Year 8 CAIE Further Mathematics: Unit Test Mock Paper Analysis | 8年级 CAIE 进阶数学：单元测试模拟卷解析

📚 Year 8 CAIE Further Mathematics: Unit Test Mock Paper Analysis | 8年级 CAIE 进阶数学：单元测试模拟卷解析

This article provides a complete walkthrough of a mock unit test designed for Year 8 CAIE Further Mathematics. Each question is solved step by step, with clear explanations in both English and Chinese, highlighting essential techniques, common pitfalls and checking strategies to build exam confidence.

本文为8年级CAIE进阶数学的一套单元模拟测试卷提供完整解析。每道题均含逐步求解、清晰的中英双语解释，并强调关键技巧、常见错误以及验算方法，帮助同学们建立应考信心。

1. Question 1: Solving Linear Equations | 第1题：解线性方程

Question: Solve the equation 3x – 7 = 2x + 5.

题目：解方程 3x – 7 = 2x + 5。

Step 1 – Collect x terms on one side. Subtract 2x from both sides of the equation to keep it balanced:

步骤1 – 将含x的项集中到一边。方程两边同时减去2x，保持等式的平衡：

3x – 2x – 7 = 5 → x – 7 = 5

Step 2 – Isolate x. Add 7 to both sides to undo the subtraction of 7:

步骤2 – 分离x。两边加7，以消去减7：

x – 7 + 7 = 5 + 7 → x = 12

Check: substitute x = 12 back into the original equation: 3(12) – 7 = 36 – 7 = 29; 2(12) + 5 = 24 + 5 = 29. Both sides equal, so the solution is correct.

验算：将x=12代回原方程：3(12)–7=29；2(12)+5=29，两边相等，答案正确。

Common mistake: Forgetting to change the sign when moving terms. Always do the same operation on both sides.

常见错误：移项时忘记变号。务必在等式两边同时进行相同的运算。

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2. Question 2: Expanding and Simplifying Brackets | 第2题：去括号并化简

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

题目：展开并化简 (2x + 3)(x – 4) – x(x + 2)。

Step 1 – Multiply the first pair of brackets using FOIL (First, Outer, Inner, Last):

步骤1 – 使用乘法分配律展开第一组括号：

(2x)(x) + (2x)(–4) + (3)(x) + (3)(–4) = 2x² – 8x + 3x – 12

Combine like terms for the first group: –8x + 3x = –5x, so we get 2x² – 5x – 12.

合并同类项：–8x+3x=–5x，得 2x² – 5x – 12。

Step 2 – Expand the second part: –x(x + 2) means multiply –x by both terms inside. Be careful with the negative sign:

步骤2 – 展开第二部分：–x(x+2) 即用–x乘以括号内的每一项，注意负号：

–x · x = –x² and –x · 2 = –2x

Step 3 – Write the full expression and simplify by collecting like terms:

步骤3 – 写出整体表达式并合并同类项：

(2x² – 5x – 12) – x² – 2x = 2x² – x² – 5x – 2x – 12 = x² – 7x – 12

Check: You can verify with a particular value, e.g. x=1: original (2+3)(1–4) – 1(1+2) = 5×(–3) – 3 = –15 – 3 = –18. Simplified: 1² – 7(1) – 12 = 1 – 7 – 12 = –18. Correct.

验算：代入x=1，原式：(5)(–3)–1(3)= –15–3= –18；化简式：1–7–12= –18。一致。

Common mistake: Missing the negative sign when subtracting x(x+2). Remember – × + = –.

常见错误：减去x(x+2)时漏掉负号。记住负号乘正项仍为负。

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3. Question 3: Factorising Quadratic Expressions | 第3题：二次三项式的因式分解

Question: Factorise x² – 5x – 24.

题目：对 x² – 5x – 24 进行因式分解。

Strategy: Look for two numbers whose product is –24 and sum is –5 (the coefficient of x). Think of pairs of factors of –24: (1, –24), (2, –12), (3, –8), (4, –6), etc. We need the pair that sums to –5.

策略：寻找两个数，积为–24，和为–5（x的系数）。列出–24的因子对：(1, –24), (2, –12), (3, –8), (4, –6)等，找到和为–5的一组。

The correct pair is 3 and –8 because 3 + (–8) = –5 and 3 × (–8) = –24.

正确的一组是3和–8，因为3+(–8)= –5，且3×(–8)= –24。

Hence the factorised form is:

因此因式分解的结果为：

x² – 5x – 24 = (x + 3)(x – 8)

Check by expanding: (x+3)(x–8) = x² – 8x + 3x – 24 = x² – 5x – 24. Correct.

验算：展开(x+3)(x–8) = x²–8x+3x–24 = x²–5x–24。正确。

Common mistake: Getting the signs wrong; always check that the product of the two numbers gives the constant term and their sum gives the middle coefficient.

常见错误：符号出错；一定要确保两个数的积等于常数项，和等于一次项系数。

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4. Question 4: Solving Simultaneous Equations | 第4题：解联立方程组

Question: Solve the simultaneous equations: 2x + y = 8 and x – 3y = –10.

题目：解联立方程组：2x + y = 8 和 x – 3y = –10。

Method – Substitution. From the second equation, express x in terms of y:

方法 – 代入法。由第二个方程用y表示x：

x = 3y – 10

Step 1 – Substitute this into the first equation:

步骤1 – 代入第一个方程：

2(3y – 10) + y = 8

Step 2 – Expand and solve for y:

步骤2 – 展开并解y：

6y – 20 + y = 8 → 7y = 28 → y = 4

Step 3 – Now substitute y = 4 back into x = 3y – 10:

步骤3 – 将y=4代回 x = 3y – 10：

x = 3(4) – 10 = 12 – 10 = 2

So the solution is (x, y) = (2, 4).

