📚 Mock Unit Test Analysis for Year 9 CAIE Mathematics | Year 9 CAIE 数学单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test designed for the Year 9 CAIE Mathematics syllabus. Each section presents a typical exam-style question, followed by a step-by-step solution, common pitfalls and revision tips. Working through these problems will strengthen your understanding of number properties, algebra, geometry and data handling, and help you prepare effectively for your real assessment.
本文为 Year 9 CAIE 数学大纲设计了一份模拟单元测试，并逐题进行详细解析。每个小节先给出一道典型考题，然后逐步展示解题过程，同时指出常见错误和复习要点。通过完成这些题目，你可以巩固对数的性质、代数、几何和统计等内容的理解，为真正的测验做好充分准备。
1. Prime Factorisation, HCF and LCM | 质因数分解与最大公因数/最小公倍数
Write 48 and 60 as products of prime factors. Hence find the highest common factor (HCF) and lowest common multiple (LCM) of 48 and 60.
将 48 和 60 写成质因数的乘积。由此求出 48 和 60 的最大公因数 (HCF) 和最小公倍数 (LCM)。
Start by drawing factor trees. For 48, divide by the smallest prime repeatedly: 48 ÷ 2 = 24, 24 ÷ 2 = 12, 12 ÷ 2 = 6, 6 ÷ 2 = 3. So 48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3. For 60, divide: 60 ÷ 2 = 30, 30 ÷ 2 = 15, 15 ÷ 3 = 5. Therefore 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5. To find the HCF, take the lowest power of each common prime factor: common primes are 2 and 3. Lowest power of 2 is 2², lowest power of 3 is 3¹. So HCF = 2² × 3 = 4 × 3 = 12. To find the LCM, take the highest power of all primes present: 2⁴, 3¹ and 5¹. So LCM = 2⁴ × 3 × 5 = 16 × 15 = 240. Always check that HCF × LCM = 48 × 60 = 2880, and 12 × 240 = 2880.
先画因数树。对于 48，不断用最小的质数去分解：48 ÷ 2 = 24，24 ÷ 2 = 12，12 ÷ 2 = 6，6 ÷ 2 = 3。所以 48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3。对于 60：60 ÷ 2 = 30，30 ÷ 2 = 15，15 ÷ 3 = 5。因此 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5。求 HCF 时，取每个公共质因数的最低次幂：公共质因数是 2 和 3，2 的最低次幂是 2²，3 的最低次幂是 3¹。HCF = 2² × 3 = 12。求 LCM 时，取所有出现质因数的最高次幂：2⁴、3¹ 和 5¹。LCM = 2⁴ × 3 × 5 = 16 × 15 = 240。一定要检查 HCF × LCM 是否等于 48 × 60 = 2880，而 12 × 240 = 2880，结果一致。
2. Operations with Fractions | 分数的运算
Evaluate 3/4 + 5/6 – 1/3. Give your answer as a fraction in its simplest form.
计算 3/4 + 5/6 – 1/3，将答案写成最简分数。
Find a common denominator for 4, 6 and 3. The LCM is 12. Rewrite each fraction: 3/4 = 9/12, 5/6 = 10/12, 1/3 = 4/12. Now perform the operations: 9/12 + 10/12 = 19/12, then 19/12 – 4/12 = 15/12. Simplify by dividing numerator and denominator by their HCF, 3: 15 ÷ 3 = 5, 12 ÷ 3 = 4, giving 5/4 or 1 1/4. You may write it as an improper fraction or mixed number; both are acceptable. A common mistake is to forget to convert all fractions to the same denominator before adding or subtracting, or to simplify only at the very end.
找出 4、6 和 3 的公分母。它们的最小公倍数是 12。把每个分数改写为以 12 为分母的分数：3/4 = 9/12，5/6 = 10/12，1/3 = 4/12。然后按顺序计算：9/12 + 10/12 = 19/12，19/12 – 4/12 = 15/12。再化简：分子分母同时除以它们的最大公因数 3，得到 5/4 或 1 1/4。可以写成假分数或带分数，两者均可。常见错误是在加减前忘记通分，或是只在最后才想起来约分。
3. Percentage Increase and Decrease | 百分数的增减
The original price of a jacket is £80. In a sale, the price is reduced by 15%. The following week, the sale price is increased by 10%. Calculate the final price of the jacket.
