📚 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 covers a full unit test mock paper for Year 10 Edexcel Mathematics. We will work through ten typical questions ranging from algebra to geometry and probability, with detailed step-by-step solutions. Use this revision resource to sharpen your exam technique and build confidence before your real end-of-unit assessment.
本解析涵盖一份 Year 10 Edexcel 数学单元测试模拟卷，从代数到几何、概率共十道典型题目，配有详细的分步解答。利用这份复习资料打磨考试技巧，在真正的单元测验前树立信心。
1. Mock Paper Overview | 模拟卷概览
This mock paper is designed to mirror the style and difficulty of an Edexcel Year 10 unit test. It includes collecting like terms, solving equations, inequalities, sequences, straight line graphs, simultaneous equations, expanding and factorising, Pythagoras, and probability. The total marks would be around 50, to be completed in 45 minutes.
本模拟卷仿照 Edexcel Year 10 单元测试的风格和难度设计。内容包括合并同类项、解方程、不等式、数列、直线图像、联立方程组、展开与因式分解、毕达哥拉斯定理和概率。总分约 50 分，建议在 45 分钟内完成。
2. Q1: Collecting Like Terms | 问题1：合并同类项
Simplify 7a - 3b + 2a + 9b - 4a.
化简 7a - 3b + 2a + 9b - 4a。
Identify like terms: 7a, 2a and -4a are a-terms; -3b and 9b are b-terms.
识别同类项：7a、2a 和 -4a 是 a 项；-3b 和 9b 是 b 项。
Combine a-coefficients: 7 + 2 - 4 = 5, giving 5a.
合并 a 的系数：7 + 2 - 4 = 5，得到 5a。
Combine b-coefficients: -3 + 9 = 6, giving 6b.
合并 b 的系数：-3 + 9 = 6，得到 6b。
Final answer: 5a + 6b.
最终答案：5a + 6b。
3. Q2: Solving a Linear Equation | 问题2：解线性方程
Solve 4(x - 3) = 2x + 8.
解方程 4(x - 3) = 2x + 8。
First, expand the bracket: 4x - 12 = 2x + 8.
首先，展开括号：4x - 12 = 2x + 8。
Subtract 2x from both sides: 2x - 12 = 8.
方程两边同时减去 2x：2x - 12 = 8。
Add 12 to both sides: 2x = 20.
两边加 12：2x = 20。
Divide by 2: x = 10.
除以 2：x = 10。
Check: LHS = 4(10 - 3) = 4 × 7 = 28, RHS = 2×10 + 8 = 20 + 8 = 28.
检验：左边 = 4(10 - 3) = 4 × 7 = 28，右边 = 2×10 + 8 = 20 + 8 = 28。
4. Q3: Representing Inequalities | 问题3：不等式表示
Solve 3x - 5 < 7 and represent the solution on a number line.
解不等式 3x - 5 < 7，并在数轴上表示解集。
Add 5 to both sides: 3x < 12.
两边加 5：3x < 12。
Divide by 3: x < 4.
除以 3：x < 4。
On a number line, draw an open circle at 4 and shade to the left, showing all values less than 4.
在数轴上，在 4 处画空心圆圈并向左画阴影，表示所有小于 4 的值。
This solution can be written as {x : x < 4} or in interval notation (-∞, 4).
解集可以写成 {x : x < 4} 或用区间符号表示为 (-∞, 4)。
5. Q4: nth Term of an Arithmetic Sequence | 问题4：等差数列的第 n 项
Find the nth term of the sequence: 5, 9, 13, 17, ...
求数列 5, 9, 13, 17, ... 的第 n 项。
First, find the common difference d = 9 - 5 = 4.
先求公差 d = 9 - 5 = 4。
The nth term formula for an arithmetic sequence is: a + (n-1)d, where a is the first term.
等差数列的通项公式为：a + (n-1)d，其中 a 为首项。
Here a = 5, d = 4, so nth term = 5 + (n-1)×4.
这里 a = 5，d = 4，所以第 n 项 = 5 + (n-1)×4。
Simplify: 5 + 4n - 4 = 4n + 1.
化简：5 + 4n - 4 = 4n + 1。
Therefore, the nth term is 4n + 1.
因此，第 n 项是 4n + 1。
6. Q5: Equation of a Straight Line | 问题5：直线方程
gradient 3 and passes through the point (2, 5). Find its equation in the form y = mx + c.
一条直线的斜率为 3，且经过点 (2, 5)。求该直线以 y = mx + c 形式表示的方程。
Use y = mx + c with m = 3 and the point (2,5): 5 = 3 × 2 + c.
将 m = 3 与点 (2,5) 代入 y = mx + c：5 = 3 × 2 + c。
So 5 = 6 + c, giving c = -1.
即 5 = 6 + c，得 c = -1。
Equation: y = 3x - 1.
