Year 10 CAIE Maths: Mock Unit Test Walkthrough & Solutions | CAIE 十年级数学单元测试模拟卷解析

📚 Year 10 CAIE Maths: Mock Unit Test Walkthrough & Solutions | CAIE 十年级数学单元测试模拟卷解析

This article provides a full walkthrough of a mock unit test designed for Year 10 CAIE Mathematics. Each question targets essential skills from the syllabus, including algebra, graphs, geometry, data handling, and probability. Work through the questions, then check the step-by-step solutions to strengthen your understanding.

本文为 CAIE 十年级数学单元测试模拟卷提供全程解析。每道题目紧扣大纲核心技能，涵盖代数、图像、几何、数据处理和概率。先尝试作答，再跟随分步解析查漏补缺，巩固理解。

1. Expanding and Simplifying | 代数展开与化简

Question: Expand and fully simplify: 3(x + 2) − 2(x − 4).

题目：展开并完全化简：3(x + 2) − 2(x − 4)。

Step 1: Distribute the coefficients. Multiply 3 by each term inside the first bracket: 3 × x = 3x, 3 × 2 = 6. Multiply −2 by each term inside the second bracket: −2 × x = −2x, −2 × (−4) = +8.

步骤一：分配系数。用 3 乘第一个括号内每一项：3×x = 3x，3×2 = 6。用 −2 乘第二个括号内每一项：−2×x = −2x，−2×(−4) = +8。

Step 2: Combine the results: 3x + 6 − 2x + 8. Collect like terms: (3x − 2x) + (6 + 8) = x + 14.

步骤二：合并结果：3x + 6 − 2x + 8。合并同类项：(3x − 2x) + (6 + 8) = x + 14。

Final answer: x + 14.

最终答案：x + 14。

2. Factorising Quadratics | 二次三项式因式分解

Question: Factorise completely: x² + 7x + 10.

题目：完全因式分解：x² + 7x + 10。

We need two numbers that multiply to +10 and add to +7. The pairs of factors of 10 are (1,10) and (2,5). The pair (2,5) gives a sum of 7.

我们需要两个数，乘积为 +10，和为 +7。10 的因数对有 (1,10) 和 (2,5)。(2,5) 的和为 7。

Therefore, x² + 7x + 10 = (x + 2)(x + 5).

因此，x² + 7x + 10 = (x + 2)(x + 5)。

3. Solving Linear Equations | 解一元一次方程

Question: Solve for x: 2(x − 3) + 4 = 3x − 5.

题目：解 x：2(x − 3) + 4 = 3x − 5。

Step 1: Expand the left side: 2x − 6 + 4 = 2x − 2. The equation becomes 2x − 2 = 3x − 5.

步骤一：展开左边：2x − 6 + 4 = 2x − 2。方程变为 2x − 2 = 3x − 5。

Step 2: Bring variable terms to one side. Subtract 2x from both sides: −2 = x − 5. Then add 5 to both sides: 3 = x.

步骤二：将含变量项移到同侧。两边同时减 2x：−2 = x − 5。然后两边加 5：3 = x。

x = 3.

x = 3。

4. Solving Quadratic Equations | 解二次方程

Question: Solve x² − 5x + 6 = 0 by factorisation.

题目：用因式分解法解 x² − 5x + 6 = 0。

Find two numbers that multiply to +6 and add to −5. The pair (−2, −3) works because (−2)×(−3)=6 and (−2)+(−3)=−5. So, (x − 2)(x − 3) = 0.

寻找乘积为 +6、和为 −5 的两个数。数对 (−2, −3) 满足 (−2)×(−3)=6 且 (−2)+(−3)=−5。因此 (x − 2)(x − 3) = 0。

Set each factor equal to zero: x − 2 = 0 or x − 3 = 0. Hence, x = 2 or x = 3.

令每个因式等于零：x − 2 = 0 或 x − 3 = 0。因此 x = 2 或 x = 3。

5. Straight Line Graphs and Gradient | 直线图像与斜率

Question: Find the gradient of the line passing through points A(1, 2) and B(3, 8). Then write the equation of the line in the form y = mx + c.

题目：求经过 A(1, 2) 和 B(3, 8) 两点的直线的斜率，并写出该直线的 y = mx + c 形式方程。

Gradient m = (y₂ − y₁)/(x₂ − x₁) = (8 − 2)/(3 − 1) = 6/2 = 3.

