📚 Year 8 OCR Advanced Maths: In-depth Analysis of Past Papers | Year 8 OCR 进阶数学：历年真题深度解析
Past papers are the most powerful tool for mastering the Year 8 OCR Advanced Mathematics curriculum. By working through real exam questions, students not only familiarise themselves with the format and style of questions but also identify recurring topics and common pitfalls. This article provides a comprehensive breakdown of key question types from recent years, offering step-by-step solutions and bilingual commentary to strengthen understanding and exam technique.
历年真题是掌握 Year 8 OCR 进阶数学课程的最有力工具。通过练习真实考题，学生不仅能熟悉题型与出题风格，还能识别反复出现的主题和常见陷阱。本文对近年来的典型题目进行了全面解析，提供分步解答和双语点评，以强化理解和应试技巧。
1. Algebraic Expansion and Factorisation | 代数展开与因式分解
Expanding brackets and factorising expressions appear in almost every OCR Advanced Maths paper. A strong grasp of the distributive law and sign rules is essential.
括号展开与因式分解几乎出现在每一份 OCR 进阶数学试卷中。牢固掌握分配律和符号规则至关重要。
Past Paper Example (OCR, 2019): Expand and simplify 5(2x − 3) − 3(4 − x).
真题示例（OCR，2019）：展开并化简 5(2x − 3) − 3(4 − x)。
rst bracket: 5 × 2x = 10x, and 5 × (−3) = −15. The expression becomes 10x − 15 − 3(4 − x).
步骤1：去掉第一个括号：5×2x=10x，5×(−3)=−15。表达式变为 10x − 15 − 3(4 − x)。
the second bracket, paying attention to the minus sign: −3 × 4 = −12, and −3 × (−x) = +3x.
步骤2：展开第二个括号，注意负号：−3×4=−12，−3×(−x)=+3x。
erms together: 10x − 15 − 12 + 3x. Combine like terms: 10x + 3x = 13x, and −15 − 12 = −27.
步骤3：写出所有项：10x − 15 − 12 + 3x。合并同类项：10x+3x=13x，−15−12=−27。
Final Answer: 13x − 27
Common mistakes include forgetting to distribute the negative sign to both terms inside the bracket, or mishandling the product of two negatives. The table below highlights these pitfalls.
常见错误包括忘记将负号分配给括号内的两项，或错误处理两个负数相乘。下表说明了这些常见陷阱。
| Mistake | Correction |
|---|---|
| −3(4 − x) → −12 − 3x | −3 × (−x) = +3x, so it should be −12 + 3x |
| Forgetting to simplify 10x + 3x | Always collect like terms to reach the simplest form |
2. Linear Equations and Inequalities | 线性方程与不等式
Solving linear equations requires balancing both sides and systematically isolating the variable. Inequalities extend this skill with an additional rule about reversing the sign when multiplying or dividing by a negative number.
解线性方程需要平衡等式两边并系统性地分离变量。不等式则在此基础上增加了一条规则：当乘以或除以负数时需要反转符号。
Past Paper Example (OCR, 2020): Solve the equation 4x + 7 = 3x − 2. Hence solve the inequality 4x + 7 < 3x − 2.
真题示例（OCR，2020）：解方程 4x + 7 = 3x − 2。并由此解不等式 4x + 7 < 3x − 2。
For the equation: Subtract 3x from both sides → x + 7 = −2. Then subtract 7 → x = −9.
对于方程：两边减去 3x → x + 7 = −2。然后减去 7 → x = −9。
+ 7 < −2 → x < −9. Notice the sign did not reverse because we never divided by a negative.
对于不等式：4x + 7 < 3x − 2 → x + 7 < −2 → x < −9。注意符号未反转，因为我们从未除以负数。
A typical exam trap is an inequality like −2x > 6. Dividing by −2 gives x < −3. Remember to flip the inequality symbol.
典型的考试陷阱是不等式如 −2x > 6。除以 −2 得到 x < −3。记得反转不等号。
3. Introduction to Quadratic Equations | 二次方程入门
nd can be solved by factorising into two binomials. Recognising factor pairs of the constant term is key.
在这个阶段，二次方程通常以 x² + bx + c = 0 的形式出现，并可通过因式分解为两个二项式求解。识别常数项的因子对是关键。
Past Paper Example (OCR, 2017): Solve x² + 5x + 6 = 0 by factorisation.
真题示例（OCR，2017）：用因式分解法解 x² + 5x + 6 = 0。
to +5. The pairs for 6 are (1,6) and (2,3). 2 + 3 = 5, so the factors are (x + 2)(x + 3).
寻找两个数，乘积为 +6，和为 +5。6 的因子对有 (1,6) 和 (2,3)。2+3=5，因此因式为 (x+2)(x+3)。
r to zero: x + 2 = 0 → x = −2, and x + 3 = 0 → x = −3. The solutions are x = −2 or x = −3.