方程：y = 3x - 1。
7. Q6: Solving Simultaneous Equations | 问题6：解联立方程组
Solve the simultaneous equations: y = 2x + 1 and 3x + 2y = 16.
解联立方程组：y = 2x + 1 与 3x + 2y = 16。
Substitute y = 2x + 1 into the second equation: 3x + 2(2x + 1) = 16.
将 y = 2x + 1 代入第二个方程：3x + 2(2x + 1) = 16。
Expand: 3x + 4x + 2 = 16 → 7x + 2 = 16.
展开：3x + 4x + 2 = 16 → 7x + 2 = 16。
Subtract 2: 7x = 14, so x = 2.
减去 2：7x = 14，故 x = 2。
Substitute x = 2 into y = 2x + 1: y = 2×2 + 1 = 5.
将 x = 2 代入 y = 2x + 1：y = 2×2 + 1 = 5。
Solution: x = 2, y = 5.
解为：x = 2, y = 5。
8. Q7: Expanding Double Brackets | 问题7：展开双括号
Expand and simplify (x + 4)(x - 3).
展开并化简 (x + 4)(x - 3)。
Use the FOIL method: First, Outer, Inner, Last.
使用 FOIL 方法：首项、外项、内项、末项。
First: x × x = x². Outer: x × (-3) = -3x. Inner: 4 × x = 4x. Last: 4 × (-3) = -12.
首项：x × x = x²。外项：x × (-3) = -3x。内项：4 × x = 4x。末项：4 × (-3) = -12。
Combine like terms: x² - 3x + 4x - 12 = x² + x - 12.
合并同类项：x² - 3x + 4x - 12 = x² + x - 12。
Expanded form is x² + x - 12.
展开式为 x² + x - 12。
9. Q8: Factorising Quadratic Expressions | 问题8：因式分解二次表达式
Factorise x² + 7x + 12.
因式分解 x² + 7x + 12。
We need two numbers that multiply to +12 and add to +7. These numbers are 3 and 4.
我们需要两个数，乘积为 +12，和为 +7。这两个数是 3 和 4。
Write as (x + 3)(x + 4).
写成 (x + 3)(x + 4)。
Check by expanding: (x+3)(x+4) = x² + 4x + 3x + 12 = x² + 7x + 12.
通过展开检验：(x+3)(x+4) = x² + 4x + 3x + 12 = x² + 7x + 12。
Factorised form: (x + 3)(x + 4).
因式分解结果：(x + 3)(x + 4)。
10. Q9: Applying Pythagoras' Theorem | 问题9：应用毕达哥拉斯定理
A right-angled triangle has shorter sides of 6 cm and 8 cm. Find the length of the hypotenuse.
一个直角三角形的两条直角边分别为 6 cm 和 8 cm。求斜边长度。
Pythagoras' theorem: a² + b² = c², where c is the hypotenuse.
毕达哥拉斯定理：a² + b² = c²，其中 c 为斜边。
Substitute: 6² + 8² = c² → 36 + 64 = c² → 100 = c².
代入：6² + 8² = c² → 36 + 64 = c² → 100 = c²。
Take the square root: c = √100 = 10.
开平方：c = √100 = 10。
Hypotenuse is 10 cm.
斜边长度为 10 cm。
11. Q10: Probability Tree Diagrams | 问题10：概率树状图
A bag contains 3 red and 5 blue counters. Two counters are taken at random without replacement. Find the probability that both are red.
一个袋子里有 3 个红色和 5 个蓝色筹码。随机取出两个且不放回。求两个都是红色的概率。
= 3/8. After taking one red, 2 red remain out of 7 total, so probability second red = 2/7.
第一次取出红色的概率 = 3/8。取出一个红色后，剩下 2 个红色共 7 个，所以第二次红色的概率 = 2/7。
Multiply along the branch: P(both red) = 3/8 × 2/7 = 6/56.
沿着树枝相乘：P(两个红) = 3/8 × 2/7 = 6/56。
Simplify the fraction: 6/56 = 3/28.
化简分数：6/56 = 3/28。
Probability both are red is 3/28.
两个都是红色的概率为 3/28。
12. Exam Tips and Common Pitfalls | 考试技巧与常见陷阱
Always show all steps of your working – method marks are awarded even if the final answer is wrong. Double-check signs when collecting like terms; a common error is to mishandle negative coefficients. For inequalities, remember to flip the direction if you multiply or divide by a negative number. When factorising, verify your factors by expanding back. In probability, note whether the question is with or without replacement. Finally, manage your time: roughly 1 minute per mark, and leave a few minutes to check your answers.
务必写出所有解题步骤——即便最终答案有误，过程分也能得分。合并同类项时要再三检查符号；常见错误是处理负数系数不当。解不等式时，若乘以或除以负数，记得改变不等号方向。因式分解后要展开验证。在概率题中，看清是有放回还是无放回抽取。最后，合理分配时间：大约每分题一分钟，留出几分钟检查答案。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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