📚 Year 9 WJEC Maths: Case Study Practice Drills | Year 9 WJEC 数学：案例分析实战演练
Real-world maths problems are the best way to see how the skills you learn in the classroom apply to everyday life. In this article, we walk through a series of case study drills designed for Year 9 WJEC students, covering percentages, geometry, algebra, statistics, ratio, graphs, probability and more. Each case is broken down step by step so you can understand the reasoning and gain confidence in tackling similar exam questions.
真实世界的数学问题是检验课堂所学技能如何应用于日常生活的最佳方式。本文将通过一系列专为 Year 9 WJEC 学生设计的案例演练，涵盖百分比、几何、代数、统计、比和比例、图像、概率等内容。每个案例都会逐步分解，让你理解推理过程，增强应对类似考试题的信心。
1. Comparing Mobile Phone Contracts | 比较手机合约
Sarah wants a new smartphone and is considering two 24‑month contracts for the same handset. Plan A charges £25 per month with zero upfront cost. Plan B charges £18 per month but requires a £120 upfront payment. Which plan is cheaper overall?
莎拉想买一部新智能手机，正在比较两款同款手机的 24 个月合约。套餐 A 每月 £25，无需预付费；套餐 B 每月 £18，但需要预付 £120。哪个套餐总费用更少？
To find the total cost, multiply the monthly fee by 24 and add any upfront charge.
计算总费用，将月费乘以 24 再加上任何预付费。
Plan A: 24 × £25 = £600
Plan B: 24 × £18 = £432; Total = £432 + £120 = £552
Plan B costs £552, which is £48 less than Plan A. Therefore, even with the upfront cost, Plan B is the better deal. This case uses multiplication, addition and comparison of money – a common percentage and finance skill.
套餐 B 总费用为 £552，比套餐 A 少 £48。因此，尽管有预付费用，套餐 B 更划算。该案例使用了乘法、加法和金额比较——这是常见的百分比和理财技能。
2. Designing a Rectangular Garden Path | 设计矩形花园小径
Ali wants to lay a gravel path around the edge of his rectangular lawn. The lawn measures 8 m by 6 m, and the path will be 1 m wide on all four sides. He needs to know the area of the path to order enough gravel. Gravel costs £4.50 per square metre.
阿里想在长方形草坪四周铺设一条碎石小径。草坪长 8 米，宽 6 米，小径四周均宽 1 米。他需要知道小径的面积以便订购足量的碎石。碎石每平方米 £4.50。
angle is 10 × 8 = 80 m². The lawn area is 8 × 6 = 48 m². So the path area = 80 − 48 = 32 m².
首先，计算草坪加小径的总尺寸。长度变为 8 + 2×1 = 10 米，宽度变为 6 + 2×1 = 8 米。大矩形面积为 10 × 8 = 80 平方米。草坪面积为 8 × 6 = 48 平方米。因此，小径面积 = 80 − 48 = 32 平方米。
Cost = 32 × £4.50 = £144. Ali will need £ to buy the gravel. This problem combines perimeter, area and multiplication with decimals.
费用 = 32 × £4.50 = £144。阿里需要 £144 购买碎石。这道题结合了周长、面积和小数乘法。
3. Solving a Savings Equation | 解储蓄方程
Emily has £120 in her savings account and plans to save £15 each month. She wants to buy a laptop that costs £420. How many months will it take to reach her goal? Write an equation and solve it.
艾米丽储蓄账户里有 £120，计划每月再存 £15。她想买一台 £420 的笔记本电脑。需要几个月才能达到目标？请写出方程并求解。
Let m be the number of months. The total savings after m months is 120 + 15m. Set this equal to 420.
设 m 为月数。m 个月后的总储蓄为 120 + 15m。令其等于 420。
120 + 15m = 420
15m = 300 → m = 20
eck: 120 + 15 × 20 = 420. This is a simple linear equation, a key algebra skill in Year 9.
