📚 SQA National 5 Mathematics Unit Test Mock Exam Analysis | SQA 数学单元测试模拟卷解析
This article provides a step-by-step analysis of a mock unit test for SQA National 5 Mathematics, designed for Year 11 learners. Each section breaks down a typical exam-style question, covering essential techniques in algebra, geometry, trigonometry, statistics and more. Use these worked examples to consolidate your understanding and sharpen your problem-solving skills for the real assessment.
本文为 SQA National 5 数学单元测试模拟卷提供逐题解析，专为 Year 11 学生设计。每个小节剖析一道典型考题，涵盖代数、几何、三角学、统计等核心技巧。通过这些详细解答，你可以巩固概念理解，提高应试能力，为正式测评做好充分准备。
1. Simplifying Algebraic Expressions | 代数化简
Question: Simplify 3(2x - 4) + 5(x + 2).
题目：化简 3(2x - 4) + 5(x + 2)。
Expand the first bracket: 3 × 2x = 6x, and 3 × (-4) = -12, so we get 6x - 12.
展开第一个括号：3 × 2x = 6x，3 × (-4) = -12，得到 6x - 12。
Expand the second bracket: 5 × x = 5x, and 5 × 2 = 10, giving 5x + 10.
展开第二个括号：5 × x = 5x，5 × 2 = 10，得到 5x + 10。
Combine the expanded parts: 6x - 12 + 5x + 10.
合并展开后的部分：6x - 12 + 5x + 10。
Collect like terms: 6x + 5x = 11x; -12 + 10 = -2.
合并同类项：6x + 5x = 11x；-12 + 10 = -2。
The simplified expression is 11x - 2. Always re-check signs when expanding.
化简后的表达式为 11x - 2。展开时务必重新检查符号。
2. Solving Linear Equations with Brackets | 解含括号的线性方程
Question: Solve 4(2y - 1) = 3(y + 5) + 2.
题目：解方程 4(2y - 1) = 3(y + 5) + 2。
Expand the left side: 4 × 2y = 8y; 4 × (-1) = -4, so left becomes 8y - 4.
展开左边：4 × 2y = 8y；4 × (-1) = -4，左边变为 8y - 4。
Expand the right side: 3 × y = 3y; 3 × 5 = 15; plus the extra +2 gives 3y + 17.
展开右边：3 × y = 3y；3 × 5 = 15；再加上后面的 +2，得 3y + 17。
Set the equation: 8y - 4 = 3y + 17.
列出方程：8y - 4 = 3y + 17。
Collect y terms on one side: subtract 3y from both sides: 8y - 3y - 4 = 17 → 5y - 4 = 17.
将含 y 的项移到同侧：两边同时减去 3y：8y - 3y - 4 = 17 → 5y - 4 = 17。
Add 4 to both sides: 5y = 21. Divide by 5: y = 21/5 or 4.2.
两边同时加 4：5y = 21。除以 5：y = 21/5 或 4.2。
(2×4.2 -1) = 4(8.4 -1) = 4×7.4 = 29.6; right: 3(4.2+5)+2 = 3×9.2+2 = 27.6+2=29.6, correct.
检验：4(2×4.2 -1) = 4(8.4 -1) = 4×7.4 = 29.6；右边：3(4.2+5)+2 = 3×9.2+2 = 27.6+2=29.6，正确。
3. Factorising Quadratic Expressions | 二次表达式因式分解
Question: Factorise x² + 7x + 10.
题目：因式分解 x² + 7x + 10。
Look for two numbers that multiply to +10 (constant term) and add to +7 (coefficient of x).
寻找两个数，其积为 +10（常数项），其和为 +7（x 的系数）。
The pairs for 10: (1,10), (2,5), (-1,-10), (-2,-5). Only 2 + 5 = 7 works.
10 的因数对有：(1,10), (2,5), (-1,-10), (-2,-5)。只有 2 + 5 = 7 满足要求。
Thus the expression factorises to (x + 2)(x + 5).
因此原式分解为 (x + 2)(x + 5)。
Expand to verify: (x+2)(x+5) = x² + 5x + 2x + 10 = x² + 7x + 10.
展开验证：(x+2)(x+5) = x² + 5x + 2x + 10 = x² + 7x + 10。
or x² - x - 12: need numbers multiplying to -12, sum to -1. Use 3 and -4 → (x + 3)(x - 4).
如果是 x² - x - 12：需要积为 -12，和为 -1。用 3 和 -4 → (x + 3)(x - 4)。
4. Straight Line: Finding Equation from Two Points | 直线：由两点求方程
Question: Find the equation of the straight line passing through A(2,5) and B(6,13).
题目：求经过 A(2,5) 和 B(6,13) 的直线方程。
Calculate the gradient m: m = (y₂ - y₁) / (x₂ - x₁) = (13 - 5) / (6 - 2) = 8 / 4 = 2.
