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

This mock paper walkthrough is designed to help Year 8 students consolidate key topics from the Cambridge advanced mathematics curriculum. Each question is modelled on typical unit test items covering algebra, sequences, ratio, geometry, Pythagoras' theorem, and statistics. Worked solutions are provided step by step, with common pitfalls explained so you can identify where marks are often lost.

这份模拟卷解析旨在帮助 Year 8 学生巩固剑桥进阶数学课程的核心主题。每道题都参照典型单元测试题设计，涵盖代数、数列、比与比例、几何、勾股定理和统计。我们提供分步解答，并解释常见错误，让你清楚哪些地方容易丢分。

1. Simplifying Expressions and Substitution | 化简表达式与代入求值

Question: Simplify 2a − 3b + 4a + b. Hence evaluate the expression when a = 5 and b = −2.

问题：化简 2a − 3b + 4a + b。并求当 a = 5 和 b = −2 时该表达式的值。

Solution: Identify like terms – re 2a and 4a are like terms, and −3b and b (which means +1b) are like terms. Combine the a‑terms: 2a + 4a = 6a. Combine the b‑terms: −3b + 1b = −2b. The simplified expression is 6a − 2b.

解：先找出同类项——即变量和指数完全相同的项。这里 2a 与 4a 是同类项，−3b 与 b（即 +1b）是同类项。合并 a 项：2a + 4a = 6a。合并 b 项：−3b + 1b = −2b。化简后的表达式为 6a − 2b。

carefully: 30 − (−4). Subtracting a negative is the same as adding a positive, so 30 + 4 = 34. Always double‑check your signs when substituting negative values.

现在把 a = 5 和 b = −2 代入 6a − 2b。用括号把负值括起来：6 × 5 − 2 × (−2)。先算乘法：30 − (−4)。减去一个负数等于加上它的相反数，所以 30 + 4 = 34。代入负值时一定要反复检查符号。

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2. Solving Linear Equations | 解一元一次方程

Question: Solve 3(x − 4) = 2x + 5.

问题：解方程 3(x − 4) = 2x + 5。

Solution: First expand the bracket on the left‑ ly each term inside the bracket by 3: 3 × x = 3x, and 3 × (−4) = −12. The equation becomes 3x − 12 = 2x + 5.

解：先去括号。把括号里的每一项都乘以 3：3 × x = 3x，3 × (−4) = −12。方程化为 3x − 12 = 2x + 5。

Next, collect x‑ ide and constant terms on the other. Subtract 2x from both sides: 3x − 2x − 12 = 5, giving x − 12 = 5. Then add 12 to both sides to isolate x: x − 12 + 12 = 5 + 12, so x = 17.

接下来把含 x 的项移到一边，常数移到另一边。两边同时减 2x：3x − 2x − 12 = 5，得 x − 12 = 5。然后两边同时加 12，得到 x = 17。

= 39. Right side: 2 × 17 + 5 = 34 + 5 = 39. Both sides match, so the solution is correct.

验算：把 x = 17 代回原方程。左边：3(17 − 4) = 3 × 13 = 39。右边：2 × 17 + 5 = 34 + 5 = 39。两边相等，说明解正确。

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3. Finding the nth Term of a Sequence | 求数列的第 n 项

Question: Consider the sequence 5, 9, 13, 17, … Find an expression for the nth term and then calculate the 50th term.

问题：观察数列 5, 9, 13, 17, … 写出第 n 项的表达式，并计算第 50 项。

= 4, so this is an arithmetic sequence. The nth term formula for an arithmetic sequence is a + (n − 1)d, where a is the first term. Here a = 5 and d = 4.

解：先看相邻两项的差。9 − 5 = 4，13 − 9 = 4，17 − 13 = 4。公差 d = 4，所以这是一个等差数列。等差数列第 n 项公式为 a + (n − 1)d，其中 a 为首项。这里 a = 5，d = 4。

Substitute the values: nth term = 5 + (n − 1) × 4. Expand: 5 + 4n − 4 = 4n + 1. To find the 50th term, let n = 50: 4 × 50 + 1 = 200 + 1 = 201.

代入数值：第 n 项 = 5 + (n − 1) × 4。展开得 5 + 4n − 4 = 4n + 1。要求第 50 项，令 n = 50：4 × 50 + 1 = 200 + 1 = 201。

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4. Ratio and Proportion | 比与比例

Question: The ratio of boys to girls in a class is 3 : 5. There are 24 boys. Work out the number of girls and the total number of students.

问题：某班级男生与女生的人数比为 3 : 5。已知男生有 24 人，求女生人数和全班总人数。

k. Then number of boys = 3k, number of girls = 5k. We are told 3k = 24, so k = 24 ÷ 3 = 8.

解：比 3 : 5 表示每 3 名男生对应 5 名女生。设比例因子为 k，那么男生人数 = 3k，女生人数 = 5k。已知 3k = 24，所以 k = 24 ÷ 3 = 8。

Now girls = 5k = 5 × 8 = 40. Total students = boys + girls = 24 + 40 = 64. Alternatively, total parts = 3 + 5 = 8, so 3 parts = 24 ⇒ 1 part = 8, then total = 8 × 8 = 64. Both methods are valid.

