📚 Year 7 WJEC Further Mathematics: Case Study Practical Exercises | Year 7 WJEC 进阶数学：案例分析实战演练
In further mathematics, real understanding comes not only from learning rules but from applying them in unfamiliar contexts. This article presents a series of case study challenges designed for Year 7 WJEC students. Each case mimics a real‑world or puzzle‑type scenario, requiring logical reasoning, number fluency, algebraic thinking, and data handling. Work through each example step by step – the aim is not just to find an answer, but to explain your process clearly.
在进阶数学中，真正的理解不仅来自学习规则，更来自于在不熟悉的情境中应用它们。本文为 Year 7 WJEC 学生设计了一系列案例研究挑战。每个案例都模拟了一个现实世界或谜题型的情境，需要逻辑推理、数字流利度、代数思维和数据处理能力。请逐步完成每个例子——目标不仅仅是找到答案，而是清晰地解释你的解题过程。
1. Getting Started with Case Study Thinking | 开启案例分析思维
Before you jump into a problem, read it twice. Underline what you know, what you need to find, and any conditions that apply. In a case study, there is often more than one part – treat each part as a stepping stone. Use diagrams, tables, or trial and error where helpful. Always write down your reasoning, because in further mathematics 'how' you solve matters as much as the answer.
在着手解题之前，请把题目读两遍。在已知条件、需要求解的内容以及任何限制条件下划线。在案例分析中，通常有多个部分——把每个部分都当作垫脚石。在适当的地方使用图表、表格或试错法。始终写下你的推理过程，因为在进阶数学中，“如何”解题与答案本身同样重要。
2. Case Study 1: The Prime Spiral Mystery | 案例 1：素数螺旋之谜
sing by 1 each step. After filling a 7 × 7 grid, you highlight all the prime numbers. Curiously, many primes appear on diagonal lines. Your task is to create a 7 × 7 spiral grid, identify the primes, and describe the pattern you see.
想象一个数字网格，你按螺旋模式写下数字，从中心的 1 开始。然后向右、向上、向左、向下移动，依次加 1。在填满一个 7 × 7 的网格后，你高亮所有的素数。奇怪的是，许多素数出现在对角线上。你的任务是创建一个 7 × 7 的螺旋网格，标出素数，并描述你所看到的模式。
3. Building the Spiral Grid | 构建螺旋网格
n the central cell (row 4, column 4). Then move right to place 2, up to place 3, left for 4 and 5, down for 6 and 7, right for 8, 9, 10, and so on. Remember that each straight segment increases in length: 1 step, 1 step, 2 steps, 2 steps, 3 steps, 3 steps, etc. Complete the grid up to 49. Now circle all prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
从一个空白的 7 × 7 表格开始。将 1 放在中心单元格（第 4 行第 4 列）。然后向右移动填入 2，向上填入 3，向左填入 4 和 5，向下填入 6 和 7，向右填入 8、9、10，依此类推。记住，每个直线段的长度递增：1 步、1 步、2 步、2 步、3 步、3 步，等等。完成网格直到 49。现在圈出所有素数：2、3、5、7、11、13、17、19、23、29、31、37、41、43、47。
Look carefully at the positions of the primes. Do they cluster along certain lines? You might notice many primes lie on the diagonal running from the top‑left to bottom‑right, and also on the diagonal from top‑right to bottom‑left. Some primes also sit one step away from these diagonals. Describe your findings in two sentences: "Most circled numbers sit on two main diagonals. The pattern is not perfect, but there is a clear tendency."
仔细观察素数的位置。它们是否沿着某些直线聚集？你可能会注意到，许多素数位于从左上到右下的对角线上，以及从右上到左下的对角线上。有些素数也位于这些对角线的一步之外。用两句话描述你的发现：“大多数带圈的数字位于两条主要的对角线上。这个规律并不完美，但存在明显的趋势。”
4. Reasoning Behind the Pattern | 模式背后的推理
The prime spiral shows that primes are not randomly scattered. Certain diagonal lines contain numbers generated by quadratic expressions, such as n² + n + 41 (which produces many primes for small n). In the spiral, numbers on the same diagonal often share similar algebraic forms. While full understanding of these formulas requires higher‑level algebra, Year 7 students can appreciate that number patterns can reveal hidden structure. This is why further mathematics values exploration – patterns lead to formulas, and formulas lead to predictions.
