📚 Year 9 CIE Maths: Unit Mock Test Walkthrough | Year 9 CIE 数学：单元测试模拟卷解析
This article walks through a representative Year 9 CIE Mathematics unit mock test, covering number, algebra, geometry, measurement, and data handling. Each selected question is broken down with clear reasoning, common pitfalls, and bilingual explanations to strengthen your revision. Use this as a self‑check before your real assessment – the thinking process matters as much as the final answer.
本文解析一份典型的 Year 9 CIE 数学单元模拟卷，覆盖数、代数、几何、测量与数据处理五大板块。每个精选题目都配有清晰的推理过程、常见错误提示和双语讲解，帮助你巩固复习。在正式评估前，你可以将这份解析当作自检工具——思考过程和最终答案同样重要。
1. Rational Number Operations | 有理数运算
Question: Evaluate 3½ ÷ ( −1¼ ) + 2. Give your answer as a mixed number in simplest form.
题目：计算 3½ ÷ ( −1¼ ) + 2。用最简带分数表示答案。
cal: 7/2 ÷ ( −5/4 ) = 7/2 × ( −4/5 ) = −28/10 = −14/5. Then add 2, which is 10/5: −14/5 + 10/5 = −4/5. As a mixed number this is simply −4/5 (no whole part). Watch the sign – many forget that dividing by a negative flips the sign.
先把带分数化为假分数：3½ = 7/2，1¼ = 5/4，除数是 −5/4。除以一个分数等于乘以它的倒数：7/2 ÷ ( −5/4 ) = 7/2 × ( −4/5 ) = −28/10 = −14/5。然后加上 2（即 10/5）：−14/5 + 10/5 = −4/5。作为带分数就是 −4/5。注意符号——很多人忘记除以负数会改变符号。
2. Laws of Indices and Standard Form | 指数律与标准形式
Question: Simplify (2 × 10⁴) × (3 × 10⁻²). Give your answer in standard form.
题目：化简 (2 × 10⁴) × (3 × 10⁻²)。用标准形式表示结果。
s is already in standard form because the coefficient 6 is between 1 and 10. Common errors include adding exponents incorrectly or leaving the answer as 600, which is not standard form.
系数相乘：2 × 3 = 6。运用指数律：10⁴ × 10⁻² = 10⁴⁻² = 10²。所以乘积是 6 × 10²。这已经是标准形式，因为系数 6 在 1 到 10 之间。常见错误包括指数运算出错，或者答案写成 600 而不使用标准形式。
3. Solving Linear Equations | 解一元一次方程
Question: Solve 2(3x − 1) = 4x + 10.
题目：解方程 2(3x − 1) = 4x + 10。
Expand the left side: 6x − 2 = 4x + 10. Collect x‑ ht side 4×6+10=24+10=34. An easy mistake is mishandling the minus when expanding or moving terms.
展开左边：6x − 2 = 4x + 10。把含 x 的项移到一边：两边减去 4x 得 2x − 2 = 10。两边加 2：2x = 12，所以 x = 6。必须代入检验：左边 2(18−1)=2×17=34，右边 4×6+10=24+10=34。常见错误是展开或移项时处理负号不当。
4. Factorising and Expanding Quadratics | 二次式的因式分解与展开
Question: Expand and simplify (x + 5)(x − 3). Then factorise x² + x − 12.
题目：展开并化简 (x + 5)(x − 3)。再对 x² + x − 12 进行因式分解。
x − 15. For the factorisation, find two numbers that multiply to −12 and add to +1. Those are +4 and −3, so the factors are (x + 4)(x − 3). Reversing the logic shows the connection between expansion and factorisation – a key skill in algebra.
使用展开法则：首项 x·x = x²，外项 x·(−3) = −3x，内项 5·x = 5x，尾项 5·(−3) = −15。合并同类项：x² + 2x − 15。因式分解时，寻找两个数，乘积为 −12 且和为 +1，这两个数是 +4 和 −3，所以分解结果为 (x + 4)(x − 3)。逆向思考体现了展开与因式分解的联系——这是代数的核心技能。
5. Inequalities and the Number Line | 不等式与数轴表示
Question: Solve the inequality 5 − 2x > 3x + 10 and represent the solution on a number line.
