📚 GCSE Edexcel Mathematics: Worked Examples of Typical Questions | GCSE Edexcel 数学：典型例题详解
In GCSE Edexcel Mathematics, mastering a broad range of question types is essential for success. This article presents detailed, step‑by‑step solutions to typical problems across the main topic areas: algebra, number, ratio, geometry, probability, and statistics. Each worked example is broken down into clear stages, with English and Chinese explanations given side by side to help you follow the reasoning and strengthen your exam technique.
在 GCSE Edexcel 数学考试中，掌握各类典型题型是取得好成绩的关键。本文精选了代数、数、比与比例、几何、概率和统计等主要领域的经典例题，逐步详细拆解，并提供中英文对照讲解，帮助你理清思路、提升解题技巧。
Use these models to check your own method, identify common pitfalls, and build confidence before the exam. Every solution reflects the precise, logical steps expected by Edexcel examiners.
用这些范例来检验自己的方法，识别常见错误脉络，并在考前建立信心。每一道题的解答都体现了 Edexcel 考官所期望的精准、逻辑清晰的步骤。
1. Solving Linear Equations with Brackets | 解含括号的一元线性方程
Example: Solve 3(x – 2) = 2x + 5.
例题：解方程 3(x – 2) = 2x + 5。
Step 1: Expand the bracket on the left‑hand side. Multiply 3 by each term inside: 3 × x = 3x, 3 × (-2) = -6, giving 3x – 6 = 2x + 5.
步骤 1：展开左边括号。3 乘以括号内每一项：3 × x = 3x，3 × (-2) = -6，得到 3x – 6 = 2x + 5。
Step 2: Collect the x terms on one side. Subtract 2x from both sides: 3x – 2x – 6 = 2x – 2x + 5, which simplifies to x – 6 = 5.
步骤 2：使含 x 的项集中到一边。两边同时减去 2x：3x – 2x – 6 = 2x – 2x + 5，化简得 x – 6 = 5。
Step 3: Isolate x by adding 6 to both sides: x – 6 + 6 = 5 + 6, so x = 11.
步骤 3：两边同时加 6 以解出 x：x – 6 + 6 = 5 + 6，得 x = 11。
Step 4: Check the solution in the original equation. Left: 3(11 – 2) = 3 × 9 = 27; Right: 2 × 11 + 5 = 22 + 5 = 27. The solution is correct.
步骤 4：代入原方程检验。左边：3(11 – 2) = 3 × 9 = 27；右边：2 × 11 + 5 = 22 + 5 = 27。解正确。
2. Factorising Quadratic Expressions | 二次式的因式分解
Example: Factorise x² + 7x + 10.
例题：因式分解 x² + 7x + 10。
Step 1: Identify the product and sum. The constant term is 10, the coefficient of x is 7. Look for two numbers that multiply to +10 and add to +7.
步骤 1：确定乘积与和。常数项为 10，x 的系数为 7。寻找两个数，它们的乘积为 +10，和为 +7。
Step 2: List factor pairs of 10: (1, 10), (2, 5). Check sums: 1 + 10 = 11, 2 + 5 = 7. The required numbers are 2 and 5.
步骤 2：列出 10 的因数对：(1, 10)、(2, 5)。检查其和：1 + 10 = 11，2 + 5 = 7。所需数字为 2 和 5。
Step 3: Write the expression as (x + 2)(x + 5). You can expand to verify: (x + 2)(x + 5) = x² + 5x + 2x + 10 = x² + 7x + 10, which matches.
步骤 3：将表达式写成 (x + 2)(x + 5)。可展开验证：(x + 2)(x + 5) = x² + 5x + 2x + 10 = x² + 7x + 10，与原式一致。
3. Solving Quadratic Equations by Factorising | 用因式分解解二次方程
Example: Solve x² – 4x – 12 = 0.
例题：解方程 x² – 4x – 12 = 0。
Step 1: Factorise the quadratic. Find two numbers with product -12 and sum -4. Pairs: (-6, 2) gives sum -4. So x² – 4x – 12 = (x – 6)(x + 2).
