportions: Action 14/40 = 0.35, Comedy 0.25, Sci‑fi 0.20, Horror 0.125, Romance 0.075.
众数类型是动作片（频数最高）。频数的极差 = 14 − 3 = 11，显示出离散程度。接下来，我们可以计算相对频率（小数）以比较比例：动作片 14/40 = 0.35，喜剧 0.25，科幻 0.20，恐怖 0.125，爱情 0.075。
7. Case Study 3: Presenting Data – Averages and Charts | 案例3：数据展示 – 平均数与图表
Since the data is categorical, the mean cannot be calculated directly, but we can present a bar chart. Draw axes: x‑axis for genre, y‑axis for frequency. The bars should have equal width and gaps. From this chart, you can instantly see which genre is most popular. If the council wants to choose a film that satisfies at least half the surveyed students, could they combine two genres? Discuss: combining Action and Comedy would give 24/40 = 60%, above half. This could be a compromise decision.
由于数据是分类变量，无法直接计算平均数，但我们可以绘制条形图。画出坐标轴：x 轴表示类型，y 轴表示频数。条形宽度相等且留有间隙。从图中可立刻看出哪种类型最受欢迎。如果学生会希望选择一部能让至少半数受访学生满意的电影，他们可以合并两种类型吗？讨论：将动作片和喜剧合并可得到 24/40 = 60%，超过半数，这可以成为一个折中方案。
If we were dealing with numerical data, such as spending money brought by students, we could compute mean, median, and mode. For example, the amounts (in £) from 10 students: 5, 8, 5, 12, 7, 5, 10, 8, 6, 24. The mean = (5+8+5+12+7+5+10+8+6+24) ÷ 10 = 90 ÷ 10 = £9.00. Ordering the data: 5,5,5,6,7,8,8,10,12,24. The median = (7+8) ÷ 2 = £7.50. The mode = £5. The outlier £24 pulls the mean above the median, showing why analysing all three averages matters.
当处理数值型数据时，比如学生携带的零花钱，我们可以计算平均数、中位数和众数。例如，10 名学生的金额（£）：5, 8, 5, 12, 7, 5, 10, 8, 6, 24。平均数 = (5+8+5+12+7+5+10+8+6+24) ÷ 10 = 90 ÷ 10 = £9.00。将数据排序：5,5,5,6,7,8,8,10,12,24。中位数 = (7+8) ÷ 2 = £7.50。众数 = £5。极端值 £24 将平均数拉高到中位数以上，反映了同时分析三个平均指标的重要性。
8. Case Study 4: Designing a Birdhouse – Volume and Surface Area | 案例4：设计鸟屋 – 体积与表面积
In Design & Technology, you are making a birdhouse shaped like a rectangular prism (without a floor) with a sloped roof. The internal base measures 20 cm by 15 cm and the height is 25 cm. First, find the volume of the interior space available for nesting: Volume = length × width × height = 20 cm × 15 cm × 25 cm.
在设计与技术课上，你需要制作一个长方体形状（无底板）并带有斜屋顶的鸟屋。内部底面尺寸为 20 cm × 15 cm，高度为 25 cm。首先，求出可供筑巢的内部空间体积：体积 = 长 × 宽 × 高 = 20 cm × 15 cm × 25 cm。
Volume = 20 × 15 × 25 = 7500 cm³
体积 = 20 × 15 × 25 = 7500 cm³
To build the birdhouse you need to cut wood for four vertical sides (ignoring the door for now). Find the total surface area of these four rectangles: two faces of size 20 cm × 25 cm and two faces of 15 cm × 25 cm. Total wall area = 2(20×25) + 2(15×25).
