GCSE CIE Mathematics: Past Paper Exam Solutions | GCSE CIE 数学：历年真题解析

📚 GCSE CIE Mathematics: Past Paper Exam Solutions | GCSE CIE 数学：历年真题解析

Working through actual CIE GCSE Mathematics past papers is the most effective way to understand the exam format, question styles and mark schemes. This article selects ten representative questions from recent years, providing step‑by‑step solutions, examiner tips and common pitfalls. Use these walkthroughs to sharpen your problem‑solving skills and boost your confidence before the final assessment.

刷透 CIE GCSE 数学历年真题是熟悉考试题型、评分标准和出题套路的最直接途径。本文精选近十年真题中的十道经典题型，逐一给出分步解析、阅卷人点评与易错提醒。用这些精讲练好解题肌肉，你一定能在考场上更稳更准。

1. Percentage Increase and Decrease | 百分数增减

Past Paper Example: A laptop is priced at £480. In a sale, the price is reduced by 15%. Find the sale price.

真题示例：一台笔记本电脑原价 £480，商场打八五折（降价 15%）。求折扣后的售价。

Step 1: Identify the multiplier for a 15% decrease. A reduction of 15% means the new price is 100% – 15% = 85% of the original. The decimal multiplier is 0.85.

第 1 步：计算降价的乘数。降价 15% 意味着现价是原价的 100% – 15% = 85%，因此乘数为 0.85。

Step 2: Multiply the original price by the multiplier. 480 × 0.85 = £408.

第 2 步：用原价乘这个乘数。480 × 0.85 = £408。

Examiner tip: If the question asks for the amount of discount, calculate 480 × 0.15 = £72 first and subtract. Both methods are accepted. Always write the unit (£) in your final answer.

阅卷人提示：如果题目要求降价金额，可以先算 480 × 0.15 = £72，再相减。两种方法都得分，但最后答案一定要带上单位 (£)。

2. Solving Linear Equations | 解一元一次方程

Past Paper Example: Solve 5x – 3 = 2x + 9.

真题示例：解方程 5x – 3 = 2x + 9。

Step 1: Collect the x‑terms on one side. Subtract 2x from both sides: 5x – 3 – 2x = 2x + 9 – 2x → 3x – 3 = 9.

第 1 步：把含 x 的项移到一边。两边同时减去 2x：5x – 3 – 2x = 2x + 9 – 2x → 3x – 3 = 9。

Step 2: Isolate the x‑term. Add 3 to both sides: 3x = 12.

第 2 步：分离含 x 项。两边加 3：3x = 12。

Step 3: Solve for x. Divide both sides by 3: x = 4.

第 3 步：解出 x。两边除以 3：x = 4。

x = 4

Check: Substitute x = 4 into the original equation: 5(4) – 3 = 20 – 3 = 17, and 2(4) + 9 = 8 + 9 = 17. Both sides match.

检验：把 x = 4 代回原方程：5(4) – 3 = 17，2(4) + 9 = 17，两边相等。

3. Factorising Quadratic Expressions | 因式分解二次式

Past Paper Example: Solve x² – 8x + 15 = 0 by factorisation.

真题示例：用因式分解法解方程 x² – 8x + 15 = 0。

Step 1: Identify two numbers that multiply to +15 and add to -8. These are -3 and -5 because (-3) × (-5) = 15 and (-3) + (-5) = -8.

第 1 步：找出两个数，它们的积为 +15，和为 -8。这两个数是 -3 和 -5，因为 (-3) × (-5) = 15，(-3) + (-5) = -8。

Step 2: Write the factorised form: (x – 3)(x – 5) = 0.

第 2 步：写出因式分解形式：(x – 3)(x – 5) = 0。

Step 3: Set each factor equal to zero. x – 3 = 0 → x = 3; x – 5 = 0 → x = 5.

第 3 步：令每个因式等于零。x – 3 = 0 → x = 3；x – 5 = 0 → x = 5。

x = 3 or x = 5

Common mistake: Forgetting the signs. If the constant term is positive and the middle term is negative, both factors will have negative signs.

常见错误：搞错符号。常数项为正且中间项为负时，两个因式都是减号。

4. Simultaneous Equations by Elimination | 消元法解联立方程

Past Paper Example: Solve the simultaneous equations: 2x + y = 10 and x – y = 2.

