Year 9 Cambridge Advanced Mathematics: Unit Test Mock Paper Walkthrough | Year 9 Cambridge 进阶数学：单元测试模拟卷解析

📚 Year 9 Cambridge Advanced Mathematics: Unit Test Mock Paper Walkthrough | Year 9 Cambridge 进阶数学：单元测试模拟卷解析

This article provides a detailed, step-by-step walkthrough of a Year 9 Cambridge Advanced Mathematics unit test mock paper. The paper covers essential topics including quadratic equations, simultaneous equations, inequalities, linear sequences, straight‑line graphs, laws of indices, factorising quadratics with a≠1, right‑angled triangle trigonometry, probability without replacement, and algebraic proof. Each solution is broken down into clear reasoning and common exam tips, helping you master the techniques required for top marks.

本文为 Year 9 Cambridge 进阶数学的一份单元测试模拟卷提供详细的分步解析。试卷涵盖二次方程、联立方程组、不等式、线性数列、直线图像、指数运算法则、二次三项式因式分解（a≠1）、直角三角形三角比、无放回概率以及代数证明等核心主题。每道题的解答都分解为清晰的推理过程和常见考试技巧，助你掌握夺取高分所需的方法。

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1. Solving Quadratic Equations by Factorising | 因式分解解二次方程

Question: Solve x² − 7x + 12 = 0.

题目：解方程 x² − 7x + 12 = 0。

Step 1: Identify two numbers whose product is +12 and whose sum is −7.

步骤1：找出两个数，使其乘积为 +12，且和为 −7。

The correct pair is −3 and −4, because (−3) × (−4) = 12 and (−3) + (−4) = −7.

正确的数对是 −3 和 −4，因为 (−3)×(−4)=12 且 (−3)+(−4)=−7。

Step 2: Write the quadratic in its factorised form: (x − 3)(x − 4) = 0.

步骤2：将二次方程写成因式分解形式：(x − 3)(x − 4) = 0。

Step 3: Apply the zero‑product property. Set each bracket equal to zero: x − 3 = 0 or x − 4 = 0.

步骤3：应用零乘积性质。令每个括号等于零：x − 3 = 0 或 x − 4 = 0。

Step 4: Solve the two simple linear equations to obtain x = 3 or x = 4.

步骤4：解这两个简单的一次方程，得到 x = 3 或 x = 4。

tion to check them. 3² − 7×3 + 12 = 9 − 21 + 12 = 0, and 4² − 7×4 + 12 = 16 − 28 + 12 = 0.

考试技巧：务必把你的答案代回原方程进行检验。3² − 7×3 + 12 = 9 − 21 + 12 = 0，且 4² − 7×4 + 12 = 16 − 28 + 12 = 0。

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2. Simultaneous Equations | 联立方程组

Question: Solve the system 2x + y = 10 and x − y = 2.

题目：解方程组 2x + y = 10 和 x − y = 2。

Method: Elimination works nicely here because the coefficients of y are +1 and −1.

解法：这里适宜用消元法，因为 y 的系数分别为 +1 和 −1。

Add the two equations directly: (2x + y) + (x − y) = 10 + 2, which simplifies to 3x = 12.

将两个方程直接相加：(2x + y) + (x − y) = 10 + 2，化简得 3x = 12。

Divide both sides by 3 to give x = 4.

两边除以 3，得到 x = 4。

Substitute x = 4 into the second equation (x − y = 2): 4 − y = 2, so y = 2.

将 x = 4 代入第二个方程 (x − y = 2)：4 − y = 2，故 y = 2。

The solution is (4, 2). Check in the first equation: 2×4 + 2 = 10, which is correct.

方程组的解为 (4, 2)。代入第一个方程检验：2×4 + 2 = 10，正确。

Alternative check: The point (4,2) satisfies both equations, so it is the unique intersection of the two lines.

备选检验：点 (4,2) 同时满足两个方程，因此是两条直线唯一的交点。

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3. Inequalities on a Number Line | 数轴上的不等式

Question: Solve 3x − 5 ≤ 7 and represent the solution on a number line.

