📚 IGCSE Maths: Step-by-Step Example Problems | IGCSE 数学：典型例题详解
Mastering IGCSE Mathematics requires more than just memorising formulas — it demands a clear understanding of how to apply concepts across a wide range of question types. In this article, we carefully work through twelve typical problems covering number, algebra, geometry, trigonometry, vectors, probability, statistics and sequences. Each example is broken down into logical steps, with bilingual explanations to strengthen both your technical vocabulary and mathematical reasoning. Use this as a revision companion to boost your confidence and exam readiness.
掌握 IGCSE 数学不仅需要记忆公式，更需要透彻理解如何在各种题型中灵活运用概念。本文精心选取了十二道典型例题，涵盖数、代数、几何、三角学、向量、概率、统计和数列。每个示例都被分解为清晰的步骤，并配有中英双语讲解，帮助你巩固专业词汇和数学推理能力。把本文作为复习伙伴，提升你的信心和应考状态。
1. Number and Powers | 数与幂
lify the expression (2⁻³ × 4²) / 8⁻¹. Give your answer as a fraction in its simplest form.
化简表达式 (2⁻³ × 4²) / 8⁻¹。将答案化为最简分数。
powers of 2. 4² = (2²)² = 2⁴; 8⁻¹ = (2³)⁻¹ = 2⁻³. The expression becomes (2⁻³ × 2⁴) / 2⁻³.
首先，将所有数字写成 2 的幂。4² = (2²)² = 2⁴；8⁻¹ = (2³)⁻¹ = 2⁻³。原式变为 (2⁻³ × 2⁴) / 2⁻³。
aw aᵐ × aⁿ = aᵐ⁺ⁿ on the numerator: 2⁻³ × 2⁴ = 2⁻³⁺⁴ = 2¹. The expression is now 2¹ / 2⁻³.
对分子使用指数法则 aᵐ × aⁿ = aᵐ⁺ⁿ：2⁻³ × 2⁴ = 2⁻³⁺⁴ = 2¹。现在式子为 2¹ / 2⁻³。
Apply the division law aᵐ / aⁿ = aᵐ⁻ⁿ: 2¹ / 2⁻³ = 2¹⁻⁽⁻³⁾ = 2¹⁺³ = 2⁴ = 16.
运用除法法则 aᵐ / aⁿ = aᵐ⁻ⁿ：2¹ / 2⁻³ = 2¹⁻⁽⁻³⁾ = 2¹⁺³ = 2⁴ = 16。
Since the answer is an integer, we can write it as 16/1 if a fraction is required. Thus the simplified value is 16.
因为答案是整数，若要求分数形式可写作 16/1。因此化简结果为 16。
2. Expanding and Factorising | 展开与因式分解
Expand and simplify (2x - 3)(x + 5) - (x - 2)².
展开并化简 (2x - 3)(x + 5) - (x - 2)²。
product: (2x - 3)(x + 5) = 2x·x + 2x·5 - 3·x - 3·5 = 2x² + 10x - 3x - 15 = 2x² + 7x - 15.
展开第一个乘积：(2x - 3)(x + 5) = 2x·x + 2x·5 - 3·x - 3·5 = 2x² + 10x - 3x - 15 = 2x² + 7x - 15。
Expand the square: (x - 2)² = (x - 2)(x - 2) = x² - 2x - 2x + 4 = x² - 4x + 4.
展开平方：(x - 2)² = (x - 2)(x - 2) = x² - 2x - 2x + 4 = x² - 4x + 4。
ond expansion from the first: 2x² + 7x - 15 - (x² - 4x + 4) = 2x² + 7x - 15 - x² + 4x - 4.
现在从第一个展开式中减去第二个：2x² + 7x - 15 - (x² - 4x + 4) = 2x² + 7x - 15 - x² + 4x - 4。
: 2x² - x² = x²; 7x + 4x = 11x; -15 - 4 = -19. The simplified expression is x² + 11x - 19.
合并同类项：2x² - x² = x²；7x + 4x = 11x；-15 - 4 = -19。化简后的表达式为 x² + 11x - 19。
3. Solving Linear Equations | 解一元一次方程
Solve for x: 3(2x - 1) - 5 = 4x + 7.
解方程：3(2x - 1) - 5 = 4x + 7。
6x - 3. So the equation becomes 6x - 3 - 5 = 4x + 7, which simplifies to 6x - 8 = 4x + 7.
展开左边：3(2x - 1) = 6x - 3。因此方程变为 6x - 3 - 5 = 4x + 7，化简得 6x - 8 = 4x + 7。
ide and constants on the other: subtract 4x from both sides: 6x - 4x - 8 = 7 → 2x - 8 = 7.
