📚 Year 10 Edexcel Maths: Unit Test Mock Paper Walkthrough | Year 10 Edexcel 数学：单元测试模拟卷解析
This walkthrough breaks down a full Unit Test mock paper designed for Year 10 Edexcel Maths. Each question targets core topics from Number, Algebra, Ratio, Geometry, Probability and Statistics, reflecting the demands of the 9‑1 GCSE specification. Follow the step‑by‑step solutions to strengthen your exam technique and avoid common errors.
本篇解析逐题拆解一份为 Year 10 Edexcel 数学设计的完整单元测试模拟卷。题目涵盖数、代数、比与比例、几何、概率与统计等核心板块，贴合 9‑1 GCSE 考纲要求。通过逐步推演，帮助你巩固解题方法，避开高频陷阱。
1. Question 1: Multi‑step Arithmetic with Fractions | 第1题：分数四则混合运算
Question: Evaluate (3/4 + 1/5) × 2/3 - 1/10. Give your answer in simplest form.
题目： 计算 (3/4 + 1/5) × 2/3 - 1/10，结果化为最简形式。
Start inside the brackets. Find a common denominator for 3/4 and 1/5. The lowest common denominator is 20. Convert:
先处理括号内。3/4 与 1/5 的最小公分母是 20。转换分数：
3/4 = 15/20, 1/5 = 4/20
Add them: 15/20 + 4/20 = 19/20.
相加得: 19/20。
2/3: (19/20) × (2/3) = 38/60. Simplify by dividing numerator and denominator by 2: 19/30.
乘以 2/3: (19/20) × (2/3) = 38/60，约去公因数 2 得 19/30。
Subtract 1/10: 19/30 - 1/10 = 19/30 - 3/30 = 16/30. Simplify: divide by 2 → 8/15.
最后减去 1/10: 19/30 - 3/30 = 16/30，约简为 8/15。
Final answer: 8/15. Always check you fully simplify.
最终答案：8/15。务必确认已化简到底。
2. Question 2: Standard Form and Significant Figures | 第2题：科学记数法与有效数字
Question: (4.2 × 10³) × (6.0 × 10⁻⁴) and give your answer in standard form to 2 significant figures.
题目： 将 0.00743 写成科学记数法。再计算 (4.2 × 10³) × (6.0 × 10⁻⁴)，并将结果用标准形式表示，保留 2 位有效数字。
3: Move the decimal 3 places right to get 7.43, so the power is 10⁻³. Answer: 7.43 × 10⁻³.
0.00743: 小数点右移 3 位得 7.43，因此指数为 -3，结果为 7.43 × 10⁻³。
Multiplication: (4.2 × 6.0) × 103-4 10⁰. Since 2.52 moves decimal one place right, it becomes 2.52 × 10⁰ (which is just 2.52).
乘法：系数相乘 4.2 × 6.0 = 25.2；幂相加 3 + (-4) = -1，得 25.2 × 10⁻¹ = 2.52 × 10⁰。
To 2 significant figures, 2.52 becomes 2.5, so answer: 2.5 × 10⁰ (or simply 2.5).
保留 2 位有效数字，2.52 四舍五入为 2.5，故答案: 2.5 × 10⁰（或直接写 2.5）。
3. Question 3: Proportion and Recipe Scaling | 第3题：比例与食谱份量调整
Question: A recipe for 8 people uses 300 g flour, 120 g sugar and 2 eggs. Jen wants to make the dessert for 10 people using the same ratios. Find the mass of flour and sugar she needs, and decide how many whole eggs to use.
题目： 一份 8 人量的食谱需要 300 g 面粉、120 g 糖和 2 枚鸡蛋。Jen 想按相同比例为 10 人制作甜点。求她需要的面粉和糖的质量，并说明应使用多少枚完整的鸡蛋。
Scaling factor: 10 ÷ 8 = 1.25.
比例因子: 10 ÷ 8 = 1.25。
Flour: 300 × 1.25 = 375 g. Sugar: 120 × 1.25 = 150 g.
面粉: 300 × 1.25 = 375 g。糖: 120 × 1.25 = 150 g。
gs. Since eggs cannot be split practically, she should use 3 whole eggs and adjust other ingredients proportionally, or accept a slightly richer mixture. In an exact mathematical answer, state 2.5 eggs; in a real kitchen, 3.
鸡蛋: 2 × 1.25 = 2.5，数学上需要 2.5 枚。实际备料时无法切分，需用 3 枚整蛋并相应微调其他配料，或接受略微湿润的质地。考试中写出 2.5 即可。
4. Question 4: Percentage Profit and Depreciation | 第4题：百分比利润与折旧
Question: A shop buys a laptop for £480 and sells it at a 15% profit. Later the customer resells it after one year at a depreciation of 20% from the buying price. Calculate both selling prices.
