📚 Year 10 WJEC Maths: Case Study Practice & Problem Solving | WJEC十年级数学：案例分析实战演练
WJEC GCSE Mathematics in Year 10 often presents real problems that go beyond straightforward exercises. These case study questions test your ability to read a scenario, pick out the relevant numbers, choose the right mathematical tools, and then communicate your reasoning clearly. This article walks you through several practical examples covering geometry, money, travel, statistics and probability, exactly as you would meet in a WJEC paper. Each section will build your confidence to tackle the kind of structured problem solving that can appear in both non‑calculator and calculator papers.
WJEC Year 10 数学考试中，你经常会遇到一些需要结合实际情境来回答的问题，而不仅仅是简单的计算题。这些案例分析题考查你阅读场景、提取关键数字、选择正确数学工具以及清晰表达解题思路的能力。本文通过几何、财务、行程、统计和概率等多个实例，带你一步步练习 WJEC 试卷中常见的结构化问题，帮助你从容应对这类挑战。
1. The Approach to Math Case Studies | 数学案例分析方法
When you face a wordy case study, do not panic. Start by reading the whole text once to get the big picture, then read again with a highlighter to mark all quantities, units and the final question. Identify the mathematical topic – is it about percentages, area, graphs or probability? Draw a quick diagram or make a list of what you know and what you need to find. This planning step often reveals the correct method before you start any number crunching.
当遇到文字很多的案例分析题时，不要慌张。先通读一遍了解大意，再用高亮笔标出所有数字、单位和题目最终要求。判断这属于哪个数学主题——百分比、面积、图表还是概率？随手画个简图或列出已知量和待求量，这一步规划往往能在你动笔计算前就指明正确方向。
Always show your working step by step, even when you use a calculator. WJEC examiners award method marks for a correct process, so a clear layout can save your grade if you slip on the final arithmetic. Remember to write the units with every measurement and to check whether your final answer makes sense in the context – for example, a garden fence that costs £13.78 is probably correct; one that costs £1378 might mean you used centimetres instead of metres.
每一步都要写出计算过程，即使用了计算器也不能省略。WJEC 阅卷老师会按正确的解题思路给方法分，清晰的步骤可以在最后计算失误时帮你保住分数。同时记得每个测量值都要带单位，并检查最终答案是否合理——比如，花园围栏花费13.78英镑可能是对的，但1378英镑就可能意味着你把单位从米错用成了厘米。
2. Case 1: Designing a Rectangular Garden | 案例一：设计矩形花园
A homeowner wants to lay new turf in a rectangular garden measuring 12 m by 8 m. Turf costs £4.50 per square metre. He also plans to put a single row of brick edging along all four sides. Each brick is 0.5 m long and costs £1.20. The garden gate is 1 m wide and will not have bricks along that section. Calculate the total cost of turf and bricks.
一位房主想在长12米、宽8米的矩形花园里铺设新草皮。草皮每平方米售价4.50英镑。他还打算沿四边砌一圈砖做边饰，每块砖长0.5米，单价1.20英镑。花园门宽1米，门下不铺砖。请计算草皮和砖的总费用。
Area of garden = length × width = 12 m × 8 m = 96 m². Cost of turf = 96 × £4.50 = £432.
花园面积 = 长 × 宽 = 12 × 8 = 96 平方米。草皮费用 = 96 × 4.50 = 432 英镑。
Perimeter of the garden = 2 × (12 + 8) = 40 m. The gate opening takes away 1 m, so length to edge = 40 – 1 = 39 m. Number of bricks needed = 39 ÷ 0.5 = 78 bricks. Cost of bricks = 78 × £1.20 = £93.60.
花园周长 = 2 × (12 + 8) = 40 米。减去门宽1米后需要砌边饰的长度为 40 – 1 = 39 米。所需砖块数 = 39 ÷ 0.5 = 78 块。砖的费用 = 78 × 1.20 = 93.60 英镑。
Total cost = £432 + £93.60 = £525.60. This answer is sensible because each component is priced in the same unit (£) and the quantities are realistic.
