📚 Year 11 WJEC Maths: Case Study Practical Workout | WJEC 11年级数学：案例分析实战演练
Case studies in WJEC Mathematics bring together multiple topics under one realistic scenario. This article walks you through four typical case studies covering percentages, geometry, travel, and data handling. You will learn how to break down complex problems, apply the right formulae, and present solutions clearly – exactly as required in your GCSE exam.
WJEC 数学中的案例分析将多个知识点融合在一个真实场景中。本文将带你演练四个典型案例，涵盖百分比、几何、旅行以及数据处理。你将学会如何拆解复杂问题、应用正确公式，并以考试要求的方式清晰地呈现解答过程。
1. Understanding Case Study Questions | 理解案例分析题
Case study questions in WJEC exams often describe a real-life situation, such as running a small business, planning a journey, or refurbishing a room. They expect you to extract numerical information, identify the mathematics required, and carry out multi-step calculations. Marks are awarded for method as well as the final answer, so always show your working step by step.
WJEC 考试中的案例分析题通常描述一个现实生活情境，比如经营小生意、规划旅程或装修房间。题目要求你提取数字信息，识别所需的数学知识，并进行多步计算。评分既看最终答案也看解题过程，因此务必逐步展示你的计算步骤。
Key skills assessed include percentages, ratio, algebra, Pythagoras’ theorem, trigonometry, speed-distance-time, interpretation of charts, and measures of central tendency. You may also need to convert units or work with compound measures like density or pressure. The best approach is to read the whole case study first, highlight key numbers and units, then tackle each question in the order it appears.
考查的核心技能包括百分比、比率、代数、毕达哥拉斯定理、三角学、速度-距离-时间关系、图表解读以及集中趋势的度量。你可能还需要进行单位换算或处理密度、压强等复合量。最佳做法是首先通读整个案例，圈出关键数字和单位，然后按顺序逐一解答每个小问。
2. Case Study 1: Monthly Budget Planning | 案例1：月度预算规划
Anna earns a net monthly salary of £1850. She plans to save 12% of her income each month. Her fixed monthly expenses are rent £650, council tax £120, and utilities £85. She also spends £240 on groceries and allows £150 for leisure. Any remaining money goes into an emergency fund. Calculate the amount she saves each month, the total fixed expenses, and the amount left for the emergency fund.
安娜每月净收入为 1850 英镑。她计划每月储蓄收入的 12%。她的固定月支出为房租 650 英镑、市政税 120 英镑和水电杂费 85 英镑。她还在食品上花费 240 英镑，并预留 150 英镑用于休闲。剩余的钱全部存入应急基金。请计算她每月的储蓄金额、总固定支出以及留给应急基金的金额。
First, find 12% of £1850: 0.12 × 1850 = £222. This is the monthly saving. Next, total fixed expenses: rent (£650) + council tax (£120) + utilities (£85) = £855. Then add groceries £240 and leisure £150, giving total outgoings of £855 + £240 + £150 = £1245. The emergency fund contribution is net income minus savings minus total outgoings: £1850 − £222 − £1245 = £383.
首先，计算 1850 英镑的 12%：0.12 × 1850 = 222 英镑。这是每月储蓄额。接着，总固定开支：房租（650）+ 市政税（120）+ 水电费（85）= 855 英镑。再加上食品 240 英镑和休闲 150 英镑，总支出为 855 + 240 + 150 = 1245 英镑。应急基金缴款为净收入减去储蓄再减去总支出：1850 − 222 − 1245 = 383 英镑。
Now suppose Anna expects a 3% pay rise next year. Her new net monthly salary becomes £1850 × 1.03 = £1905.50. If she keeps the same percentage saving and all expenses rise by 5% tes how small percentage changes affect disposable income, a typical WJEC style extension.