素数螺旋表明，素数并不是随机分布的。某些对角线上的数是由二次表达式生成的，例如 n² + n + 41（对小数值 n 能产生许多素数）。在螺旋中，同一条对角线上的数往往具有相似的代数形式。虽然完全理解这些公式需要更高级的代数知识，但 Year 7 学生可以体会到数字模式能够揭示隐藏的结构。这就是为什么进阶数学重视探索——模式引出公式，公式引出预测。
5. Case Study 2: Designing a Rectangular Enclosure | 案例 2：设计一个矩形围栏
A school gardening club has 24 metres of fencing to create a rectangular vegetable plot against a wall. The wall forms one side of the rectangle, so fencing is needed only for the other three sides. The club wants the maximum possible area for growing vegetables. Investigate what length and width give the largest area. Use algebra and tables to support your reasoning.
学校园艺社团有 24 米长的围栏，用来靠墙建一个矩形的蔬菜地。墙构成矩形的一条边，因此只需要为另外三条边使用围栏。社团想要尽可能大的种植面积。研究哪个长度和宽度能给出最大面积。运用代数和表格来支持你的推理。
6. Setting Up Variables and Testing Values | 设定变量并测试数值
Let the side parallel to the wall be L metres, and each of the two perpendicular sides be W metres. The total fencing used is L + 2W = 24. So L = 24 − 2W. The area A = L × W = (24 − 2W) × W. Now try different whole‑number values of W, calculate L and A.
设与墙平行的一边为 L 米，两条垂直的边各为 W 米。所用的围栏总长度为 L + 2W = 24。因此 L = 24 − 2W。面积 A = L × W = (24 − 2W) × W。现在尝试 W 的不同整数值，计算 L 和 A。
| W (m) | L = 24 − 2W (m) | Area A (m²) |
|---|---|---|
| 1 | 22 | 22 |
| 2 | 20 | 40 |
| 3 | 18 | 54 |
| 4 | 16 | 64 |
| 5 | 14 | 70 |
| 6 | 12 | 72 |
| 7 | 10 | 70 |
| 8 | 8 | 64 |
From the table, the maximum area occurs when W = 6 m and L = 12 m, giving an area of 72 m². Notice that the maximum area arises when L is exactly twice W (a rectangle that is half as wide as it is long). This is a special property of three‑sided enclosures.
从表格中可以看出，最大面积出现在 W = 6 米、L = 12 米时，面积为 72 平方米。注意，最大面积出现在 L 正好是 W 的两倍时（这种矩形的宽是长的一半）。这是三边围栏的一个特殊性质。
7. Extending the Garden Problem | 拓展花园问题
If the fencing length changes to a different total, you can still use the same method. In general, for a fixed perimeter against a wall, the maximum area for a rectangle is achieved when the side parallel to the wall is twice the width. To prove this algebraically, you can rewrite the area expression as A = 24W − 2W² and complete the square, but that is explored in later years. For now, systematic testing gives you confidence and a strong intuition for optimisation.
如果围栏的总长度改变，你仍然可以使用相同的方法。一般来说，对于靠墙的固定周长，当与墙平行的一边是宽度的两倍时，矩形的面积最大。为了用代数证明这一点，你可以将面积表达式改写为 A = 24W − 2W² 并进行配方，不过这将在高年级探索。目前，系统的测试能让你对优化问题充满信心并建立起强烈的直觉。
8. Case Study 3: Interpreting a School Survey | 案例 3：解读学校调查
A survey asked 60 Year 7 pupils which of three activities they preferred: chess club, drama club, or coding club. The results are: chess 15, drama 25, coding 20. Your case study asks you to display this data in a pie chart, calculate the mode, and work out the fraction of pupils who did not choose coding. Then explain what the data suggests about the popularity of activities.
一项调查询问了 60 名 Year 7 学生他们最喜欢三种活动中的哪一种：国际象棋社、戏剧社或编程社。结果如下：象棋 15 人，戏剧 25 人，编程 20 人。你的案例研究要求你用饼图展示这些数据，计算众数，并计算出没有选择编程的学生所占的比例。然后解释数据对活动受欢迎程度的启示。
9. Pie Chart Construction and Fractions | 饼图绘制与分数
25 ÷ 60) × 360° = 150°. Coding = (20 ÷ 60) × 360° = 120°. Check: 90° + 150° + 120° = 360°.
