📚 Year 8 WJEC Further Maths: Case Study Practical Exercises | 威尔士八年级进阶数学：案例分析实战演练
Mathematics comes alive when you apply it to real-world problems. This article takes you through ten practical case studies designed to stretch your thinking, sharpen your problem-solving skills and prepare you for the WJEC Year 8 Further Mathematics curriculum. Each scenario requires you to combine several topics – from percentages and perimeter to algebra, sequences and transformations – just as you would in an investigation or a project. Work through the examples, check the reasoning, and then have a go at similar challenges yourself.
当数学应用于现实世界的问题时，它才真正鲜活起来。本文将带你完成十个精心设计的案例分析实战演练，旨在拓展你的思维、磨炼解决问题的能力，并为 WJEC 八年级进阶数学课程做好准备。每个情景都需要你将多个知识点结合起来——从百分数、周长到代数、数列以及变换——就像你在探究或项目中所做的那样。跟随例题理解推理过程，然后自己尝试解决类似的挑战。
1. Budgeting a School Trip | 规划学校旅行预算
The school is organising a trip to a science museum for 120 students. Each entry ticket costs £8.50, but the museum offers a special deal: for every 10 tickets bought, the school receives 1 free ticket. Three coaches are needed, and each coach costs £450 and can carry up to 40 passengers.
学校正为 120 名学生组织一次科学博物馆之旅。每张门票 £8.50，但博物馆提供特别优惠：每购买 10 张门票，学校可获赠 1 张免费门票。需要租用三辆大巴，每辆大巴费用 £450，最多可载 40 人。
since 110 ÷ 10 = 11 free tickets, and 110 + 11 = 121 entries). The cost of paid tickets is 110 × £8.50 = £935. Total coach cost is 3 × £450 = £1350, making the overall trip cost £935 + £1350 = £2285. If the school subsidises 20% of the total, the amount parents must pay is 80% of £2285, which is £1828. Divided equally among 120 students, each student pays £1828 ÷ 120 = £15.23 (rounded to the nearest penny).
首先计算学校可获得多少免费门票。购买 10 张送 1 张，意味着花 10 张票的钱可获得 11 个入场名额，因此 120 名学生只需购买 110 张票（110 ÷ 10 = 11 张免费票，110 + 11 = 121 个名额，足够）。付费门票花费 110 × £8.50 = £935。大巴总费用 3 × £450 = £1350，旅行总成本为 £935 + £1350 = £2285。若学校补贴总费用的 20%，家长需支付的金额为 £2285 的 80%，即 £1828。平均分给 120 名学生，每人应付 £1828 ÷ 120 = £15.23（四舍五入到分）。
This case used multiplication, division, percentage decrease and rounding. Further Maths students should also consider alternative scenarios: what if the group size changes or the coach company offers a discount for booking more than two coaches? Extend the model to include a profit from a cake sale that reduces the final cost.
该案例运用了乘法、除法、百分数减少和四舍五入。进阶数学的学生还应考虑其他情景：如果团队人数变化，或者大巴公司对预订超过两辆给予折扣该怎么办？可以扩展模型，将义卖蛋糕的利润纳入，用以降低最终费用。
2. Designing a Half Basketball Court | 设计半场篮球场
A community project wants to mark out a half basketball court on a tarmac rectangle 15 m long and 14 m wide. The free-throw lane, known as 'the key', is a trapezoid: it measures 5.8 m at the baseline, 3.6 m at the free-throw line, and the distance from baseline to free-throw line is 4.9 m.
社区项目需要在一块长 15 m、宽 14 m 的沥青矩形场地上标出半场篮球场。罚球区（也叫“钥匙区”）是一个梯形：底线处长 5.8 m，罚球线处长 3.6 m，底线到罚球线的距离为 4.9 m。
tire half court: 15 m × 14 m = 210 m². The area of the trapezoidal key is calculated using ½ × (a + b) × height ntage of the court that is the key, divide 23.03 by 210 and multiply by 100, giving roughly 11.0% (to 1 decimal place).
首先计算整个半场的面积：15 m × 14 m = 210 m²。梯形钥匙区的面积用公式 ½ × (a + b) × 高 计算：½ × (5.8 + 3.6) × 4.9 = ½ × 9.4 × 4.9 = 23.03 m²。剩余活动区域面积为 210 - 23.03 = 186.97 m²。要计算钥匙区占整个球场的比例，用 23.03 ÷ 210 × 100，结果约为 11.0%（保留一位小数）。
1 = 13 m, so perimeter = 2 × (14 + 13) = 54 m. If each tin of line paint covers 20 m, how many tins are needed? erimeter and rounding up in a real-life context, which is typical of WJEC problem solving.
