📚 Year 8 WJEC Maths Unit Test Mock Paper Walkthrough | Year 8 WJEC 数学：单元测试模拟卷解析
Welcome to this in-depth walkthrough of a Year 8 WJEC Mathematics unit test mock paper. Whether you are revising for an end-of-topic assessment or preparing for a school exam, this guide breaks down every question step by step. We cover key topics from the WJEC Year 8 curriculum including decimals, fractions, percentages, algebra, geometry, and data handling. Each solution is clearly explained so you can understand the method and avoid common mistakes.
欢迎阅读这篇 Year 8 WJEC 数学单元测试模拟卷的详细解析。无论你是在为单元测验复习，还是为校内考试做准备，这篇指南都会逐步拆解每一道题目。我们涵盖了 WJEC 八年级课程的核心内容，包括小数、分数、百分数、代数、几何和数据处理。每道题都提供了清晰的解析，帮助你理解方法，规避常见错误。
1. Mock Paper Overview | 模拟卷概览
This mock paper contains ten questions designed to reflect the style and difficulty of a typical WJEC Year 8 unit test. Topics tested include arithmetic with decimals, fractions, percentages, simplifying algebraic expressions, solving linear equations, nth term sequences, angle reasoning, area and perimeter, mean and median, and ratio. Try to work through each question on your own before reading the walkthrough.
这份模拟卷包含十道题目，旨在反映典型 WJEC 八年级单元测试的风格和难度。考查的主题包括小数运算、分数、百分数、代数式化简、解一元一次方程、找第 n 项通项公式、角度推理、面积与周长、平均数与中位数，以及比和比例。在阅读解析之前，建议你先独立尝试完成每一道题。
2. Arithmetic with Decimals | 小数运算
Question: Work out 3.7 × (2.4 + 1.6) – 5.2 ÷ 0.4. Remember the order of operations: brackets first, then multiplication and division from left to right, and finally addition and subtraction.
题目：计算 3.7 × (2.4 + 1.6) – 5.2 ÷ 0.4。请记住运算顺序：先算括号，再算乘除（从左到右），最后算加减。
Step 1: Inside the brackets, 2.4 + 1.6 = 4.0. Step 2: Multiply 3.7 × 4.0 = 14.8. Step 3: Divide 5.2 ÷ 0.4. You can multiply both numbers by 10 to get 52 ÷ 4 = 13. Step 4: Subtract: 14.8 – 13 = 1.8. The answer is 1.8.
第一步：括号内，2.4 + 1.6 = 4.0。第二步：乘法，3.7 × 4.0 = 14.8。第三步：除法，5.2 ÷ 0.4，可以将两个数同时乘以 10，变为 52 ÷ 4 = 13。第四步：减法，14.8 – 13 = 1.8。答案是 1.8。
3. Fractions Combined Operations | 分数混合运算
Question: Evaluate 2 ⅓ + 1 ⅚ ÷ ¾ . Give your answer as a mixed number in its simplest form.
题目：计算 2 ⅓ + 1 ⅚ ÷ ¾ ，将结果写成最简带分数。
Division takes priority over addition. First, convert mixed numbers to improper fractions: 1 ⅚ = 11/6. Division by ¾ is the same as multiplying by 4/3. So 11/6 × 4/3 = 44/18. Simplify by dividing numerator and denominator by 2 to get 22/9. Now add to 2 ⅓: 2 ⅓ = 7/3. Convert 7/3 to 21/9 and add 21/9 + 22/9 = 43/9. As a mixed number, 43 ÷ 9 is 4 remainder 7, so 4 ⁷/₉. The answer is 4 ⁷/₉.
除法优先于加法。首先把带分数化为假分数：1 ⅚ = 11/6。除以 ¾ 等于乘以 4/3。所以 11/6 × 4/3 = 44/18。分子分母同时除以 2 化简得 22/9。现在加上 2 ⅓：2 ⅓ = 7/3。将 7/3 化为 21/9，然后 21/9 + 22/9 = 43/9。化为带分数，43 ÷ 9 = 4 余 7，所以答案是 4 ⁷/₉。
4. Percentage Change in a Real-Life Context | 百分数在实际情境中的应用
Question: A jacket originally costs £45. In a sale, the price is reduced by 20%. Then a 15% VAT is added to the sale price. What is the final price?
