📚 GCSE Mathematics: Typical Example Problems Explained in Detail | GCSE 数学：典型例题详解
In your GCSE Mathematics revision, nothing builds confidence like working through a range of typical exam-style questions. This article selects 10 classic problems from key topics, showing clear step-by-step solutions with paired English and Chinese explanations. Use these examples to sharpen your problem-solving technique and avoid common pitfalls.
在 GCSE 数学复习中，没有什么比演练一系列典型考试题更能建立信心。本文从核心主题中精选 10 道经典题目，以中英对照的方式展示清晰的逐步解题过程。通过这些例题，你可以磨练解题技巧，避开常见失分点。
1. Solving Linear Equations | 解一元一次方程
Solve: 3(2x − 5) = x + 8
解方程：3(2x − 5) = x + 8
First expand the bracket: 3 × 2x gives 6x, and 3 × (−5) gives −15. The equation becomes 6x − 15 = x + 8.
首先展开括号：3 × 2x 得到 6x，3 × (−5) 得到 −15。方程变为 6x − 15 = x + 8。
Subtract x from both sides to collect like terms: 6x − x − 15 = 8, which simplifies to 5x − 15 = 8.
两边同时减去 x，合并同类项：6x − x − 15 = 8，化简得 5x − 15 = 8。
Add 15 to both sides: 5x = 23. Finally divide by 5: x = 23/5 or 4.6.
两边加 15：5x = 23。最后除以 5：x = 23/5 即 4.6。
Always check by substitution: 3(2×4.6 − 5) = 3(9.2 − 5) = 3×4.2 = 12.6, and 4.6 + 8 = 12.6. Both sides equal.
务必代入检验：3(2×4.6 − 5) = 3(9.2 − 5) = 3×4.2 = 12.6，而 4.6 + 8 = 12.6。两边相等。
2. Factorising Quadratic Expressions | 因式分解二次表达式
Factorise fully: x² − 5x − 24
因式分解：x² − 5x − 24
We look for two numbers that multiply to give −24 and add to give −5. Pairs of factors of 24 are (1,24), (2,12), (3,8), (4,6). Because the product is negative, one number must be positive and the other negative.
我们要找两个数，乘积为 −24，和为 −5。24 的因数对有 (1,24)、(2,12)、(3,8)、(4,6)。由于乘积为负，两数必须一正一负。
Testing pairs: 3 and −8 work because 3 × (−8) = −24 and 3 + (−8) = −5. So we write x² − 5x − 24 = (x + 3)(x − 8).
试出 3 和 −8 符合，因为 3 × (−8) = −24，3 + (−8) = −5。因此 x² − 5x − 24 = (x + 3)(x − 8)。
Check by expanding: (x + 3)(x − 8) = x² − 8x + 3x − 24 = x² − 5x − 24. Correct.
验证：展开得 (x + 3)(x − 8) = x² − 8x + 3x − 24 = x² − 5x − 24。正确。
3. Pythagoras’ Theorem | 勾股定理
In a right-angled triangle, the two shorter sides are 9 cm and 12 cm. Find the length of the hypotenuse.
在一个直角三角形中，两条较短的边分别为 9 cm 和 12 cm。求斜边的长度。
Pythagoras’ theorem states: a² + b² = c², where c is the hypotenuse. Substitute: 9² + 12² = c².
勾股定理：a² + b² = c²，其中 c 为斜边。代入：9² + 12² = c²。
Calculate: 81 + 144 = 225. So c² = 225. Taking the square root gives c = √225 = 15 cm.
计算：81 + 144 = 225。所以 c² = 225。开平方得 c = √225 = 15 cm。
Always remember to take the positive square root because length is positive. The hypotenuse is 15 cm.
注意取正平方根，因为长度为正。斜边长度为 15 cm。
4. Trigonometry (SOHCAHTOA) | 三角函数 (SOHCAHTOA)
A right-angled triangle has an angle of 35° and the adjacent side is 8 cm. Find the length of the opposite side to the nearest mm.
一直角三角形中，一个角为 35°，邻边为 8 cm。求对边的长度，精确到毫米。
Use tan because tan θ = opposite/adjacent. So tan 35° = opposite / 8.
使用正切，因为 tan θ = 对边/邻边。故 tan 35° = 对边 / 8。
Rearrange: opposite = 8 × tan 35°. Using a calculator, tan 35° ≈ 0.700208. Hence opposite ≈ 8 × 0.700208 = 5.601664 cm.
整理得：对边 = 8 × tan 35°。用计算器，tan 35° ≈ 0.700208。因此对边 ≈ 8 × 0.700208 = 5.601664 cm。
To nearest mm (one decimal place in cm): 5.6 cm or 56 mm. Remember to state units clearly.
