📚 Case Study Practicals for Year 8 WJEC Further Mathematics | Year 8 WJEC 进阶数学案例分析实战演练
In Year 8 WJEC Further Mathematics, you are expected to move beyond straightforward textbook questions and apply your knowledge to realistic, multi-step problems. These case studies combine topics such as algebra, geometry, statistics and ratio, challenging you to think like a mathematician. This article presents a series of worked examples and strategies to help you feel confident when tackling extended scenarios.
在 WJEC 八年级进阶数学中，你需要超越简单的课本习题，将知识应用于现实的、多步骤的问题。这些案例分析结合了代数、几何、统计和比例等主题，要求你像数学家一样思考。本文提供了一系列详细的示例和策略，帮助你自信地应对扩展场景。
1. Understanding Case Studies in Further Mathematics | 理解进阶数学中的案例分析
A case study sets you a problem in a real-life context – you might be designing a garden, planning a school trip or analysing game probabilities. The goal is not just to find one answer but to show your method clearly and justify your decisions. You must be able to choose the right mathematical tools and explain why they work.
案例分析将问题置于真实情境中——你可能需要设计一个花园、规划一次学校旅行或分析游戏中的概率。目标不仅仅是找出一个答案，而是清晰地展示你的方法并证明你的决定。你必须能够选择合适的数学工具，并解释它们为何有效。
In WJEC Further Maths, examiners look for evidence of reasoning, accurate calculations and the ability to check your solution. Common themes include linear relationships, area and perimeter, statistical averages, ratio and proportion, and transformations on a coordinate grid. Each case study will link two or more of these ideas.
在 WJEC 进阶数学中，考官寻找的是推理过程、精确计算以及检查答案的能力。常见主题包括线性关系、面积与周长、统计平均值、比例以及坐标网格上的变换。每个案例分析会将两个或更多这样的想法联系起来。
2. Case Study 1: Garden Design – Area and Perimeter | 案例1：花园设计——面积与周长
You are helping a family design a rectangular front garden that measures 12 m by 8 m. They want a lawn covering the whole area except for a semicircular flower bed of radius 2 m placed against one of the shorter sides. You need to calculate the area of turf required and the length of fencing needed to enclose the garden (except along the house wall, which is one of the 12 m sides).
你正在帮助一个家庭设计一个长 12 m、宽 8 m 的矩形前花园。他们希望除了一个半径为 2 m 且紧靠一条短边的半圆形花坛外，整个区域都铺上草坪。你需要计算所需的草皮面积，以及围住花园（沿着房屋墙壁的一侧不围，该侧是 12 m 的边）所需的围栏长度。
This case study requires you to recall the area of a rectangle and the area of a circle, then adapt the latter for a semicircle. You will also work with perimeter and understand which sides need fencing. The challenge is to combine these steps correctly and use the appropriate value for π.
这个案例分析要求你回想矩形的面积公式和圆的面积公式，然后将后者调整为半圆的面积。你还需要处理周长，并理解哪些边需要围栏。挑战在于正确地组合这些步骤并使用合适的 π 值。
3. Case Study 2: Budgeting a School Trip – Linear Equations | 案例2：学校旅行预算——线性方程
A school hires a coach for a day trip. The fixed hire cost is £120, and each student is charged £15 to cover the coach and entry fees. The trip organiser has a total budget of £360. How many students can go on the trip? After finding the number, check that the total cost does not exceed the budget.
学校为一日游租用一辆大巴。固定租车费用为 120 英镑，每位学生支付 15 英镑以涵盖大巴和门票费用。旅行组织者的总预算为 360 英镑。有多少学生可以参加这次旅行？在得出人数后，检查总费用是否未超出预算。
s to get 15n = 240, then divide by 15 to obtain n = 16 students. To check: 120 + 15(16) = 120 + 240 = 360, which exactly matches the budget.
建立一个线性方程，总费用 C = 120 + 15n，其中 n 表示学生人数。解方程 120 + 15n = 360。两边同时减去 120 得到 15n = 240，再除以 15 得出 n = 16 名学生。检查：120 + 15(16) = 120 + 240 = 360，恰好与预算相符。
If the budget were increased to £ ve a fraction of a student, the maximum number would be 18 students, leaving some money unspent. Always interpret your solution in the context of the problem.
如果预算提高到 400 英镑，方程 120 + 15n = 400 的解为 n = 280/15 ≈ 18.67。因为不能有分数个学生，所以最多人数为 18 名，这样会剩一些钱。永远要在题目情境中解读你的解。
4. Case Study 3: Sports Statistics – Mean, Median, Mode and Range | 案例3：体育统计数据——平均数、中位数、众数与极差
A basketball team recorded the points scored by their players in five matches: 12, 15, 12, 20, 14. The coach wants to know the mean, median, mode and range of these scores to analyse consistency. Calculate each measure and explain which gives the most useful information for the coach.
