📚 Year 8 WJEC Maths: Case Study Practice | 八年级WJEC数学：案例分析实战演练
Case studies bring mathematics to life by showing how the topics you learn in class apply to real-world situations. In this article, we will work through a series of practical problems that cover the key areas of the Year 8 WJEC maths curriculum – number, algebra, geometry and statistics. Each case is designed to strengthen your problem-solving skills and build confidence for assessments.
案例研究将数学带进生活，展示课堂上学到的知识如何应用于现实场景。本文将带领大家完成一系列实际问题，这些问题涵盖了八年级 WJEC 数学课程的核心领域——数、代数、几何与统计。每个案例都旨在强化你的问题解决能力，并为你应对考试建立信心。
1. Budgeting for a School Trip | 学校旅行预算
Scenario: A group of 96 Year 8 pupils are going on a geography field trip. The school will cover the cost of 4 accompanying teachers. The coach hire is £280, and the outdoor activity centre charges £5.40 per student. The school also allocates £48 for refreshments. Find the total cost and the amount each student needs to pay.
情景：96 名八年级学生将参加一次地理实地考察。学校将承担 4 位跟队老师的费用。大巴租金为 280 英镑，户外活动中心向每位学生收取 5.40 英镑。学校还拨出 48 英镑用于茶点。计算总费用以及每位学生需要支付的金额。
Step 1: Calculate the activity centre cost for all students: 96 × £5.40 = £518.40.
步骤 1：计算所有学生的活动中心费用：96 × £5.40 = £518.40。
Step 2: Add the fixed costs: £280 (coach) + £518.40 + £48 = £846.40.
步骤 2：加上固定费用：£280（大巴）+ £518.40 + £48 = £846.40。
Step 3: Divide the total by the number of students: £846.40 ÷ 96 = £8.81666… which rounds to £8.82 per student.
步骤 3：用总费用除以学生人数：£846.40 ÷ 96 = £8.81666…，四舍五入后为每位学生 £8.82。
Therefore, each student should contribute £8.82. The school may round it up to £8.85 to build a small contingency fund.
因此，每位学生应交 £8.82。学校可能将其向上取整为 £8.85，以建立一个小额应急基金。
2. Travel Time and Average Speed | 行程时间与平均速度
Scenario: The coach leaves school at 08:15 and arrives at the activity centre at 10:45. The distance travelled is 84 miles. Work out the average speed of the coach in miles per hour (mph). On the way back, the driver expects the journey to take 1 hour 50 minutes. Will the average speed be higher or lower? Explain.
情景：大巴早上 08:15 离开学校，10:45 到达活动中心。行驶距离为 84 英里。计算大巴的平均速度（英里/小时）。返程时，司机预计路程将花费 1 小时 50 分钟。回程的平均速度会更高还是更低？请解释。
First, find the journey time: from 08:15 to 10:45 is 2 hours 30 minutes, which is 2.5 hours.
首先，计算行程时间：从 08:15 到 10:45 是 2 小时 30 分钟，即 2.5 小时。
Average speed = distance ÷ time = 84 ÷ 2.5 = 33.6 mph.
平均速度 = 距离 ÷ 时间 = 84 ÷ 2.5 = 33.6 英里/小时。
The return journey time is 1 hour 50 minutes = 1 + ⁵⁰⁄₆₀ = 1⅚ hours (approximately 1.833 hours). Using the same distance of 84 miles, the speed would be 84 ÷ 1⅚ = 84 × ⁶⁄₁₁ = 45.8 mph (to 1 d.p.).
返程时间为 1 小时 50 分钟 = 1 + ⁵⁰⁄₆₀ = 1⅚ 小时（约 1.833 小时）。使用相同的 84 英里距离，速度为 84 ÷ 1⅚ = 84 × ⁶⁄₁₁ = 45.8 英里/小时（保留一位小数）。
The return average speed is higher. This could be due to lighter traffic or a more direct route.
返程平均速度更高。这可能是由于交通较顺畅或路线更直接。
3. Percentage Discounts in a School Shop | 学校商店的折扣百分比
Scenario: The school shop is having a sale. A water bottle originally costs £6.50 and is now reduced by 25%. A notebook set costs £4.80 with a 15% discount. A pupil buys one bottle and two notebook sets. How much does he pay in total? What is the overall percentage saving compared to the original total price?
