📚 Year 10 CIE Maths: Unit Test Mock Paper Analysis | 10年级 CIE 数学：单元测试模拟卷解析
This article walks you through a Year 10 CIE Mathematics unit test mock paper, covering algebra, geometry, statistics and more. Each question is analysed step by step, showing the correct method, common pitfalls and revision strategies. Use this as a guided review to strengthen your understanding and boost your confidence before the real assessment.
本文带你逐题解析一份10年级 CIE 数学单元测试模拟卷，内容涵盖代数、几何、统计等领域。每道题目都给出了详细的解题步骤、常见错误分析和复习策略。你可以把这篇解析作为一次有指导的复习，用来巩固知识、提升信心，为真正的评估做好准备。
1. Mock Paper Overview | 模拟卷概览
This mock paper consists of 10 questions designed to reflect the style and difficulty of a typical CIE IGCSE core/extended unit test. Topics include linear equations, quadratics, inequalities, graphs, simultaneous equations, Pythagoras, trigonometry, statistics, probability and compound measures. Most questions are worth 3 to 6 marks, and working must be shown clearly.
这份模拟卷包含10道题目，难度和风格都与典型的 CIE IGCSE 核心/拓展单元测试一致。考查内容涉及线性方程、二次表达式、不等式、直线图像、联立方程、勾股定理、三角函数、统计量、概率以及复合度量。大部分题目分值为3至6分，解题过程必须书写清晰。
The full paper is designed to be completed in 45 minutes. A calculator is allowed for some questions, but you should be prepared to work without it for algebraic manipulation and exact values. The mark scheme values method marks, so even if you make a numerical slip, you can still gain credit.
全卷设计完成时间为45分钟。部分题目允许使用计算器，但在代数变形和精确值的计算中，你仍需具备笔算能力。评分方案重视方法分，因此即使出现数值错误，你仍然可以获得相应步骤的分数。
2. Question 1: Solving a Linear Equation | 题目1：解线性方程
Problem: Solve 4(2x - 3) + 5 = 3x + 7.
问题：解方程 4(2x - 3) + 5 = 3x + 7。
ep 1: Expand the brackets. 4 × 2x = 8x and 4 × (-3) = -12, so we get 8x - 12 + 5 = 3x + 7.
步骤1：展开括号。4 × 2x = 8x，4 × (-3) = -12，因此得到 8x - 12 + 5 = 3x + 7。
Step 2: Simplify the left-hand side. -12 + 5 = -7, which gives 8x - 7 = 3x + 7.
步骤2：化简左边。-12 + 5 = -7，得到 8x - 7 = 3x + 7。
Step 3: Collect like terms. Subtract 3x from both sides: 5x - 7 = 7.
步骤3：移项合并。两边同时减去 3x：5x - 7 = 7。
Step 4: Add 7 to both sides: 5x = 14.
步骤4：两边加7：5x = 14。
Step 5: Divide by 5: x = 14/5 or 2.8.
步骤5：除以5：x = 14/5，即 2.8。
= 4×2.6 + 5 = 10.4 + 5 = 15.4; RHS = 3×2.8 + 7 = 8.4 + 7 = 15.4. The solution is correct.
检验：左边 = 4(2×2.8 - 3) + 5 = 15.4；右边 = 3×2.8 + 7 = 15.4，答案正确。
Common mistake: forgetting to distribute the 4 to both terms inside the bracket. Always multiply each term separately.
常见错误：忘记将括号外的 4 与括号内的每一项相乘。务必对每一项分别进行乘法运算。
3. Question 2: Factorising a Quadratic Expression | 题目2：二次三项式因式分解
Problem: Factorise x² + 8x + 15 completely.
问题：将 x² + 8x + 15 彻底因式分解。
We need two numbers that multiply to 15 and add to 8. The pairs of factors for 15 are (1,15) and (3,5). Only 3 and 5 sum to 8.
我们需要找到两个数，乘积为15，且和为8。15的因数对有 (1,15) 和 (3,5)。只有 3 和 5 的和为8。
Therefore, x² + 8x + 15 = (x + 3)(x + 5).
因此，x² + 8x + 15 = (x + 3)(x + 5)。
and: (x+3)(x+5) = x² + 5x + 3x + 15 = x² + 8x + 15, which matches the original expression.
验证：展开 (x+3)(x+5) = x² + 5x + 3x + 15 = x² + 8x + 15，与原式相符。
Note: If the constant term is positive and the coefficient of x is negative, you would need two negative numbers. If the constant is negative, the factors have opposite signs.
注意：当常数项为正而 x 的系数为负时，需要寻找两个负数；当常数项为负时，两个因数异号。
4. Question 3: Solving a Linear Inequality | 题目3：解一元一次不等式
Problem: Solve the inequality 3x - 2 ≤ 4x + 3 and represent the solution on a number line.
问题：解不等式 3x - 2 ≤ 4x + 3，并在数轴上表示解集。
Step 1: Subtract 3x from both sides to gather the variable on one side: -2 ≤ x + 3.
