📚 CAIE Year 8 Further Mathematics: Unit Test Mock Paper Walkthrough | CAIE 8年级进阶数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test designed for CAIE Year 8 Further Mathematics. The paper covers essential topics including indices, algebraic expansion, factorisation, linear equations, simultaneous equations, coordinate geometry, sequences, polygon angles, basic probability, and statistical mean. Each section breaks down a representative question, explains the key concepts, and presents a step-by-step solution to consolidate understanding and exam technique.
本文详细解析了一套为CAIE 8年级进阶数学设计的单元模拟测试卷。试卷涵盖指数、代数展开、因式分解、一次方程、联立方程、坐标几何、数列、多边形角度、基础概率以及统计平均数等核心主题。每个部分拆解一道典型题目，讲解关键概念，并逐步展示解题过程，以巩固理解和应试技巧。
1. Simplifying Expressions with Negative Indices | 化简含负指数的表达式
stion: Simplify (2⁴ × 2⁻³) ÷ (2²)², giving your answer as a fraction in its simplest form.
题目：化简 (2⁴ × 2⁻³) ÷ (2²)²，并将答案表示为最简分数。
1: Apply the multiplication law aᵐ × aⁿ = aᵐ⁺ⁿ to the numerator: 2⁴ × 2⁻³ = 2⁴⁺⁽⁻³⁾ = 2¹.
步骤1：对分子使用乘法法则 aᵐ × aⁿ = aᵐ⁺ⁿ：2⁴ × 2⁻³ = 2⁴⁻³ = 2¹。
Step 2: Simplify the denominator using the power of a power rule (aᵐ)ⁿ = aᵐⁿ: (2²)² = 2⁴.
步骤2：运用幂的乘方法则化简分母：(2²)² = 2⁴。
Divide the numerator by the denominator, keeping the base the same: 2¹ ÷ 2⁴ = 2¹⁻⁴ = 2⁻³.
步骤3：分子除以分母，底数保持不变：2¹ ÷ 2⁴ = 2¹⁻⁴ = 2⁻³。
Step 4: Rewrite with a positive index: 2⁻³ = 1 / 2³ = 1/8.
步骤4：改写为正指数形式：2⁻³ = 1 / 2³ = 1/8。
2. Expanding and Simplifying Brackets | 去括号展开并化简
Question: Expand and simplify fully: 2x(3x − 4) − (x + 2)(x − 5).
题目：展开并彻底化简：2x(3x − 4) − (x + 2)(x − 5)。
each term inside the bracket: 2x × 3x = 6x² and 2x × (−4) = −8x, so 2x(3x − 4) = 6x² − 8x.
第一部分：将 2x 乘以括号内的每一项：2x × 3x = 6x²，2x × (−4) = −8x，因此 2x(3x − 4) = 6x² − 8x。
) = −5x, 2 × x = 2x, 2 × (−5) = −10. Combine like terms: x² − 5x + 2x − 10 = x² − 3x − 10.
第二部分：用乘法分配律展开 (x + 2)(x − 5)：x·x = x²，x·(−5) = −5x，2·x = 2x，2·(−5) = −10。合并同类项：x² − 5x + 2x − 10 = x² − 3x − 10。
te the minus sign: 6x² − 8x − x² + 3x + 10 = (6x² − x²) + (−8x + 3x) + 10 = 5x² − 5x + 10.
从第一部分的结果中减去第二部分：6x² − 8x − (x² − 3x − 10)。去括号时注意变号：6x² − 8x − x² + 3x + 10。将其合并：(6x² − x²) + (−8x + 3x) + 10 = 5x² − 5x + 10。
3. Factorising Quadratic Trinomials | 二次三项式的因式分解
Question: Factorise fully: 6x² + 11x − 10.
题目：因式分解：6x² + 11x − 10。
and add to b = 11. These numbers are 15 and −4 because 15 × (−4) = −60 and 15 + (−4) = 11.
寻找两个数，使其乘积为 a × c = 6 × (−10) = −60，且和为 b = 11。这两个数是 15 和 −4，因为 15 × (−4) = −60，15 + (−4) = 11。
Split the middle term: 6x² + 15x − 4x − 10. Group the first two terms and the last two terms: (6x² + 15x) + (−4x − 10). Factor each group: 3x(2x + 5) − 2(2x + 5).
拆项：6x² + 15x − 4x − 10。将前两项与后两项分别分组：(6x² + 15x) + (−4x − 10)。对每组提取公因式：3x(2x + 5) − 2(2x + 5)。
Factor out the common binomial (2x + 5): (2x + 5)(3x − 2).
提取公因子 (2x + 5)，得到 (2x + 5)(3x − 2)。
4. Solving Multistep Linear Equations | 解多步一次方程
Question: Solve for x: 3(2x − 1) − 4(x + 3) = 7x − 5.
题目：解方程求 x：3(2x − 1) − 4(x + 3) = 7x − 5。
ving −4x − 12. Combine: 6x − 3 − 4x − 12 = 2x − 15. The equation becomes 2x − 15 = 7x − 5.