总费用 = 432 + 93.60 = 525.60 英镑。这个结果是合理的，因为所有组成部分都以英镑计价，且用量真实可信。
3. Case 2: Choosing a Mobile Phone Plan | 案例二：选择手机套餐
A mobile company offers two plans. Plan A has a monthly fee of £10 and charges £0.10 per minute for calls. Plan B has a monthly fee of £15 and includes 200 free minutes; after that, each additional minute costs £0.08. Sam expects to use t minutes per month. Write a cost expression for each plan and find the number of minutes where the two plans cost the same.
一家移动运营商提供两种套餐。套餐A月租10英镑，通话每分钟0.10英镑。套餐B月租15英镑，包含200分钟免费通话，超出部分每分钟0.08英镑。Sam 预计每月通话 t 分钟。写出每种套餐的费用表达式，并求两种套餐费用相同时的通话分钟数。
Plan A cost = 10 + 0.10t. Plan B cost: if t ≤ 200, cost = 15; if t > 200, cost = 15 + 0.08(t – 200). We suspect the break-even point will be above 200 minutes, so we set 10 + 0.10t = 15 + 0.08(t – 200).
套餐A费用 = 10 + 0.10t。套餐B费用：当 t ≤ 200 时，费用 = 15；当 t > 200 时，费用 = 15 + 0.08(t – 200)。我们估计平衡点会超过200分钟，因此设方程：10 + 0.10t = 15 + 0.08(t – 200)。
Simplify the right side: 15 + 0.08t – 16 = 0.08t – 1. Now equate: 10 + 0.10t = 0.08t – 1 → 0.10t – 0.08t = –1 – 10 → 0.02t = –11. This gives t = –550, which is impossible. Re-examine: the right side should be 15 + 0.08(t – 200) = 15 + 0.08t – 16 = 0.08t – 1, correct. The mistake is in the sign: 10 + 0.10t = 0.08t – 1 → 0.02t = –11 is indeed negative, meaning Plan A is never cheaper above 200 minutes? Let us check values: at t=300, A=10+30=40; B=15+8=23. B is cheaper. At t=100, A=20, B=15, B cheaper. So Plan B is always cheaper for any t, because even with zero minutes B is £15 vs A £10 + 0? A is £10 at t=0, so B is more expensive there. Let us correct: at t=0 A=£10, B=£15. So A cheaper initially. The breaker must be below 200 minutes. For t ≤ 200, B cost = 15. Equation: 10 + 0.10t = 15 → 0.10t = 5 → t = 50 minutes. At t=50, A = 10 + 5 = £15, B = £15. For t < 50, A is cheaper; for 50 < t ≤ 200, B is cheaper. Above 200 minutes B is even cheaper. So the break-even is 50 minutes. The original set‑up with t > 200 was unnecessary.
化简右边：15 + 0.08t – 16 = 0.08t – 1。两边相等：10 + 0.10t = 0.08t – 1，移项得 0.02t = –11，t = –550，不可能。检查发现在 t ≤ 200 时 B 费用固定为15，设 10 + 0.10t = 15 即可得 t = 50 分钟。所以当通话时间为50分钟时两套餐费用相等；少于50分钟时A更划算，超过50分钟时B更划算。
A quick table helps visualize:
| t (min) | Plan A (£) | Plan B (£) |
|---|---|---|
| 0 | 10.00 | 15.00 |
| 50 | 15.00 | 15.00 |
| 100 | 20.00 | 15.00 |
| 250 | 35.00 | 19.00 |
This case study shows why you must always check which part of a piecewise function is relevant before solving an equation.
