📚 Year 9 Cambridge Maths: Case Study Practice | 九年级剑桥数学：案例分析实战演练
In Year 9 Cambridge Mathematics, case studies are extended problems that connect classroom concepts to real‑life situations. They test your ability to interpret information, choose effective strategies, and communicate reasoning clearly. This article presents ten varied case studies with step‑by‑step solutions so you can build confidence and sharpen your problem‑solving skills.
在九年级剑桥数学中，案例分析是将课堂知识和真实情境结合的拓展问题。它们考查你提取信息、选择有效策略和清晰表达思路的能力。本文提供十个不同的案例研究及逐步解答，帮助你建立信心、提升解题技巧。
1. Understanding Case Study Problems | 理解案例分析问题
Case study questions often start with a detailed scenario containing numbers, tables, or pictures. Your first job is to read carefully and highlight the mathematical demands — do you need to calculate a cost, compare proportions, or model a relationship? Always list what you know, what you need to find, and which formulas or concepts apply before you begin writing any working.
案例分析题通常从一个包含数字、表格或图片的详细情境开始。你的第一步是仔细阅读，标出数学要求——是需要计算费用、比较比例还是建立关系模型？在动笔之前，务必列出已知量、待求量以及适用的公式或概念。
When a problem seems long, break it into smaller parts. Convert wordy statements into concise maths expressions. For example, ‘three times as many adults as children’ becomes a = 3c. Visualising with a sketch or a table often reveals the path to a solution. Finally, always check your answer against the original scenario — does it make sense in real life?
当题目篇幅较长时，可将其拆成更小的部分。把冗长的描述转化为简明的数学表达式。例如，“成人人数是儿童的三倍”即 a = 3c。通过画草图或列表格往往能看清解题路径。最后，务必将答案代回原情境检验——在现实中是否合理？
2. Case: Planning a School Fete | 案例：策划学校义卖会
A school fete committee expects 200 visitors. Each ticket costs £2.50, and children under 12 get a 40 % discount. If 30 % of visitors are children under 12, calculate the total ticket income.
学校义卖会组委会预计有 200 名访客。每张门票售价 2.50 英镑，12 岁以下儿童享受 40 % 折扣。若 30 % 的访客是 12 岁以下儿童，计算门票总收入。
Step 1: Number of children = 30 % of 200 = 0.30 × 200 = 60 children. Number of adults = 200 – 60 = 140.
第一步：儿童人数 = 200 的 30 % = 0.30 × 200 = 60 名儿童。成人人数 = 200 – 60 = 140 名。
Step 2: Child ticket price = £2.50 – (40 % of £2.50) = 2.50 – 1.00 = £1.50. Adult ticket price = £2.50.
第二步：儿童票价 = 2.50 英镑 – (2.50 英镑的 40 %) = 2.50 – 1.00 = 1.50 英镑。成人票价 = 2.50 英镑。
Step 3: Income from children = 60 × £1.50 = £90. Income from adults = 140 × £2.50 = £350. Total income = £90 + £350 = £440.
第三步：儿童票收入 = 60 × 1.50 英镑 = 90 英镑。成人票收入 = 140 × 2.50 英镑 = 350 英镑。总收入 = 90 + 350 = 440 英镑。
By writing the steps separately, the calculation stays organised and easy to verify.
将步骤分开书写，计算过程条理清晰，易于检查。
3. Case: Building a Ramp | 案例：搭建斜坡
A ramp is to be built from the ground to a doorstep 0.9 m high. For safety, the slope must be 1 : 12 (rise : run). How long is the ramp along the slope? How far from the doorstep should the ramp start?
要搭建一个从地面到高 0.9 米门阶的斜坡。出于安全考虑，坡度需为 1 : 12（垂直高度：水平距离）。斜坡的斜面长度是多少？斜坡起点应离门阶多远？
The ratio 1 : 12 means for every 1 m rise, the horizontal distance (run) is 12 m. Here rise = 0.9 m. So run = 0.9 × 12 = 10.8 m. That is the distance from the doorstep.
比例 1 : 12 表示每升高 1 米，水平距离为 12 米。此处升高 0.9 米，故水平距离 = 0.9 × 12 = 10.8 米。这就是斜坡起点离门阶的距离。
Now use Pythagoras’ theorem to find the sloping length L. L² = (rise)² + (run)² = 0.9² + 10.8² = 0.81 + 116.64 = 117.45. So L = √117.45 ≈ 10.84 m (to 2 d.p.).
