IGCSE CIE Mathematics: Detailed Example Problem Walkthrough | IGCSE CIE 数学：典型例题详解

📚 IGCSE CIE Mathematics: Detailed Example Problem Walkthrough | IGCSE CIE 数学：典型例题详解

This article presents a carefully selected set of typical IGCSE CIE Mathematics problems, each solved step by step with clear reasoning. Working through these examples will strengthen your problem-solving skills and build confidence for the examination. Every solution is explained in both English and Chinese to support bilingual learners and deepen conceptual understanding.

本文精选了一组典型的 IGCSE CIE 数学题目，并对每道题进行逐步解析和清晰推理。通过研习这些例题，你将提升解题能力，并为考试树立信心。所有解答均以中英双语呈现，以便双语学习者理解并深化概念掌握。

1. Solving Linear Equations | 解一元一次方程

Problem: Solve 4x − 7 = 2x + 9.

问题：解方程 4x − 7 = 2x + 9。

Step 1: Collect the x‑terms on one side. Subtract 2x from both sides: 4x − 2x − 7 = 9, which gives 2x − 7 = 9.

第1步：将含 x 的项移到一边。两边同时减去 2x：4x − 2x − 7 = 9，得到 2x − 7 = 9。

Step 2: Isolate the term with x. Add 7 to both sides: 2x = 16.

第2步：分离含 x 的项。两边加 7：2x = 16。

Step 3: Divide both sides by 2: x = 8.

第3步：两边除以 2：x = 8。

x = 8

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2. Quadratic Equations by Factoring | 因式分解法解二次方程

Problem: Solve x² − 6x + 8 = 0.

问题：解方程 x² − 6x + 8 = 0。

y to +8 and add to −6: −2 and −4. Therefore, the expression factors as (x − 2)(x − 4) = 0.

寻找两个数，乘积为 +8，和为 −6：−2 与 −4。因此原式可分解为 (x − 2)(x − 4) = 0。

Set each factor equal to zero: x − 2 = 0 → x = 2, or x − 4 = 0 → x = 4.

令每个因式为零：x − 2 = 0 ⇒ x = 2，或 x − 4 = 0 ⇒ x = 4。

x = 2 or x = 4

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3. Simultaneous Linear Equations | 联立一次方程组

Problem: Solve the system: 3x + 2y = 11, 2x − y = 3.

问题：解方程组：3x + 2y = 11，2x − y = 3。

Method: elimination. Multiply the second equation by 2 to align the y‑coefficients: 4x − 2y = 6.

解法：消元法。将第二个方程乘以 2，使 y 的系数对齐：4x − 2y = 6。

Add the new equation to the first: (3x + 2y) + (4x − 2y) = 11 + 6 → 7x = 17, so x = 17/7.

将新方程与第一个方程相加：(3x + 2y) + (4x − 2y) = 11 + 6 ⇒ 7x = 17，因此 x = 17/7。

tute x = 17/7 into 2x − y = 3: 2(17/7) − y = 3 → 34/7 − y = 21/7 → y = 34/7 − 21/7 = 13/7.

将 x = 17/7 代入 2x − y = 3：2×(17/7) − y = 3 ⇒ 34/7 − y = 21/7 ⇒ y = 34/7 − 21/7 = 13/7。

x = 17/7, y = 13/7

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4. Sequences and the nth Term | 数列与第 n 项

Problem: Find the nth term of the linear sequence: 7, 11, 15, 19, …

问题：求一次序列的第 n 项：7，11，15，19，…

ence d = 11 − 7 = 4. The nth term formula for an arithmetic sequence is uₙ = a + (n − 1)d.

首项 a = 7。公差 d = 11 − 7 = 4。等差数列第 n 项公式为 uₙ = a + (n − 1)d。

Substitute: uₙ = 7 + (n − 1)×4 = 7 + 4n − 4 = 4n + 3.