题目：解不等式 5 − 2x > 3x + 10，并在数轴上表示解集。
: −5 > 5x. Divide by positive 5: −1 > x, which is the same as x < −1. On a number line draw an open circle at −1 and shade to the left. Remember: dividing or multiplying by a negative flips the inequality sign, but here we divided by a positive so no flip needed. However, when reading the final inequality, it's convenient to write the variable on the left: x < −1.
把含变量项移到一边：两边加 2x 得 5 > 5x + 10。减去 10：−5 > 5x。除以正数 5：−1 > x，也即 x < −1。在数轴上用空心圆表示 −1，并向左画阴影。记住：乘以或除以负数要反转不等号，但这里除以的是正数，所以不等号方向不变。最后习惯将变量写在左边：x < −1。
6. Sequences and the nth Term | 数列与第 n 项公式
Question: The first three terms of a linear sequence are 7, 11, 15. Find an expression for the nth term and hence find the 50th term.
题目：一个线性数列的前三项是 7, 11, 15。求第 n 项表达式，并由此计算第 50 项。
4n + 3. Check: n=1 → 7, n=2 → 11, n=3 → 15. For n=50, term = 4×50 + 3 = 203. A common mistake is mixing up n and n−1; always test your formula.
求公差：11 − 7 = 4，15 − 11 = 4，所以 d = 4。等差数列第 n 项公式为 a + (n−1)d，其中 a = 7。因此第 n 项 = 7 + (n−1)×4 = 7 + 4n − 4 = 4n + 3。检验：n=1 → 7，n=2 → 11，n=3 → 15。第 50 项为 4×50 + 3 = 203。常见错误是混淆 n 和 n−1，务必用已知项检验公式。
7. Angles in Polygons and Parallel Lines | 多边形内角与平行线
Question: A regular polygon has interior angle 140°. How many sides does it have? Two parallel lines are cut by a transversal; one co‑interior angle is 2x° and the other is (x + 30)°. Find x.
题目：一个正多边形的内角为 140°。它有几条边？两条平行线被一条横截线所截，一对同旁内角分别为 2x° 和 (x+30)°。求 x。
0°/n = 40° → n = 360/40 = 9. So the polygon is a nonagon (9 sides). For parallel lines, co‑ = 150 → x = 50. Geometry questions require both formula recall and clear angle reasoning.
内角 = 180° − 360°/n = 140° → 360°/n = 40° → n = 360/40 = 9，所以是九边形。对于平行线，同旁内角之和为 180°。因此 2x + (x+30) = 180 → 3x + 30 = 180 → 3x = 150 → x = 50。几何题需要既熟记公式，又能清晰推理角的关系。
8. Pythagoras and Right‑Angled Triangles | 勾股定理与直角三角形
Question: A ladder 5 m long leans against a vertical wall. The foot of the ladder is 1.4 m from the wall. How high up the wall does the ladder reach? Round to one decimal place.
题目：一把 5 米长的梯子靠在竖直墙上。梯脚离墙 1.4 米。梯子顶端能达到墙多高？结果保留一位小数。
Use Pythagoras' e hypotenuse. Let the height be h. Then 1.4² + h² = 5² → 1.96 + h² = 25 → h² = 23.04 → h = √23 So the ladder reaches 4.8 m up the wall. Watch the labelling: the ladder is the hypotenuse, not a leg.
运用勾股定理：a² + b² = c²，其中 c 为斜边。设高度为 h，则 1.4² + h² = 5² → 1.96 + h² = 25 → h² = 23.04 → h = √23.04 = 4.8（取正平方根）。梯子顶端达到 4.8 米高。注意标注：梯子本身是斜边，而不是直角边。
9. Area, Perimeter and Volume | 面积、周长与体积
Question: A compound shape consists of a rectangle 8 cm by 5 cm and a right‑angled triangle sitting on one 8 cm side, with height 3 cm. Find the total area. Later, the shape is revolved around the rectangle’s 5 cm side to form a cylinder with a cone on top. Calculate the total volume in terms of π.
题目：一个组合图形由一个 8 cm × 5 cm 的矩形和一个以 8 cm 边为底、高 3 cm 的直角三角形组成。求总面积。之后，该图形绕矩形 5 cm 边旋转，形成一个圆柱体顶部加一个圆锥。用 π 表示总体积。
he axis), height = 5 cm → volume = π × 8² × 5 = 320π. Cone radius = 8 cm, height = 3 cm → volume = ⅓ × π × 8² × 3 = 64π. Total volume = 384π cm³. Always visualise the rotation axis correctly; students often swap radius and height.
