📚 Year 8 SQA Advanced Mathematics: Mock Unit Test Walkthrough | Year 8 SQA 进阶数学：单元测试模拟卷解析
This mock unit test walkthrough is designed for Year 8 (S2) pupils following the SQA Curriculum for Excellence in Advanced Mathematics. It covers linear equations, angle properties with parallel lines, perimeter and area of composite shapes, ratio and probability, grouped frequency statistics, and speed problems. Each question is modelled on typical Scottish assessment items, helping you build confidence for class tests or end-of-unit checks.
本模拟卷解析针对遵循 SQA“卓越课程”进阶数学的 Year 8（S2）学生编写。题目涵盖一元一次方程、平行线角度性质、组合图形的周长与面积、比例与概率、分组频数统计以及速度问题。每道题都参考了苏格兰典型的测评风格，旨在帮助你巩固知识，从容应对课堂测验或单元考试。
1. Question 1: Solving Linear Equations with Brackets | 题目1：含括号的一元一次方程
The first problem asks you to solve the equation 3(2x - 1) - 4(x + 3) = 10.
第一题要求解方程 3(2x - 1) - 4(x + 3) = 10。
- 1) = 6x - 3. Next, multiply -4 by each term in the second bracket: -4(x + 3) = -4x - 12.
首先展开两组括号。将 3 与第一个括号内的每一项相乘：3(2x - 1) = 6x - 3。接着将 -4 与第二个括号内的每一项相乘：-4(x + 3) = -4x - 12。
Rewrite the left-hand side combining all terms: 6x - 3 - 4x - 12. Group the x‑terms and the constants separately: (6x - 4x) + (-3 - 12) = 2x - 15.
将左边重写为 6x - 3 - 4x - 12。把含 x 的项和常数项分别合并：(6x - 4x) + (-3 - 12) = 2x - 15。
e equation becomes 2x - 15 = 10. Add 15 to both sides to isolate the term with x: 2x = 25.
此时方程变为 2x - 15 = 10。两边同时加 15，使含 x 的项单独在一边：2x = 25。
both sides by 2 to find x = 12.5. You may also express the answer as a fraction: x = 25/2.
最后，两边除以 2 得到 x = 12.5。你也可以将答案写成分数形式：x = 25/2。
2.5 + 3) = 3(25 - 1) - 4(15.5) = 3 × 24 - 62 = 72 - 62 = 10, which matches the right side.
务必把解代回原方程进行检验：左边 3(2 × 12.5 - 1) - 4(12.5 + 3) = 3(25 - 1) - 4(15.5) = 3 × 24 - 62 = 72 - 62 = 10，与右边相等。
Watch out for sign errors when expanding brackets preceded by a minus sign. A common mistake is to write -4(x + 3) as -4x + 12; remember that -4 multiplied by +3 gives -12.
展开括号时要留意符号，尤其是括号前带负号的情况。常见错误是把 -4(x + 3) 误写为 -4x + 12；请记住，-4 乘以 +3 得到 -12。
In an exam, it is a good idea to write each expansion step clearly and avoid doing too much mental arithmetic at once.
在考试中，最好将每一步展开的过程清晰写出，尽量避免过多心算。
2. Question 2: Angles with Parallel Lines | 题目2：平行线中的角度
E. You are given that angle ABE = 40° and angle CDE = 110°. The task is to find angle BED.
本题中，直线 AB 与 CD 平行。一条截线穿过它们并交于点 E。已知角 ABE = 40°，角 CDE = 110°。要求求出角 BED。
l to AB, angle BEF is equal to angle ABE as alternate interior angles, so angle BEF = 40°.
解题时，过点 E 作一条辅助线 EF，使其平行于 AB 与 CD。由于 EF ∥ AB，根据内错角相等，角 BEF = 角 ABE = 40°。
on the same side of the transversal. Their sum is 180°, so angle FED = 180° - 110° = 70°.
再考虑平行线 CD 与 EF。角 CDE 和角 FED 是同旁内角，它们的和为 180°，因此角 FED = 180° - 110° = 70°。
Angle BED is the sum of the two angles BEF and FED: 40° + 70° = 110°.
