Year 7 CIE Advanced Mathematics: Unit Test Mock Paper Analysis | Year 7 CIE 进阶数学：单元测试模拟卷解析

📚 Year 7 CIE Advanced Mathematics: Unit Test Mock Paper Analysis | Year 7 CIE 进阶数学：单元测试模拟卷解析

This article presents a complete unit test mock paper designed for Year 7 CIE Advanced Mathematics, along with detailed step‑by‑step solutions. Each question targets key topics such as number theory, fractions, algebra, geometry, ratio and statistics, helping students consolidate their understanding and check their readiness for real assessments.

本文提供了一份为 Year 7 CIE 进阶数学设计的单元测试模拟卷，并附有详细的逐步解析。每道题都直指数论、分数、代数、几何、比和统计等核心主题，帮助学生巩固知识并检查自己的备考情况。

1. Mock Paper Overview | 模拟试卷概览

The mock paper contains 10 questions covering prime factorisation, fraction and decimal operations, solving linear equations, simplifying algebraic expressions, properties of triangles, perimeter and area of a square, map scales, sharing in a given ratio, calculating mean and median, and unit conversion with rate. Each question is worth equal marks, with a total of 50 marks. Students should aim to complete the paper within 45 minutes.

本模拟卷共10道题，涵盖质因数分解、分数与小数运算、求解一元一次方程、化简代数式、三角形的性质、正方形的周长与面积、地图比例尺、按比例分配、均值与中位数的计算以及单位换算与速率问题。每题分值相同，满分50分，建议学生45分钟内完成。

2. Question 1: HCF, LCM and Prime Factorisation | 第1题：最大公约数、最小公倍数与质因数分解

Question: (a) Find the highest common factor (HCF) and the lowest common multiple (LCM) of 72 and 90. (b) Express 120 as a product of its prime factors.

题目：(a) 求72和90的最大公约数(HCF)和最小公倍数(LCM)。(b) 将120分解为质因数的乘积。

To determine the HCF, first write both numbers as products of prime factors. 72 = 2³ × 3² and 90 = 2 × 3² × 5. The prime factors common to both numbers are 2 and 3. Use the lowest power of each common prime: for 2 the lowest power is 2¹ (or simply 2), and for 3 it is 3². Multiplying these gives HCF = 2 × 3² = 2 × 9 = 18.

求最大公约数时，先将两数写成质因数乘积的形式。72 = 2³ × 3²，90 = 2 × 3² × 5。公共的质因数是2和3。对每一个公共质因数取最低次幂：2的最低次幂是2¹（即2），3的最低次幂是3²。相乘即得最大公约数：HCF = 2 × 3² = 2 × 9 = 18。

The LCM is found by taking the highest power of every prime factor that appears in either number. The prime factors are 2, 3 and 5. Highest power of 2 is 2³, of 3 is 3², and of 5 is 5¹. Therefore, LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360.

求最小公倍数时，对两个数中出现的每一个质因数，都取最高次幂。质因数有2、3、5。2的最高次幂是2³，3的最高次幂是3²，5的最高次幂是5¹。因此，LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360。

For part (b), break 120 down into prime factors using a factor tree: 120 = 12 × 10, then 12 = 2² × 3 and 10 = 2 × 5. Combining gives 120 = 2³ × 3 × 5, which is the product of prime factors.

第(b)部分，利用因数树将120分解：120 = 12 × 10，而12 = 2² × 3，10 = 2 × 5。合并后得到120 = 2³ × 3 × 5，这就是质因数乘积形式。

HCF = 18, LCM = 360, 120 = 2³ × 3 × 5

HCF = 18，LCM = 360，120 = 2³ × 3 × 5

3. Question 2: Fractions and Decimals | 第2题：分数与小数运算

Question: Evaluate (a) 3/5 + 2/3 – 1/2, giving your answer as a fraction in its simplest form. (b) 0.75 × 1.2 ÷ 0.3, giving your answer as a decimal.

题目：计算 (a) 3/5 + 2/3 – 1/2，结果用最简分数表示。(b) 0.75 × 1.2 ÷ 0.3，结果用小数表示。

For (a), find a common denominator for 5, 3 and 2, which is 30. Convert each fraction: 3/5 = 18/30, 2/3 = 20/30, 1/2 = 15/30. Then evaluate: 18/30 + 20/30 – 15/30 = (18 + 20 – 15)/30 = 23/30. The fraction 23/30 is already in its simplest form since 23 is prime and does not divide 30.

对于(a)题，找出分母5、3、2的公分母30。转换为同分母分数：3/5 = 18/30，2/3 = 20/30，1/2 = 15/30。然后计算：18/30 + 20/30 – 15/30 = (18 + 20 – 15)/30 = 23/30。23/30已是最简分数，因为23是质数且不能整除30。

For (b), perform the operations from left to right. 0.75 × 1.2 = 0.90 (or 0.9). Then divide 0.9 by 0.3: 0.9 ÷ 0.3 = 3. Alternatively, multiply both dividend and divisor by 10 to obtain whole numbers: 9 ÷ 3 = 3. The final answer is 3.