真题示例：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.

第 1 步：两式相加消去 y。(2x + y) + (x – y) = 10 + 2 → 3x = 12。

Step 2: Solve for x. 3x = 12 → x = 4.

第 2 步：解出 x。3x = 12 → x = 4。

Step 3: Substitute x = 4 into the simpler equation (x – y = 2). 4 – y = 2 → y = 2.

第 3 步：将 x = 4 代入较简单的方程 (x – y = 2)。4 – y = 2 → y = 2。

x = 4, y = 2

Check: 2(4) + 2 = 10, and 4 – 2 = 2. Both equations are satisfied.

检验：2(4) + 2 = 10，4 – 2 = 2。两个方程都成立。

5. Pythagoras’ Theorem in 2D | 二维空间勾股定理

Past Paper Example: A right‑angled triangle has legs of length 6 cm and 8 cm. Find the length of the hypotenuse.

真题示例：一直角三角形两直角边分别为 6 cm 和 8 cm，求斜边长度。

Step 1: Recall Pythagoras’ theorem: c² = a² + b², where c is the hypotenuse.

第 1 步：回忆勾股定理：c² = a² + b²，其中 c 为斜边。

Step 2: Substitute the given lengths. a = 6 cm, b = 8 cm → c² = 6² + 8² = 36 + 64 = 100.

第 2 步：代入已知边长。a = 6 cm, b = 8 cm → c² = 6² + 8² = 36 + 64 = 100。

Step 3: Take the square root. c = √100 = 10 cm.

第 3 步：开平方。c = √100 = 10 cm。

The hypotenuse is 10 cm.

Pitfall: Do not forget the units. If the question gives mm, answer in mm. Make sure your final answer is labelled.

易错提醒：别忘了单位。题目给的是 cm，答案也必须是 cm。所有最终结果都要写单位。

6. Trigonometric Ratios (SOH CAH TOA) | 三角函数比（SOH CAH TOA）

Past Paper Example: In a right‑angled triangle, the angle at A is 35° and the adjacent side is 12 m. Find the length of the opposite side.

真题示例：一直角三角形中，∠A = 35°，邻边长 12 m，求对边长度。

Step 1: Identify the relevant ratio. We have the adjacent side and want the opposite side, so use tan θ = opposite / adjacent.

第 1 步：选择适用的三角比。已知邻边，要求对边，因此用正切：tan θ = 对边 / 邻边。

Step 2: Set up the equation. tan 35° = opposite / 12.

第 2 步：列出方程。tan 35° = 对边 / 12。

Step 3: Rearrange and calculate. opposite = 12 × tan 35°. Using a calculator, tan 35° ≈ 0.7002, so opposite ≈ 12 × 0.7002 = 8.4024 m ≈ 8.40 m (3 significant figures).

第 3 步：移项求解。对边 = 12 × tan 35°。用计算器，tan 35° ≈ 0.7002，故对边 ≈ 8.40 m（保留三位有效数字）。

opposite side ≈ 8.40 m

Examiner advice: Always check your calculator is in degree mode, not radian mode. Round your final answer as specified (usually 3 s.f. unless the question says otherwise).

阅卷建议：务必确认计算器处于角度模式（DEG）而非弧度模式。最终答案按题目要求取有效数字（通常三位有效数字）。

7. Probability Tree Diagrams (Independent Events) | 独立事件树状图

Past Paper Example: A bag contains 4 red and 5 blue counters. One counter is taken at random, its colour noted, and then returned to the bag. A second counter is taken. Draw a tree diagram and find the probability that both counters are red.

真题示例：袋中有 4 个红筹码和 5 个蓝筹码。随机取一个，记录颜色后放回，再取第二个。画出树状图，并求两次都取到红筹码的概率。

Step 1: Note the probabilities for the first pick: P(red) = 4/9, P(blue) = 5/9. Because the counter is replaced, the probabilities for the second pick are identical.

第 1 步：第一次抽取的概率：P(红) = 4/9，P(蓝) = 5/9。由于放回，第二次抽取的概率完全相同。

Step 2: The path for “both red” is the first red branch followed by the second red branch. Multiply along the branch: 4/9 × 4/9 = 16/81.

