📚 Year 10 CAIE Maths: Unit Test Paper Walkthrough | 十年级CAIE数学单元测试模拟卷解析
Mock unit tests are an excellent way to consolidate your understanding and prepare for the real IGCSE Mathematics examination. This walkthrough presents a carefully compiled Year 10 CAIE-style test paper, covering all the major topics from algebra and geometry to statistics. Each question is broken down step by step, with reasoning explained in both English and Chinese. Use this resource to identify weak spots, strengthen problem-solving skills, and build confidence before your next class test.
模拟单元测试是巩固知识、备战 IGCSE 数学考试的绝佳方式。本文提供一份精心编制的十年级 CAIE 风格测试卷，涵盖代数、几何、统计等核心主题。每道题目均按步骤详细解析，并以中英双语阐释解题思路。利用这份材料排查薄弱环节、提升解题技巧，为下一次课堂测验积累信心。
1. Solving Linear Equations | 解线性方程
Example Question: Solve 3(x − 2) = 2x + 1.
例题： 解方程 3(x − 2) = 2x + 1。
Step 1: Expand the bracket on the left-hand side. Multiply 3 by each term inside: 3 × x = 3x and 3 × (−2) = −6, giving 3x − 6 = 2x + 1.
步骤1：展开左边括号。3 乘以括号内每一项：3 × x = 3x，3 × (−2) = −6，得到 3x − 6 = 2x + 1。
Step 2: Collect like terms. Subtract 2x from both sides to get x − 6 = 1.
步骤2：移项合并。两边同时减去 2x，得 x − 6 = 1。
Step 3: Isolate x by adding 6 to both sides: x = 7.
步骤3：两边加 6，求得 x = 7。
Check by substituting x = 7 into the original equation: LHS = 3(5) = 15, RHS = 14+1 = 15. The solution is correct.
将 x = 7 代入原方程检验：左边 = 3×5 = 15，右边 = 14+1 = 15，成立。
2. Factorising Quadratic Expressions | 因式分解二次表达式
Example Question: Factorise x² + 7x + 10.
例题： 因式分解 x² + 7x + 10。
We need two numbers that multiply to +10 and add up to +7. The pair 2 and 5 satisfy both conditions: 2 × 5 = 10 and 2 + 5 = 7.
我们需要两个数，乘积为 +10，和为 +7。数字 2 和 5 满足：2 × 5 = 10，2 + 5 = 7。
Thus the factorised form is (x + 2)(x + 5).
因此分解结果为 (x + 2)(x + 5)。
Always expand to verify: (x + 2)(x + 5) = x² + 5x + 2x + 10 = x² + 7x + 10, which matches the original expression.
务必展开验证：(x + 2)(x + 5) = x² + 5x + 2x + 10 = x² + 7x + 10，与原式一致。
3. Simultaneous Equations | 联立方程组
Example Question: Solve the system: 2x + y = 7, x − y = −1.
例题： 解方程组：2x + y = 7，x − y = −1。
Use the elimination method. Add the two equations directly: (2x + y) + (x − y) = 7 + (−1). The y terms cancel, leaving 3x = 6, so x = 2.
使用加减消元法。两式相加：(2x + y) + (x − y) = 7 + (−1)。y 项抵消，得 3x = 6，因此 x = 2。
Substitute x = 2 into the second equation: 2 − y = −1. Solve for y: −y = −3, so y = 3.
将 x = 2 代入第二个方程：2 − y = −1。解得 −y = −3，所以 y = 3。
The solution is x = 2, y = 3. Check in the first equation: 2(2) + 3 = 7, which holds.
方程组的解为 x = 2, y = 3。代入第一个方程检验：2×2 + 3 = 7，成立。
4. Percentages and Compound Interest | 百分比与复利
Example Question: Jack invests $800 in an account offering 4% compound interest per annum. How much will the investment be worth after 3 years?
例题： Jack 将 800 美元存入一个年利率为 4% 的复利账户。3 年后这笔投资的价值是多少？
The compound interest formula is A = P(1 + r/100)ⁿ, where P is principal, r is the annual interest rate, and n is the number of years.
A = P(1 + r/100)ⁿ
复利计算公式为 A = P(1 + r/100)ⁿ，其中 P 为本金，r 为年利率，n 为年数。
Here, P = 800, r = 4, n = 3. So A = 800 × (1 + 0.04)³ = 800 × (1.04)³.
此题中 P = 800，r = 4，n = 3。由此 A = 800 × (1 + 0.04)³ = 800 × (1.04)³。
Calculate (1.04)³ stepwise: 1.04² = 1.0816; then 1.0816 × 1.04 = 1.124864. Multiply by 800: A = 899.8912, which rounds to $899.89.