要绘制饼图，先计算每个类别的角度。总频数 = 60。象棋的角度 = (15 ÷ 60) × 360° = 90°。戏剧 = (25 ÷ 60) × 360° = 150°。编程 = (20 ÷ 60) × 360° = 120°。检查：90° + 150° + 120° = 360°。
e number who chose chess or drama divided by the total: (15 + 25)/60 = 40/60 = 2/3. So two‑ tell the school that more pupils enjoy performance and strategic board games over computer‑based clubs, but further investigation would be needed to understand why.
众数（选择最多的）是戏剧，有 25 名学生。没有选择编程的学生所占比例为选择象棋或戏剧的人数除以总人数：(15 + 25)/60 = 40/60 = 2/3。因此三分之二的学生没有选择编程。这可能告诉学校，比起基于电脑的社团，更多的学生喜欢表演和策略棋盘游戏，但需要进一步调查来理解原因。
10. Case Study 4: Fair Game Design | 案例 4：公平游戏设计
You are inventing a simple game using a spinner divided into four equal sections coloured red, blue, green, and yellow. A player wins if the spinner lands on red. Your case study asks: is this a fair game? If not, how can you change the rules to make the expected winnings equal for both the player and the stall holder? Assume a win pays £1 and a loss costs £1.
你正在发明一个简单的游戏，使用一个被分为红、蓝、绿、黄四个相等扇区的转盘。如果转盘停在红色上，玩家获胜。你的案例研究询问：这是一个公平的游戏吗？如果不是，你如何改变规则，使玩家和摊主的期望收益相等？假设赢一次获得 £1，输一次付出 £1。
11. Probability and Expected Value | 概率与期望值
s, the probability of winning (red) is 1/4, and the probability of losing is 3/4. A player's expected net gain per play = (1/4 × £1) + (3/4 × −£1) = £0.25 − £0.75 = −£0.50. On average, the player loses 50p per spin, so the game is not fair – it favours the stall holder.
由于四个扇区相等，获胜（红色）的概率为 1/4，失败的概率为 3/4。玩家每次游戏的期望净收益 = (1/4 × £1) + (3/4 × −£1) = £0.25 − £0.75 = −£0.50。平均而言，玩家每次转盘损失 50 便士，因此这个游戏不公平——它偏向于摊主。
To make it fair, we can adjust the payout. Let the win amount be £3 (while the loss still costs £1). Then expected net gain = (1/4 × £3) + (3/4 × −£1) = £0.75 − £0.75 = £ ely, you can change the spinner: make half the spinner red (probability 1/2), and keep win £1, loss £ a further mathematician, you explain that fairness means the expected value must be zero.
为了让它变得公平，我们可以调整奖金数额。假设获胜获得 £3（而失败仍然付出 £1）。那么期望净收益 = (1/4 × £3) + (3/4 × −£1) = £0.75 − £0.75 = £0。这个游戏就公平了。或者，你可以改变转盘：让一半的转盘是红色（概率为 1/2），保持赢 £1、输 £1，此时期望收益 = (1/2 × 1) + (1/2 × −1) = 0。作为进阶数学学习者，你要解释公平意味着期望值必须为零。
12. Bringing It All Together: Tips for Case Study Success | 总结：案例分析成功秘诀
Across all four case studies, you have used tables, diagrams, algebra, data interpretation, and probability reasoning. These are core skills in WJEC further mathematics. When tackling an unseen case study in an assessment, remember: read and annotate, break the problem into smaller steps, show your working clearly, and write a short conclusion. Reflection on why your solution makes sense will earn you top marks for reasoning.
在这四个案例研究中，你使用了表格、图表、代数、数据解释和概率推理。这些都是 WJEC 进阶数学的核心技能。当你在评估中遇到全新的案例研究时，请记住：阅读并做好标注、将问题分解成更小的步骤、清晰地展示你的解题过程，并写出简短的结论。反思你的解决方案为何合理，将为你赢得推理方面的最高分数。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
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