📚 Year 8 Cambridge Advanced Maths: Case Study Practice | Year 8 剑桥进阶数学：案例分析实战演练
Real-world problem-solving lies at the heart of Cambridge advanced mathematics. This article presents a collection of case study exercises carefully designed for Year 8 learners, blending reasoning, modelling, and multi-step analysis across topics such as proportional reasoning, linear equations, data handling, geometry, and probability. Each case mimics the style of open-ended tasks found in extension papers, encouraging you to interpret information, make assumptions, and justify conclusions with mathematical evidence. Working through these scenarios will sharpen your ability to apply mathematics beyond routine exercises and is excellent preparation for higher-level challenges.
现实世界的问题解决是剑桥进阶数学的核心。本文为 Year 8 学生精心设计了一组案例分析练习题，融合了比例推理、线性方程、数据处理、几何与概率等主题中的逻辑推理、建模和多步骤分析。每个案例都模仿了进阶试卷中常见的开放式任务风格，鼓励你解读信息、作出假设，并用数学证据论证结论。通过这些场景训练，你将提升在常规练习之外运用数学的能力，为更高层次的挑战做好充分准备。
1. Case Study: Charity Fun Run – Proportional Reasoning | 案例一：慈善趣味跑 – 比例推理
A school organises a charity fun run. Participants can choose a 3 km, 5 km, or 10 km route. Last year, the ratio of participants in the three distances was 5 : 3 : 2, and the total sponsorship raised was £4 500. The average sponsorship per kilometre was the same across all three groups. This year, the event attracted 40% more runners in the 10 km category, while the numbers for 3 km and 5 km stayed exactly the same. The sponsorship rate per kilometre increased by 25%.
某学校组织慈善趣味跑。参与者可选择 3 公里、5 公里或 10 公里路线。去年，三个距离的参赛人数比为 5 : 3 : 2，共筹集赞助金 £4 500，且三组的每公里平均赞助金额相同。今年，10 公里组人数增加了 40%，而 3 公里和 5 公里组人数与去年完全相同。每公里的赞助费率上升了 25%。
k, 3k, and 2k respectively. Because sponsorship/km is the same, total distance run was (5k × 3) + (3k × 5) + (2k × 10) = 15k + 15k + 20k = 50k km. Let the rate be r £/km, so 50k × r = 4 express total funds this year using the new quantities. The new 10 km runners count is 2k × 1.4 = 2.8k, while 5k and 3k remain. New total distance = 15k + 15k + (2.8k × 10) = 15k + 15k + 28k = 58k km. New rate = 1.25 × (90/k) = 112.5/k. Total raised = 58k × (112.5/k) = £6 525. The percentage increase is (6 525 – 4 500) ÷ 4 500 × 100% = 45%.
设去年参赛人数分别为 5k、3k 和 2k。因每公里赞助金额相同，总跑步里程为 (5k × 3) + (3k × 5) + (2k × 10) = 50k 公里。设费率为 r 英镑/公里，则 50k × r = 4 500，故 r = 90/k。今年 10 公里组人数变为 2k × 1.4 = 2.8k，其余不变。新的总里程 = 15k + 15k + 28k = 58k 公里。新费率 = 1.25 × (90/k) = 112.5/k。筹集总额 = 58k × (112.5/k) = £6 525。增幅为 (6 525 – 4 500) ÷ 4 500 × 100% = 45%。
This case highlights how proportional reasoning removes the need to know the exact number of participants. Representing quantities with a variable k is a powerful technique.
此案例表明，比例推理可以免去确定参赛者具体人数的需要。用变量 k 表示数量是一种强有力的技巧。
2. Case Study: Mobile Phone Battery – Linear Model | 案例二：手机电池 – 线性模型
A mobile phone battery discharges at a steady rate when video is streamed continuously. At 1:00 pm the battery is 85% full, and at 3:00 pm it drops to 45%. Assume a linear relationship between time and remaining charge.
一部手机在连续播放视频时电池以恒定速率放电。下午 1:00 电量 85%，下午 3:00 降至 45%。假设时间与剩余电量之间为线性关系。
and B(t) the battery percentage. From the data: B(1) = 85, B(3) = 45. The gradient m = (45 – 85) ÷ (3 – 1) = –40 ÷ 2 = –20% per hour. The equation is B(t) = 85 – 20(t – 1) = 105 – 20t. Check: at t=3, 105 – 60 = 45.