假设安娜明年预计加薪 3%，她新的净月收入变为 1850 × 1.03 = 1905.50 英镑。如果她保持相同的储蓄比例，而所有支出上涨 5%，你可以重新计算她的应急基金。这展示了微小的百分比变化如何影响可支配收入，是典型的 WJEC 风格延伸题。
3. Break-Even Analysis | 盈亏平衡分析
A local bakery sells cakes for £12 each. The cost to make one cake is £5. The monthly fixed costs for the shop are £1400. Let x be the number of cakes sold in a month. Write an expression for total revenue and total cost, and find the break-even point.
一家本地蛋糕店每块蛋糕售价 12 英镑。制作一块蛋糕的成本为 5 英镑。店铺每月固定成本为 1400 英镑。设 x 为每月销售的蛋糕数量。写出总收入与总成本的表达式，并求出盈亏平衡点。
akery must sell 200 cakes each month to cover all costs. If they sell 250 cakes, profit is 12(250) − [5(250) + 1400] = 3000 − 2650 = £350.
收入 R = 12x。总成本 C = 5x + 1400。在盈亏平衡点，收入等于成本：12x = 5x + 1400。两边减去 5x：7x = 1400，得 x = 200 块蛋糕。该店每月必须售出 200 块蛋糕才能覆盖所有成本。若售出 250 块，利润为 12(250) − [5(250) + 1400] = 3000 − 2650 = 350 英镑。
This case study blends algebra with percentages if we later include VAT or discount offers. For example, if a 10% service charge is added to the selling price, the revenue per cake becomes £13.20, and the break-even point shifts. Always define your variable clearly and interpret the result in the context of the problem.
如果后续引入增值税或折扣优惠，这个案例就会将代数与百分比结合起来。例如，若售价加收 10% 服务费，每块蛋糕收入变为 13.20 英镑，盈亏平衡点也会随之改变。务必清晰定义变量，并在问题情境中解释结果。
4. Case Study 2: Home Improvement – Paint and Flooring | 案例2：家居装修——油漆与地板
Tom wants to paint the four walls of a rectangular room measuring 4.5 m by 3.2 m, with a height of 2.4 m. The room has a door of area 2.0 m² and a window of area 1.5 m². One litre of paint covers 12 m². Paint is sold in 2.5-litre tins costing £18 each. Calculate the total wall area to be painted, the number of tins required, and the total cost.
汤姆想给一个长 4.5 米、宽 3.2 米、高 2.4 米的长方形房间的四壁刷漆。房间有一扇面积为 2.0 平方米的门和一扇面积为 1.5 平方米的窗户。一升油漆可刷 12 平方米。油漆以 2.5 升一桶出售，每桶 18 英镑。计算需要刷漆的总墙面面积、所需桶数以及总费用。
.5 = 33.46 m². Paint needed = 33.46 ÷ 12 = 2.7883 litres. Since tins are 2.5 litres each, two tins give 5 litres, which is sufficient. Number of tins = 2. Total cost = 2 × £18 = £36.
房间周长 = 2 × (4.5 + 3.2) = 2 × 7.7 = 15.4 米。包括门窗洞口的墙面总面积 = 15.4 × 2.4 = 36.96 平方米。减去门和窗：36.96 − 2.0 − 1.5 = 33.46 平方米。所需油漆量 = 33.46 ÷ 12 = 2.7883 升。由于每桶 2.5 升，两桶共 5 升，足够使用。桶数 = 2。总费用 = 2 × 18 = 36 英镑。
m. Floor area = 4.5 × 3.2 = 14.4 m². He must buy packs that cover 2.1 m² per pack, costing £ quired: 14.4 ÷ 2.1 ≈ 6.857, so he must purchase 7 packs. Total flooring cost = 7 × 23.50 = £164.50. This task involves rounding up, a real-life application of integer constraints.
