📚 IGCSE Edexcel Maths: Typical Example Questions Explained | IGCSE Edexcel 数学：典型例题详解
Mastering IGCSE Edexcel Mathematics requires not only knowing the formulas but also understanding how to apply them in typical exam-style questions. In this article, we will walk through twelve carefully selected example questions that cover the key topics: Number, Algebra, Graphs, Geometry, Trigonometry, Statistics, and Probability. Each question is broken down step by step, with paired explanations in both English and Chinese to help you build confidence and accuracy for the real exam.
要掌握 IGCSE Edexcel 数学，不仅要熟记公式，更要懂得如何将其应用到典型的考试题型中。本文精选了 12 道典型例题，涵盖数、代数、图像、几何、三角学、统计和概率等核心板块。每道题都配有逐步解析，并以中英文对照的方式呈现，帮助你扎实提升解题信心与准确性。
1. Number: Prime Factors, HCF & LCM | 数：质因数、最大公因数与最小公倍数
Question: Find the Highest Common Factor (HCF) and Lowest Common Multiple (LCM) of 60 and 84 using prime factorisation.
问题：利用质因数分解求 60 和 84 的最大公因数（HCF）和最小公倍数（LCM）。
First, list the prime factors of each number. 60 = 2 × 2 × 3 × 5 = 22 × 3 × 5. 84 = 2 × 2 × 3 × 7 = 22 × 3 × 7.
首先，写出每个数的质因数分解。60 = 2 × 2 × 3 × 5 = 22 × 3 × 5。84 = 2 × 2 × 3 × 7 = 22 × 3 × 7。
For HCF, take the common prime factors with the lowest power. Both numbers have 22 and 3. So HCF = 22 × 3 = 4 × 3 = 12.
求 HCF 时，取共同质因数的最低次幂。两数都有 22 和 3，因此 HCF = 22 × 3 = 4 × 3 = 12。
For LCM, take all prime factors with the highest power. We have 22, 3, 5, and 7. LCM = 22 × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420.
求 LCM 时，取所有质因数的最高次幂。共有 22、3、5 和 7。LCM = 22 × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420。
Thus, HCF(60,84) = 12 and LCM(60,84) = 420. Always check that HCF × LCM = 60 × 84 = 5040, indeed 12 × 420 = 5040.
因此，HCF(60,84) = 12，LCM(60,84) = 420。记住验证 HCF × LCM = 60 × 84 = 5040，确实 12 × 420 = 5040。
2. Algebra: Solving Linear Equations | 代数：解一次方程
Question: Solve the equation 3(2x − 1) = 2(x + 5) − 7.
问题：解方程 3(2x − 1) = 2(x + 5) − 7。
Expand both sides: LHS = 6x − 3, RHS = 2x + 10 − 7 = 2x + 3. Now the equation is 6x − 3 = 2x + 3.
先去括号：左边 = 6x − 3，右边 = 2x + 10 − 7 = 2x + 3。现在方程为 6x − 3 = 2x + 3。
Collect x terms on one side: subtract 2x from both sides → 4x − 3 = 3. Then add 3 to both sides → 4x = 6.
将 x 项移到一边：两边减去 2x → 4x − 3 = 3。然后两边加 3 → 4x = 6。
Divide by 4: x = 6 ÷ 4 = 1.5. The solution is x = 1.5 or x = 3/2.
除以 4 得 x = 6 ÷ 4 = 1.5。解为 x = 1.5（或 x = 3/2）。
3. Inequalities: Solving and Graphing | 不等式：求解与数轴表示
Question: Solve the inequality 4x − 3 < 2x + 7 and illustrate the solution on a number line.
问题：解不等式 4x − 3 < 2x + 7，并将解在数轴上表示。
Subtract 2x from both sides: 2x − 3 < 7. Add 3 to both sides: 2x < 10. Divide by 2: x < 5.
两边减去 2x：2x − 3 < 7。两边加 3：2x < 10。除以 2：x < 5。
On a number line, draw an open circle at 5 and shade to the left, indicating all values less than 5.
在数轴上，在 5 处画空心圆并向左画阴影，表示所有小于 5 的值。
4. Sequences: nth Term of an Arithmetic Sequence | 数列：等差数列的第 n 项
Question: Find the nth term of the sequence: 5, 8, 11, 14, … Then calculate the 50th term.
