📚 GCSE CCEA Maths: Worked Examples Explained | GCSE CCEA 数学：典型例题精讲
This revision guide provides a set of carefully chosen worked examples covering the main topic areas of the CCEA GCSE Mathematics specification. Each problem is broken down into clear steps, helping you to understand the methods and build confidence for the exam. Both Foundation and Higher tier students will find useful practice here.
本复习指南提供一组精心挑选的典型例题，涵盖 CCEA GCSE 数学考试大纲的主要知识点。每道题都分解为清晰的步骤，帮助你理解方法，树立考试信心。无论是基础卷还是高级卷的学生，都能在这里找到有用的练习。
1. Fractions, Decimals and Percentages | 分数、小数与百分数
Problem: Evaluate 2/3 + 4/5 − 1/2, giving your answer as a fraction in its simplest form.
题目：计算 2/3 + 4/5 − 1/2，将答案写成最简分数。
Step 1: Find a common denominator for 3, 5 and 2. The LCM is 30.
第 1 步：找出 3、5 和 2 的公分母。最小公倍数是 30。
ep 2: Convert each fraction to have denominator 30. 2/3 = 20/30, 4/5 = 24/30, 1/2 = 15/30.
第 2 步：将每个分数转换为分母为 30。2/3 = 20/30，4/5 = 24/30，1/2 = 15/30。
3: Perform the operations. 20/30 + 24/30 − 15/30 = (20 + 24 − 15)/30 = 29/30.
第 3 步：进行运算。20/30 + 24/30 − 15/30 = (20 + 24 − 15)/30 = 29/30。
Step 4: Check if the fraction can be simplified. 29 and 30 have no common factors, so 29/30 is the simplest form.
第 4 步：检查分数是否可以化简。29 和 30 没有公因数，所以 29/30 就是最简形式。
2. Ratio and Proportion | 比与比例
Problem: A prize of £250 is shared between Chloe and David in the ratio 4:6. (a) How much money does David receive? (b) Chloe spends 30% of her share. How much does she have left?
题目：一笔 250 英镑的奖金由克洛伊和大卫按 4:6 的比例分配。(a) 大卫得到多少钱？(b) 克洛伊花掉她份额的 30%。她还剩多少钱？
Step 1: Find the total number of parts. Ratio 4:6 gives total parts = 4 + 6 = 10.
第 1 步：找出总份数。比例 4:6 得到总份数 = 4 + 6 = 10。
Step 2: Calculate the value of one part. £250 ÷ 10 = £25 per part.
第 2 步：计算每份的价值。£250 ÷ 10 = 每份 £25。
Step 3: David’s share = 6 parts. 6 × £25 = £150. Answer (a): £150.
第 3 步：大卫的份额 = 6 份。6 × £25 = £150。答案 (a)：£150。
Step 4: Chloe’s share = 4 parts, 4 × £25 = £100.
第 4 步：克洛伊的份额 = 4 份，4 × £25 = £100。
Step 5: She spends 30%, so she keeps 70%. 70% of £100 = 0.70 × 100 = £70. Answer (b): £70.
第 5 步：她花掉 30%，因此还剩 70%。£100 的 70% = 0.70 × 100 = £70。答案 (b)：£70。
3. Algebraic Expressions and Equations | 代数表达式与方程
Problem: ) Expand and simplify (2x + 3)(x − 5). (b) Solve the equation 3x − 7 = 2x + 8.
题目：(a) 展开并化简 (2x + 3)(x − 5)。(b) 解方程 3x − 7 = 2x + 8。
(−5) = −10x. Inner: 3 × x = 3x. Last: 3 × (−5) = −15.
第 1 步 (a)：使用 FOIL 方法。首项：2x × x = 2x²。外项：2x × (−5) = −10x。内项：3 × x = 3x。尾项：3 × (−5) = −15。
bine like terms. −10x + 3x = −7x. Final expression: 2x² − 7x − 15.
第 2 步 (a)：合并同类项。−10x + 3x = −7x。最终表达式：2x² − 7x − 15。
ct 2x from both sides to collect x terms. 3x − 2x − 7 = 8, so x − 7 = 8.
第 1 步 (b)：两边同时减去 2x，把含 x 的项移到一边。3x − 2x − 7 = 8，得到 x − 7 = 8。
Step 2 (b): Add 7 to both sides to isolate x. x = 8 + 7, so x = 15.
第 2 步 (b)：两边同时加 7，求出 x。x = 8 + 7，所以 x = 15。
4. Sequences and nth Term | 数列与第 n 项
Problem: Here are the first four terms of a linear sequence: 5, 9, 13, 17. (a) Write down the next term. (b) Find an expression for the nth term. (c) Is 121 a term in this sequence? Explain.