设 t 为中午 12:00 后的小时数，B(t) 为电量百分比。由数据得：B(1) = 85，B(3) = 45。斜率 m = (45 – 85) ÷ (3 – 1) = –20% 每小时。方程为 B(t) = 85 – 20(t – 1) = 105 – 20t。验证：t=3 时，105 – 60 = 45。
When will the battery be completely drained? Set 105 – s 5:15 pm. If a power-saving mode is activated at 5:00 pm, slowing the discharge rate to 8% per hour, how long can the phone last from 5:00 pm? At 5:00 pm, t=5, B(5) = 105 – 100 = 5%. Remaining time at new rate = 5 ÷ 0 pm. Alternatively, using the linear model before 5:00 pm and a separate model afterward.
电池何时完全耗尽？令 105 – 20t = 0 → t = 5.25，即下午 5:15。若在下午 5:00 启用省电模式，放电速率降至每小时 8%，手机还能用多久？下午 5:00 时 t=5，B(5)=5%。按新速率剩余时间 = 5 ÷ 8 = 0.625 小时 = 37.5 分钟，故将在 5:37:30 没电。此方法在下午 5:00 之前使用原线性模型，之后换用另一模型。
This case demonstrates how constructing a linear function from two data points allows predictions, and how piecewise linear models can handle changes in behaviour.
此案例示范了如何从两个数据点构建线性函数进行预测，以及分段线性模型如何处理行为变化。
3. Case Study: Designing a Water Tank – Surface Area and Volume | 案例三：水箱设计 – 表面积与体积
A manufacturer wants to produce a cylindrical water tank with a lid. The volume must be exactly 1 000 litres (1 000 000 cm³). The material for the curved side costs £0.02 per cm², while the circular top and bottom cost £0.03 per cm² due to rigid reinforcement. Find the dimensions (radius r and height h) that minimise the total material cost.
制造商想生产带盖的圆柱形水箱。容积必须精确为 1 000 升（1 000 000 立方厘米）。侧面材料费用为每平方厘米 £0.02，而圆形顶盖和底盖因加固需要每平方厘米 £0.03。求使材料总成本最小的尺寸（半径 r 和高 h）。
Volume constraint: π r² h = 1 000 000 → h = 1 000 000 ÷ (π r²). Surface area: side = 2π r h; top + bottom = 2π r². Cost function C = 0.02 × (2π r h) + 0.03 × (2π r²) = 0.04π r h + 0.06π r². Substitute h: C(r) = 0.04π r × [1 000 000 ÷ (π r²)] + 0.06π r² = 40 000/r + 0.06π r².
容积约束：π r² h = 1 000 000 → h = 1 000 000 ÷ (π r²)。表面积：侧面 = 2π r h；顶底 = 2π r²。成本函数 C = 0.02 × (2π r h) + 0.03 × (2π r²) = 0.04π r h + 0.06π r²。代入 h：C(r) = 0.04π r × [1 000 000 ÷ (π r²)] + 0.06π r² = 40 000/r + 0.06π r²。
e minimum, we can use numerical exploration (trial and improvement). Try r = 20 cm: C ≈ 40 ≈ 1271.2. r = 55 cm: 727.3 + 570.2 ≈ 1297.5. Minimum is around r = 50 cm. At r = 50, h = 1 000 000 ÷ (π × 2500) ≈ 127.3 cm. The minimum cost is approximately £1 271. This problem builds intuition before formal calculus.
为寻找最小值，可用数值探索（试错法）。尝试 r = 20 cm：C ≈ 40 000/20 + 0.06π(400) = 2000 + 75.4 ≈ 2075.4。r = 30 cm：1333.3 + 169.6 ≈ 1502.9。r = 40 cm：1000 + 301.6 ≈ 1301.6。r = 50 cm：800 + 471.2 ≈ 1271.2。r = 55 cm：727.3 + 570.2 ≈ 1297.5。最小值约在 r = 50 cm 处。此时 h = 1 000 000 ÷ (π × 2500) ≈ 127.3 cm。最低成本约为 £1 271。此问题在正式微积分前培养了直觉。
4. Case Study: Traffic Survey – Data Handling and Averages | 案例四：交通调查 – 数据处理与平均值
A student records the number of vehicles passing her house in 5-minute intervals over one hour: 8, 12, 9, 15, 10, 14, 11, 13, 16, 10, 12, 9. She wants to estimate the average number of vehicles per hour and comment on variability.
