📚 Year 9 WJEC Mathematics: Case Study Practical Exercises | Year 9 WJEC 数学：案例分析实战演练
In Year 9 WJEC Mathematics, applying mathematical skills to real-world scenarios is a vital part of developing problem-solving abilities. This article presents a series of practical case studies covering topics such as percentages, algebra, geometry, statistics, and probability. Each case provides a problem, step-by-step solutions, and explanations to help you master WJEC-style applied questions.
在九年级 WJEC 数学中，将数学技能应用于现实场景是培养解决问题能力的重要部分。本文提供一系列涵盖百分比、代数、几何、统计和概率的实战案例分析。每个案例都包含问题、分步解答和解释，帮助你掌握 WJEC 风格的应用题。
1. Supermarket Discounts | 超市折扣
A supermarket offers a 25% discount on all clothing. A jacket originally costs £48. What is the sale price, and what fraction of the original price do you pay?
一家超市所有服装打七五折。一件夹克原价 48 英镑。售价是多少？你支付原价的几分之几？
25% means 25 out of 100, which simplifies to the fraction 1/4. As a decimal, it is 0.25. The discount amount is found by multiplying: 0.25 × 48 = £12.
25% 表示 100 份中的 25 份，化简为分数 1/4。化为小数是 0.25。折扣金额通过乘法求得：0.25 × 48 = 12 英镑。
The sale price is the original price minus the discount: £48 − £12 = £36. You pay 3/4 of the original price.
售价是原价减去折扣：48 英镑 − 12 英镑 = 36 英镑。你支付原价的 3/4。
Imagine a different item has a sale price of £30 after a 25% discount. To find the original price, let it be x. Since you pay 75%, we have 0.75x = 30, so x = 30 ÷ 0.75 = £40.
假设另一件商品在七五折后售价为 30 英镑。设原价为 x。由于支付了 75%，有 0.75x = 30，因此 x = 30 ÷ 0.75 = 40 英镑。
2. Planning a School Trip | 规划学校旅行
A school needs to transport 150 students to a museum. Each coach can carry 50 students and costs £120 to hire. How many coaches are needed and what is the total cost? What is the cost per student?
一所学校需要将 150 名学生运送到博物馆。每辆大巴可载 50 名学生，租金为 120 英镑。需要多少辆大巴？总费用是多少？每名学生成本是多少？
Number of coaches required: 150 ÷ 50 = 3. Total cost: 3 × £120 = £360. Cost per student: £360 ÷ 150 = £2.40.
所需大巴数量：150 ÷ 50 = 3。总费用：3 × 120 英镑 = 360 英镑。每名学生成本：360 英镑 ÷ 150 = 2.40 英镑。
The distance to the museum is 45 miles. If 1 mile ≈ 1.6 km, convert this distance to kilometres. Then, if the coach travels at an average speed of 60 km/h, find the travel time in hours and minutes.
到博物馆的距离是 45 英里。如果 1 英里 ≈ 1.6 公里，请将距离转换为公里。如果大巴平均速度为 60 km/h，求出行时间（小时和分钟）。
Distance in km: 45 × 1.6 = 72 km. Time = distance ÷ speed = 72 ÷ 60 = 1.2 hours. 0.2 hours = 0.2 × 60 = 12 minutes, so the journey takes 1 hour 12 minutes.
距离（公里）：45 × 1.6 = 72 公里。时间 = 距离 ÷ 速度 = 72 ÷ 60 = 1.2 小时。0.2 小时 = 0.2 × 60 = 12 分钟，所以行程需要 1 小时 12 分钟。
3. Designing a Rectangular Garden | 设计矩形花园
A gardener plans a rectangular lawn with length 15 m and width 10 m. Calculate the perimeter for fencing and the area for turf. Fencing costs £8 per metre and turf costs £3 per square metre. Find the total cost.
一位园丁规划一块长 15 米、宽 10 米的矩形草坪。计算围栏所需的周长和铺设草皮所需的面积。围栏每米 8 英镑，草皮每平方米 3 英镑。求总费用。
Perimeter: 2 × (15 + 10) = 2 × 25 = 50 m. Fencing cost: 50 × £8 = £400. Area: 15 × 10 = 150 m². Turf cost: 150 × £3 = £450. Total cost: £400 + £450 = £850.