一件夹克的原价为 £80。促销期间降价 15%，接下来的一周，促销价又上涨 10%。求夹克的最终价格。
First, find the sale price after the 15% decrease. 15% of £80 = 0.15 × 80 = £12, so the reduced price is 80 – 12 = £68. Alternatively, use a multiplier: a 15% decrease means 100% – 15% = 85%, multiplier 0.85. £80 × 0.85 = £68. Next, increase £68 by 10%. 10% of £68 = 6.80, so the new price is 68 + 6.80 = £74.80. The multiplier method: 10% increase means 100% + 10% = 110%, multiplier 1.10. £68 × 1.10 = £74.80. Many students mistakenly apply the second percentage to the original price instead of the new price, so always read the question carefully.
首先计算 15% 降价后的促销价。£80 的 15% 是 0.15 × 80 = £12，降价后的价格为 80 – 12 = £68。也可以使用乘数：降价 15% 意味着剩下 100% – 15% = 85%，乘数为 0.85。£80 × 0.85 = £68。接下来，£68 再上涨 10%。£68 的 10% 是 6.80，所以新价格为 68 + 6.80 = £74.80。乘数法：上涨 10% 即 100% + 10% = 110%，乘数为 1.10。£68 × 1.10 = £74.80。很多学生容易错误地将第二次百分比作用在原价上，而不是作用在第一次变动后的价格上，因此务必仔细读题。
4. Simplifying Algebraic Expressions | 代数式的化简
Simplify: (a) 2x + 3y – x + 5y (b) 3(2x – 4) + 2(x + 1)
化简：(a) 2x + 3y – x + 5y (b) 3(2x – 4) + 2(x + 1)
For part (a), collect like terms. The terms in x are 2x and -x, which combine to x. The terms in y are 3y and 5y, which give 8y. So the expression simplifies to x + 8y. For part (b), first expand the brackets using the distributive law. 3(2x – 4) = 6x – 12. 2(x + 1) = 2x + 2. Then collect like terms: 6x + 2x = 8x, and -12 + 2 = -10. The simplified expression is 8x – 10. Always be careful with the signs when expanding, especially when a negative sign is outside the bracket.
对于 (a) 部分，合并同类项。包含 x 的项是 2x 和 -x，合并为 x。包含 y 的项是 3y 和 5y，合并为 8y。因此代数式化简为 x + 8y。对于 (b) 部分，首先利用分配律展开括号。3(2x – 4) = 6x – 12。2(x + 1) = 2x + 2。接着合并同类项：6x + 2x = 8x，-12 + 2 = -10。化简后的式子为 8x – 10。展开时务必注意符号，尤其当括号前是负号时更要小心。
5. Solving Linear Equations | 解一次方程
Solve the equation 5x – 7 = 3x + 9.
解方程 5x – 7 = 3x + 9。
To isolate x, collect the x terms on one side and the constant terms on the other. Subtract 3x from both sides: 5x – 3x – 7 = 3x – 3x + 9, which gives 2x – 7 = 9. Then add 7 to both sides: 2x = 16. Finally, divide both sides by 2: x = 8. You should check the solution by substituting x = 8 back into the original equation. Left side: 5(8) – 7 = 40 – 7 = 33. Right side: 3(8) + 9 = 24 + 9 = 33. Both sides match, so the solution is correct. A frequent error is to move terms without changing their sign, e.g. writing 5x + 3x instead of subtracting.
为了把 x 分离出来，先把含 x 的项移到一边，常数项移到另一边。方程两边同时减去 3x：5x – 3x – 7 = 3x – 3x + 9，得到 2x – 7 = 9。然后两边同时加 7：2x = 16。最后两边同时除以 2：x = 8。应把 x = 8 代入原方程检验。左边：5(8) – 7 = 40 – 7 = 33。右边：3(8) + 9 = 24 + 9 = 33。两边相等，所以解正确。常见的错误是在移项时忘记变号，例如写成 5x + 3x 而不是相减。
6. Coordinates and Midpoint/Distance | 坐标、中点与距离
Point A is (2, 5) and point B is (6, 9). Find: (a) the coordinates of the midpoint of AB, (b) the length of AB, leaving your answer in simplified surd form.