代入：uₙ = 7 + (n − 1)×4 = 7 + 4n − 4 = 4n + 3。

nth term = 4n + 3

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5. Functions and Inverse Functions | 函数与反函数

Problem: f(x) = (3x − 6)/2. Find f⁻¹(x).

问题：f(x) = (3x − 6)/2，求 f⁻¹(x)。

Write y = (3x − 6)/2. To find the inverse, swap x and y: x = (3y − 6)/2.

记 y = (3x − 6)/2。求反函数时，交换 x 与 y：x = (3y − 6)/2。

multiply both sides by 2 → 2x = 3y − 6. Add 6 → 3y = 2x + 6. Divide by 3 → y = (2x + 6)/3.

解出 y：两边乘 2 ⇒ 2x = 3y − 6。加 6 ⇒ 3y = 2x + 6。除以 3 ⇒ y = (2x + 6)/3。

Thus f⁻¹(x) = (2x + 6)/3, which simplifies to 2(x + 3)/3.

因此 f⁻¹(x) = (2x + 6)/3，可简化为 2(x + 3)/3。

f⁻¹(x) = (2x + 6)/3

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6. Circle Theorems: Angle at the Centre | 圆定理：圆心角定理

Problem: Points A, B and C lie on a circle with centre O. ∠ABC = 35°. Find ∠AOC, giving a reason.

问题：A、B、C 三点在圆心为 O 的圆上，∠ABC = 35°。求 ∠AOC，并说明理由。

The angle at the centre is twice the angle at the circumference when both subtend the same arc AC. Here ∠ABC is at the circumference, subtending arc AC, and ∠AOC is the corresponding angle at the centre.

圆心角等于同弧所对圆周角的两倍。此处 ∠ABC 为圆周角，对应弧 AC；∠AOC 为同弧所对的圆心角。

Therefore, ∠AOC = 2 × ∠ABC = 2 × 35° = 70°.

因此，∠AOC = 2 × ∠ABC = 2 × 35° = 70°。

∠AOC = 70°

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7. Transformations: Enlargement | 变换：放大

Problem: Enlarge triangle PQR with vertices P(1,2), Q(3,1), R(2,4) by scale factor 2, centre (0,0). Give the coordinates of the image.

问题：以原点为中心、放大因子 2 放大 △PQR，顶点为 P(1,2)，Q(3,1)，R(2,4)。写出像的坐标。

For an enlargement centre (0,0), multiply each coordinate by the scale factor. P' = (1×2, 2×2) = (2,4). Q' = (3×2, 1×2) = (6,2). R' = (2×2, 4×2) = (4,8).

以原点为放大中心时，将各坐标乘以放大因子。P' = (1×2, 2×2) = (2,4)；Q' = (3×2, 1×2) = (6,2)；R' = (2×2, 4×2) = (4,8)。

P'(2,4), Q'(6,2), R'(4,8)

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8. Probability Tree Diagrams | 概率树图

Problem: A bag contains 4 red and 6 blue balls. Two balls are drawn without replacement. Find the probability that both balls are red.

问题：袋中有 4 个红球和 6 个蓝球。不放回地抽取两次。求两次都抽到红球的概率。

(red) = 4/10 = 2/5. If red is drawn first, 3 red remain out of 9 balls, so P(red on second | red first) = 3/9 = 1/3.

画树图。第一次抽球：P(红) = 4/10 = 2/5。若第一次抽到红球，剩下 3 红共 9 球，故 P(第二次红 | 第一次红) = 3/9 = 1/3。

Multiply along the branches: P(both red) = (2/5) × (1/3) = 2/15.

沿树枝相乘：P(两红) = (2/5) × (1/3) = 2/15。

P(both red) = 2/15

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9. Statistics: Central Tendency and Spread | 统计：集中趋势与离散程度

Problem: The ages in a club are: 14, 15, 12, 15, 16, 14, 14, 13. Find the mean, median, mode, and range.