情景：学校商店正在进行促销。一个水瓶原价 £6.50，现在降价 25%。一套笔记本原价 £4.80，享受 15% 折扣。一名学生购买一个水瓶和两套笔记本。他一共需要支付多少钱？与原始总价相比，整体节省了百分之几？
Water bottle discount: 25% of £6.50 = 0.25 × 6.50 = £1.625. Sale price = £6.50 – £1.625 = £4.875, which rounds to £4.88.
水瓶折扣：£6.50 的 25% = 0.25 × 6.50 = £1.625。销售价 = £6.50 – £1.625 = £4.875，四舍五入为 £4.88。
Notebook set discount: 15% of £4.80 = 0.15 × 4.80 = £0.72. Sale price per set = £4.80 – £0.72 = £4.08.
笔记本折扣：£4.80 的 15% = 0.15 × 4.80 = £0.72。每套销售价 = £4.80 – £0.72 = £4.08。
Total paid = £4.88 + (2 × £4.08) = £4.88 + £8.16 = £13.04.
支付总额 = £4.88 + (2 × £4.08) = £4.88 + £8.16 = £13.04。
Original total = £6.50 + (2 × £4.80) = £6.50 + £9.60 = £16.10. Saving = £16.10 – £13.04 = £3.06. Percentage saving = (3.06 ÷ 16.10) × 100% ≈ 19.0%.
原始总价 = £6.50 + (2 × £4.80) = £6.50 + £9.60 = £16.10。节省 = £16.10 – £13.04 = £3.06。节省百分比 = (3.06 ÷ 16.10) × 100% ≈ 19.0%。
4. Ratio and Proportion in Cooking | 烹饪中的比与比例
Scenario: A recipe for 8 portions of pasta bake requires 320 g of pasta, 2 eggs, 160 g of cheese and 400 ml of milk. A canteen needs to make 50 portions for lunch. How much of each ingredient is needed? Express the ratio of pasta to cheese to milk in its simplest form for the original recipe.
情景：一份 8 人份的烤意面食谱需要 320 克意面、2 个鸡蛋、160 克奶酪和 400 毫升牛奶。食堂需要制作 50 人份的午餐。每种食材各需要多少？请将原食谱中意面、奶酪与牛奶的比例化为最简形式。
Find the amount per portion: Pasta = 320 ÷ 8 = 40 g per portion; Eggs = 2 ÷ 8 = ¼ egg per portion; Cheese = 160 ÷ 8 = 20 g per portion; Milk = 400 ÷ 8 = 50 ml per portion.
计算每人份量：意面 = 320 ÷ 8 = 40 克/份；鸡蛋 = 2 ÷ 8 = ¼ 个/份；奶酪 = 160 ÷ 8 = 20 克/份；牛奶 = 400 ÷ 8 = 50 毫升/份。
For 50 portions: Pasta = 50 × 40 g = 2000 g (2 kg); Eggs = 50 × ¼ = 12.5, so use 13 eggs; Cheese = 50 × 20 g = 1000 g (1 kg); Milk = 50 × 50 ml = 2500 ml (2.5 litres).
制作 50 份：意面 = 50 × 40 克 = 2000 克（2 千克）；鸡蛋 = 50 × ¼ = 12.5，因此使用 13 个鸡蛋；奶酪 = 50 × 20 克 = 1000 克（1 千克）；牛奶 = 50 × 50 毫升 = 2500 毫升（2.5 升）。
Original ratio pasta : cheese : milk = 320 : 160 : 400. Divide each by 80: 4 : 2 : 5. This is the simplest form.
原食谱中意面 : 奶酪 : 牛奶 = 320 : 160 : 400。每项除以 80：4 : 2 : 5。此为最简形式。
5. Area and Perimeter of a Rectangular Garden | 矩形花园的面积与周长
Scenario: A school garden is rectangular, measuring 12.5 m in length and 8.4 m in width. It needs new turf (grass) and a fence around it. Turf costs £5.20 per square metre. Fencing panels are sold in 1.8 m lengths, each costing £14.99, and a gate costs £35. How much will the total project cost?
