📚 SQA Year 9 Advanced Maths Mock Exam Walkthrough | SQA 九年级进阶数学模拟卷解析
Mock exams are one of the most effective tools for consolidating your understanding of Advanced Mathematics. In this walkthrough, we will carefully go through a full set of typical SQA Year 9 questions, showing step‑by‑step solutions and explaining the reasoning behind each method. Whether you are preparing for an end‑of‑unit test or simply reviewing key skills, this guide will help you build confidence and accuracy.
模拟考试是巩固进阶数学知识最有效的方法之一。本文将对一套典型的 SQA 九年级单元测试题进行详细解析，逐步展示解题过程，并解释每一步背后的逻辑。无论你是在准备单元测验，还是想复习关键技能，这篇解析都能帮助你提升信心与准确率。
1. Simplifying Expressions and Solving Equations | 化简表达式与解方程
Question: Simplify 2(3x − 4) − (x + 5) and then solve 4x − 7 = 2x + 9.
题目：化简 2(3x − 4) − (x + 5)，然后解方程 4x − 7 = 2x + 9。
irst bracket by 2: 2 × 3x = 6x and 2 × (−4) = −8. The expression becomes 6x − 8 − (x + 5).
首先展开表达式中的括号。用 2 乘以第一个括号内的每一项：2 × 3x = 6x，2 × (−4) = −8。表达式变为 6x − 8 − (x + 5)。
Combine like terms: 6x − x = 5x and −8 − 5 = −13. So the simplified expression is 5x − 13.
现在去掉第二个括号，注意分配负号：−(x + 5) = −x − 5。合并同类项：6x − x = 5x，−8 − 5 = −13。因此化简后的表达式为 5x − 13。
For the equation 4x − 7 = 2x + 9, start by gathering the x‑terms on one side. Subtract 2x from both sides: 4x − 2x − 7 = 9, which gives 2x − 7 = 9.
对于方程 4x − 7 = 2x + 9，先将含 x 的项移到一边。两边同时减 2x：4x − 2x − 7 = 9，得到 2x − 7 = 9。
sides to isolate the term with x: 2x = 16. Finally, divide both sides by 2 to find x = 8.
接着，两边同时加 7，分离含 x 的项：2x = 16。最后两边同除以 2，解得 x = 8。
2. Factorising Quadratic Expressions | 因式分解二次三项式
Question: Factorise x² − 5x + 6.
题目：对 x² − 5x + 6 进行因式分解。
3). We need both numbers to be negative since their sum is negative and product positive: (−2) and (−3) work perfectly because (−2) × (−3) = 6 and (−2) + (−3) = −5.
我们寻找两个数，它们的乘积为常数项 +6，和为 x 的系数 −5。6 的因数对有 (1,6) 和 (2,3)。因为两数之和为负且乘积为正，这两个数都必须是负数：−2 和 −3 符合条件，因为 (−2) × (−3) = 6，(−2) + (−3) = −5。
− 2)(x − 3). You can check by expanding: (x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6.
因此，因式分解的结果为 (x − 2)(x − 3)。可以通过展开来检验：(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6。
3. Straight Line Equations | 直线方程
Question: For the line 3x + 2y = 6, find the gradient and the y‑intercept.
题目：对于直线 3x + 2y = 6，求其斜率和 y 轴截距。
Rearrange the equation into the slope–intercept form y = mx + c. Subtract 3x from both sides: 2y = −3x + 6.
将方程改写为斜截式 y = mx + c。两边同时减 3x：2y = −3x + 6。
x + 3. Now the coefficient of x is the gradient m, so m = −3/2. The constant term is the y‑intercept c, so c = 3. The line crosses the y‑axis at (0, 3).
每一项除以 2：y = (−3/2)x + 3。此时 x 的系数即为斜率 m，因此 m = −3/2。常数项为 y 轴截距 c，所以 c = 3。该直线与 y 轴交于点 (0, 3)。
4. Solving Simultaneous Equations | 解联立方程组
Question: Solve the system: 2x + y = 10, x − y = 2.
题目：解方程组：2x + y = 10，x − y = 2。
We can use the elimination method because the y‑ ions together: (2x + y) + (x − y) = 10 + 2. The y terms cancel, leaving 3x = 12, so x = 4.
我们可以使用加减消元法，因为两个方程中 y 的系数互为相反数。将两个方程相加：(2x + y) + (x − y) = 10 + 2。y 项抵消，得到 3x = 12，因此 x = 4。
o y = 2. The solution is x = 4, y = 2. Check in the first equation: 2(4) + 2 = 8 + 2 = 10.
将 x = 4 代入第二个方程：4 − y = 2。两边减 4：−y = −2，因此 y = 2。方程组的解为 x = 4, y = 2。代入第一个方程检验：2(4) + 2 = 8 + 2 = 10，结果正确。
5. Solving Inequalities | 解不等式
Question: Solve 5 − 3x ≤ 11 and represent the solution on a number line.
题目：解不等式 5 − 3x ≤ 11，并把解表示在数轴上。
Begin by subtracting 5 from both sides: −3x ≤ 11 − 5, so −3x ≤ 6.
首先两边同时减 5：−3x ≤ 11 − 5，得到 −3x ≤ 6。
iding by a negative number reverses the inequality sign: x ≥ 6 ÷ (−3), which gives x ≥ −2.
现在两边同时除以 −3。注意除以负数时不等号方向要改变：x ≥ 6 ÷ (−3)，得到 x ≥ −2。
On a number line, you would draw a closed circle at −2 and shade the line to the right, indicating all values greater than or equal to −2.
