📚 Cambridge IGCSE Maths: Case Study Problem Solving | 剑桥IGCSE数学：案例分析实战演练
In Cambridge IGCSE Mathematics, case studies bridge the gap between textbook theory and real-world application. These problems require you to select the right mathematical tools, interpret data, and communicate solutions clearly. This article presents a series of practical case studies covering percentages, ratio, speed, geometry, statistics, algebra, probability, and trigonometry. Work through each scenario to strengthen your problem-solving skills and prepare for exam-style contextual questions.
在剑桥IGCSE数学中，案例研究在课本理论与现实应用之间架起了桥梁。这些问题要求你选择合适的数学工具、解读数据并清晰地表达解决方案。本文呈现一系列实践案例研究，涵盖百分比、比率、速度、几何、统计、代数、概率和三角学。逐一练习这些场景，提升你的问题解决能力，为考试中的情境化题目做好准备。
1. Percentage Discounts and Profit | 百分比折扣与利润
A clothing store buys a jacket for £60. The manager decides to apply a 25% markup on the cost price to set the selling price. During a seasonal sale, a 10% discount is offered on the selling price. Determine the final price a customer pays.
一家服装店以60英镑购进一件夹克。经理决定在成本价上加价25%来设定售价。在季末促销中，售价再打九折。请计算顾客最终支付的价格。
First calculate the profit amount: 25% of £60 = 0.25 × 60 = £15. So the selling price before discount is £60 + £15 = £75.
首先计算利润额：60英镑的25% = 0.25 × 60 = 15英镑。因此折前售价为60 + 15 = 75英镑。
Now apply the 10% discount on £75. Discount amount = 0.10 × 75 = £7.50. The final price = £75 - £7.50 = £67.50.
现在对75英镑应用10%折扣。折扣额 = 0.10 × 75 = 7.50英镑。最终价格 = 75 - 7.50 = 67.50英镑。
Always check if the discount is applied to the original cost or the selling price. Here the sequence matters – percentage increase followed by a percentage decrease does not bring you back to the original value.
务必检查折扣是针对成本价还是售价。这里顺序很重要——先增百分率再减百分率并不会让你回到初始值。
2. Ratio and Proportion | 比率与比例
An interior designer needs 12 litres of violet paint. The shade is created by mixing blue and red paint in the ratio 5:3. How many litres of each colour are required?
一位室内设计师需要12升紫罗兰色油漆。这种色调由蓝色和红色油漆按5:3的比例混合而成。请问各需要多少升？
The total number of parts in the ratio is 5 + 3 = 8 parts. One part equals 12 litres ÷ 8 = 1.5 litres.
比率中的总份数为5 + 3 = 8份。一份等于12升 ÷ 8 = 1.5升。
Blue paint: 5 parts × 1.5 = 7.5 litres. Red paint: 3 parts × 1.5 = 4.5 litres.
蓝色油漆：5份 × 1.5 = 7.5升。红色油漆：3份 × 1.5 = 4.5升。
adding 7.5 + 4.5 = 12 litres, and the ratio 7.5:4.5 simplifies to 5:3 when divided by 1.5.
你可以验证：7.5 + 4.5 = 12升，并且7.5:4.5同除以1.5后化简为5:3。
3. Speed, Distance and Time | 速度、距离与时间
A train travels a distance of 240 km at an average speed of 80 km/h. However, it makes a scheduled stop of 20 minutes in the middle of the journey. Find the total time taken for the whole trip in hours and minutes.
一列火车以平均速度80公里/小时行驶240公里。然而，它在行程中途计划停车20分钟。求整个旅程花费的总时间，以小时和分钟表示。
Travel time (moving) = Distance ÷ Speed = 240 ÷ 80 = 3 hours.
行驶时间 = 距离 ÷ 速度 = 240 ÷ 80 = 3小时。
Add the stop time: 20 minutes = 20/60 = 1/3 hour. So total time = 3 + 1/3 = 3 1/3 hours, which is 3 hours 20 minutes.
加上停车时间：20分钟 = 20/60 = 1/3小时。因此总时间 = 3 + 1/3 = 3又1/3小时，即3小时20分钟。
and minutes, remember that 1 hour = 60 minutes. Keeping units consistent prevents errors.
在小时和分钟之间转换时，记住1小时 = 60分钟。保持单位一致能避免错误。
4. Geometry and Measurement | 几何与测量
A school wants to build a rectangular playground with length 30 m and width 20 m. They need to install a fence around it and lay turf on the entire area. Fencing costs £12 per metre, and turf costs £5 per square metre. Calculate the total cost of the project.
一所学校想建一个长30米、宽20米的矩形操场。他们需要围上篱笆并铺设草坪。篱笆成本为每米12英镑，草皮为每平方米5英镑。计算项目总成本。
Perimeter for fencing = 2 × (30 + 20) = 2 × 50 = 100 m. Area for turf = 30 × 20 = 600 m².
围栏所需周长 = 2 × (30 + 20) = 2 × 50 = 100米。草坪面积 = 30 × 20 = 600平方米。
Fencing cost = 100 × 12 = £1200. Turf cost = 600 × 5 = £3000. Total cost = 1200 + 3000 = £4200.
篱笆费用 = 100 × 12 = 1200英镑。草皮费用 = 600 × 5 = 3000英镑。总费用 = 1200 + 3000 = 4200英镑。
This case shows how perimeter and area formulas serve different purposes: perimeter for boundaries and area for covering surfaces.
