📚 Case Study Practice for Year 8 Edexcel Further Maths | Year 8 Edexcel 进阶数学：案例分析实战演练
This article presents a series of practical case studies designed to strengthen problem-solving skills for Year 8 Edexcel Further Maths. Each case study reflects real-world scenarios that require students to apply algebra, geometry, statistics, and proportional reasoning. By working through these examples, learners will develop the confidence to analyse complex problems, break them into manageable steps, and justify their reasoning clearly.
本文提供一系列实际案例分析，旨在帮助 Year 8 Edexcel 进阶数学的学生提升解题能力。每个案例都取材于真实情境，要求学生综合运用代数、几何、统计和比例推理。通过演练这些题目，学生将能更有信心地分析复杂问题，将其分解为可执行步骤，并清晰地阐述推理过程。
1. Designing a School Garden | 设计学校花园
A school plans to build a rectangular garden with a perimeter of 36 metres. The length is twice the width. Students must calculate the dimensions and then determine the cost of fencing if each metre of fencing costs £8.50.
学校计划建造一个周长为 36 米的长方形花园，长度是宽度的两倍。学生需要计算出长和宽，并确定围栏的总成本，假设每米围栏的价格为 8.50 英镑。
Let the width be w metres. Then the length is 2w metres. The perimeter is 2(length + width) = 2(2w + w) = 6w. Given 6w = 36, we find w = 6 m and length = 12 m. Total fencing needed = perimeter = 36 m. Cost = 36 × 8.50 = £306.
设宽度为 w 米，则长度为 2w 米。周长为 2（长 + 宽）= 2(2w + w) = 6w。已知 6w = 36，解得 w = 6 米，长度 = 12 米。所需围栏总长度即周长 36 米。总成本 = 36 × 8.50 = 306 英镑。
Extension: What would be the area of the garden? How many square-metre flower beds could fit if each bed occupies 1.5 m²?
拓展：花园的面积是多少？如果每个花坛占地 1.5 平方米，可以容纳多少个花坛？
2. Mobile Phone Contract Comparison | 手机套餐比较
A student wants to choose between two mobile phone contracts. Contract A charges a fixed monthly fee of £10 plus £0.05 per minute of calls. Contract B has no monthly fee but charges £0.12 per minute. Determine which contract is cheaper for 200 minutes of calls per month and find the break-even point where both contracts cost the same.
一位学生想在两种手机套餐中做出选择。套餐 A 每月固定费用为 10 英镑，外加每分钟 0.05 英镑的通话费。套餐 B 没有月租，但每分钟通话费为 0.12 英镑。请判断每月通话 200 分钟时哪个套餐更便宜，并找出两种套餐费用相同时的通话分钟数。
For Contract A: Cost = 10 + 0.05 × minutes. For 200 minutes: 10 + 0.05×200 = 10 + 10 = £20. Contract B: 0.12 × 200 = £24. So A is cheaper. Let the break-even point be m minutes. 10 + 0.05m = 0.12m → 10 = 0.07m → m = 10 ÷ 0.07 ≈ 142.86 minutes. At 143 minutes, costs are nearly equal; for fewer minutes, B is cheaper.
套餐 A 费用 = 10 + 0.05 × 分钟数。200 分钟时：10 + 0.05×200 = 10 + 10 = 20 英镑。套餐 B：0.12 × 200 = 24 英镑。因此 A 更便宜。设费用相等时的分钟数为 m，则 10 + 0.05m = 0.12m → 10 = 0.07m → m = 10 ÷ 0.07 ≈ 142.86 分钟。当通话约 143 分钟时两者费用几乎相同；少于该时长则 B 更便宜。
3. Statistical Analysis of Bus Arrival Times | 公交车到达时间的统计分析
A student records the arrival delays (in minutes) of the school bus over 20 days: 2, 0, 5, 1, -1 (early), 3, 7, 0, 4, 2, 6, -2, 1, 0, 8, 3, 2, 5, 1, 4. Construct a frequency table, calculate the mean delay, and find the median and mode. Discuss whether the bus is generally punctual.
一名学生记录了校车在 20 天内的到达延迟时间（分钟）：2、0、5、1、-1（提前）、3、7、0、4、2、6、-2、1、0、8、3、2、5、1、4。要求构建频数表，计算平均延迟时间，并求出中位数和众数。讨论该校车总体上是否准时。
Sort data: -2, -1, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 7, 8. Frequency: -2(1), -1(1), 0(3), 1(3), 2(3), 3(2), 4(2), 5(2), 6(1), 7(1), 8(1). Total sum = (-2)+(-1)+0+0+0+1+1+1+2+2+2+3+3+4+4+5+5+6+7+8 = 54. Mean = 54 ÷ 20 = 2.7 minutes. Median (10th and 11th values): both are 2 → median = 2. Mode = 0, 1, and 2 (all appear 3 times). The mean positive delay indicates the bus tends to be late on average, but the mode of 0 shows it is often on time.