步骤 1：因式分解二次式。找出乘积为 -12、和为 -4 的两个数字。因数对：(-6, 2) 的和为 -4。因此 x² – 4x – 12 = (x – 6)(x + 2)。
Step 2: Set each bracket equal to zero. (x – 6) = 0 or (x + 2) = 0.
步骤 2：令每个因式等于零。(x – 6) = 0 或 (x + 2) = 0。
Step 3: Solve the two simple equations. x – 6 = 0 ⇒ x = 6; x + 2 = 0 ⇒ x = -2.
步骤 3：解两个一元一次方程。x – 6 = 0 ⇒ x = 6；x + 2 = 0 ⇒ x = -2。
Step 4: Check by substitution. For x = 6: 36 – 24 – 12 = 0. For x = -2: 4 + 8 – 12 = 0. Both are correct.
步骤 4：代入检验。x = 6 时：36 – 24 – 12 = 0。x = -2 时：4 + 8 – 12 = 0。均正确。
4. Percentage Increase and Compound Interest | 百分比增长与复利
Example: A bicycle costs £240. In a sale, the price is increased by 15%. After one month, the new price is reduced by 10% for a clearance deal. Find the final price.
例题：一辆自行车价格为 £240。在促销调整中，价格先上涨 15%，一个月后进行清仓处理，再降价 10%。求最终价格。
Step 1: Calculate the 15% increase. 15% of £240 = 0.15 × 240 = £36. New price = 240 + 36 = £276.
步骤 1：计算 15% 的涨幅。£240 的 15% = 0.15 × 240 = £36。新价格 = 240 + 36 = £276。
Step 2: Now apply a 10% decrease to the increased price. 10% of £276 = 0.10 × 276 = £27.60.
步骤 2：将 10% 的折扣应用于涨价后的价格。£276 的 10% = 0.10 × 276 = £27.60。
Step 3: Final price = 276 – 27.60 = £248.40.
步骤 3：最终价格 = 276 – 27.60 = £248.40。
Step 4: Alternatively, use multipliers. Increase by 15% is ×1.15, decrease by 10% is ×0.90. Combined multiplier = 1.15 × 0.90 = 1.035. Final price = 240 × 1.035 = £248.40.
步骤 4：亦可用乘数法。上涨 15% 对应乘数 ×1.15，下降 10% 对应 ×0.90。综合乘数 = 1.15 × 0.90 = 1.035。最终价格 = 240 × 1.035 = £248.40。
5. Sharing in a Ratio | 按比例分配
Example: Alice, Bob and Charlie share £360 in the ratio 2 : 3 : 5. How much does each person receive?
例题：Alice、Bob 和 Charlie 按 2 : 3 : 5 的比例分配 £360。每人各得多少？
Step 1: Find the total number of parts. 2 + 3 + 5 = 10 parts.
步骤 1：求总份数。2 + 3 + 5 = 10 份。
Step 2: Calculate the value of one part. £360 ÷ 10 = £36 per part.
步骤 2：计算每份的价值。£360 ÷ 10 = £36/份。
Step 3: Multiply each person’s share by the value per part. Alice: 2 × 36 = £72; Bob: 3 × 36 = £108; Charlie: 5 × 36 = £180.
步骤 3：用每人份数乘以每份价值。Alice：2 × 36 = £72；Bob：3 × 36 = £108；Charlie：5 × 36 = £180。
Step 4: Check the total: 72 + 108 + 180 = 360. The distribution is correct.
步骤 4：检验总和：72 + 108 + 180 = 360。分配正确。
6. Pythagoras’ Theorem | 毕达哥拉斯定理
Example: In a right‑angled triangle, the two shorter sides measure 8 cm and 15 cm. Find the length of the hypotenuse.
例题：在一个直角三角形中，两条直角边分别为 8 cm 和 15 cm。求斜边长。
Step 1: Write Pythagoras’ theorem: a² + b² = c², where c is the hypotenuse. Let a = 8, b = 15.
