📚 Year 7 WJEC Maths: Case Study Challenge – Planning a School Sports Day | 七年级 WJEC 数学：案例分析实战演练 – 策划学校运动会
Welcome to a real-world Year 7 WJEC Maths challenge. In this case study, you will take on the role of a sports day organiser. You will use arithmetic, fractions, geometry, data handling, and simple algebra to plan a successful school sports event. Solve each task step by step, showing your working clearly.
欢迎参加七年级 WJEC 数学真实案例挑战。在这个案例中，你将扮演运动会策划者，运用算术、分数、几何、数据处理和简单代数来策划一次成功的校园运动会。请逐步解决每个任务，清楚地展示你的解题过程。
1. The Scenario: Sports Day Planning | 场景：运动会策划
Your school is hosting a sports day for all Year 7 students. There are 120 students, split into four houses: Red, Blue, Green, and Yellow. The event will run from 09:00 to 15:00 with a one-hour lunch break. The available field is a rectangle measuring 100 metres by 60 metres. You need to plan the track, set up events, manage budgets, and collect results.
你的学校要为所有七年级学生举办运动会。共有 120 名学生，分为四个学院：红、蓝、绿、黄。活动从 09:00 到 15:00 举行，中间有一小时午餐休息。可用场地为长方形，长 100 米，宽 60 米。你需要规划跑道、安排比赛项目、管理预算并收集比赛结果。
2. Estimating and Measuring the Track | 估计与测量跑道
You decide to mark a 200-metre oval track using cones on the grass. Each straight side is 60 metres long. The remaining distance consists of two semi-circular ends. First, find the total length of the two curved sections:
你决定用圆锥在草地上标出一个 200 米长的椭圆形跑道。两条直道各长 60 米，剩余距离由两个半圆弯道组成。先求两个弯道总长：
200 − 2 × 60 = 80 m
ameter d = (40 × 2) ÷ π ≈ 80 ÷ 3.14 ≈ 25.5 m. You round up to 26 m to ensure enough space.
每个半圆长 40 米。整圆的周长是 π × d，所以直径 d = (40 × 2) ÷ π ≈ 80 ÷ 3.14 ≈ 25.5 米。向上取整为 26 米，确保有足够空间。
The track, including a 2-metre border each side, has a total width of about 30 m, which fits comfortably within the 60 m width of the field. You measure and mark the corners using a trundle wheel and cones.
跑道加上每边 2 米边界，总宽度约 30 米，能够舒适地放入 60 米宽的场地中。你用测距轮和圆锥测量并标记弯道位置。
3. Managing the Timetable | 管理时间表
The school day runs from 09:00 to 15:00. Subtract a one-hour lunch (12:00– want to run 5 events: 100 m sprint, 200 m race, 4 × 50 m relay, long jump, and tug-of-war.
上学日从 09:00 到 15:00。扣除一小时午餐（12:00–13:00），剩下 5 小时用于比赛。你要安排 5 个项目：100 米短跑、200 米赛跑、4×50 米接力、跳远和拔河。
Each event needs 45 minutes, plus 15 minutes for transitioning between events. That means each block is exactly 1 hour. The timetable fits perfectly:
每个项目需要 45 分钟，外加 15 分钟转换时间，因此每个时段正好 1 小时。时间表完美契合：
- 09:00–10:00 100 m sprint
- 10:00–11:00 200 m race
- 11:00–12:00 Relay
- 12:00–13:00 Lunch
- 13:00–14:00 Long jump
- 14:00–15:00 Tug-of-war
中文时间表：
- 09:00–10:00 100 米短跑
- 10:00–11:00 200 米赛跑
- 11:00–12:00 接力赛
- 12:00–13:00 午餐
- 13:00–14:00 跳远
- 14:00–15:00 拔河
the available time. No overlap occurs, and each house can rotate through events smoothly.
你再次确认总时间：5 × 1 小时 = 5 小时，与可用时间一致。没有重叠，每个学院可以顺畅地轮换项目。
4. Fractions and Grouping Participants | 分数与参赛者分组
or the relay, each team needs 4 runners. How many complete relay teams can each house field?
每个学院人数相等：120 ÷ 4 = 30 人。接力赛中每队需要 4 名跑者。每个学院能派出多少支完整的接力队？
30 ÷ 4 = 7 remainder 2
So each house forms 7 teams, with 2 students left over. Those extra students can act as substitutes. Now, the PE department rules that ⅓ of all participants must take part in field events (long jump or tug-of-war). Calculate the number of students for track and field:
因此每个学院组成 7 支队，剩余 2 名学生。多余的学生可作替补。现在，体育部规定所有参赛者中必须有 ⅓ 参加田赛（跳远或拔河）。计算径赛和田赛的人数：
Field events: ⅓ × 120 = 40 students
Track events: ⅔ × 120 = 80 students
The 80 track athletes are divided among the sprint, 200 m, and relay. The 40 field athletes split equally between long jump and tug-of-war, with 20 each.