问题：求数列 5, 8, 11, 14, … 的第 n 项表达式，并求出第 50 项。
The first term a = 5 and the common difference d = 3. The formula for the nth term is a + (n−1)d.
首项 a = 5，公差 d = 3。第 n 项公式为 a + (n−1)d。
So nth term = 5 + (n−1)×3 = 5 + 3n − 3 = 3n + 2.
因此，第 n 项 = 5 + (n−1)×3 = 5 + 3n − 3 = 3n + 2。
For the 50th term, substitute n = 50: 3×50 + 2 = 150 + 2 = 152.
求第 50 项时，代入 n = 50：3×50 + 2 = 150 + 2 = 152。
5. Straight Line Graphs: Equation from Two Points | 直线图像：由两点求方程
Question: Find the equation of the straight line passing through the points (1, 3) and (4, 9). Give your answer in the form y = mx + c.
问题：求经过点 (1, 3) 和 (4, 9) 的直线方程，以 y = mx + c 的形式给出。
Calculate the gradient m = (y2 − y1) / (x2 − x1) = (9 − 3) / (4 − 1) = 6 / 3 = 2.
计算斜率 m = (y2 − y1) / (x2 − x1) = (9 − 3) / (4 − 1) = 6 / 3 = 2。
Using y = mx + c with point (1, 3): 3 = 2×1 + c → 3 = 2 + c → c = 1. Equation: y = 2x + 1.
利用点 (1, 3) 代入 y = mx + c：3 = 2×1 + c → 3 = 2 + c → c = 1。方程为 y = 2x + 1。
You can check with the other point: when x = 4, y = 2×4 + 1 = 9, which matches.
可用另一点检验：当 x = 4 时，y = 2×4 + 1 = 9，吻合。
6. Quadratic Equations: Factorisation and Solutions | 二次方程：因式分解法求解
Question: Solve x2 − 5x + 6 = 0 by factorisation.
问题：用因式分解法解 x2 − 5x + 6 = 0。
We need two numbers that multiply to +6 and add to −5. The numbers are −2 and −3.
我们需要两个数，乘积为 +6，和为 −5。这两个数是 −2 和 −3。
Factorise: (x − 2)(x − 3) = 0. Therefore, either x − 2 = 0 or x − 3 = 0. So x = 2 or x = 3.
因式分解得 (x − 2)(x − 3) = 0。因此，x − 2 = 0 或 x − 3 = 0，故 x = 2 或 x = 3。
The graph of y = x2 − 5x + 6 is a U-shaped parabola crossing the x-axis at (2,0) and (3,0).
y = x2 − 5x + 6 的图像是一条 U 形抛物线，与 x 轴交于 (2,0) 和 (3,0)。
7. Simultaneous Equations: Elimination Method | 联立方程：加减消元法
Question: Solve the simultaneous equations: 2x + y = 7 and x − y = 2.
问题：解联立方程组：2x + y = 7 与 x − y = 2。
Add the two equations to eliminate y: (2x + y) + (x − y) = 7 + 2 → 3x = 9 → x = 3.
将两式相加消去 y：(2x + y) + (x − y) = 7 + 2 → 3x = 9 → x = 3。
Substitute x = 3 into x − y = 2: 3 − y = 2 → y = 1. Or into the first: 2(3) + y = 7 → 6 + y = 7 → y = 1.
把 x = 3 代入 x − y = 2：3 − y = 2 → y = 1。或代入第一式：2(3) + y = 7 → 6 + y = 7 → y = 1。
Solution: x = 3, y = 1. Always check both original equations.
解为 x = 3, y = 1。务必代入原方程组检验。
8. Angles in Polygons: Interior and Exterior Angles | 多边形中的角：内角与外角
Question: Each interior angle of a regular polygon is 156°. Find the number of sides.
问题：一个正多边形的每个内角为 156°，求其边数。
The exterior angle = 180° − interior angle = 180° − 156° = 24°.
外角 = 180° − 内角 = 180° − 156° = 24°。
For any polygon, the sum of exterior angles is 360°. Number of sides n = 360° ÷ exterior angle = 360° ÷ 24° = 15.
任何多边形的外角和为 360°。边数 n = 360° ÷ 外角 = 360° ÷ 24° = 15。
Thus the polygon has 15 sides. The sum of interior angles would be (15 − 2) × 180° = 2340°, and each interior angle 2340° ÷ 15 = 156°, confirming the result.