问题：某俱乐部成员年龄为：14, 15, 12, 15, 16, 14, 14, 13。求平均数、中位数、众数和极差。

n = sum ÷ count. Sum = 14+15+12+15+16+14+14+13 = 113, count = 8, so mean = 113/8 = 14.125.

平均数 = 总和 ÷ 个数。总和 = 113，个数 = 8，故平均数 = 113/8 = 14.125。

he 4th and 5th values: (14+14)/2 = 14. Mode is the most frequent: 14. Range = 16 − 12 = 4.

排序：12, 13, 14, 14, 14, 15, 15, 16。中位数是第4与第5个值的平均：(14+14)/2 = 14。众数为出现最多的值：14。极差 = 16 − 12 = 4。

Mean = 14.125, Median = 14, Mode = 14, Range = 4

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10. Trigonometry in Right‑Angled Triangles | 直角三角形的三角比

Problem: In a right‑angled triangle, the hypotenuse is 15 cm and one acute angle is 40°. Find the length of the side opposite the 40° angle to 1 decimal place.

问题：一直角三角形斜边为 15 cm，一锐角为 40°。求 40° 角所对的边长，保留一位小数。

site the 40° angle and the hypotenuse are linked by sine: sin 40° = opposite / hypotenuse.

使用 SOH CAH TOA。40° 角的对边与斜边用正弦联系：sin 40° = 对边 / 斜边。

15 × sin 40°. Using a calculator, sin 40° ≈ 0.6428, so opposite ≈ 15 × 0.6428 = 9.642 cm.

故对边 = 15 × sin 40°。计算器得 sin 40° ≈ 0.6428，对边 ≈ 15 × 0.6428 = 9.642 cm。

To 1 decimal place, the side length is 9.6 cm.

取一位小数，边长为 9.6 cm。

Opposite length ≈ 9.6 cm

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11. Vectors: Addition and Magnitude | 向量：加法与模

Problem: b = (−1, 6), find the vector a + 2b and the magnitude of a + 2b to 3 significant figures.

问题：已知向量 a = (3, −4)，b = (−1, 6)。求向量 a + 2b 及其模（保留三位有效数字）。

First compute 2b: 2 × (−1, 6) = (−2, 12). Then a + 2b = (3 + (−2), −4 + 12) = (1, 8).

先计算 2b：2 × (−1, 6) = (−2, 12)。然后 a + 2b = (3 + (−2), −4 + 12) = (1, 8)。

Magnitude of (1, 8) = √1² + 8² = √1 + 64 = √65 ≈ 8.0623. To 3 s.f., this is 8.06.

向量 (1, 8) 的模 = √1² + 8² = √1 + 64 = √65 ≈ 8.0623。四舍五入至三位有效数字，为 8.06。

a + 2b = (1, 8), magnitude ≈ 8.06

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12. Mensuration: Cylinder Volume and Surface Area | 测量：圆柱体积与表面积

Problem: = 5 cm and height h = 12 cm. Calculate its volume and total surface area. (Use π ≈ 3.142)

问题：一圆柱底面半径 r = 5 cm，高 h = 12 cm。计算其体积与总表面积。（取 π ≈ 3.142）

. Substitute: V = 3.142 × 5² × 12 = 3.142 × 25 × 12 = 3.142 × 300 = 942.6 cm³ (to 1 d.p.).

体积 V = πr²h。代入得 V = 3.142 × 5² × 12 = 3.142 × 25 × 12 = 3.142 × 300 = 942.6 cm³（保留一位小数）。

h + r). A = 2 × 3.142 × 5 × (12 + 5) = 31.42 × 17 = 534.14 cm², which rounds to 534.1 cm².

总表面积 A = 2πrh + 2πr² = 2πr(h + r)。A = 2 × 3.142 × 5 × (12 + 5) = 31.42 × 17 = 534.14 cm²，四舍五入为 534.1 cm²。

Volume ≈ 942.6 cm³, Surface area ≈ 534.1 cm²

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