📚 Year 8 Edexcel Mathematics: Unit Test Mock Paper Analysis | Year 8 Edexcel 数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test designed for Year 8 Edexcel Mathematics. We will go through typical questions covering number, algebra, geometry, and statistics, explaining each step and highlighting common pitfalls to help you prepare effectively.
本文详细解析一份为 Year 8 Edexcel 数学设计的单元模拟试卷。我们将逐一讲解涵盖数、代数、几何和统计的典型题目，解释每个步骤并指出常见错误，帮助你有效备考。
1. Order of Operations and Integers | 整数与运算顺序
Question: Evaluate –4² + 30 ÷ (2 – 5) × 2.
题目：计算 –4² + 30 ÷ (2 – 5) × 2.
Step 1: Always start with brackets. Here, (2 – 5) = –3. The expression becomes –4² + 30 ÷ (–3) × 2.
第1步：始终先计算括号。这里 (2 – 5) = –3。表达式变为 –4² + 30 ÷ (–3) × 2.
Step 2: Next, handle the exponent. 4² = 16, but note the leading minus applies after the exponent, so –4² = –16.
第2步：接下来处理指数。4² = 16，但注意前面的负号是在指数计算之后才作用的，因此 –4² = –16。
Step 3: Division and multiplication are at the same level, so we work from left to right. 30 ÷ (–3) = –10, then –10 × 2 = –20.
第3步：除法和乘法同级，按从左到右的顺序计算。30 ÷ (–3) = –10，然后 –10 × 2 = –20。
Step 4: Finally, add the results: –16 + (–20) = –36. The correct answer is –36.
第4步：最后做加法：–16 + (–20) = –36。正确答案是 –36。
A common mistake is to read –4² as (–4)² = 16, forgetting that the exponent only applies to the 4 unless brackets are used.
常见错误是把 –4² 当作 (–4)² = 16 来计算，忘记了除非使用括号，否则指数仅作用于数字4。
2. Fractions and Mixed Numbers | 分数与带分数
Question: Evaluate 2½ ÷ (1⅓ – ¼). Give your answer as a mixed number in its simplest form.
题目：计算 2½ ÷ (1⅓ – ¼)，并将答案写成最简带分数形式。
Step 1: Convert mixed numbers to improper fractions. 2½ = 5/2, 1⅓ = 4/3, and ¼ = 1/4.
第1步：将带分数化为假分数。2½ = 5/2，1⅓ = 4/3，¼ = 1/4。
Step 2: Solve inside the bracket: 4/3 – 1/4. The common denominator is 12: 4/3 = 16/12, 1/4 = 3/12. So 16/12 – 3/12 = 13/12.
第2步：计算括号内的式子：4/3 – 1/4。公分母为12：4/3 = 16/12，1/4 = 3/12。因此 16/12 – 3/12 = 13/12。
Step 3: Now divide: 5/2 ÷ 13/12. Change division to multiplication by flipping the second fraction: 5/2 × 12/13.
第3步：现在做除法：5/2 ÷ 13/12。将除法变为乘法，并取第二个分数的倒数：5/2 × 12/13。
Step 4: Multiply numerators and denominators: (5 × 12) / (2 × 13) = 60 / 26.
第4步：分子乘分子，分母乘分母：(5 × 12) / (2 × 13) = 60 / 26。
Step 5: Simplify: divide top and bottom by 2 to get 30/13. Convert to a mixed number: 30 ÷ 13 = 2 remainder 4, so 2 4/13.
第5步：化简：分子分母同时除以2得到 30/13。化为带分数：30 ÷ 13 = 2 余 4，因此答案是 2 4/13。
3. Decimals, Percentages and Conversions | 小数、百分数及其互化
Question: A jacket costs £40 before a 15% discount. After the discount, an additional 8% sales tax is applied to the reduced price. What is the final price?
题目：一件夹克原价 £40，先享受15%的折扣。折扣后，按折后价格加收8%的消费税。最终价格是多少？
Step 1: Find the discount amount: 15% of £40 = 0.15 × 40 = £6. The price after discount is £40 – £6 = £34.
第1步：计算折扣金额：£40 的 15% = 0.15 × 40 = £6。折扣后价格为 £40 – £6 = £34。
Step 2: Work out the tax: 8% of £34 = 0.08 × 34 = £2.72.
