📚 Year 10 SQA Advanced Mathematics: Unit Test Mock Paper Analysis | Year 10 SQA 进阶数学：单元测试模拟卷解析
This article walks you through a typical Year 10 SQA Advanced Mathematics unit test, dissecting each question type with clear, step‑by‑step solutions. We will cover algebra, quadratic equations, coordinate geometry, trigonometry, similarity, and statistics. Use this analysis to deepen your understanding and boost your exam confidence.
本文为你详细解析一套典型的 Year 10 SQA 进阶数学单元测试卷，逐步拆解每类题型并给出清晰解答。内容涵盖代数、二次方程、坐标几何、三角学、相似形和统计。通过这篇解析，你能够加深理解、提升应试信心。
1. Simplifying Algebraic Fractions | 化简代数分式
Mock question: Simplify fully (3x² - 12) / (6x + 12).
模拟题：完全化简 (3x² - 12) / (6x + 12)。
Factorise the numerator: 3x² - 12 = 3(x² - 4) = 3(x - 2)(x + 2).
分子因式分解：3x² - 12 = 3(x² - 4) = 3(x - 2)(x + 2)。
Factorise the denominator: 6x + 12 = 6(x + 2).
分母因式分解：6x + 12 = 6(x + 2)。
The common factor (x + 2) cancels, leaving 3(x - 2) / 6, which simplifies to (x - 2) / 2.
公因子 (x + 2) 约去，剩下 3(x - 2) / 6，进一步化简为 (x - 2) / 2。
Always state restrictions: x ≠ -2, because the denominator would be zero.
必须注明限制条件：x ≠ -2，否则分母为零。
2. Completing the Square for a Quadratic | 二次式的配方法
Mock question: Express x² + 6x - 5 in the form (x + p)² + q, and state the coordinates of the turning point.
模拟题：将 x² + 6x - 5 写成 (x + p)² + q 的形式，并写出转折点坐标。
ficient of x, which is 6. Half of it is 3, and 3² = 9. So we write x² + 6x = (x + 3)² - 9.
x 的系数是 6。它的一半是 3，3² = 9。因此写为 x² + 6x = (x + 3)² - 9。
Now incorporate the constant term: x² + 6x - 5 = (x + 3)² - 9 - 5 = (x + 3)² - 14.
现在并入常数项：x² + 6x - 5 = (x + 3)² - 9 - 5 = (x + 3)² - 14。
Thus p = 3 and q = -14. The turning point (minimum) is at (-3, -14).
因此 p = 3，q = -14。转折点（最小值）在 (-3, -14)。
3. Solving a Quadratic Equation by Factorising | 因式分解法解二次方程
Mock question: Solve x² - 7x + 12 = 0.
模拟题：解方程 x² - 7x + 12 = 0。
We need two numbers that multiply to +12 and add to -7. They are -3 and -4.
需要两个数，乘积为 +12，和为 -7。它们是 -3 和 -4。
Therefore the factorised form is (x - 3)(x - 4) = 0.
因此因式分解形式为 (x - 3)(x - 4) = 0。
Set each bracket to zero: x - 3 = 0 ⇒ x = 3; x - 4 = 0 ⇒ x = 4.
令每个括号为零：x - 3 = 0 ⇒ x = 3；x - 4 = 0 ⇒ x = 4。
4. Using the Quadratic Formula | 使用求根公式
Mock question: Solve 2x² - 5x - 3 = 0, giving your answers correct to two decimal places where necessary.
模拟题：解 2x² - 5x - 3 = 0，如有需要，答案保留两位小数。
Here a = 2, b = -5, c = -3. The quadratic formula: x = [-b ± √b² - 4ac] / (2a).
这里 a = 2，b = -5，c = -3。求根公式：x = [-b ± √b² - 4ac] / (2a)。
Calculate the discriminant: b² - 4ac = (-5)² - 4(2)(-3) = 25 + 24 = 49.