步骤 1：写出毕达哥拉斯定理：a² + b² = c²，其中 c 为斜边。令 a = 8，b = 15。
Step 2: Substitute values: 8² + 15² = c² → 64 + 225 = c² → 289 = c².
步骤 2：代入数值：8² + 15² = c² → 64 + 225 = c² → 289 = c²。
Step 3: Solve for c by taking the positive square root: c = √289 = 17 cm.
步骤 3：开方求 c（取正值）：c = √289 = 17 cm。
Step 4: Check the triangle inequality: the hypotenuse is indeed the longest side, and 17 > 15 and 17 > 8.
步骤 4：检验三角形边长关系：斜边应最长，且 17 > 15、17 > 8，符合要求。
7. Area of a Trapezium | 梯形的面积
Example: A trapezium has parallel sides of length 12 cm and 18 cm. The perpendicular distance between them (height) is 7 cm. Calculate its area.
例题：一梯形，两底长分别为 12 cm 和 18 cm，它们之间的垂直距离（高）为 7 cm。求其面积。
Step 1: Recall the formula: Area = ½ × (a + b) × h, where a and b are the parallel sides.
步骤 1：回忆公式：面积 = ½ × (a + b) × h，其中 a 和 b 为两底边长。
Step 2: Substitute the numbers: a = 12, b = 18, h = 7. Sum of parallel sides = 12 + 18 = 30 cm.
步骤 2：代入数值：a = 12，b = 18，h = 7。两底之和 = 12 + 18 = 30 cm。
Step 3: Multiply ½ × 30 × 7 = 15 × 7 = 105 cm².
步骤 3：计算 ½ × 30 × 7 = 15 × 7 = 105 cm²。
Step 4: State the answer with correct units: The area is 105 cm².
步骤 4：给出正确答案和单位：面积为 105 cm²。
8. Cumulative Frequency and Box Plots | 累积频率与箱线图
Example: The table shows the masses of 40 apples, grouped. Draw a cumulative frequency graph and use it to find the median and interquartile range.
例题：下表显示了 40 个苹果的质量分组情况。绘制累积频率图，并利用它求中位数和四分位距。
Mass (g): 100–110 (freq 5), 110–120 (8), 120–130 (12), 130–140 (10), 140–150 (5).
质量（克）：100–110（频数 5），110–120（8），120–130（12），130–140（10），140–150（5）。
Step 1: Add an extra column for cumulative frequency. For 100–110: 5; 110–120: 5+8=13; 120–130: 13+12=25; 130–140: 25+10=35; 140–150: 35+5=40.
步骤 1：增加累积频率列。100–110：5；110–120：5+8=13；120–130：13+12=25；130–140：25+10=35；140–150：35+5=40。
Step 2: Plot cumulative frequency (vertical axis) against the upper class boundary (horizontal axis): 110, 120, 130, 140, 150. Draw a smooth curve through the points.
步骤 2：以累积频率为纵轴，以上组界为横轴（110, 120, 130, 140, 150）描点，连成平滑曲线。
Step 3: Median is the value at half the total frequency: total = 40, halfway = 20th value. From the graph, read the mass at cumulative frequency 20 (approximately 126 g).
步骤 3：中位数是总频数一半处的值：总数 40，一半为第 20 个值。从图中累积频率 20 处读得质量约为 126 g。
Step 4: Lower quartile (Q1) at ¼ × 40 = 10th value ≈ 118 g; upper quartile (Q3) at ¾ × 40 = 30th value ≈ 133 g. Interquartile range = Q3 – Q1 = 133 – 118 = 15 g.
步骤 4：下四分位数 (Q1) 位于 ¼ × 40 = 第 10 个值 ≈ 118 g；上四分位数 (Q3) 位于 ¾ × 40 = 第 30 个值 ≈ 133 g。四分位距 = Q3 – Q1 = 133 – 118 = 15 g。
9. Probability Tree Diagrams (Without Replacement) | 概率树形图（不放回）
Example: A bag contains 5 red sweets and 3 green sweets. Two sweets are picked at random without replacement. Find the probability that both sweets are the same colour.