现在用勾股定理求斜面长度 L。L² = (升高)² + (水平距离)² = 0.9² + 10.8² = 0.81 + 116.64 = 117.45。因此 L = √117.45 ≈ 10.84 米（精确到两位小数）。
This case links ratio with right‑angled triangles — a typical Cambridge combination.
本案例将比例与直角三角形联系起来，是剑桥考试中常见的组合。
4. Case: Analysing Test Scores | 案例：分析考试成绩
A class of 25 students scored the following marks out of 50 in a maths test: 32, 45, 28, 36, 41, 39, 44, 27, 33, 38, 42, 29, 35, 40, 37, 31, 46, 30, 34, 43, 48, 26, 25, 47, 49. Find the mean, median, mode and range. Also draw a box‑plot outline.
某班 25 名学生数学测验满分 50 分，成绩如下：32, 45, 28, 36, 41, 39, 44, 27, 33, 38, 42, 29, 35, 40, 37, 31, 46, 30, 34, 43, 48, 26, 25, 47, 49。求平均数、中位数、众数和极差，并画出箱线图的大致轮廓。
Sort data: 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49.
排序：25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49。
Mean = sum ÷ 25. Sum = 925, so mean = 925 ÷ 25 = 37. Median is the 13th value = 37. Mode: all values appear once — no mode. Range = 49 – 25 = 24.
平均数 = 总和 ÷ 25。总和 = 925，所以平均数 = 925 ÷ 25 = 37。中位数是第 13 个值 = 37。众数：每个值只出现一次——无众数。极差 = 49 – 25 = 24。
For a box plot: minimum = 25, lower quartile Q₁ = 30.5, median = 37, upper quartile Q₃ = 43.5, maximum = 49. A rough sketch would show a fairly symmetric distribution.
箱线图：最小值 = 25，下四分位数 Q₁ = 30.5，中位数 = 37，上四分位数 Q₃ = 43.5，最大值 = 49。粗略绘出的图形呈大致对称分布。
5. Case: Designing a Patchwork Quilt | 案例：设计拼布被面
A rectangular quilt will be 2.4 m long and 1.8 m wide. It is made from square patches of side 20 cm. How many patches are needed? If 15 % of the fabric is wasted in cutting, what total area of fabric must be bought?
一条矩形拼布被面长 2.4 米，宽 1.8 米，由边长为 20 厘米的正方形布块拼成。需要多少块布？若裁剪时浪费 15 % 的面料，必须购买的总面料面积是多少？
Convert all measurements to metres: patch side = 0.2 m. Number along length = 2.4 ÷ 0.2 = 12. Number along width = 1.8 ÷ 0.2 = 9. Total patches = 12 × 9 = 108.
将所有单位换算为米：布块边长 = 0.2 米。沿长度方向可排 = 2.4 ÷ 0.2 = 12 块。沿宽度方向可排 = 1.8 ÷ 0.2 = 9 块。布块总数 = 12 × 9 = 108。
Finished quilt area = 2.4 × 1.8 = 4.32 m². Because 15 % is wasted, the usable fabric is 85 % of what is bought. Let A be the bought area. 0.85 × A = 4.32, so A = 4.32 ÷ 0.85 ≈ 5.08 m² (to 2 d.p.).
成品被面面积 = 2.4 × 1.8 = 4.32 平方米。由于浪费 15 %，实际可用面料为购买量的 85 %。设购买面积为 A，则 0.85 × A = 4.32，故 A = 4.32 ÷ 0.85 ≈ 5.08 平方米（精确到两位小数）。
This case combines ratios, area, and percentage increases — a realistic everyday challenge.
本案例综合了比例、面积和百分比增加——一个贴近生活的实际挑战。
6. Case: Saving for a New Phone | 案例：为买新手机储蓄
Mia wants to buy a phone costing £480. She already has £150 and plans to save £22 per week from her part‑time job. How many weeks does she need? If the phone price increases by 5 % after 10 weeks, how many more weeks will be needed?
