Year 8 OCR Advanced Mathematics: Unit Test Mock Paper Breakdown | Year 8 OCR 进阶数学：单元测试模拟卷解析

📚 Year 8 OCR Advanced Mathematics: Unit Test Mock Paper Breakdown | Year 8 OCR 进阶数学：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test designed for Year 8 OCR Advanced Mathematics. Each section features a typical exam question, followed by a step-by-step solution and examiner tips to help you master key concepts.

本文详细解析了一套为 Year 8 OCR 进阶数学设计的单元测试模拟卷。每个小节包含一道典型考题，并配有逐步解答和考官提示，帮助你掌握核心概念。

1. Simplifying Algebraic Expressions | 代数表达式化简

Question: Simplify 3a + 5b – a + 2b – 4a.

（题目）化简：3a + 5b – a + 2b – 4a。

Combine like terms for a: 3a – a – 4a = –2a. Combine like terms for b: 5b + 2b = 7b. The simplified expression is –2a + 7b.

合并 a 的同类项：3a – a – 4a = –2a。合并 b 的同类项：5b + 2b = 7b。化简后的表达式为 –2a + 7b。

Tip: Always group the coefficients of identical variables before adding or subtracting. Check that you haven't lost any signs.

提示：先将相同字母的系数归为一组，再进行加减运算。仔细检查正负号，避免遗漏。

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2. Solving Linear Equations | 解线性方程

Question: Solve 2(x + 3) = 14 – x.

（题目）解方程：2(x + 3) = 14 – x。

Expand the left-hand side: 2x + 6 = 14 – both sides: 3x + 6 = 14. Subtract 6 from both sides: 3x = 8. Divide by 3: x = 8/3 or 2 ⅔.

左边展开：2x + 6 = 14 – x。两边同时加 x：3x + 6 = 14。两边同时减 6：3x = 8。两边除以 3：x = 8/3 或 2 ⅔。

Trick: Eliminate the brackets first, then collect x‑terms on one side and constants on the other. Always perform the same operation on both sides.

窍门：先去掉括号，然后将含 x 的项移到一边、常数项移到另一边。每一步都必须在等式两边同时进行相同运算。

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3. Operations with Fractions | 分数运算

Question: Work out 2 ⅓ ÷ 1 ¼. Give your answer as a mixed number in its simplest form.

（题目）计算 2 ⅓ ÷ 1 ¼，将答案写成最简带分数。

28/15. Convert back to a mixed number: 28/15 = 1 13/15. That is already in simplest form.

将带分数化为假分数：2 ⅓ = 7/3，1 ¼ = 5/4。除法即乘以其倒数：(7/3) × (4/5) = 28/15。再化回带分数：28/15 = 1 13/15，已是最简形式。

Common pitfall: raction when dividing. Always keep the first fraction, change ÷ to ×, and flip the second.

常见错误：在除法时忘记将第二个分数翻转。一定要牢记：保留第一个分数，变 ÷ 为 ×，再将第二个分数的分子分母颠倒。

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4. Percentage Increase and Decrease | 百分数增减

Question: A jacket priced at £80 is reduced by 15% in a sale. The sale price is then increased by 15%. Is the final price back to £80? Explain.

（题目）一件夹克原价 £80，先降价 15%，然后在此降价新价格基础上再涨价 15%。最终价格是否回到 £80？请解释。

Step 1: 15% of £80 = £12, so sale price = £80 – £12 = £68. Step 2: 15% of £68 = £10.20, so final price = £68 + £10.20 = £78.20. The final price is not £80; it is lower because the 15% increase is calculated on a smaller amount.

第一步：£80 的 15% 是 £12，降价后为 £68。第二步：£68 的 15% 是 £10.20，最终价格 £78.20。最终价格不是 £80，比原价低，因为第二次的 15% 是按更小的基数计算的。

Key idea: Percentage changes are not symmetric. A decrease of 15% followed by an increase of the same percentage always results in a smaller number.

