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

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

This article provides a step-by-step walkthrough of a typical Year 8 Cambridge Advanced Mathematics unit test mock paper. It covers core topics including algebraic techniques, fractions and percentages, ratio, indices, sequences, geometry of polygons and composite shapes, statistical measures, and probability. Each worked example is carefully explained to strengthen conceptual understanding and boost exam confidence.

本文逐步解析一套典型的 Year 8 剑桥进阶数学单元测试模拟卷，涵盖代数方法、分数与百分数、比与比例、指数、数列、多边形与复合图形几何、统计量以及概率等核心主题。每个范例都经过细致讲解，旨在加深概念理解并提升应试信心。

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1. Simplifying and Expanding Expressions | 化简与展开表达式

Question: Simplify 3a + 2b – a + 4b and expand 2(x – 3) + 3(4 – x).

题目：化简 3a + 2b – a + 4b 并展开 2(x – 3) + 3(4 – x)。

Step 1: For the first part, identify like terms. The a‑terms are 3a and –a; the b‑terms are 2b and 4b. Combine them: 3a – a = 2a and 2b + 4b = 6b. The simplified expression is 2a + 6b.

步骤 1：第一部分先找出同类项。含 a 的项为 3a 和 –a；含 b 的项为 2b 和 4b。合并：3a – a = 2a，2b + 4b = 6b。化简后的表达式为 2a + 6b。

Step 2: For the expansion, distribute the numbers outside the brackets. 2(x – 3) becomes 2x – 6, and 3(4 – x) becomes 12 – 3x. Write the sum: 2x – 6 + 12 – 3x. Now collect like terms: 2x – 3x = –x and –6 + 12 = 6. The result is –x + 6 or 6 – x.

步骤 2：展开部分需将括号外的数分配进去。2(x – 3) 得到 2x – 6，3(4 – x) 得到 12 – 3x。写成加法：2x – 6 + 12 – 3x。再合并同类项：2x – 3x = –x，–6 + 12 = 6。结果为 –x + 6 或 6 – x。

Final answers: 2a + 6b and 6 – x

最终答案：2a + 6b 与 6 – x

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2. Solving Linear Equations with Fractions | 解含分数的一元一次方程

Question: Solve (2x – 5) / 3 = 7.

题目：解方程 (2x – 5) / 3 = 7。

Step 1: Eliminate the denominator by multiplying both sides by 3. This gives 2x – 5 = 21.

步骤 1：两边同时乘以 3，消去分母，得到 2x – 5 = 21。

Step 2: Add 5 to both sides to isolate the term with x: 2x = 26.

步骤 2：两边加 5，使含 x 的项单独在一边：2x = 26。

Step 3: Divide by 2 to find x = 13. Always substitute back to verify: (2 × 13 – 5) / 3 = (26 – 5) / 3 = 21 / 3 = 7. Correct.

步骤 3：两边除以 2，得 x = 13。务必代回检验：(2 × 13 – 5) / 3 = (26 – 5) / 3 = 21 / 3 = 7，正确。

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3. Fractions, Decimals and Percentages | 分数、小数与百分数

Question: (a) Convert 5/8 to a decimal and a percentage. (b) Calculate 35% of 240.

题目：(a) 将 5/8 转换为小数和百分数。(b) 计算 240 的 35%。

For (a), divide 5 by 8: 5 ÷ 8 = 0.625. To get a percentage, multiply by 100%: 0.625 × 100% = 62.5%. So 5/8 = 0.625 = 62.5%.

(a) 5 除以 8：5 ÷ 8 = 0.625。转换为百分数则乘以 100%：0.625 × 100% = 62.5%。因此 5/8 = 0.625 = 62.5%。

For (b), 35% means 35/100. Multiply: 35/100 × 240 = 0.35 × 240 = 84. Alternatively, find 10% (24) and 5% (12), then 3 × 24 + 12 = 84.

(b) 35% 即 35/100。相乘：35/100 × 240 = 0.35 × 240 = 84。也可先求 10%（24）和 5%（12），然后 3 × 24 + 12 = 84。

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4. Ratio and Proportion | 比与比例

Question: (a) Divide £120 in the ratio 3 : 5. (b) A recipe uses 200 g of flour for 12 biscuits. How much flour is needed for 30 biscuits?