现在假设在外缘内侧 0.5 m 处画一条边界线。求这个内部矩形的周长。其长度变为 15 - 1 = 14 m，宽度 14 - 1 = 13 m，因此周长 = 2 × (14 + 13) = 54 m。若每罐边界漆可画 20 m，需要多少罐？54 ÷ 20 = 2.7，因此必须购买 3 罐。这涉及现实情境中的减法、周长计算以及向上取整，是 WJEC 问题解决的典型特征。
3. Analysing Football Match Data | 足球比赛数据分析
A striker played 5 matches and recorded the following goal tallies: 2, 0, 3, 1, 4. The player also took 25 shots on target across these games. We want to find the mean, median and range of goals, as well as the shot conversion rate.
一名前锋参加了 5 场比赛，进球数据如下：2, 0, 3, 1, 4。该球员在这几场比赛中共有 25 脚射正。我们需要计算进球的平均数、中位数和极差，以及射门转化率。
ange is 4 - 0 = 4. Shot conversion rate uses the total goals scored (10) divided by total shots on target (25) then multiplied by 100: (10 ÷ 25) × 100% = 40%. A 40% conversion rate is excellent in football.
首先排序进球数据：0, 1, 2, 3, 4。中位数（中间值）为 2。平均数为总进球（2+0+3+1+4=10）除以 5，即每场 2 球。极差为 4 - 0 = 4。射门转化率用总进球数（10）除以总射正次数（25）再乘以 100%：(10 ÷ 25) × 100% = 40%。在足球中，40% 的转化率非常出色。
rsion rate: total goals become 10, total shots become 25 + 2 = 27, so rate = 10 ÷ 27 × 100% ≈ 37.0% weighted averages and updating statistics, an important concept in further data handling.
假设下一场比赛该前锋未能进球，但有 2 脚射正。重新计算转化率：总进球仍为 10，总射正次数变为 25 + 2 = 27，转化率 = 10 ÷ 27 × 100% ≈ 37.0%（一位小数）。随着数据的增加，比率会相应调整。这就引出了加权平均和动态更新统计的概念，这是进阶数据处理中的一个重要思想。
4. Constructing a Birdhouse | 搭建鸟屋
A birdhouse consists of a rectangular base box with a triangular prism roof. The base box is 20 cm wide, 20 cm deep and 25 cm tall. The roof overhangs by 2 cm on each side and the ridge is 30 cm above the ground. The roof forms an isosceles triangle in cross-section.
一个鸟屋由一个矩形底箱和一个三棱柱屋顶组成。底箱宽 20 cm，深 20 cm，高 25 cm。屋顶每边悬挑 2 cm，屋脊距地面 30 cm。屋顶的横截面为等腰三角形。
ise of the roof triangle is 30 - 25 = 5 cm. The base of the roof triangle equals the box depth plus two overhangs: 20 + 2 + 2 = 24 cm. But each sloping side's horizontal run is half of 24 = 12 cm. Applying Pythagoras: slope length = √5² + 12² = √25 + 144 = √169 = 13 cm.
要计算屋顶斜面板的长度，我们需要使用勾股定理。从地面到屋脊的总高为 30 cm，底箱高 25 cm，因此屋顶三角形的垂直升高为 30 - 25 = 5 cm。屋顶三角形的底边等于箱体深度加两个悬挑：20 + 2 + 2 = 24 cm。但每条斜边的水平投影为 24 的一半 = 12 cm。应用勾股定理：斜边长 = √5² + 12² = √25 + 144 = √169 = 13 cm。
o 20×25), ignoring the base. Roof area: 2 × (13 × 20) = 520 cm². Box sides: front/back: 2 × (20 × 25) = 1000 cm²; left/right: 2 × (20 × 25) = 1000 cm². Total wood = 520 + 1000 + 1000 = 2520 cm², or 0.252 m². Volume of the entire birdhouse (box volume + prism volume) can also be found: box volume = 20×20×25 = 10000 cm³, prism volume = (½ × 24 × 5) × 20 = 60 × 20 = 1200 cm³, total = 11200 cm³.
