Year 10 SQA Mathematics: Unit Test Mock Paper Walkthrough | SQA 数学单元测试模拟卷解析

📚 Year 10 SQA Mathematics: Unit Test Mock Paper Walkthrough | SQA 数学单元测试模拟卷解析

Welcome to this detailed walkthrough of a mock unit test designed for Year 10 SQA Mathematics, covering key topics from the National 5 Relationships unit. This resource breaks down typical exam-style questions step by step, helping you understand the methods and reasoning required to achieve full marks. We will explore algebraic fractions, quadratic equations, straight lines, simultaneous equations, trigonometry, vectors, and contextual problem-solving. Use this guide to consolidate your knowledge and build confidence for your actual assessment.

欢迎阅读这篇为 Year 10 SQA 数学设计的单元测试模拟卷详细解析，内容涵盖 National 5 关系中单元的核心主题。本资源逐步拆解典型的考试风格题目，帮助你理解取得满分所需的方法和推理。我们将探讨代数分式、二次方程、直线、联立方程组、三角函数、向量以及情境问题解决。使用本指南巩固你的知识，并为实际评估建立信心。

1. Simplifying Algebraic Fractions | 化简代数分式

Question: Simplify the rational expression (x² − 9) / (x² − 2x − 3) fully.

题目：完全化简有理式 (x² − 9) / (x² − 2x − 3)。

Step 1: Factorise the numerator as a difference of squares. x² − 9 = (x − 3)(x + 3).

步骤1：将分子因式分解为平方差。x² − 9 = (x − 3)(x + 3)。

s that multiply to −3 and add to −2. These are −3 and +1, so x² − 2x − 3 = (x − 3)(x + 1).

步骤2：将分母分解为两个二项式。寻找乘积为 −3 且和为 −2 的两个数，分别是 −3 和 +1，因此 x² − 2x − 3 = (x − 3)(x + 1)。

Step 3: Cancel the common factor (x − 3), provided x ≠ 3 to avoid division by zero.

步骤3：约去公因式 (x − 3)，注意 x ≠ 3 以避免分母为零。

Simplified result:

化简结果：

(x + 3) / (x + 1), x ≠ 3, x ≠ −1

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2. Solving Quadratic Equations by Factorisation | 用因式分解法解二次方程

Question: Solve x² − 5x + 6 = 0.

题目：解方程 x² − 5x + 6 = 0。

Step 1: Identify two numbers with product 6 and sum −5. These are −2 and −3.

步骤1：找出乘积为 6 且和为 −5 的两个数，即 −2 和 −3。

Step 2: Write the factorised form: (x − 2)(x − 3) = 0.

步骤2：写出因式分解形式：(x − 2)(x − 3) = 0。

Step 3: Apply the zero product property: set each factor equal to zero.

步骤3：应用零乘积性质：令每个因式等于零。

Step 4: x − 2 = 0 gives x = 2; x − 3 = 0 gives x = 3.

步骤4：由 x − 2 = 0 得 x = 2；由 x − 3 = 0 得 x = 3。

Solutions: x = 2, x = 3

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3. Using the Quadratic Formula | 使用二次公式

Question: Solve 2x² + 3x − 2 = 0 using the quadratic formula.

题目：使用二次公式解方程 2x² + 3x − 2 = 0。

Step 1: Identify coefficients a = 2, b = 3, c = −2. The quadratic formula is x = [−b ± √b² − 4ac] / (2a).

步骤1：确定系数 a = 2，b = 3，c = −2。二次公式为 x = [−b ± √b² − 4ac] / (2a)。

Step 2: Compute the discriminant Δ = b² − 4ac = 3² − 4(2)(−2) = 9 + 16 = 25.

步骤2：计算判别式 Δ = b² − 4ac = 3² − 4(2)(−2) = 9 + 16 = 25。

Step 3: Substitute into the formula: x = [−3 ± √25] / (2 × 2) = [−3 ± 5] / 4.

步骤3：代入公式：x = [−3 ± √25] / (2 × 2) = [−3 ± 5] / 4。

ep 4: Calculate the two roots: x = (−3 + 5)/4 = 2/4 = 1/2, and x = (−3 − 5)/4 = −8/4 = −2.

步骤4：计算两个根：x = (−3 + 5)/4 = 2/4 = 1/2，以及 x = (−3 − 5)/4 = −8/4 = −2。

x = ½, x = −2

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4. Completing the Square | 配方法

Question: Express x² + 6x + 5 in the form (x + p)² + q.

题目：将 x² + 6x + 5 写成 (x + p)² + q 的形式。

Step 1: Take the coefficient of x, which is 6, halve it to get 3, and square it to obtain 9.

