📚 Year 8 CIE Additional Mathematics: Unit Test Mock Paper Walkthrough | 八年级 CIE 进阶数学：单元测试模拟卷解析
This article unpacks a full mock unit test designed for the Year 8 CIE Additional Mathematics course. We work through each problem systematically, highlighting key techniques in algebra, equations, functions and proportional reasoning. Use this walkthrough to identify common pitfalls and strengthen your problem-solving toolkit.
本文全面解析一份为八年级 CIE 进阶数学课程设计的单元测试模拟卷。我们系统地讲解每道题目，突出代数、方程、函数和比例推理中的关键技巧。通过这份解析，你可以发现常见错误，强化解题能力。
1. Expanding and Simplifying Expressions | 展开并化简表达式
Question: Simplify (3x – 2)(x + 5) – 4(x – 1)².
题目：化简 (3x – 2)(x + 5) – 4(x – 1)²。
Step 1: Expand the first pair of brackets. Multiply (3x – 2) by (x + 5) to get 3x² + 15x – 2x – 10, which simplifies to 3x² + 13x – 10.
步骤1：展开第一组括号。将 (3x – 2) 乘以 (x + 5) 得到 3x² + 15x – 2x – 10，化简为 3x² + 13x – 10。
Step 2: Expand the square. (x – 1)² = x² – 2x + 1, then multiply by -4 to obtain -4x² + 8x – 4.
步骤2：展开完全平方。(x – 1)² = x² – 2x + 1，然后乘以 -4 得到 -4x² + 8x – 4。
Step 3: Combine all terms. 3x² + 13x – 10 – 4x² + 8x – 4 = (3x² – 4x²) + (13x + 8x) + (-10 – 4) = -x² + 21x – 14.
步骤3：合并所有同类项。3x² + 13x – 10 – 4x² + 8x – 4 = (3x² – 4x²) + (13x + 8x) + (-10 – 4) = -x² + 21x – 14。
Final answer: -x² + 21x – 14
2. Factorising Simple Quadratics | 因式分解简单二次式
Question: Factorise 2x² – 8x completely.
题目：将 2x² – 8x 完全因式分解。
Look for the greatest common factor (GCF) first. Both terms contain 2x. Taking out 2x leaves x – 4 inside the bracket.
首先寻找最大公因式 (GCF)。两项都含有 2x。提取 2x 后，括号内剩下 x – 4。
2x² – 8x = 2x(x – 4)
Always check by expanding: 2x × x = 2x² and 2x × (-4) = -8x. The expression is now fully factorised.
务必通过展开来检查：2x × x = 2x²，2x × (-4) = -8x。表达式已完全分解。
3. Solving Linear Equations | 解一元线性方程
Question: Solve 5(x – 3) + 2 = 3x – 7.
题目：解方程 5(x – 3) + 2 = 3x – 7。
Step 1: Expand the left-hand side. 5(x – 3) + 2 = 5x – 15 + 2 = 5x – 13.
步骤1：展开左边。5(x – 3) + 2 = 5x – 15 + 2 = 5x – 13。
Step 2: Bring variable terms to one side. 5x – 13 = 3x – 7 → 5x – 3x = -7 + 13 → 2x = 6.
步骤2：将含变量项移到同一边。5x – 13 = 3x – 7 → 5x – 3x = -7 + 13 → 2x = 6。
Step 3: Divide by the coefficient. x = 6 ÷ 2 = 3.
步骤3：除以系数。x = 6 ÷ 2 = 3。
x = 3
4. Solving an Inequality and Number Line Representation | 解不等式及数轴表示
Question: Solve the inequality 4x – 3 > 5 and represent the solution on a number line, given also that x ≤ 6.
题目：解不等式 4x – 3 > 5，并结合 x ≤ 6 在数轴上表示解集。
Solve the inequality: 4x > 8 → x > 2. Combine with x ≤ 6 to get 2 < x ≤ 6.
解不等式：4x > 8 → x > 2。结合条件 x ≤ 6，得到 2 < x ≤ 6。
On a number line, place an open circle at 2 (not included), a closed circle at 6 (included), and shade the region between them.
在数轴上，在 2 处画空心圆（不包含），在 6 处画实心圆（包含），并将它们之间的区域涂上阴影。
The solution in set notation: {x : 2 < x ≤ 6}.
用集合记号表示解集：{x : 2 < x ≤ 6}。
5. Applying Laws of Indices | 指数律的应用
Question: Simplify (2x³y⁻¹)² × 3x⁻²y⁴.
题目：化简 (2x³y⁻¹)² × 3x⁻²y⁴。
Step 1: Apply the power to the first bracket. (2²)(x³ˣ²)(y⁻¹ˣ²) = 4x⁶y⁻².
步骤1：将幂作用到第一个括号上。(2²)(x³ˣ²)(y⁻¹ˣ²) = 4x⁶y⁻²。
Step 2: Multiply by the second term. 4x⁶y⁻² × 3x⁻²y⁴ = 12 x⁶⁺⁽⁻²⁾ y⁻²⁺⁴ = 12x⁴y².
步骤2：乘以第二项。4x⁶y⁻² × 3x⁻²y⁴ = 12 x⁶⁺⁽⁻²⁾ y⁻²⁺⁴ = 12x⁴y²。
Simplified expression: 12x⁴y²
Remember: when multiplying same bases, add exponents. Negative exponents indicate reciprocals, so any variable with a negative exponent ends up in the denominator, but here all exponents become positive.
