KS3 CAIE Further Maths: Case Study Practical Exercises | KS3 CAIE 进阶数学：案例分析实战演练

📚 KS3 CAIE Further Maths: Case Study Practical Exercises | KS3 CAIE 进阶数学：案例分析实战演练

Case studies are a powerful way to develop problem-solving skills in further mathematics. Instead of memorising formulas, you will learn to apply them to real-world puzzles.

案例分析是培养进阶数学解题能力的有力方法。你不必死记硬背公式，而是学会将它们应用于现实世界的谜题中。

1. Setting the Scene: Decoding the Problem | 场景设定：解码问题

Before tackling any case, read the scenario carefully. Identify the known quantities, the unknown you need to find, and which mathematical tools (algebra, geometry, probability, etc.) are relevant. Break the problem into manageable steps.

在处理任何案例之前，仔细阅读场景。识别已知量、需要求解的未知量，以及哪些数学工具（代数、几何、概率等）是相关的。将问题分解成可管理的步骤。

In a case study, there is often more than one path to the answer. Your goal is to choose a logical route, show your working clearly, and interpret the result in the context of the problem.

在案例分析中，通常不止一条路径通向答案。你的目标是选择一条合乎逻辑的路径，清晰地展示你的运算过程，并在问题背景下解释结果。

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2. Case 1: The Staircase of Blocks | 案例1：积木楼梯

A staircase is built with square blocks. The first step uses 1 block, the second step uses 2 blocks, the third step uses 3 blocks, and so on. How many blocks are needed for a staircase with n steps?

用正方形积木搭建楼梯。第一级台阶用1块，第二级用2块，第三级用3块，以此类推。搭建一个有n级台阶的楼梯需要多少块积木？

Observe the totals: for 1 step: 1 block; for 2 steps: 1+2 = 3; for 3 steps: 1+2+3 = 6; for 4 steps: 1+2+3+4 = 10. These are the triangular numbers. To find a general formula, write S = 1 + 2 + … + n. Write the sum backwards: S = n + (n−1) + … + 1. Adding the two gives 2S = n(n+1), therefore S = n(n+1)/2.

观察总数：1级：1块；2级：1+2=3块；3级：1+2+3=6块；4级：1+2+3+4=10块。这些是三角数。为找到一般公式，写出 S = 1 + 2 + … + n。再倒序写出和：S = n + (n−1) + … + 1。两式相加得 2S = n(n+1)，因此 S = n(n+1)/2。

Sₙ = n(n+1) ÷ 2

For a 5-step staircase, n = 5, so S₅ = 5×6/2 = 15 blocks. This formula allows you to compute the total for any number of steps instantly, turning a pattern into a powerful algebraic tool.

对于5级楼梯，n=5，故 S₅ = 5×6/2 = 15块。该公式使你能立即计算任意级数的总数，将一个模式转化为强大的代数工具。

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3. Case 2: The Age Puzzle | 案例2：年龄谜题

A father is currently four times as old as his son. In five years, the father will be three times as old as his son. Find their current ages.

一位父亲目前的年龄是儿子的四倍。五年后，父亲的年龄将是儿子的三倍。求他们当前的年龄。

Let the son’s current age be x. Then the father’s current age is 4x. In five years, son’s age = x + 5, father’s age = 4x + 5. The condition becomes 4x + 5 = 3(x + 5).

设儿子当前年龄为x。那么父亲当前年龄为4x。五年后，儿子年龄 = x + 5，父亲年龄 = 4x + 5。条件为 4x + 5 = 3(x + 5)。

Solve the equation: 4x + 5 = 3x + 15 → 4x − 3x = 15 − 5 → x = 10. Hence the son is 10 years old, and the father is 40 years old. Verify: in 5 years, son 15, father 45, and 45 = 3 × 15.

解方程：4x + 5 = 3x + 15 → 4x − 3x = 15 − 5 → x = 10。因此儿子10岁，父亲40岁。验证：5年后，儿子15岁，父亲45岁，45 = 3 × 15。

This case shows how setting up a variable and forming an equation can unlock problems that seem confusing when tackled by guesswork.

这个案例展示了如何通过设定变量并建立方程，来解开凭猜测似乎混乱的问题。

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4. Case 3: Maximising Garden Area | 案例3：最大化花园面积

A gardener has 40 metres of fencing to enclose a rectangular flower bed against an existing wall. Only three sides need fencing. What dimensions give the maximum area, and what is that area?

一位园丁有40米长的围栏，用来靠着一堵现有的墙围出一个矩形花坛。只有三边需要围栏。怎样的尺寸能获得最大面积？最大面积是多少？

Let the width of the bed perpendicular to the wall be w metres. Then the length parallel to the wall is 40 − 2w metres. The area A = w × (40 − 2w) = 40w − 2w².

设花坛垂直于墙的宽度为w米。那么平行于墙的长度为 40 − 2w 米。面积 A = w × (40 − 2w) = 40w − 2w²。

A = −2w² + 40w

This is a quadratic function opening downwards. The maximum occurs at the vertex. Using the axis of symmetry w = −b/(2a) with a = −2, b = 40 gives w = −40/(−4) = 10 m.

这是一个开口向下的二次函数。最大值出现在顶点处。利用对称轴 w = −b/(2a)，其中 a = −2, b = 40，得 w = −40/(−4) = 10 m。

Then length = 40 − 2×10 = 20 m, and maximum area = 10 × 20 = 200 m². The optimal shape is twice as long as it is wide.

于是长度 =

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