📚 Worked Examples for IGCSE AQA Mathematics | IGCSE AQA 数学：典型例题详解
This article presents a selection of carefully chosen worked examples from the IGCSE AQA Mathematics specification. Each problem is solved step by step in both English and Chinese, giving you clear insights into key methods and common question types.
本文精选了 IGCSE AQA 数学考试中的典型例题，以中英双语、分步详解的方式呈现，帮助你清晰掌握核心解题方法与常见题型。
1. Fractions and Decimals | 分数与小数
Example: Evaluate 2/3 + 1/4. Write your answer as a mixed number.
例题： 计算 2/3 + 1/4，将答案写成带分数形式。
Step 1: Find a common denominator. The least common multiple of 3 and 4 is 12. Convert each fraction: 2/3 = (2×4)/(3×4) = 8/12, and 1/4 = (1×3)/(4×3) = 3/12.
步骤1：寻找公分母。3 和 4 的最小公倍数是 12。转换两个分数：2/3 = 8/12，1/4 = 3/12。
Step 2: Add the fractions: 8/12 + 3/12 = 11/12.
步骤2：将分数相加：8/12 + 3/12 = 11/12。
Step 3: The result 11/12 is already a proper fraction, so as a mixed number it is simply 11/12 (there is no whole number part).
步骤3：结果 11/12 已经是一个真分数，因此写成带分数就是 11/12（没有整数部分）。
2. Percentages | 百分比
Example: A jacket originally costs £80. In a sale, the price is reduced by 25%. Find the sale price.
例题： 一件夹克原价 £80。打折促销时降价 25%。求促销价。
Step 1: Calculate 25% of £80. 25% = 25/100 = 0.25, so 0.25 × 80 = £20.
步骤1：计算 £80 的 25%。25% = 0.25，0.25 × 80 = £20。
Step 2: Subtract the discount from the original price: £80 − £20 = £60.
步骤2：从原价中减去折扣：£80 − £20 = £60。
Step 3: The sale price is £60. Alternatively, you could multiply by 0.75 directly.
步骤3：促销价为 £60。也可以直接乘以 0.75 得出结果。
3. Ratio and Proportion | 比与比例
Example: Divide £90 in the ratio 2 : 3.
例题： 将 £90 按 2 : 3 的比例分配。
Step 1: Find the total number of parts: 2 + 3 = 5 parts.
步骤1：计算总份数：2 + 3 = 5 份。
Step 2: Determine the value of one part: £90 ÷ 5 = £18.
步骤2：求出一份的价值：£90 ÷ 5 = £18。
Step 3: Multiply: the first share is 2 × £18 = £36, the second share is 3 × £18 = £54.
步骤3：相乘：第一份额为 2 × £18 = £36，第二份额为 3 × £18 = £54。
4. Algebraic Expressions | 代数表达式
Example: Simplify 5a − 3b + 2a + 7b − 4a.
例题： 化简 5a − 3b + 2a + 7b − 4a。
Step 1: Group like terms for a: 5a + 2a − 4a = 3a.
步骤1：合并含 a 的同类项：5a + 2a − 4a = 3a。
Step 2: Group like terms for b: −3b + 7b = 4b.
步骤2：合并含 b 的同类项：−3b + 7b = 4b。
Step 3: Write the simplified expression: 3a + 4b.
步骤3：写出化简结果：3a + 4b。
Additional example: Expand and simplify 2(x + 4) − 3(x − 1).
附加例题：展开并化简 2(x + 4) − 3(x − 1)。
Step: 2(x + 4) = 2x + 8; −3(x − 1) = −3x + 3. Combine: (2x − 3x) + (8 + 3) = −x + 11.
步骤：2(x + 4) = 2x + 8；−3(x − 1) = −3x + 3。合并得 −x + 11。
5. Linear Equations | 线性方程
Example: Solve the equation (2x)/3 − 1 = 5.
例题： 解方程 (2x)/3 − 1 = 5。
Step 1: Add 1 to both sides: (2x)/3 = 6.
步骤1：两边同时加 1：(2x)/3 = 6。
Step 2: Multiply both sides by 3: 2x = 18.
步骤2：两边同时乘 3：2x = 18。
Step 3: Divide by 2: x = 9.
步骤3：两边除以 2：x = 9。
Check: (2 × 9)/3 − 1 = 18/3 − 1 = 6 − 1 = 5. Correct.
检验：(2×9)/3 − 1 = 18/3 − 1 = 6 − 1 = 5。正确。
6. Simultaneous Equations | 联立方程
Example: Solve the simultaneous equations: 2x + y = 10, x − y = 2.
例题： 解联立方程组：2x + y = 10，x − y = 2。
Step 1: Add the two equations to eliminate y: (2x + y) + (x − y) = 10 + 2 ⇒ 3x = 12.
步骤1：两式相加消去 y：(2x + y) + (x − y) = 10 + 2，得 3x = 12。
Step 2: Solve for x: x = 12 ÷ 3 = 4.
步骤2：求出 x：x = 4。
Step 3: Substitute x = 4 into the second equation: 4 − y = 2 ⇒ y = 4 − 2 = 2.
步骤3：将 x = 4 代入第二个方程：4 − y = 2，得 y = 2。
The solution is x = 4, y = 2.
解为 x = 4，y = 2。
7. Inequalities | 不等式
Example: Solve the inequality 5x − 7 ≤ 2x + 8 and represent the solution on a number line.
例题： 解不等式 5x − 7 ≤ 2x + 8，并在数轴上表示解集。
Step 1: Subtract 2x from both sides: 3x − 7 ≤ 8.