将含 x 的项移到一边，常数项移到另一边：两边同减 4x，得 6x - 4x - 8 = 7 → 2x - 8 = 7。
Add 8 to both sides: 2x = 15. Divide by 2: x = 15/2 or 7.5.
两边同加 8：2x = 15。除以 2：x = 15/2 或 7.5。
- 1) - 5 = 3(15 - 1) - 5 = 3×14 - 5 = 42 - 5 = 37; RHS = 4×7.5 + 7 = 30 + 7 = 37. Correct.
检验：左边 = 3(2×7.5 - 1) - 5 = 3(15 - 1) - 5 = 3×14 - 5 = 42 - 5 = 37；右边 = 4×7.5 + 7 = 30 + 7 = 37。正确。
4. Quadratic Equations by Factorisation | 因式分解解二次方程
Solve x² - 5x + 6 = 0.
解方程 x² - 5x + 6 = 0。
tiply to +6 and add to -5. The numbers are -2 and -3 because (-2)×(-3)=6 and (-2)+(-3)=-5.
我们需要两个数，它们的积为 +6，和为 -5。这两个数是 -2 和 -3，因为 (-2)×(-3)=6 且 (-2)+(-3)=-5。
Factorise the quadratic: (x - 2)(x - 3) = 0.
因式分解二次式：(x - 2)(x - 3) = 0。
Set each factor equal to zero: x - 2 = 0 or x - 3 = 0. Therefore x = 2 or x = 3.
令每个因式等于零：x - 2 = 0 或 x - 3 = 0。因此 x = 2 或 x = 3。
5. Linear Inequalities | 一元一次不等式
Solve the inequality 2x - 5 < 5x + 4 and represent the solution on a number line.
解不等式 2x - 5 < 5x + 4，并在数轴上表示解集。
Subtract 2x from both sides: -5 < 3x + 4. Then subtract 4: -9 < 3x. Divide by 3: -3 < x, which is equivalent to x > -3.
两边同减 2x：-5 < 3x + 4。再减 4：-9 < 3x。除以 3：-3 < x，等价于 x > -3。
The solution is all real numbers greater than -3. On a number line, we draw an open circle at -3 and shade the line to the right, with an arrow indicating all values greater than -3.
解集为所有大于 -3 的实数。在数轴上，在 -3 处画空心圆，并向右画阴影和箭头，表示所有大于 -3 的值。
6. Straight Line Graphs | 直线图像
Find the gradient and the y-intercept of the line 2y = 4x + 3.
求直线 2y = 4x + 3 的斜率和 y 截距。
c, where m is the gradient and c is the y-intercept. Divide both sides by 2: y = 2x + 3/2.
将方程化为 y = mx + c 的形式，其中 m 是斜率，c 是 y 截距。两边同除以 2：y = 2x + 3/2。
+ c: gradient m = 2; y-intercept c = 3/2 or 1.5. The line crosses the y-axis at (0, 1.5).
与 y = mx + c 对比：斜率 m = 2；y 截距 c = 3/2 即 1.5。该直线与 y 轴交于点 (0, 1.5)。
A larger gradient means the line is steeper. Here, for every 1 unit increase in x, y increases by 2 units.
斜率越大，直线越陡。此处，x 每增加 1 单位，y 增加 2 单位。
7. Basic Trigonometry | 基础三角学
In a right-angled triangle, sin θ = 0.6. Find the exact value of cos θ.
在一个直角三角形中，sin θ = 0.6。求 cos θ 的精确值。
identity sin²θ + cos²θ = 1. Substitute sin θ = 0.6: (0.6)² + cos²θ = 1 → 0.36 + cos²θ = 1.
利用恒等式 sin²θ + cos²θ = 1。代入 sin θ = 0.6：(0.6)² + cos²θ = 1 → 0.36 + cos²θ = 1。
So cos²θ = 1 - 0.36 = 0.64. Therefore cos θ = √0.64 = 0.8 (taking the positive root as θ is acute in a right-angled triangle).
所以 cos²θ = 1 - 0.36 = 0.64。因此 cos θ = √0.64 = 0.8（取正根，因为在直角三角形中 θ 为锐角）。
riangle: if sin θ = opposite/hypotenuse = 3/5 = 0.6, then adjacent/hypotenuse = 4/5 = 0.8.
或者，联想 3-4-5 三角形：若 sin θ = 对边/斜边 = 3/5 = 0.6，则邻边/斜边 = 4/5 = 0.8。
8. Vector Arithmetic | 向量运算
Given a = (3, -2) and b = (-1, 4), find the vector 2a - b and its magnitude.
已知 a = (3, -2)，b = (-1, 4)，求向量 2a - b 及其模长。
-2) = (6, -4). Then subtract b: 2a - b = (6, -4) - (-1, 4) = (6 - (-1), -4 - 4) = (7, -8).
先计算 2a：2a = 2×(3, -2) = (6, -4)。然后减去 b：2a - b = (6, -4) - (-1, 4) = (6 - (-1), -4 - 4) = (7, -8)。
The magnitude of a vector (x, y) is √x² + y². So |2a - b| = √(7² + (-8)²) = √49 + 64 = √113. This cannot be simplified further and is left in surd form.
向量 (x, y) 的模长为 √x² + y²。故 |2a - b| = √(7² + (-8)²) = √49 + 64 = √113。这已经是最简根式形式。
9. Probability | 概率
A bag contains 3 red, 5 blue and 2 green balls. One ball is picked at random. Find the probability that it is red or blue.
一个袋子里有 3 个红球、5 个蓝球和 2 个绿球。随机抽取一个球，求抽到红球或蓝球的概率。
Total number of balls = 3 + 5 + 2 = 10. The event 'red or blue' includes 3 + 5 = 8 favourable outcomes.
球的总数 = 3 + 5 + 2 = 10。事件“红或蓝”包含 3 + 5 = 8 个有利结果。
Probability = (number of favourable outcomes) / (total outcomes) = 8/10 = 4/5 or 0.8.
概率 = (有利结果数) / (总结果数) = 8/10 = 4/5 即 0.8。
This is a simple 'or' probability where the events are mutually exclusive: P(red or blue) = P(red) + P(blue) = 3/10 + 5/10 = 8/10.
这是一个简单的“或”概率问题，且事件互斥：P(红或蓝) = P(红) + P(蓝) = 3/10 + 5/10 = 8/10。
10. Descriptive Statistics | 描述性统计
For the data set 4, 7, 2, 9, 7, 3, 7, find the mean, median, mode and range.
对于数据集 4, 7, 2, 9, 7, 3, 7，求平均数、中位数、众数和极差。
Sum of values = 4+7+2+9+7+3+7 = 39. Number of values = 7. Mean = 39/7 ≈ 5.57 (to 2 d.p.).
总和 = 4+7+2+9+7+3+7 = 39。数据个数 = 7。平均数 = 39/7 ≈ 5.57（保留两位小数）。
Arrange in order: 2, 3, 4, 7, 7, 7, 9. Median is the middle value, the 4th number: 7.
按顺序排列：2, 3, 4, 7, 7, 7, 9。中位数是中间的值，即第 4 个数：7。
Mode is the most frequent value: 7 appears three times, so mode = 7. Range = maximum – minimum = 9 – 2 = 7.
众数是出现次数最多的值：7 出现了三次，因此众数 = 7。极差 = 最大值 – 最小值 = 9 – 2 = 7。
Notice how the mean is lower than the median and mode because of the low value 2 pulling it down.
注意到平均数低于中位数和众数，这是因为较小的值 2 拉低了平均数。
11. Transformations | 图形变换
Triangle A has vertices (1,2), (3,2), (1,5). Triangle A is reflected in the y-axis. Find the coordinates of the vertices of the image.
三角形 A 的顶点为 (1,2)、(3,2)、(1,5)。将三角形 A 关于 y 轴反射，求像的顶点坐标。
nges the sign of the x-coordinate while the y-coordinate stays the same: (x, y) → (-x, y).
关于 y 轴反射会改变 x 坐标的符号，y 坐标保持不变：(x, y) → (-x, y)。
; (3,2) → (-3,2); (1,5) → (-1,5). Therefore the image has vertices (-1,2), (-3,2), (-1,5).
对每个顶点进行变换：(1,2) → (-1,2)；(3,2) → (-3,2)；(1,5) → (-1,5)。因此像的顶点坐标为 (-1,2)、(-3,2)、(-1,5)。
is preserved but orientation is reversed. The right-angled corner moves from x=1 to x=-1.
检验：图形全等，到 y 轴的距离保持不变，但朝向相反。直角顶点从 x=1 移至 x=-1。
12. Arithmetic Sequences | 等差数列
Find the nth term of the sequence: 5, 9, 13, 17, 21, …
求数列 5, 9, 13, 17, 21, … 的第 n 项。
Identify the first term a = 5 and the common difference d = 9 - 5 = 4.
确定首项 a = 5，公差 d = 9 - 5 = 4。
uence is given by aₙ = a + (n - 1)d. Substitute: aₙ = 5 + (n - 1)×4 = 5 + 4n - 4 = 4n + 1.
等差数列的第 n 项公式为 aₙ = a + (n - 1)d。代入：aₙ = 5 + (n - 1)×4 = 5 + 4n - 4 = 4n + 1。
5; n=2: 4×2+1=9; n=3: 13. The formula works. Use it to find e.g. the 20th term: 4×20+1=81.
检验：n=1 时 a₁ = 4×1 + 1 = 5；n=2：4×2+1=9；n=3：13。公式正确。可用其求第 20 项：4×20+1=81。
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