第 2 步：“两次红”的路径是第一次红分支接着第二次红分支。沿分支相乘：4/9 × 4/9 = 16/81。

P(both red) = 16/81

Reminder: If the question had been “without replacement”, the probabilities on the second stage would change. Always read the question carefully to see if counters are replaced or not.

提醒：如果题目改成“不放回”，第二阶段概率会变化。审题时务必看清是否放回。

8. Cumulative Frequency and Quartiles | 累积频数及四分位数

Past Paper Example: The table shows the heights of 40 students. Construct a cumulative frequency table and estimate the median height.

Height (cm)	Frequency
140 < h ≤ 150	6
150 < h ≤ 160	14
160 < h ≤ 170	12
170 < h ≤ 180	8

真题示例：下表给出 40 名学生的身高。构建累积频数表，并估计中位数身高。

Step 1: Compute cumulative frequencies: 140‑150 cm: 6; 150‑160 cm: 6+14=20; 160‑170 cm: 20+12=32; 170‑180 cm: 32+8=40.

第 1 步：计算累积频数：140‑150 cm: 6；150‑160 cm: 20；160‑170 cm: 32；170‑180 cm: 40。

Step 2: The median position is at n/2 = 40/2 = 20th value. From the cumulative table, the 20th value lies in the 150‑160 cm class.

第 2 步：中位数位置在 n/2 = 20。从累积频数表可知第 20 个数据落在 150‑160 cm 组。

Step 3: Use linear interpolation (or simply read from a cumulative frequency graph). For quick estimate: median ≈ lower boundary + (group frequency) × (position within group) = 150 + (10 / 14) × 10 ≈ 150 + 7.1 = 157.1 cm.

第 3 步：用线性插值（或从累积频数图上读取）。中位数 ≈ 150 + (10/14)×10 ≈ 157.1 cm。

Median ≈ 157 cm (to 3 s.f.)

9. Straight Line Graphs | 直线图形

Past Paper Example: A straight line passes through the points (2, 7) and (6, 15). Find the equation of the line in the form y = mx + c.

真题示例：一条直线经过 (2, 7) 和 (6, 15) 两点，求该直线的方程，写成 y = mx + c 的形式。

Step 1: Calculate the gradient m. m = (y₂ – y₁) / (x₂ – x₁) = (15 – 7) / (6 – 2) = 8 / 4 = 2.

第 1 步：求斜率 m。m = (15 – 7) / (6 – 2) = 8 / 4 = 2。

Step 2: Substitute m and one point into y = mx + c. Using (2,7): 7 = 2(2) + c → 7 = 4 + c → c = 3.

第 2 步：把 m 和一个点的坐标代入 y = mx + c。用 (2,7)：7 = 2×2 + c → 7 = 4 + c → c = 3。

Step 3: Write the final equation: y = 2x + 3.

第 3 步：写出最终方程：y = 2x + 3。

Equation: y = 2x + 3

10. Circle Theorem – Angles in the Same Segment | 圆定理——同弓形内的圆周角

Past Paper Example: Points A, B, C and D lie on a circle. The chord AB subtends angle ACB = 42° at point C and angle ADB at point D. Find angle ADB.

真题示例：点 A, B, C, D 在同一个圆上。弦 AB 在点 C 所对的圆周角 ∠ACB = 42°，在点 D 所对的圆周角为 ∠ADB。求 ∠ADB。

Step 1: Recognise that angles in the same segment are equal. Both ∠ACB and ∠ADB stand on the same arc AB (the arc not containing C and D).

第 1 步：识别出同弓形内的圆周角相等。∠ACB 和 ∠ADB 都对同一条弧 AB（不含 C 与 D 的那段弧），因此它们相等。

Step 2: Therefore, ∠ADB = ∠ACB = 42°.

第 2 步：所以 ∠ADB = 42°。

∠ADB = 42°

Tip: Always refer to the specific arc when explaining your reasoning. Examiners expect the phrase “angles in the same segment” or “angles subtended by the same chord in the same segment are equal”.

技巧：解释理由时一定要指明具体的弧。阅卷人希望看到“同弓形内的圆周角相等”或“同一弦在同侧所对的圆周角相等”这样的表述。

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GCSE CIE Mathematics: Past Paper Exam Solutions | GCSE CIE 数学：历年真题解析