接着，汤姆计划在同一房间铺强化木地板。地板面积 = 4.5 × 3.2 = 14.4 平方米。他必须购买每包覆盖 2.1 平方米的地板，每包价格 23.50 英镑。所需包数：14.4 ÷ 2.1 ≈ 6.857，因此必须购买 7 包。地板总费用 = 7 × 23.50 = 164.50 英镑。这个任务涉及向上取整，是整数约束的现实应用。
5. Geometry of a Ramp | 坡道的几何
A wheelchair ramp is to be built against a step of height 0.45 m. Safety regulations state the ramp must have a maximum slope of 1:12, meaning for every 1 m of vertical rise there must be at least 12 m of horizontal run. Using Pythagoras’ theorem, calculate the minimum ramp length and check if the angle of incline is within the 5° guideline.
需要为一段 0.45 米高的台阶修建轮椅坡道。安全规章规定坡道最大坡度为 1:12，即垂直高度每升高 1 米，水平距离至少需 12 米。利用毕达哥拉斯定理，计算最短坡道长度，并检验倾斜角是否在 5° 的建议范围内。
The ramp is the hypotenuse of a right triangle with legs 0.45 m and 5.4 m. Ramp length L = √0.45² + 5.4² = √0.2025 + 29.16 = √29 0.08333. θ = tan⁻¹(0.08333) ≈ 4.76°, which is under 5°, so the design meets the guideline.
最短水平距离 = 12 × 0.45 = 5.4 米。坡道是直角三角形斜边，两直角边分别为 0.45 米和 5.4 米。坡道长度 L = √0.45² + 5.4² = √0.2025 + 29.16 = √29.3625 ≈ 5.42 米。求角度，tan θ = 对边/邻边 = 0.45 ÷ 5.4 = 0.08333。θ = tan⁻¹(0.08333) ≈ 4.76°，低于 5°，因此设计符合指引。
If the ramp were built with a 1:10 slope instead, the horizontal run would be 4.5 m, the length √0.45² + 4.5² .71°. This would not meet the 5° recommendation. WJEC questions often ask you to decide between options by calculating both and comparing.
如果改用 1:10 的坡度修建，水平距离将为 4.5 米，坡道长度约为 √0.45² + 4.5² ≈ 4.52 米，角度约 5.71°，这就不符合 5° 的建议。WJEC 题目经常要求你通过计算两种方案并进行比较来做出选择。
6. Case Study 3: Travel and Timetables | 案例3：旅行与时间表
Megan catches a train from Cardiff to London. The timetable shows departure at 07:52 and arrival at 10:34. She walks 12 minutes to the station and needs to arrive 5 minutes before departure. What time must she leave home? Also, calculate the journey time on the train and the average speed if the distance is 234 km.
梅根从卡迪夫乘火车去伦敦。时刻表显示出发时间为 07:52，到达时间为 10:34。她需要步行 12 分钟到车站，并提前 5 分钟到达。她最迟必须什么时间离开家？同时计算火车行程时间和平均速度，已知距离为 234 公里。
0:34 is 2 hours and 42 minutes. In hours: 42 min = 0.7 h, so time = 2.7 h. Average speed = distance ÷ time = 234 ÷ 2.7 ≈ 86.7 km/h.
发车时间为 07:52，因此她必须在 07:47 前到达车站。步行需 12 分钟，故应于 07:47 − 12 分钟 = 07:35 离开家。在火车上，从 07:52 到 10:34 为 2 小时 42 分钟。换算为小时：42 分钟 = 0.7 小时，因此时间 = 2.7 小时。平均速度 = 距离 ÷ 时间 = 234 ÷ 2.7 ≈ 86.7 公里/小时。
l travel time, excluding the wait. The elapsed time is 2 hours 33 minutes (from 17:15 to 19:48). Subtract the 8-minute wait: 2 hours 25 minutes = 2.4167 h. The average speed then becomes 234 ÷ 2.4167 ≈ 96.8 km/h. Tabulate your results for clarity.