在数轴上，在 −2 处画实心圆点，并向右画阴影线，表示所有大于或等于 −2 的数值。
6. Pythagoras’ Theorem | 勾股定理
Question: A right‑angled triangle has legs of 5 cm and 12 cm. Calculate the length of the hypotenuse.
题目：一直角三角形的两条直角边分别为 5 cm 和 12 cm。计算斜边的长度。
se c satisfies c² = a² + b². Substitute the known lengths: c² = 5² + 12² = 25 + 144 = 169.
根据勾股定理，斜边 c 满足 c² = a² + b²。代入已知边长：c² = 5² + 12² = 25 + 144 = 169。
e the positive square root because length is positive: c = √169 = 13. The hypotenuse is 13 cm.
由于长度为正值，取正平方根：c = √169 = 13。斜边长为 13 cm。
7. Basic Trigonometry – Finding an Angle | 基础三角函数——求角
Question: In a right‑angled triangle, the side opposite angle θ is 7 cm and the hypotenuse is 25 cm. Find the size of θ to one decimal place.
题目：在一个直角三角形中，角 θ 的对边长为 7 cm，斜边长为 25 cm。求 θ 的大小，精确到一位小数。
The ratio opposite/hypotenuse corresponds to the sine function: sin θ = opposite / hypotenuse = 7/25.
对边与斜边的比值对应正弦函数：sin θ = 对边 / 斜边 = 7/25。
in⁻¹(7/25). Using a calculator, sin⁻¹(0.28) ≈ 16.26°, so θ ≈ 16.3° (to one decimal place).
要求 θ，需使用反正弦函数：θ = sin⁻¹(7/25)。用计算器计算，sin⁻¹(0.28) ≈ 16.26°，因此 θ ≈ 16.3°（精确到一位小数）。
8. Area of Composite Shapes | 复合图形的面积
Question: A shape is made from a semi‑circle of radius 4 cm sitting on top of a rectangle 8 cm by 4 cm. Calculate the total area. Give your answer in terms of π.
题目：一个图形由一个半径为 4 cm 的半圆放置在一个长 8 cm、宽 4 cm 的长方形上组成。计算总面积，答案保留 π。
Area of the rectangle = length × width = 8 × 4 = 32 cm².
长方形的面积 = 长 × 宽 = 8 × 4 = 32 cm²。
Area of a full circle with radius 4 cm would be π × 4² = 16π. The semi‑circle is half of that, so its area is 8π cm².
半径为 4 cm 的完整圆的面积是 π × 4² = 16π。半圆的面积是它的一半，即 8π cm²。
Total area = area of rectangle + area of semi‑circle = 32 + 8π. Expressing as 8(4 + π) cm² is also acceptable.
总面积 = 长方形面积 + 半圆面积 = 32 + 8π。写成 8(4 + π) cm² 也可以。
9. Statistics – Averages and Box Plots | 统计——平均数与箱线图
Question: For the data set 5, 7, 7, 8, 10, 12, calculate the mean, median and mode. Then determine the five‑number summary needed for a box plot.
题目：对于数据集 5, 7, 7, 8, 10, 12，计算平均数、中位数和众数。然后确定绘制箱线图所需的五数概括。
Mean: Add all values (5+7+7+8+10+12 = 49) and divide by 6. Mean = 49 ÷ 6 ≈ 8.17 (to 2 d.p.).
平均数：将所有数值相加（5+7+7+8+10+12 = 49）再除以 6。平均数 = 49 ÷ 6 ≈ 8.17（保留两位小数）。
dy in order. With six values, the median is the average of the 3rd and 4th: (7+8)/2 = 7.5.
中位数：数据已按顺序排列。有六个数，中位数是第 3 和第 4 个数的平均值：(7+8)/2 = 7.5。
Mode: The value that appears most often is 7 (it occurs twice).
众数：出现次数最多的数值是 7（出现了两次）。
For the box plot, find the five‑ edian of upper half {8,10,12} = 10, Maximum = 12. Interquartile range (IQR) = Q₃ − Q₁ = 3.
绘制箱线图需要五数概括：最小值 = 5，下四分位数 Q₁ = 下半部分 {5,7,7} 的中位数 = 7，中位数 Q₂ = 7.5，上四分位数 Q₃ = 上半部分 {8,10,12} 的中位数 = 10，最大值 = 12。四分位距 (IQR) = Q₃ − Q₁ = 3。
10. Probability – Without Replacement | 概率——不放回抽取
Question: A bag contains 3 red and 2 blue marbles. Two marbles are drawn at random without replacement. Find the probability that both marbles are red.
题目：一个袋子里有 3 个红球和 2 个蓝球。随机不放回地抽取两个球。求两个球都是红色的概率。
Total marbles initially = 5. Probability that the first marble is red = 3/5.
初始时总球数为 5。第一个球是红色的概率 = 3/5。
eft, so total 4. Probability the second marble is red given the first was red = 2/4 = 1/2.
抽出一个红球后，剩下 2 个红球和 2 个蓝球，总数为 4。在第一个球是红色的条件下，第二个球是红色的概率 = 2/4 = 1/2。
Multiply along the branch: P(both red) = (3/5) × (1/2) = 3/10.
沿着树状图分支相乘：P(两个都是红球) = (3/5) × (1/2) = 3/10。
d from 3 is ³C₂ = 3; total ways to choose any 2 from 5 is ⁵C₂ = 10. So probability = 3/10.
也可以用组合来解：从 3 个红球中选 2 个的方法数为 ³C₂ = 3；从 5 个球中任选 2 个的方法数为 ⁵C₂ = 10。因此概率为 3/10。
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