📚 Year 7 SQA Advanced Mathematics: Unit Test Mock Paper Walkthrough | Year 7 SQA 进阶数学：单元测试模拟卷解析
Welcome to this detailed walkthrough of a mock unit test designed for Year 7 pupils following the SQA Advanced Mathematics pathway in Scotland. The problems here reflect the breadth of skills expected at this level: algebraic manipulation, proportional reasoning, geometry, statistics, and basic Pythagoras or trigonometry. Each question is first presented in full, then broken down step by step in both English and Chinese, so you can build confidence and accuracy whether you are revising in Glasgow or Guangzhou.
欢迎阅读这份为苏格兰 SQA 进阶数学课程 Year 7 学生设计的单元测试模拟卷详细解析。此卷涵盖了该阶段需要掌握的核心技能：代数运算、比例推理、几何、统计以及基础的毕达哥拉斯定理或三角函数。每道题先给出完整题干，再以中英双语逐步讲解，无论你是在格拉斯哥还是广州复习，都能从中建立信心并提升解题准确率。
1. Question 1: Simplifying Algebraic Expressions | 第1题：化简代数表达式
Question: Simplify 3a + 5b – 2a + 7b – 4b.
题目：化简 3a + 5b – 2a + 7b – 4b。
Collect like terms: the ‘a’ terms are 3a and –2a, giving a. The ‘b’ terms are 5b + 7b – 4b, which is 8b. So the simplified expression is a + 8b.
合并同类项：含 a 的项是 3a 和 –2a，得到 a。含 b 的项是 5b + 7b – 4b，得到 8b。因此化简结果为 a + 8b。
2. Question 2: Solving a Linear Equation | 第2题：解一元一次方程
Question: Solve 4x – 7 = 13.
题目：解方程 4x – 7 = 13。
Add 7 to both sides: 4x = 20. Divide both sides by 4: x = 5. Check: 4(5) – 7 = 20 – 7 = 13, so the solution is correct.
两边同时加 7：4x = 20。两边同时除以 4：x = 5。检验：4×5 – 7 = 20 – 7 = 13，所以解正确。
3. Question 3: Ratio and Proportion | 第3题：比与比例
Question: The ratio of boys to girls in a Year 7 class is 3 : 5. If there are 24 boys, how many girls are there?
题目：一个 Year 7 班级中男生与女生的比例是 3 : 5。如果男生有 24 人，女生有多少人？
The ratio 3 : 5 means for every 3 boys there are 5 girls. The number of boys is 24, which is 3 parts scaled up by a factor of 8 (since 3 × 8 = 24). Multiply the girls’ part by the same factor: 5 × 8 = 40 girls. Alternatively, set up the proportion 3/5 = 24/g, giving g = (24 × 5) ÷ 3 = 40.
比例 3 : 5 表示每 3 名男生对应 5 名女生。男生人数为 24，即 3 份乘以 8（因为 3 × 8 = 24）。将女生的份数乘以相同因子：5 × 8 = 40 名女生。或者用比例式 3/5 = 24/g，解得 g = (24 × 5) ÷ 3 = 40。
4. Question 4: Area of a Compound Shape | 第4题：组合图形的面积
Question: The diagram shows an L-shaped figure formed by two rectangles. One rectangle is 10 cm by 4 cm, attached to another rectangle 6 cm by 3 cm along the shorter side. Calculate the total area.
题目：下图是一个由两个矩形组成的 L 形图形。一个矩形长 10 厘米、宽 4 厘米，与另一个长 6 厘米、宽 3 厘米的矩形沿短边拼接。计算总面积。
Area of larger rectangle = 10 × 4 = 40 cm². Area of smaller rectangle = 6 × 3 = 18 cm². Since they do not overlap, total area = 40 + 18 = 58 cm².
大矩形面积 = 10 × 4 = 40 平方厘米。小矩形面积 = 6 × 3 = 18 平方厘米。因为两个矩形不重叠，总面积 = 40 + 18 = 58 平方厘米。
5. Question 5: Calculating the Mean | 第5题：计算平均数
Question: The test scores of five pupils are: 7, 9, 6, 10, 8. Work out the mean score.
题目：五位学生的测验成绩为：7、9、6、10、8。计算平均分。
Sum the scores: 7 + 9 + 6 + 10 + 8 = 40. Number of scores = 5. Mean = 40 ÷ 5 = 8.
将成绩相加：7 + 9 + 6 + 10 + 8 = 40。成绩个数为 5。平均数 = 40 ÷ 5 = 8。
6. Question 6: Pythagoras’ Theorem | 第6题：毕达哥拉斯定理
Question: A right-angled triangle has legs of length 5 cm and 12 cm. Find the length of the hypotenuse.
题目：一个直角三角形的两条直角边长分别为 5 厘米和 12 厘米。求斜边长。
Use Pythagoras’ theorem: c² = a² + b². So c² = 5² + 12² = 25 + 144 = 169. Taking the square root, c = √169 = 13 cm. The hypotenuse is 13 cm.
使用毕达哥拉斯定理：c² = a² + b²。因此 c² = 5² + 12² = 25 + 144 = 169。开平方，c = √169 = 13 厘米。斜边长为 13 厘米。
7. Question 7: Percentage Increase | 第7题：百分数增加
Question: A jacket originally costs £40. In a sale, the price is increased by 15%. What is the new price?
题目：一件夹克原价 40 英镑。打折季中，价格反而上涨了 15%。新价格是多少？
Calculate 15% of £40: 0.15 × 40 = £6. Add the increase to the original: £40 + £6 = £46. So the new price is £46.
计算 40 英镑的 15%：0.15 × 40 = 6 英镑。将上涨额加回原价：40 + 6 = 46 英镑。新价格为 46 英镑。
8. Question 8: Probability with Fractions | 第8题：分数概率
Question: A bag contains 3 red balls, 5 blue balls and 2 green balls. One ball is taken at random. What is the probability that the ball is not blue?
题目：一个袋子里有 3 个红球、5 个蓝球和 2 个绿球。随机取出一个球，这个球不是蓝色的概率是多少？
Total balls = 3 + 5 + 2 = 10. Number of balls that are not blue = red + green = 3 + 2 = 5. Probability = 5/10, which simplifies to 1/2.
球的总数 = 3 + 5 + 2 = 10。不是蓝色的球数 = 红球 + 绿球 = 3 + 2 = 5。概率 = 5/10，约分为 1/2。
9. Question 9: Coordinates and Midpoint | 第9题：坐标与中点
Question: Points A and B have coordinates (2, 7) and (8, 3). Find the coordinates of the midpoint of AB.
题目：点 A 的坐标为 (2, 7)，点 B 的坐标为 (8, 3)。求线段 AB 的中点坐标。
The midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2). Substitute: ((2 + 8)/2, (7 + 3)/2) = (10/2, 10/2) = (5, 5).
中点公式：((x₁ + x₂)/2, (y₁ + y₂)/2)。代入数值：((2 + 8)/2, (7 + 3)/2) = (10/2, 10/2) = (5, 5)。
10. Question 10: Angles in a Triangle | 第10题：三角形的内角
Question: Two angles of a triangle are 37° and 84°. Calculate the size of the third angle.
题目：三角形的两个内角分别是 37° 和 84°。计算第三个角的大小。
The sum of angles in a triangle is 180°. So the third angle = 180° – (37° + 84°) = 180° – 121° = 59°.
三角形内角和为 180°。因此第三个角 = 180° – (37° + 84°) = 180° – 121° = 59°。
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