记住：同底数幂相乘，指数相加。负指数表示倒数，因此任何带负指数的变量最终会移至分母，但本题中所有指数最终都为正。
6. Finding the Equation of a Straight Line | 求直线方程
Question: A line passes through the points (2, 5) and (4, 9). Find its gradient and give the equation in slope-intercept form.
题目：一条直线经过点 (2, 5) 和 (4, 9)。求其斜率，并用斜截式写出方程。
Gradient (m) = (y₂ – y₁)/(x₂ – x₁) = (9 – 5)/(4 – 2) = 4/2 = 2.
斜率 (m) = (y₂ – y₁)/(x₂ – x₁) = (9 – 5)/(4 – 2) = 4/2 = 2。
Using point-slope form with (2, 5): y – 5 = 2(x – 2) → y – 5 = 2x – 4 → y = 2x + 1.
用点斜式，以 (2, 5) 代入：y – 5 = 2(x – 2) → y – 5 = 2x – 4 → y = 2x + 1。
Equation: y = 2x + 1
Check with the other point: when x = 4, y = 2×4 + 1 = 9. Correct.
用另一个点检验：当 x = 4 时，y = 2×4 + 1 = 9。正确。
7. Angles in a Triangle | 三角形内角
Question: The angles of a triangle are (2x + 10)°, (3x – 20)° and (x + 30)°. Find x and determine whether the triangle is scalene, isosceles or right-angled.
题目：三角形的三个内角分别为 (2x + 10)°、(3x – 20)° 和 (x + 30)°。求 x 的值，并判断该三角形是不等边三角形、等腰三角形还是直角三角形。
Sum of angles in a triangle is 180°: (2x + 10) + (3x – 20) + (x + 30) = 180.
三角形内角和为 180°：(2x + 10) + (3x – 20) + (x + 30) = 180。
Simplify: 2x + 3x + x + 10 – 20 + 30 = 180 → 6x + 20 = 180 → 6x = 160 → x = 160/6 = 80/3 ≈ 26.67.
化简：2x + 3x + x + 10 – 20 + 30 = 180 → 6x + 20 = 180 → 6x = 160 → x = 160/6 = 80/3 ≈ 26.67。
Now find the angles: 2x+10 = 2×(80/3)+10 = 160/3+30/3 = 190/3° ≈ 63.33°, 3x-20 = 3×(80/3)-20 = 80-20 = 60°, x+30 = 80/3+90/3 = 170/3° ≈ 56.67°.
计算三个角：2x+10 = 2×(80/3)+10 = 160/3+30/3 = 190/3° ≈ 63.33°，3x-20 = 3×(80/3)-20 = 80-20 = 60°，x+30 = 80/3+90/3 = 170/3° ≈ 56.67°。
All angles are different, so the triangle is scalene. None equals 90°, so not right-angled.
三个角都不相等，因此是不等边三角形。没有角等于90°，不是直角三角形。
8. Word Problem: Rectangle Perimeter | 应用题：长方形周长
Question: The length of a rectangle is 3 cm more than its width. The perimeter is 26 cm. Find the length and width.
题目：一个长方形的长比宽多 3 cm，周长为 26 cm。求长和宽。
Let width = x cm, then length = x + 3 cm. Perimeter = 2(length + width) = 2(x + x + 3) = 2(2x + 3) = 4x + 6.
设宽为 x cm，则长为 x + 3 cm。周长 = 2(长+宽) = 2(x + x + 3) = 2(2x + 3) = 4x + 6。
Set equal to 26: 4x + 6 = 26 → 4x = 20 → x = 5. Width = 5 cm, length = 8 cm.
令其等于 26：4x + 6 = 26 → 4x = 20 → x = 5。宽 = 5 cm，长 = 8 cm。
Dimensions: 5 cm by 8 cm
9. Arithmetic Sequences | 等差数列
Question: Consider the sequence 3, 7, 11, 15, … . Find the formula for the nth term and use it to calculate the 20th term.
题目：给定数列 3, 7, 11, 15, … 。写出第 n 项的公式，并计算第 20 项。
The first term a₁ = 3, common difference d = 7 – 3 = 4. The nth term of an arithmetic sequence is given by aₙ = a₁ + (n-1)d.
首项 a₁ = 3，公差 d = 7 – 3 = 4。等差数列的第 n 项公式为 aₙ = a₁ + (n-1)d。
Substitute: aₙ = 3 + (n-1)×4 = 3 + 4n – 4 = 4n – 1.
代入：aₙ = 3 + (n-1)×4 = 3 + 4n – 4 = 4n – 1。
For the 20th term, n = 20: a₂₀ = 4×20 – 1 = 80 – 1 = 79.
第 20 项，n = 20：a₂₀ = 4×20 – 1 = 80 – 1 = 79。
nth term: 4n – 1; 20th term: 79
10. Direct Proportion with a Square | 平方正比例
Question: y is directly proportional to x². When x = 2, y = 12. Find y when x = 5.
题目：y 与 x² 成正比。当 x = 2 时，y = 12。求当 x = 5 时 y 的值。
Write the proportion as an equation: y = kx², where k is the constant of proportionality.
将正比例关系写成等式：y = kx²，其中 k 为比例常数。
Find k using the given values: 12 = k × (2)² → 12 = 4k → k = 3.
用已知数值求 k：12 = k × (2)² → 12 = 4k → k = 3。
Now substitute x = 5: y = 3 × (5)² = 3 × 25 = 75.
现在代入 x = 5：y = 3 × (5)² = 3 × 25 = 75。
y = 75
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