计算判别式：b² - 4ac = (-5)² - 4×2×(-3) = 25 + 24 = 49。
Then x = [5 ± √49] / 4 = [5 ± 7] / 4. The two solutions are x = (5+7)/4 = 3 and x = (5-7)/4 = -0.5.
于是 x = [5 ± √49] / 4 = [5 ± 7] / 4。两个解为 x = (5+7)/4 = 3 和 x = (5-7)/4 = -0.5。
5. Determining the Equation of a Straight Line | 确定直线方程
Mock question: ne passes through the points (2, 5) and (6, 17). Find its equation in the form y = mx + c.
模拟题：一条直线经过点 (2, 5) 和 (6, 17)。求其方程，写成 y = mx + c 形式。
Gradient m = (y₂ - y₁) / (x₂ - x₁) = (17 - 5)/(6 - 2) = 12/4 = 3.
斜率 m = (y₂ - y₁) / (x₂ - x₁) = (17 - 5)/(6 - 2) = 12/4 = 3。
Using point (2, 5) with y = mx + c: 5 = 3(2) + c ⇒ 5 = 6 + c ⇒ c = -1.
将点 (2, 5) 代入 y = mx + c：5 = 3×2 + c ⇒ 5 = 6 + c ⇒ c = -1。
The equation is y = 3x - 1.
方程为 y = 3x - 1。
6. Applying Trigonometric Ratios in Right‑Angled Triangles | 直角三角形中的三角比应用
Mock question: In right‑ , angle B = 90°, AB = 8 cm, and angle C = 35°. Calculate the length of AC, the hypotenuse.
模拟题：在直角三角形 ABC 中，角 B = 90°，AB = 8 cm，角 C = 35°。计算斜边 AC 的长度。
ve to angle C, side AB is the opposite. Use the sine ratio: sin C = opposite / hypotenuse.
相对于角 C，边 AB 是对边。使用正弦比：sin C = 对边 / 斜边。
sin 35° = 8 / AC ⇒ AC = 8 / sin 35° ≈ 8 / 0.573576 ≈ 13.95 cm (to 2 d.p.).
sin 35° = 8 / AC ⇒ AC = 8 / sin 35° ≈ 8 / 0.573576 ≈ 13.95 cm（保留两位小数）。
7. Using the Cosine Rule to Find an Angle | 用余弦定理求角度
Mock question: In triangle PQR, PQ = 7 m, QR = 9 m, and PR = 12 m. Calculate angle PQR.
模拟题：在三角形 PQR 中，PQ = 7 m，QR = 9 m，PR = 12 m。计算角 PQR。
The cosine rule states: cos PQR = (PQ² + QR² - PR²) / (2 × PQ × QR).
余弦定理：cos PQR = (PQ² + QR² - PR²) / (2 × PQ × QR)。
stitute: cos PQR = (7² + 9² - 12²) / (2×7×9) = (49 + 81 - 144) / 126 = (-14) / 126 = -1/9.
代入：cos PQR = (7² + 9² - 12²) / (2×7×9) = (49 + 81 - 144) / 126 = (-14) / 126 = -1/9。
Angle PQR = cos⁻¹(-1/9) ≈ 96.38° (to 2 d.p.).
角 PQR = cos⁻¹(-1/9) ≈ 96.38°（保留两位小数）。
8. Area of a Triangle Using Two Sides and the Included Angle | 用两边及其夹角求三角形面积
Mock question: Find the area of triangle XYZ where XY = 15 cm, XZ = 10 cm, and angle YXZ = 54°.
模拟题：求三角形 XYZ 的面积，已知 XY = 15 cm，XZ = 10 cm，角 YXZ = 54°。
Area = ½ × XY × XZ × sin(YXZ) = ½ × 15 × 10 × sin 54°.