这个案例说明周长和面积公式的不同用途：周长用于边界，面积用于覆盖表面。
5. Statistics: Analysing Survey Data | 统计：分析调查数据
A teacher recorded the scores of 10 students on a maths test: 45, 52, 78, 66, 52, 89, 52, 71, 66, 90. Determine the mean, median, mode, and range of these scores.
一位老师记录了10名学生的数学测试成绩：45, 52, 78, 66, 52, 89, 52, 71, 66, 90。求这些分数的平均数、中位数、众数和极差。
First sort the data in ascending order: 45, 52, 52, 52, 66, 66, 71, 78, 89, 90.
首先将数据从小到大排序：45, 52, 52, 52, 66, 66, 71, 78, 89, 90。
Mode = 52 (it appears three times). Range = 90 - 45 = 45.
众数 = 52（出现三次）。极差 = 90 - 45 = 45。
middle lies between the 5th and 6th values. 5th = 66, 6th = 66, so median = (66 + 66)/2 = 66.
中位数：10个数值，中间位置在第5和第6个之间。第5个是66，第6个是66，因此中位数 = (66+66)/2 = 66。
Mean = sum of scores ÷ 10. Sum = 45+52+52+52+66+66+71+78+89+90 = 661. Mean = 661 ÷ 10 = 66.1.
平均数 = 总分 ÷ 10。总分 = 45+52+52+52+66+66+71+78+89+90 = 661。平均数 = 661 ÷ 10 = 66.1。
Even though the mean and median are close here, the mode reveals the most frequent score. All three averages give different insights.
尽管这里的平均数和中位数很接近，但众数揭示了最常见的分数。三个平均数提供了不同角度的信息。
6. Algebra: Mobile Phone Data Plan | 代数：手机流量套餐
A mobile network offers two plans: Plan A has a monthly fee of £10 and charges £3 per GB of data used. Plan B has no monthly fee but charges £5 per GB. Write a linear expression for the monthly cost of each plan and determine the data usage at which both plans cost the same.
一家移动网络提供两种套餐：A套餐月租10英镑，每使用1 GB数据收费3英镑。B套餐无月租，但每GB收费5英镑。写出每种套餐月费用的线性表达式，并求出使两个套餐费用相同的数据使用量。
Let g be the number of GB used. Cost of Plan A: CA = 10 + 3g. Cost of Plan B: CB = 5g.
设 g 为使用的GB数。A套餐费用：CA = 10 + 3g。B套餐费用：CB = 5g。
To find when they are equal, solve 10 + 3g = 5g ⇒ 10 = 2g ⇒ g = 5 GB. At 5 GB, both plans cost £25.
为求相等时的用量，解方程 10 + 3g = 5g ⇒ 10 = 2g ⇒ g = 5 GB。在5 GB时，两个套餐费用均为25英镑。
This break-even point helps customers choose: below 5 GB, Plan B is cheaper; above 5 GB, Plan A is more economical. Linear equations model real financial decisions.
这个盈亏平衡点帮助客户选择：低于5 GB时B套餐更便宜，高于5 GB时A套餐更划算。线性方程能为真实的财务决策建模。
7. Probability: Carnival Game | 概率：嘉年华游戏
A carnival spinner has 8 equal sectors numbered 1 to 8. To win a prize, a player must spin an even number greater than 4. Calculate the probability of winning on a single spin.
一个嘉年华转盘有8个相等的扇形区域，编号1到8。玩家必须转到大于4的偶数才能获奖。计算单次旋转的获奖概率。
Even numbers greater than 4 from 1 to 8 are 6 and 8. That gives 2 favourable outcomes.
1到8中大于4的偶数是6和8。共有2种有利结果。
Total possible outcomes = 8. Probability = favourable / total = 2/8 = 1/4 or 0.25.
总可能结果 = 8。概率 = 有利结果 / 总数 = 2/8 = 1/4 即0.25。
If the carnival charges £1 per spin, what is the expected number of prizes in 200 spins? Expected wins = 200 × 1/4 = 50 prizes. Probability helps predict long-term results.
如果嘉年华每次旋转收费1英镑，旋转200次的预期获奖次数是多少？期望获奖次数 = 200 × 1/4 = 50次。概率有助于预测长期结果。
8. Trigonometry: Ramp Accessibility | 三角学：坡道可访问性
angle of inclination must not exceed 4°. What is the minimum horizontal length of the ramp? Use tan θ = opposite/adjacent.
一个轮椅坡道需要升高1.2米。建筑规范要求倾斜角不得超过4°。求坡道的最小水平长度。使用 tan θ = 对边/邻边。
horizontal length (adjacent side). tan 4° = 1.2 / L. Using a calculator, tan 4° ≈ 0.0699.
设 L 为水平长度（邻边）。tan 4° = 1.2 / L。用计算器得 tan 4° ≈ 0.0699。
Then L = 1.2 / 0.0699 ≈ 17.17 metres. The ramp length (hypotenuse) can be found by Pythagoras: √17.17² + 1.2² ≈ 17.21 m, but the horizontal length is the key design constraint.
则 L = 1.2 / 0.0699 ≈ 17.17米。坡道长度（斜边）可用勾股定理求得：√17.17² + 1.2² ≈ 17.21米，但水平长度是关键的设计约束。
Always set your calculator to degree mode. A small angle requires a much longer ramp to keep the slope gentle, illustrating how trigonometry ensures safety in construction.
务必将计算器设置为度数模式。较小的角度需要更长的坡道来保持坡度平缓，这体现了三角学如何确保建筑安全。
Published by TutorHao | Maths Revision Series | aleveler.com
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