Year 11 CIE Maths: Unit Test Mock Paper Analysis | Year 11 CIE 数学：单元测试模拟卷解析

📚 Year 11 CIE Maths: Unit Test Mock Paper Analysis | Year 11 CIE 数学：单元测试模拟卷解析

This article provides a detailed walkthrough of a typical Year 11 CIE IGCSE Mathematics unit test mock paper. We cover a wide range of core topics, including algebra, equations, geometry, trigonometry, probability, and functions. Each section presents a representative question with step-by-step solutions, designed to help you revise effectively and master exam‑style problem solving.

本文详细解析了一份典型的 Year 11 CIE IGCSE 数学单元测试模拟卷。内容涵盖代数、方程、几何、三角学、概率与函数等多个核心主题。每个部分均提供一道代表性题目并逐步解答，旨在帮助你高效复习，掌握考试风格的解题技巧。

1. Algebraic Simplification and Factorisation | 代数化简与因式分解

Question: Simplify 3x(2x – 5) – 2(x – 3) and factorise x² – 5x + 6.

题目：化简 3x(2x – 5) – 2(x – 3)，并对 x² – 5x + 6 进行因式分解。

First, expand 3x(2x – 5): 3x × 2x = 6x², and 3x × (–5) = –15x, giving 6x² – 15x.

首先展开 3x(2x – 5)：3x × 2x = 6x²，3x × (–5) = –15x，得到 6x² – 15x。

Next, expand –2(x – 3): –2 × x = –2x, and –2 × (–3) = +6, so we obtain –2x + 6.

接着展开 –2(x – 3)：–2 × x = –2x，–2 × (–3) = +6，因此得到 –2x + 6。

Combine like terms from both expansions: 6x² – 15x – 2x + 6 = 6x² – 17x + 6.

合并两个展开式中的同类项：6x² – 15x – 2x + 6 = 6x² – 17x + 6。

Simplified expression: 6x² – 17x + 6

化简结果：6x² – 17x + 6

To factorise x² – 5x + 6, look for two numbers that multiply to +6 and add to –5. Those numbers are –2 and –3. Hence, x² – 5x + 6 = (x – 2)(x – 3).

对 x² – 5x + 6 因式分解，需找到相乘得 +6、相加得 –5 的两个数。这两个数是 –2 和 –3。因此，x² – 5x + 6 = (x – 2)(x – 3)。

Factorised form: (x – 2)(x – 3)

因式分解形式：(x – 2)(x – 3)

2. Solving Linear and Quadratic Equations | 解一次与二次方程

Question: Solve 2(3x – 1) = 5x + 4 and x² + 2x – 8 = 0.

题目：解方程 2(3x – 1) = 5x + 4 和 x² + 2x – 8 = 0。

For the linear equation, expand the left-hand side: 2 × 3x = 6x, and 2 × (–1) = –2, so the equation becomes 6x – 2 = 5x + 4.

对于一次方程，先展开左边：2 × 3x = 6x，2 × (–1) = –2，方程化为 6x – 2 = 5x + 4。

Collect x‑terms on one side: 6x – 5x = 4 + 2, which simplifies to x = 6.

将含 x 的项移至同一边：6x – 5x = 4 + 2，化简得 x = 6。

Solution: x = 6

解：x = 6

For the quadratic, identify factors of –8 that sum to +2. They are +4 and –2. Therefore, x² + 2x – 8 = (x + 4)(x – 2) = 0.

对于二次方程，找出相乘得 –8、相加得 +2 的两个因数：+4 和 –2。因此 x² + 2x – 8 = (x + 4)(x – 2) = 0。

Setting each factor to zero gives x + 4 = 0 or x – 2 = 0, so x = –4 or x = 2.

令每个因式为零，得 x + 4 = 0 或 x – 2 = 0，所以 x = –4 或 x = 2。

Solutions: x = –4, x = 2

解：x = –4, x = 2

3. Inequalities and Number Line Representation | 不等式及其在数轴上的表示

Question: Solve 4x – 7 > 5 – 2x and represent the solution on a number line.

题目：解不等式 4x – 7 > 5 – 2x，并在数轴上表示解集。

Bring x‑terms to one side and constants to the other: 4x + 2x > 5 + 7, giving 6x > 12.

将含 x 项与常数项分别移到不等式两边：4x + 2x > 5 + 7，得 6x > 12。

Divide both sides by 6: x > 2.

两边同时除以 6：x > 2。

On a number line, draw an open circle at 2 (since 2 is not included) and shade the line to the right, indicating all values greater than 2.

在数轴上，在 2 处画一个空心圆圈（因为不包含 2），并将右侧部分涂色，表示所有大于 2 的数。

Solution set: {x : x > 2}

解集：{x : x > 2}

4. Equation of a Straight Line | 直线方程

Question: Find the equation of the line passing through (2, 5) with gradient –3. Write your answer in the form y = mx + c.

题目：求经过点 (2, 5) 且梯度为 –3 的直线方程，结果写成 y = mx + c 的形式。

Use the point‑slope form: y – y₁ = m(x – x₁), where m = –3 and (x₁, y₁) = (2, 5).

使用点斜式：y – y₁ = m(x – x₁)，其中 m = –3，(x₁, y₁) = (2, 5)。

Substituting gives y – 5 = –3(x – 2). Expand the right side: –3x + 6.

代入得 y – 5 = –3(x – 2)。展开右边：–3x + 6。

Add 5 to both sides: y = –3x + 11.

两边同时加 5：y = –3x + 11。

Equation: y = –3x + 11

直线方程：y = –3x + 11

5. Geometry – Pythagoras’ Theorem and Area of a Triangle | 几何 – 勾股定理与三角形面积

Question: In triangle ABC, angle C = 90°, AC = 6 cm, BC = 8 cm. Calculate the length AB and the area of the triangle.