题目：一件夹克原价 45 英镑。打折季降价 20%，之后又加上 15% 的增值税。求最终价格。
Find the sale price first. 20% of £45 is 0.20 × 45 = £9. Subtract to get the sale price: 45 – 9 = £36. Next, add 15% VAT. 15% of £36 = 0.15 × 36 = £5.40. Final price = 36 + 5.40 = £41.40. So the jacket costs £41.40 after discount and VAT.
先求打折后价格。45 的 20% 是 0.20 × 45 = 9 英镑。相减得到打折价：45 – 9 = 36 英镑。接着加上 15% 的增值税。36 的 15% = 0.15 × 36 = 5.40 英镑。最终价格 = 36 + 5.40 = 41.40 英镑。所以这件夹克最终价格为 41.40 英镑。
5. Simplifying Algebraic Expressions and Substitution | 代数式化简与代入求值
Question: Simplify 3a + 2b – a + 5b. Then find the value of the expression when a = 2 and b = –1.
题目：化简 3a + 2b – a + 5b，并求当 a = 2，b = –1 时该表达式的值。
Collect like terms: for a, 3a – a = 2a. For b, 2b + 5b = 7b. The simplified expression is 2a + 7b. Now substitute a = 2 and b = –1: 2(2) + 7(–1) = 4 – 7 = –3. The value is –3.
合并同类项：含 a 的项，3a – a = 2a；含 b 的项，2b + 5b = 7b。化简后的表达式为 2a + 7b。代入 a = 2，b = –1：2(2) + 7(–1) = 4 – 7 = –3。值为 –3。
6. Solving a Linear Equation with Brackets | 解含括号的一元一次方程
Question: Solve 4(x – 3) = 2x + 10.
题目：解方程 4(x – 3) = 2x + 10。
First expand the left side: 4x – 12 = 2x + 10. Bring variable terms to one side: subtract 2x from both sides gives 2x – 12 = 10. Add 12 to both sides: 2x = 22. Divide by 2: x = 11.
首先展开左边：4x – 12 = 2x + 10。将所有含 x 的项移到一边：两边同时减去 2x 得 2x – 12 = 10。两边同时加 12：2x = 22。两边除以 2：x = 11。
4x – 12 = 2x + 10 → 2x = 22 → x = 11
7. Finding the nth Term of a Linear Sequence | 线性序列的第 n 项通项公式
Question: Here are the first four terms of a sequence: 3, 10, 17, 24. Find an expression for the nth term and use it to find the 20th term.
题目：一个序列的前四项为：3，10，17，24。写出第 n 项的通项公式，并利用它求出第 20 项。
The difference between consecutive terms is constant: 10 – 3 = 7, 17 – 10 = 7, 24 – 17 = 7. The common difference is 7, so the nth term rule is of the form 7n + c. For n = 1, 7(1) + c = 3, so c = –4. The nth term is 7n – 4. The 20th term: 7(20) – 4 = 140 – 4 = 136.
相邻项的差是常数：10 – 3 = 7，17 – 10 = 7，24 – 17 = 7。公差为 7，因此通项公式形如 7n + c。当 n = 1 时，7(1) + c = 3，所以 c = –4。第 n 项公式为 7n – 4。第 20 项：7(20) – 4 = 140 – 4 = 136。
8. Angle Reasoning in a Triangle | 三角形中的角度推理
Question: In a triangle, the largest angle is equal to the sum of the other two angles. What is the size of the largest angle?
题目：在一个三角形中，最大的角等于另外两个角之和。求最大角的度数。
Let the angles be x, y, and z, with z being the largest. According to the problem, z = x + y. Since the angles in a triangle sum to 180°, we have x + y + z = 180. Substitute z with x + y: (x + y) + (x + y) = 180, or 2(x + y) = 180. Therefore x + y = 90. Then z = 90°. The largest angle is a right angle, 90°.
