KS3 Cambridge Advanced Mathematics: Mock Test Walkthrough | KS3 剑桥进阶数学：模拟卷解析

📚 KS3 Cambridge Advanced Mathematics: Mock Test Walkthrough | KS3 剑桥进阶数学：模拟卷解析

Welcome to our detailed walkthrough of a KS3 Cambridge Advanced Mathematics unit test mock paper. This resource is designed to help you consolidate key skills across algebra, geometry, number, statistics and more. Each question is explained step by step in both English and Chinese, so you can follow the logic, learn essential techniques, and identify common pitfalls. Use these explanations to refine your exam strategy and build confidence.

欢迎阅读我们针对 KS3 剑桥进阶数学单元测试模拟卷的详细解析。本资料旨在帮助您巩固代数、几何、数、统计等核心技能。每道题都以英文和中文逐步讲解，让您能够跟上逻辑、学习关键技巧并识别常见易错点。请利用这些解析优化您的应考策略并建立信心。

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1. Algebra: Expanding and Simplifying | 代数：展开与化简

Question: Simplify the expression 4(2x – 3) + 5(x + 1) – 2x.

题目：化简表达式 4(2x – 3) + 5(x + 1) – 2x。

Step 1: Expand the brackets by multiplying each term inside. For 4(2x – 3), we get 8x – 12. For 5(x + 1), we get 5x + 5.

步骤 1：展开括号，将括号外的系数乘以括号内的每一项。4(2x – 3) 得到 8x – 12。5(x + 1) 得到 5x + 5。

Step 2: Rewrite the expression without brackets: 8x – 12 + 5x + 5 – 2x.

步骤 2：将表达式重写为无括号形式：8x – 12 + 5x + 5 – 2x。

Step 3: Group like terms. Collect all terms with x: 8x + 5x – 2x = 11x. Combine constants: -12 + 5 = -7.

步骤 3：合并同类项。集中所有含 x 的项：8x + 5x – 2x = 11x。合并常数：-12 + 5 = -7。

Final answer: 11x – 7. Always check that no further simplification is possible.

最终答案：11x – 7。务必检查是否还能继续化简。

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2. Solving Equations: Fractions and Inverse Operations | 解方程：分数与逆运算

Question: Solve for y: (3y – 4)/2 = 7.

题目：解方程求 y：(3y – 4)/2 = 7。

Step 1: Eliminate the denominator by multiplying both sides by 2. This gives 3y – 4 = 14.

步骤 1：将方程两边同时乘以 2 以消去分母。得到 3y – 4 = 14。

Step 2: Isolate the term containing y by adding 4 to both sides. 3y – 4 + 4 = 14 + 4, so 3y = 18.

步骤 2：通过两边同时加 4 使含 y 的项独立。3y – 4 + 4 = 14 + 4，即 3y = 18。

Step 3: Divide both sides by 3 to solve for y. y = 18 ÷ 3 = 6.

步骤 3：两边同时除以 3 求出 y。y = 18 ÷ 3 = 6。

Verification: Substitute y = 6 back into (3y – 4)/2: (18 – 4)/2 = 14/2 = 7, which checks out.

验证：将 y = 6 代回 (3y – 4)/2： (18 – 4)/2 = 14/2 = 7，结果正确。

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3. Geometry: Area and Circumference of Circles | 几何：圆的面积与周长

Question: A circle has a radius of 7 cm. Find its area, giving the answer in terms of π and as an approximate value using π ≈ 3.14.

题目：一个圆的半径为 7 cm。求其面积，答案分别用 π 表示和使用 π ≈ 3.14 的近似值。

Recall the area formula: A = πr², where r is the radius.

回顾面积公式：A = πr²，其中 r 是半径。

Substitute r = 7 cm: A = π × 7² = 49π cm².

代入 r = 7 cm：A = π × 7² = 49π cm²。

For the approximate value, replace π with 3.14: A ≈ 49 × 3.14 = 153.86 cm².

求近似值时，用 3.14 替换 π：A ≈ 49 × 3.14 = 153.86 cm²。

Note: In this topic you might also be asked for circumference using C = 2πr, which would be 14π cm.

注意：本主题也可能会要求求周长，公式为 C = 2πr，结果为 14π cm。

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4. Number: Standard Form and Significant Figures | 数：标准形式与有效数字

Question: Write 0.00045 in standard form and round 36721 to 3 significant figures.

题目：将 0.00045 写成标准形式，并将 36721 四舍五入至 3 位有效数字。

For standard form, move the decimal point to create a number between 1 and 10. 0.00045 becomes 4.5, and we moved 4 places to the right, so the exponent is -4: 4.5 × 10⁻⁴.

对于标准形式，移动小数点以得到一个介于 1 和 10 之间的数。0.00045 变为 4.5，我们向右移动了 4 位，因此指数为 -4：4.5 × 10⁻⁴。

For 36721 to 3 significant figures, look at the first 3 digits: 3,6,7. The next digit is 2, so we round down to 36700. In standard form: 3.67 × 10⁴.

对于 36721 保留 3 位有效数字，看前三位数字：3、6、7。下一位数字是 2，因此我们舍去为 36700。写成标准形式：3.67 × 10⁴。

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5. Ratio and Proportion in Context | 情境中的比与比例

Question: A concrete mix uses cement and sand in the ratio 2 : 5. If 14 kg of cement is available, how much sand is needed?