核心思想：百分数变化不具有对称性。先降 15% 再涨相同百分数，结果总是小于原数。

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5. Ratio and Proportion | 比率与比例

Question: The ratio of boys to girls in a school is 5 : 3. There are 320 more boys than girls. How many students are there altogether?

（题目）某学校男生与女生的比是 5 : 3。男生比女生多 320 人。全校共有多少名学生？

Let the number of boys be 5x and girls be 3x. The difference is 5x – 3x = 2x = 320, so x = 160. Total students = 5x + 3x = 8x = 8 × 160 = 1280.

设男生人数为 5x，女生为 3x。差值为 5x – 3x = 2x = 320，得 x = 160。总人数 = 5x + 3x = 8x = 8 × 160 = 1280 人。

Strategy: Use a multiplier x to turn a ratio into algebraic expressions. Find x from the given difference or sum, then answer the question.

策略：用乘数 x 将比率转化为代数表达式。根据已知的差值或和求出 x，再回答问题。

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6. Angles in Polygons | 多边形内角

Question: A regular polygon has an interior angle of 156°. How many sides does it have?

（题目）一个正多边形的每个内角为 156°。它有几条边？

Interior + exterior angle = 180°, so exterior angle = 180° – °. The sum of exterior angles of any polygon is 360°. Number of sides n = 360° ÷ 24° = 15.

内角 + 外角 = 180°，所以外角 = 180° – 156° = 24°。任意多边形的外角和为 360°。边数 n = 360° ÷ 24° = 15。

Use fact: For a regular polygon, exterior angle = 180° – interior angle. The number of sides is 360° divided by the exterior angle.

牢记：正多边形的外角 = 180° – 内角。边数等于 360° 除以外角。

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7. Area and Perimeter of Compound Shapes | 复合图形的面积与周长

Question: The shape consists of a rectangle 10 cm by 4 cm with a right-angled triangle cut out. The triangle’s base is 3 cm and height 4 cm. Find the shaded area.

（题目）图形由一个 10 cm × 4 cm 的矩形挖去一个直角三角形组成。三角形底为 3 cm，高为 4 cm。求阴影部分面积。

0 × 4 = 40 cm². Area of triangle = ½ × base × height = ½ × 3 × 4 = 6 cm². Shaded area = 40 – 6 = 34 cm².

矩形面积 = 10 × 4 = 40 cm²。三角形面积 = ½ × 底 × 高 = ½ × 3 × 4 = 6 cm²。阴影面积 = 40 – 6 = 34 cm²。

Watch out: Make sure the triangle's dimensions are correct even when the shape is rotated. Subtract the unshaded region from the whole.

注意：即使图形被旋转，也要确保三角形的底和高正确。从总面积中减去非阴影部分的面积。

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8. Interpreting Statistical Diagrams | 统计图解读

Question: From the frequency table: Score 0–9: freq 4; 10–19: 6; 20–29: 12; 30–39: 8. Which class contains the median?

（题目）根据频数表：分数 0–9：频数 4；10–19：6；20–29：12；30–39：8。中位数位于哪一组？

6 + 12 + 8 = 30. Median position = (30 + 1) ÷ 2 = 15.5th value. Cumulative frequencies: 0–9: 4; 10–19: 10; 20–29: 22. The 15.5th value falls in the 20–29 class.

总频数 = 4 + 6 + 12 + 8 = 30。中位数位置 = (30 + 1) ÷ 2 = 第 15.5 个。累计频数：0–9: 4；10–19: 10；20–29: 22。第 15.5 个值位于 20–29 组。

Method: Find the cumulative frequencies; the class where the cumulative count first reaches or exceeds the median position is the median class.

方法：算累计频数；累计频数首次达到或超过中位位置的那个组即为中位数组。

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9. Basic Probability | 基础概率

Question: A bag contains 5 red, 3 blue and 2 green marbles. One marble is taken at random and not replaced. A second marble is taken. Find the probability both are red.