角 BED 是角 BEF 与角 FED 之和：40° + 70° = 110°。
Always label your auxiliary line clearly on the diagram. This helps the examiner follow your reasoning. Do not assume that BED is simply the difference between the two given angles; drawing the auxiliary line reveals the correct structure.
务必将辅助线在图上标注清楚。这有助于阅卷人理解你的思路。不要想当然地认为角 BED 就是所给两角的和或差；画出辅助线才能揭示正确的角度关系。
3. Question 3: Perimeter and Area of Composite Shapes | 题目3：组合图形的周长与面积
The composite shape is made up of a rectangle and a semicircle. The rectangle has length 12 cm and width 6 cm. The semicircle has a diameter equal to the width, meaning its radius is 3 cm. The semicircle is attached to one of the longer sides, so the diameter covers the full 6 cm of that side.
组合图形由一个矩形和一个半圆构成。矩形的长和宽分别为 12 cm 与 6 cm。半圆的直径等于宽，即半径为 3 cm。半圆附着在矩形的一条长边上，其直径恰好覆盖该边 6 cm 的长度。
. The top length is replaced by the semicircular arc. Thus the perimeter = 6 + 12 + 6 + (π × 3).
求周长时，注意哪些边是外部的。外边界包括两条垂直的宽边（各 6 cm）和底部长边（12 cm）。顶部长边被半圆弧取代。因此周长 = 6 + 12 + 6 + (π × 3)。
The semicircle' , so the total perimeter is about 30 + 9.42 = 39.42 cm. In exact form, it is (30 + 3π) cm.
半圆弧长是整圆周长的一半：(1/2) × 2πr = π × 3。取 π ≈ 3.14，弧长约为 9.42 cm，因此周长约为 30 + 9.42 = 39.42 cm。精确形式为 (30 + 3π) cm。
For the area, sum the rectangle's area (12 × 6 = 72 cm²) and the semicircle's area (½ × π × 3² = 4.5π cm²). Expressed fully, the total area = (72 + 4.5π) cm².
求面积时，将矩形面积（12 × 6 = 72 cm²）与半圆面积（½ × π × 3² = 4.5π cm²）相加。总面积 = (72 + 4.5π) cm²。
A frequent error is to include the shared side in the perimeter or to use the diameter instead of the radius. Always check which edges are actually exposed on the outside of the compound figure.
常见错误是将重合边也计入周长，或用直径代替半径计算。一定要检查哪些边是组合图形的实际外部轮廓。
4. Question 4: Ratio and Probability | 题目4：比例与概率
A bag contains red, blue and green marbles in the ratio 3 : 4 : 5. There are 36 blue marbles. First, determine the total number of marbles, then find the probability that a marble drawn at random is not green.
一个袋子里装有红、蓝、绿三色弹珠，数量比为 3 : 4 : 5。已知蓝色弹珠有 36 颗。首先求出弹珠总数，然后计算随机抽取一颗弹珠不是绿色的概率。
green. The blue share corresponds to 4 parts. If 4 parts = 36, then one part = 36 ÷ 4 = 9.
比例 3 : 4 : 5 表示每 3 颗红弹珠对应 4 颗蓝弹珠和 5 颗绿弹珠。蓝色部分对应 4 份。如果 4 份 = 36，那么每份为 36 ÷ 4 = 9。
es = 3 × 9 = 27. Number of green marbles = 5 × 9 = 45. Total marbles = 27 + 36 + 45 = 108.
红弹珠数量 = 3 × 9 = 27。绿弹珠数量 = 5 × 9 = 45。弹珠总数 = 27 + 36 + 45 = 108。
63 / 108. Simplify the fraction by dividing numerator and denominator by 9: 63/108 = 7/12.
抽到的不是绿弹珠的概率 = (红珠数 + 蓝珠数) / 总数 = (27 + 36) / 108 = 63 / 108。约简分数：分子分母同除以 9，得到 7/12。
en) = 45/108 = 5/12, so P(not green) = 1 - 5/12 = 7/12. Both methods give the same answer.