对于(b)题，按从左到右的顺序计算。0.75 × 1.2 = 0.90（即0.9）。再用0.9除以0.3：0.9 ÷ 0.3 = 3。另一种方法是把被除数和除数同时乘10，变成整数：9 ÷ 3 = 3。最终结果为3。

(a) 23/30 (b) 3

4. Question 3: Solving a Linear Equation | 第3题：解一元一次方程

Question: Solve for x: 4(x – 3) = 2x + 10

题目：解方程：4(x – 3) = 2x + 10

Start by expanding the left‑hand side: 4(x – 3) = 4x – 12. The equation becomes 4x – 12 = 2x + 10. To isolate the variable, subtract 2x from both sides: 4x – 2x – 12 = 10, which simplifies to 2x – 12 = 10.

首先展开等号左边：4(x – 3) = 4x – 12。原方程化为 4x – 12 = 2x + 10。为了分离未知数，两边同时减去2x，得到 4x – 2x – 12 = 10，即 2x – 12 = 10。

Next, add 12 to both sides: 2x = 22. Finally, divide by 2: x = 11. Always check by substituting back: 4(11 – 3) = 4 × 8 = 32; 2 × 11 + 10 = 22 + 10 = 32. Both sides match, confirming the solution.

接着，两边同时加上12：2x = 22。最后，两边除以2：x = 11。将x = 11代回原方程验算：4(11 – 3) = 4 × 8 = 32；2 × 11 + 10 = 22 + 10 = 32。两边相等，答案正确。

x = 11

5. Question 4: Simplifying Algebraic Expressions | 第4题：化简代数表达式

Question: Simplify the expression: 5a – 3b + 2a + 7b – 4a

题目：化简表达式：5a – 3b + 2a + 7b – 4a

Identify and group the like terms. Terms containing a: 5a, 2a and –4a. Terms containing b: –3b and 7b. Combine the a‑terms: 5 + 2 – 4 = 3, so the a‑terms simplify to 3a. Combine the b‑terms: –3 + 7 = 4, giving 4b.

找出同类项（即含有相同字母的项）。含有a的项：5a、2a和–4a。含有b的项：–3b和7b。合并a项：5 + 2 – 4 = 3，因此a项合并为3a。合并b项：–3 + 7 = 4，得到4b。

The simplified expression is 3a + 4b. Since a and b are different variables, the terms cannot be combined further. If numerical values for a and b were given, we could then evaluate the expression, but for now the simplest form is just 3a + 4b.

化简后的表达式为3a + 4b。由于a和b是不同字母，无法继续合并。如果题目给出a和b的具体数值，可以代入求值；但就化简而言，最终结果就是3a + 4b。

3a + 4b

6. Question 5: Angles in a Triangle | 第5题：三角形内角

Question: Two angles of a triangle measure 48° and 67°. Find the size of the third angle and classify the triangle by its angles.

题目：一个三角形的两个内角分别是48°和67°，求第三个角的度数，并根据角度对三角形进行分类。

The sum of the interior angles of any triangle is 180°. Add the two known angles: 48° + 67° = 115°. Subtract this sum from 180°: 180° – 115° = 65°. Therefore, the third angle measures 65°.

任意三角形的三个内角之和为180°。将已知两个角相加：48° + 67° = 115°。用180°减去这个和：180° – 115° = 65°。所以第三个角为65°。

The three angles are 48°, 67° and 65°, each of which is less than 90°. A triangle in which all three angles are acute (less than 90°) is called an acute‑angled triangle. Hence, the triangle is an acute triangle.

三角形的三个角分别是48°、67°和65°，每一个角都小于90°。三个角都是锐角（小于90°）的三角形叫作锐角三角形。因此，这是一个锐角三角形。

Third angle = 65°, triangle is acute‑angled

第三个角 = 65°，三角形为锐角三角形

7. Question 6: Perimeter and Area of a Square | 第6题：正方形的周长与面积

Question: A square has a perimeter of 32 cm. Find the length of one side and then calculate its area.

题目：一个正方形的周长是32 cm，求边长并计算面积。

For a square, all four sides are equal. Perimeter = 4 × side length. Set up the equation: 4 × side = 32 cm. Divide both sides by 4 to find side length = 32 ÷ 4 = 8 cm.

正方形的四条边长度相等。周长 = 4 × 边长。列出等式：4 × 边长 = 32 cm。两边除以4，得到边长 = 32 ÷ 4 = 8 cm。

The area of a square is found by squaring the side length: Area = side × side = 8 cm × 8 cm = 64 cm². Always remember to include the correct unit; area is given in square centimetres (cm²).

正方形面积等于边长的平方：面积 = 边长 × 边长 = 8 cm × 8 cm = 64 cm²。务必带上正确的单位，面积以平方厘米（cm²）表示。

Side = 8 cm, Area = 64 cm²

8. Question 7: Map Scale | 第7题：地图比例尺

Question: A map is drawn to a scale of 1 : 25 000. Two towns are 8 cm apart on the map. Calculate the actual distance between the towns in kilometres.