计算斜率 m：m = (y₂ - y₁) / (x₂ - x₁) = (13 - 5) / (6 - 2) = 8 / 4 = 2。
Use point A(2,5) with gradient m = 2 in y - y₁ = m(x - x₁): y - 5 = 2(x - 2).
将点 A(2,5) 和斜率 m = 2 代入点斜式 y - y₁ = m(x - x₁)：y - 5 = 2(x - 2)。
Expand and simplify: y - 5 = 2x - 4 → y = 2x + 1.
展开并整理：y - 5 = 2x - 4 → y = 2x + 1。
The equation of the line is y = 2x + 1. Check with point B: when x=6, y=2×6+1=13, correct.
直线方程为 y = 2x + 1。用 B 点检验：x=6 时 y=2×6+1=13，正确。
The y-intercept is 1 and the gradient 2 means for every 1 unit right, go up 2 units.
y 轴截距为 1，斜率为 2，表示每向右移动 1 个单位，向上移动 2 个单位。
5. Pythagoras' Theorem and Converse | 勾股定理及其逆定理
Question: A triangle has sides of lengths 9 cm, 12 cm and 15 cm. Show whether it is right-angled.
题目：三角形三边长分别为 9 cm、12 cm 和 15 cm。判断它是否为直角三角形。
Pythagoras' a right-angled triangle, hypotenuse² = leg₁² + leg₂². Here suspected hypotenuse is 15 cm.
勾股定理：直角三角形中，斜边² = 直角边₁² + 直角边₂²。怀疑斜边为 15 cm。
Calculate leg squares: 9² = 81, 12² = 144. Sum = 81 + 144 = 225.
计算直角边的平方：9² = 81，12² = 144。和 = 81 + 144 = 225。
Hypotenuse square: 15² = 225. Since the sum matches, the triangle is right-angled.
斜边的平方：15² = 225。由于等式成立，该三角形是直角三角形。
triangle is right-angled with c as the longest side. This is a 3-4-5 triangle scaled by 3.
逆定理：若 a² + b² = c²，且 c 为最长边，则三角形为直角三角形。此为 3-4-5 三角形按比例放大 3 倍。
In a problem where the hypotenuse is unknown, use c = √a² + b². For a=6, b=8, c = √36+64=√100=10.
在未知斜边的问题中，用 c = √a² + b²。若 a=6, b=8，则 c = √36+64=√100=10。
6. Trigonometry: Sine, Cosine, Tangent | 三角学：正弦、余弦、正切
Question: In a right-angled triangle, one angle is 35°, the hypotenuse is 20 cm. Find the length of the side opposite the 35° angle.
题目：一直角三角形中，一个锐角为 35°，斜边长为 20 cm。求 35° 角所对的边长。
Identify ratio: opposite and hypotenuse involve sine: sin θ = opposite / hypotenuse.
确定比值：对边与斜边使用正弦：sin θ = 对边 / 斜边。
Rearrange: opposite = hypotenuse × sin θ = 20 × sin 35°.
变形：对边 = 斜边 × sin θ = 20 × sin 35°。
g a calculator, sin 35° ≈ 0.5736. Therefore opposite ≈ 20 × 0.5736 = 11.47 cm (to 2 d.p.).
用计算器，sin 35° ≈ 0.5736。因此对边 ≈ 20 × 0.5736 = 11.47 cm（保留两位小数）。
If the adjacent were needed, use cos 35°; for the angle itself, use inverse trig after ratio is found. Always label sides relative to the angle.
若要求邻边，则用 cos 35°；若已知比值求角度，则使用反三角函数。务必相对给定的角标记各边。
7. Area of a Triangle and Composite Shapes | 三角形及组合图形的面积
Question: Find the area of a triangle with base 8.4 cm and height 5.6 cm. Then find the area of a shape made by attaching a semicircle of diameter 8.4 cm to that base.
题目：求底边 8.4 cm、高 5.6 cm 的三角形面积；再求将该底边作为一个直径 8.4 cm 的半圆的组合图形面积。
Triangle area: A = ½ × base × height = ½ × 8.4 × 5.6.
三角形面积：A = ½ × 底 × 高 = ½ × 8.4 × 5.6。
Calculate: 8.4 × 5.6 = 47.04; half is 23.52 cm².
计算：8.4 × 5.6 = 47.04；一半为 23.52 cm²。
Semicircle area: radius = 8.4 ÷ 2 = 4.2 cm. Area = ½ × π × r² = ½ × π × (4.2)².
半圆面积：半径 = 8.4 ÷ 2 = 4.2 cm。面积 = ½ × π × r² = ½ × π × (4.2)²。
us: 4.2² = 17.64. Multiply: ½ × π × 17.64 ≈ 0.5 × 3.1416 × 17.64 ≈ 27.708 cm² (to 3 d.p.).