步骤1：取 x 的系数 6，除以 2 得 3，再平方得到 9。

Step 2: Rewrite the expression as (x² + 6x + 9) − 9 + 5.

步骤2：将表达式重写为 (x² + 6x + 9) − 9 + 5。

p 3: The first three terms form a perfect square: (x + 3)². Combine constants −9 + 5 = −4.

步骤3：前三项构成完全平方：(x + 3)²。合并常数 −9 + 5 = −4。

(x + 3)² − 4

Here p = 3 and q = −4.

这里 p = 3，q = −4。

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5. Finding the Equation of a Straight Line | 求直线方程

Find the equation of the line passing through (2, 5) that is perpendicular to y = 2x + 1.

题目：求经过点 (2, 5) 且垂直于直线 y = 2x + 1 的直线方程。

1: The given line has gradient m₁ = 2. For perpendicular lines, m₁ × m₂ = −1, so m₂ = −½.

步骤1：所给直线的斜率为 m₁ = 2。对于垂直线，m₁ × m₂ = −1，因此 m₂ = −½。

Step 2: Use the point-gradient form y − y₁ = m(x − x₁) with (x₁, y₁) = (2, 5) and m = −½.

步骤2：使用点斜式 y − y₁ = m(x − x₁)，代入 (x₁, y₁) = (2, 5) 和 m = −½。

tep 3: y − 5 = −½(x − 2). Multiply both sides by 2: 2y − 10 = −(x − 2) → 2y − 10 = −x + 2.

步骤3：y − 5 = −½(x − 2)。两边乘以 2：2y − 10 = −(x − 2) → 2y − 10 = −x + 2。

Step 4: Rearrange to the standard form: x + 2y − 12 = 0, or y = −½x + 6.

步骤4：整理成标准形式：x + 2y − 12 = 0，或 y = −½x + 6。

Equation: y = −½x + 6

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6. Solving Simultaneous Equations | 解联立方程组

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: Use the elimination method by adding the two equations to cancel y.

方法：使用加减消元法，将两方程相加消去 y。

Step 1: (2x + y) + (x − y) = 10 + 2 → 3x = 12 → x = 4.

步骤1：(2x + y) + (x − y) = 10 + 2 → 3x = 12 → x = 4。

Step 2: Substitute x = 4 into the second equation: 4 − y = 2 → y = 2.

步骤2：将 x = 4 代入第二个方程：4 − y = 2 → y = 2。

Check in the first equation: 2(4) + 2 = 8 + 2 = 10, which is true.

代入第一个方程检验：2(4) + 2 = 8 + 2 = 10，成立。

x = 4, y = 2

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7. Solving a Trigonometric Equation | 解三角方程

Question: Solve sin x = 0.5 for 0° ≤ x ≤ 360°.

题目：在 0° ≤ x ≤ 360° 范围内解 sin x = 0.5。

1: Determine the first acute angle using the inverse sine function: x = sin⁻¹(0.5) = 30°.

步骤1：用反正弦函数求出第一个锐角：x = sin⁻¹(0.5) = 30°。

e is positive in the first and second quadrants, the second solution is 180° − 30° = 150°.

步骤2：因为正弦在第一和第二象限为正，第二个解为 180° − 30° = 150°。

Step 3: No further solutions in the given interval. Both 30° and 150° satisfy the equation.

步骤3：在给定区间内没有其他解。30° 和 150° 均满足方程。

x = 30°, 150°

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8. Applying the Cosine Rule | 应用余弦定理

ion: In triangle ABC, side AB = 7 cm, BC = 8 cm, and angle B = 60°. Find the length of AC.

题目：在三角形 ABC 中，边 AB = 7 cm，BC = 8 cm，∠B = 60°。求 AC 的长度。

2ac cos B. Here we need side AC opposite angle B, so label AC = b, AB = c = 7, BC = a = 8.

步骤1：余弦定理为 b² = a² + c² − 2ac cos B。我们需要求角 B 的对边 AC，故设 AC = b，AB = c = 7，BC = a = 8。

Step 2: Substitute values: b² = 8² + 7² − 2 × 8 × 7 × cos 60°.

步骤2：代入数值：b² = 8² + 7² − 2 × 8 × 7 × cos 60°。

Step 3: cos 60° = 0.5, so b² = 64 + 49 − 2 × 56 × 0.5 = 113 − 56 = 57.

步骤3：cos 60° = 0.5，因此 b² = 64 + 49 − 2 × 56 × 0.5 = 113 − 56 = 57。

Step 4: Take the positive square root: b = √57 cm (exact value ≈ 7.55 cm).