步骤1：两边同时减去 3x，将未知数移到一边：-2 ≤ x + 3。
Step 2: Subtract 3 from both sides: -5 ≤ x. This can also be written as x ≥ -5.
步骤2：两边再减去 3：-5 ≤ x。也可以写作 x ≥ -5。
The solution set includes all numbers greater than or equal to -5. On a number line, we mark a closed circle at -5 and shade to the right.
解集包括所有大于或等于 -5 的数。在数轴上，我们在 -5 处画一个实心圆点，并向右侧画阴影线。
Important: When multiplying or dividing by a negative number, the inequality sign must be reversed. In this problem, we did not need to flip the sign because we subtracted only.
重要提醒：当乘以或除以一个负数时，不等号方向必须改变。本题仅进行了减法运算，因此不等号方向不变。
5. Question 4: Straight Line Graphs | 题目4：直线图像
ite down the gradient and the y-intercept. Then find the coordinates of the point where the line crosses the x-axis.
问题：对于方程 y = 2x + 3，写出它的斜率和 y 轴截距。然后求出该直线与 x 轴交点的坐标。
where m is the gradient and c is the y-intercept. So m = 2, and the y-intercept is (0, 3).
方程形如 y = mx + c，其中 m 为斜率，c 为 y 轴截距。因此斜率 m = 2，y 轴截距为 (0, 3)。
the x-intercept, set y = 0: 0 = 2x + 3 → 2x = -3 → x = -3/2. The x-intercept is (-1.5, 0).
要求 x 轴截距，令 y = 0：0 = 2x + 3 → 2x = -3 → x = -3/2。因此 x 轴截距为 (-1.5, 0)。
Sketching tip: Plot the y-intercept (0,3) first. From that point, use the gradient 'rise over run' of 2 (up 2 units for every 1 unit right) to find a second point, then draw the line.
绘图技巧：先标出 y 轴截距 (0,3)。从该点出发，利用斜率 2（每向右移动1个单位，向上移动2个单位）找到第二个点，然后画出直线。
6. Question 5: Simultaneous Equations | 题目5：联立方程
Problem: Solve the simultaneous equations: 2x + y = 10 and x - y = 2.
问题：解联立方程组：2x + y = 10，x - y = 2。
elimination method, add the two equations: (2x + y) + (x - y) = 10 + 2 → 3x = 12 → x = 4.
用消元法，将两式相加：(2x + y) + (x - y) = 10 + 2 → 3x = 12 → x = 4。
Substitute x = 4 into the second equation: 4 - y = 2 → y = 2.
将 x = 4 代入第二个方程：4 - y = 2 → y = 2。
Check with the first equation: 2(4) + 2 = 8 + 2 = 10. The solution is x = 4, y = 2.
用第一个方程检验：2(4) + 2 = 8 + 2 = 10。解为 x = 4，y = 2。
If the coefficients do not easily cancel, multiply one or both equations before adding. For substitution method, rearrange one equation to make a variable the subject first.
如果系数不能直接抵消，可先将一个或两个方程乘以适当倍数再进行加减。使用代入法时，需要先把一个方程变形，用其中一个未知数表示另一个。
7. Question 6: Pythagoras’ Theorem | 题目6：勾股定理
Problem: A right-angled triangle has legs of length 6 cm and 8 cm. Find the length of the hypotenuse.
问题：一个直角三角形两条直角边长分别为 6 cm 和 8 cm，求斜边的长度。
ythagoras’ theorem states that a² + b² = c², where c is the hypotenuse. Here a = 6, b = 8.
勾股定理指出 a² + b² = c²，其中 c 为斜边。此处 a = 6，b = 8。
Calculate: 6² + 8² = 36 + 64 = 100. Then c = √100 = 10 cm.
计算：6² + 8² = 36 + 64 = 100。因此 c = √100 = 10 cm。
This is a classic 3-4-5 Pythagorean triple scaled by factor 2. Recognising such triples can save time in exams.
这是一个经典的3-4-5勾股数乘以因子2得到的。识别这类勾股数可以在考试中节省时间。
ge to a² = c² - b² before square rooting. Never forget to take the square root at the end.
如果题目给出斜边和一条直角边，则需先用 a² = c² - b² 变形再开平方。切勿忘记最后一步要开平方根。
8. Question 7: Trigonometry in Right-Angled Triangles | 题目7：直角三角形中的三角比
Problem: In a right-angled triangle, the side opposite angle θ is 5 cm and the adjacent side is 12 cm. Find the size of θ to 1 decimal place.
问题：在一个直角三角形中，角 θ 的对边长为 5 cm，邻边长为 12 cm。求 θ 的大小，精确到小数点后1位。
We use the tangent ratio: tan θ = opposite / adjacent = 5/12.
我们使用正切比：tan θ = 对边 / 邻边 = 5/12。
Therefore θ = tan⁻¹ (5/12). Using a calculator, θ ≈ 22.6° (to 1 d.p.).
因此 θ = tan⁻¹ (5/12)。用计算器求得 θ ≈ 22.6°（精确到1位小数）。
Always check whether the question asks for degrees or radians. In CIE IGCSE, angles are typically given in degrees. Label the sides (opposite, adjacent, hypotenuse) carefully relative to the angle you are working with.