例题：一个袋子里有 5 颗红色糖和 3 颗绿色糖。随机无放回地取两次，每次取一颗。求两颗糖颜色相同的概率。
Step 1: Draw the tree diagram. First pick: P(red) = 5/8, P(green) = 3/8.
步骤 1：画树状图。第一次取：P(红) = 5/8，P(绿) = 3/8。
Step 2: Second pick probabilities change. After red first, 4 red and 3 green remain (total 7). So P(red|red) = 4/7, P(green|red) = 3/7. After green first, 5 red and 2 green remain (total 7). So P(red|green) = 5/7, P(green|green) = 2/7.
步骤 2：第二次取的概率发生变化。先取红后，剩下 4 红 3 绿（共 7 颗），P(红|红) = 4/7，P(绿|红) = 3/7。先取绿后，剩下 5 红 2 绿（共 7 颗），P(红|绿) = 5/7，P(绿|绿) = 2/7。
Step 3: Same colour means (red, red) or (green, green). Multiply along branches: P(RR) = 5/8 × 4/7 = 20/56 = 5/14. P(GG) = 3/8 × 2/7 = 6/56 = 3/28.
步骤 3：颜色相同即（红，红）或（绿，绿）。沿树枝相乘：P(RR) = 5/8 × 4/7 = 20/56 = 5/14。P(GG) = 3/8 × 2/7 = 6/56 = 3/28。
Step 4: Add the probabilities: P(same) = 5/14 + 3/28. Convert to common denominator 28: 10/28 + 3/28 = 13/28.
步骤 4：概率相加：P(同色) = 5/14 + 3/28。通分分母 28：10/28 + 3/28 = 13/28。
10. Algebraic Fractions – Simplifying and Solving | 代数分式——化简与解方程
Example: Simplify (2x² + 5x – 3) / (x² + 4x + 3) and hence solve the equation (2x² + 5x – 3)/(x² + 4x + 3) = 2.
例题：化简 (2x² + 5x – 3)/(x² + 4x + 3)，并由此解方程 (2x² + 5x – 3)/(x² + 4x + 3) = 2。
Step 1: Factorise numerator and denominator. Numerator: 2x² + 5x – 3. Find two numbers that multiply to 2 × (-3) = -6 and add to 5 → 6 and -1. Write 2x² + 6x – x – 3 = 2x(x + 3) – 1(x + 3) = (2x – 1)(x + 3).
步骤 1：因式分解分子和分母。分子：2x² + 5x – 3。找两数乘积为 2 × (-3) = -6，和为 5 → 6 和 -1。写成 2x² + 6x – x – 3 = 2x(x + 3) – 1(x + 3) = (2x – 1)(x + 3)。
Step 2: Denominator: x² + 4x + 3 = (x + 1)(x + 3).
步骤 2：分母：x² + 4x + 3 = (x + 1)(x + 3)。
Step 3: Cancel the common factor (x + 3), provided x ≠ -3. The simplified expression is (2x – 1)/(x + 1).
步骤 3：约去公因式 (x + 3)，须 x ≠ -3。化简后为 (2x – 1)/(x + 1)。
Step 4: To solve (2x² + 5x – 3)/(x² + 4x + 3) = 2, first use the simplified form: (2x – 1)/(x + 1) = 2. Multiply both sides by (x + 1): 2x – 1 = 2(x + 1) → 2x – 1 = 2x + 2. Subtract 2x from both sides: -1 = 2, which is a contradiction. Therefore there is no solution (and x = -3 is excluded by the domain). The equation has no valid solutions.
