📚 Year 10 WJEC Maths: Unit Test Mock Paper Analysis | Year 10 WJEC 数学：单元测试模拟卷解析
This mock paper analysis is designed to help Year 10 students prepare for their WJEC Mathematics unit test. It covers a range of core topics – from algebra and equations to geometry, statistics and probability – with fully worked solutions and key exam tips. Use this walkthrough to sharpen your problem‑solving skills and avoid common pitfalls.
本模拟卷解析旨在帮助 Year 10 学生备战 WJEC 数学单元测试。内容涵盖代数、方程、几何、统计与概率等核心主题，每道题都配有完整解答和重要考试技巧。通过本解析，你可以提升解题能力并避开典型失分陷阱。
1. Simplifying Algebraic Expressions | 第1题：化简代数表达式
Question: Simplify 4a − 3b + 2a + 7b.
题目：化简 4a − 3b + 2a + 7b。
Solution: Identify and group like terms. The terms containing a are 4a and +2a, which sum to 6a. The terms containing b are −3b and +7b, which sum to +4b. Hence the simplified expression is 6a + 4b.
解：先找出同类项并分组。含 a 的项是 4a 和 +2a，相加得 6a。含 b 的项是 −3b 和 +7b，合并得 +4b。因此化简结果为 6a + 4b。
Common mistake: Forgetting to take the sign in front of each term when grouping. Always move the term with its sign: −3b + 7b, not 3b + 7b.
常见错误：分组时忽略了各项前面的正负号。务必带着符号移动项：−3b + 7b，而不是 3b + 7b。
2. Solving Linear Equations | 第2题：解一元一次方程
Question: Solve 5(x − 2) = 3x + 4.
题目：解方程 5(x − 2) = 3x + 4。
Solution: First expand the bracket: 5x − 10 = 3x + 4. Next, collect x‑terms on one side by subtracting 3x from both sides: 2x − 10 = 4. Add 10 to both sides: 2x = 14. Finally, divide by 2 to obtain x = 7.
解：先去括号：5x − 10 = 3x + 4。接着将含 x 的项移到一边，两边同时减去 3x：2x − 10 = 4。两边加 10：2x = 14。最后除以 2，得 x = 7。
Tip: Always check your solution by substituting back into the original equation: 5(7−2)=5×5=25 and 3×7+4=21+4=25, so it is correct.
技巧：务必把解代入原方程检验：5(7−2)=5×5=25，而 3×7+4=21+4=25，结果一致，答案正确。
3. Factorising Quadratics | 第3题：二次三项式的因式分解
Question: Factorise x² + 5x + 6.
题目：将 x² + 5x + 6 因式分解。
Solution: We need two numbers that multiply to +6 (the constant term) and add to +5 (the coefficient of x). Those numbers are +2 and +3. Thus, x² + 5x + 6 = (x + 2)(x + 3).
解：需要找到两个数，它们的乘积等于常数项 +6，且和等于 x 的系数 +5。这两个数是 +2 和 +3。因此 x² + 5x + 6 = (x + 2)(x + 3)。
Verification: Expanding (x+2)(x+3) gives x²+3x+2x+6 = x²+5x+6, confirming the factorisation.
验证：展开 (x+2)(x+3) 得 x²+3x+2x+6 = x²+5x+6，确认分解无误。
4. Equation of a Straight Line | 第4题：直线方程
Question: Find the equation of the straight line passing through (1, 3) and (3, 7).
题目：求经过点 (1, 3) 和 (3, 7) 的直线方程。
Solution: First calculate the gradient m using the formula: m = (y₂ − y₁)/(x₂ − x₁). Substituting the coordinates gives m = (7 − 3)/(3 − 1) = 4/2 = 2. The line has equation y = 2x + c. Substitute (1, 3): 3 = 2(1) + c ⇒ c = 1. Hence the equation is y = 2x + 1.
解：先用公式求斜率 m：m = (y₂ − y₁)/(x₂ − x₁)。代入坐标得 m = (7 − 3)/(3 − 1) = 4/2 = 2。直线方程为 y = 2x + c。代入 (1, 3)：3 = 2(1) + c ⇒ c = 1。因此方程为 y = 2x + 1。
Check: For x = 3, y = 2×3 + 1 = 7, matching the second point.
检验：当 x = 3 时，y = 2×3 + 1 = 7，与第二个点吻合。
5. Solving Quadratic Equations | 第5题：解二次方程
Question: Solve x² − 4x − 5 = 0.
题目：解方程 x² − 4x − 5 = 0。
Solution by factorisation: Look for two numbers that multiply to −5 and add to −4. They are −5 and +1. So the equation factorises as (x − 5)(x + 1) = 0. Setting each factor to zero gives x − 5 = 0 ⇒ x = 5, or x + 1 = 0 ⇒ x = −1.
因式分解法解：寻找乘积为 −5 且和为 −4 的两个数，它们分别是 −5 和 +1。因此方程可分解为 (x − 5)(x + 1) = 0。令每个因式为零，得 x − 5 = 0 ⇒ x = 5，或 x + 1 = 0 ⇒ x = −1。
Alternative using the quadratic formula: a = 1, b = −4, c = −5. The formula is x = [−b ± √(b² − 4ac)]/(2a). Substituting gives x = [4 ± √(16 + 20)]/2 = [4 ± √36]/2 = [4 ± 6]/2. Thus x = 5 or x = −1.