题目：求解不等式 3x − 5 ≤ 7，并在数轴上表示解集。

Step 1: Isolate the term containing x. Add 5 to both sides: 3x ≤ 12.

步骤1：分离含 x 的项。两边同时加 5：3x ≤ 12。

Divide both sides by 3. Since 3 is positive, the inequality sign remains unchanged: x ≤ 4.

步骤2：两边除以 3。因为 3 是正数，不等号方向不变：x ≤ 4。

To draw the number line: place a closed (filled) circle at 4 to show that 4 is included, then shade the line to the left of 4.

数轴画法：在 4 上画一个实心圆点，表示 4 包含在内，然后向 4 的左侧画阴影线。

sed circle for ≤ or ≥ and an open circle for < or >. Always label the scale clearly.

请牢记：对于 ≤ 或 ≥ 使用实心圆点，对于 < 或 > 使用空心圆圈。始终清晰地标注刻度。

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4. nth Term of a Linear Sequence | 线性数列的第n项

Question: Find the nth term of the sequence 5, 8, 11, 14, 17, …

题目：求数列 5, 8, 11, 14, 17, … 的第 n 项公式。

First, identify the common difference d. 8 − 5 = 3, 11 − 8 = 3, so d = 3.

首先，找出公差 d。8 − 5 = 3，11 − 8 = 3，因此 d = 3。

c) sequence, the nth term is given by a + (n − 1)d, where a is the first term. Here a = 5.

对于线性（等差）数列，第 n 项公式为 a + (n − 1)d，其中 a 为首项。这里 a = 5。

Substitute the values: nth term = 5 + (n − 1) × 3 = 5 + 3n − 3 = 3n + 2.

代入数值：第 n 项 = 5 + (n − 1) × 3 = 5 + 3n − 3 = 3n + 2。

st the formula: when n = 1, 3×1 + 2 = 5; when n = 2, 3×2 + 2 = 8. Both match the sequence.

检验公式：当 n = 1 时，3×1 + 2 = 5；当 n = 2 时，3×2 + 2 = 8。均符合数列。

Quick tip: The nth term of a linear sequence can also be found by writing the term as ‘dn + (a − d)’ which gives 3n + (5 − 3) = 3n + 2.

快记技巧：线性数列的第 n 项也可以写成 ‘dn + (a − d)’，即 3n + (5 − 3) = 3n + 2。

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5. Straight Line Graphs | 直线图像

ete a table of values for x from −1 to 2, draw the graph, and find the coordinates where the line crosses the axes.

题目：对于直线 y = 2x − 1，完成 x 从 −1 到 2 的取值表，画出图像，并求出直线与坐标轴的交点坐标。

able of values: x = −1 → y = 2(−1) − 1 = −3; x = 0 → y = −1; x = 1 → y = 1; x = 2 → y = 3.

取值表：x = −1 → y = 2(−1) − 1 = −3；x = 0 → y = −1；x = 1 → y = 1；x = 2 → y = 3。

Plot the points (−1, −3), (0, −1), (1, 1) and (2, 3) on a coordinate grid, then join them with a straight line.

在坐标网格上描出点 (−1, −3), (0, −1), (1, 1) 和 (2, 3)，然后用直尺将它们连成一条直线。

The y‑intercept is where x = 0; from the table this is (0, −1).

y 轴截距是 x = 0 时的点；由表可知为 (0, −1)。

For the x‑intercept, set y = 0: 0 = 2x − 1 → 2x = 1 → x = 0.5, giving the point (0.5, 0).

要求 x 轴截距，令 y = 0：0 = 2x − 1 → 2x = 1 → x = 0.5，得到点 (0.5, 0)。

Always label both intercepts clearly on your sketch.

在你的草图上务必清晰地标出这两个截距。

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6. Laws of Indices | 指数运算法则

Question: Simplify (3a²b)³ × (2a⁻¹b²)².