再考虑返程。她于 17:15 离开伦敦，19:48 抵达卡迪夫。列车在中途站等候 8 分钟。计算除去等候时间的实际行驶时间。总经过时间为 2 小时 33 分钟（从 17:15 到 19:48）。减去 8 分钟等候：2 小时 25 分钟 = 2.4167 小时。此时平均速度为 234 ÷ 2.4167 ≈ 96.8 公里/小时。用表格汇总结果会更清晰。
| Trip | Time | Distance | Avg Speed |
|---|---|---|---|
| Outward | 2.7 h | 234 km | 86.7 km/h |
| Return (excl. wait) | 2.4167 h | 234 km | 96.8 km/h |
7. Speed, Distance and Time with Unit Conversion | 速度、距离、时间与单位转换
m). Calculate the time to cover 300 km at this speed, giving your answer in hours and minutes.
许多旅行案例分析要求你转换单位。例如，一辆车以 50 英里每小时的速度行驶。将此转换为米每秒（1 英里 = 1609 米）。并计算以此速度行驶 300 公里所需的时间，答案以小时和分钟表示。
ce 1 mile = 1.609 km, 50 mph = 50 × 1.609 = 80.45 km/h. Time = distance ÷ speed = 300 ÷ 80.45 ≈ 3.729 hours. 0.729 h × 60 = 43.7 min, so about 3 hours 44 minutes.
首先，50 mph 转换为 m/s：50 × 1609 米每小时 = 80,450 米/小时。除以 3600 秒：80,450 ÷ 3600 ≈ 22.35 米/秒。再求时间：先将速度用 km/h 表示。因为 1 英里 = 1.609 公里，50 mph = 50 × 1.609 = 80.45 km/h。时间 = 距离 ÷ 速度 = 300 ÷ 80.45 ≈ 3.729 小时。0.729 小时 × 60 = 43.7 分钟，所以大约 3 小时 44 分钟。
Always check whether you need to convert lengths, masses, or volumes within the same case study. For example, fuel consumption might be given in miles per gallon, while distances are in kilometres. You will be expected to multiply by appropriate conversion factors like 4.546 litres per gallon or 1.609 km per mile. Practise two-step conversions to avoid errors.
务必检查是否需要在同一案例中转换长度、质量或体积单位。例如，燃油消耗可能以英里每加仑给出，而距离却是公里。你需要乘以适当的换算系数，如 1 加仑 = 4.546 升或 1 英里 = 1.609 公里。练习两步转换以避免出错。
8. Case Study 4: Data Analysis – Comparing Mobile Plans | 案例4：数据分析——手机套餐比较
A company offers two mobile phone plans. Plan A has a fixed monthly fee of £15 and charges 5p per minute of calls. Plan B has no fixed fee but charges 8p per minute. Write cost equations for each plan and find the number of minutes where both plans cost the same. Recommend which plan is cheaper for 400 minutes per month.
一家公司提供两种手机套餐。套餐 A 每月固定费用 15 英镑，通话每分钟 5 便士。套餐 B 无固定费用，但通话每分钟 8 便士。写出每种套餐的费用方程，并找出两种套餐费用相同时的通话分钟数。推荐每月通话 400 分钟时哪种套餐更便宜。
m = 0.08m => 15 = 0.03m => m = 500 minutes. At 400 minutes, A = 15 + 0.05(400) = 15 + 20 = £35. B = 0.08(400) = £32. So Plan B is cheaper by £ simple linear model can be extended by adding data allowances or international call rates.
设 m 为分钟数。套餐 A 费用（英镑）：A = 15 + 0.05m。套餐 B：B = 0.08m。令两者相等：15 + 0.05m = 0.08m，得 15 = 0.03m，m = 500 分钟。通话 400 分钟时，A = 15 + 0.05(400) = 15 + 20 = 35 英镑。B = 0.08(400) = 32 英镑。因此套餐 B 便宜 3 英镑。这个简单的线性模型可以通过增加数据流量或国际通话费率来扩展。
ean = (320+480+510+390+450+470) ÷ 6 = 2620 ÷ 6 ≈ 436.7 min. Ordered: 320, 390, 450, 470, 480, 510. Median = (450+470)/2 = 460 min. Range = 510 − 320 = 190 min. With these statistics, she can decide which plan suits her typical usage.