精确到毫米（厘米保留一位小数）：5.6 cm 或 56 mm。记得清晰注明单位。
5. Area and Circumference of a Circle | 圆的面积和周长
A circle has a diameter of 14 cm. Calculate its area and circumference. Give answers in terms of π and rounded to 2 decimal places.
一个圆的直径为 14 cm。计算它的面积和周长。答案分别用 π 表示，并四舍五入到两位小数。
Radius r = diameter / 2 = 7 cm. Circumference formula: C = 2πr = 2 × π × 7 = 14π cm. Area formula: A = πr² = π × 7² = 49π cm².
半径 r = 直径 / 2 = 7 cm。周长公式：C = 2πr = 2 × π × 7 = 14π cm。面积公式：A = πr² = π × 7² = 49π cm²。
Using π ≈ 3.14159, C ≈ 14 × 3.14159 = 43.98 cm (2 d.p.). A ≈ 49 × 3.14159 = 153.94 cm² (2 d.p.).
使用 π ≈ 3.14159，C ≈ 14 × 3.14159 = 43.98 cm（两位小数）。A ≈ 49 × 3.14159 = 153.94 cm²（两位小数）。
Always check whether the question asks for exact form (in terms of π) or a decimal approximation.
务必看清题目要求的是精确值（用 π 表示）还是小数近似值。
6. Ratios and Proportion | 比例和比例关系
Flour, sugar and butter are mixed in the ratio 5:2:3 to make pastry. How much sugar is needed if 900 g of flour is used?
面粉、糖和黄油按 5:2:3 的比例混合制作糕点。如果用 900 g 面粉，需要多少糖？
The ratio tells us flour corresponds to 5 parts. So 5 parts = 900 g. Find the value of one part: 1 part = 900 ÷ 5 = 180 g.
比例表明面粉对应 5 份。因此 5 份 = 900 g。求一份的值：1 份 = 900 ÷ 5 = 180 g。
Sugar corresponds to 2 parts, so sugar needed = 2 × 180 = 360 g.
糖对应 2 份，所以糖的量 = 2 × 180 = 360 g。
You can use the same method to find butter: 3 × 180 = 540 g. Always keep the order of the ratio consistent.
你可以用同样方法求黄油：3 × 180 = 540 g。始终保持比例顺序一致。
7. Mean, Median, Mode and Range | 平均数、中位数、众数和极差
Find the mean, median, mode and range of this data set: 7, 9, 12, 8, 9, 15, 6, 9, 11.
求以下数据的平均数、中位数、众数和极差：7, 9, 12, 8, 9, 15, 6, 9, 11。
First arrange in order: 6, 7, 8, 9, 9, 9, 11, 12, 15. The mode (most frequent) is 9. The range = max − min = 15 − 6 = 9.
先排序：6, 7, 8, 9, 9, 9, 11, 12, 15。众数（出现次数最多）是 9。极差 = 最大值 − 最小值 = 15 − 6 = 9。
There are 9 values, so median is the 5th value: 9. Mean = sum ÷ 9. Sum = 6+7+8+9+9+9+11+12+15 = 86. Mean = 86/9 ≈ 9.56 (2 d.p.).
有 9 个数据，中位数是第 5 个值：9。平均数 = 总和 ÷ 9。总和 = 6+7+8+9+9+9+11+12+15 = 86。平均数 = 86/9 ≈ 9.56（两位小数）。
Be careful: if the number of values is even, the median is the mean of the two middle numbers.
注意：若数据个数为偶数，中位数是中间两个数的平均数。
8. Probability Tree Diagrams | 概率树状图
A bag contains 4 red and 6 blue counters. Two counters are drawn randomly without replacement. Draw a tree diagram and find the probability that both are red.
一个袋子里有 4 个红色筹码和 6 个蓝色筹码。随机抽取两个筹码，不放回。画出树状图，并求两个都是红色的概率。
First draw: P(Red) = 4/10 = 2/5, P(Blue) = 6/10 = 3/5. If first is red, remaining: 3 red, 6 blue, total 9. So P(Red|Red) = 3/9 = 1/3. P(Blue|Red) = 6/9 = 2/3.
第一次抽取：P(红) = 4/10 = 2/5，P(蓝) = 6/10 = 3/5。若第一次是红色，剩余：3 红 6 蓝，共 9 个。故 P(红|红) = 3/9 = 1/3。P(蓝|红) = 6/9 = 2/3。
If first is blue, remaining: 4 red, 5 blue, total 9. Then P(Red|Blue) = 4/9, P(Blue|Blue) = 5/9.