步骤4：取正平方根：b = √57 cm（精确值约 7.55 cm）。

AC = √57 cm

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9. Vector Operations and Magnitude | 向量运算与模

Given vectors a = (3, −2) and b = (−1, 4), calculate the magnitude of the vector 3a + 2b.

题目：给定向量 a = (3, −2) 和 b = (−1, 4)，计算向量 3a + 2b 的模。

ultiply each vector by the scalar: 3a = (3×3, 3×−2) = (9, −6); 2b = (2×−1, 2×4) = (−2, 8).

步骤1：用标量乘以各向量：3a = (3×3, 3×−2) = (9, −6)；2b = (2×−1, 2×4) = (−2, 8)。

Step 2: Add the corresponding components: 3a + 2b = (9 + (−2), −6 + 8) = (7, 2).

步骤2：对应分量相加：3a + 2b = (9 + (−2), −6 + 8) = (7, 2)。

Step 3: Find the magnitude using |v| = √x² + y²: |(7, 2)| = √7² + 2² = √49 + 4 = √53.

步骤3：用公式 |v| = √x² + y² 求模：|(7, 2)| = √7² + 2² = √49 + 4 = √53。

Magnitude = √53

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10. Quadratic Modelling and Maximum Height | 二次函数建模与最大高度

ds after launch is given by h(t) = −2t² + 12t + 3 metres. Find the maximum height reached.

题目：烟花弹发射 t 秒后的高度由 h(t) = −2t² + 12t + 3 米给出。求到达的最大高度。

so the maximum occurs at the vertex. Use t = −b/(2a) from the standard form at² + bt + c.

步骤1：抛物线开口向下（t² 系数为负），因此最大值出现在顶点。使用标准式 at² + bt + c 的 t = −b/(2a)。

Step 2: Here a = −2, b = 12. t = −12 / (2 × −2) = −12 / −4 = 3 seconds.

步骤2：这里 a = −2，b = 12。t = −12 / (2 × −2) = −12 / −4 = 3 秒。

3: Substitute t = 3 into h(t): h(3) = −2(3)² + 12(3) + 3 = −2(9) + 36 + 3 = −18 + 39 = 21.

步骤3：将 t = 3 代入 h(t)：h(3) = −2(3)² + 12(3) + 3 = −2(9) + 36 + 3 = −18 + 39 = 21。

Maximum height = 21 metres

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11. Working with Scientific Notation and Significant Figures | 科学记数法与有效数字

Question: The mass of a particle is 0.0000457 kg. Write this in scientific notation correct to three significant figures.

题目：一个粒子的质量为 0.0000457 kg。用科学记数法表示该数，并精确到三位有效数字。

decimal point 5 places to the right: 4.57, so the exponent is n = −5. Thus 4.57 × 10⁻⁵ kg.

步骤1：用科学记数法表示为 a × 10ⁿ，其中 1 ≤ a < 10。将小数点右移5位得 4.57，因此指数 n = −5，即 4.57 × 10⁻⁵ kg。

actly), so it remains 4.57 × 10⁻⁵ kg. If it had been 4.578, we would round to 4.58 × 10⁻⁵.

步骤2：四舍五入到三位有效数字。第三位是 7，下一位是 0（因为原数为确切的 4.57），所以保持 4.57 × 10⁻⁵ kg。若为 4.578，则进位为 4.58 × 10⁻⁵。

4.57 × 10⁻⁵ kg

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12. Interpreting Graphs and Gradient in Context | 情境图的解读与斜率

Question: A distance-time graph shows a straight line from (0,0) to (5, 200) where distance is in metres and time in seconds. Calculate the speed and interpret what a steeper gradient would mean.

题目：一张距离-时间图显示从 (0,0) 到 (5, 200) 的直线，距离单位为米，时间单位为秒。计算速度，并解释斜率更陡代表什么。

time graph. Gradient = change in distance / change in time = (200 − 0) / (5 − 0) = 40 m/s.

步骤1：速度是距离-时间图的斜率。斜率 = 距离变化量 / 时间变化量 = (200 − 0) / (5 − 0) = 40 m/s。

Step 2: A steeper gradient indicates a higher speed, meaning more distance is covered in the same time interval.

步骤2：更陡的斜率表示更高的速度，意味着在相同时间间隔内移动了更远的距离。

Step 3: In real-life terms, the object would be moving faster. If the graph becomes a horizontal line, the gradient is zero, indicating the object is stationary.

步骤3：在实际情况下，物体运动得更快。如果图像变为水平线，斜率为零，则表明物体静止。

Speed = 40 m/s; steeper gradient = higher speed

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