步骤 4：解方程 (2x² + 5x – 3)/(x² + 4x + 3) = 2，先用化简式：(2x – 1)/(x + 1) = 2。两边同乘 (x + 1)：2x – 1 = 2(x + 1) → 2x – 1 = 2x + 2。两边减 2x 得 -1 = 2，矛盾。故方程无解（且 x = -3 已被排除在定义域之外）。方程无有效解。
11. Indices and Standard Index Form | 指数与标准指数形式
Example: Simplify (3⁻² × (3⁴)²) ÷ 3³ and write the answer as a power of 3. Then evaluate (2 × 10³) × (5 × 10⁻²) giving your answer in standard form.
例题：化简 (3⁻² × (3⁴)²) ÷ 3³ 并将结果写成 3 的幂。然后计算 (2 × 10³) × (5 × 10⁻²)，结果以标准形式表示。
Step 1: Simplify the power expression. First, (3⁴)² = 3⁴ ˣ ² = 3⁸. Then 3⁻² × 3⁸ = 3⁽⁻²⁺⁸⁾ = 3⁶. Now divide by 3³: 3⁶ ÷ 3³ = 3⁽⁶⁻³⁾ = 3³.
步骤 1：化简幂的运算。首先 (3⁴)² = 3⁴ ˣ ² = 3⁸。然后 3⁻² × 3⁸ = 3⁽⁻²⁺⁸⁾ = 3⁶。再除以 3³：3⁶ ÷ 3³ = 3⁽⁶⁻³⁾ = 3³。
Step 2: For the standard form multiplication: (2 × 10³) × (5 × 10⁻²). Multiply the number parts: 2 × 5 = 10. Multiply the powers of 10: 10³ × 10⁻² = 10⁽³⁻²⁾ = 10¹. So we have 10 × 10¹ = 10².
步骤 2：对于标准形式乘法：(2 × 10³) × (5 × 10⁻²)。数字部分相乘：2 × 5 = 10。10 的幂相乘：10³ × 10⁻² = 10⁽³⁻²⁾ = 10¹。得到 10 × 10¹ = 10²。
Step 3: Express 10² in standard form: it is already 1 × 10². Therefore the final answer is 1 × 10².
步骤 3：将 10² 写成标准形式：它本身就是 1 × 10²。因此最终答案为 1 × 10²。
12. Surds – Simplifying and Rationalising | 根式——化简与分母有理化
Example: Write √48 in its simplest form. Hence simplify (√48 + √27) / √75, and rationalise the denominator of 5/(4 + √3).
例题：将 √48 化为最简形式。由此化简 (√48 + √27) / √75，并将 5/(4 + √3) 分母有理化。
Step 1: Simplify √48. Look for the largest square factor: 48 = 16 × 3. So √48 = √(16 × 3) = √16 × √3 = 4√3.
步骤 1：化简 √48。寻找最大的平方因子：48 = 16 × 3。所以 √48 = √(16 × 3) = √16 × √3 = 4√3。
Step 2: Simplify √27 and √75. 27 = 9 × 3 → √27 = 3√3. 75 = 25 × 3 → √75 = 5√3.
步骤 2：化简 √27 和 √75。27 = 9 × 3 → √27 = 3√3。75 = 25 × 3 → √75 = 5√3。
Step 3: Then (√48 + √27) / √75 = (4√3 + 3√3) / (5√3) = (7√3) / (5√3) = 7/5.
步骤 3：因此 (√48 + √27) / √75 = (4√3 + 3√3) / (5√3) = (7√3) / (5√3) = 7/5。
Step 4: Rationalise 5/(4 + √3). Multiply numerator and denominator by the conjugate 4 – √3: = 5(4 – √3) / [(4 + √3)(4 – √3)]. Denominator: 4² – (√3)² = 16 – 3 = 13. So the expression becomes (20 – 5√3)/13.
步骤 4：分母有理化 5/(4 + √3)。分子分母同乘共轭式 4 – √3：= 5(4 – √3) / [(4 + √3)(4 – √3)]。分母：4² – (√3)² = 16 – 3 = 13。因此表达式变为 (20 – 5√3)/13。
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