情景：学校花园为矩形，长 12.5 米，宽 8.4 米。需要铺设新草皮并在周围安装围栏。草皮每平方米 £5.20。围栏板按 1.8 米长的片段出售，每片 £14.99，一扇门售价 £35。整个工程将花费多少？
Area of garden = length × width = 12.5 × 8.4 = 105 m².
花园面积 = 长 × 宽 = 12.5 × 8.4 = 105 平方米。
Cost of turf = 105 × £5.20 = £546.00.
草皮费用 = 105 × £5.20 = £546.00。
Perimeter of garden = 2 × (12.5 + 8.4) = 2 × 20.9 = 41.8 m. Subtract the gate width (e.g. 1 m) to get fencing length needed: 40.8 m. (We assume the gate fits in a panel gap.)
花园周长 = 2 × (12.5 + 8.4) = 2 × 20.9 = 41.8 米。减去门宽（假设为 1 米）后所需围栏长度为 40.8 米。（我们假定门安装在围栏板间隙中。）
Number of panels = 40.8 ÷ 1.8 = 22.666…, so 23 panels must be bought. Cost of panels = 23 × £14.99 = £344.77. Add the gate: £344.77 + £35 = £379.77.
围栏板数量 = 40.8 ÷ 1.8 = 22.666…，因此必须购买 23 片。围栏板费用 = 23 × £14.99 = £344.77。加上门：£344.77 + £35 = £379.77。
Total project cost = £546.00 + £379.77 = £925.77.
工程总费用 = £546.00 + £379.77 = £925.77。
6. Mean, Median and Mode from a Data Set | 一组数据的平均数、中位数与众数
Scenario: The PE teacher records the number of laps 11 students complete in 12 minutes: 7, 9, 8, 10, 7, 8, 9, 11, 8, 7, 12. Calculate the mean, median and mode. Which average best represents the typical performance? Explain.
情景：体育老师记录了 11 名学生在 12 分钟内完成的圈数：7, 9, 8, 10, 7, 8, 9, 11, 8, 7, 12。计算平均数、中位数和众数。哪个平均数最能代表典型表现？请解释。
First, sort the data: 7, 7, 7, 8, 8, 8, 9, 9, 10, 11, 12.
首先，将数据排序：7, 7, 7, 8, 8, 8, 9, 9, 10, 11, 12。
Mean = sum ÷ number of values = (7+7+7+8+8+8+9+9+10+11+12) ÷ 11 = 96 ÷ 11 ≈ 8.73 laps (to 2 d.p.).
平均数 = 总和 ÷ 数据个数 = (7+7+7+8+8+8+9+9+10+11+12) ÷ 11 = 96 ÷ 11 ≈ 8.73 圈（保留两位小数）。
Median: the 6th value in the ordered list is 8, so median = 8 laps.
中位数：排序后第 6 个数值为 8，因此中位数 = 8 圈。
Mode: both 7 and 8 appear three times, so the data is bimodal with modes 7 and 8. However, if a single mode is required, we could say there are two modal values.
众数：7 和 8 均出现三次，因此数据为双峰，众数为 7 和 8。但如果需要单一众数，则可说明存在两个众数值。
The median of 8 is a good representation because it is not affected by the higher value of 12. The mean (8.73) is slightly pulled upwards by the 12.
中位数 8 是一个不错的代表性数值，因为它不受较高数值 12 的影响。平均数 8.73 则被 12 略微拉高。
7. Solving Linear Equations: Mobile Phone Plans | 解一次方程：手机套餐
Scenario: An electronics shop offers two contract deals for a phone. Plan A: £15 monthly fee plus £0.08 per minute of calls. Plan B: £10 monthly fee plus £0.12 per minute. For how many minutes of calls per month do both plans cost the same? What is that cost?
情景：一家电子产品商店为一部手机提供两种合约套餐。套餐 A：月费 £15，通话每分钟 £0.08。套餐 B：月费 £10，通话每分钟 £0.12。每月通话多少分钟时，两种套餐费用相同？该费用是多少？
Let m be the number of minutes. Cost A = 15 + 0.08m; Cost B = 10 + 0.12m.