若第一次是蓝色，剩余：4 红 5 蓝，共 9。则 P(红|蓝) = 4/9，P(蓝|蓝) = 5/9。
The branch for both red: multiply along branches: (2/5) × (1/3) = 2/15. So probability both are red is 2/15.
两红分支：沿分支相乘：(2/5) × (1/3) = 2/15。因此两个都是红色的概率为 2/15。
Tree diagrams are extremely useful for sequential events.
树状图对于连续事件非常有用。
9. Solving Simultaneous Equations | 解联立方程组
Solve simultaneously: 2x + 3y = 16 and 5x − 2y = 2.
解联立方程组：2x + 3y = 16，5x − 2y = 2。
We can use elimination. Multiply the first equation by 2 and the second by 3 to make y-coefficients opposite: 4x + 6y = 32 and 15x − 6y = 6.
可以采用消元法。第一式乘 2，第二式乘 3，使 y 系数互为相反数：4x + 6y = 32 和 15x − 6y = 6。
Add the two equations: (4x+6y)+(15x−6y) = 32+6 → 19x = 38. So x = 2.
两式相加：(4x+6y)+(15x−6y) = 32+6 → 19x = 38。得 x = 2。
Substitute x = 2 into the first equation: 2(2) + 3y = 16 → 4 + 3y = 16 → 3y = 12 → y = 4. Check in second: 5(2) − 2(4) = 10 − 8 = 2. Correct.
将 x = 2 代入第一式：2(2) + 3y = 16 → 4 + 3y = 16 → 3y = 12 → y = 4。代入第二式检验：5(2) − 2(4) = 10 − 8 = 2，正确。
10. Nth Term of a Linear Sequence | 线性数列的第 n 项
Find the nth term of the sequence: 7, 11, 15, 19, 23, …
求数列 7, 11, 15, 19, 23, … 的第 n 项公式。
The difference between consecutive terms is 4. So the nth term is of the form 4n + c. To find c, substitute n = 1: 4(1) + c = 7 → c = 3.
相邻项的差为 4。所以第 n 项的形式为 4n + c。代入 n = 1，求 c：4(1) + c = 7 → c = 3。
Thus the nth term is 4n + 3. Check n = 2: 4(2) + 3 = 11, correct. n = 5: 4(5) + 3 = 23, correct.
因此第 n 项为 4n + 3。验证 n = 2：4(2) + 3 = 11，正确。n = 5：4(5) + 3 = 23，正确。
For linear sequences, always find the common difference first, then adjust with the zero term.
对于线性数列，总是先找到公差，再用第零项调整。
11. Compound Interest | 复利计算
£2000 is invested at a compound interest rate of 3% per annum. How much will the investment be worth after 4 years?
2000 英镑以年利率 3% 的复利投资。4 年后投资价值是多少？
Use the formula: A = P(1 + r/100)ⁿ, where P is principal, r is rate, n is number of years. Here P = 2000, r = 3, n = 4.
使用公式：A = P(1 + r/100)ⁿ，其中 P 为本金，r 为利率，n 为年数。这里 P = 2000，r = 3，n = 4。
Substitute: A = 2000 × (1 + 3/100)⁴ = 2000 × (1.03)⁴. Calculate (1.03)⁴: 1.03² = 1.0609, then 1.0609² = 1.12550881. So A ≈ 2000 × 1.12550881 = 2251.01762.
代入：A = 2000 × (1 + 3/100)⁴ = 2000 × (1.03)⁴。计算 (1.03)⁴：1.03² = 1.0609，再 1.0609² = 1.12550881。因此 A ≈ 2000 × 1.12550881 = 2251.01762。
Round to the nearest penny: £2251.02. The interest earned is £251.02.
四舍五入到便士：2251.02 英镑。获得利息 251.02 英镑。
12. Volume of a Prism | 棱柱的体积
A triangular prism has a cross-sectional area of 24 cm² and a length of 15 cm. Calculate its volume.
一个三棱柱的横截面积为 24 cm²，长度为 15 cm。计算它的体积。
Volume of a prism = area of cross-section × length. So volume = 24 × 15 = 360 cm³.
棱柱体积 = 横截面积 × 长度。因此体积 = 24 × 15 = 360 cm³。
You must ensure the cross-section is perpendicular to the length. This formula works for any uniform prism.
必须确保横截面与长度垂直。该公式适用于任何直棱柱。
If the cross-section area is not given, you may need to calculate it first from the shape’s dimensions.
如果没有直接给出横截面积，你可能需要先根据图形尺寸计算出面积。
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