数据排序：-2、-1、0、0、0、1、1、1、2、2、2、3、3、4、4、5、5、6、7、8。频数：-2(1), -1(1), 0(3), 1(3), 2(3), 3(2), 4(2), 5(2), 6(1), 7(1), 8(1)。总和 = (-2)+(-1)+0+0+0+1+1+1+2+2+2+3+3+4+4+5+5+6+7+8 = 54。平均值 = 54 ÷ 20 = 2.7 分钟。中位数（第 10 和 11 个数）均为 2 → 中位数 = 2。众数为 0、1、2（均出现 3 次）。正的平均延迟显示校车平均有晚点趋势，但众数为 0 表明经常准时。
4. Ratio in a Cake Recipe | 蛋糕配方中的比例
A Victoria sponge recipe uses flour, butter, and sugar in the ratio 3:2:2 by weight. If a baker wants to make a cake using 450 g of flour, how much butter and sugar are needed? What is the total mass of the mixture?
一款维多利亚海绵蛋糕的配方要求面粉、黄油和糖的重量比为 3:2:2。如果烘焙师想用 450 克面粉制作蛋糕，需要多少黄油和糖？混合物的总质量是多少？
The ratio 3:2:2 means for every 3 parts flour, there are 2 parts butter and 2 parts sugar. Total parts = 7. If 3 parts = 450 g, then 1 part = 150 g. Butter = 2 × 150 = 300 g, sugar = 2 × 150 = 300 g. Total mass = 450 + 300 + 300 = 1050 g. Alternatively, using the multiplier: 450 ÷ 3 = 150 g per part.
比例 3:2:2 表示每 3 份面粉对应 2 份黄油和 2 份糖。总份数 = 7。如果 3 份对应 450 克，那么 1 份 = 150 克。黄油 = 2 × 150 = 300 克，糖 = 2 × 150 = 300 克。总质量 = 450 + 300 + 300 = 1050 克。也可以使用乘数：450 ÷ 3 = 150 克/份。
5. Speed, Distance, and Time in a Cycling Trip | 骑行旅行中的速度、距离和时间
Two friends cycle towards each other from towns 42 km apart. One cycles at 12 km/h, the other at 15 km/h. How long will it take for them to meet? What distance has each travelled when they meet?
两个朋友从相距 42 公里的两个城镇相对骑行。一人速度为 12 公里/小时，另一人为 15 公里/小时。他们多长时间后相遇？相遇时各自骑行了多远？
Combined speed = 12 + 15 = 27 km/h. Time to cover 42 km = 42 ÷ 27 = 14/9 hours = 1 hour 33 minutes 20 seconds (approximately). Distance travelled by first cyclist = 12 × 14/9 = 56/3 ≈ 18.67 km; second cyclist = 15 × 14/9 = 70/3 ≈ 23.33 km. Check: 56/3 + 70/3 = 126/3 = 42 km.
相对速度 = 12 + 15 = 27 公里/小时。覆盖 42 公里所需时间 = 42 ÷ 27 = 14/9 小时 ≈ 1 小时 33 分 20 秒。第一位骑行的距离 = 12 × 14/9 = 56/3 ≈ 18.67 公里；第二位 = 15 × 14/9 = 70/3 ≈ 23.33 公里。检验：56/3 + 70/3 = 126/3 = 42 公里。
6. Volume and Surface Area of a Water Tank | 水箱的体积与表面积
A cuboid water tank has length 80 cm, width 50 cm, and height 60 cm. Find its capacity in litres (1 litre = 1000 cm³). The tank is to be painted on the outside (excluding the base). Calculate the area to be painted.
一个长方体水箱长 80 厘米，宽 50 厘米，高 60 厘米。求其容量（升）（1 升 = 1000 立方厘米）。现在要给水箱的外部刷漆（不包括底部），计算需要刷漆的面积。
Volume = length × width × height = 80 × 50 × 60 = 240,000 cm³. Capacity in litres = 240,000 ÷ 1000 = 240 litres. Surface area to paint: four vertical walls + top. Two side walls: 2 × (80 × 60) = 9600 cm²; front and back walls: 2 × (50 × 60) = 6000 cm²; top: 80 × 50 = 4000 cm². Total area = 9600 + 6000 + 4000 = 19,600 cm².