艾米丽需要 20 个月才能存够钱。可以验证：120 + 15 × 20 = 420。这是一个简单的一元一次方程，属于 Year 9 代数核心技能。
4. Analysing Survey Results | 分析调查结果
A school surveyed 200 students about their favourite sport. The results are: Football 80, Netball 50, Tennis 30, Athletics 40. Display the data in a table and calculate each sport as a percentage of the total. Then suggest the most popular sport.
某学校对 200 名学生进行了最喜爱运动的调查。结果为：足球 80 人，无挡板篮球 50 人，网球 30 人，田径 40 人。用表格展示数据，计算每项运动占总数的百分比，并指出最受欢迎的运动。
We can organise the data in a table:
我们可以用表格整理数据：
| Sport | Frequency | Percentage |
|---|---|---|
| Football | 80 | 40% |
| Netball | 50 | 25% |
| Tennis | 30 | 15% |
| Athletics | 40 | 20% |
age is calculated by (frequency ÷ 200) × 100. For example, Football: (80 ÷ 200) × 100 = 40%. Football is the most popular with 40% of votes. This case practises frequency tables, percentages and interpreting data.
每个百分比的计算方法是 (频数 ÷ 200) × 100。例如，足球：(80 ÷ 200) × 100 = 40%。足球最受欢迎，占 40%。该案例练习了频数表、百分比和数据分析。
5. Scaling a Recipe | 按比例调整食谱
A recipe for 4 people requires 300 g of flour, 200 g of sugar, 100 g of butter and 2 eggs. Mariam needs to make the same recipe for 10 people. How much of each ingredient does she need? Give the quantities in grams and state the number of eggs.
一份 4 人份的食谱需要 300 克面粉、200 克糖、100 克黄油和 2 个鸡蛋。玛丽亚姆需要做 10 人份的相同食谱。每种原料需要多少？以克为单位给出数量，并说明鸡蛋个数。
250 g; Eggs: 2 × 2.5 = 5 eggs. This is a direct proportion problem, a common ratio skill.
比例因子为 10 ÷ 4 = 2.5。将每种原料乘以 2.5。面粉：300 × 2.5 = 750 克；糖：200 × 2.5 = 500 克；黄油：100 × 2.5 = 250 克；鸡蛋：2 × 2.5 = 5 个。这是一个直接比例问题，属于常见的比和比例技能。
6. Interpreting a Distance–Time Graph | 解读距离–时间图
A cyclist’s journey is shown on a distance–time graph. From the graph: she travels 20 km in the first 1.5 hours, then stops for 0.5 hours, and finally cycles another 10 km in 1 hour. Calculate her average speed for the whole journey.
一位骑自行车者的行程在距离–时间图中呈现。从图中可知：她在最初 1.5 小时内骑行 20 公里，然后休息 0.5 小时，最后在 1 小时内再骑行 10 公里。计算她整个行程的平均速度。
= 1.5 + 0.5 + 1 = 3 hours. Average speed = total distance ÷ total time = 30 ÷ 3 = 10 km/h.
总距离 = 20 + 10 = 30 公里。总时间 = 1.5 + 0.5 + 1 = 3 小时。平均速度 = 总距离 ÷ 总时间 = 30 ÷ 3 = 10 公里/小时。
Even though there was a stop, the average speed is over the entire duration. Interpreting graphs and calculating speed are essential for WJEC exams.
尽管中途有停顿，平均速度仍是整个时段的平均值。解读图表和计算速度是 WJEC 考试的基本要求。
7. Probability of a Board Game | 棋盘游戏的概率
In a board game, you spin a spinner divided into 8 equal sectors numbered 1 to 8. What is the probability of landing on a prime number? List the prime numbers from 1 to 8.
在一个棋盘游戏中，你需要转动一个被分成 8 等分的转盘，编号为 1 至 8。转到质数的概率是多少？列出 1 到 8 中的质数。
Prime numbers up to 8: 2, 3, 5, 7. (1 is not prime.) There are 4 favourable outcomes out of 8.
8 以内的质数有：2、3、5、7。（1 不是质数。）共有 4 个有利结果，总共有
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