📚 Year 8 WJEC Further Mathematics: Unit Test Mock Paper Walkthrough | Year 8 WJEC 进阶数学：单元测试模拟卷解析
Preparing for a WJEC Further Mathematics unit test can be challenging, but working through a mock paper is one of the best ways to revise. In this walkthrough, we will go through a full set of 8 questions covering key topics such as algebra, geometry, statistics and probability. Each question is explained step by step, highlighting common pitfalls and efficient methods. Follow along to boost your confidence and exam technique.
准备 WJEC 进阶数学单元测试很有挑战，但做模拟卷是最高效的复习方式之一。在本文中，我们将完整解析一套包含 8 道题的模拟卷，涵盖代数、几何、统计和概率等核心内容。每道题都会逐步讲解，点明常见错误与快捷方法。跟着一起练习，提升你的信心和应试技巧。
1. Simplifying Algebraic Expressions | 化简代数表达式
Question 1: Simplify (3x² + 5x – 2) – (x² – 2x + 4).
问题1：化简 (3x² + 5x – 2) – (x² – 2x + 4)。
Step 1: Remove the brackets, paying close attention to the minus sign before the second bracket. That sign changes all the terms inside: – (x² – 2x + 4) becomes –x² + 2x – 4. The expression now is 3x² + 5x – 2 – x² + 2x – 4.
步骤1：去掉括号，特别注意第二个括号前的减号。这个减号会改变括号内每一项的符号：– (x² – 2x + 4) 变为 –x² + 2x – 4。此时表达式为 3x² + 5x – 2 – x² + 2x – 4。
Step 2: Identify and collect like terms. Group the x² terms: 3x² – x² = 2x². Group the x terms: 5x + 2x = 7x. Group the constant terms: –2 – 4 = –6.
步骤2：找出并合并同类项。x² 项：3x² – x² = 2x²。x 项：5x + 2x = 7x。常数项：–2 – 4 = –6。
Step 3: Write the final simplified expression: 2x² + 7x – 6.
步骤3：写出最终化简结果：2x² + 7x – 6。
bstitute x = 1 into both the original and the simplified expression. Original: (3·1² + 5·1 – 2) – (1² – 2·1 + 4) = (3+5–2) – (1–2+4) = 6 – 3 = 3. Simplified: 2·1² + 7·1 – 6 = 2+7–6 = 3. Answers match, confirming the simplification is correct.
快速检验：将 x = 1 分别代入原式和化简式。原式：(3·1² + 5·1 – 2) – (1² – 2·1 + 4) = (3+5–2) – (1–2+4) = 6 – 3 = 3。化简式：2·1² + 7·1 – 6 = 2+7–6 = 3。结果一致，说明化简正确。
Common mistake: forgetting to change the sign of every term inside the subtracted bracket. Many students write –x² – 2x + 4 by only flipping the first sign, which is wrong. Always apply the negative sign to all three terms.
常见错误：忘记对括号内每一项变号。不少同学只改变第一项的符号，写成 –x² – 2x + 4，这是错误的。务必把减号作用到括号内所有三项。
2. Solving Linear Equations | 解一次方程
Question 2: Solve 2(3x – 4) = 5x + 8.
问题2：解方程 2(3x – 4) = 5x + 8。
Step 1: Expand the left-hand side: 2 × 3x – 2 × 4 = 6x – 8. The equation becomes 6x – 8 = 5x + 8.
步骤1：展开左边：2 × 3x – 2 × 4 = 6x – 8。方程变为 6x – 8 = 5x + 8。
Step 2: Collect the x terms on one side. Subtract 5x from both sides: 6x – 5x – 8 = 5x – 5x + 8, giving x – 8 = 8.
步骤2：将含 x 的项移到一边。两边同时减去 5x：6x – 5x – 8 = 5x – 5x + 8，得到 x – 8 = 8。
Step 3: Isolate x by adding 8 to both sides: x – 8 + 8 = 8 + 8, so x = 16.
步骤3：通过两边加 8 将 x 分离出来：x – 8 + 8 = 8 + 8，因此 x = 16。
he solution by substituting x = 16 back into the original equation. Left-hand side: 2(3×16 – 4) = 2(48 – Right-hand side: 5×16 + 8 = 80 + 8 = 88. Both sides equal 88, so the solution is correct.