令每个因式为零：x+2=0 → x=−2，以及 x+3=0 → x=−3。解为 x=−2 或 x=−3。
In more challenging questions, the coefficient of x² may not be 1. For example, 2x² + 7x + 3 factorises to (2x + 1)(x + 3). Always expand to verify.
在更具挑战性的题目中，x² 的系数可能不是 1。例如，2x²+7x+3 分解为 (2x+1)(x+3)。务必展开验证。
4. Sequences and the nth Term Rule | 数列与第 n 项公式
Linear sequences are tested frequently. Students must find the nth term and use it to determine whether a specific number belongs to the sequence.
线性数列经常被考查。学生必须找到第 n 项公式，并利用它判断某个特定数字是否属于该数列。
Past Paper Example (OCR, 2021): The sequence begins 5, 9, 13, 17, ... (a) Find the nth term. (b) Is 101 a term in the sequence? Justify your answer.
真题示例（OCR，2021）：数列开始为 5, 9, 13, 17, ... (a) 求第 n 项公式。 (b) 101 是数列中的一项吗？证明你的答案。
hen n = 1, 4 × 1 = 4, but the first term is 5, so we need +1. The nth term is Tₙ = 4n + 1.
项之间的差为 +4，因此第 n 项以 4n 开始。当 n=1 时，4×1=4，但第一项为 5，故需加 1。第 n 项公式为 Tₙ = 4n + 1。
et 4n + 1 = 101 → 4n = 100 → n = 25, which is a positive integer. So 101 is the 25th term.
对于 (b)，设 4n+1=101 → 4n=100 → n=25，为正整数。因此 101 是第 25 项。
Common exam request: 'Find the 50th term' – substitute n = 50 into Tₙ = 4n + 1 to get 201.
常见考题：“求第 50 项”——将 n=50 代入 Tₙ=4n+1，得到 201。
5. Angles in Parallel Lines and Polygons | 平行线与多边形的角
Angle reasoning questions test knowledge of alternate, corresponding and co-interior angles, often combined with interior angles of regular polygons.
角度推理题考查内错角、同位角、同旁内角的知识，常常结合正多边形的内角一起出现。
Past Paper Example (OCR, 2018): In the diagram, a pair of parallel lines is cut by a transversal. One angle is given as 65°. Find the sizes of angles a, b and c, giving reasons.
真题示例（OCR，2018）：如图，一对平行线被一条截线所截。已知一角为 65°。求角 a、b 和 c 的度数，并说明理由。
ther parallel line is also 65° (angle a). Angle b is vertically opposite to a, so b = 65°.
假设 65° 是一个锐角，另一条平行线上的同位角也是 65°（角 a）。角 b 是 a 的对顶角，因此 b=65°。
o they sum to 180°: c = 180° − 65° = 115°. Reasons: angles on a straight line sum to 180°.
角 c 与 65° 角在一条直线上，因此它们之和为 180°：c=180°−65°=115°。理由：直线上的角之和为 180°。
interior angle of a regular n-gon: (n−2)×180° / n. A regular hexagon: (6−2)×180°/6 = 120°.
涉及多边形时，使用正 n 边形的内角公式：(n−2)×180°/n。正六边形：(6−2)×180°/6=120°。
6. Pythagoras' Theorem Applications | 勾股定理的应用
Questions on Pythagoras' theorem often move beyond simple triangles to real-life contexts, such as finding the length of a ladder leaning against a wall or the diagonal of a rectangle.
关于勾股定理的题目常常超越简单的三角形，延伸到现实生活情境，比如求靠墙梯子的长度或矩形的对角线。
Past Paper Example (OCR, 2019): A ladder of length 5 m leans against a vertical wall. The foot of the ladder is 1.5 m from the base of the wall. How high up the wall does the ladder reach?
真题示例（OCR，2019）：一架长 5 m 的梯子斜靠在竖直墙壁上。梯脚距墙基 1.5 m。梯子能到达墙上的多高处？
Let the height be h. By Pythagoras: h² + (1.5)² = 5² → h² + 2.25 = 25 → h² = 22.75 → h = √22.75 ≈ 4.77 m (to 3 s.f.).
设高度为 h。根据勾股定理：h²+(1.5)²=5² → h²+2.25=25 → h²=22.75 → h=√22.75≈4.77 m（保留 3 位有效数字）。
Always check whether the side to be found is the hypotenuse or a shorter side. If the hypotenuse is missing, add the squares; if a leg is missing, subtract.
始终检查需求的是斜边还是直角边。若求斜边，则平方相加；若求直角边，则相减。
7. Area, Perimeter and Volume of Compound Shapes | 复合图形的面积、周长与体积
Compound shapes require splitting the figure into standard forms, calculating individual areas or volumes, and combining them correctly. Common elements include semicircles, triangles and cuboids.
复合图形需要将图形分割为标准形状，计算各个面积或体积，然后正确组合。常见元素包括半圆、三角形和长方体。
Past Paper Example (OCR, 2020): A shape is made from a rectangle of width 6 m and height 4 m, with a semicircle of diameter 6 m attached on one of the longer sides. Find the total area to 3 significant figures.