注意确认题目要求使用角度制还是弧度制。在 CIE IGCSE 中，角度通常以度为单位。解题时务必相对于目标角正确标注对边、邻边和斜边。
9. Question 8: Statistics – Mean and Median | 题目8：统计量 – 平均数与中位数
Problem: Here are the scores of 7 students: 14, 18, 12, 15, 12, 20, 14. Calculate the mean and the median.
问题：7名学生的成绩如下：14, 18, 12, 15, 12, 20, 14。计算平均数和中位数。
+12+20+14 = 105. Divide by the number of students (7): 105 ÷ 7 = 15. The mean score is 15.
求平均数：先将数值相加：14+18+12+15+12+20+14 = 105。再除以学生人数7：105 ÷ 7 = 15。平均成绩为15。
For the median, arrange the data in ascending order: 12, 12, 14, 14, 15, 18, 20. The middle value is the 4th term, which is 14. Thus the median is 14.
求中位数：将数据按升序排列：12, 12, 14, 14, 15, 18, 20。中间的数是第4项，即14。因此中位数为14。
If there is an even number of data points, take the mean of the two middle values. Always reorder the list; a common mistake is to pick the middle of the unsorted data.
如果数据个数为偶数，则取中间两个数的平均数。务必先将数据排序，常见错误是直接在未排序的列表中选取中间值。
10. Question 9: Probability – Tree Diagram | 题目9：概率 – 树状图
Problem: A bag contains 4 red balls and 3 blue balls. Two balls are drawn at random without replacement. Find the probability that both balls are red.
问题：一个袋子里有4个红球和3个蓝球。随机从袋中不重复地抽取两个球。求抽到两个球都是红色的概率。
Total balls: 4 + 3 = 7. Probability first ball is red = 4/7.
总球数：4 + 3 = 7。第一个球是红色的概率为 4/7。
3 red left out of 6 balls. Probability second ball is red given first is red = 3/6 = 1/2.
在抽走一个红球后，袋中剩下3个红球，总共6个球。因此在第一个是红球的条件下，第二个是红球的概率为 3/6 = 1/2。
Multiply along the branch: P(both red) = (4/7) × (1/2) = 4/14 = 2/7.
沿树状图分支相乘：P(两个红球) = (4/7) × (1/2) = 4/14 = 2/7。
Drawing a tree diagram clearly shows the without-replacement situation: the second probabilities change. Remember to label branches and write final probability as a simplified fraction.
画出树状图可以清晰地展示不重复抽取的情况：第二次的概率会发生变化。记得标注每个分支，并将最终概率化为最简分数。
11. Question 10: Compound Measures – Speed | 题目10：复合度量 – 速度
Problem: A cyclist travels 45 km in 1 hour 15 minutes. Calculate the average speed in km/h and in m/s.
问题：一位自行车手在1小时15分钟内行驶了45 km。计算平均速度，以 km/h 和 m/s 为单位。
Convert time to hours: 1 hour 15 minutes = 1 + 15/60 = 1.25 hours.
将时间转换为小时：1小时15分钟 = 1 + 15/60 = 1.25 小时。
Speed in km/h = distance / time = 45 km / 1.25 h = 36 km/h.
速度 (km/h) = 路程 / 时间 = 45 km / 1.25 h = 36 km/h。
s in a km) and divide by 3600 (seconds in an hour): 36 × 1000 / 3600 = 36 × 5/18 = 10 m/s.
要将 km/h 转化为 m/s，需乘以1000（每千米的米数）再除以3600（每小时秒数）：36 × 1000 / 3600 = 36 × 5/18 = 10 m/s。
ch confirms the result. Always pay attention to unit conversions, especially when minutes or other units are involved.
利用换算关系 1 m/s = 3.6 km/h 进行验证：10 m/s × 3.6 = 36 km/h，结果一致。要特别注意单位换算，尤其是当题目中涉及分钟或其他单位时。
12. Key Takeaways and Revision Tips | 关键要点与复习建议
After working through this mock paper, reflect on your errors. Did you misapply a sign? Forget to square a number? Or perhaps you did not read the question carefully? Targeting these small mistakes can lift your grade significantly.
完成这份模拟卷后，请反思自己的错误。是不是符号弄错了？忘记给某个数字平方？还是审题不够仔细？针对这些小失误进行改进，可以显著提升成绩。
For algebra topics, master expanding, factorising and rearranging. In geometry, practise drawing clear diagrams and labelling sides. For data handling, focus on ordering data and understanding 'without replacement' scenarios.
在代数部分，要熟练掌握展开、因式分解和变形技巧；几何部分，养成画清晰示意图并标注边长的习惯；数据处理部分，请重点关注数据排序以及对“不重复抽取”场景的理解。
Regular timed practice with mock papers helps you manage exam pressure. Always show your working; even if the final answer is wrong, you can earn method marks. When stuck, read the question several times and underline key values.
定期进行限时模拟训练有助于应对考试压力。一定要写出解题过程；即使最终答案错误，你仍可获得步骤分。遇到难题时，反复读题并圈出关键数值。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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