一支篮球队记录了他们球员在五场比赛中的得分：12, 15, 12, 20, 14。教练想知道这些分数的平均数、中位数、众数和极差，以分析表现的一致性。请计算每个指标，并解释哪一个能给教练提供最有用的信息。
dle value, 14. The mode is 12, as it appears twice. The range is maximum minus minimum: 20 – 12 = 8.
首先，将数据排序：12, 12, 14, 15, 20。平均数为 (12+15+12+20+14)/5 = 73/5 = 14.6 分。中位数是中间的值，14。众数是 12，它出现了两次。极差为最大值减最小值：20 – 12 = 8。
The mean (14.6) gives a typical score but is pulled up slightly by the high score of 20. The median (14) is less affected by that extreme. The mode (12) shows a common low score, and the range (8) indicates variability. For consistency, the coach might look at a small range and a median close to the target.
平均数（14.6）给出了一个典型分数，但被高分 20 稍稍拉高。中位数（14）受极端值影响较小。众数（12）显示了一个常见的低分，而极差（8）表明得分有波动。对于稳定性来说，教练可能会期望较小的极差和接近目标的中位数。
5. Case Study 4: Recipe Scaling – Direct Proportion | 案例4：食谱配料缩放——正比例
A recipe for 8 people requires 300 g of flour, 200 g of sugar and 4 eggs. You need to adapt the recipe for 10 people. All quantities must be scaled up proportionally. Calculate the new amounts, keeping your answers exact where possible, and round to a sensible precision when necessary.
一份供 8 人食用的食谱需要 300 克面粉、200 克糖和 4 个鸡蛋。你需要将其调整为 10 人份。所有用量必须按比例增加。计算新的用量，尽可能保持精确，必要时按合理精度取整。
nts are exact. If you needed to scale for 7 people, the factor would be 7/8 = 0.875, and you would get 262.5 g flour, 175 g sugar and 3.5 eggs – but you would have to interpret half an egg realistically!
比例因子为 10 / 8 = 1.25。将每种原料乘以 1.25：面粉变为 300 × 1.25 = 375 克；糖 200 × 1.25 = 250 克；鸡蛋 4 × 1.25 = 5 个。所有用量都很精确。如果是要调整为 7 人份，因子为 7/8 = 0.875，你将得到 262.5 克面粉、175 克糖和 3.5 个鸡蛋——但你必须现实地处理半个鸡蛋！
e flour-to-sugar ratio should remain 300:200 = 3:2, and 375:250 simplifies to 3:2 as well.
比例推理是进阶数学的关键。始终检查新的用量是否保持了相同比例。例如，面粉与糖的比例应保持 300:200 = 3:2，而 375:250 简化后也是 3:2。
6. Case Study 5: Coordinates and Transformations – Creating a Logo | 案例5：坐标与变换——设计徽标
A graphic designer wants to create a simple logo by transforming a basic shape on a coordinate grid. The original triangle has vertices A(1, 2), B(3, 5) and C(1, 5). First, reflect the triangle in the y-axis. Then translate the reflected image by the vector (4, -3). Write down the coordinates of the final image and describe the single transformation that would map the original triangle to the final one.
一位平面设计师想通过在坐标网格上变换基本图形来设计一个简单徽标。原始三角形的顶点为 A(1, 2)、B(3, 5) 和 C(1, 5)。首先，将三角形关于 y 轴反射。然后将反射后的图形按照向量 (4, -3) 平移。写出最终图形的顶点坐标，并描述能够将原始三角形一次映射到最终图形的单一变换。
Reflecting in the y-axis changes the sign of the x-coordinate: A'(−1, 2), B'(−3, 5), C'(−1, 5). Then translating by (4, −3) adds 4 to each x-coordinate and subtracts 3 from each y-coordinate: A''(3, −1), B''(1, 2), C''(3, 2). The final triangle has vertices (3, −1), (1, 2), (3, 2).
关于 y 轴反射会改变 x 坐标的符号：A'(−1, 2)、B'(−3, 5)、C'(−1, 5)。然后按向量 (4, −3) 平移，即每个 x 坐标加 4，每个 y 坐标减 3：A''(3, −1)、B''(1, 2)、C''(3, 2)。最终三角形的顶点为 (3, −1)、(1, 2)、(3, 2)。
Is there a single transformation? Compare original A(1,2) with A''(3,−1). The movement is 2 units right and 3 units down, but that translation alone would not map the other points because the reflection reversed the triangle's orientation. No single translation or reflection can achieve the same effect; a reflection followed by a translation is a glide reflection, which is a combination.