题目： 商店以 £480 购入一台笔记本电脑，加价 15% 售出。顾客一年后以买入价贬值 20% 再次转卖。分别计算两次的售价。
Profit sale: 15% of £480 = 0.15 × 480 = £72. Shop selling price = 480 + 72 = £552.
利润售价: 成本的 15% = 0.15 × 480 = £72。商店售价 = 480 + 72 = £552。
Resale: depreciation is 20% of the customer's buying price (£552). Depreciation amount = 0.20 × 552 = £110.40. Resale price = 552 - 110.40 = £441.60.
再次转卖: 折旧为顾客购入价 £552 的 20%。折旧额 = 0.20 × 552 = £110.40。转卖价 = 552 - 110.40 = £441.60。
5. Question 5: Solving Quadratic Equations by Factorising | 第5题：因式分解法解二次方程
Question: Solve x² + x - 12 = 0.
题目： 解方程 x² + x - 12 = 0。
Find two numbers that multiply to -12 and add to +1. They are +4 and -3.
寻找乘积为 -12、和为 +1 的两数。符合条件的是 +4 和 -3。
Factorise: (x + 4)(x - 3) = 0.
因式分解: (x + 4)(x - 3) = 0。
Set each bracket to zero: x + 4 = 0 → x = -4; x - 3 = 0 → x = 3.
令每个括号为 0: x + 4 = 0 → x = -4; x - 3 = 0 → x = 3。
Solutions: x = -4 or x = 3. Always check by substituting back.
解：x = -4 或 x = 3。须代回原方程检验。
6. Question 6: Simultaneous Linear Equations | 第6题：二元一次联立方程组
Question: Solve the simultaneous equations: 3x + 2y = 16, 5x - y = 5.
题目： 解联立方程组: 3x + 2y = 16, 5x - y = 5。
Label: (1) 3x + 2y = 16, (2) 5x - y = 5. Multiply equation (2) by 2 to align y‑coefficients: 10x - 2y = 10.
标号: (1) 3x + 2y = 16, (2) 5x - y = 5。将 (2) 式乘以 2: 10x - 2y = 10。
Add to equation (1): (3x + 2y) + (10x - 2y) = 16 + 10 → 13x = 26 → x = 2.
与 (1) 式相加: 13x = 26，解得 x = 2。
Substitute x = 2 into (2): 5(2) - y = 5 → 10 - y = 5 → y = 5.
将 x = 2 代入 (2): 10 - y = 5，得 y = 5。
Solution: x = 2, y = 5.
解：x = 2, y = 5。
7. Question 7: Compound Units – Speed and Density | 第7题：复合单位 – 速度与密度
Question: A lorry travels 135 km in 1 hour 30 minutes. Calculate its average speed in km/h. Then the fuel tank holds 150 litres of diesel with density 0.85 kg/litre. Find the mass of fuel.
题目： 一辆卡车 1 小时 30 分钟行驶 135 km。计算平均速度（km/h）。若油箱装有 150 升柴油，密度为 0.85 kg/litre，求燃油质量。
Time: 1 h 30 min = 1.5 hours. Speed = distance ÷ time = 135 ÷ 1.5 = 90 km/h.
时间: 1 小时 30 分 = 1.5 小时。速度 = 135 ÷ 1.5 = 90 km/h。
Mass = volume × density = 150 × 0.85 = 127.5 kg.
质量 = 体积 × 密度 = 150 × 0.85 = 127.5 kg。
8. Question 8: Angles in Polygons and Parallel Lines | 第8题：多边形内角与平行线角度
Question: A regular hexagon has one exterior angle marked x°. Find x. Then two parallel lines are cut by a transversal, giving an alternate angle of 65° and a co‑interior angle of y°. Find y.
题目： 一个正六边形的一个外角为 x°，求 x。另外，两条平行线被一条截线所切，给出内错角 65° 和同旁内角 y°，求 y。
xterior angles of any convex polygon = 360°. Regular hexagon: 6 sides → x = 360 ÷ 6 = 60°.
任意凸多边形外角和为 360°。正六边形有 6 边，x = 360 ÷ 6 = 60°。
Co‑ As the alternate angle is 65°, the corresponding supplementary angle is 115°, but using co‑interior pairs: if one interior angle is 65°, y = 180 - 65 = 115°.