也可以先求抽到绿弹珠的概率，再用 1 减去它。P(绿) = 45/108 = 5/12，因此 P(非绿) = 1 - 5/12 = 7/12。两种方法结果一致。
When working with ratios, always identify the value of one part before calculating the other quantities. Keep probabilities as simplified fractions unless the question specifies otherwise.
处理比例问题时，一定要先求出一份的量，再计算其他数量。除非题目另有要求，概率应保留为最简分数。
5. Question 5: Statistics – Frequency Polygon and Mean | 题目5：统计——频数多边形与平均数
The table below shows the distances run by a group of 28 students. Use it to draw a frequency polygon and estimate the mean distance.
下表列出 28 名学生跑步的距离。请根据表格绘制频数多边形并估算平均距离。
| Distance (km) | Frequency |
|---|---|
| 0 – 2 | 5 |
| 2 – 4 | 12 |
| 4 – 6 | 8 |
| 6 – 8 | 3 |
To plot a frequency polygon, use the midpoint of each class interval. The midpoints are 1, 3, 5 and 7. Plot these against the corresponding frequencies and join the points with straight lines. Remember to anchor the polygon at the midpoints of the next empty intervals (e.g., coordinate -1,0 and 9,0) to close the shape.
绘制频数多边形时，使用每个组区间的组中值。组中值分别为 1、3、5 和 7。将它们在坐标系中标出，与对应频数配对，并用直线连接各点。注意要将多边形两端连接到下一个空组的中值（例如坐标 (-1,0) 和 (9,0)），使图形封闭。
each midpoint by its frequency and find the total: (1 × 5) + (3 × 12) + (5 × 8) + (7 × 3).
估算平均数，将每个组中值乘以对应频数并求和：(1 × 5) + (3 × 12) + (5 × 8) + (7 × 3)。
Sum of these products = 5 + 36 + 40 + 21 = 102. Total frequency Σf = 5 + 12 + 8 + 3 = 28.
逐步计算：1×5 = 5，3×12 = 36，5×8 = 40，7×3 = 21。乘积之和 = 5 + 36 + 40 + 21 = 102。总频数 Σf = 5 + 12 + 8 + 3 = 28。
Estimated mean = Σ(f × midpoint) / Σf = 102 / 28 ≈ 3.64 km (to 2 decimal places).
估算平均数 = Σ(f × 组中值) / Σf = 102 / 28 ≈ 3.64 km（保留两位小数）。
Mean = Σfx / Σf
平均数 = Σfx / Σf
Do not use class boundaries when plotting the polygon; always rely on midpoints. For the mean, accept that it is an estimate because we assume all data within an interval lies at the midpoint.
绘图时切勿使用组边界，必须使用组中值。这里的平均数是估计值，因为我们假设各组数据都集中在中点。
6. Question 6: Speed and Average Speed | 题目6：速度与平均速度
A car travels at a constant speed of 60 km/h for 2.5 hours, and then at 80 km/h for the next 1.5 hours. Find the total distance travelled and the average speed for the whole journey.
一辆汽车先以 60 km/h 的恒速行驶 2.5 小时，接着以 80 km/h 行驶 1.5 小时。求总路程以及全程的平均速度。
ance = 60 km/h × 2.5 h = 150 km. For the second part, distance = 80 km/h × 1.5 h = 120 km.
运用公式 路程 = 速度 × 时间。第一段路程 = 60 km/h × 2.5 h = 150 km。第二段路程 = 80 km/h × 1.5 h = 120 km。
Total distance = 150 + 120 = 270 km. Total time = 2.5 + 1.5 = 4 hours.
总路程 = 150 + 120 = 270 km。总时间 = 2.5 + 1.5 = 4 小时。
Average speed = total distance ÷ total time = 270 ÷ 4 = 67.5 km/h.
平均速度 = 总路程 ÷ 总时间 = 270 ÷ 4 = 67.5 km/h。
his is incorrect because the car spent more time at 60 km/h than at 80 km/h, pulling the weighted average below 70 km/h.
常见错误是直接对两个速度求算术平均：(60 + 80) ÷ 2 = 70 km/h。这种算法是错误的，因为汽车以 60 km/h 行驶的时间更长，加权平均会被拉低到 70 km/h 以下。
Always treat average speed
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