备选方法–公式法：a = 1, b = −4, c = −5。求根公式为 x = [−b ± √(b² − 4ac)]/(2a)。代入得 x = [4 ± √(16 + 20)]/2 = [4 ± √36]/2 = [4 ± 6]/2，故 x = 5 或 x = −1。
6. Area of a Triangle Using Trigonometry | 第6题：利用三角函数求三角形面积
Question: Two sides of a triangle measure 8 cm and 6 cm, and the included angle is 30°. Calculate the area of the triangle.
题目：三角形两边长分别为 8 cm 和 6 cm，夹角为 30°。求该三角形的面积。
Solution: Use the formula Area = ½ × a × b × sin C. Substituting the values: Area = ½ × 8 × 6 × sin 30°. Since sin 30° = 0.5, Area = ½ × 8 × 6 × 0.5 = (48 × 0.5)/2 = 24/2 = 12 cm².
解：使用公式 面积 = ½ × a × b × sin C。代入已知量：面积 = ½ × 8 × 6 × sin 30°。因为 sin 30° = 0.5，面积 = ½ × 8 × 6 × 0.5 = (48 × 0.5)/2 = 24/2 = 12 cm²。
Note: Remember that the angle must be the included angle between the two given sides.
注意：所给角必须是已知两边的夹角才能使用该公式。
7. Mean from a Frequency Table | 第7题：根据频数表求平均数
Question: The table shows the scores of 20 students: score 0 (frequency 2), score 1 (5), score 2 (7), score 3 (4), score 4 (2). Calculate the mean score.
题目：下表给出了20名学生的得分情况：得分0（频数2），得分1（5），得分2（7），得分3（4），得分4（2）。计算平均得分。
Solution: First find the total sum: (0×2) + (1×5) + (2×7) + (3×4) + (4×2) = 0 + 5 + 14 + 12 + 8 = 39. The total frequency is 2+5+7+4+2 = 20. Mean = total sum ÷ total frequency = 39 ÷ 20 = 1.95.
解：先求总分：(0×2) + (1×5) + (2×7) + (3×4) + (4×2) = 0 + 5 + 14 + 12 + 8 = 39。总频数为 2+5+7+4+2 = 20。平均分 = 总分 ÷ 总频数 = 39 ÷ 20 = 1.95。
Tip: Always include the zero scores in both the sum and the frequency count – omitting them is a common mistake.
技巧：计算总分和总频数时务必把0分项纳入，漏算是常见错误。
8. Probability Without Replacement | 第8题：不放回情况下的概率
Question: A bag contains 3 red balls and 5 blue balls. Two balls are drawn at random without replacement. Find the probability that both balls are blue.
题目：一个袋子里有3个红球和5个蓝球。随机取出两个球且不放回。求两个球都是蓝色的概率。
Solution: Probability the first ball is blue = 5/8. After removing one blue, there are 4 blue balls left out of 7. So probability the second is blue given the first was blue = 4/7. Multiply: P(both blue) = (5/8) × (4/7) = 20/56, which simplifies to 5/14.
解：第一个球是蓝色的概率 = 5/8。取走一个蓝球后，袋中剩4个蓝球，总球数7个，故第二球为蓝色的条件概率为 4/7。相乘：P(两球皆蓝) = (5/8) × (4/7) = 20/56，约分得 5/14。
Alternative tree diagram: Drawing a tree with branches ‘Blue’ and ‘Red’ on each pick helps visualise the dependence and ensures the fractions are multiplied correctly.
树状图辅助：画树状图，每次抽取分“蓝”和“红”两支，可直观体现事件相依性，并确保分数相乘无误。
9. Percentage Decrease | 第9题：百分数减少
Question: The original price of a jacket is £80. In a sale, the price is reduced by 15%. Find the sale price.
题目：一件夹克的原价为 £80。打折活动中价格降低了 15%。求打折后的价格。
Solution: A 15% decrease means the new price is 100% − 15% = 85% of the original. 85% as a decimal is 0.85. Sale price = £80 × 0.85 = £68.
解：降价 15% 意味着新价格为原价的 100% − 15% = 85%。85% 化为小数是 0.85。打折价 = £80 × 0.85 = £68。
Alternative method: Find 15% of £80 = £12, then subtract from £80: £80 − £12 = £68. Both methods give the same result.
另解：先算 £80 的 15% 为 £12，再从原价中扣除：£80 − £12 = £68。两种方法结果一致。
10. Translation of a Point | 第10题：点的平移变换
Question: Point A has coordinates (2, 3). It is translated by the vector (−4, 1). Find the coordinates of the image point A′.
题目：点 A 的坐标为 (2, 3)，经向量 (−4, 1) 平移后，求像点 A′ 的坐标。
Solution: To translate a point, add the vector components to the original coordinates. New x‑coordinate = 2 + (−4) = −2. New y‑coordinate = 3 + 1 = 4. Therefore A′ is (−2, 4).
解：平移一个点时，将向量的分量分别加到原坐标上。新横坐标 = 2 + (−4) = −2，新纵坐标 = 3 + 1 = 4。故 A′ 坐标为 (−2, 4)。
Check: Plotting the original and image points on a grid confirms the movement 4 units left and 1 unit up.
验证：在坐标网格中标出原点和像点，可见点向左移动4个单位、向上移动1个单位，与向量一致。
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