题目：一种混凝土混合物中水泥与沙子的比为 2 : 5。如果有 14 kg 水泥，需要多少沙子？

The ratio tells us that for every 2 parts cement, there are 5 parts sand. Total parts = 7, but we use direct proportion.

这个比告诉我们，每 2 份水泥对应 5 份沙子。总份数为 7，但我们使用正比例。

Scale factor for cement: 14 kg ÷ 2 = 7. So the amount of sand is 5 × 7 = 35 kg.

水泥的缩放因子：14 kg ÷ 2 = 7。因此沙子的用量为 5 × 7 = 35 kg。

Alternative method: Set up the proportion 2/5 = 14/x, cross-multiply: 2x = 70, x = 35.

替代方法：建立比例式 2/5 = 14/x，交叉相乘：2x = 70，x = 35。

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6. Sequences: Finding the nth Term | 数列：求第 n 项通项

Question: For the arithmetic sequence 5, 9, 13, 17, …, find an expression for the nth term.

题目：对于等差数列 5, 9, 13, 17, …，写出第 n 项的表达式。

Find the common difference: 9 – 5 = 4, 13 – 9 = 4, so d = 4.

求公差：9 – 5 = 4，13 – 9 = 4，因此 d = 4。

The nth term of an arithmetic sequence has the form an + b, where a = common difference. So we start with 4n.

等差数列的第 n 项形式为 an + b，其中 a 为公差。因此我们从 4n 开始。

When n=1, 4n = 4, but the first term is 5, so b = 1. The nth term is 4n + 1.

当 n=1 时，4n = 4，但首项是 5，所以 b = 1。通项为 4n + 1。

Check with n=3: 4×3 + 1 = 13, which matches the third term.

用 n=3 验证：4×3 + 1 = 13，与第三项一致。

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7. Statistics: Mean, Median from a Frequency Table | 统计：根据频数表求平均数与中位数

Question: The table below shows test scores and their frequencies. Find the mean and median score.

Score	10	20	30
Frequency	4	6	2

题目：下表显示了测试分数及其频数。求平均分和中位数。

Total number of students = 4 + 6 + 2 = 12.

学生总人数 = 4 + 6 + 2 = 12。

Mean = (sum of scores × frequency) / total = (10×4 + 20×6 + 30×2) / 12 = (40 + 120 + 60) / 12 = 220 ÷ 12 ≈ 18.33.

平均数 = (分数 × 频数) 之和 / 总人数 = (10×4 + 20×6 + 30×2) / 12 = (40 + 120 + 60) / 12 = 220 ÷ 12 ≈ 18.33。

To find the median, list all 12 data points in order. The 6th and 7th data points both fall in the score 20, so median = 20.

求中位数时，将全部 12 个数据按顺序排列。第 6 和第 7 个数据均落在分数 20 上，因此中位数 = 20。

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8. Probability: Single Events and Complements | 概率：单一事件与补集

Question: A bag contains 3 red, 5 blue and 2 green marbles. One marble is drawn at random. Find the probability it is not red.

题目：一个袋子里有 3 颗红球、5 颗蓝球和 2 颗绿球。随机抽取一粒，求抽出的不是红球的概率。

Total marbles = 3 + 5 + 2 = 10.

总球数 = 3 + 5 + 2 = 10。

Probability of red = 3/10. Using the complement rule, probability of not red = 1 – 3/10 = 7/10.

抽到红球的概率 = 3/10。利用补集规则，非红球的概率 = 1 – 3/10 = 7/10。

Alternatively, count favourable outcomes directly: (5 blue + 2 green) = 7, so 7/10.

也可直接计算有利结果：(5 颗蓝 + 2 颗绿) = 7，因此 7/10。

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9. Transformations: Reflection in Coordinate Plane | 变换：坐标平面内的反射

Question: The point A has coordinates (3, 4). Give the coordinates of its image after a reflection in the y-axis.

题目：点 A 的坐标为 (3, 4)。求其关于 y 轴反射后的像点坐标。

When reflecting in the y-axis, the x-coordinate changes sign, while the y-coordinate stays the same. So (3, 4) → (-3, 4).

当关于 y 轴反射时，x 坐标变号，y 坐标保持不变。因此 (3, 4) → (-3, 4)。

A helpful way is to picture the mirror line x=0; the point moves to the opposite side at equal distance.

一个有用的方法是想象镜面线 x=0；点移动到另一侧，距离相等。

Similarly, reflecting in the x-axis would give (3, -4).

类似地，关于 x 轴反射将得到 (3, -4)。

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10. Pythagoras’ Theorem: Finding the Hypotenuse | 勾股定理：求斜边

Question: In a right-angled triangle, the two shorter sides are 6 cm and 8 cm. Calculate the length of the hypotenuse.

题目：在一个直角三角形中，两条较短的边分别为 6 cm 和 8 cm。计算斜边的长度。

Pythagoras’ theorem: a² + b² = c², where c is the hypotenuse. Substitute a = 6, b = 8.

勾股定理：a² + b² = c²，其中 c 是斜边。代入 a = 6, b = 8。

6² + 8² = c²

36 + 64 = c²

100 = c²

Taking the square root of both sides gives c = √100 = 10 cm.

两边开平方根得到 c = √100 = 10 cm。

Always check: 6² + 8² = 100, and 10² = 100, so the result is correct.

务必检验：6² + 8² = 100，而 10² = 100，结果正确。

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