（题目）袋中有 5 红、3 蓝、2 绿弹珠。随机取出一颗不放回，再取第二颗。求两颗都是红色的概率。

en, 4 reds remain out of 9. P(second red) = 4/9. P(both red) = (1/2) × (4/9) = 4/18 = 2/9.

总数 10。P(第一颗红) = 5/10 = 1/2。取走一颗红后，剩余 4 红，共 9 颗。P(第二颗红) = 4/9。P(两红) = (1/2) × (4/9) = 4/18 = 2/9。

Remember: Without replacement changes the denominator for the second event. Multiply the probabilities along the branch of the tree diagram.

切记：不放回会改变第二次事件的样本空间大小。沿树状图分支路线的概率相乘。

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10. Finding the nth Term of Sequences | 求数列的通项

Question: Here are the first four terms of a sequence: 7, 13, 19, 25. Find an expression for the nth term.

（题目）数列前四项为：7, 13, 19, 25。求第 n 项的通项公式。

Find the difference between consecutive terms: 13 – 7 = 6, 19 – 13 = 6, 25 – r with common difference 6. So the nth term is of the form 6n + c. For n = 1: 6(1) + c = 7 ⇒ c = 1. Therefore the nth term is 6n + 1.

计算连续两项的差：13 – 7 = 6，19 – 13 = 6，25 – 19 = 6。公差为 6，是线性数列。通项形式为 6n + c。代入 n=1：6(1) + c = 7 ⇒ c = 1。所以通项为 6n + 1。

Quick check: Substitute n = 3: 6(3) + 1 = 19, which matches the third term. The rule works.

快速检验：代入 n=3，6(3)+1=19，与第三项相符。公式正确。

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11. Applying Pythagoras’ Theorem | 勾股定理应用

Question: A ladder 5 m long leans against a wall. The foot of the ladder is 1.5 m from the wall. How high up the wall does the ladder reach? (Give your answer to 2 d.p.)

（题目）一根 5 m 长的梯子斜靠在墙上，梯脚离墙 1.5 m。梯子顶端能达到墙多高？（保留两位小数）

a² + b² = c², where c is the hypotenuse (ladder). Let the height be h. Then h² + 1.5² = 5² ⇒ h² + 2.25 = 25 ⇒ h² = 22.75 ⇒ h = √22.75 ≈ 4.77 m (2 d.p.).

利用 a² + b² = c²，其中 c 为斜边（梯子）。设高度为 h。则 h² + 1.5² = 5² ⇒ h² + 2.25 = 25 ⇒ h² = 22.75 ⇒ h = √22.75 ≈ 4.77 m（保留两位小数）。

Tip: Always identify the hypotenuse correctly – it is the longest side, opposite the right angle. In context, the ladder is the hypotenuse.

提示：务必正确识别斜边——最长边，对直角。在此情境中，梯子为斜边。

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12. Mock Test Strategy | 模拟测试应试技巧

Before the test, revise core formulae and practise questions under timed conditions. Use the mark scheme to understand how working marks are allocated. Read each question carefully, show every step, and check units.

考前复习核心公式，并限时练习。利用评分标准了解步骤分的分配。仔细审题，写出每一步解答，检查单位。

During the test, start with the questions you find easiest to build confidence. For multi‑step problems, write down given information and what you need to find. Manage time wisely, and if stuck, move on and return later.

考试时先做最拿手的题目以建立信心。遇到多步骤问题，先列出已知条件和求解目标。合理分配时间，如果卡住了先跳过去，之后再回头。

Finally, always leave a few minutes to review your answers, especially common errors such as sign mistakes in algebra or misreading percentage problems.

最后，务必留出几分钟检查答案，尤其注意常见错误，如代数里的正负号错误，或对百分数问题的误读。

Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com

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