半径平方：4.2² = 17.64。相乘：½ × π × 17.64 ≈ 0.5 × 3.1416 × 17.64 ≈ 27.708 cm²（保留三位小数）。
e area = triangle area + semicircle area ≈ 23.52 + 27.708 = 51.228 cm², so about 51.2 cm².
组合图形总面积 = 三角形面积 + 半圆面积 ≈ 23.52 + 27.708 = 51.228 cm²，约为 51.2 cm²。
8. Simultaneous Equations: Elimination Method | 联立方程：消元法
Question: Solve the system: 2x + 3y = 13, 4x - y = 5.
题目：解方程组：2x + 3y = 13，4x - y = 5。
Label equations: (1) 2x + 3y = 13; (2) 4x - y = 5.
标号方程：(1) 2x + 3y = 13；(2) 4x - y = 5。
Multiply equation (2) by 3 to align y coefficients: 12x - 3y = 15.
将方程 (2) 乘以 3 使 y 系数匹配：12x - 3y = 15。
Add this to equation (1): (2x + 3y) + (12x - 3y) = 13 + 15 → 14x = 28.
将此式与方程 (1) 相加：(2x + 3y) + (12x - 3y) = 13 + 15 → 14x = 28。
Solve for x: x = 28 / 14 = 2.
解出 x：x = 28 / 14 = 2。
Substitute x = 2 into (2): 4(2) - y = 5 → 8 - y = 5 → -y = -3 → y = 3.
将 x = 2 代入 (2)：4(2) - y = 5 → 8 - y = 5 → -y = -3 → y = 3。
Check in (1): 2(2)+3(3)=4+9=13. Solution: x=2, y=3.
代入 (1) 检验：2(2)+3(3)=4+9=13。解为 x=2, y=3。
For systems with fractions, clear denominators first. Always label and show the multiplier step clearly.
对于包含分数的方程组，先去分母。务必清晰标注并展示乘数步骤。
9. Indices and Surds Simplification | 指数与根式的化简
Question: Simplify (a) 3p²q × 4p⁵q³; (b) √50 - √18.
题目：化简 (a) 3p²q × 4p⁵q³；(b) √50 - √18。
tiply p terms: p² × p⁵ = p²⁺⁵ = p⁷. Multiply q terms: q¹ × q³ = q¹⁺³ = q⁴. Result: 12p⁷q⁴.
(a) 部分：系数相乘：3 × 4 = 12。p 项相乘：p² × p⁵ = p²⁺⁵ = p⁷。q 项相乘：q¹ × q³ = q¹⁺³ = q⁴。结果为 12p⁷q⁴。
Part (b): Simplify each surd. √50 = √25×2 = 5√2; √18 = √9×2 = 3√2.
(b) 部分：分别化简根式。√50 = √25×2 = 5√2；√18 = √9×2 = 3√2。
Then 5√2 - 3√2 = (5-3)√2 = 2√2.
所以 5√2 - 3√2 = (5-3)√2 = 2√2。
ng indices, subtract powers: aᵐ ÷ aⁿ = aᵐ⁻ⁿ. Negative indices mean reciprocal: a⁻ⁿ = 1/aⁿ.
指数相除时，指数相减：aᵐ ÷ aⁿ = aᵐ⁻ⁿ。负指数表示倒数：a⁻ⁿ = 1/aⁿ。
10. Statistics: Mean and Range from a Frequency Table | 统计：由频数表求平均数与极差
Question: The table shows the number of books read by 20 students: 0 books: 3 students, 1 book: 5, 2 books: 7, 3 books: 4, 4 books: 1. Find the mean and the range.
题目：下表为 20 名学生阅读的书籍数量：0本:3人，1本:5人，2本:7人，3本:4人，4本:1人。求平均数与极差。
→2×7=14; 3→3×4=12; 4→4×1=4. Sum fx = 0+5+14+12+4 = 35. Σf = 20. Mean = 35/20 = 1.75 books.
平均数 = (Σ fx) / Σf。扩展表格：fx 列：0→0×3=0；1→1×5=5；2→2×7=14；3→3×4=12；4→4×1=4。总和 Σ fx = 0+5+14+12+4 = 35。Σf = 20。平均数 = 35/20 = 1.75 本。
difference between maximum and minimum values: max = 4, min = 0, range = 4 - 0 = 4 books.
极差是最大值与最小值之差：max = 4, min = 0，极差 = 4 - 0 = 4 本。
When data are grouped, use midpoints. Always check the total frequency matches the sum given. Median from a frequency table requires cumulative frequency.
当数据分组时，使用组中点。务必检查总频数是否与所给人数相符。由频数表求中位数需要先计算累积频数。
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