验证解：将 x = 16 代回原方程。左边：2(3×16 – 4) = 2(48 – 4) = 2×44 = 88。右边：5×16 + 8 = 80 + 8 = 88。两边都等于 88，因此解正确。
Tip: When you have a multiplier outside brackets, always expand first before moving terms. Also, if the equation had fractions, you would multiply every term by the common denominator to clear them.
提示：当括号外有乘数时，一定要先展开再移项。另外，如果方程中含有分数，可以先给每一项乘以公分母以去掉分母。
3. Factorising Quadratic Expressions | 因式分解二次三项式
Question 3: Factorise x² – 7x + 12.
问题3：因式分解 x² – 7x + 12。
Step 1: Look for two numbers that multiply to the constant term +12 and add to the coefficient of x, which is –7.
步骤1：寻找两个数，乘积为常数项 +12，同时相加等于 x 的系数 –7。
, (3,4). Since we need a negative sum and positive product, both numbers must be negative: –3 and –4 fit because (–3) × (–4) = 12 and (–3) + (–4) = –7.
步骤2：列出 12 的因数对：(1,12)，(2,6)，(3,4)。因为我们需要负的和与正的积，所以两个数都必须为负：–3 和 –4 符合条件，因为 (–3) × (–4) = 12 且 (–3) + (–4) = –7。
Step 3: Write the factorised form using these numbers: (x – 3)(x – 4).
步骤3：用这两个数写出因式分解形式：(x – 3)(x – 4)。
Check by expanding: (x – 3)(x – 4) = x² – 4x – 3x + 12 = x² – 7x + 12, which matches the original expression.
通过展开检验：(x – 3)(x – 4) = x² – 4x – 3x + 12 = x² – 7x + 12，与原式一致。
Watch out: a frequent error is to write (x + 3)(x + 4) or (x + 3)(x – 4). Always double-check the signs by considering the middle term. If both numbers were positive, the middle term would be +7x, not –7x.
注意：常见错误是写成 (x + 3)(x + 4) 或 (x + 3)(x – 4)。一定要通过中间项来检查符号。如果两数都为正，中间项就会是 +7x 而不是 –7x。
4. Angles in Parallel Lines | 平行线中的角度
Question 4: Two parallel lines AB and CD are cut by a transversal EF. One angle formed is 125°. Find the related angles a, b and c as described below.
问题4：两条平行线 AB 和 CD 被一条截线 EF 所截。已知形成的其中一个角为 125°。求相关角 a、b 和 c，描述如下。
on the other parallel line. Corresponding angles on parallel lines are equal, so a = 125°.
角 a 是另一条平行线上与 125° 角对应的同位角。平行线上的同位角相等，所以 a = 125°。
posite to the 125° angle. Vertically opposite angles are always equal, therefore b = 125°.
角 b 是 125° 角的对顶角。对顶角始终相等，因此 b = 125°。
gle. Interior angles between parallel lines are supplementary (sum to 180°). Thus c = 180° – 125° = 55°.
角 c 是与 125° 角位于截线同侧的同旁内角。平行线之间的同旁内角互补（和为 180°）。所以 c = 180° – 125° = 55°。
These relationships help you find missing angles without measuring. Always state which angle fact you used, just as in the steps above, to earn full marks in a test.
这些角的关系可以帮助你不出错地找出未知角。在考试中，像上面步骤一样，务必说明你使用的角性质，以获得全部分数。
Diagram summary in words: if the 125° angle is at the top intersection, then a is in the matching position at the bottom intersection, b is opposite the 125° angle at the top, and c is next to a inside the parallels, making a C-shaped pair with 125°.
文字示意：若 125° 角位于上方的交点，则 a 在下方交点与之匹配的位置，b 在上方交点的 125° 对角，c 在 a 旁边、平行线内侧，与 125° 形成同旁内角对。
5. Pythagoras’ Theorem | 勾股定理
Question 5: In a right-angled triangle, the two shorter sides are 5 cm and 12 cm. Calculate the length of the hypotenuse.
问题5：在一个直角三角形中，两条直角边分别为 5 cm 和 12 cm。计算斜边的长度。 Published by TutorHao | Year 8 进阶数学 Revision Series | aleveler.com
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