逐步计算 (1.04)³：1.04² = 1.0816；再乘以 1.04 得 1.124864。乘以 800：A = 899.8912，四舍五入为 899.89 美元。
5. Ratio and Proportion | 比例与比率
Example Question: Divide £360 between Alice and Bob in the ratio 2 : 3.
例题： 将 360 英镑按 2 : 3 的比例分给 Alice 和 Bob。
The total number of parts in the ratio is 2 + 3 = 5. Each part is worth £360 ÷ 5 = £72.
比例的总份数为 2 + 3 = 5。每一份的价值为 £360 ÷ 5 = £72。
Alice receives 2 parts: 2 × £72 = £144. Bob receives 3 parts: 3 × £72 = £216.
Alice 得到 2 份：2 × £72 = £144；Bob 得到 3 份：3 × £72 = £216。
Check that the sum is £144 + £216 = £360, and the ratio simplifies to 144:216 = 2:3 when dividing both by 72.
检验：总和为 £144 + £216 = £360，且 144:216 同时除以 72 化简为 2:3。
6. Angles in Polygons | 多边形中的角度
Example Question: Find the size of one interior angle of a regular pentagon.
例题： 求正五边形的一个内角大小。
The formula for the sum of interior angles of an n-sided polygon is (n − 2) × 180°. For a pentagon, n = 5, so the sum is (5 − 2) × 180° = 3 × 180° = 540°.
n 边形内角和公式为 (n − 2) × 180°。五边形 n = 5，内角和为 (5 − 2) × 180° = 3 × 180° = 540°。
Since the pentagon is regular, all 5 interior angles are equal. Divide the sum by 5: 540° ÷ 5 = 108°.
因为是正五边形，所有内角相等，将总和除以 5：540° ÷ 5 = 108°。
Each interior angle is 108°. Alternatively, you can use the exterior angle: 360° ÷ 5 = 72°, then interior angle = 180° − 72° = 108°.
每个内角为 108°。也可用外角方法：360° ÷ 5 = 72°，内角 = 180° − 72° = 108°。
7. Pythagoras’ Theorem | 毕达哥拉斯定理
Example Question: A right-angled triangle has legs of length 6 cm and 8 cm. Calculate the length of the hypotenuse.
例题： 一直角三角形的两条直角边分别为 6 cm 和 8 cm，求斜边长。
Pythagoras’ theorem states that in a right-angled triangle, c² = a² + b², where c is the hypotenuse.
c² = a² + b²
毕达哥拉斯定理：在直角三角形中，c² = a² + b²，c 为斜边。
Substitute a = 6 and b = 8: c² = 6² + 8² = 36 + 64 = 100.
代入 a = 6, b = 8：c² = 6² + 8² = 36 + 64 = 100。
Take the square root of both sides: c = √100 = 10 cm. The hypotenuse is 10 cm.
两边开平方：c = √100 = 10 cm。斜边长为 10 cm。
8. Basic Trigonometry | 基础三角学
Example Question: In a right-angled triangle, the side opposite angle θ is 5 cm and the hypotenuse is 13 cm. Find sin θ and the angle θ (to the nearest degree).
例题： 在直角三角形中，角 θ 的对边为 5 cm，斜边为 13 cm。求 sin θ 及角 θ（精确到度）。
Using SOH CAH TOA, sin θ = Opposite / Hypotenuse. So sin θ = 5/13.
根据 SOH CAH TOA，sin θ = 对边 / 斜边。因此 sin θ = 5/13。
Calculate the decimal: sin θ ≈ 0.3846. To find θ, use the inverse sine function: θ = sin⁻¹(0.3846).
计算小数：sin θ ≈ 0.3846。通过反正弦函数求角：θ = sin⁻¹(0.3846)。
Using a calculator, θ ≈ 22.6°, which rounds to 23° (to the nearest degree).
用计算器得 θ ≈ 22.6°，四舍五入为 23°。
9. Data Handling: Mean, Median, Mode | 数据处理：平均数、中位数、众数
Example Question: The scores of eight students in a test are: 7, 9, 5, 7, 6, 10, 7, 5. Find the mean, median and mode.
例题： 八名学生的测验成绩为：7, 9, 5, 7, 6, 10, 7, 5。求平均数、中位数和众数。
First, list the data in ascending order: 5, 5, 6, 7, 7, 7, 9, 10.