题目：化简 (3a²b)³ × (2a⁻¹b²)²。

brackets. For (3a²b)³: 3³ = 27, (a²)³ = a⁶, b³ = b³. So the first bracket becomes 27a⁶b³.

步骤1：将指数分别应用于括号内的每个因式。(3a²b)³：3³ = 27，(a²)³ = a⁶，b³ = b³。第一个括号变为 27a⁶b³。

For (2a⁻¹b²)²: 2² = 4, (a⁻¹)² = a⁻², (b²)² = b⁴. The second bracket simplifies to 4a⁻²b⁴.

对于 (2a⁻¹b²)²：2² = 4，(a⁻¹)² = a⁻²，(b²)² = b⁴。第二个括号化简为 4a⁻²b⁴。

Step 2: Multiply the simplified parts. Multiply the coefficients: 27 × 4 = 108.

步骤2：将化简后的部分相乘。系数相乘：27 × 4 = 108。

For the base a, add the exponents: 6 + (−2) = 4, so we get a⁴.

对于底数 a，指数相加：6 + (−2) = 4，故得 a⁴。

For base b, add the exponents: 3 + 4 = 7, so b⁷.

对于底数 b，指数相加：3 + 4 = 7，因此 b⁷。

Final answer: 108a⁴b⁷. Remember that a negative index means a reciprocal, but here the addition resolved it to a positive exponent.

最终答案：108a⁴b⁷。牢记负指数表示倒数，但这里通过加法得到了正指数。

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7. Factorising Quadratics (a ≠ 1) | 二次三项式因式分解 (a≠1)

Question: Factorise 2x² − 5x − 3 completely.

题目：将 2x² − 5x − 3 完全因式分解。

Here a = 2, b = −5, c = −3. Multiply a and c: 2 × (−3) = −6.

这里 a = 2，b = −5，c = −3。将 a 和 c 相乘：2 × (−3) = −6。

We need two numbers that multiply to −6 and add to −5. The numbers are −6 and 1.

我们需要两个数，乘积为 −6，和为 −5。这两个数是 −6 和 1。

Split the middle term −5x into −6x + 1x: 2x² − 6x + x − 3.

将中间项 −5x 拆分为 −6x + 1x：2x² − 6x + x − 3。

Group the first two terms and the last two terms: (2x² − 6x) + (x − 3).

将前两项和后两项分别分组：(2x² − 6x) + (x − 3)。

Factor each group: 2x(x − 3) + 1(x − 3). Now factor out the common binomial (x − 3): (2x + 1)(x − 3).

分解各组：2x(x − 3) + 1(x − 3)。现在提取公因式 (x − 3)：(2x + 1)(x − 3)。

Expand to verify: (2x + 1)(x − 3) = 2x² − 6x + x − 3 = 2x² − 5x − 3. Correct.

展开检验：(2x + 1)(x − 3) = 2x² − 6x + x − 3 = 2x² − 5x − 3。正确。

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8. Trigonometry in Right-Angled Triangles | 直角三角形中的三角比

, angle B = 90°, AB = 6 cm, BC = 8 cm. Calculate the size of angle A to one decimal place.

题目：在三角形 ABC 中，∠B = 90°，AB = 6 cm，BC = 8 cm。求 ∠A 的度数，精确到一位小数。

Sketch the triangle: right angle at B, so side AB is adjacent to angle A, and side BC is opposite angle A.

画出草图：直角在 B，因此对于角 A 而言，AB 是邻边，BC 是对边。

se we know the opposite and adjacent sides: tan A = opposite/adjacent = BC/AB = 8/6 = 4/3.

由于已知对边和邻边，使用正切比：tan A = 对边/邻边 = BC/AB = 8/6 = 4/3。

Take the inverse tangent: A = tan⁻¹(4/3). Use a calculator to obtain A ≈ 53.1301...°

取反正切：A = tan⁻¹(4/3)。用计算器求得 A ≈ 53.1301...°

Round to one decimal place: A ≈ 53.1°.

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