步骤1：两边同时减去 2x：3x − 7 ≤ 8。
Step 2: Add 7 to both sides: 3x ≤ 15.
步骤2：两边同时加 7：3x ≤ 15。
Step 3: Divide by 3: x ≤ 5.
步骤3：两边除以 3：x ≤ 5。
Number line: a solid circle at 5 with an arrow pointing to the left, showing all numbers less than or equal to 5.
数轴表示：在 5 处画实心圆点，并向左画箭头，表示所有小于或等于 5 的数。
8. Sequences | 数列
Example: The first four terms of a linear sequence are 7, 11, 15, 19. Find the nth term and the 50th term.
例题： 一个线性数列的前四项为 7, 11, 15, 19。求第 n 项公式和第 50 项的值。
Step 1: Find the common difference: 11 − 7 = 4, so the term-to-term rule is +4.
步骤1：计算公差：11 − 7 = 4，即相邻项相差 4。
Step 2: The nth term of an arithmetic sequence has the form an + b. Here a = 4, so nth term = 4n + b.
步骤2：等差数列第 n 项可写为 an + b，这里 a = 4，所以第 n 项 = 4n + b。
Step 3: Use the first term (n = 1) to find b: 4(1) + b = 7 ⇒ b = 3. The nth term is 4n + 3.
步骤3：代入第一项 (n=1) 求 b：4 + b = 7 ⇒ b = 3。第 n 项公式为 4n + 3。
Step 4: For the 50th term, substitute n = 50: 4(50) + 3 = 200 + 3 = 203.
步骤4：第 50 项：代入 n = 50，4(50) + 3 = 203。
9. Graphs of Functions | 函数图像
Example: Complete the table of values for y = 2x − 3 from x = −2 to x = 3, then draw the graph.
例题： 为函数 y = 2x − 3 完成从 x = −2 到 x = 3 的数值表，并绘制图像。
Step 1: Build the table:
步骤1：制作表格：
| x | −2 | −1 | 0 | 1 | 2 | 3 |
| y | −7 | −5 | −3 | −1 | 1 | 3 |
Step 2: Plot the coordinates (−2, −7), (−1, −5), (0, −3), (1, −1), (2, 1), (3, 3) on a grid.
步骤2：在坐标网格上描点：(−2,−7)、(−1,−5)、(0,−3)、(1,−1)、(2,1)、(3,3)。
Step 3: Join the points with a straight line extending beyond the plotted range. The graph is a straight line with gradient 2 and y-intercept at (0, −3).
步骤3：用直线连接各点并延长。该图像是一条斜率为 2、与 y 轴交于 (0, −3) 的直线。
10. Geometry: Angles and Polygons | 几何：角与多边形
Example: Find the sum of the interior angles of a regular octagon, and hence find the size of one interior angle.
例题： 求正八边形的内角和，并求出一个内角的大小。
Step 1: Use the formula for the sum of interior angles of an n-sided polygon: (n − 2) × 180°. For an octagon, n = 8.
步骤1：使用 n 边形内角和公式：(n − 2) × 180°。八边形 n = 8。
Step 2: Substitute: (8 − 2) × 180° = 6 × 180° = 1080°.
步骤2：代入计算：(8 − 2) × 180° = 6 × 180° = 1080°。
Step 3: Since the octagon is regular, all interior angles are equal. One interior angle = 1080° ÷ 8 = 135°.
步骤3：因为是正八边形，所有内角相等。一个内角 = 1080° ÷ 8 = 135°。
11. Pythagoras’ Theorem | 勾股定理
Example: A right-angled triangle has shorter sides of length 5 cm and 12 cm. Calculate the length of the hypotenuse.
例题： 一个直角三角形的两条直角边分别为 5 cm 和 12 cm。求斜边的长度。
Step 1: Pythagoras’ theorem states that in a right-angled triangle, a2 + b2 = c2, where c is the hypotenuse.
步骤1：勾股定理指出，在直角三角形中，a2 + b2 = c2，其中 c 为斜边。
Step 2: Substitute the given lengths: 52 + 122 = c2.
步骤2：代入已知边长：52 + 122 = c2。
Step 3: Compute: 25 + 144 = c2 ⇒ c2 = 169.
步骤3：计算：25 + 144 = c2，得 c2 = 169。
Step 4: Take the square root: c = √169 = 13 cm.
步骤4：求平方根：c = √169 = 13 cm。
12. Probability | 概率
Example: A box contains 4 red sweets, 3 green sweets and 5 blue sweets. One sweet is taken at random. Find the probability that it is not red.
例题： 一个盒子里有 4 颗红色糖果、3 颗绿色糖果和 5 颗蓝色糖果。随机取出一颗糖果，求取出的不是红色糖果的概率。
Step 1: Total number of sweets = 4 + 3 + 5 = 12.
步骤1：糖果总数 = 4 + 3 + 5 = 12。
Step 2: Number of sweets that are not red = 3 + 5 = 8.
步骤2：非红色糖果的数量 = 3 + 5 = 8。
Step 3: Probability = (number of favourable outcomes) / (total outcomes) = 8/12 = 2/3.
步骤3：概率 = （有利结果数） / （所有可能结果数）= 8/12 = 2/3。
Alternative method: P(not red) = 1 − P(red) = 1 − 4/12 = 1 − 1/3 = 2/3.
另一种解法：P(非红色) = 1 − P(红色) = 1 − 4/12 = 1 − 1/3 = 2/3。
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