题目：在三角形 ABC 中，∠C = 90°，AC = 6 cm，BC = 8 cm。计算 AB 的长度和三角形的面积。

By Pythagoras’ theorem, AB² = AC² + BC² = 6² + 8² = 36 + 64 = 100. Hence AB = √100 = 10 cm.

由勾股定理，AB² = AC² + BC² = 6² + 8² = 36 + 64 = 100。因此 AB = √100 = 10 cm。

Since the triangle is right‑angled at C, its area = ½ × base × height = ½ × 6 × 8 = 24 cm².

因为三角形在 C 处为直角，其面积 = ½ × 底 × 高 = ½ × 6 × 8 = 24 cm²。

AB = 10 cm, Area = 24 cm²

AB = 10 cm，面积 = 24 cm²

6. Similarity and Area Scale Factor | 相似形与面积比例因子

Question: Two similar triangles have side lengths in the ratio 3 : 5. If the area of the smaller triangle is 18 cm², find the area of the larger triangle.

题目：两个相似三角形的边长比为 3 : 5。若小三角形的面积为 18 cm²，求大三角形的面积。

For similar figures, the ratio of areas is the square of the linear scale factor. Here the scale factor is 3 : 5, so the area ratio = (3/5)² = 9/25.

对于相似图形，面积比等于线性比例因子的平方。此处比例因子为 3 : 5，因此面积比 = (3/5)² = 9/25。

The smaller area corresponds to 9 parts, the larger to 25 parts. Let A be the larger area: 18 / A = 9 / 25, thus A = 18 × (25/9) = 50 cm².

小长方形对应 9 份，大长方形对应 25 份。设大面积为 A：18 / A = 9 / 25，因此 A = 18 × (25/9) = 50 cm²。

Area of larger triangle = 50 cm²

大三角形面积 = 50 cm²

7. Trigonometry – The Cosine Rule | 三角学 – 余弦定理

Question: In triangle ABC, angle A = 35°, AB = 12 cm, AC = 15 cm. Use the cosine rule to calculate side BC to 3 significant figures.

题目：在三角形 ABC 中，∠A = 35°，AB = 12 cm，AC = 15 cm。用余弦定理计算边 BC，结果保留 3 位有效数字。

The cosine rule states: a² = b² + c² – 2bc cos A, where a = BC, b = 15 (AC), c = 12 (AB).

余弦定理为：a² = b² + c² – 2bc cos A，其中 a = BC，b = 15 (AC)，c = 12 (AB)。

Substitute the values: BC² = 15² + 12² – 2 × 15 × 12 × cos 35°.

代入数值：BC² = 15² + 12² – 2 × 15 × 12 × cos 35°。

Calculate: 15² = 225, 12² = 144, sum = 369. 2 × 15 × 12 = 360. cos 35° ≈ 0.81915. So 360 × 0.81915 ≈ 294.894.

计算：15² = 225，12² = 144，和为 369。2 × 15 × 12 = 360。cos 35° ≈ 0.81915。因此 360 × 0.81915 ≈ 294.894。

BC² ≈ 369 – 294.894 = 74.106. Taking the square root: BC ≈ √74.106 ≈ 8.61 cm (3 s.f.).

BC² ≈ 369 – 294.894 = 74.106。开平方根：BC ≈ √74.106 ≈ 8.61 cm（3 位有效数字）。

BC ≈ 8.61 cm

8. Probability – Combined Events | 概率 – 复合事件

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

题目：一个袋子中有 5 个红色、3 个蓝色和 2 个绿色弹珠。随机抽取一个弹珠，求抽到的不是蓝色的概率。

Total number of marbles = 5 + 3 + 2 = 10. Number of marbles that are not blue = 5 (red) + 2 (green) = 7.

弹珠总数 = 5 + 3 + 2 = 10。不是蓝色的弹珠数 = 5（红）+ 2（绿）= 7。

Probability(not blue) = number of favourable outcomes / total outcomes = 7/10 = 0.7.

概率（非蓝）= 有利结果数 / 总结果数 = 7/10 = 0.7。

P(not blue) = 7/10 or 0.7

P（非蓝）= 7/10 或 0.7

9. Statistics – Mean and Median | 统计 – 平均数与中位数

Question: Find the mean and median of the data set: 8, 5, 12, 9, 7, 10, 14.

题目：计算以下数据集的中位数和平均数：8, 5, 12, 9, 7, 10, 14。

Sum of values = 8 + 5 + 12 + 9 + 7 + 10 + 14 = 65. Number of values = 7. Mean = 65 ÷ 7 ≈ 9.29 (2 d.p.).

数值总和 = 8 + 5 + 12 + 9 + 7 + 10 + 14 = 65。数据个数 = 7。平均数 = 65 ÷ 7 ≈ 9.29（保留两位小数）。

To find the median, arrange the data in ascending order: 5, 7, 8, 9, 10, 12, 14. The median is the middle value, which is the 4th value: 9.

求中位数时，先将数据按升序排列：5, 7, 8, 9, 10, 12, 14。中位数为中间值，即第 4 个值：9。

Mean ≈ 9.29, Median = 9

平均数 ≈ 9.29，中位数 = 9

10. Functions and Composite Functions | 函数与复合函数

Question: Given f(x) = 2x – 5 and g(x) = x², evaluate f(3) and fg(2).

题目：已知 f(x) = 2x – 5，g(x) = x²，求 f(3) 和 fg(2)。

f(3) = 2(3) – 5 =

Published by TutorHao | Year 11 Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Year 11 CIE Maths: Unit Test Mock Paper Analysis | Year 11 CIE 数学：单元测试模拟卷解析