一名学生记录了每小时按 5 分钟间隔统计经过她家门前车辆的数据：8, 12, 9, 15, 10, 14, 11, 13, 16, 10, 12, 9。她想估算平均每小时车流量并讨论变异性。
al vehicles in 12 intervals = 8+12+9+15+10+14+11+13+16+10+12+9 = 139. Mean per 5 min = 139 ÷ 12 ≈ 11.58 vehicles. Hence mean per hour = 11.58 × )/2 = 11.5 vehicles per 5 min. Mode = multi-modal: 9, 10, 12 appear twice each. Range = 16 – 8 = 8.
先求 12 个间隔的总车辆数 = 8+12+9+15+10+14+11+13+16+10+12+9 = 139。每 5 分钟平均值 = 139 ÷ 12 ≈ 11.58 辆。因此每小时平均值 = 11.58 × 12 = 139 辆/小时，与预期一致。中位数：排序后为 8,9,9,10,10,11,12,12,13,14,15,16，中位数 = (11+12)/2 = 11.5 辆/5 分钟。众数：9、10、12 各出现两次，为多众数。极差 = 16 – 8 = 8。
(8,9,9,10,10,11) = (9+10)/2 = 9.5. Q3 = median of upper half (12,12,13,14,15,16) = (13+14)/2 = 13.5. IQR = 13.5 – 9.5 = 4 vehicles. This means the middle 50% each other. Such analysis helps assess traffic consistency and plan road safety measures.
数据表明分布较为分散。她可以计算四分位距（IQR）来更好地描述离散程度：Q1 为下半部分的中位数 (8,9,9,10,10,11) = (9+10)/2 = 9.5。Q3 为上半部分中位数 (12,12,13,14,15,16) = (13+14)/2 = 13.5。IQR = 13.5 – 9.5 = 4 辆。这意味着中间 50% 的 5 分钟车流量彼此相差不超过 4 辆。此类分析有助于评估交通稳定性并规划道路安全措施。
5. Case Study: School Trip – Algebraic Modelling | 案例五：学校旅行 – 代数建模
arry up to 52 students. The total cost of hiring c coaches is given by the formula: Cost = £200c + £350 (fixed booking fee). The total amount collected from n students is £ s required to break even, assuming as many coaches are hired as needed (no empty coaches).
学校为旅行租用大巴。每辆大巴最多可载 52 名学生。租用 c 辆大巴的总费用公式为：费用 = £200c + £350（固定预订费）。从 n 名学生处收取的总金额为每人 £15。求利润的表达式，并确定在租用恰好足够、不留空车的情况下，达到收支平衡所需的最低学生人数。
×52=780, cost = 550 → profit 230 (always profit). But breakeven occurs when cost exceeds revenue at low numbers. Try c=1 with minimum n? ed per coach. So breakeven for c=1: 15n = 550 → n=36.67, so minimum n=37 (since 37 students fit in 1 coach). Check: cost=550, revenue=555, profit=£ 2 coaches. However, 50 is less than 52, so actually you could use 1 coach for 50 students! The condition is you must use enough coaches for n without empty coaches? The problem says 'as many coaches are hired as needed' st respect the coach requirement. The true minimum n for overall operation: if n ≤ 52, c=1, breakeven n=37. For 53 ≤ n ≤ 104, c=2, breakeven 15n = 750 → n=50, but n=50 is in c=1 region. So for c=2, the smallest n is 53, at which profit = 15×53 - 750 = 795-750=45 positive. So the business is always profitable beyond n=36? Let's check n=36, c=1, revenue 540, cost 550, loss £10. So minimum n for breakeven is indeed 37.
设 c 为满足 52c ≥ n 的最小整数。收支平衡要求 15n = 200c + 350。需探索整数 c 值。若 c=1，最大 n=52，收入 = 15×52=780，成本 550 → 利润 230（始终盈利）。但收支平衡点出现在低人数成本超过收入时。c=1 时，最小 n=1，收入 15，成本 550 → 亏损。随着 n 增加，收入上升，但每车成本固定。c=1 的平衡点：15n=550 → n=36.67，故最小整数 n=37（37 人刚好用 1 辆车）。验证：成本 550，收入 555，利润 £5。c=2 时，平衡点：15n=400+350=750 → n=50。此人数低于 52×2=104，故可用 2 辆车，但 50 ≤ 52，完全可用 1 辆车。题意是“根据需要租用车辆”，因此若 n=50，只需 1 辆车，成本 550，利润 200。所以平衡点必须符合车辆使用规则。真正的全局最小 n：若 n ≤ 52，c=1，平衡 n=37。对 53 ≤ n ≤ 104，c=2，平衡方程得 n=50。但 n=50 在 c=1 区间内。实际上 c=2 的最小 n=53，此时利润 = 15×53 – 750 = 45 为正。因此只要 n > 36 就始终盈利。最小保本人数为 37。
6. Case Study: Probability in a Dice Game – Expected Value | 案例六：骰子游戏中的概率 – 期望值
In a game, you roll two fair six-sided dice. If the sum is 7 or 11, you win £5. If the sum is 2, 3, or 12, you lose £3. For any other sum, you win nothing and lose nothing. Calculate the expected gain per game and discuss whether the game is fair.