首先将数据从小到大排列：5, 5, 6, 7, 7, 7, 9, 10。
Mean: Sum = 5+5+6+7+7+7+9+10 = 56. There are 8 values. Mean = 56 ÷ 8 = 7.
平均数： 总和 = 5+5+6+7+7+7+9+10 = 56。共 8 个值，平均数 = 56 ÷ 8 = 7。
Median: With an even number of values, the median is the average of the 4th and 5th values. The 4th is 7 and the 5th is 7, so median = 7.
中位数： 数据个数为偶数，中位数为第 4 和第 5 个数值的平均数。第 4 个为 7，第 5 个为 7，因此中位数 = 7。
Mode: The value that appears most often. Here, 7 appears three times, so the mode is 7.
众数： 出现次数最多的数据。此处 7 出现三次，众数为 7。
10. Probability | 概率
Example Question: A fair six-sided die is rolled once. What is the probability of rolling a number greater than 4?
例题： 掷一枚均匀的六面骰子一次。掷出点数大于 4 的概率是多少？
Probability = (Number of favourable outcomes) / (Total number of possible outcomes). The possible outcomes are {1, 2, 3, 4, 5, 6}, total 6.
概率 = 有利结果数 / 所有可能结果数。可能结果为 {1, 2, 3, 4, 5, 6}，共 6 种。
Numbers greater than 4 are 5 and 6, so there are 2 favourable outcomes.
点数大于 4 的有 5 和 6，所以有利结果数为 2。
Thus, P(greater than 4) = 2/6 = 1/3. The probability is 1/3.
因此，P(大于 4) = 2/6 = 1/3。概率为 1/3。
11. Sequences | 数列
Example Question: The first four terms of a sequence are 5, 8, 11, 14. Find an expression for the nth term and the 50th term.
例题： 某数列前四项为 5, 8, 11, 14。求第 n 项表达式及第 50 项。
First, check the common difference: 8 − 5 = 3, 11 − 8 = 3, 14 − 11 = 3. This is an arithmetic sequence with first term a = 5 and common difference d = 3.
首先检查公差：8−5=3，11−8=3，14−11=3。这是一个等差数列，首项 a = 5，公差 d = 3。
The formula for the nth term of an arithmetic sequence is: aₙ = a + (n − 1)d.
aₙ = a + (n − 1)d
等差数列第 n 项公式为：aₙ = a + (n − 1)d。
Substitute: aₙ = 5 + (n − 1)×3 = 5 + 3n − 3 = 3n + 2.
代入：aₙ = 5 + (n − 1)×3 = 5 + 3n − 3 = 3n + 2。
To find the 50th term, let n = 50: a₅₀ = 3(50) + 2 = 150 + 2 = 152.
求第 50 项，令 n = 50：a₅₀ = 3×50 + 2 = 150 + 2 = 152。
12. Graphs of Functions | 函数图像
Example Question: Sketch the graph of y = x² − 4x + 3 and label the coordinates of the turning point.
例题： 画出 y = x² − 4x + 3 的图像，并标出顶点坐标。
First, find where the curve crosses the x-axis by setting y = 0: x² − 4x + 3 = 0. Factorise to (x − 1)(x − 3) = 0, giving x = 1 and x = 3. The x-intercepts are (1,0) and (3,0).
首先，令 y = 0 求与 x 轴交点：x² − 4x + 3 = 0。因式分解得 (x − 1)(x − 3) = 0，解得 x = 1 和 x = 3。x 轴截点为 (1,0) 和 (3,0)。
The y-intercept is found by setting x = 0: y = 0² − 0 + 3 = 3, so the point is (0,3).
令 x = 0 求 y 轴截点：y = 0² − 0 + 3 = 3，即点 (0,3)。
To locate the turning point, use the symmetry of the parabola. The x-coordinate of the vertex lies midway between the roots: x = (1+3)/2 = 2.
利用抛物线对称性求顶点。顶点 x 坐标在两根的中点：x = (1+3)/2 = 2。
Substitute x = 2 into the equation: y = 2² − 4(2) + 3 = 4 − 8 + 3 = −1. The turning point is (2, −1). Since the coefficient of x² is positive, it is a minimum point.
将 x = 2 代入方程：y = 2² − 4×2 + 3 = 4 − 8 + 3 = −1。顶点为 (2, −1)。由于 x² 系数为正，此为最小值点。
The graph is a U-shaped parabola passing through (0,3), (1,0), (3,0) with a vertex at (2, −1).
图像为开口向上的 U 形抛物线，经过 (0,3), (1,0), (3,0)，顶点在 (2, −1)。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply