📚 GCSE CIE Maths: Worked Examples | GCSE CIE 数学：典型例题详解
This article presents a collection of carefully chosen worked examples for the CIE IGCSE Mathematics (0580) syllabus. Each problem is solved step by step, with clear reasoning in both English and Chinese to help you master the techniques required in the exam. Topics covered include algebra, geometry, probability, statistics, and more.
本文精选了 CIE IGCSE 数学（0580）课程中的典型例题，逐一进行详细解析。每道题都配有清晰的步骤和逻辑说明，英中双语对照，帮助你掌握考试所需的核心方法。内容涵盖代数、几何、概率、统计等主要板块。
1. Solving Linear Equations | 解一元一次方程
Example: Solve 3(2x − 5) + 4 = 2x + 7.
例题：解方程 3(2x − 5) + 4 = 2x + 7。
First, expand the bracket: 6x − 15 + 4 = 2x + 7.
首先展开括号：6x − 15 + 4 = 2x + 7。
Simplify the left side: 6x − 11 = 2x + 7.
化简左边：6x − 11 = 2x + 7。
Subtract 2x from both sides: 4x − 11 = 7. Then add 11 to both sides: 4x = 18, so x = 4.5.
两边减去 2x：4x − 11 = 7。两边再加 11：4x = 18，因此 x = 4.5。
Always check by substituting back: 3(2×4.5−5)+4 = 3(9−5)+4 = 3×4+4=12+4=16; right side: 2×4.5+7=9+7=16. Correct.
务必代回检验：左式 3(2×4.5−5)+4 = 3×4+4=16，右式 2×4.5+7=16，正确。
2. Factorising Quadratic Expressions | 二次三项式的因式分解
Example: Factorise x² − 7x + 12.
例题：分解因式 x² − 7x + 12。
We need two numbers that multiply to +12 and add to −7. The pairs for 12 are (1,12), (2,6), (3,4). To get a negative sum, both must be negative: (−3) + (−4) = −7 and (−3)×(−4) = 12.
我们需要找到两个数，乘积为 +12，和为 −7。12 的因数对有 (1,12)、(2,6)、(3,4)。要使和为负数，两个数都应为负：(−3) + (−4) = −7，且 (−3)×(−4) = 12。
Therefore, x² − 7x + 12 = (x − 3)(x − 4).
因此，x² − 7x + 12 = (x − 3)(x − 4)。
You can verify by expanding: (x−3)(x−4) = x² −4x −3x +12 = x² −7x +12.
可通过展开验证：(x−3)(x−4) = x² −4x −3x +12 = x² −7x +12。
3. Angle Properties in Parallel Lines | 平行线中的角度性质
Example: In the diagram below, AB is parallel to CD. Angle EFB = 72°. Find angle FGD.
例题：下图中，AB∥CD，∠EFB = 72°。求 ∠FGD。
Since AB ∥ CD, the alternate interior angles are equal. Angle EFB and angle FGD are alternate angles. Thus, ∠FGD = 72°.
由于 AB∥CD，内错角相等。∠EFB 和 ∠FGD 是一对内错角，因此 ∠FGD = 72°。
If a question involves corresponding or co-interior angles, apply the same principle: corresponding angles are equal; co-interior angles sum to 180°.
如果题目涉及同位角或同旁内角，应用同样的原理：同位角相等；同旁内角互补，和为 180°。
4. Circle Theorem: Angle at the Centre | 圆定理：圆心角与圆周角
Example: Points A, B, C lie on a circle with centre O. ∠AOB = 116°. Find ∠ACB.
例题：A、B、C 在圆心为 O 的圆上，∠AOB = 116°。求 ∠ACB。
The angle at the centre is twice any angle at the circumference subtended by the same arc. Here, arc AB subtends ∠AOB at the centre and ∠ACB at the circumference.
圆心角等于同弧所对的圆周角的两倍。此处，弧 AB 对圆心角 ∠AOB，对圆周角 ∠ACB。
So ∠ACB = ½ × ∠AOB = ½ × 116° = 58°.
因此 ∠ACB = ½ × 116° = 58°。
Remember this theorem applies only when the angle at the circumference is subtended by the same arc as the central angle.
记住，这一定理仅在圆周角与圆心角为同一弧所对时适用。
5. Mean, Median, Mode, and Range | 平均数、中位数、众数和极差
Example: Find the mean, median, mode, and range for the data set: 8, 12, 9, 12, 7, 14, 10.
例题：求数据集 8, 12, 9, 12, 7, 14, 10 的平均数、中位数、众数和极差。
First, order the data: 7, 8, 9, 10, 12, 12, 14. There are 7 values.
首先将数据排序：7, 8, 9, 10, 12, 12, 14，共 7 个数值。
Mean = (7+8+9+10+12+12+14) ÷ 7 = 72 ÷ 7 ≈ 10.3.
平均数 = (7+8+9+10+12+12+14) ÷ 7 = 72 ÷ 7 ≈ 10.3。
Median is the 4th value: 10.
中位数为第 4 个值：10。
Mode is the most frequent value: 12.
众数为出现最频繁的值：12。
Range = maximum − minimum = 14 − 7 = 7.
极差 = 最大值 − 最小值 = 14 − 7 = 7。
6. Probability with Tree Diagrams | 树状图与概率
Example: A bag contains 4 red and 6 blue balls. Two balls are drawn without replacement. Find the probability that both are red.
例题：一个袋子里有 4 个红球和 6 个蓝球。依次摸出两个球，不放回。求两个都是红球的概率。
First draw: P(red) = 4/10 = 2/5. Second draw: if first was red, remaining are 3 red out of 9 balls, so P(red second | red first) = 3/9 = 1/3.
第一次摸球：P(红) = 4/10 = 2/5。第二次摸球：如果第一次是红球，则剩下 9 个球中有 3 个红球，P(第二次红|第一次红) = 3/9 = 1/3。
Multiply along the branch: P(both red) = (2/5) × (1/3) = 2/15.
沿分支相乘：P(两红) = (2/5) × (1/3) = 2/15。
Always draw a tree diagram to visualise the outcomes, especially for ‘without replacement’ scenarios.
遇到不放回情形，务必画出树状图以理清所有结果。
7. Direct and Inverse Proportion | 正比例与反比例
Example: y is directly proportional to x². When x = 4, y = 80. Find y when x = 7.
例题：y 与 x² 成正比。当 x = 4 时，y = 80。求 x = 7 时的 y 值。
Write the equation: y = k x². Substitute given values: 80 = k × 4² ⇒ 80 = 16k ⇒ k = 5.
写出方程：y = k x²。代入已知值：80 = k × 4² ⇒ 80 = 16k ⇒ k = 5。
Now use k to find y when x = 7: y = 5 × 7² = 5 × 49 = 245.
利用 k 求 x = 7 时的 y 值：y = 5 × 7² = 5 × 49 = 245。
If y is inversely proportional to x, use y = k/x. Always find k first.
若 y 与 x 成反比，则用 y = k/x。解题时先求出常数 k。
8. Nth Term of a Sequence | 数列的第 n 项
Example: Find the nth term of the sequence: 5, 9, 13, 17, 21, …
例题：求数列 5, 9, 13, 17, 21, … 的通项公式。
This is an arithmetic sequence. Find the common difference: 9 − 5 = 4, 13 − 9 = 4, etc. So d = 4.
这是一个等差数列。找出公差：9 − 5 = 4，13 − 9 = 4，因此 d = 4。
The nth term formula is aₙ = a₁ + (n−1)d. Here a₁ = 5, so aₙ = 5 + (n−1)×4 = 5 + 4n − 4 = 4n + 1.
通项公式为 aₙ = a₁ + (n−1)d。这里 a₁ = 5，所以 aₙ = 5 + (n−1)×4 = 5 + 4n − 4 = 4n + 1。
Check: when n=1, 4×1+1=5; n=2, 4×2+1=9, correct.
检验：n=1 时，4×1+1=5；n=2 时，4×2+1=9，正确。
9. Inequalities and Number Lines | 不等式与数轴表示
Example: Solve 3x + 4 > 2x − 1 and represent the solution on a number line.
例题：解不等式 3x + 4 > 2x − 1，并在数轴上表示解集。
Subtract 2x from both sides: x + 4 > −1. Then subtract 4: x > −5.
两边减去 2x：x + 4 > −1。再减去 4：x > −5。
On a number line, draw an open circle at −5 and an arrow pointing to the right, indicating all numbers greater than −5.
在数轴上，在 −5 处画一个空心圆，并画指向右边的箭头，表示所有大于 −5 的数。
Remember: open circle for < or >; closed circle for ≤ or ≥.
记住：< 或 > 用空心圆；≤ 或 ≥ 用实心圆。
10. Area and Perimeter of Compound Shapes | 组合图形的面积与周长
Example: A shape is made from a rectangle 8 cm by 5 cm and a semicircle on one of the shorter sides. Calculate the total area and perimeter. Take π = 3.14.
例题：一个图形由一个 8 cm × 5 cm 的矩形和一个以较短边为直径的半圆组成。计算总面积和总周长。（π 取 3.14）
The shorter side is 5 cm, so the semicircle’s diameter is 5 cm, radius r = 2.5 cm.
较短边为 5 cm，因此半圆的直径为 5 cm，半径 r = 2.5 cm。
Area of rectangle = 8 × 5 = 40 cm². Area of semicircle = ½ × π × r² = ½ × 3.14 × 2.5² = ½ × 3.14 × 6.25 = 9.8125 cm². Total area ≈ 49.81 cm².
矩形面积 = 8 × 5 = 40 cm²。半圆面积 = ½ × π × r² = ½ × 3.14 × 2.5² = 9.8125 cm²。总面积约为 49.81 cm²。
Perimeter: three sides of rectangle (two lengths and one width) + arc length of semicircle. Arc length = π × r = 3.14 × 2.5 = 7.85 cm. Perimeter = 8 + 8 + 5 + 7.85 = 28.85 cm.
周长：矩形的三条边（两条长边和一条宽边）加上半圆弧长。弧长 = π × r = 3.14 × 2.5 = 7.85 cm。周长 = 8 + 8 + 5 + 7.85 = 28.85 cm。
Always separate compound shapes into basic parts and then add areas, being careful not to double-count edges for perimeter.
处理组合图形时，先拆分为基本形状再求和；计算周长时注意不要重复计算公共边。
11. Transformations: Enlargement | 变换：位似放大
Example: Triangle ABC has coordinates A(2,1), B(4,1), C(3,4). Enlarge by scale factor 3 with centre (0,0). Find the image coordinates.
例题：三角形 ABC 的顶点坐标为 A(2,1), B(4,1), C(3,4)。以原点为位似中心、比例因子 3 进行放大。求像点坐标。
When the centre is the origin, multiply each coordinate by the scale factor. A’ = (2×3, 1×3) = (6,3); B’ = (12,3); C’ = (9,12).
当位似中心为原点时，各坐标乘以比例因子。A’ = (6,3)；B’ = (12,3)；C’ = (9,12)。
If the centre is not the origin, draw rays from the centre and use the scale factor to find distances.
若位似中心不是原点，则从中心向各点引射线，按比例因子计算距离。
12. Simultaneous Equations | 联立方程组
Example: Solve the simultaneous equations: 2x + y = 10 and x − y = 2.
例题：解方程组 2x + y = 10 和 x − y = 2。
Add the two equations to eliminate y: (2x + y) + (x − y) = 10 + 2 ⇒ 3x = 12 ⇒ x = 4.
将两式相加以消去 y：(2x + y) + (x − y) = 12 ⇒ 3x = 12 ⇒ x = 4。
Substitute x = 4 into the second equation: 4 − y = 2 ⇒ y = 2.
将 x = 4 代入第二个方程：4 − y = 2 ⇒ y = 2。
Check with the first equation: 2×4 + 2 = 8 + 2 = 10. Solution: x = 4, y = 2.
用第一个方程检验：2×4 + 2 = 10。解为 x = 4，y = 2。
You can also use substitution or matrix methods, but elimination is often fastest for linear equations.
也可使用代入法或矩阵解法，但对于一次方程组，消元法通常最快捷。
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