于是女生人数 = 5k = 5 × 8 = 40。总人数 = 24 + 40 = 64。也可以这样想：总份数 = 3 + 5 = 8，3 份对应 24 人，所以 1 份为 8 人，总人数 = 8 × 8 = 64。两种方法都正确。

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5. Area of a Composite Shape | 组合图形的面积

Question: The diagram shows a composite shape made of a rectangle and a right‑angled triangle on top. The rectangle has length 8 cm and width 5 cm. The triangle shares the top 8 cm side as its base and has a height of 3 cm. Find the total area.

问题：下图是由一个矩形和其上方的一个直角三角形组成的组合图形。矩形长 8 cm，宽 5 cm。三角形以矩形上方的 8 cm 边为底，高为 3 cm。求总面积。

Solution: Area of rectangle = length × width = 8 × 5 = 40 cm². Area of triangle = ½ × base × height = ½ × 8 × 3 = 12 × 1? Let's compute step by step: ½ of 8 is 4, then 4 × 3 = 12, so triangle area = 12 cm².

解：矩形面积 = 长 × 宽 = 8 × 5 = 40 cm²。三角形面积 = ½ × 底 × 高 = ½ × 8 × 3。先算 8 的一半是 4，再乘 3 得 12，所以三角形面积 = 12 cm²。

Total area = rectangle area + triangle area = 40 + 12 = 52 cm². Always include the correct units (cm² for area). Remember that the height of the triangle must be perpendicular to its base.

总面积 = 矩形面积 + 三角形面积 = 40 + 12 = 52 cm²。务必带上正确的单位（面积用 cm²）。切记三角形的高必须与底垂直。

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6. Volume of a Triangular Prism | 三棱柱的体积

Question: A triangular prism has a uniform cross‑section in the shape of a right‑angled triangle. The triangle's base is 6 cm and its height is 4 cm. The length of the prism is 10 cm. Calculate the volume of the prism.

问题：一个三棱柱的横截面是直角三角形，底为 6 cm，高为 4 cm。棱柱长 10 cm。求该棱柱的体积。

Solution: Volume of a prism = area of cross‑section × length. First find the area of the triangular cross‑section. Area of a right‑angled triangle = ½ × base × height = ½ × 6 × 4 = ½ × 24 = 12 cm².

解：棱柱体积 = 横截面积 × 长。先计算三角形横截面的面积。直角三角形面积 = ½ × 底 × 高 = ½ × 6 × 4 = ½ × 24 = 12 cm²。

Now multiply by the length of the prism: volume = 12 × 10 = 120 cm³. Volume is measured in cubic centimetres (cm³), not square units. A common mistake is forgetting to halve the product of base and height for the triangle.

再乘以棱柱长度：体积 = 12 × 10 = 120 cm³。体积的单位是立方厘米（cm³），而不是平方单位。常见错误是忘记三角形面积需要除以 2。

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7. Pythagoras' Theorem | 勾股定理

Question: A right‑angled triangle has legs of length 9 cm and 12 cm. Find the length of the hypotenuse.

问题：一个直角三角形的两条直角边分别为 9 cm 和 12 cm，求斜边的长度。

Solution: Pythagoras' theorem states that in a right‑angled triangle, c² = a² + b² egs. Here a = 9, b = 12. Calculate a² = 9² = 81, b² = 12² = 144. Then c² = 81 + 144 = 225.

解：勾股定理指出，直角三角形中，c² = a² + b²，其中 c 为斜边（直角对边，最长边），a、b 为直角边。这里 a = 9，b = 12。计算 a² = 9² = 81，b² = 12² = 144。因此 c² = 81 + 144 = 225。

To find c, take the positive square root: c = √225 = 15 cm you were finding a shorter leg, you would subtract: a² = c² − b². Remember that Pythagoras' theorem only applies to right‑angled triangles.

要求 c，取正的平方根：c = √225 = 15 cm。记得带上单位。如果求的是直角边，则要用减法：a² = c² − b²。请注意勾股定理只适用于直角三角形。

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8. Statistics: Mean, Median and Mode from a Frequency Table | 统计：从频数表求平均数、中位数和众数

Question: The frequency table shows the scores of 20 students on a short quiz. Calculate the mean, median and mode.

Score	1	2	3	4	5
Frequency	2	5	8	4	1

问题：频数表显示了 20 名学生一次小测验的成绩。求平均数、中位数和众数。

=5. Sum of fx = 2+10+24+16+5 = 57. Sum of frequencies n = 2+5+8+4+1 = 20. Mean = Σfx / n = 57 ÷ 20 = 2.85.

平均数解：扩展表格加入 fx 列（成绩 × 频数）。各乘积：1×2=2，2×5=10，3×8=24，4×4=16，5×1=5。fx 总和 = 2+10+24+16+5 = 57。频数总和 n =

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