在一款游戏中，你掷两枚公平的六面骰子。若和为 7 或 11，赢得 £5；若和为 2、3 或 12，损失 £3；其他和数既不赢也不亏。计算每局游戏的期望收益，并讨论游戏是否公平。
7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 ways. Sum of 11: (5,6),(6,5) → 2 ways. So P(win £ → 1 way. sum 3: (1,2),(2,1) → 2 ways. sum 12: (6,6) → 1 way. Total losing = 4 ways, P(lose £3) = 4/36 = 1/9. Other sums: 36 – 8 – 4 = 24 ways, gain £0. Expected gain = (5 × 8/36) + (–3 × 4/36) + (0 × 24/36) = (40 – 12)/36 = 28/36 = £ and favours the player. To make it fair, the loss amount or win amount could be adjusted.
总可能结果 = 36。有利结果：和为 7：(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 种。和为 11：(5,6),(6,5) → 2 种。所以 P(赢 £5) = (6+2)/36 = 8/36 = 2/9。损失结果：和 2：(1,1) → 1 种。和 3：(1,2),(2,1) → 2 种。和 12：(6,6) → 1 种。损失总计 4 种，P(亏 £3) = 4/36 = 1/9。其他和数 24 种，收益 £0。期望收益 = (5 × 8/36) + (–3 × 4/36) + (0 × 24/36) = (40 – 12)/36 = 28/36 ≈ £0.78。正期望值意味着长期来看玩家每局平均赢约 78 便士，所以游戏不公平，有利于玩家。为使游戏公平，可调整损失金额或赢得金额。
7. Case Study: Park Refurbishment – Area and Cost | 案例七：公园翻新 – 面积与成本
A rectangular park measures 120 m by 80 m. A uniform path of width x metres is constructed along the inside perimeter, leaving a central grassy area. The cost of paving the path is £25 per m², and the remaining grass area must be reseeded at £4 per m². The total budget for both tasks is £50 000. Find the possible widths of the path.
一个矩形公园长 120 米，宽 80 米。沿内侧周边修建一条宽度为 x 米的均匀小径，中央为草坪区域。铺路费用为每平方米 £25，剩余草地需重新播种，费用为每平方米 £4。两项任务的总预算为 £50 000。求小径的可能宽度。
Inner grassy rectangle dimensions: (120 – 2x) by (80 – 2x). Total area = 9600 m². Path area = total area – grass area = 9600 – (120 – 2x)(80 – 2x). Expand: (120 – 2x)(80 – 2x) = 9600 – 240x – 160x + 4x² = 9600 – 400x + 4x². So path area = 9600 – (9600 – 400x + 4x²) = 400x – 4x². Total cost = 25(400x – 4x²) + 4(9600 – 400x + 4x²) = 10 000x – 100x² + 38 400 – 1 600x + 16x². Simplify: (10 000x – 1 600x) + (–100x² + 16x²) + 38 400 = 8 400x – 84x² + 38 400. Set equal to 50 000: –84x² + 8 400x + 38 400 = 50 000 → –84x² + 8 400x – 11 600 = 0. Divide by –4: 21x² – 2 100x + 2 900 = 0. Use the quadratic formula: x = [2 100 ± √2 100² – 4×21×2 900] / (2×21). Discriminant = 4 410 000 – 243 600 = 4 166 400. √4 166 400 = 2 040 (since 2 040² = 4 161 inant is 4,166,400, difference 4,800. 2041² = 2040²+2040+2041 = 4,161,600+4,081=4,165,681. 2042² = 4,165,681+2041+2042 = 4,169,764. So sqrt is not integer. 2041.xx? Let's compute precisely later with fractions). Actually recalculate discriminant: 2 100² = 4 410 000. 4×21×2 900 = 84×2 900 = 243 600. Difference = 4 166 664 = 16 × 2604 = 64 × 651. 651=3×7×31. Not a perfect square. So sqrt(4,166,400) = 2040.88? But we can keep exact: sqrt(4,166,400) = 20√10,416? Maybe simplify: 4,166,400 = 100 × 41,664, not helpful. We' (4,140.94)/42 ≈ 98.6 m (not valid, width too large, 2x would exceed 80). The other: (2,100 – ible width. This modelling task mixes geometry, quadratic equations, and real constraints.