题目：一个线性数列的前四项为：5, 9, 13, 17。(a) 写出下一项。(b) 求出第 n 项的表达式。(c) 121 是这个数列的一项吗？请解释。
een terms is constant: 9 − 5 = 4, 13 − 9 = 4. So the next term is 17 + 4 = 21.
第 1 步 (a)：各项之间的差是常数：9 − 5 = 4，13 − 9 = 4。所以下一项是 17 + 4 = 21。
rst term, d = common difference. Here a = 5, d = 4. Expression: 5 + (n − 1)×4.
第 2 步 (b)：对于线性数列，第 n 项公式为：a + (n − 1)d，其中 a 为首项，d 为公差。这里 a = 5，d = 4。表达式：5 + (n − 1)×4。
Step 3 (b): Simplify: 5 + 4n − 4 = 4n + 1. So nth term = 4n + 1.
第 3 步 (b)：化简：5 + 4n − 4 = 4n + 1。所以第 n 项 = 4n + 1。
4n = 120, n = 30. Since n is a positive integer, 121 is the 30th term. Yes, it is a term.
第 4 步 (c)：令 4n + 1 = 121。求解：4n = 120，n = 30。因为 n 是正整数，121 是第 30 项。是的，它是数列的一项。
5. Linear Graphs and Equations | 线性图像与方程
Problem: a) the gradient, (b) the equation of the line in the form y = mx + c, (c) the x-intercept.
题目：一条直线经过点 (2, 5) 和 (6, 1)。求 (a) 斜率，(b) 直线方程，形式为 y = mx + c，(c) x 轴截距。
m = (y₂ − y₁)/(x₂ − x₁) = (1 − 5)/(6 − 2) = −4/4 = −1.
第 1 步 (a)：斜率 m = (y₂ − y₁)/(x₂ − x₁) = (1 − 5)/(6 − 2) = −4/4 = −1。
orm y − y₁ = m(x − x₁) with point (2, 5): y − 5 = −1(x − 2).
第 2 步 (b)：使用点斜式 y − y₁ = m(x − x₁)，代入点 (2, 5)：y − 5 = −1(x − 2)。
b): Simplify: y − 5 = −x + 2. Add 5 to both sides: y = −x + 7. So c = 7.
第 3 步 (b)：化简：y − 5 = −x + 2。两边加 5：y = −x + 7。所以 c = 7。
4 (c): For x-intercept, set y = 0. 0 = −x + 7. Solve: x = 7. x-intercept is (7, 0).
第 4 步 (c)：求 x 轴截距，令 y = 0。0 = −x + 7。解得 x = 7。x 轴截距为 (7, 0)。
6. Angles in Polygons | 多边形角度
Problem: A regular polygon has an exterior angle of 24°. (a) Work out the number of sides of this polygon. (b) Calculate the size of its interior angle.
题目：一个正多边形的一个外角为 24°。(a) 求该多边形的边数。(b) 计算其内角的大小。
Step 1 (a): The sum of exterior angles of any polygon is 360°. For a regular polygon, all exterior angles are equal.
第 1 步 (a)：任何多边形外角和均为 360°。对正多边形，所有外角相等。
n = 360° ÷ exterior angle = 360 ÷ 24 = 15. So the polygon has 15 sides.
第 2 步 (a)：边数 n = 360° ÷ 外角 = 360 ÷ 24 = 15。所以该多边形有 15 条边。
exterior angle = 180°. Therefore interior angle = 180° − 24° = 156°.
第 3 步 (b)：内角 + 外角 = 180°。因此内角 = 180° − 24° = 156°。
× 180° ÷ n = (13 × 180) ÷ 15 = 2340 ÷ 15 = 156°.
或者，使用公式：内角 ) × 180° ÷ n = (13 × 180) ÷ 15 = 2340 ÷ 15 = 156°。
7. Area and Volume | 面积与体积
Problem: A cylinder has radius 6 cm and height 10 cm. (a) Calculate the volume, giving your answer in terms of π. (b) Find the curved surface area, also in terms of π.
题目：一个圆柱体的半径为 6 cm，高为 10 cm。(a) 计算体积，答案用 π 表示。(b) 求其侧面积，也用 π 表示。
r = 6, h = 10: V = π × 6² × 10 = π × 36 × 10 = 360π cm³.
第 1 步 (a)：体积公式：V = πr²h。代入 r = 6，h = 10：V = π × 6² × 10 = π × 36 × 10 = 360π cm³。
a formula: A = 2πrh. Substitute: A = 2 × π × 6 × 10 = 120π cm².
第 2 步 (b)：侧面积公式：A = 2πrh。代入：A = 2 × π × 6 × 10 = 120π cm²。
Step 3: Ensure units are correct and the question asks for answers in terms of π — leave π in the expression.
