📚 Year 7 CAIE Advanced Mathematics: Case Study Practical Exercises | Year 7 CAIE 进阶数学：案例分析实战演练
Advanced Mathematics at Year 7 level goes far beyond routine calculation. It asks you to reason logically, to recognise patterns, and to apply familiar ideas inside unfamiliar situations. In this article we work through ten carefully selected case studies written in the style of CAIE problem-solving papers, together with two supporting sections on technique. Every case study is broken down the problem step by step, so that you can rehearse the exact skills that higher-level mathematics demands: interpreting the question, choosing a method, showing clear working, and checking whether the answer is sensible.
Year 7 阶段的进阶数学远远超越了常规计算。它要求你进行逻辑推理、识别模式，并把熟悉的概念应用到陌生的情境中。本文精选了十个符合 CAIE 解题风格的案例，并附加两个技巧专题小节。每个案例都一步步拆解，帮助你演练高阶数学所必需的核心技能：读懂题意、选择方法、清晰书写过程，以及检验答案是否合理。
1. Number Theory Puzzle: The Missing Digit | 数论谜题：缺失的数字
Consider the five-digit number 7A2B8, where A and B are single digits. Three facts are given: . It is known that the number is divisible by 9, the sum of its digits equals 18, and when the number is divided by 5 the remainder is 3. Your task is to find the digits A and B.
请看五位数 7A2B8，其中 A 和 B 都是单个数字。题目给出三个条件：这个数能被 9 整除；它的各位数字之和等于 18；这个数除以 5 时余数为 3。你的任务是求出 A 和 B。
nder on division by 5 is 3 whatever A and B happen to be. This condition is therefore satisfied automatically and tells us nothing new.
第一步——看个位数字。最后一位是 8，而 8 = 5 × 1 + 3，所以无论 A 和 B 取什么值，除以 5 的余数都是 3。这个条件自动满足，不能提供新的信息。
its: 7 + A + 2 + B + 8 = 17 + A + B, and the question states this equals 18, so A + B = 1.
第二步——利用数字和。把已知数字相加：7 + A + 2 + B + 8 = 17 + A + B，题目说明它等于 18，因此 A + B = 1。
ltiple of 9. Here the digit sum is 18 = 2 × 9, so this condition also holds automatically.
第三步——检验 9 的整除性。一个数能被 9 整除，当且仅当它的各位数字之和是 9 的倍数。这里数字和为 18 = 2 × 9，所以这个条件同样自动成立。
7 + A + 2 + B + 8 = 18 ⇒ A + B = 1
er 70218 or 71208. Check: 70218 ÷ 9 = 7802 and 71208 ÷ 9 = 7912, with no remainder in either case.
第四步——列举可能性。因为 A 和 B 是满足 A + B = 1 的数字，数对只能是 (A, B) = (0, 1) 或 (1, 0)。两者都成立，所以原数可能是 70218 或 71208。检验：70218 ÷ 9 = 7802，71208 ÷ 9 = 7912，都没有余数。
2. Divisibility Rules and Digit Sums: The Toolkit | 整除规则与数字和：工具包
Case study 1 becomes much faster once you own the standard divisibility tests. Learn them as a toolkit rather than as separate tricks, because CAIE questions often combine two or three of them in a single part.
一旦掌握了标准的整除判定法则，案例 1 就会变得非常快。要把它们当作一套工具来记忆，而不是孤立的技巧，因为 CAIE 题目经常在一个小问中同时用到两三条。
| Divisor 除数 | Test 判定方法 | Example 例子 |
| 2 | last digit even 末位是偶数 | 4316 ✓ |
| 3 | digit sum is a multiple of 3 数字和是 3 的倍数 | 417 → 12 ✓ |
| 4 | last two digits form a multiple of 4 末两位是 4 的倍数 | 7316 → 16 ✓ |
| 5 | last digit 0 or 5 末位是 0 或 5 | 7295 ✓ |
| 8 | last three digits form a multiple of 8 末三位是 8 的倍数 | 5128 → 128 ✓ |
| 9 | digit sum is a multiple of 9 数字和是 9 的倍数 | 70218 → 18 ✓ |
| 11 | alternating digit sum is 0 or a multiple of 11 交错数字和为 0 或 11 的倍数 | 2728 → 2 − 7 + 2 − 8 = −11 ✓ |
emember that 6 = 2 × 3, so the number must pass both the even test and the digit-sum test.
常见的追问是：让某个数能被 6 整除时，缺失数字的最小值是多少。请记住 6 = 2 × 3，所以这个数必须同时通过"末位为偶数"和"数字和为 3 的倍数"两项检验。
3. Fractions, Ratio and Proportion in Context | 分数、比与比例的实际应用
A prize fund of £ en three students in the ratio 3 : 4 : 5. Add the parts: 3 + 4 + 5 = 12. One part is worth £180 ÷ 12 = £15, so the shares are 3 × £15 = £45, 4 × £15 = £60 and 5 × £15 = £75. Always check that the shares add back to the total: 45 + 60 + 75 = 180.
一笔 180 英镑的奖金按 3 : 4 : 5 的比例分给三位学生。先求份数总和：3 + 4 + 5 = 12。每份价值 180 ÷ 12 = 15 英镑，所以三人分别得到 3 × 15 = 45、4 × 15 = 60、5 × 15 = 75 英镑。一定要检验各份之和是否等于总数：45 + 60 + 75 = 180。
g. Fraction-of-amount questions follow the same logic: three eighths of 240 is 240 ÷ 8 × 3 = 90.
比例也常以配方缩放的形式出现。若一份供 4 人食用的配方需要 250 克面粉，那么供 10 人食用时的倍数为 10 ÷ 4 = 2.5，得到 250 × 2.5 = 625 克。分数求值类题目逻辑相同：240 的八分之三是 240 ÷ 8 × 3 = 90。
proportional to x, then y = kx, and the constant k is found from one known pair of values.
