📚 Year 11 WJEC Maths: Unit Test Mock Paper Walkthrough | Year 11 WJEC 数学：单元测试模拟卷解析
This walkthrough takes you through a typical WJEC Year 11 unit test, covering non‑calculator and calculator skills. Each section dissects a common question type, unpacks the method step by step, and highlights the key marking points so you can approach your exam with confidence.
本篇解析带你完整演练一份 WJEC Year 11 单元测试模拟卷，涵盖非计算器与计算器技巧。每个小节剖析一类典型考题，逐步拆解方法并点出关键得分点，帮助你自信应考。
1. Simplifying Algebraic Expressions | 代数式化简
Question: Simplify 3x + 5y − 2x + 7y − 4. Identify the like terms first: terms in x, terms in y, and constant numbers. For x: 3x − 2x gives x. For y: 5y + 7y gives 12y. The constant is −4. The simplified expression is x + 12y − 4.
题目：化简 3x + 5y − 2x + 7y − 4。首先找出同类项：含 x 的项、含 y 的项和常数项。x 部分：3x − 2x 得到 x。y 部分：5y + 7y 得到 12y。常数项为 −4。化简后的表达式为 x + 12y − 4。
Common mistake: mixing unlike terms. Remember that x and x² are not like terms – they cannot be combined. Also, always keep the sign in front of each term when rearranging.
常见错误：混淆不同类项。切记 x 和 x² 不是同类项，不能合并。另外，重新排列时务必保留每一项前面的符号。
2. Solving Linear Equations | 解线性方程
Question: Solve 2(3x − 1) = 4x + 10. Expand the bracket first: 6x − 2 = 4x + 10. Collect x terms on one side by subtracting 4x: 2x − 2 = 10. Add 2 to both sides: 2x = 12. Divide by 2: x = 6.
题目：解方程 2(3x − 1) = 4x + 10。首先展开括号：6x − 2 = 4x + 10。将含 x 的项移到一边，两边同时减去 4x：2x − 2 = 10。两边加 2：2x = 12。除以 2 得 x = 6。
Always check your solution by substituting back into the original equation. Left side: 2(3×6 − 1) = 2(18 − 1) = 34. Right side: 4×6 + 10 = 34. Both sides match, so x = 6 is correct.
务必代入原方程检验解。左边：2(3×6 − 1) = 2(18 − 1) = 34。右边：4×6 + 10 = 34。两边相等，因此 x = 6 正确。
3. Factorising and Solving Quadratics | 因式分解与解二次方程
Question: Factorise x² − 5x − 14. Look for two numbers that multiply to −14 and add to −5. The pair −7 and +2 works because (−7) × 2 = −14 and (−7) + 2 = −5. So the factorised form is (x − 7)(x + 2). To solve x² − 5x − 14 = 0, set each bracket to zero: x − 7 = 0 gives x = 7, x + 2 = 0 gives x = −2.
题目：因式分解 x² − 5x − 14。寻找两个数，乘积为 −14，和为 −5。 −7 和 +2 满足条件，因为 (−7) × 2 = −14 且 (−7) + 2 = −5。因式分解结果为 (x − 7)(x + 2)。解方程 x² − 5x − 14 = 0 时，令每个括号为零：x − 7 = 0 得 x = 7，x + 2 = 0 得 x = −2。
When the quadratic has a coefficient greater than 1 for x², such as 2x² + 7x + 3, you factor by splitting the middle term or using trial and error. The pair for ac = 2×3 = 6 that adds to 7 is 6 and 1, so rewrite as 2x² + 6x + x + 3, factor in pairs, giving (2x + 1)(x + 3).
当二次项系数大于 1，例如 2x² + 7x + 3，可通过拆分中项或试错法因式分解。ac = 2×3 = 6，和为 7 的数是 6 和 1，重写为 2x² + 6x + x + 3，分组提取公因式得 (2x + 1)(x + 3)。
4. Straight Line Graphs | 直线图形
Question: Find the gradient of the line passing through (2, 5) and (6, 13). Use the formula gradient = (y₂ − y₁) ÷ (x₂ − x₁). Here (13 − 5) ÷ (6 − 2) = 8 ÷ 4 = 2. The gradient is 2.