接下来计算所需木材的表面积：屋顶（两块矩形，每块 13 cm × 20 cm）以及底箱四壁（忽略底板）。屋顶面积：2 × (13 × 20) = 520 cm²。箱体四壁：前后两面 2 × (20 × 25) = 1000 cm²；左右两面 2 × (20 × 25) = 1000 cm²。木材总面积 = 520 + 1000 + 1000 = 2520 cm²，即 0.252 m²。此外还可以计算整个鸟屋的体积（箱体体积 + 棱柱体积）：箱体体积 = 20×20×25 = 10000 cm³，棱柱体积 = (½ × 24 × 5) × 20 = 60 × 20 = 1200 cm³，总体积 = 11200 cm³。
5. Mapping a Garden with Coordinates | 用坐标规划花园
A garden is plotted on a grid where 1 unit represents 1 m. The corners are at A(0,0), B(12,0), C(12,8), D(6,8), E(6,4) and F(0,4). The shape is an L-shaped polygon. The owner wants to lay turf and fence the entire perimeter.
一座花园绘制在网格上，其中 1 单位代表 1 m。各角点坐标为 A(0,0), B(12,0), C(12,8), D(6,8), E(6,4) 和 F(0,4)。图形是一个 L 型多边形。主人想铺设草坪并围起整个周界。
e area, split the shape into two rectangles. Rectangle 1: from A to the line x=6 up to y=8? imple method is to extend the vertical line at x=6 upwards, creating a large rectangle ABC'F'? from x=6 to 12 and y=4 to 8, which is 6 m × 4 m = 24 m². So garden area = 96 - 24 = 72 m².
要计算面积，可将形状分割为两个矩形。或者采用完整矩形减去空缺的方法。完整矩形为 12 m × 8 m = 96 m²。空缺矩形位于 x 从 6 到 12，y 从 4 到 8，大小为 6 m × 4 m = 24 m²。因此花园面积 = 96 - 24 = 72 m²。
aced with its bottom-left corner at E(6,4), find its top-right coordinate. Since it extends 2 m in x and 1 m in y, the top-right is at (6+2, 4+1) = (8,5). Translate the shed 3 units right and 2 units up: new bottom-left becomes (9,6), top-right (11,7). This connects to transformation skills taught in Further Maths.
周长需要将所有边长相加：AB = 12 m, BC = 8 m, CD = 6 m, DE = 4 m, EF = 6 m, FA = 4 m。总和 = 12+8+6+4+6+4 = 40 m。如果将一个 2 m × 1 m 的工具棚放置在其左下角位于 E(6,4) 的位置，求其右上角坐标。因其在 x 方向延伸 2 m，y 方向延伸 1 m，所以右上角坐标为 (6+2, 4+1) = (8,5)。将工具棚向右平移 3 个单位，向上平移 2 个单位：新左下角变为 (9,6)，右上角 (11,7)。这对应了进阶数学中教授的变换技能。
6. Comparing Mobile Phone Tariffs | 手机话费套餐对比
Two mobile phone tariffs are available. Tariff X costs £12 per month and includes 150 free minutes; additional minutes are charged at 8p each. Tariff Y has no monthly fee but charges 10p per minute for all calls. Determine which tariff is cheaper for 100, 200 and 300 minutes.
现有两种手机话费套餐。套餐 X 月租 £12，含 150 分钟免费通话；超出部分每分钟收费 8p。套餐 Y 无月租，但所有通话每分钟收费 10p。分别确定通话 100、200 和 300 分钟时哪种套餐更便宜。
For 100 minutes, Tariff X costs £12 (no extra charges) because 100 ≤ 150. Tariff Y costs 100 × £0.10 = £10. So Y is cheaper. For 200 minutes, X costs £12 + (200 - 150) × £0.08 = £12 + £4 = £16. Y costs 200 × £0.10 = £20. So X is cheaper. For 300 minutes, X costs £12 + (150 × £0.08) = £12 + £12 = £24. Y costs 300 × £0.10 = £30. X remains cheaper.