设三个角分别为 x、y、z，其中 z 最大。根据题意，z = x + y。由于三角形内角和为 180°，有 x + y + z = 180。将 z 替换为 x + y：(x + y) + (x + y) = 180，即 2(x + y) = 180。因此 x + y = 90。于是 z = 90°。最大角是直角，度数为 90°。
9. Area and Perimeter of a Rectangle | 长方形的面积与周长
Question: The length of a rectangle is twice its width. The perimeter is 36 cm. Find the area of the rectangle.
题目：一个长方形的长是宽的两倍，周长为 36 cm。求这个长方形的面积。
Let the width be w cm, so the length is 2w cm. Perimeter = 2(length + width) = 2(2w + w) = 6w. Set 6w = 36, giving w = 6 cm. Length = 12 cm. Area = length × width = 12 × 6 = 72 cm².
设宽为 w cm，则长为 2w cm。周长 = 2(长 + 宽) = 2(2w + w) = 6w。令 6w = 36，得 w = 6 cm。长为 12 cm。面积 = 长 × 宽 = 12 × 6 = 72 cm²。
10. Mean and Median from a Frequency Table | 从频数表求平均数和中位数
Question: The table shows the number of books read by 11 students in a month: 0 books (2 students), 1 book (4 students), 2 books (3 students), 3 books (2 students). Calculate the mean and median number of books read.
题目：下表显示了 11 名学生一个月内阅读的书籍数量：0 本（2 人），1 本（4 人），2 本（3 人），3 本（2 人）。计算阅读本数的平均数和中位数。
Mean: total books = (0×2) + (1×4) + (2×3) + (3×2) = 0 + 4 + 6 + 6 = 16. Total students = 11. Mean = 16 ÷ 11 ≈ 1.45 books. Median: list all 11 data points in order: 0,0, 1,1,1,1, 2,2,2, 3,3. The middle value is the 6th term. Counting: 6th term is 1. So median = 1 book.
平均数：总本数 = (0×2) + (1×4) + (2×3) + (3×2) = 0 + 4 + 6 + 6 = 16。总人数 11。平均数 = 16 ÷ 11 ≈ 1.45 本。中位数：将 11 个数据从小到大排列：0,0, 1,1,1,1, 2,2,2, 3,3。中间位置是第 6 个数据。数到第 6 个是 1。所以中位数 = 1 本。
11. Ratio in Sharing a Quantity | 按比例分配
Question: Share £120 between two friends in the ratio 3:5. How much does each receive?
题目：将 120 英镑按 3:5 的比例分给两位朋友。每人各得多少？
Total number of parts = 3 + 5 = 8. Value of one part = £120 ÷ 8 = £15. First friend gets 3 parts: 3 × 15 = £45. Second friend gets 5 parts: 5 × 15 = £75. So the amounts are £45 and £75.
总份数 = 3 + 5 = 8。每份金额 = 120 ÷ 8 = 15 英镑。第一位朋友得 3 份：3 × 15 = 45 英镑。第二位朋友得 5 份：5 × 15 = 75 英镑。因此分配金额为 45 英镑和 75 英镑。
12. Final Tips and Recap | 总结与备考建议
You have now seen a full set of typical Year 8 WJEC Maths questions with step-by-step solutions. The key to success is practising each type of problem until the method becomes automatic. Always show your working clearly, check your answers, and memorise the essential formulas for area, perimeter, and angle facts. Use this mock paper as a revision tool, and try similar questions to build confidence before your real test.
你现在已经看完了一套完整的 Year 8 WJEC 数学典型题目以及详细解答步骤。成功的关键在于反复练习每一种题型，直到方法变得熟练自动化。一定要清晰地展示解题过程，检查答案，并牢记面积、周长和角度基本公式。把这份模拟卷当作复习工具，尝试类似的题目，在实际考试前建立信心。
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