📚 Year 11 SQA Mathematics: Unit Test Mock Paper Solutions | Year 11 SQA 数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock Unit Test for SQA National 5 Mathematics, covering key skills in Expressions, Formulae, Relationships and Applications. Each question is presented with a clear solution and commentary, helping Year 11 learners consolidate their understanding and prepare effectively for the final assessment.
本文细致解析了一套 SQA National 5 数学单元测试模拟卷，涵盖表达式、公式、关系及应用等重要考点。每道题都配有清晰的解答与点评，旨在帮助 Year 11 学生夯实基础，高效备战最终考试。
1. Scientific Notation and Significant Figures | 科学记数法与有效数字
Question: Write 0.000045 in scientific notation, giving your answer to 2 significant figures.
题目：将 0.000045 用科学记数法表示，并保留两位有效数字。
Solution: Move the decimal point 5 places to the right to get 4.5 × 10⁻⁵. The digits ‘4’ and ‘5’ are already the first two significant figures, so the answer remains 4.5 × 10⁻⁵.
解答：将小数点向右移动 5 位得到 4.5 × 10⁻⁵。数字‘4’和‘5’本身就是前两位有效数字，因此答案为 4.5 × 10⁻⁵。
2. Simplifying Expressions with Indices | 指数表达式化简
Question: Simplify (m³)² × m⁴ ÷ m⁵.
题目：化简 (m³)² × m⁴ ÷ m⁵。
Solution: First compute the power of a power: (m³)² = m⁶. Then multiply by m⁴ to obtain m⁶⁺⁴ = m¹⁰. Finally divide by m⁵: m¹⁰⁻⁵ = m⁵.
解答：先计算幂的乘方：(m³)² = m⁶。再乘以 m⁴ 得 m⁶⁺⁴ = m¹⁰。最后除以 m⁵：m¹⁰⁻⁵ = m⁵。
3. Expanding Brackets and Factorising | 括号展开与因式分解
Question (a): Expand 3(x + 2) − 2(x − 1). (b): Factorise fully x² − 9.
题目 (a)：展开 3(x + 2) − 2(x − 1)。(b)：将 x² − 9 彻底因式分解。
Solution (a): Distribute: 3x + 6 − 2x + 2. Combine like terms: x + 8. (b): Recognise the difference of two squares: x² − 9 = (x + 3)(x − 3).
解答 (a)：去括号得 3x + 6 − 2x + 2，合并同类项为 x + 8。(b)：利用平方差公式，x² − 9 = (x + 3)(x − 3)。
4. Solving Linear Equations | 解一元一次方程
Question: Solve 2(x + 3) = 10 − (x − 2).
题目：解方程 2(x + 3) = 10 − (x − 2)。
Solution: Expand both sides: 2x + 6 = 10 − x + 2, which simplifies to 2x + 6 = 12 − x. Add x to both sides: 3x + 6 = 12. Subtract 6: 3x = 6. Divide by 3: x = 2.
解答：两边展开：2x + 6 = 10 − x + 2，化简得 2x + 6 = 12 − x。两边同时加 x 得 3x + 6 = 12。减 6 得 3x = 6。除以 3 得 x = 2。
5. Rearranging Formulae | 公式变形
Question: Make r the subject of the formula V = (4/3)πr³.
题目：将公式 V = (4/3)πr³ 变形，使 r 作为主语。
Solution: Multiply both sides by 3: 3V = 4πr³. Divide by 4π: r³ = 3V / (4π). Take the cube root: r = ³√(3V / (4π)).
r = ³√(3V / (4π))
解答：两边同乘 3：3V = 4πr³。再同除以 4π 得 r³ = 3V / (4π)。开立方根即得 r = ³√(3V / (4π))。
6. Exact Trigonometric Values | 精确三角函数值
Question: Without using a calculator, evaluate sin 30° + cos 60°.