解为 (x, y) = (2, 4)。

Check in both equations: 2(2)+4=8, and 2–3(4)=2–12= –10. Both hold true.

代入两方程验证：2(2)+4=8，2–3(4)=–10。均成立。

Common mistake: Misapplying distribution, e.g. forgetting to multiply both terms inside the bracket when substituting.

常见错误：乘法分配律运用错误，例如代入时忘记乘以括号内的每一项。

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5. Question 5: Pythagoras’ Theorem | 第5题：毕达哥拉斯定理

Question: A right-angled triangle has legs of lengths 6 cm and 8 cm. Find the length of the hypotenuse.

题目：一个直角三角形的两条直角边分别为6 cm和8 cm。求斜边的长度。

Use Pythagoras' theorem: c² = a² + b², where c is the hypotenuse.

使用毕达哥拉斯定理：c² = a² + b²，其中c为斜边。

c² = 6² + 8² = 36 + 64 = 100

Take the square root of both sides (length must be positive):

两边开平方（长度为正值）：

c = √100 = 10 cm

Check: 10² = 100, and 6²+8² = 36+64 = 100. So the hypotenuse is 10 cm.

验算：10²=100，6²+8²=100。斜边长度为10 cm。

Common mistake: Mixing up the hypotenuse with a leg. Always identify the longest side opposite the right angle.

常见错误：混淆斜边与直角边。斜边总是直角所对的最长边。

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6. Question 6: Angles with Parallel Lines | 第6题：平行线的角度问题

Question: In the diagram, lines AB and CD are parallel. Line EF intersects them. Angle AEF is 55°. Find angle EFD, giving a reason.

题目：如图，AB∥CD，直线EF与它们相交。∠AEF = 55°。求∠EFD，并说明理由。

Since AB // CD, the alternate interior angles formed by the transversal EF are equal. Angle AEF and angle EFD are on alternate sides of the transversal and inside the parallel lines.

由于AB∥CD，截线EF形成的交替内角相等。∠AEF和∠EFD位于截线两侧且在平行线之间，是一对交替内角。

∠EFD = 55° (alternate angles are equal)

Common mistake: Confusing alternate angles with corresponding angles. Alternate angles form a "Z" shape, while corresponding angles form an "F" shape.

常见错误：混淆交替内角与同位角。交替内角呈“Z”型，同位角呈“F”型。

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7. Question 7: Sequences and the nth Term | 第7题：数列与第n项

Question: The nth term of a sequence is given by 3n + 2. Write down the first three terms and the 10th term.

题目：某数列的第n项公式为 3n+2。写出该数列的前三项及第10项。

To find the first term, substitute n = 1: 3(1) + 2 = 5.

求第一项，代入n=1：3(1)+2=5。

Second term (n=2): 3(2) + 2 = 8.

第二项(n=2)：3(2)+2=8。

Third term (n=3): 3(3) + 2 = 11.

第三项(n=3)：3(3)+2=11。

The sequence starts 5, 8, 11, ... (an arithmetic sequence with common difference 3).

数列开头为5, 8, 11, …（公差为3的等差数列）。

Tenth term (n=10): 3(10) + 2 = 32.

第十项(n=10)：3(10)+2=32。

Check: Using the pattern, 11th term = 35, 12th = 38, etc., confirming the rule.

验证：按规律，第11项为35，第12项为38，符合公式。

Common mistake: Forgetting the order of operations; always multiply n by 3 before adding 2.

常见错误：运算顺序出错；必须先乘3再加2。

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8. Question 8: Simplifying with Index Laws | 第8题：指数律化简

Question: Simplify (2a²b)³ × (ab²)² ÷ 4a⁶b⁴.

题目：化简 (2a²b)³ × (ab²)² ÷ 4a⁶b⁴。

Step 1 – Apply the power to each factor inside the brackets. Remember (xy)ⁿ = xⁿyⁿ.

步骤1 – 将指数分配给括号内的每个因式。牢记 (xy)ⁿ = xⁿyⁿ。

(2a²b)³ = 2³ · (a²)³ · b³ = 8 a⁶ b³

(ab²)² = a² · (b²)² = a² b⁴

Step 2 – Multiply the two results together. When multiplying, add the exponents of like bases:

步骤2 – 将两式相乘。同底数幂相乘，指数相加：

8 a⁶ b³ × a² b⁴ = 8 a⁶⁺² b³⁺⁴ = 8 a⁸ b⁷

Step 3 – Divide by 4a⁶b⁴, subtracting exponents for each base:

步骤3 – 除以4a⁶b⁴，同底数幂相除，指数相减：

8 \ 4 = 2, a⁸⁻⁶ = a², b⁷⁻⁴ = b³ → 2 a² b³

Final simplified expression: 2a²b³.

最终化简结果：2a²b³。

Check: Pick a=1, b=1. Original: (2·1·1)³×(1·1)² ÷4 = 8×1÷4=2. Simplified: 2·1·1=2. Works.

验算：取a=1, b=1。原式：(2)³×(1)²÷4=8÷4=2；化简式：2×1×1=2。一致。

Common mistake: Forgetting to raise the coefficient to the power as well,

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