矩形面积 = 8 × 5 = 40 cm²。三角形面积 = ½ × 8 × 3 = 12 cm²，总面积 = 52 cm²。旋转时：圆柱半径 = 8 cm（垂直于旋转轴的边），高 = 5 cm → 体积 = π × 8² × 5 = 320π。圆锥半径 = 8 cm，高 = 3 cm → 体积 = ⅓ × π × 8² × 3 = 64π。总体积 = 384π cm³。务必正确设想旋转轴；学生常把半径和高度弄反。
10. Statistics: Mean, Median and Range | 统计：平均数、中位数与极差
Question: The marks of 7 students: 12, 15, 9, 18, 12, 10, 15. Find the mean, median and range. A new student joins with mark x. If the mean becomes 13, find x.
题目：7 名学生的分数：12, 15, 9, 18, 12, 10, 15。求平均数、中位数和极差。一名新同学加入，分数为 x，使平均数变为 13，求 x。
dents = 8, total sum = 8 × 13 = 104. Original sum was 91, so x = 104 − 91 = 13. Note: the median might change with the new value, but the question focuses on mean.
总和 = 12+15+9+18+12+10+15 = 91，平均数 = 91÷7 = 13。按大小排列：9, 10, 12, 12, 15, 15, 18；中位数 = 12。极差 = 18 − 9 = 9。加入新生后，总人数 8，总和为 8 × 13 = 104。原总和 91，所以 x = 104 − 91 = 13。注意：中位数可能因新值改变，但本题重点在平均数。
11. Probability with Fractions and Two‑Way Tables | 分数概率与双向表
Question: A bag contains 3 red, 2 blue and 5 green marbles. Two marbles are drawn without replacement. Find the probability that both are blue. Then, given a two‑way table of favourite subjects by gender, find P(Boy ∩ likes Maths).
题目：一个袋子里有 3 红、2 蓝和 5 绿弹珠。不放回抽取两颗。求两颗都是蓝色的概率。再根据一张性别与喜爱科目的双向表，求 P(男生 ∩ 喜欢数学)。
1 blue out of 9 total, P = 1/9. Multiply: 1/5 × 1/9 = 1/45. For the table, locate the cell "Boys who like Maths"; divide that frequency by the grand total. Two‑ minator (total of all entries). Combine probabilities carefully when events are dependent.
第一次抽取：总数 10，P(蓝) = 2/10 = 1/5。第二次，若第一次抽到蓝球：剩余 1 蓝，总数 9，P = 1/9。相乘：1/5 × 1/9 = 1/45。对于双向表，找到“喜欢数学的男生”对应的频数，除以总人数即可。只要确认正确的分母（所有调查对象总数），双向表概率计算很简单。注意区分独立与相关事件，正确相乘。
12. Ratio, Proportion and Real‑Life Graphs | 比率、比例与实际应用图像
Question: A recipe uses flour and sugar in the ratio 5:3. If 400 g of flour is used, how much sugar is needed? Then interpret a distance‑time graph: a car travels at constant speed for 2 hours covering 140 km, stops for 30 minutes, then returns at 70 km/h. Draw the graph and find the total time for the round trip.
题目：一份食谱中面粉和糖的比例为 5:3。如果用了 400 克面粉，需要多少糖？再解读一幅距离‑时间图：一辆汽车以恒定速度行驶 2 小时，行程 140 公里，停车 30 分钟，然后以 70 公里/小时返回。画出图像并求往返总时间。
d = 140 km ÷ 2 h = 70 km/h. Return distance is also 140 km, at 70 km/h, so return takes 2 hours. Total moving time = 2 h + 0.5 h + 2 h = 4.5 hours. Graph: rising line from (0,0) to (2,140), horizontal to (2.5,140), descending to (4.5,0). Ratio reasoning often simplifies by finding the value of one part.
比例 5:3 意味着每 5 份面粉对应 3 份糖。面粉 5 份为 400 g，所以 1 份 = 400 ÷ 5 = 80 g。糖需要 3 × 80 = 240 g。图像方面：去程速度 = 140 km ÷ 2 h = 70 km/h。返程也是 140 km，速度 70 km/h，所以用时 2 小时。行驶总时间 = 2 + 0.5 + 2 = 4.5 小时。图像：从 (0,0) 上升至 (2,140)，水平至 (2.5,140)，下降至 (4.5,0)。比例计算常可通过求出一份的量简化。
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