要建造这个鸟屋，你需要为四面垂直侧板（暂且忽略门洞）切割木板。求这四块矩形的总表面积：两块 20 cm × 25 cm 的面，两块 15 cm × 25 cm 的面。侧板总面积 = 2(20×25) + 2(15×25)。
Wall area = 2×500 + 2×375 = 1000 + 750 = 1750 cm²
侧板面积 = 2×500 + 2×375 = 1000 + 750 = 1750 cm²
This helps you buy the right amount of wood sheet. If wood costs £0.05 per 10 cm², calculate the cost. It’s another opportunity to use proportion.
这能帮你购买准确数量的木板。如果木板每 10 cm² 售价 £0.05，请计算费用，这是再次运用比例的好机会。
9. Case Study 4: Optimising Material Usage – Substitution | 案例4：优化材料使用 – 代入
The technology teacher provides a formula linking the base length L and the amount of wood saved if a smaller birdhouse is made. Suppose the formula for total surface area (walls only) is S = 2L×25 + 2W×25, but W is always 3/4 of L. Write S in terms of L: W = ¾ L. Then S = 2L×25 + 2(¾ L)×25 = 50L + 50 × ¾ L = 50L + 37.5L = 87.5L. If you want the surface area to be exactly 2000 cm², solve 87.5L = 2000 → L = 2000 ÷ 87.5 ≈ 22.86 cm. Then W ≈ 17.14 cm. Substitution allows you to test different designs quickly.
技术课老师提供了一个公式，将底面长度 L 与制作较小鸟屋所节省的木材量关联起来。假设侧板总表面积公式为 S = 2L×25 + 2W×25，但 W 始终是 L 的 3/4。用 L 表示 S： W = ¾ L。那么 S = 2L×25 + 2(¾ L)×25 = 50L + 50 × ¾ L = 50L + 37.5L = 87.5L。如果你希望表面积恰好为 2000 cm²，求解 87.5L = 2000 → L = 2000 ÷ 87.5 ≈ 22.86 cm，于是 W ≈ 17.14 cm。代入法让你可以快速测试不同设计方案。
Check the volume with these dimensions: Volume = L × W × 25 = 22.86 × 17.14 × 25 ≈ 9793 cm³, a larger volume than the original design but using less material because the base is narrower. This demonstrates optimisation—balancing different factors.
用这些尺寸验算体积：体积 = L × W × 25 = 22.86 × 17.14 × 25 ≈ 9793 cm³，比原设计体积更大，但由于底部更窄，用料更少。这体现了优化思想——平衡不同因素。
10. Bringing It All Together: Reflection and Extension | 总结归纳：反思与拓展
Throughout these case studies, you have used area and perimeter for planning spaces, ratio and proportion for mixing and costing, linear equations and inequalities for financial decisions, data collection and analysis for surveys, and volume and surface area for designs. The most successful problem‑solvers are those who ask: “What mathematics do I see here?” and draw a simple model—a sketch, a table, or a graph—before diving into calculations. Extend your thinking: what if the garden had two different rectangular beds? What if the film night also sold drinks? What if the birdhouse had a circular entrance hole and you needed to find its circumference? These variations deepen your grasp of further mathematics.
在这些案例研究中，你运用了面积和周长规划空间、比率和比例进行配比与费用计算、线性方程和不等式做出财务决策、数据收集与分析进行问卷调查，以及体积和表面积完成设计。最擅长解决问题的人总会先问自己：“这里用到什么数学？”并在动手计算前画出简单模型——一张草图、一个表格或一幅图表。进一步拓展思维：如果菜园有两个不同的矩形苗床呢？如果电影之夜还售卖饮料呢？如果鸟屋带有一个圆形入口，你需要计算其周长呢？这些变式练习能加深你对进阶数学的掌握。
Remember to always check your answers: does the profit make sense in the context? Are the units consistent? Could you solve the problem another way? Keep practising, and soon you will spot the mathematics behind everyday situations automatically.
记得时常检查答案：在情境中这个利润数字合理吗？单位是否统一？能用另一种方法求解吗？不断练习，很快你就能自动识别日常情境背后的数学原理。
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