点 A 的坐标为 (2, 5)，点 B 的坐标为 (6, 9)。求：(a) 线段 AB 的中点坐标，(b) AB 的长度，答案保留最简根式。
For part (a), use the midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2). Substitute: ((2 + 6)/2, (5 + 9)/2) = (8/2, 14/2) = (4, 7). So the midpoint is (4, 7). For part (b), apply the distance formula: √[(x₂ – x₁)² + (y₂ – y₁)²]. Substitute: √[(6 – 2)² + (9 – 5)²] = √(4² + 4²) = √(16 + 16) = √32. Simplify √32: 32 = 16 × 2, so √32 = √16 × √2 = 4√2. The length of AB is 4√2 units. Remember that the difference in x and the difference in y are both positive once squared.
对于 (a)，使用中点公式：((x₁ + x₂)/2, (y₁ + y₂)/2)。代入：((2 + 6)/2, (5 + 9)/2) = (8/2, 14/2) = (4, 7)。中点为 (4, 7)。对于 (b)，应用距离公式：√[(x₂ – x₁)² + (y₂ – y₁)²]。代入：√[(6 – 2)² + (9 – 5)²] = √(4² + 4²) = √(16 + 16) = √32。化简 √32：32 = 16 × 2，√32 = √16 × √2 = 4√2。AB 的长度为 4√2 单位。记住两个坐标的差值在平方后都是正值。
AB = √[(6−2)² + (9−5)²] = √(4² + 4²) = √32 = 4√2
7. Angles in Parallel Lines | 平行线中的角
In the diagram, two parallel lines are cut by a transversal. One angle is labelled 70°. State the sizes of the alternate angle, the corresponding angle and the interior angle on the same side of the transversal. (Assume the given 70° is an acute angle between one parallel line and the transversal.)
在图示中，两条平行线被一条横截线所截。已知其中一个角为 70°。说出它的内错角、同位角以及同旁内角的度数。（假设已知的 70° 是一条平行线与横截线之间的锐角。）
When a transversal intersects parallel lines, several angle relationships are always true. The alternate angle is equal to the given angle, so the alternate interior angle is 70°. The corresponding angle is also equal, so the corresponding angle is 70°. The interior angles on the same side of the transversal are supplementary, meaning they add up to 180°. Therefore the interior angle on the same side is 180° – 70° = 110°. If the given angle were an exterior angle, the relationships still hold for corresponding and alternate pairs, but always identify the position carefully in a diagram.
当一条横截线与两条平行线相交时，存在一些恒成立的角关系。内错角与已知角相等，因此内错角为 70°。同位角也相等，所以同位角也为 70°。同旁内角互补，即两角之和为 180°，因此同旁内角为 180° – 70° = 110°。如果已知角是外角，同位角和内错角的关系仍然成立，但在看图时务必仔细辨认每个角的位置。
8. Perimeter, Area and Problem Solving | 周长、面积与问题解决
A rectangle has a perimeter of 30 cm. Its length is 3 cm longer than its width. Find the area of the rectangle.
一个长方形的周长为 30 cm，且长比宽多 3 cm。求该长方形的面积。
Let the width be w cm. Then the length is w + 3 cm. The perimeter of a rectangle is 2(length + width), so 2(w + 3 + w) = 30. Simplify inside the brackets: 2w + 3. Equation: 2(2w + 3) = 30. Divide both sides by 2: 2w + 3 = 15. Subtract 3: 2w = 12, so w = 6. The width is 6 cm, and the length is 6 + 3 = 9 cm. Area = length × width = 9 × 6 = 54 cm². Always check the perimeter: 2(9+6)=30, correct. A common mistake is to confuse perimeter with area, or to set up the equation incorrectly by forgetting to double the sum.