80 名径赛运动员被分配到短跑、200 米和接力项目。40 名田赛运动员平均分配到跳远和拔河，各 20 人。
5. Percentages and Discounts on Equipment | 器材的百分数与折扣
You have a budget of £150 for equipment. You need cones (£20 for a pack of 10), marking tape (£15 per roll), and medals (£2 each). A local shop offers a 15% discount on orders over £100. You order 2 packs of cones, 3 rolls of tape, and 50 medals. Calculate the original total cost:
你的器材预算为 150 英镑。你需要圆锥（每包 10 个，20 英镑）、标记胶带（每卷 15 英镑）和奖牌（每枚 2 英镑）。本地商店对超过 100 英镑的订单提供 15% 折扣。你订购 2 包圆锥、3 卷胶带和 50 枚奖牌。计算原始总费用：
Cones: 2 × £20 = £40
Tape: 3 × £15 = £45
Medals: 50 × £2 = £100
Total before discount = £40 + £45 + £100 = £185
Since £185 > £100, the discount applies. Calculate 15% off:
由于 185 英镑 > 100 英镑，可享受折扣。计算 85 折：
Discount = 0.15 × £185 = £27.75
Final cost = £185 − £27.75 = £157.25
You are over budget by £7.25. To stay within £150, you could reduce medals to 46 (cost £92) and keep other items, giving a new pre-discount total of £40+£45+£92=£177. After 15% discount: £177 − £26.55 = £.45, almost on budget. Or buy one fewer roll of tape. Always check different combinations.
你超出预算 7.25 英镑。要控制在 150 英镑内，可将奖牌减至 46 枚（费用 92 英镑），其他不变，新折前总额为 40+45+92=177 英镑。85 折后：177 − 26.55 = 150.45 英镑，几乎符合预算。或者少买一卷胶带。务必核对不同方案。
6. Geometry of the Field Layout | 场地的几何布局
The main field is 100 m by 60 m. You must fit the 200 m track, a long jump pit, and a circular tug-of-war safety zone. The straight sections of the track run along the length of the field (100 m). With 60 m straights and two semi-circles of diameter 26 m, the track occupies a rectangle of length 60 m and width 26 m, plus extra boundary space. The remaining area can host other events.
主场地长 100 米，宽 60 米。你必须容纳 200 米跑道、一个跳远沙坑和一个圆形拔河安全区。跑道直道沿场地长边布置（100 米）。包含 60 米直道和两个直径为 26 米的半圆，跑道占据长约 60 米、宽约 26 米的矩形，外加边界空间。剩余面积可用于其他项目。
The long jump pit measures 1.5 m by 8 m. Calculate its area:
跳远沙坑尺寸为 1.5 米 × 8 米。计算其面积：
Area of pit = 1.5 m × 8 m = 12 m²
The tug-of-war rope is 10 m long, and you need a safety zone of at least 5 m around. This creates a circle with a radius of 10 m (centre to boundary). Find the area required:
拔河绳长 10 米，你需设置至少 5 米的安全区域，形成一个半径为 10 米的圆形（中心至边界）。求所需面积：
Area = π × r² ≈ 3.14 × (10 m)² = 3.14 × 100 = 314 m²
You verify that the field' × 60 = 6000 m²) is far larger than the combined event areas, so all activities fit safely.
你核实场地总面积（100 × 60 = 6000 平方米）远大于各项目面积之和，因此所有活动均能安全容纳。
7. Collecting and Analysing Race Data | 收集与分析比赛数据
After the 100 m sprint, the times in seconds for 8 finalists are recorded: 12.3, 11.8, 13.1, 12.0, 11.5, 12.7, 13.5, 12.9. You need to find the mean, median, and range.
100 米短跑结束后，记录了 8 名决赛选手的时间（秒）：12.3、11.8、13.1、12.0、11.5、12.7、13.5、12.9。你需要计算平均数、中位数和极差。
First, sum the times:
首先，求时间总和：
12.3 + 11.8 + 13.1 + 12.0 + 11.5 + 12.7 + 13.5 + 12.9 = 99.8 s
Mean (average):
平均数：
Mean = 99.8 ÷ 8 = 12.475 s (≈ 12.5 s to 1 decimal place)
To find the median, order the data from fastest to slowest:
为求中位数，将数据从最快到最慢排序：
11.5, 11.8, 12.0, 12.3, 12.7, 12.9, 13.1, 13.5
With 8 values (even number), the median is the average of the 4th and 5th values:
共有 8 个数值（偶数），中位数为第 4 和第 5 个值的平均：
Median = (12.3 + 12.7) ÷ 2 = 12.5 s
Range = fastest time subtracted from slowest:
极差 = 最慢时间 − 最快时间：