是否存在单一变换？比较原始点 A(1,2) 和 A''(3,−1)。移动为向右 2 个单位、向下 3 个单位，但仅靠平移无法映射其他点，因为反射翻转了三角形的方向。没有单一的平移或反射能实现相同效果；反射后再平移是一种滑动反射，属于组合变换。
7. Case Study 6: Probability in Games – Fairness and Expected Outcomes | 案例6：游戏中的概率——公平性与期望结果
Sasha designs a game where two fair six-sided dice are thrown. If the sum is 7, she wins a point; otherwise her friend wins a point. Is the game fair? Use a sample space diagram to calculate the probability of getting a sum of 7, and compare it with the probability of not getting 7.
萨莎设计了一个游戏：投掷两个公平的六面骰子。如果点数之和为 7，她得一分；否则她的朋友得一分。这个游戏公平吗？使用样本空间图计算得到和为 7 的概率，并与得到非 7 的概率进行比较。
ally likely outcomes. The pairs that sum to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) – for Sasha is 1/6, while her friend wins with probability 5/6. The game is heavily biased against Sasha, so it is not fair.
共有 36 种等可能的结果。和为 7 的组合有 (1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1)——共 6 种结果。因此 P(和为 7) = 6/36 = 1/6。萨莎获胜的概率为 1/6，而她的朋友获胜的概率为 5/6。该游戏严重偏向于朋友一方，所以不公平。
ave 6+2 = 8 favourable outcomes, with probability 8/36 = 2/9, still not 1/2. Alternatively, they could take turns with different winning conditions.
为了使游戏公平，萨莎可以调整规则，例如，当和为 7 或 11 时她赢。得到 11 的方式有两种：(5,6) 和 (6,5)，这样她就有 6+2 = 8 种有利结果，概率为 8/36 = 2/9，仍不是 1/2。或者，他们可以轮流设置不同的获胜条件。
8. Problem-Solving Strategies – Working Backwards and Drawing Diagrams | 解题策略——倒推法和绘图法
When a case study feels overwhelming, break it down using diagrams, tables or flowcharts. For the garden case study, start by drawing the rectangle and the semicircle. Label the dimensions and mark which sides need fencing. This visual step often clarifies which formulas to use and prevents simple mistakes like mixing up radius and diameter.
当案例分析让你感到棘手时，可以用图表、表格或流程图将其分解。以花园案例为例，首先画出矩形和半圆。标注尺寸，并标明哪些边需要围栏。这个视觉步骤通常能帮你理清该用哪些公式，并避免混淆半径和直径等简单错误。
Another powerful strategy is working backwards from the answer you need. If you know the budget and the cost per student, set up an equation as in Case Study 2. If a problem involves unknown angles, label them and use angle facts in reverse. Always write down what you know and what you need to find.
另一个强大的策略是从所需的答案倒推。如果你知道预算和每名学生的费用，就像案例 2 那样建立方程。如果问题涉及未知角度，可以为其标注，并逆向使用角度定理。始终把你已知的和需要求的东西写下来。
9. Step-by-Step Worked Example – Solving the Garden Case Study | 分步骤详解——解决花园案例分析
We now return to the garden design from Section 2. Let's go through the solution carefully.
现在我们回到第 2 节的花园设计。让我们仔细过一遍解答。
Step 1 – Understand the layout. The garden is a 12 m by 8 m rectangle. One of the 8 m sides is against the house and does not need fencing. A semicircular flower bed of radius 2 m sits against the other 8 m side (or the inner part – assume it is a half-circle with diameter along the 8 m side, but the exact position does not affect area). We need to find the turf area and the length of fencing.
第 1 步——理解布局。 花园是一个 12 m × 8 m 的矩形。其中一条 8 m 的边靠房，不需要围栏。一个半径为 2 m 的半圆形花坛位于另一条 8 m 边上（假设是半圆，其直径沿 8 m 边，但确切位置不影响面积）。我们需要求草皮面积和围栏长度。
Step 2 – Calculate total garden area. Area_rectangle = length × width = 12 × 8 = 96 m².
第 2 步——计算花园总面积。 矩形面积 = 长 × 宽 = 12 × 8 = 96 m²。
Step 3 – Calculate area of the flower bed. ². Using π ≈ 3.14 gives 2 × 3.14 = 6.28 m². (Leave as 2π for an exact answer if required.)
第 3 步——计算花坛面积
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