第 3 步：确保单位正确，题意要求答案保留 π，所以表达式里保留 π。
8. Pythagoras and Trigonometry | 勾股定理与三角学
Problem: In a right-angled triangle, the hypotenuse is 13 cm and one shorter side is 5 cm. (a) Calculate the length of the other side. (b) Find the smallest angle in the triangle.
题目：在一个直角三角形中，斜边长为 13 cm，一条直角边为 5 cm。(a) 计算另一条直角边的长度。(b) 求该三角形最小的角。
Let a = 5, c = 13. So b² = 13² − 5² = 169 − 25 = 144. b = √144 = 12 cm.
第 1 步 (a)：使用勾股定理。a² + b² = c²，其中 c 为斜边。设 a = 5，c = 13。则 b² = 13² − 5² = 169 − 25 = 144。b = √144 = 12 cm。
nt. Let θ be the angle opposite side 5 cm. sin θ = opposite/hypotenuse = 5/13.
第 2 步 (b)：最小角对着最短边，即 5 cm 的边。使用正弦、余弦或正切。设 θ 为 5 cm 边所对的角。sin θ = 对边/斜边 = 5/13。
13). Using a calculator, θ ≈ 22.6° (to 1 d.p.). Thus the smallest angle is about 22.6°.
第 3 步 (b)：θ = sin⁻¹(5/13)。用计算器计算，θ ≈ 22.6°（精确到小数点后一位）。因此最小角约为 22.6°。
9. Statistics: Averages and Charts | 统计：平均数与图表
Problem: The frequency table shows the number of pets owned by 30 pupils. Number of pets: 0, 1, 2, 3, 4. Frequency: 5, 12, 8, 3, 2. Calculate (a) the mean number of pets, (b) the median, (c) the mode.
题目：频数表显示了 30 名学生拥有的宠物数量。宠物数量：0, 1, 2, 3, 4。频数：5, 12, 8, 3, 2。计算 (a) 平均宠物数量，(b) 中位数，(c) 众数。
es;5) + (1×12) + (2×8) + (3×3) + (4×2) = 0 + 12 + 16 + 9 + 8 = 45.
第 1 步 (a)：将每个宠物数量乘以对应的频数，并求和：mes;5) + (1×12) + (2×8) + (3×3) + (4×2) = 0 + 12 + 16 + 9 + 8 = 45。
5 + 12 + 8 + 3 + 2 = 30. Mean = sum ÷ total frequency = 45 ÷ 30 = 1.5 pets.
第 2 步 (a)：总频数 = 5 + 12 + 8 + 3 + 2 = 30。平均数 = 总和 ÷ 总频数 = 45 ÷ 30 = 1.5 只宠物。
h values lie in the group with 1 pet (since cumulative up to 1 is 5+12=17). So median = 1.
第 3 步 (b)：求中位数，列出累积频数。第 15 和第 16 个数值落在“1 只宠物”这一组（因为累积到 1 的频数为 5+12=17）。所以中位数 = 1。
Step 4 (c): The mode is the number with the highest frequency, which is 1 pet (frequency 12).
第 4 步 (c)：众数是频数最高的数值，即 1 只宠物（频数为 12）。
10. Probability | 概率
Problem: A bag contains 4 red, 3 blue and 2 green marbles. One marble is taken at random and then replaced. A second marble is then taken. (a) Draw a tree diagram to show the probabilities. (b) Calculate the probability that both marbles are blue. (c) What is the probability of drawing at least one green?
题目：一个袋子装有 4 个红色、3 个蓝色和 2 个绿色弹珠。随机抽取一个弹珠后放回，然后再次抽取第二个弹珠。(a) 画出树状图表示概率。(b) 计算两次都抽到蓝色的概率。(c) 至少抽到一个绿色的概率是多少？
(Green) = 2/9. The tree diagram has two sets of identical branches because of replacement.
第 1 步 (a)：弹珠总数 = 4+3+2 = 9。P(红) = 4/9，P(蓝) = 3/9 = 1/3，P(绿) = 2/9。由于是放回抽取，树状图有两组完全相同的分支。
ue) = P(blue on 1st) × P(blue on 2nd) = (3/9) × (3/9) = 1/3 × 1/3 = 1/9.
第 2 步 (b)：两次都抽到蓝的概率 = 第一次抽到蓝 × 第二次抽到蓝 = (3/9) × (3/9) = 1/3 × 1/3 = 1/9。
Step 3 (c): 'At least one green' on 2nd) = (7/9) × (7/9) = 49/81. So P(at least one green) = 1 − 49/81 = 32/81.
第 3 步 (c)：计算“至少一个绿色”用补集更简单：1 − P(没有绿色)。P(没有绿色) = 第一次没抽到绿 × 第二次没抽到绿 = (7/9) × (7/9) = 49/81。所以至少一个绿色的概率 = 1 − 49/81 = 32/81。
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