题目：(a) 将 £120 按 3 : 5 分配。(b) 一份食谱用 200 g 面粉做 12 块饼干，做 30 块饼干需要多少面粉？

For (a), total parts = 3 + 5 = 8. One part = £120 ÷ 8 = £15. First share = 3 × £15 = £45; second share = 5 × £15 = £75. Check: £45 + £75 = £120.

(a) 总份数 = 3 + 5 = 8。一份金额 = £120 ÷ 8 = £15。第一份为 3 × £15 = £45，第二份为 5 × £15 = £75。检验：£45 + £75 = £120。

12) = 30 × 16.67 = 500 g. Or find the scale factor: 30 / 12 = 2.5, so 200 g × 2.5 = 500 g.

(b) 建立比例关系。每块饼干用面粉 200 g / 12 ≈ 16.67 g。30 块饼干需要 30 × (200 / 12) = 500 g。也可用倍率法：30 / 12 = 2.5，则 200 g × 2.5 = 500 g。

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5. Laws of Indices | 指数律

Question: Simplify (2³ × 2⁴) / (2²)⁴.

题目：化简 (2³ × 2⁴) / (2²)⁴。

Step 1: Apply the multiplication law: aᵐ × aⁿ = aᵐ⁺ⁿ. Here, 2³ × 2⁴ = 2³⁺⁴ = 2⁷.

步骤 1：运用同底数幂的乘法法则：aᵐ × aⁿ = aᵐ⁺ⁿ。此处 2³ × 2⁴ = 2³⁺⁴ = 2⁷。

Step 2: Use the power of a power law: (aᵐ)ⁿ = aᵐⁿ. So (2²)⁴ = 2²ˣ⁴ = 2⁸.

步骤 2：运用幂的乘方法则：(aᵐ)ⁿ = aᵐⁿ。所以 (2²)⁴ = 2²ˣ⁴ = 2⁸。

2⁷ / 2⁸. Apply the division law: aᵐ / aⁿ = aᵐ⁻ⁿ, giving 2⁷⁻⁸ = 2⁻¹. This is equal to 1/2.

步骤 3：现在得到 2⁷ / 2⁸。运用同底数幂的除法法则：aᵐ / aⁿ = aᵐ⁻ⁿ，得 2⁷⁻⁸ = 2⁻¹。这等于 1/2。

Simplified expression: 2⁻¹ or ½

化简结果：2⁻¹ 或 ½

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6. Sequences and the nth Term | 数列与第 n 项

Question: The sequence begins 3, 8, 13, 18, ... Find the nth term and the 50th term.

题目：数列前几项为 3, 8, 13, 18, ... 求第 n 项及第 50 项。

thmetic, with first term a = 3 and common difference d = 5. The nth term formula is a + (n – 1)d. Substituting gives 3 + (n – 1) × 5 = 3 + 5n – 5 = 5n – 2.

相邻两项的差为 +5，这是一个等差数列，首项 a = 3，公差 d = 5。第 n 项公式为 a + (n – 1)d。代入得 3 + (n – 1) × 5 = 3 + 5n – 5 = 5n – 2。

For the 50th term, let n = 50: 5 × 50 – 2 = 250 – 2 = 248.

第 50 项将 n = 50 代入：5 × 50 – 2 = 250 – 2 = 248。

nth term = 5n – 2; 50th term = 248

第 n 项：5n – 2；第 50 项：248

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7. Interior Angles of Regular Polygons | 正多边形的内角

Question: A regular polygon has an interior angle of 140°. How many sides does it have?

题目：一个正多边形的每个内角为 140°，它有多少条边？

or angle and exterior angle at a vertex sum to 180°. Therefore, each exterior angle = 180° – 60° ÷ exterior angle = 360° ÷ 40° = 9. Alternatively, use the formula: interior angle = (n – 2) × 180° / n. Set this equal to 140° and solve for n: (n – 2) × 180° = 140° n → 180° n – 360° = 140° n → 40° n = 360° → n = 9.

多边形的一个顶点处内角与外角之和为 180°。因此每个外角 = 180° – 140° = 40°。任意多边形的外角和为 360°。正多边形各外角相等，故边数 n = 360° ÷ 40° = 9。也可用公式：内角 = (n – 2) × 180° / n。令其等于 140° 并求解 n：(n – 2) × 180° = 140° n → 180° n – 360° = 140° n → 40° n = 360° → n = 9。该正多边形有 9 条边，是一个正九边形。

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8. Perimeter and Area of a Composite Shape | 复合图形的周长与面积

Question: A shape consists of a rectangle of length 10 m and width 4 m attached to a semicircle of diameter 4 m along one width. Find the perimeter and area of the whole shape.

