📚 Case Study Practice for AQA GCSE Maths | AQA 数学案例分析实战演练
Welcome to this focused case study practice for AQA GCSE Mathematics. In this article, you will work through ten real-world scenarios that test your ability to apply mathematical concepts such as number, algebra, ratio, geometry, statistics and probability. Each case study presents a problem, guides you through the reasoning and provides a full worked solution. Use these examples to sharpen your problem-solving skills and build confidence for the exam.
欢迎参加本次 AQA GCSE 数学专项案例分析实战演练。本文将带你演练十个真实场景，检验你运用数字、代数、比例、几何、统计和概率等数学概念的能力。每个案例均呈现问题，引导推理过程，并给出完整解题步骤。通过这些实例，你可以提升解题技巧，为考试增添信心。
1. Savings Plan: Compound Interest | 储蓄计划：复利
Mia deposits £2000 into a savings account that pays 2.4% compound interest per year. She plans to leave the money untouched for 5 years. Calculate the total amount in the account after 5 years, giving your answer to the nearest penny. Explain why the interest earned in the fifth year is greater than the interest earned in the first year.
米娅将 2000 英镑存入一个年利率 2.4% 的复利储蓄账户。她计划在 5 年内不动用这笔钱。请计算 5 年后账户中的总金额，结果精确到便士。并解释为什么第五年获得的利息比第一年多。
Compound interest formula: A = P(1 + r/100)n. Here P = 2000, r = 2.4, n = 5. So A = 2000 × (1.024)5. Using a calculator, (1.024)5 ;48.00. In year 5, interest is calculated on a larger balance (approximately £2198.83 at the start of year 5), so the interest is about £52.77. Compound interest earns 'interest on interest', making later years produce more growth.
复利公式：A = P(1 + r/100)n。此处 P = 2000，r = 2.4，n = 5。因此 A = 2000 × (1.024)5。用计算器可得 (1.024)5 ≈ 1.1259，A ≈ 2251.80 英镑。第一年的利息为 2000 × 0.024 = 48.00 英镑。第五年时，利息基于更高的本金（第五年初约 2198.83 英镑）计算，所以利息约为 52.77 英镑。复利产生“利滚利”效应，使后期利息更高。
2. Home Improvement: Area and Cost | 家居装修：面积与成本
Sam wants to cover a rectangular wall measuring 3.6 m by 2.8 m with wooden panels. Each panel covers an area of 0.56 m2 and costs £12.40. Panels are sold in whole numbers. He also needs adhesive at a cost of £2.80 per panel used. Calculate the total area of the wall, the minimum number of panels required, and the total cost including adhesive.
萨姆想用木板覆盖一面长 3.6 米、宽 2.8 米的矩形墙壁。每块木板覆盖面积 0.56 平方米，单价 12.40 英镑，且必须整块购买。每块木板还需 2.80 英镑的粘合剂。请计算墙壁总面积、最少需要购买的木板数量，以及包含粘合剂在内的总费用。
Wall area = 3.6 × 2.8 = 10.08 m2 £223.20. Adhesive cost = 18 × 2.80 = £50.40. Total cost = £273.60.
墙壁面积 = 3.6 × 2.8 = 10.08 m2。所需木板数量 = 10.08 ÷ 0.56 = 18，恰好为整数。因此萨姆需购买 18 块木板。木板费用 = 18 × 12.40 = 223.20 英镑。粘合剂费用 = 18 × 2.80 = 50.40 英镑。总费用 = 273.60 英镑。
3. Recipe Scaling: Proportion | 食谱缩放：比例
A recipe for 6 people uses 250 g of flour, 120 g of sugar and 90 g of butter. Jess needs to make the same recipe for 15 people. Find the amount of each ingredient she needs. She has only 550 g of flour. Does she have enough? Explain.
一份供 6 人食用的食谱需用 250 克面粉、120 克糖和 90 克黄油。杰丝需要为 15 人制作同样食谱。请计算她所需每种食材的用量。她家里只有 550 克面粉，是否足够？请说明理由。
50 g of flour, which is less than 625 g, so she does not have enough; she needs 75 g more.