设 m 为通话分钟数。费用 A = 15 + 0.08m；费用 B = 10 + 0.12m。
Set them equal: 15 + 0.08m = 10 + 0.12m.
令两者相等：15 + 0.08m = 10 + 0.12m。
Subtract 10 from both sides: 5 + 0.08m = 0.12m. Subtract 0.08m: 5 = 0.04m. Therefore, m = 5 ÷ 0.04 = 125 minutes.
两边同时减去 10：5 + 0.08m = 0.12m。减去 0.08m：5 = 0.04m。因此，m = 5 ÷ 0.04 = 125 分钟。
Cost at 125 minutes: Plan A = 15 + 0.08 × 125 = 15 + 10 = £25. Plan B also gives £25.
125 分钟时的费用：套餐 A = 15 + 0.08 × 125 = 15 + 10 = £25。套餐 B 同样得出 £25。
If a customer talks less than 125 minutes, Plan B is cheaper; if more than 125 minutes, Plan A becomes better value.
如果用户通话少于 125 分钟，套餐 B 更便宜；如果多于 125 分钟，套餐 A 性价比更高。
8. Angles in Triangles and on a Straight Line | 三角形与平角中的角度计算
Scenario: In a triangle, two of the angles are 48° and 73°. Calculate the third angle. Then, one side of the triangle is extended to form an exterior angle. Find the size of that exterior angle and show that it equals the sum of the two opposite interior angles.
情景：一个三角形中，两个角分别为 48° 和 73°。计算第三个角。然后，将三角形的一边延长形成一个外角。计算该外角的大小，并证明它等于两个不相邻内角之和。
Sum of angles in a triangle = 180°. So third angle = 180° – (48° + 73°) = 180° – 121° = 59°.
三角形内角和为 180°。因此第三个角 = 180° – (48° + 73°) = 180° – 121° = 59°。
When one side is extended, the exterior angle is supplementary to the adjacent interior angle. If we extend the side at the 48° vertex, the interior angle is 48°. Then exterior angle = 180° – 48° = 132°.
当延长一边时，外角与其相邻内角互补。若我们在 48° 角的顶点处延长边，内角为 48°。则外角 = 180° – 48° = 132°。
The two opposite interior angles are 73° and 59°. Their sum = 73° + 59° = 132°, which matches the exterior angle. This confirms the exterior angle theorem.
两个不相邻内角为 73° 和 59°。它们的和 = 73° + 59° = 132°，与外角相等。这验证了外角定理。
9. Interpreting a Pie Chart from Survey Data | 根据调查数据解读饼状图
Scenario: 180 students were asked their favourite school subject. The results: 50 chose Maths, 45 Science, 35 English, 30 History and 20 Art. Calculate the angle for each sector in a pie chart and explain what the chart tells you about students’ preferences.
情景：180 名学生被问及他们最喜欢的学校科目。结果为：50 人选择数学，45 人科学，35 人英语，30 人历史，20 人艺术。计算饼状图中每个扇形的角度，并说明该图表揭示了学生偏好的哪些信息。
Angle for a sector = (frequency ÷ total) × 360°.
扇形的角度 = (频数 ÷ 总数) × 360°。
Maths: (50 ÷ 180) × 360° = 100°. Science: (45 ÷ 180) × 360° = 90°. English: (35 ÷ 180) × 360° = 70°. History: (30 ÷ 180) × 360° = 60°. Art: (20 ÷ 180) × 360° = 40°.
数学：(50 ÷ 180) × 360° = 100°。科学：(45 ÷ 180) × 360° = 90°。英语：(35 ÷ 180) × 360° = 70°。历史：(30 ÷ 180) × 360° = 60°。艺术：(20 ÷ 180) × 360° = 40°。
Check: 100° + 90° + 70° + 60° + 40° = 360° – correct. The pie chart would show that Maths and Science together account for more than half the total, indicating a strong preference for STEM subjects among these students.
检验：100° + 90° + 70° + 60° + 40° = 360°——正确。饼状图将显示数学和科学合计超过总数的一半，表明这些学生对理工科有强烈偏好。
Published by TutorHao | Maths Revision Series | aleveler.com
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