这个案例说明，在解方程之前一定要先判断分段函数的哪个部分适用于当前情景。
4. Case 3: Savings with Compound Interest | 案例三：复利储蓄
Megan invests £800 in a savings account that pays 2.5% compound interest per annum. How much will she have after 3 years? Give your answer to the nearest penny. Then compare with an account paying 3.1% simple interest over the same period.
Megan 将 800 英镑存入一个年复利为2.5%的储蓄账户。3年后她将有多少钱？答案精确到便士。然后与同样期限、年单利3.1%的账户进行比较。
Amount = P × (1 + r/100)ⁿ
复利总额 = 本金 × (1 + 年利率/100)ⁿ
Compound: 800 × (1 + 2.5/100)³ = 800 × (1.025)³ = 800 × 1.076890625 ≈ £861.51 (to nearest penny).
复利计算：800 × (1.025)³ = 800 × 1.076890625 ≈ 861.51 英镑（精确到便士）。
Simple interest: Interest = P × r × t = 800 × 3.1% × 3 = 800 × 0.031 × 3 = 800 × 0.093 = £74.40. Total = £874.40. The simple interest account gives more in this instance because the rate is higher, but over longer periods compound interest usually wins if rates are similar.
单利计算：利息 = 800 × 3.1% × 3 = 800 × 0.093 = 74.40 英镑，总额 874.40 英镑。此时单利账户收益更高，因为利率更高；但在利率相近时，长期来看复利通常胜出。
This case reminds you to read the interest type carefully – WJEC questions often mix simple and compound interest to see if you spot the difference.
这个案例提醒你仔细阅读利息类型——WJEC 常将单利和复利混合出题，看你能否分辨两者的区别。
5. Case 4: Journey Planning with Speed and Time | 案例四：旅行计划与速度时间
A car travels 150 km from town X to town Y. The first 60 km are on country roads at an average speed of 40 km/h. The remaining distance is on a motorway. If the total journey must not exceed 2.5 hours, find the minimum average speed required on the motorway.
一辆汽车从 X 镇驶往 Y 镇，全程 150 公里。前 60 公里在乡间道路上行驶，平均速度为 40 公里/小时。剩余路程在高速公路上。如果总行程不得超过 2.5 小时，求在高速公路上需要的最低平均速度。
Time on country roads = distance ÷ speed = 60 ÷ 40 = 1.5 hours. Remaining time for motorway = 2.5 – 1.5 = 1.0 hour. Remaining distance = 150 – 60 = 90 km. Minimum motorway speed = 90 km ÷ 1 h = 90 km/h.
乡间道路用时 = 60 ÷ 40 = 1.5 小时。高速公路可用时间 = 2.5 – 1.5 = 1.0 小时。剩余距离 = 150 – 60 = 90 公里。高速公路最低平均速度 = 90 公里/1 小时 = 90 公里/小时。
Be careful with unit conversions – if parts of a question give distances in metres or times in minutes, change them to km and hours (or keep consistent). Also, always state your final speed with its unit, km/h.
注意单位换算——如果题目里部分距离给的是米或时间给的是分钟，要统一转换为公里或小时。最后作答时一定要注明速度单位 km/h。
6. Case 5: School Canteen Survey Analysis | 案例五：学校食堂调查分析
A school asked 200 students to name their favourite fruit. The results were: Apple 45, Banana 35, Orange 55, Grapes 40, Strawberry 25. Draw a frequency table with a percentage column. Which fruit is the mode? Explain whether a pie chart or a bar chart would be more suitable to display this data.
一所学校调查了 200 名学生最喜欢哪种水果。结果为：苹果45人，香蕉35人，橘子55人，葡萄40人，草莓25人。请绘制包含百分比列的频数表，指出众数是哪种水果，并说明用饼图还是条形图展示这组数据更合适。
| Fruit | Frequency | Percentage |
|---|---|---|
| Apple | 45 | 22.5% |
| Banana | 35 | 17.5% |
| Orange | 55 | 27.5% |
| Grapes | 40 | 20% |
| Strawberry | 25 | 12.5% |
The mode is Orange because it has the highest frequency (55). A pie chart is suitable because the categories add up to a whole (200 students), and it clearly shows the share of each fruit. A bar chart would also work, but pie charts are especially useful when showing proportions of a total.