斜率 m = (y₂ − y₁)/(x₂ − x₁) = (8 − 2)/(3 − 1) = 6/2 = 3。

Equation: y = 3x + c. Substitute point A(1,2): 2 = 3(1) + c → c = −1. Therefore, the line equation is y = 3x − 1.

方程：y = 3x + c。代入 A(1,2)：2 = 3(1) + c → c = −1。因此直线方程为 y = 3x − 1。

6. Evaluating Functions | 函数求值

Question: Given the function f(x) = 2x² − 3x + 1, find f(−2).

题目：已知函数 f(x) = 2x² − 3x + 1，求 f(−2)。

Substitute x = −2 into the expression: f(−2) = 2(−2)² − 3(−2) + 1.

将 x = −2 代入表达式：f(−2) = 2(−2)² − 3(−2) + 1。

Calculate: (−2)² = 4, so 2×4 = 8. −3×(−2) = +6. Then 8 + 6 + 1 = 15.

计算：(−2)² = 4，2×4 = 8；−3×(−2) = +6；然后 8 + 6 + 1 = 15。

f(−2) = 15.

f(−2) = 15。

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

Question: A shape is formed by a rectangle of length 6 cm and width 4 cm, with a semicircle of diameter 4 cm attached along one of the shorter sides. Calculate the total area of the shape. Leave π in your answer.

题目：一个图形由一个长 6 cm、宽 4 cm 的矩形和一个直径 4 cm 的半圆沿一条短边拼接而成。计算该图形的总面积，答案保留 π。

Area of rectangle = length × width = 6 × 4 = 24 cm².

矩形面积 = 长 × 宽 = 6 × 4 = 24 cm²。

Radius of semicircle = diameter/2 = 4/2 = 2 cm. Area of a full circle = πr² = π×(2)² = 4π. Semicircle area = 4π/2 = 2π cm².

半圆的半径 = 直径/2 = 4/2 = 2 cm。整圆面积 = πr² = π×(2)² = 4π。半圆面积 = 4π/2 = 2π cm²。

Total area = 24 + 2π cm².

总面积 = 24 + 2π cm²。

8. Pythagoras’ Theorem | 勾股定理

Question: In a right-angled triangle, the legs are 5 cm and 12 cm. Find the length of the hypotenuse. Determine whether the triangle is a Pythagorean triple.

题目：在一个直角三角形中，两条直角边分别为 5 cm 和 12 cm。求斜边长度，并判断该三角形是否为勾股数三角形。

By Pythagoras’ theorem: hypotenuse² = 5² + 12² = 25 + 144 = 169. So, hypotenuse = √169 = 13 cm.

由勾股定理：斜边² = 5² + 12² = 25 + 144 = 169。因此斜边 = √169 = 13 cm。

Since all three sides are integers (5, 12, 13), it is a Pythagorean triple.

由于三边长度均为整数 (5, 12, 13)，所以这是一个勾股数三角形。

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

Question: The following data set represents test scores: 4, 7, 9, 12, 15. Calculate the mean and the median.

题目：以下数据集代表测验得分：4, 7, 9, 12, 15。计算平均数和中位数。

Mean = (sum of values) ÷ (number of values) = (4+7+9+12+15) ÷ 5 = 47 ÷ 5 = 9.4.

平均数 = (数值总和) ÷ (数值个数) = (4+7+9+12+15) ÷ 5 = 47 ÷ 5 = 9.4。

The data set is already in ascending order. The median is the middle value, the 3rd term: 9.

数据集已按升序排列。中位数为中间的值，即第 3 项：9。

10. Probability | 概率

Question: A bag contains 3 red marbles and 5 blue marbles. One marble is drawn at random. What is the probability that it is red? After drawing a red marble (and not replacing it), what is the probability that a second marble drawn is blue?

题目：一个袋子里有 3 个红球和 5 个蓝球。随机摸出一个球，它是红色的概率是多少？如果第一次摸出红球且不放回，第二次摸出蓝球的概率是多少？

Total marbles initially = 3 + 5 = 8. Probability of red first = 3/8.

初始总球数 = 3 + 5 = 8。第一次摸出红球的概率 = 3/8。

After drawing a red marble, marbles left: 2 red, 5 blue; total 7. Probability that the second is blue = number of blue marbles left ÷ total remaining = 5/7.

第一次摸出红球后，剩余球数：2 个红球、5 个蓝球，共 7 个。第二次摸出蓝球的概率 = 剩余蓝球数 ÷ 剩余总数 = 5/7。

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Year 10 CAIE Maths: Mock Unit Test Walkthrough & Solutions | CAIE 十年级数学单元测试模拟卷解析