周长：2 × (15 + 10) = 2 × 25 = 50 米。围栏费用：50 × 8 英镑 = 400 英镑。面积：15 × 10 = 150 平方米。草皮费用：150 × 3 英镑 = 450 英镑。总费用：400 英镑 + 450 英镑 = 850 英镑。
If the gardener decides to add a semicircular flower bed at one end with diameter equal to the width (10 m), find the area of the flower bed (π ≈ 3.14). The area is ½ × π × r² = 0.5 × 3.14 × 5² = 0.5 × 3.14 × 25 = 39.25 m².
如果园丁决定在一端添加一个直径等于宽度（10 米）的半圆形花坛，求花坛面积（π ≈ 3.14）。面积是 ½ × π × r² = 0.5 × 3.14 × 5² = 0.5 × 3.14 × 25 = 39.25 平方米。
4. Mobile Phone Tariffs | 手机资费比较
Two mobile networks offer monthly plans: Network A charges a £10 monthly fee plus 5p per minute of calls. Network B charges a £15 monthly fee plus 3p per minute. Write expressions for the total monthly cost for x minutes.
两家移动网络提供月度套餐：网络 A 收取 10 英镑月租外加每分钟 5 便士。网络 B 收取 15 英镑月租外加每分钟 3 便士。写出 x 分钟通话的总月费表达式。
Network A: CostA = 10 + 0.05x. Network B: CostB = 15 + 0.03x. To find when the costs are equal, solve 10 + 0.05x = 15 + 0.03x.
网络 A：费用A = 10 + 0.05x。网络 B：费用B = 15 + 0.03x。为求费用相等时的通话时长，解方程 10 + 0.05x = 15 + 0.03x。
Subtract 10 from both sides: 0.05x = 5 + 0.03x. Subtract 0.03x: 0.02x = 5. Divide by 0.02: x = 250 minutes. The equal cost is 10 + 0.05×250 = £22.50.
两边减去 10：0.05x = 5 + 0.03x。减去 0.03x：0.02x = 5。除以 0.02：x = 250 分钟。费用相等时为 10 + 0.05×250 = 22.50 英镑。
If you talk for 300 minutes, Network A costs £10+0.05×300=£25, Network B costs £15+0.03×300=£24, so B is cheaper. Use algebra to decide based on usage.
如果你通话 300 分钟，网络 A 费用为 10+0.05×300=25 英镑，网络 B 为 15+0.03×300=24 英镑，因此 B 更便宜。根据使用量用代数做决定。
5. Sports Statistics: Basketball Scores | 运动统计：篮球得分
A basketball team recorded the following points in 8 matches: 12, 15, 10, 18, 20, 15, 14, 16. Find the mean, median, mode, and range of this data set.
一支篮球队在 8 场比赛中记录了以下得分：12, 15, 10, 18, 20, 15, 14, 16。求该数据集的平均数、中位数、众数和极差。
Mean: sum of scores = 12+15+10+18+20+15+14+16 = 120, divided by 8 matches gives 120 ÷ 8 = 15.
平均数：总分 = 12+15+10+18+20+15+14+16 = 120，除以 8 场比赛得到 120 ÷ 8 = 15。
To find the median, order the data: 10, 12, 14, 15, 15, 16, 18, 20. With 8 values, the median is the average of the 4th and 5th terms: (15+15) ÷ 2 = 15.
为求中位数，排序：10, 12, 14, 15, 15, 16, 18, 20。共 8 个值，中位数是第 4 和第 5 项的平均数：(15+15) ÷ 2 = 15。
Mode: the most frequent score is 15 (appears twice). Range: maximum minus minimum = 20 − 10 = 10.
众数：出现最多的得分是 15（出现两次）。极差：最大值减最小值 = 20 − 10 = 10。
| Measure | Value |
|---|---|
| Mean | 15 |
| Median | 15 |
| Mode | 15 |
| Range | 10 |
6. Train Timetable and Speed | 火车时刻表与速度
A train departs from Cardiff Central at 09:15 and arrives at London Paddington at 10:45. The distance travelled is 120 km. Calculate the journey time in hours and the average speed in km/h.
一列火车于 09:15 从卡迪夫中央站出发，于 10:45 抵达伦敦帕丁顿站。行驶距离为 120 公里。计算行程时间（小时）和平均速度（km/h）。
Journey time: 09:15 to 10:45 is 1 hour 30 minutes = 1.5 hours. Average speed = distance ÷ time = 120 ÷ 1.5 = 80 km/h.
行程时间：09:15 到 10:45 是 1 小时 30 分钟 = 1.5 小时。平均速度 = 距离 ÷
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