真题示例（OCR，2020）：一个图形由一个宽 6 m、高 4 m 的矩形，以及一个连接在较长边上、直径为 6 m 的半圆组成。求总面积，保留 3 位有效数字。
emicircle = ½ × π × r² = ½ × π × 9 = 4.5π ≈ 14.137 m². Total area ≈ 24 + 14.137 = 38.1 m².
矩形面积 = 6×4=24 m²。半圆半径 r=3 m。半圆面积=½×π×r² = ½×π×9 = 4.5π ≈ 14.137 m²。总面积≈24+14.137=38.1 m²。
For volume, a common question is a prism with a compound cross-section. Multiply the area of the cross-section by the length of the prism.
对于体积，常见题目是一个具有复合截面的棱柱。用截面面积乘以棱柱的长度即可。
8. Ratio, Proportion and Rates of Change | 比、比例与变化率
Ratio problems often involve sharing an amount in a given ratio or scaling recipes. Proportional reasoning is also tested through conversion graphs and speed calculations.
比例问题常常涉及按给定比例分配金额或按比例调整食谱。比例推理也会通过转换图表和速度计算进行考查。
Past Paper Example (OCR, 2016): Anna, Ben and Cindy share £120 in the ratio 3 : 5 : 2. How much does Ben receive?
真题示例（OCR，2016）：Anna、Ben 和 Cindy 按 3 : 5 : 2 的比例分享 £120。Ben 得到多少？
Total parts = 3 + 5 + 2 = 10. Each part is worth £120 ÷ 10 = £12. Ben gets 5 parts → 5 × £12 = £60.
总份数 = 3+5+2=10。每份价值 £120÷10=£12。Ben 得 5 份 → 5×£12=£60。
Rates of change may appear as distance-time graphs. The gradient represents speed; constant speed implies a straight line. Always label units.
变化率可能以距离-时间图的形式出现。斜率表示速度；匀速运动意味着直线。务必标注单位。
9. Statistical Graphs and Averages | 统计图表与平均数
OCR papers frequently include bar charts, pie charts and stem-and-leaf diagrams. Students are expected to interpret data and calculate mean, median, mode and range.
OCR 试卷中经常包含条形图、饼图和茎叶图。要求学生解读数据，并计算平均数、中位数、众数和极差。
Past Paper Example (OCR, 2022): The stem-and-leaf diagram shows the test scores of 20 students. Key: 3|7 means 37. 2|2 5 8, 3|0 3 4 4 7 9, 4|1 2 2 5 6 8 8, 5|0 1 3. Find the median score.
真题示例（OCR，2022）：茎叶图显示了 20 名学生的测验分数。图例：3|7 表示 37。茎叶图：2|2 5 8, 3|0 3 4 4 7 9, 4|1 2 2 5 6 8 8, 5|0 1 3。求中位数分数。
2,25,28,30,33,34,34,37,39,41,42,... The 10th is 41, 11th is 42. Median = (41+42)/2 = 41.5.
首先按顺序列出所有分数。学生总数=20。中位数位于第 10 和第 11 个值之间。计数：22,25,28,30,33,34,34,37,39,41,42,... 第 10 个是 41，第 11 个是 42。中位数 = (41+42)/2 = 41.5。
r. For example, if 12 out of 30 students walk to school, the angle is (12/30)×360° = 144°.
饼图题目需要求出扇形的角度。例如，若 30 名学生中有 12 人走路上学，则扇形的角度为 (12/30)×360° = 144°。
10. Probability with Two-Way Tables | 概率与双向表
Probability questions at this level often involve two-way tables for combined events, requiring students to understand 'and' and 'or' rules, as well as conditional language.
这一阶段的概率题目常常涉及组合事件的双向表，要求学生理解“且”和“或”的规则，以及条件概率的语言表达。
Past Paper Example (OCR, 2017): A survey of 50 students asks whether they like tea (T) or coffee (C). The results: 20 like both, 12 like only tea, 8 like only coffee, and the rest like neither. (a) Draw a two-way table. (b) Find the probability a randomly selected student likes tea. (c) Given a student likes coffee, find the probability they also like tea.
真题示例（OCR，2017）：对 50 名学生进行调查，询问是否喜欢茶 (T) 或咖啡 (C)。结果：20 人两者都喜欢，12 人只喜欢茶，8 人只喜欢咖啡，其余人都不喜欢。(a) 绘制双向表。(b) 求随机选取的学生喜欢茶的概率。(c) 已知一位学生喜欢咖啡，求其也喜欢茶的概率。
a lovers = 20 + 12 = 32. Total coffee lovers = 20 + 8 = 28. Neither = 50 − (20+12+8) = 10.
表格：喜欢茶的总数 = 20+12=32。喜欢咖啡的总数 = 20+8=28。都不喜欢 = 50−(20+12+8)=10。
Published by TutorHao | Year 8 进阶数学 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图