体积 = 长 × 宽 × 高 = 80 × 50 × 60 = 240,000 立方厘米。容量 = 240,000 ÷ 1000 = 240 升。需刷漆面积：四个垂直面 + 顶部。两个侧面：2 × (80 × 60) = 9600 平方厘米；正面和背面：2 × (50 × 60) = 6000 平方厘米；顶部：80 × 50 = 4000 平方厘米。总面积 = 9600 + 6000 + 4000 = 19,600 平方厘米。
7. Interpreting a Travel Graph | 解读行程图
The distance-time graph of a train journey shows a steady line from (0,0) to (30, 60), then a horizontal segment from (30,60) to (45,60), followed by a straight line from (45,60) to (75,0). Describe the journey, calculate speeds for each segment, and find the average speed for the entire trip.
某火车行程的距离-时间图像显示：从 (0,0) 到 (30,60) 的稳定上升直线，然后从 (30,60) 到 (45,60) 的水平线段，最后从 (45,60) 到 (75,0) 的直线下降段。请描述这段行程，计算各段的速率，并求出全程平均速率。
First segment: travels 60 km in 30 min → speed = 60 / 0.5 h = 120 km/h. Second segment: stationary for 15 minutes. Third segment: returns 60 km in 30 min → speed = 120 km/h towards start. Total distance = 120 km, total time = 75 min = 1.25 hours. Average speed = 120 ÷ 1.25 = 96 km/h.
第一段：30 分钟行驶 60 公里 → 速率 = 60 / 0.5 小时 = 120 公里/小时。第二段：停留 15 分钟。第三段：30 分钟返回 60 公里 → 速率 = 120 公里/小时。总路程 = 120 公里，总时间 = 75 分钟 = 1.25 小时。平均速率 = 120 ÷ 1.25 = 96 公里/小时。
8. Probability in a Dice Game | 骰子游戏中的概率
In a game, two fair six-sided dice are rolled. A player wins if the sum is 7 or 11. If the sum is 2, 3, or 12, the player loses immediately. Otherwise, the game continues. Calculate the probability of winning on the first roll.
在一个游戏中，投掷两枚公平的六面骰子。如果点数之和为 7 或 11，玩家获胜；如果点数之和为 2、3 或 12，玩家立即输掉；其余情况游戏继续。计算第一次投掷即获胜的概率。
Total outcomes = 6 × 6 = 36. Ways to get 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways. Ways to get 11: (5,6), (6,5) → 2 ways. Winning outcomes = 8. P(win) = 8/36 = 2/9. Losing outcomes: sum 2 (1 way), sum 3 (2 ways), sum 12 (1 way) → 4 ways; P(lose) = 4/36 = 1/9. P(continue) = 24/36 = 2/3.
总结果数 = 6 × 6 = 36。得到和为 7 的方式：(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 种。得到 11 的方式：(5,6), (6,5) → 2 种。获胜结果数 = 8。获胜概率 = 8/36 = 2/9。失败结果：和 2（1 种），和 3（2 种），和 12（1 种）→ 4 种；失败概率 = 4/36 = 1/9。继续游戏概率 = 24/36 = 2/3。
9. Optimising a Rectangular Enclosure | 优化矩形围栏面积
A farmer has 40 metres of fencing and wants to enclose a rectangular area against an existing wall (so only three sides need fencing). What dimensions give the maximum area? What is that area?
一位农民有 40 米长的围栏，想靠着一面已有的墙围出一个矩形区域（因此只需围三面）。怎样的尺寸可以获得最大面积？最大面积是多少？
Let the side perpendicular to the wall be x metres, and the side parallel to the wall be y metres. Then 2x + y = 40 → y = 40 – 2x. Area A = x × y = x(40 – 2x) = 40x – 2x². This is a quadratic, maximum at vertex. Vertex x = -b/(2a) with a = -2, b = 40 → x = -40/(2×-2) = 10. Then y = 40 – 20 = 20. Maximum area = 10 × 20 = 200 m².
设垂直于墙的边长为 x 米，平行于墙的边长为 y 米。则 2x + y = 40 → y = 40 – 2x。面积 A = x × y = x(40 – 2x) = 40x – 2x²。这是一个二次函数，顶点处最大。顶点 x = -b/(2a)，a = -2，b = 40 → x = -40/(2×-2) = 10。则 y = 40 – 20 = 20。最大面积 = 10 × 20 = 200 平方米。
10. Currency Conversion and Best Buy | 货币兑换与最优购买
An online game accessory costs $24 in the US store and €19 in the EU store. The exchange rates are £1 = $1.32 and £1 = €1.15. Which store offers the cheaper price in pounds? Also, a UK store sells it for £17.50. Which is the best buy?