众数是橘子，因为它的频数最高（55）。饼图很适合这组数据，因为各类别合计构成一个整体（200名学生），它能直观显示每种水果的占比。条形图也可用，但饼图在展示整体中各部分比例时更直观。
When calculating percentages, work out (frequency ÷ total) × 100. Round to one decimal place if necessary and double‑check that the percentages sum close to 100% (here 22.5+17.5+27.5+20+12.5 = 100). Small rounding errors are acceptable.
计算百分比时，用（频数÷总数）×100，可保留一位小数，最后验证百分比之和是否接近100%（本题各项相加正好100%）。允许存在微小的舍入误差。
7. Case 6: Probability in Games | 案例六：游戏中的概率
Two fair six‑sided dice are rolled. List all the outcomes where the sum of the two numbers equals 7. Hence find the probability that the sum is 7. In a game, a player wins £5 if the sum is 7 and loses £1 otherwise. Calculate the expected gain per game.
同时掷两枚公平的六面骰子。列出所有点数之和为7的结果，并据此求点数和为7的概率。在一个游戏中，玩家若掷出和为7则赢5英镑，否则输1英镑。计算每局游戏的期望收益。
Favourable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes. Total possible outcomes = 6 × 6 = 36. Probability(sum=7) = 6/36 = 1/6. Probability(not 7) = 5/6.
有利结果：(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) 共6种。总结果数 = 6 × 6 = 36。概率(和为7) = 6/36 = 1/6。非7的概率 = 5/6。
Expected gain = (1/6) × 5 + (5/6) × (–1) = 5/6 – 5/6 = 0
期望收益 = (1/6) × 5 + (5/6) × (–1) = 5/6 – 5/6 = 0
An expected gain of £0 means the game is fair in the long run – neither the player nor the organiser can expect to profit. WJEC questions often ask you to comment on whether a game is fair; always support your answer with the expected value calculation.
期望收益为0意味着长期来看游戏是公平的——玩家和游戏组织者均不期望获利。WJEC 常要求你评价游戏是否公平，作答时一定要用期望值计算作为依据。
8. Interpreting Diagrams and Scale Drawings | 案例七：图表与比例图的解读
A scale drawing shows a playground with a scale of 1 : 200. On the drawing, the length is 7.5 cm and the width is 4.2 cm. Find the real dimensions in metres. Then calculate the area of the playground. The council wants to cover the surface with rubber tiles costing £22 per square metre. What is the total cost?
一幅比例图展示了一个操场，比例为 1 : 200。图上长度为 7.5 cm，宽度为 4.2 cm。求实际尺寸（单位：米）。然后计算操场面积。市政部门想用每平方米22英镑的橡胶地砖铺设地面，问总费用是多少？
Real length = 7.5 cm × 200 = 1500 cm = 15.00 m. Real width = 4.2 cm × 200 = 840 cm = 8.40 m. Real area = 15 × 8.4 = 126 m². Total cost = 126 × £22 = £2772.
实际长度 = 7.5 × 200 = 1500 cm = 15.00 m。实际宽度 = 4.2 × 200 = 840 cm = 8.40 m。实际面积 = 15 × 8.4 = 126 m²。总费用 = 126 × 22 = 2772 英镑。
When working with scales, always convert the scaled measurement into the real measurement in the same units first, then change to the required unit – here cm to m. Writing the conversion factor (1 cm = 200 cm = 2 m) can help speed up the process.
处理比例时，始终先将图上尺寸换算为实际尺寸（单位统一），再转换为题目要求的单位——这里由厘米转为米。记住 1 cm 代表 200 cm 即 2 m，能帮助你快速心算。
9. Combining Percentages with Real Data | 案例八：百分比与实际数据结合
In a sale, a jacket that originally cost £85 is reduced by 20%. The store then offers a further 10% off the reduced price for loyalty card holders. Calculate the final price a loyalty card holder pays. Show that a single reduction of 28% would give the same final price.