正比例可以写成一个等式。若 y 与 x 成正比，则 y = kx，其中常数 k 可由一组已知数值确定。
4. Algebra: Forming and Solving Equations | 代数：列方程与解方程
hich simplifies to 4w + 6 = 46 and then to 4w = 40, so w = 10. The rectangle is 10 cm by 13 cm and its area is 10 × 13 = 130 cm².
一个长方形的长比宽多 3 厘米，周长是 46 厘米。设宽为 w 厘米，则长为 (w + 3) 厘米。由周长得 2(w + w + 3) = 46，化简为 4w + 6 = 46，即 4w = 40，所以 w = 10。长方形为 10 厘米 × 13 厘米，面积为 10 × 13 = 130 平方厘米。
The discipline here is to define the variable in words before writing any algebra, and to answer the question actually asked. Many candidates solve for w and then forget that the question wanted the area.
这里的关键在于：动笔写代数式之前，先用文字定义变量；并且一定要回答题目真正问的东西。很多考生解出 w 之后就忘了题目要求的是面积。
× 4 × (−2) = 48 + 16 = 64. Watch the signs and apply the index before the multiplication.
代入求值题是相反的过程。若 a = 4，b = −2，则 3a² − 2ab = 3 × 16 − 2 × 4 × (−2) = 48 + 16 = 64。注意符号，并先算乘方再算乘法。
5. Sequences and Pattern Recognition | 数列与模式识别
Consider the sequence 5, 9, 13, 17, ... The term-to-term rule is "add 4" erm 5 shows that c = 1, so the nth term is 4n + 1. Check with n = 3: 4 × 3 + 1 = 13, which matches.
看数列 5, 9, 13, 17, …。相邻项关系是"每次加 4"，公差为 4。对于线性数列，第 n 项形如 dn + c，其中 d 是公差。把 dn = 4n 与首项 5 比较，可得 c = 1，所以第 n 项是 4n + 1。用 n = 3 检验：4 × 3 + 1 = 13，与数列相符。
Special sequences are worth memorising. Square numbers are n², giving 1, 4, 9, 16, ... Triangular numbers are n(n + 1) ÷ 2, giving 1, 3, 6, 10, ..., and the Fibonacci sequence starts 1, 1, 2, 3, 5, 8, 13, each term being the sum of the two before it.
特殊数列值得牢记。平方数是 n²，即 1, 4, 9, 16, …；三角形数是 n(n + 1) ÷ 2，即 1, 3, 6, 10, …；斐波那契数列以 1, 1, 2, 3, 5, 8, 13 开始，每一项都等于前两项之和。
then asks how many matches are needed for figure 20. Substituting n = 20 gives 61 matches.
典型的案例题会给出火柴棒图形，第 n 个图形用 3n + 1 根火柴，然后问第 20 个图形需要多少根。代入 n = 20 得 61 根。
6. Geometry: Angles, Triangles and Bearings | 几何：角、三角形与方位角
0° and 80°. This is a right scalene triangle only if one angle is 90°, which it is not - so resist the temptation to over-claim.
三角形的三个内角分别是 2x、3x 和 4x。由于三角形内角和为 180°，得 2x + 3x + 4x = 9x = 180，所以 x = 20，三个角为 40°、60° 和 80°。只有当其中一个角为 90° 时它才是直角三角形，而这里不是——所以不要过度断言。
Useful angle facts to keep at your fingertips: angles on a straight line sum to 180°; angles around a point sum to 360°; vertically opposite angles are equal; the exterior angle of a triangle equals the sum of the two opposite interior angles.
需要熟记的角度结论：同一直线上的角之和为 180°；围绕一点的角之和为 360°；对顶角相等；三角形的外角等于与它不相邻的两个内角之和。
en the bearing of A from B is 065° + 180° = 245°. Reverse bearings differ by exactly 180°.
方位角从正北方向顺时针测量，并且总写成三位数。若从 A 看 B 的方位角是 065°，那么从 B 看 A 的方位角是 065° + 180° = 245°。互反方位角恰好相差 180°。
7. Area, Perimeter and Composite Shapes | 面积、周长与复合图形
ius 4 cm, so its arc length is half the circumference: π × 4 = 12.568 cm, and its area is half of πr²: ½ × 3.142 × 16 = 25.136 cm².
一条跑道由一个 12 厘米 × 8 厘米的长方形，以及附在一条短边上的直径 8 厘米的半圆组成。取 π = 3.142。半圆半径为 4 厘米，因此弧长为圆周长的一半：π × 4 = 12.568 厘米；面积为 πr² 的一半：½ × 3.142 × 16 = 25.136 平方厘米。
Perimeter = 12 + 8 + 12 + πr = 32 + 12.568 ≈ 44.57 cm
Area = 12 × 8 + ½πr² = 96 + 25.136 ≈ 121.14 cm²
The exam skill is deciding which edges belong to the outside of the shape. The 8 cm side is covered by the semicircle, so it must not be counted in the perimeter, even though it is still inside the diagram.
考试技巧在于判断哪些边属于图形外边界。那条 8 厘米的边被半圆覆盖，因此不能计入周长，尽管它在图中依然画着。
Always state the value of π you used, give units, and round only at the end of the calculation. Premature rounding is one of the most common sources of lost marks.
一定要写明所用的 π 值、带上单位，并且只在计算最后一步取近似。过早取近似是失分最常见的原因之一。
8. Coordinates, Transformations and Straight-Line Graphs | 坐标、变换与直线图像
Find the equation of the line through A(1, 2) and B(5,
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