设宽为 w cm，则长为 w + 3 cm。长方形的周长公式为 2 × (长 + 宽)，因此 2(w + 3 + w) = 30。化简括号内：2w + 3。方程：2(2w + 3) = 30。两边同除以 2：2w + 3 = 15。再减去 3：2w = 12，得到 w = 6。宽为 6 cm，长为 6 + 3 = 9 cm。面积 = 长 × 宽 = 9 × 6 = 54 cm²。务必验算周长：2 × (9 + 6) = 30，正确。常见的错误是把周长和面积搞混，或者在列方程时忘记把长宽之和乘以 2。
9. Averages and Range | 平均数与极差
The marks of seven students in a test are: 12, 15, 11, 18, 14, 16, 13. Calculate the mean, median, mode and range.
七名学生的测验成绩为：12, 15, 11, 18, 14, 16, 13。求平均数、中位数、众数和极差。
First, sort the data in ascending order: 11, 12, 13, 14, 15, 16, 18. Range = highest value – lowest value = 18 – 11 = 7. Mean = sum of all values divided by the number of values. Sum = 11 + 12 + 13 + 14 + 15 + 16 + 18 = 99. Number of values = 7, so mean = 99 ÷ 7 ≈ 14.14 (or exactly 99/7). Median is the middle value in an ordered list. Since there are 7 values, the 4th value is the median: 14. Mode is the value that appears most frequently; each mark appears only once, so there is no mode (or the data set is ‘bimodal’ if two values repeat, but here none do). Range tells you the spread of the data. Always remember to order the data before finding the median and range.
首先将数据从小到大排序：11, 12, 13, 14, 15, 16, 18。极差 = 最大值 – 最小值 = 18 – 11 = 7。平均数 = 所有数据之和除以数据个数。总和 = 11 + 12 + 13 + 14 + 15 + 16 + 18 = 99。数据个数为 7，所以平均数 = 99 ÷ 7 ≈ 14.14（或精确为 99/7）。中位数是排序后位于中间位置的数，因为有 7 个数据，第 4 个数值就是中位数：14。众数是出现次数最多的值；这里每个分数只出现一次，所以没有众数。极差反映了数据的分散程度。在求中位数和极差之前，一定要先把数据排序。
10. Ratio and Proportion | 比与比例
A recipe for shortbread uses flour and sugar in the ratio 5:2. How much sugar is needed if 300 g of flour is used? What mass of flour would be needed to make 420 g of the mixture?
一份黄油饼干的配方中，面粉和糖的比例为 5:2。如果使用 300 g 面粉，需要多少糖？若要制作 420 g 的混合料，需要多少面粉？
For the first part, set up an equivalent ratio. Flour:sugar = 5:2, so for every 5 parts flour there are 2 parts sugar. 300 g flour corresponds to 5 parts, so one part = 300 ÷ 5 = 60 g. Sugar is 2 parts, so sugar = 2 × 60 = 120 g. Alternatively, write a proportion: 5/2 = 300/x, cross-multiply: 5x = 600, x = 120 g. For the second part, the total mixture is flour + sugar = 5 + 2 = 7 parts. 420 g of mixture corresponds to 7 parts, so one part = 420 ÷ 7 = 60 g. Flour is 5 parts, so flour needed = 5 × 60 = 300 g. This shows scaling by finding the value of one part, which is a powerful strategy for all ratio problems.
第一部分，建立等效比例。面粉:糖 = 5:2，即每 5 份面粉对应 2 份糖。300 g 面粉对应 5 份，所以 1 份 = 300 ÷ 5 = 60 g。糖为 2 份，因此糖 = 2 × 60 = 120 g。也可以列比例式：5/2 = 300/x，交叉相乘得 5x = 600，x = 120 g。第二部分，混合料总量 = 面粉 + 糖 = 5 + 2 = 7 份。420 g 混合料对应 7 份，所以 1 份 = 420 ÷ 7 = 60 g。面粉为 5 份，因此面粉需要 5 × 60 = 300 g。这种先求出一份量的方法，对所有比例问题都非常有用。
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