题目：一个图形由长 10 m、宽 4 m 的矩形，和在其一条宽边上拼接一个直径为 4 m 的半圆组成。求整个图形的周长和面积。

diameter of the semicircle equals the width of the rectangle, 4 m, so its radius r = 2 m.

该复合图形如同矩形的一个短边外侧附加一个半圆。半圆的直径等于矩形宽 4 m，因此半径 r = 2 m。

ference of a full circle is 2πr, so the semicircular arc length is πr = π × 2 ≈ 6.28 m. The straight rectangle edges are 10 m + 10 m + 4 m = 24 m. Total perimeter ≈ 24 m + 6.28 m = 30.28 m (using π ≈ 3.14).

周长：外围边界包括矩形的三条边（两条长和一条对边宽）以及半圆弧长。整圆周长是 2πr，因此半圆弧长 = πr = π × 2 ≈ 6.28 m。矩形直边部分为 10 m + 10 m + 4 m = 24 m。总周长 ≈ 24 m + 6.28 m = 30.28 m（取 π ≈ 3.14）。

f semicircle = (1/2) × πr² = ½ × π × 2² = 2π ≈ 6.28 m². Total area = 40 + 6.28 = 46.28 m².

面积：矩形面积 = 长 × 宽 = 10 × 4 = 40 m²。半圆面积 = ½ × πr² = ½ × π × 2² = 2π ≈ 6.28 m²。总面积 = 40 + 6.28 = 46.28 m²。

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9. Statistics: Mean, Median, Mode and Range | 统计：平均数、中位数、众数与极差

Question: The data set shows test scores: 12, 14, 13, 12, 16, 11, 10, 12, 15. Find the mean, median, mode and range.

题目：数据为测验分数：12, 14, 13, 12, 16, 11, 10, 12, 15。求平均数、中位数、众数和极差。

First, order the data: 10, 11, 12, 12, 12, 13, 14, 15, 16. There are 9 values.

首先排序：10, 11, 12, 12, 12, 13, 14, 15, 16。共有 9 个数值。

um = 10 + 11 + 12 + 12 + 12 + 13 + 14 + 15 + 16 = 115. Mean = 115 ÷ 9 ≈ 12.78 (to 2 d.p.).

平均数：总和 = 10 + 11 + 12 + 12 + 12 + 13 + 14 + 15 + 16 = 115。平均数 = 115 ÷ 9 ≈ 12.78（保留两位小数）。

Median: the median is the 5th value in the ordered list, which is 12.

中位数：排序后第 5 个数据为 12。

Mode: the most frequent score is 12 (appears 3 times).

众数：出现次数最多的成绩是 12（出现 3 次）。

Range = maximum – minimum = 16 – 10 = 6.

极差 = 最大值 – 最小值 = 16 – 10 = 6。

A potential pitfall: mixing up the median with the mean or forgetting to order the data before finding the median. Always order first.

易错点：将中位数与平均数混淆，或在未排序的情况下找中位数。务必先排序。

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10. Probability of Combined Events | 复合事件的概率

Question: A fair coin is flipped and a fair six-sided die is rolled. List all possible outcomes and find the probability of getting a head and an even number.

题目：同时抛一枚均匀硬币和掷一个均匀六面骰子。列出所有可能结果，并求得到正面且点数为偶数的概率。

Sample space: coin outcomes H, T; die outcomes 1, 2, 3, 4, 5, 6. Pairs: (H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6). Total 12 equally likely outcomes.

样本空间：硬币为 H（正面）、T（反面）；骰子点数为 1 至 6。配对：(H,1) 到 (H,6) 以及 (T,1) 到 (T,6)，共 12 个等可能结果。

Favourable outcomes: head and even number → (H,2), (H,4), (H,6) → 3 outcomes.

有利结果：正面且偶数 → (H,2), (H,4), (H,6)，共 3 个。

Probability = number of favourable outcomes / total number of outcomes = 3 / 12 = 1/4.

概率 = 有利结果数 / 总结果数 = 3 / 12 = 1/4。

ication rule: P(H and even) = P(H) × P(even) = ½ × ½ = ¼. Both methods confirm the answer.

也可用树状图或乘法法则：P(正面且偶数) = P(正面) × P(偶数) = ½ × ½ = ¼。两种方法答案一致。

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