一件夹克原价85英镑，促销时降价20%。商场再对会员卡用户提供折上折，再减10%。计算会员最终支付的价格，并证明一次性的28%减价能得到同样的最终价格。
After 20% off: price = 85 × (1 – 20/100) = 85 × 0.80 = £68. After extra 10% off: final price = 68 × 0.90 = £61.20. Single 28% reduction: 85 × (1 – 28/100) = 85 × 0.72 = £61.20. Indeed they give the same result.
降价20%后：85 × 0.80 = 68 英镑。再减10%：68 × 0.90 = 61.20 英镑。一次性减28%：85 × 0.72 = 61.20 英镑，结果相同。
This example highlights that successive percentage decreases do not add directly: a 20% then 10% reduction is equivalent to a multiplier of 0.80 × 0.90 = 0.72, i.e. a 28% overall decrease. Always multiply the decimal multipliers rather than adding percentages.
这个例子说明连续的百分数减少不能直接相加：减20%再减10%相当于乘以 0.80 × 0.90 = 0.72，即总共减少28%。处理复合百分数时应始终将小数相乘，而非将百分数简单相加。
10. Key Skills and Common Pitfalls | 关键技能与常见错误
Case studies demand that you draw together skills across different topics. Make sure you are confident with unit conversions (e.g. cm² to m², minutes to hours), reading scales, and using formulas such as speed = distance/time. A very common mistake is to forget to use the same units throughout, leading to answers that are 100 times too large or too small. Before you finalise an answer, re‑read the question to check the unit required.
案例分析题要求你综合运用不同主题的技能。务必熟练掌握单位换算（如平方厘米到平方米、分钟到小时）、读数刻度以及使用公式（如速度 = 距离/时间）。最常见的一个错误就是前后单位不一致，导致答案偏大或偏小100倍。落笔前务必回看题目所要求的单位。
Another pitfall is misreading a graph axis or missing a key in a statistical chart. Take a few seconds to label axes or write what each bar represents. In probability, double‑check that your probabilities sum to 1 and that you have accounted for all possible outcomes. Structured working is the best defence against careless errors.
另一个易错点是误读图表坐标轴或忽略统计图中的图例。花几秒钟标出轴的含义或者写下每个条形代表什么。在概率题中，要检查概率之和是否为1，是否已考虑所有可能的结果。有条理的书写步骤是预防粗心错误的最佳手段。
11. Exam Tips for Case Study Questions | 案例分析题的考试技巧
Start by skimming the whole question – how many marks is it worth? If it is a high‑mark question, expect several sub‑calculations. Jot down the given information in a structured way. Show every small step; even a simple multiplication should be written. In WJEC exams, a correct method with a small arithmetic slip can still earn most of the marks. Finally, present your answer in a sentence where appropriate, for example “The total cost is £525.60”, as this often satisfies the communication mark.
先快速浏览整道题——它占多少分？如果是高分题，往往包含多个计算步骤。有条理地记下已知信息，写出每一步，即使简单的乘法也不要跳过。在 WJEC 考试中，方法正确但计算有小误仍可获得大部分分数。最后，适当用完整句子表述答案，例如“总费用为525.60英镑”，这样通常能拿到表达分。
Time management is crucial. A complex case study might appear towards the end of the paper; leave enough time by practising similar exercises at home. Use the mark allocation as a guide: a 6‑mark case study might need about 8‑10 minutes. If you get stuck, write what you know and move on – you can return later.
时间管理很关键。复杂的案例分析题可能会出现在试卷末尾；在家多练习同类题目，确保留出充足时间。以分值为指引：一道6分的案例分析题大约需要8–10分钟。如果卡住了，
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