题目：求过点 (2, 5) 和 (6, 13) 的直线的斜率。使用公式 斜率 = (y₂ − y₁) ÷ (x₂ − x₁)。计算 (13 − 5) ÷ (6 − 2) = 8 ÷ 4 = 2。斜率为 2。
To find the equation of the line, substitute the gradient and one point into y = mx + c. Using (2, 5): 5 = 2×2 + c → 5 = 4 + c → c = 1. The equation is y = 2x + 1. On a WJEC unit test, you may be asked to draw the line or read the y‑intercept directly from a graph.
要求出直线方程，将斜率和一点代入 y = mx + c。用 (2, 5)：5 = 2×2 + c → 5 = 4 + c → c = 1。方程为 y = 2x + 1。在 WJEC 单元测试中，你可能会被要求画出直线或直接从图中读出 y 轴截距。
5. Ratio and Proportion | 比与比例
Question: Divide £450 in the ratio 2:3:5. Add the parts: 2 + 3 + 5 = 10 parts in total. One part is £450 ÷ 10 = £45. The shares are 2×£45 = £90, 3×£45 = £135, and 5×£45 = £225. Always check that the amounts sum to the original total: £90 + £135 + £225 = £450.
题目：将 £450 按 2:3:5 的比例分配。先求出总份数：2 + 3 + 5 = 10 份。一份为 £450 ÷ 10 = £45。各份额为 2×£45 = £90，3×£45 = £135，5×£45 = £225。务必检查各项之和等于原始总数：£90 + £135 + £225 = £450。
For direct proportion, if y is proportional to x, write y = kx. Use given values to find k. Inverse proportion is modelled as y = k/x. WJEC often embeds proportion problems in real-life contexts, such as recipes, currency exchange or speed‑time calculations.
对于正比例，若 y 与 x 成正比，写成 y = kx，利用已知值求 k。反比例模型为 y = k/x。WJEC 常将比例问题融入生活情境，如食谱、货币兑换或速度‑时间计算。
6. Area and Circumference of Circles | 圆的面积与周长
Question: A circle has radius 7 cm. Find its circumference and area, giving answers in terms of π. Circumference = 2πr = 2×π×7 = 14π cm. Area = πr² = π×7² = 49π cm². If a rounded answer is required, substitute π ≈ 3.14 or use the π button on your calculator.
题目：圆的半径为 7 cm，求周长和面积，答案保留 π。周长 = 2πr = 2×π×7 = 14π cm。面积 = πr² = π×7² = 49π cm²。若需要四舍五入的答案，代入 π ≈ 3.14 或使用计算器上的 π 键。
When given the diameter, halve it to get the radius. Be careful with semi‑circles and quarter‑circles: divide the full circle area or circumference appropriately and add straight edges where needed. WJEC questions frequently combine area of circles with area of rectangles or triangles.
当给出直径时，将其除以 2 得到半径。注意半圆和四分之一圆：适当分割整个圆的面积或周长，并加上相应的直线边。WJEC 题目常将圆的面积与矩形或三角形面积结合考查。
7. Trigonometry in Right‑Angled Triangles | 直角三角形中的三角函数
Question: In a right‑angled triangle, the hypotenuse is 12 cm and an angle of 35° is given. Find the length of the side opposite the 35° angle. Use sinθ = opposite/hypotenuse, so sin35° = opposite/12. Rearranging: opposite = 12 × sin35° ≈ 12 × 0.5736 = 6.88 cm (to 2 decimal places).
题目：在直角三角形中，斜边为 12 cm，一个角为 35°。求该角对边的长度。使用 sinθ = 对边/斜边，即 sin35° = 对边/12。变形得：对边 = 12 × sin35° ≈ 12 × 0.5736 = 6.88 cm（保留两位小数）。
Label the sides relative to the given angle: opposite (O), adjacent (A), hypotenuse (H). Choose the correct ratio: SOH CAH TOA. With WJEC unit tests, you must show the equation setup clearly to earn method marks, even if the final answer has a slight rounding error.