题目：不用计算器，求 sin 30° + cos 60° 的值。
Solution: From exact value triangles, sin 30° = 1/2 and cos 60° = 1/2. Adding them gives 1.
解答：由精确值三角形可知 sin 30° = 1/2，cos 60° = 1/2。两者相加得 1。
7. Volume of a Sphere | 球体体积
Question: Calculate the volume of a sphere with radius 5 cm. Give your answer to 3 significant figures. (Use π = 3.14)
题目：计算半径为 5 cm 的球体体积，结果保留三位有效数字。（取 π = 3.14）
Solution: Volume V = (4/3)πr³ = (4/3) × 3.14 × 5³. Since 5³ = 125, V = (4/3) × 3.14 × 125. Compute 3.14 × 125 = 392.5. Then (4 ÷ 3) ≈ 1.3333, multiply: 1.3333 × 392.5 ≈ 523.33. To three significant figures, the volume is 523 cm³.
解答：体积 V = (4/3)πr³ = (4/3) × 3.14 × 5³。5³ = 125，先求 3.14 × 125 = 392.5。再乘以 4/3：1.3333 × 392.5 ≈ 523.33。保留三位有效数字为 523 cm³。
8. Gradient and Equation of a Straight Line | 斜率与直线方程
Question: Find the gradient of the line passing through A(2, 3) and B(4, 9). Hence write the equation of the line in the form y = mx + c.
题目：求过点 A(2, 3) 和 B(4, 9) 的直线的斜率，并写出该直线的斜截式方程 y = mx + c。
Solution: Gradient m = (y₂ − y₁) / (x₂ − x₁) = (9 − 3) / (4 − 2) = 6 / 2 = 3. Using point A(2,3): y − 3 = 3(x − 2). Simplify: y − 3 = 3x − 6 → y = 3x − 3.
解答：斜率 m = (9 − 3) / (4 − 2) = 3。代入点 A(2,3) 得 y − 3 = 3(x − 2)，化简得 y = 3x − 3。
9. Quadratic Formula and Nature of Roots | 二次求根公式与根的性质
Question: Solve 2x² + 3x − 5 = 0 using the quadratic formula. Comment on the nature of the roots.
题目：用二次求根公式解方程 2x² + 3x − 5 = 0，并说明根的性质。
Solution: Here a = 2, b = 3, c = −5. Discriminant Δ = b² − 4ac = 9 − 4(2)(−5) = 9 + 40 = 49. Since Δ > 0, there are two distinct real roots. The formula gives:
x = (−b ± √Δ) / (2a) = (−3 ± √49) / (4) = (−3 ± 7) / 4
Thus x = (−3 + 7)/4 = 1, and x = (−3 − 7)/4 = −2.5. The roots are x = 1 and x = −2.5.
解答：a = 2，b = 3，c = −5。判别式 Δ = b² − 4ac = 49 > 0，因此方程有两个不相等的实数根。求根公式得 x = (−3 ± 7) / 4，两根为 x = 1 和 x = −2.5。
10. Statistics: Median and Interquartile Range | 统计：中位数与四分位距
Question: The test scores of seven pupils are: 12, 15, 14, 10, 18, 20, 22. Find the median, lower quartile, upper quartile and interquartile range.
题目：七名学生的测试分数为：12, 15, 14, 10, 18, 20, 22。求中位数、下四分位数、上四分位数和四分位距。
Solution: First order the data: 10, 12, 14, 15, 18, 20, 22. The median is the 4th value: 15. The lower half is 10, 12, 14; its median gives Q₁ = 12. The upper half is 18, 20, 22; its median gives Q₃ = 20. Interquartile range IQR = Q₃ − Q₁ = 20 − 12 = 8.
解答：数据排序：10, 12, 14, 15, 18, 20, 22。中位数为第4个值 15。下半组为 10,12,14，中位数为下四分位数 Q₁ = 12。上半组为 18,20,22，中位数为上四分位数 Q₃ = 20。四分位距 IQR = 20 − 12 = 8。
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