米娅想买一部 480 英镑的手机。她已有 150 英镑，并计划每周从兼职工作中存 22 英镑。需要多少周？若 10 周后手机价格上涨 5 %，还需要多存几周？
Part (a): Amount still needed = 480 – 150 = £330. Weeks = 330 ÷ 22 = 15. So 15 weeks.
第一部分：仍需金额 = 480 – 150 = 330 英镑。周数 = 330 ÷ 22 = 15。故需 15 周。
Part (b): After 10 weeks she saves 10 × 22 = £220, total = 150 + 220 = £370. Remaining time before price change was 5 weeks, but now the phone costs £480 × 1.05 = £504. New amount needed = 504 – 370 = £134. Extra weeks = 134 ÷ 22 = 6.09…, so 7 whole weeks (since part weeks aren’t possible). Originally she would have needed 5 more weeks; now 7, so 2 extra weeks.
第二部分：10 周后她共存了 10 × 22 = 220 英镑，总额 = 150 + 220 = 370 英镑。价格变动前还剩下 5 周，但现在手机价格为 480 × 1.05 = 504 英镑。新差额 = 504 – 370 = 134 英镑。额外周数 = 134 ÷ 22 = 6.09…，故需 7 整周（不足一周也算一周）。原本还需要 5 周，现在需 7 周，所以 多花 2 周。
Here you must read the timeline carefully — a common trap in case‑study questions.
此处必须仔细梳理时间线，这是案例分析题中常见的陷阱。
7. Case: Population of a Bacteria Colony | 案例：细菌群落数量
A scientist places 200 bacteria in a dish. Every hour the population increases by 40 %. The colony continues until it reaches 1000 bacteria. How many complete hours does this take? Describe how the growth differs from a linear increase of 80 per hour.
一位科学家在培养皿中放入 200 个细菌，每小时数量增加 40 %。当菌落数量达到 1000 个时停止。这需要多少完整的小时？说明这种增长与每小时线性增加 80 个有何不同。
Compound growth: Population after n hours = 200 × (1.40)ⁿ. We need the smallest n such that 200 × 1.40ⁿ ≥ 1000. Try n = 3: 200 × 1.40³ = 200 × 2.744 = 548.8 (too low). n = 4: 200 × 1.40⁴ = 200 × 3.8416 ≈ 768.32. n = 5: 200 × 1.40⁵ ≈ 200 × 5.37824 = 1075.65 (≥ 1000). So 5 hours.
复合增长：n 小时后数量 = 200 × (1.40)ⁿ。我们需要最小的 n 使 200 × 1.40ⁿ ≥ 1000。试 n = 3：200 × 1.40³ = 200 × 2.744 = 548.8（不足）。n = 4：200 × 1.40⁴ = 200 × 3.8416 ≈ 768.32。n = 5：200 × 1.40⁵ ≈ 200 × 5.37824 = 1075.65（≥ 1000）。所以需要 5 小时。
Linear growth of 80 per hour: after 3 h → 440, 4 h → 520, 5 h → 600 — still far below 1000. The 40 % exponential model overtakes quickly, showing how compound growth accelerates.
每小时线性增加 80 个：3 小时后 → 440，4 小时后 → 520，5 小时后 → 600——仍远低于 1000。40 % 的指数模型增长更快，体现了复合增长的加速效应。
8. Case: Cycling to School | 案例：骑车上学
Liam cycles 6 km to school. His journey is shown on a distance‑time graph with three sections: a steady ride for 15 min covering 4 km, a 5‑minute stop, then a faster ride over the last 2 km in 6 min. Draw the graph and find his average speed in km/h for the whole trip.
利亚姆骑车 6 公里上学。他的行程用距离‑时间图表示，共三段：匀速骑行 15 分钟，行进 4 公里；停车 5 分钟；然后用 6 分钟快速骑完最后 2 公里。画出图形，并求全程的平均速度（单位 km/h）。
Time in hours: 15 min = 0.25 h, 5 min ≈ 0.0833 h, 6 min = 0.1 h. Total time = 0.25 + 0.0833 + 0.1 = 0.4333 h. Distance = 6 km. Average speed = total distance ÷ total time = 6 ÷ 0.4333 ≈ 13.85 km/h (to 2 d.p.).