将左边的两组括号展开：3 × 2x = 6x，3 × (−1) = −3，得到 6x − 3；接着 −4 × x = −4x，−4 × 3 = −12，给出 −4x − 12。合并：6x − 3 − 4x − 12 = 2x − 15。方程变为 2x − 15 = 7x − 5。
ract 2x from both sides: −15 = 5x − 5. Add 5 to both sides: −10 = 5x. Divide by 5: x = −2.
将含 x 的项移到同一边：两边同时减去 2x，得 −15 = 5x − 5。两边同时加 5：−10 = 5x。两边除以 5：x = −2。
5. Solving Simultaneous Equations by Substitution | 代入法解联立方程组
Question: Solve the simultaneous equations: 2x + y = 7, x − 3y = −14.
题目：解联立方程组：2x + y = 7，x − 3y = −14。
in terms of x: y = 7 − 2x. Substitute this into the second equation: x − 3(7 − 2x) = −14.
由第一个方程将 y 用 x 表示：y = 7 − 2x。将其代入第二个方程：x − 3(7 − 2x) = −14。
e bracket: x − 21 + 6x = −14. Combine like terms: 7x − 21 = −14. Add 21: 7x = 7, so x = 1.
展开括号：x − 21 + 6x = −14。合并同类项：7x − 21 = −14。两边加 21：7x = 7，于是 x = 1。
Substitute x = 1 back into y = 7 − 2x: y = 7 − 2 = 5. The solution is x = 1, y = 5.
将 x = 1 代回 y = 7 − 2x：y = 7 − 2 = 5。方程组的解为 x = 1，y = 5。
6. Finding an Equation of a Straight Line | 求直线方程
ent −3 and passes through the point (2, 5). Determine its equation in the form y = mx + c.
题目：一条直线的斜率为 −3，并且经过点 (2, 5)。求该直线方程，写成 y = mx + c 的形式。
3. Substitute the coordinates of the point into the equation: 5 = −3 × 2 + c → 5 = −6 + c.
斜截式为 y = mx + c，其中 m = −3。将已知点的坐标代入：5 = −3 × 2 + c → 5 = −6 + c。
Solve for c: c = 5 + 6 = 11. Therefore, the equation of the line is y = −3x + 11.
解出 c：c = 5 + 6 = 11。因此，直线方程为 y = −3x + 11。
7. Quadratic Sequences and Term Calculation | 二次数列与项的计算
equence is given by T(n) = 2n² + 1. Find the 4th term and the sum of the first four terms.
题目：某数列的第 n 项公式为 T(n) = 2n² + 1。求第 4 项以及前四项的和。
To find the 4th term, substitute n = 4: T(4) = 2 × (4)² + 1 = 2 × 16 + 1 = 32 + 1 = 33.
求第 4 项，代入 n = 4：T(4) = 2 × 4² + 1 = 2 × 16 + 1 = 32 + 1 = 33。
y: T(1) = 2×1²+1=3, T(2)=2×4+1=9, T(3)=2×9+1=19, T(4)=33. The sum is 3 + 9 + 19 + 33 = 64.
逐一计算前四项：T(1)=2×1²+1=3，T(2)=2×4+1=9，T(3)=2×9+1=19，T(4)=33。它们的和为 3 + 9 + 19 + 33 = 64。
8. Interior Angles of Regular Polygons | 正多边形的内角
Question: The interior angle of a regular polygon is 156°. How many sides does the polygon have?
题目：一个正多边形的每个内角均为 156°。这个多边形有多少条边？
des: interior angle = (n − 2) × 180° / n. Set this equal to 156°: (n − 2) × 180 / n = 156.
正 n 边形的内角公式为：(n − 2) × 180° / n。令其等于 156°：(n − 2) × 180 / n = 156。
− 360 = 156n. Bring 156n to the left: 180n − 156n = 360 → 24n = 360. Divide by 24: n = 15.
两边同时乘以 n：(n − 2) × 180 = 156n。展开：180n − 360 = 156n。将 156n 移项：180n − 156n = 360 → 24n = 360。两边除以 24，得 n = 15。
9. Probability with Two Sets | 两个集合的概率
Question: In a class of 15 students, 12 play football (F), 8 play basketball (B), and 5 play both sports. One student is chosen at random. Find P(F and not B), the probability that the student plays football but not basketball.
题目：某班有 15 名学生，其中 12 人踢足球 (F)，8 人打篮球 (B)，5 人两项都参加。随机抽取一名学生，求 P(F 且非 B)，即该生踢足球但不打篮球的概率。
F − those in both = 12 − 5 = 7. Total students = 15 (given by context or from 12+8−5=15).
使用维恩图或减法：只踢足球的人数 = 集合 F 的总人数 − 两项都参加的人数 = 12 − 5 = 7。总人数为 15（由题意或由 12+8−5=15 得到）。
Probability = (number who play football only) / (total students) = 7 / 15.
概率 = (只踢足球的人数) / (总人数) = 7 / 15。
10. Calculating the Mean from a Frequency Table | 由频数表计算平均数
Question: The frequency table shows the scores of 20 students. Find the mean score.
| Score | Frequency |
|---|---|
| 0 | 3 |
| 1 | 5 |
| 2 | 7 |
| 3 | 4 |
| 4 | 1 |
题目：频数表显示了 20 名学生的得分。计算平均分。
| 得分 | 频数 |
|---|---|
| 0 | 3 |