现在考虑实际使用数据。一名学生记录了她六个月的每月通话分钟数：320、480、510、390、450、470。求平均值、中位数和极差。平均值 = (320+480+510+390+450+470) ÷ 6 = 2620 ÷ 6 ≈ 436.7 分钟。排序后：320、390、450、470、480、510。中位数 = (450+470)/2 = 460 分钟。极差 = 510 − 320 = 190 分钟。有了这些统计数据，她可以决定哪种套餐适合她的典型使用量。
9. Statistical Measures and Interpretation | 统计测量与解读
WJEC often asks you to draw conclusions from calculated statistics. In the mobile plan case, the mean usage (437 min) is below the 500-minute crossover point, so Plan B is cheaper for the average month. However, if the median is 460, more than half the months have usage above 437, suggesting Plan A might also be considered if the trend rises. Here, the median also falls below 500, reinforcing Plan B.
WJEC 经常要求你根据计算的统计量得出结论。在手机套餐案例中，平均使用量（437 分钟）低于 500 分钟的临界点，因此一般月份中套餐 B 更便宜。然而，如果中位数为 460，超过半数的月份使用量都在 437 以上，说明若使用趋势上升，套餐 A 也值得考虑。但这里中位数同样低于 500，再次印证套餐 B 更优。
Sometimes case studies provide a chart, such as a cumulative frequency graph or a box plot. You should be able to estimate the median, quartiles, and interquartile range from the graph, and then compare two distributions. When comparing, write about a typical value (median) and the spread (IQR or range), making sure to relate your statements back to the context.
有时案例分析会提供图表，例如累积频率图或箱线图。你应当能根据图表估计中位数、四分位数和四分位距，然后比较两个分布。比较时，围绕典型值（中位数）和分散程度（四分位距或极差）进行论述，并确保将陈述与情境联系起来。
For instance, comparing test scores of two classes: "The median mark for Class A was 62, while for Class B it was 58, showing that Class A typically performed higher. The IQR for A was 14 and for B was 22, indicating marks in Class B were more spread out." Such interpretation is expected in WJEC Mathematics – Numeracy papers.
例如，比较两个班级的考试成绩：“A 班的中位数分数为 62，而 B 班为 58，表明 A 班总体表现更好。A 班的四分位距为 14，B 班为 22，说明 B 班分数分布更分散。” 在 WJEC 数学——计算能力试卷中，这类解读是必考内容。
10. Review and Exam Tips | 复习与考试技巧
When revising case studies, practise with past WJEC papers and mark schemes. Focus on showing every calculation step, labelling parts (a), (b), (c), and writing conclusions in full sentences. If a question asks "Which is better?", you must compare using numbers, not just give an opinion.
复习案例分析时，请用 WJEC 历年真题和评分方案进行练习。重点关注展示每个计算步骤，标注小题 (a)、(b)、(c)，并用完整句子撰写结论。如果题目问“哪个更好？”，你必须用数字进行比较，而不仅仅是给出看法。
Common pitfalls include forgetting to subtract areas of openings in paint problems, failing to round up when buying materials, mixing units in speed questions, or not converting pence to pounds in cost equations. Always scan your working for these errors.
常见陷阱包括油漆问题中忘记减去洞口面积，购买材料时未向上取整，速度问题中混淆单位，或在费用方程中没有将便士转换为英镑。务必检查解题过程，避免此类错误。
Finally, create your own mini case studies from everyday situations, such as planning a party budget, comparing fuel efficiency, or analysing monthly screen time. The more you connect mathematics to real life, the more confident you will be in the exam.
最后，你可以根据日常生活情景自创小型案例分析，比如规划派对预算、比较燃油效率或分析月度屏幕使用时间。你与生活联系得越紧密，考试时就会越自信。
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