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

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

This article provides detailed, step-by-step solutions for a typical Year 8 CCEA Advanced Mathematics unit test mock paper. By working through these model answers, you can consolidate core skills, recognise common mistakes and build the confidence needed for the real assessment. The questions span number, algebra, geometry, measurement, statistics and probability, reflecting the full breadth of the curriculum.

本文为一份典型的 Year 8 CCEA 进阶数学单元测试模拟卷提供详细的、逐步的解答。通过学习这些范例答案，你可以巩固核心技能、识别常见错误并建立应对真实评估的信心。题目涵盖数、代数、几何、测量、统计与概率，完整体现了课程广度。

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1. Number Skills and Index Laws | 数字运算与指数律

nd aᵐ ÷ aⁿ = aᵐ⁻ⁿ, which allow us to simplify expressions involving the same base quickly.

本部分检验运算流畅度、根式及运算顺序。核心指数法则为 aᵐ × aⁿ = aᵐ⁺ⁿ 和 aᵐ ÷ aⁿ = aᵐ⁻ⁿ，它们能帮助我们快速化简同底数幂的表达式。

Example 1: Simplify (5² × 5³) ÷ 5⁴.

例题1：化简 (5² × 5³) ÷ 5⁴。

t apply the multiplication law: 5² × 5³ = 5²⁺³ = 5⁵. Then divide: 5⁵ ÷ 5⁴ = 5⁵⁻⁴ = 5¹ = 5.

首先运用乘法法则：5² × 5³ = 5²⁺³ = 5⁵。再做除法：5⁵ ÷ 5⁴ = 5⁵⁻⁴ = 5¹ = 5。

Example 2: Evaluate 3 × 4² - 10 ÷ 2.

例题2：求值 3 × 4² - 10 ÷ 2。

right), Addition/Subtraction. 4² = 16, so 3 × 16 = 48 and 10 ÷ 2 = 5. Finally 48 - 5 = 43.

遵循 BIDMAS（括号、指数、除/乘、加/减）顺序：4² = 16，因此 3 × 16 = 48，10 ÷ 2 = 5。最后 48 - 5 = 43。

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2. Simplifying Algebraic Expressions | 化简代数表达式

Simplifying algebraic expressions involves identifying like terms — terms that have exactly the same variable part — and combining them. Brackets must be expanded first using the distributive law.

化简代数表达式需要识别同类项（变量部分完全相同的项）并将其合并。如果含有括号，必须先用分配律展开。

Example 1: Simplify 6a - 2b + 3a + 5b.

例题1：化简 6a - 2b + 3a + 5b。

rms: 6a + 3a = 9a. Group the b-terms: -2b + 5b = 3b. The simplified expression is 9a + 3b.

将含 a 项合并：6a + 3a = 9a。将含 b 项合并：-2b + 5b = 3b。化简后的表达式为 9a + 3b。

Example 2: Expand and simplify 4(2x - 3) + 5x.

例题2：展开并化简 4(2x - 3) + 5x。

Multiply the bracket by 4: 8x - 12. Then add 5x: 8x + 5x - 12 = 13x - 12.

将括号乘 4：得到 8x - 12。再加上 5x：8x + 5x - 12 = 13x - 12。

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3. Solving Linear Equations | 解一元一次方程

Solving linear equations requires isolating the unknown on one side. You can add, subtract, multiply or divide both sides by the same quantity, and always collect variable terms before constant terms.

解一元一次方程需要将未知数单独移到等式一边。你可以对等式两边同时进行加、减、乘、除操作，并且始终先将变量项合并，再处理常数项。

Example 1: Solve 5x - 3 = 2x + 9.

例题1：解方程 5x - 3 = 2x + 9。

Subtract 2x from both sides: 3x - 3 = 9. Add 3 to both sides: 3x = 12. Divide by 3: x = 4.

两边减 2x：3x - 3 = 9。两边加 3：3x = 12。两边除以 3：x = 4。

Example 2: Solve 2(y + 4) = 10.

例题2：解方程 2(y + 4) = 10。

t y = 1. Method 2: Divide both sides by 2 first → y + 4 = 5, then subtract 4 to get y = 1.

方法1：展开括号 → 2y + 8 = 10，然后减 8 再除以 2 得 y = 1。方法2：先将两边除以 2 → y + 4 = 5，再减 4 得 y = 1。

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4. Sequences and the nth Term | 数列与第n项

A linear sequence has a constant difference between terms. The nth term of a sequence is an expression that lets you calculate any term directly. For a sequence with common difference d, the nth term takes the form an + b, where a equals the difference.

线性数列的各项之间具有不变的差值。数列的第 n 项是一个能让你直接求出任意项的表达式。对于公差为 d 的数列，第 n 项的形式为 an + b，其中 a 等于公差。

Example: For the sequence 3, 7, 11, 15, … find the nth term and the 30th term.

例题：对于数列 3, 7, 11, 15, …，求第 n 项与第 30 项。

we must subtract 1. Hence nth term = 4n - 1. The 30th term is 4 × 30 - 1 = 120 - 1 = 119.

公差为 4，因此第 n 项以 4n 开始。当 n = 1 时，4 × 1 = 4，但首项是 3，因此需要减 1。第 n 项为 4n - 1。第 30 项为 4 × 30 - 1 = 120 - 1 = 119。

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5. Angles and Polygons | 角度与多边形

) × 180°. Vertically opposite angles are equal, and angles on a straight line sum to 180°.

任意三角形内角和为 180°。对于 n 边形，内角和为 (n - 2) × 180°。对顶角相等，平角为 180°。

Example 1: In a triangle, two angles are 45° and 70°. Find the missing angle.

例题1：在一个三角形中，两个角分别是 45° 和 70°。求第三个角。

of the known angles from 180°: 45° + 70° = 115°, so the missing angle = 180° - 115° = 65°.

从 180° 中减去已知角和：45° + 70° = 115°，因此缺失角 = 180° - 115° = 65°。

Example 2: Four interior angles of a pentagon are 108°, 120°, 95° and 110°. Find the fifth angle.

例题2：一个五边形的四个内角分别为 108°、120°、95° 和 110°。求第五个角。

Total of given angles = 108° + 120° + 95° + 110° = 433°. Fifth angle = 540° - 433° = 107°.

五边形内角和 = (5 - 2) × 180° = 540°。已知角总和 = 108° + 120° + 95° + 110° = 433°。第五个角 = 540° - 433° = 107°。

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6. Perimeter and Area | 周长与面积

rface enclosed and is expressed in square units. Memorising standard formulas is essential — for a rectangle, area = length × width; for a triangle, area = ½ × base × height.

周长是图形外部一周的总长度，以长度单位计量。

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