题目：一幅地图的比例尺为1 : 25 000，两座城镇在地图上相距8 cm。计算实际距离（以公里为单位）。

The scale 1 : 25 000 means that 1 cm on the map corresponds to 25 000 cm in reality. Multiply the map distance by the scale factor: Actual distance = 8 cm × 25 000 = 200 000 cm.

比例尺1 : 25 000表示图上1 cm代表实际25 000 cm。用图上距离乘比例尺：实际距离 = 8 cm × 25 000 = 200 000 cm。

Convert centimetres to kilometres. Recall that 100 cm = 1 m, and 1000 m = 1 km, so 1 km = 100 000 cm. Divide the distance in centimetres by 100 000: 200 000 ÷ 100 000 = 2. Hence, the actual distance is 2 km.

将厘米换算为公里。已知100 cm = 1 m，1000 m = 1 km，所以1 km = 100 000 cm。用厘米数除以100 000：200 000 ÷ 100 000 = 2。因此，实际距离为2 km。

Actual distance = 2 km

9. Question 8: Sharing in a Given Ratio | 第8题：按比例分配

Question: Divide 45 into the ratio 2 : 3.

题目：将45按2 : 3的比例分成两部分。

The total number of parts in the ratio 2 : 3 is 2 + 3 = 5 parts. Therefore, one part is equal to the total amount divided by the total number of parts: 45 ÷ 5 = 9.

比例2 : 3的总份数为2 + 3 = 5份。因此，每一份的大小为总数量除以总份数：45 ÷ 5 = 9。

The first share corresponds to 2 parts: 2 × 9 = 18. The second share corresponds to 3 parts: 3 × 9 = 27. Check the sum: 18 + 27 = 45, which matches the original total.

第一部分占2份：2 × 9 = 18。第二部分占3份：3 × 9 = 27。验算总和：18 + 27 = 45，与原总数一致。

18 and 27

10. Question 9: Mean and Median | 第9题：均值与中位数

Question: The marks of seven students in a test are: 12, 15, 10, 13, 14, 16, 11. Find the mean and the median of this data set.

题目：七名学生的测验成绩为：12, 15, 10, 13, 14, 16, 11。求这组数据的均值和中位数。

To find the mean, add all the values together: 12 + 15 + 10 + 13 + 14 + 16 + 11 = 91. There are 7 students, so divide the total by 7: 91 ÷ 7 = 13. The mean mark is 13.

计算均值时，先把所有数据相加：12 + 15 + 10 + 13 + 14 + 16 + 11 = 91。共有7个数据，总和除以7：91 ÷ 7 = 13。均值是13分。

For the median, arrange the marks in ascending order: 10, 11, 12, 13, 14, 15, 16. With 7 numbers, the median is the middle value, the 4th term. Counting through, the 4th term is 13. Thus, the median is also 13.

求中位数时，先将成绩从小到大排列：10, 11, 12, 13, 14, 15, 16。有7个数据，中位数就是正中间的数值，即第4个。数到第4个是13，因此中位数也是13。

In this case the mean and median happen to be equal, but this is not always true. The median is often a better measure of centre when the data set contains an outlier.

本例中均值与中位数恰好相等，但这种情况并不总会出现。当数据中含有异常值时，中位数往往是描述数据中心更好的指标。

Mean = 13, Median = 13

11. Question 10: Volume Conversion and Rate | 第10题：容积换算与速率问题

Question: Convert 3.5 litres into millilitres. Water flows out of a tap at a rate of 70 ml per minute. How many minutes will it take to fill a container with a capacity of 3.5 litres?

题目：将3.5升换算为毫升。水龙头每分钟流出70毫升水，请问需要多少分钟才能灌满一个容积为3.5升的容器？

First, convert litres to millilitres using the fact that 1 L = 1000 ml. Multiply: 3.5 × 1000 = 3500 ml. So 3.5 L is exactly 3500 ml.

首先，利用1升 = 1000毫升进行换算。相乘：3.5 × 1000 = 3500 ml。所以3.5升等于3500毫升。

The flow rate is 70 ml per minute. To find the time required, divide the total volume by the rate: Time (minutes) = 3500 ml ÷ 70 ml/min = 50 minutes. The unit ‘ml’ cancels out, leaving the time in minutes.

流速为每分钟70毫升。所需时间等于总容积除以速率：时间（分钟）= 3500 ml ÷ 70 ml/min = 50分钟。单位“ml”互相抵消，得到时间单位为分钟。

It is always useful to check the answer: 70 ml/min × 50 min = 3500 ml = 3.5 L, which confirms the calculation.

验算一下：70毫升/分钟 × 50分钟 = 3500毫升 = 3.5升，计算无误。

3.5 L = 3500 ml, Time = 50 minutes

12. Tips for the Test | 考试小贴士

Read each question carefully to identify what is being asked. When working with fractions, always simplify your answer as much as possible. In algebra, show each step of expanding and isolating the variable to avoid careless errors. For geometry

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Year 7 CIE Advanced Mathematics: Unit Test Mock Paper Analysis | Year 7 CIE 进阶数学：单元测试模拟卷解析