因此，这个多边形有 15 条边。内角和为 (15 − 2) × 180° = 2340°，每个内角 2340° ÷ 15 = 156°，验证一致。
9. Circle Theorems: Angle at Centre and Circumference | 圆定理：圆心角与圆周角
Question: In a circle, an arc subtends an angle of 40° at the circumference. What is the angle subtended by the same arc at the centre?
问题：在一个圆中，某段弧所对的圆周角为 40°，求该弧所对的圆心角。
The angle at the centre is twice the angle at the circumference, standing on the same arc.
在同弧上，圆心角等于圆周角的两倍。
Therefore, centre angle = 2 × 40° = 80°.
因此，圆心角 = 2 × 40° = 80°。
This is a direct application of a key circle theorem. Make sure to cite it: ‘The angle at the centre is twice any angle at the circumference subtended by the same arc.’
这是核心圆定理的直接应用。记得写明理由：“同弧所对的圆心角等于任一圆周角的两倍”。
10. Trigonometry: SOH CAH TOA | 三角学：SOH CAH TOA 法则
Question: In a right-angled triangle, the hypotenuse is 10 cm, and one acute angle is 30°. Find the length of the side opposite the 30° angle.
问题：在一个直角三角形中，斜边长为 10 cm，一个锐角为 30°。求 30° 角对边的长度。
Using sine: sin 30° = opposite / hypotenuse. So opposite = hypotenuse × sin 30°.
利用正弦：sin 30° = 对边 / 斜边。因此对边 = 斜边 × sin 30°。
We know sin 30° = 0.5. Thus opposite = 10 × 0.5 = 5 cm.
已知 sin 30° = 0.5，故对边 = 10 × 0.5 = 5 cm。
Always label the sides relative to the given angle. Here o = opposite, h = hypotenuse.
一定要根据已知角标注各边。此处 o 为对边，h 为斜边。
11. Probability: Without Replacement | 概率：不放回抽取
Question: A bag contains 4 red, 3 blue, and 2 green balls. Two balls are drawn at random without replacement. Find the probability that both balls are the same colour.
问题：一个袋子里有 4 个红球、3 个蓝球和 2 个绿球。随机不放回地抽取两个球。求两个球颜色相同的概率。
Total balls = 4 + 3 + 2 = 9. We consider three cases: both red, both blue, both green.
总球数 = 9。考虑三种情况：两红、两蓝、两绿。
P(both red) = (4/9) × (3/8) = 12/72. P(both blue) = (3/9) × (2/8) = 6/72. P(both green) = (2/9) × (1/8) = 2/72.
P(两红) = (4/9) × (3/8) = 12/72。P(两蓝) = (3/9) × (2/8) = 6/72。P(两绿) = (2/9) × (1/8) = 2/72。
Sum = 12/72 + 6/72 + 2/72 = 20/72, which simplifies to 5/18.
总概率 = 12/72 + 6/72 + 2/72 = 20/72，约分为 5/18。
Remember that without replacement means the denominators change after the first draw.
注意，不放回意味着第一次抽取后样本空间分母会变化。
12. Statistics: Mean from a Frequency Table | 统计：根据频数表求平均值
Question: The table shows the number of goals scored by a hockey team in 20 matches. Find the mean number of goals.
问题：下表显示了某曲棍球队在 20 场比赛中的进球数。求平均进球数。
| Goals (x) | Frequency (f) |
|---|---|
| 0 | 3 |
| 1 | 7 |
| 2 | 6 |
| 3 | 4 |
First, add an fx column: 0×3 = 0, 1×7 = 7, 2×6 = 12, 3×4 = 12. Total fx = 0+7+12+12 = 31. Total frequency = 20.
首先，增加 fx 一列：0×3 = 0, 1×7 = 7, 2×6 = 12, 3×4 = 12。fx 总和 = 31。频数总和为 20。
Mean = total fx ÷ total f = 31 ÷ 20 = 1.55 goals per match.
平均数 = fx 总和 ÷ f 总和 = 31 ÷ 20 = 1.55，即平均每场进球 1.55 个。
Always check the sum of frequencies matches the given total number of matches (20 here).
记得检查频数之和是否与给定的总场次数一致（此处为 20）。
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