内部草坪矩形尺寸：(120 – 2x) × (80 – 2x)。总面积 = 9600 m²。小径面积 = 总面积 – 草地面积 = 9600 – (120 – 2x)(80 – 2x)。展开：(120 – 2x)(80 – 2x) = 9600 – 240x – 160x + 4x² = 9600 – 400x + 4x²。故小径面积 = 9600 – (9600 – 400x + 4x²) = 400x – 4x²。总费用 = 25(400x – 4x²) + 4(9600 – 400x + 4x²) = 10 000x – 100x² + 38 400 – 1 600x + 16x²。化简得 8 400x – 84x² + 38 400。令其等于 50 000：–84x² + 8 400x – 11 600 = 0。除以 –4 得 21x² – 2 100x + 2 900 = 0。代入二次公式，判别式 = 4 166 400，平方根约为 2 040.94。x = (2 100 – 2 040.94) ÷ 42 ≈ 1.406 m。另一解 x ≈ 98.6 m 无效（宽度过大）。故可行路径宽度约为 1.41 米。此题综合了几何、二次方程与现实约束。
8. Case Study: Population Projection – Compound Growth | 案例八：人口预测 – 复合增长
A small town has a population of 12 000, which is growing at a steady rate of 3.2% per year. The town council wants to know when the population will reach 18 000 and how long it takes to double. Use an exponential model and solve using trial and error or logarithms.
某小镇现有人口 12 000，正以每年 3.2% 的速度稳定增长。镇委员会想知道何时人口将达到 18 000，以及翻倍需要多长时间。请使用指数模型并通过试错法或对数求解。
Model: P(t) = 12 000 × (1.032)t, where t is years from now. To find t for 18 000: 12 000 × (1.032)t = 18 000 → (1.032)t 61 / 0.01368 ≈ 12.87 years, so around 12 years and 10 months. Doubling time: solve (1.032)t 1368 ≈ 22.0 years. Alternatively, using the rule of 70: 70/3.2 ≈ 21.875 years, very close.
模型：P(t) = 12 000 × (1.032)t，t 为距今的年数。求达 18 000 时 t：12 000 × (1.032)t = 18 000 → (1.032)t = 1.5。取对数：t = log(1.5) / log(1.032) ≈ 0.1761 / 0.01368 ≈ 12.87 年，即约 12 年 10 个月。翻倍时间：求解 (1.032)t = 2 → t = log(2)/log(1.032) = 0.3010 / 0.01368 ≈ 22.0 年。也可用 70 法则：70/3.2 ≈ 21.875 年，十分接近。
This case reinforces exponential growth concepts and logarithmic solving. The council can plan infrastructure accordingly, knowing that population pressure will significantly increase in two decades.
此案例巩固了指数增长概念与对数解法。委员会可据此规划基础设施，知晓 20 年内人口压力将显著增加。
9. Case Study: Business Profit Maximisation – Quadratic Modelling | 案例九：企业利润最大化 – 二次模型
A company sells handmade candles. The monthly profit £P is given by P = –2n² + 160n – 2 ds maximum profit and the maximum profit. Also determine the break-even production levels.
公司销售手工蜡烛。每月利润 £P 满足 P = –2n² + 160n – 2 400，其中 n 为生产和销售的数量。求实现最大利润的蜡烛数量及最大利润值。同时确定盈亏平衡的产量水平。
e quadratic is concave down (coefficient of n² is negative). Maximum occurs at vertex: n = –b/(2a) = –160/(2×–2) = –160/–4 = 40 candles. Maximum profit P(40) = –2(1600) + 160×40 – 2 400 = –3 200 + 6 400 – 2 400 = 800. So maximum profit is £800 when 40 candles are sold. Break-even: solve –2n² + 160n – 2 400 = 0 → divide by –2: n² – 80n + 1 200 = 0. Factorise: (n – 20)(n – Thus, the company makes a profit only when 20 < n < 60 and loses money outside this range.
二次项系数为
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