相对于已知角标注各边：对边 (O)、邻边 (A)、斜边 (H)。选择正确的比：SOH CAH TOA。在 WJEC 单元测试中，即使最终答案有舍入误差，也必须清楚展示方程式的建立才能得到方法分。
8. Mean, Median, Mode and Range | 平均数、中位数、众数和极差
Question: Find the mean, median, mode and range of the data set: 4, 7, 9, 7, 5, 10, 7. Mean = (4+7+9+7+5+10+7) ÷ 7 = 49 ÷ 7 = 7. To find the median, order the data: 4, 5, 7, 7, 7, 9, 10. The middle value is the fourth number, 7. Mode is the most frequent: 7 appears three times. Range = maximum − minimum = 10 − 4 = 6.
题目：对于数据集 4, 7, 9, 7, 5, 10, 7，求平均数、中位数、众数和极差。平均数 = (4+7+9+7+5+10+7) ÷ 7 = 49 ÷ 7 = 7。求中位数时先排序：4, 5, 7, 7, 7, 9, 10。中间值为第四个数 7。众数为出现次数最多的 7（出现三次）。极差 = 最大值 − 最小值 = 10 − 4 = 6。
With an even number of data points, the median is the average of the two middle values. For a grouped frequency table, estimate the mean using midpoints × frequencies. WJEC often asks students to compare two distributions using mean and range, or median and interquartile range.
当数据个数为偶数时，中位数为中间两个数的平均值。对于分组频数表，可利用组中值×频数估算平均数。WJEC 常要求考生利用平均数和极差，或中位数和四分位距来比较两个分布。
9. Probability | 概率
Question: A bag contains 5 red, 3 blue and 2 green marbles. One marble is taken at random. Find the probability it is not red. Total marbles = 5 + 3 + 2 = 10. Number not red = 3 + 2 = 5. P(not red) = 5/10 = 1/2.
题目：袋中有 5 个红色、3 个蓝色和 2 个绿色弹珠。随机取出一个，求取出不是红色的概率。总数 = 5 + 3 + 2 = 10。非红色的数量 = 3 + 2 = 5。P(非红) = 5/10 = 1/2。
For combined events, such as taking two marbles without replacement, draw a tree diagram. Multiply along the branches. For the probability of at least one, it is often easier to calculate 1 − P(none). Always leave probability as a fraction in its simplest form unless instructed otherwise.
对于组合事件，如不放回地取两个弹珠，画出树状图。沿分支相乘。求“至少一个”的概率时，通常计算 1 − P(无) 更简便。除非另有说明，概率应保留为最简分数。
10. Fractions, Decimals and Percentages | 分数、小数与百分比
Question: Convert 7/8 to a decimal and a percentage. 7 ÷ 8 = 0.875. As a percentage, multiply by 100: 0.875 × 100 = 87.5%. To find 12.5% of 240 without a calculator, first find 10% which is 24, then 2.5% is a quarter of 10% = 6, so 12.5% = 24 + 6 = 30.
题目：将 7/8 转换为小数和百分比。7 ÷ 8 = 0.875。转换为百分比乘以 100：0.875 × 100 = 87.5%。不依赖计算器求 240 的 12.5%，先求 10% 为 24，2.5% 是 10% 的四分之一为 6，因此 12.5% = 24 + 6 = 30。
In WJEC non‑calculator papers, fluently moving between fractions, decimals and percentages is essential. Know key equivalences like 1/8 = 12.5% and 3/5 = 0.6 = 60%. Percentage increase and decrease often appear in shopping or salary contexts – use the multiplier method for efficiency.
在 WJEC 非计算器试卷中，分数、小数和百分比的熟练转换至关重要。牢记关键等值关系，如 1/8 = 12.5%，3/5 = 0.6 = 60%。百分比增减常出现在购物或薪酬情境——使用乘数法更高效。
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