通话 100 分钟时，套餐 X 费用为 £12（无额外收费），因为 100 ≤ 150。套餐 Y 费用为 100 × £0.10 = £10。因此 Y 更便宜。通话 200 分钟时，X 费用为 £12 + (200 - 150) × £0.08 = £12 + £4 = £16。Y 费用为 200 × £0.10 = £20。此时 X 更便宜。通话 300 分钟时，X 费用为 £12 + (150 × £0.08) = £12 + £12 = £24。Y 费用为 300 × £0.10 = £30。X 仍然更便宜。
: 12 + 0.08(m - 150) = 0.10m. Simplify: 12 + 0.08m - 12 = 0.10m, which gives 0.08m = 0.10m? Wait: 12 - 12 cancels? Let's recalc: 12 + 0.08m - 12 = 0.08m. That can' - 12 = 0.10m → 0.08m = 0.10m, which is only true for m=0. There is an error; the constant £ need it to equal 0.10m, which only holds at m=0. So there is no break-even point after 150? Let's check: for m=150, X=12, Y=15. For m=200, X=16, Y=20; X is always £4 cheaper for m≥150? But earlier at 200 minutes X was £16, Y=£20, difference £4. At 300, X=24, Y=30, difference £ so they do not intersect for m>0. The true break-even occurs below 150 minutes: 12 = 0.10m ⇒ m = 120 minutes. So if usage is exactly 120 minutes, both cost £ sting Further Maths discussion point: exploring gradients and intercepts of linear models.
盈亏平衡点指费用相等时的通话时长。设分钟数为 m (m > 150)，列方程：12 + 0.08(m - 150) = 0.10m。化简：12 + 0.08m - 12 = 0.10m，得到 0.08m = 0.10m，这仅在 m=0 时成立，说明在超出 150 分钟后没有交点。实际上，对于 m≥150，X 始终比 Y 便宜且差值变化。真正的平衡点出现在少于 150 分钟时：方程 12 = 0.10m ⇒ m = 120 分钟。即通话恰好 120 分钟时，两者话费均为 £12。这引发了一个有趣的进阶数学讨论点：探究线性模型的斜率和截距。
7. Mountain Bike Downhill Trail | 山地车速降赛道
A mountain bike trail drops 300 m vertically over a horizontal distance of 1.2 km. Riders want to know the average slope percentage and the actual distance they travel along the trail, assuming it runs in a straight line.
一条山地自行车赛道在水平距离 1.2 km 内垂直下降 300 m。骑手想知道平均坡度百分比，以及假设赛道为直线时他们实际骑行的距离。
pe as a percentage = (vertical drop / horizontal distance) × 100 = (300 / 1200) × 100 = 25% ry steep trail. To find the actual distance (the hypotenuse), use Pythagoras: distance d = √300² + 1200² = √90000 + 1440000 about 1.237 km. Slight difference from the horizontal distance because the slope is steep.
将所有单位转换为米：1.2 km = 1200 m。坡度百分比 = (垂直下降 / 水平距离) × 100 = (300 / 1200) × 100 = 25%，这是一条非常陡峭的赛道。要计算实际骑行距离（斜边），使用勾股定理：距离 d = √300² + 1200² = √90000 + 1440000 = √1530000。估算 √1530000：1236² = 1527696，因此 d ≈ 1236.93 m，约合 1.237 km。由于坡度较陡，斜面距离与水平距离略有差异。
A rider completes the trail in 3 minutes 45 seconds. What is the average speed in km/h? , about 19.8 km/h. This requires fluent unit conversion and understanding the relationship between speed, distance and time, key skills in WJEC Further Maths.
一名骑手以 3 分 45 秒完成该赛段，平均速度是多少 km/h？转换时间：3 分 45 秒 = 3.75 分钟 = 3.75/60 小时 = 0.0625 小时。距离为 1.237 km。平均速度 = 距离 / 时间 = 1.237 / 0.0625 = 19.792 km/h，约 19.8 km/h。这需要熟练的单位换算以及对速度、距离和时间关系的理解，这些都是 WJEC 进阶数学的关键技能。
8. Scaling a Recipe for a Bake Sale | 为义卖缩放食谱
A brownie recipe for 6 people requires 200 g of plain flour, 150 g of caster sugar, 2 large eggs and 100 g of butter. We need to make enough for 15 people. Ingredient costs: flour 1.5 kg bag £1.20, sugar 1 kg bag £0.95, eggs £1.80 for 6, butter 250 g block £1.50. We want the total cost of ingredients for 15 portions and the cost per portion.
一份 6 人份的布朗尼食谱需要 200 g 中筋面粉、150 g 细砂糖、2 个大鸡蛋和 100 g 黄油。我们需要制作足够 15 人食用的份量。原料成本如下：面粉 1.5 kg 装 £1.20，糖 1 kg 装 £0.95，鸡蛋 6 枚 £1.80，黄油 250 g 装 £1.50。求 15 份所需原料的总成本以及每份的成本。
ow cost each ingredient using the packs. Flour: 500 g from a 1.5 kg bag costs (500/1500) × £1.20 = ⅓ × £1.20 = £0.40. Sugar: 375 g from 1 kg
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