第2步：计算税额：£34 的 8% = 0.08 × 34 = £2.72。
Step 3: Add the tax to the discounted price: £34 + £2.72 = £36.72. The final price is £36.72.
第3步：将税额加到折后价上：£34 + £2.72 = £36.72。最终价格为 £36.72。
This type of multi-step percentage problem is common in Edexcel exams. Remember to apply the second percentage to the new amount, not the original.
这类多步骤百分数问题在 Edexcel 考试中很常见。切记第二步的百分数要作用在折后价上，而非原价。
4. Algebraic Simplification | 代数式化简
Question: Simplify fully: 3(2a – b) – 2(a + 3b) + 4b.
题目：完全化简：3(2a – b) – 2(a + 3b) + 4b。
Step 1: Expand the brackets by multiplying out. 3(2a – b) = 6a – 3b. –2(a + 3b) = –2a – 6b.
第1步：展开括号。3(2a – b) = 6a – 3b。–2(a + 3b) = –2a – 6b。
Step 2: Write the whole expression: 6a – 3b – 2a – 6b + 4b.
第2步：写出整个式子：6a – 3b – 2a – 6b + 4b。
Step 3: Group like terms. For a: 6a – 2a = 4a. For b: –3b – 6b + 4b = –5b.
第3步：合并同类项。a 项：6a – 2a = 4a。b 项：–3b – 6b + 4b = –5b。
Step 4: The simplified expression is 4a – 5b.
第4步：化简后的式子为 4a – 5b。
Be extra careful with the minus sign in front of the second bracket — it changes the signs of all terms inside.
要特别注意第二个括号前的负号——它会使括号内所有项的符号发生改变。
5. Solving Linear Equations | 解一元一次方程
Question: Solve for x: 5x – 3 = 2(x + 6).
题目：解方程求 x：5x – 3 = 2(x + 6)。
Step 1: Expand the right-hand side: 2(x + 6) = 2x + 12. The equation becomes 5x – 3 = 2x + 12.
第1步：展开右边：2(x + 6) = 2x + 12。方程变为 5x – 3 = 2x + 12。
Step 2: Collect x terms on one side and numbers on the other. Subtract 2x from both sides: 5x – 2x – 3 = 12, which gives 3x – 3 = 12.
第2步：将含 x 的项移到一边，常数项移到另一边。两边同时减去 2x：5x – 2x – 3 = 12，得到 3x – 3 = 12。
Step 3: Add 3 to both sides: 3x = 15.
第3步：两边同时加 3：3x = 15。
Step 4: Divide both sides by 3: x = 5.
第4步：两边同时除以 3：x = 5。
Always check your solution by substituting back into the original equation: left = 5(5) – 3 = 22, right = 2(5+6) = 22. It works!
务必把解代入原方程检验：左边 = 5(5) – 3 = 22，右边 = 2(5+6) = 22。成立！
6. Angles on Parallel Lines | 平行线上的角度
Question: Two parallel lines are cut by a transversal. One acute angle formed is 72°. State the sizes of its corresponding angle, its alternate interior angle, and the co-interior angle on the same side of the transversal.
题目：两条平行线被一条截线所截。形成的一个锐角为 72°。请写出它的同位角、内错角以及位于截线同侧的同旁内角的大小。
Step 1: Corresponding angles are equal. So the corresponding angle is also 72°.
第1步：同位角相等。因此同位角也是 72°。
Step 2: Alternate interior angles are equal. Thus the alternate interior angle is 72°.
第2步：内错角相等。因此内错角为 72°。
Step 3: Co-interior (allied) angles add up to 180°. So the co-interior angle = 180° – 72° = 108°.
第3步：同旁内角（同侧内角）互补，和为 180°。因此同旁内角 = 180° – 72° = 108°。
These angle facts are fundamental in Edexcel geometry. Always identify which angle relationship is being asked about by looking at their positions relative to the parallel lines and transversal.