平行线同旁内角互补，和为 180°。已知内错角 65°，则同旁内角 y = 180° - 65° = 115°。
9. Question 9: Probability Tree Diagrams | 第9题：概率树形图
Question: A bag contains 5 red and 3 blue counters. Two counters are taken at random without replacement. Draw a tree diagram and find the probability of taking at least one red counter.
题目： 袋中有 5 个红球和 3 个蓝球。随机取出两个球且不放回。画出树形图，并求至少取到一个红球的概率。
Probabilities: 1st red = 5/8, 1st blue = 3/8.
概率: 第一次红 = 5/8，蓝 = 3/8。
2nd red = 4/7, blue = 3/7. After a blue taken: remains 5R, 2B → 2nd red = 5/7, blue = 2/7.
取出一个红后: 剩余 4 红 3 蓝 → 第二次红 = 4/7，蓝 = 3/7。取出一个蓝后: 剩余 5 红 2 蓝 → 第二次红 = 5/7，蓝 = 2/7。
At least one red means not both blue. P(both blue) = (3/8) × (2/7) = 6/56 = 3/28.
至少一个红即不全是蓝。P(两蓝) = (3/8) × (2/7) = 6/56 = 3/28。
P(at least one red) = 1 - 3/28 = 25/28.
P(至少一红) = 1 - 3/28 = 25/28。
10. Question 10: Scatter Graphs and Line of Best Fit | 第10题：散点图与最佳拟合线
Question: A scatter graph shows test scores in Maths (x) and Physics (y) for 10 students. The mean point is (65, 70). The line of best fit passes through (50, 58) and the mean point. Estimate the Physics score for a student scoring 80 in Maths.
题目： 散点图展示了 10 名学生的数学分数 (x) 与物理分数 (y)。均值点为 (65, 70)。最佳拟合线经过点 (50, 58) 与均值点。估测一位数学 80 分的学生的物理分数。
Gradient m = (70 - 58) / (65 - 50) = 12 / 15 = 0.8.
斜率 m = (70 - 58) ÷ (65 - 50) = 12 / 15 = 0.8。
Equation: y - 70 = 0.8(x - 65). Simplify: y = 0.8x + 18.
直线方程: y - 70 = 0.8(x - 65)，化简得 y = 0.8x + 18。
For x = 80: y = 0.8 × 80 + 18 = 64 + 18 = 82. Estimate: Physics score 82.
当 x = 80: y = 0.8 × 80 + 18 = 64 + 18 = 82。估测物理分数为 82。
11. Question 11: Volume and Surface Area – Cylinder | 第11题：圆柱体的体积与表面积
Question: A solid cylinder has radius 3.5 cm and height 10 cm. Calculate its volume in terms of π, and its total surface area correct to 3 significant figures.
题目： 一个实心圆柱体底面半径 3.5 cm，高 10 cm。用 π 表示体积，并计算总表面积（保留 3 位有效数字）。
Volume V = πr²h = π × (3.5)² × 10 = π × 12.25 × 10 = 122.5π cm³.
体积 V = πr²h = π × 3.5² × 10 = 122.5π cm³。
circular ends = 2 × πr² = 2 × π × 12.25 = 24.5π. Total surface area = 70π + 24.5π = 94.5π.
侧面积 = 2πrh = 70π，上下底面积 = 2 × πr² = 24.5π，总表面积 = 94.5π。
Numerical value: 94.5 × 3.14159 ≈ 296.9 (to 3 s.f.: 297 cm²).
数值结果: 94.5 × 3.14159 ≈ 297 cm²（保留 3 位有效数字）。
12. Question 12: Trigonometric Ratios in Right‑Angled Triangles | 第12题：直角三角形中的三角比
Question: the length of QR (opposite to P) and PQ (adjacent to P). Give answers to 1 decimal place.
题目： 在三角形 PQR 中，∠Q = 90°，斜边 PR = 15 cm，∠P = 40°。求边 QR（∠P 的对边）和 PQ（邻边），结果保留 1 位小数。
opposite/hypotenuse = QR / 15 → QR = 15 × sin40° ≈ 15 × 0.6428 = 9.642 → 9.6 cm (1 d.p.).
利用 SOH CAH TOA。sin40° = 对边/斜边 = QR / 15 → QR = 15 × sin40° ≈ 9.6 cm。
adjacent/hypotenuse = PQ / 15 → PQ = 15 × cos40° ≈ 15 × 0.7660 = 11.49 → 11.5 cm (1 d.p.).
cos40° = 邻边/斜边 = PQ / 15 → PQ = 15 × cos40° ≈ 11.5 cm。
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