缩放比例 = 15 ÷ 6 = 2.5。所需面粉 = 250 × 2.5 = 625 克。糖 = 120 × 2.5 = 300 克。黄油 = 90 × 2.5 = 225 克。她只有 550 克面粉，少于 625 克，因此面粉不足，尚需 75 克。
4. Map Reading: Scale and Distance | 地图阅读：比例尺与距离
On a map with scale 1 : 25 000, the distance between two villages is 8.4 cm. Calculate the actual distance in kilometres. Toby cycles this route at an average speed of 14 km/h. How long does the journey take in minutes?
在一张比例尺为 1 : 25 000 的地图上，两个村庄之间的距离为 8.4 厘米。请计算实际距离（千米）。托比以 14 公里/小时的平均速度骑行该路线，旅程需要多少分钟？
ime = distance ÷ speed = 2.1 ÷ 14 = 0.15 hours. 0.15 × 60 = 9 minutes.
实际距离 = 8.4 厘米 × 25 000 = 210 000 厘米 = 2100 米 = 2.1 千米。时间 = 距离 ÷ 速度 = 2.1 ÷ 14 = 0.15 小时。0.15 × 60 = 9 分钟。
5. Business Profit: Linear Equations | 商业利润：线性方程
A small café has fixed monthly costs of £1200. Each cup of coffee sells for £2.80, and the cost to make each cup is £0.90. Write an expression for the monthly profit if n cups are sold. Hence, find the minimum number of cups that must be sold to make a profit of at least £2000 in a month.
一家小咖啡馆每月固定成本为 1200 英镑。每杯咖啡售价 2.80 英镑，每杯制作成本 0.90 英镑。写出每月售出 n 杯时利润的表达式。由此求出每月至少需售出多少杯咖啡，才能使利润不低于 2000 英镑。
lve 1.90n - 1200 ≥ 2000 → 1.90n ≥ 3200 → n ≥ 3200 ÷ 1.90 ≈ 1684.21. Therefore, at least 1685 cups are needed.
每杯利润 = 2.80 - 0.90 = 1.90 英镑。月利润 = 1.90n - 1200。解不等式 n - 1200 ≥ 2000 → 1.90n ≥ 3200 → n ≥ 3200 ÷ 1.90 ≈ 1684.21。因此，至少需售出 1685 杯。
6. Sports Statistics: Mean and Range | 体育统计：均值与范围
The points scored by a basketball team in 8 matches are: 72, 85, 68, 91, 74, 88, 79, 83. Calculate the mean score and the range. In the ninth match, the team scores 102 points. Without calculating, explain whether the mean and range will increase, decrease or stay the same.
一支篮球队在 8 场比赛中的得分分别为：72、85、68、91、74、88、79、83。请计算平均得分和极差。在第九场比赛中，该队得到 102 分。不进行计算，解释平均分和极差将如何变化：增加、减少还是保持不变。
se the mean. The new maximum becomes 102, so the range increases from 23 to 102 - 68 = 34.
总和 = 72+85+68+91+74+88+79+83 = 640。均值 = 640 ÷ 8 = 80。极差 = 91 - 68 = 23。加入一个高于当前均值的 102 分会使均值上升。新的最大值变为 102，因此极差由 23 增大为 102 - 68 = 34。
7. Probability in Games: Expected Wins | 游戏概率：期望获胜次数
At a fair, a spinner has 3 red, 2 blue and 5 green sections, all equal in size. A player wins a prize if the spinner lands on red. The entry fee is £1.50, and a prize is worth £2.00. If 480 people play the game, how many prizes would you expect to be won? Calculate the expected profit for the organiser from these 480 players.
在游乐会上，一个转盘被均分为 3 个红色区域、2 个蓝色区域和 5 个绿色区域。转到红色即获奖。入场费为 1.50 英镑，奖品价值 2.00 英镑。若有 480 人参与游戏，预期有多少人会获奖？请计算组织者在这 480 人中的预期利润。
otal prize cost = 144 × 2.00 = £288. Expected profit = 720 - 288 = £432.