这些角度关系是 Edexcel 几何的基础。始终通过观察角度相对于平行线和截线的位置来确定题目问的是哪一种角度关系。
7. Perimeter and Area of Compound Shapes | 组合图形的周长与面积
Question: A shape is formed by a rectangle 10 cm long and 4 cm wide, with a semicircle of diameter 4 cm attached to one of the shorter sides. Calculate the total area and the perimeter. (Use π = 3.14)
题目：一个图形由一个长 10 cm、宽 4 cm 的矩形和一个直径为 4 cm 的半圆组成，半圆贴在矩形的一条短边上。计算该图形的总面积和周长。（π 取 3.14）
Area – Step 1: Area of rectangle = length × width = 10 × 4 = 40 cm².
面积 – 第1步：矩形面积 = 长 × 宽 = 10 × 4 = 40 cm²。
Area – Step 2: Radius of semicircle = diameter / 2 = 2 cm. Area of full circle = πr² = 3.14 × 2² = 3.14 × 4 = 12.56 cm². Area of semicircle = 12.56 / 2 = 6.28 cm².
面积 – 第2步：半圆的半径 = 直径 / 2 = 2 cm。完整圆的面积 = πr² = 3.14 × 2² = 3.14 × 4 = 12.56 cm²。半圆的面积 = 12.56 / 2 = 6.28 cm²。
Area – Step 3: Total area = 40 + 6.28 = 46.28 cm².
面积 – 第3步：总面积 = 40 + 6.28 = 46.28 cm²。
Perimeter – Step 1: The straight edges of the rectangle that are part of the outer perimeter: both long sides 10 cm + 10 cm, and the unoccupied short side (4 cm). The side touching the semicircle is not part of the outer edge.
周长 – 第1步：属于外部轮廓的矩形直边：两条长边 10 cm + 10 cm，以及未被半圆占据的一条短边（4 cm）。与半圆贴合的那条短边不是外边缘。
Perimeter – Step 2: The curved edge is half the circumference of a full circle: ½ × 2πr = πr = 3.14 × 2 = 6.28 cm.
周长 – 第2步：曲线边是完整圆周长的一半：½ × 2πr = πr = 3.14 × 2 = 6.28 cm。
Perimeter – Step 3: Total perimeter = 10 + 10 + 4 + 6.28 = 30.28 cm.
周长 – 第3步：总周长 = 10 + 10 + 4 + 6.28 = 30.28 cm。
8. Mean, Median, Mode and Range | 平均数、中位数、众数和极差
Question: The marks for 10 pupils in a test are: 7, 9, 5, 6, 8, 10, 9, 4, 7, 8. Work out the mean, median, mode and range.
题目：10 名学生的测验成绩为：7, 9, 5, 6, 8, 10, 9, 4, 7, 8。计算平均数、中位数、众数和极差。
Step 1: Sort the data in ascending order: 4, 5, 6, 7, 7, 8, 8, 9, 9, 10.
第1步：将数据按从小到大排序：4, 5, 6, 7, 7, 8, 8, 9, 9, 10。
Step 2 – Mean: Sum all marks: 4+5+6+7+7+8+8+9+9+10 = 73. Divide by 10: 73 ÷ 10 = 7.3.
第2步 – 平均数：求和：4+5+6+7+7+8+8+9+9+10 = 73。除以 10：73 ÷ 10 = 7.3。
Step 3 – Median: With 10 numbers, the median is the mean of the 5th and 6th values. 5th = 7, 6th = 8, so median = (7+8)/2 = 7.5.
第3步 – 中位数：共 10 个数，中位数是第 5 和第 6 个数的平均数。第5个 = 7，第6个 = 8，中位数 = (7+8)/2 = 7.5。
Step 4 – Mode: The numbers appearing most often: 7 appears twice, 8 appears twice, 9 appears twice. The data is multimodal with modes 7, 8 and 9.
第4步 – 众数：出现次数最多的数：7 出现2次，8 出现2次，9 出现2次。该组数据为多众数，众数为 7、8 和 9。
Step 5 – Range: Highest – lowest = 10 – 4 = 6.
第5步 – 极差：最大值 – 最小值 = 10 – 4 = 6。
The following table summarises the results:
下表总结了计算结果：
| Statistic | 统计量 | Value |
| Mean | 平均数 | 7.3 |
| Median | 中位数 | 7.5 |
| Mode | 众数 | 7, 8, 9 |
| Range | 极差 | 6 |
Always remember to sort the list before finding the median, and note that a data set can have more than one mode or no mode at all.
请记住在找中位数之前先排序，并且注意一组数据可以有多个众数，也可能没有众数。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导