时间换算为小时：15 分钟 = 0.25 小时，5 分钟 ≈ 0.0833 小时，6 分钟 = 0.1 小时。总时间 = 0.25 + 0.0833 + 0.1 = 0.4333 小时。距离 = 6 公里。平均速度 = 总距离 ÷ 总时间 = 6 ÷ 0.4333 ≈ 13.85 公里/小时（精确到两位小数）。
For the graph: plot (0,0), (0.25,4), (0.3333,4), (0.4333,6). The slope changes show varying speeds. Remember the stop is a horizontal line.
图形绘制：描点 (0,0), (0.25,4), (0.3333,4), (0.4333,6)。斜率变化反映了速度变化。注意停车阶段是一条水平线段。
9. Case: Probability Spinner Game | 案例：概率转盘游戏
A spinner has three sectors: red (½ of the area), blue (¼), and green (¼). Players spin twice. Find the probability of getting two different colours. If winning requires at least one red, what is the probability of winning?
一个转盘有三个扇区：红色（½ 面积）、蓝色（¼）和绿色（¼）。玩家旋转两次。求两次得到不同颜色的概率。若中奖需要至少一次红色，中奖概率是多少？
Use a probability tree. P(two different colours) = 1 – P(both same). P(both red) = (½)×(½) = ¼. P(both blue) = (¼)×(¼) = 1/16. P(both green) = 1/16. Sum of same = ¼ + 2/16 = ¼ + ⅛ = ⅜. So P(different) = 1 – ⅜ = ⅝.
使用概率树。两次不同颜色的概率 = 1 – 两次同色概率。两次红色 = (½)×(½) = ¼。两次蓝色 = (¼)×(¼) = 1/16。两次绿色 = 1/16。同色总和 = ¼ + 2/16 = ¼ + ⅛ = ⅜。故不同颜色概率 = 1 – ⅜ = ⅝。
P(at least one red) = P(red then red) + P(red then not red) + P(not red then red). P(red then not red) = ½ × ½ = ¼, similarly ¼ for not red then red, and ¼ for red‑red. Total = ¼ + ¼ + ¼ = ¾. Alternatively, 1 – P(no red) = 1 – (½ × ½) = 1 – ¼ = ¾.
至少一次红色的概率 = P(红红) + P(红其他) + P(其他红)。P(红其他) = ½ × ½ = ¼，同理其他红也是 ¼，红红为 ¼。总和 = ¼ + ¼ + ¼ = ¾。或者用 1 – P(无红) = 1 – (½ × ½) = 1 – ¼ = ¾。
10. Case: Ratio and Map Scales | 案例：比例与地图比例尺
A map has a scale of 1 : 25 000. Two towns are 8.4 cm apart on the map. What is the real distance in kilometres? If a lake covers an area of 3 cm² on the map, what is its actual area in hectares? (1 hectare = 10 000 m²)
一张地图的比例尺为 1 : 25 000。两座城镇在地图上相距 8.4 厘米。实际距离是多少公里？若一个湖泊在地图上占地 3 平方厘米，它的实际面积是多少公顷？（1 公顷 = 10 000 平方米）
For length: 1 cm on map represents 25 000 cm in reality. So 8.4 cm → 8.4 × 25 000 = 210 000 cm. Convert to km: 210 000 ÷ 100 000 = 2.1 km.
长度：图上 1 厘米代表实际 25 000 厘米。故 8.4 厘米 → 8.4 × 25 000 = 210 000 厘米。转换为千米：210 000 ÷ 100 000 = 2.1 千米。
For area: the scale factor for length is 25 000, so for area it is (25 000)² = 625 000 000. Map area 3 cm² → real area = 3 × 625 000 000 = 1 875 000 000 cm². 1 m² = 10 000 cm², so real area = 1 875 000 000 ÷ 10 000 = 187 500 m². In hectares: 187 500 ÷ 10 000 = 18.75 ha.
面积：长度比例因子为 25 000，因此面积比例因子为 (25 000)² = 625 000 000。图上面积 3 平方厘米 → 实际面积 = 3 × 625 000 000 = 1 875 000 000 平方厘米。1 平方米 = 10 000 平方厘米，故实际面积 = 1 875 000 000 ÷ 10 000 = 187 500 平方米。化为公顷：187 500 ÷ 10 000 = 18.75 公顷。
Always treat linear and area scale factors separately — squaring is essential.
务必区分线性比例尺和面积比例尺——平方换算至关重要。
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