P(红色) = 3/10。预期获奖人数 = 480 × 3/10 = 144。总入场费 = 480 × 1.50 = 720 英镑。总奖品成本 = 144 × 2.00 = 288 英镑。预期利润 = 720 - 288 = 432 英镑。
8. Travel Graph: Speed and Time | 旅行图：速度与时间
The distance-time graph of a car journey shows: from 0 to 2 hours, distance increases from 0 to 120 km; then a stop for 30 minutes; then from 2.5 hours to 4 hours, distance increases from 120 km to 220 km. Calculate the average speed for the whole journey, and the speed during the first 2 hours. During which part was the speed greatest?
一辆汽车的行程距离-时间图显示：0 至 2 小时，距离从 0 增至 120 公里；接着停留 30 分钟；然后从 2.5 小时到 4 小时，距离由 120 公里增至 220 公里。请计算全程平均速度，以及前 2 小时的速度。哪一阶段速度最快？
ent = (220 - 120) ÷ (4 - 2.5) = 100 ÷ 1.5 ≈ 66.7 km/h. The greatest speed was during the final 1.5 hours.
总距离 = 220 公里，总时间 = 4 小时。平均速度 = 220 ÷ 4 = 55 公里/小时。前 2 小时的速度 = 120 ÷ 2 = 60 公里/小时。最后阶段速度 = (220 - 120) ÷ (4 - 2.5) = 100 ÷ 1.5 ≈ 66.7 公里/小时。速度最快的是最后的 1.5 小时阶段。
9. Mobile Phone Tariffs: Simultaneous Equations | 手机套餐：联立方程
Two pay-as-you-go tariffs are offered: Tariff A charges £10 per month plus 5p per minute of calls. Tariff B charges £15 per month plus 3p per minute. For how many minutes of calls per month would the total monthly cost of both tariffs be equal? What is that equal cost?
某运营商提供两种预付费套餐：套餐 A 月租 10 英镑，通话每分钟 5 便士；套餐 B 月租 15 英镑，通话每分钟 3 便士。每月通话多少分钟时，两种套餐的月总费用相同？该相同费用是多少？
00 + 3m → 2m = 500 → m = 250 minutes. Equal cost = 1000 + 5×250 = 2250p = £22.50.
设通话分钟数为 m。费用 A（便士）= 1000 + 5m，或 10 + 0.05m 英镑。费用 B = 1500 + 3m。令两者相等：1000 + 5m = 1500 + 3m → 2m = 500 → m = 250 分钟。相等费用 = 1000 + 5×250 = 2250 便士 = 22.50 英镑。
10. Population Sampling: Data Collection | 人口抽样：数据收集
A school has 1200 students: 720 in Key Stage 3, 360 in Key Stage 4, and 120 in Key Stage 5. The student council wants to survey a stratified sample of 150 students for opinions on a new uniform. Calculate how many students from each key stage should be included in the sample. Explain one advantage of stratified sampling over simple random sampling in this context.
一所学校有 1200 名学生：关键阶段 3（KS3）720 人，关键阶段 4（KS4）360 人，关键阶段 5（KS5）120 人。学生会希望采用分层抽样调查 150 名学生对新校服的看法。请计算各关键阶段应抽取的学生人数。说明在此情境下，分层抽样相比简单随机抽样的一个优势。
Stratified sampling ensures proportional representation of each group, so opinions from all key stages are fairly reflected, reducing bias compared to simple random sampling which might accidentally miss a group.
抽样比例 = 150/1200 = 1/8。KS3：720 × 1/8 = 90。KS4：360 × 1/8 = 45。KS5：120 × 1/8 = 15。分层抽样确保各小组按比例入选，因此所有关键阶段的意见都能被公平反映，相较于简单随机抽样可能偶然遗漏某个小组，减少了偏差。
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