某款在线游戏配件在美国商店售价为 24 美元，在欧盟商店售价为 19 欧元。汇率分别为 1 英镑 = 1.32 美元，1 英镑 = 1.15 欧元。哪家商店的英镑价格更便宜？另外，一家英国商店的售价为 17.50 英镑，哪个是最优购买选择？
US price in £: 24 ÷ 1.32 ≈ £18.18. EU price in £: 19 ÷ 1.15 ≈ £16.52. So the EU store is cheaper than the US store. The UK store price is £17.50, which is more expensive than the EU price. Best buy is the EU store. However, if delivery costs apply, those must be added for a fair comparison.
美元价格换算英镑：24 ÷ 1.32 ≈ 18.18 英镑。欧元价格换算英镑：19 ÷ 1.15 ≈ 16.52 英镑。因此欧盟商店比美国商店便宜。英国商店售价 17.50 英镑，比欧盟价格贵。最优购买是欧盟商店。但如果有配送费，需将其加入进行公平比较。
11. Interpreting Pie Charts and Bar Charts | 饼图和条形图的解读
A survey of 240 students on favourite subject shows: Maths 120°, Science 90°, English 60°, History 90°. Calculate the number of students for each subject. Then represent the data on a bar chart, and explain why the pie chart might be misleading if category totals are close.
一项针对 240 名学生的最喜爱科目调查显示：数学 120°，科学 90°，英语 60°，历史 90°。计算每个科目的学生人数。然后使用条形图表示数据，并解释如果各类别总数相近，饼图为何可能具有误导性。
Total degrees = 360°. One student represents 360 ÷ 240 = 1.5°. Maths: 120 ÷ 1.5 = 80 students. Science: 90 ÷ 1.5 = 60 students. English: 60 ÷ 1.5 = 40 students. History: 90 ÷ 1.5 = 60 students. Pie charts can make similar-sized slices hard to compare precisely; bar charts show exact heights and differences more clearly.
总角度 = 360°。每名学生对应 360 ÷ 240 = 1.5°。数学：120 ÷ 1.5 = 80 人。科学：90 ÷ 1.5 = 60 人。英语：60 ÷ 1.5 = 40 人。历史：90 ÷ 1.5 = 60 人。饼图中大小相近的扇区难以精确比较；条形图则能更清晰地显示确切高度和差异。
12. Designing a Multi-Stage Problem | 设计一个多阶段问题
Create your own multi-stage problem combining percentages and geometry. Example: A shop reduces a jacket priced at £80 by 15%, then an additional 10% off the reduced price during clearance. Meanwhile, a sports ball has volume 288π cm³. Find the sphere’s radius. Decide which final price is greater: the jacket or the cost of 8 such balls at £5 per 10 cm³ of material.
设计一个结合百分数和几何的多阶段问题。例如：某商店将一件标价 80 英镑的夹克降价 15%，清仓时在降价基础上再打九折。同时，一个运动球体积为 288π 立方厘米，求该球体的半径。判断最终的夹克价格与以每 10 立方厘米 5 英镑的价格买 8 个这样的球，哪个花费更多。
Jacket final price: first reduction: 80 × 0.85 = £68. Additional 10% off: 68 × 0.90 = £61.20. Sphere volume V = 4/3 π r³ = 288π → 4/3 r³ = 288 → r³ = 216 → r = 6 cm. Material volume of one ball = 288π ≈ 904.78 cm³. Cost per ball: (904.78 / 10) × 5 = 90.478 × 5 ≈ £452.39. Eight balls cost ≈ £3619.12, far more than the jacket. The ball cost is greater. This highlights how quickly volume cost scaling works.
夹克最终价格：第一次降价：80 × 0.85 = 68 英镑。清仓再打九折：68 × 0.90 = 61.20 英镑。球体体积 V = 4/3 π r³ = 288π → 4/3 r³ = 288 → r³ = 216 → r = 6 厘米。单个球的体积 = 288π ≈ 904.78 立方厘米。单个球成本：(904.78 / 10) × 5 ≈ 90.478 × 5 = 452.39 英镑。八个球成本 ≈ 3619.12 英镑，远超夹克价格。球的成本更高。这凸显了体积成本如何快速放大。
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