面积 = ½ × XY × XZ × sin(YXZ) = ½ × 15 × 10 × sin 54°。
sin 54° ≈ 0.8090, so Area ≈ 0.5 × 150 × 0.8090 = 60.675 cm² ≈ 60.7 cm² (to 1 d.p.).
sin 54° ≈ 0.8090，所以面积 ≈ 0.5 × 150 × 0.8090 = 60.675 cm² ≈ 60.7 cm²（保留一位小数）。
9. Similar Triangles and Scale Factor | 相似三角形与比例因子
Mock question: Two similar triangles have corresponding sides in the ratio 3:7. The area of the smaller triangle is 36 cm². Find the area of the larger triangle.
模拟题：两个相似三角形对应边的比为 3:7。小三角形的面积是 36 cm²。求大三角形的面积。
scale factor from small to large is k = 7/3. The area scale factor is k² = (7/3)² = 49/9.
从小到大的长度比例因子 k = 7/3。面积的比例因子是 k² = (7/3)² = 49/9。
Area of larger triangle = 36 × (49/9) = 36 ÷ 9 × 49 = 4 × 49 = 196 cm².
大三角形的面积 = 36 × (49/9) = 36 ÷ 9 × 49 = 4 × 49 = 196 cm²。
Remember: for similar figures, if lengths are multiplied by k, areas are multiplied by k², and volumes by k³.
牢记：对于相似图形，长度乘以 k，面积乘以 k²，体积乘以 k³。
10. Interpreting Cumulative Frequency Diagrams | 解读累积频数图
Mock question: A cumulative frequency graph of test scores shows the median is at 62 marks and the lower quartile is at 48 marks. Estimate the interquartile range if the upper quartile is found at 75 marks.
模拟题：某测验成绩的累积频数图显示中位数为 62 分，下四分位数为 48 分。若上四分位数为 75 分，请估计四分位距。
The interquartile range (IQR) = upper quartile - lower quartile = 75 - 48 = 27 marks.
四分位距 (IQR) = 上四分位数 - 下四分位数 = 75 - 48 = 27 分。
The median tells us half the students scored below 62, and the IQR measures the spread of the middle 50% of scores.
中位数表明一半学生成绩低于 62 分，而四分位距衡量的是中间 50% 成绩的离散程度。
A higher IQR indicates greater variability in the central performance.
较大的四分位距表示中间部分的成绩波动更大。
11. Solving Simultaneous Equations Graphically | 图解法解联立方程
Mock question: graph to solve the system: y = 2x + 1 and x + y = 7. Confirm your solution algebraically.
模拟题：用作图法解方程组：y = 2x + 1 与 x + y = 7。并用代数方法验证。
pears at (2, 5). Check: for y = 2x+1, when x=2, y=5. For x+y=7, 2+5=7. Both are satisfied.
画出两条直线。交点似乎在 (2, 5)。验证：对 y=2x+1，当 x=2 时 y=5；对 x+y=7，2+5=7。均成立。
Algebraic solution: substitute y = 2x+1 into x + (2x+1) = 7 ⇒ 3x + 1 = 7 ⇒ 3x = 6 ⇒ x = 2, then y = 5.
代数解法：将 y = 2x+1 代入 x + (2x+1) = 7 ⇒ 3x + 1 = 7 ⇒ 3x = 6 ⇒ x = 2，进而 y = 5。
12. Key Formulae Recap | 重点公式回顾
Below is a summary of essential formulae used in the mock paper. Keep them handy for revision.
以下是模拟卷中使用到的基本公式汇总，方便你复习时查阅。
| Topic / 主题 | Formula / 公式 |
|---|---|
| Quadratic formula / 求根公式 | x = [-b ± √b² - 4ac] / 2a |
| Gradient / 斜率 | m = (y₂ - y₁) / (x₂ - x₁) |
| Trigonometric area / 三角面积 | Area = ½ ab sin C |
| Cosine rule (angle) / 余弦定理（求角） | cos A = (b² + c² - a²) / 2bc |
| Area scale factor / 面积比例因子 | k² for similar shapes / 相似形用 k² |
Practice applying these formulas in varied contexts to strengthen your problem‑solving fluency.
在不同情境中反复应用这些公式，能提升你解题的熟练度和灵活性。
Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com
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