📚 Year 7 CIE Maths: Unit Test Mock Paper Analysis | Year 7 CIE 数学：单元测试模拟卷解析
This article walks you through a typical Year 7 CIE Maths unit test mock paper, providing detailed solutions and examiner advice for every question. Use it to review core topics, spot common misconceptions and build confidence before your real assessment.
本文带你逐题解析一份典型 Year 7 CIE 数学单元测试模拟卷，提供详细解答与考官建议。通过它复习核心知识点，发现常见误区，并在考前建立信心。
1. Number Operations and Place Value | 数的运算与位值
Question: Write the number 3 046 215 in words and state the value of the digit 4.
题目：写出数字 3 046 215 的英文读法，并说明数字 4 的位值。
Solution: Three million, forty-six thousand, two hundred and fifteen. The digit 4 is in the ten-thousands place, so its value is 40 000.
解答：三百零四万六千二百一十五。数字 4 在万位，因此它的值是 40 000。
Common mistake: mixing up ‘forty’ and ‘fourteen’ when reading. Always group digits in threes from the right: millions, thousands, ones.
常见错误：读数时混淆“四十”和“十四”。始终从右向左每三位一组：百万、千、个。
2. Negative Numbers | 负数
Question: The temperature at midnight was −8 °C. By midday it had risen by 15 °C. What was the temperature at midday? Work out (−3) × 5 + (−10) ÷ 2.
题目：午夜气温是 −8 °C，到了正午上升了 15 °C。正午的气温是多少？并计算 (−3) × 5 + (−10) ÷ 2。
Solution: Temperature at midday = −8 + 15 = 7 °C. For the calculation: (−3) × 5 = −15; (−10) ÷ 2 = −5; then −15 + (−5) = −20.
解答：正午气温 = −8 + 15 = 7 °C。计算部分：(−3) × 5 = −15；(−10) ÷ 2 = −5；然后 −15 + (−5) = −20。
Remember the order of operations: multiplication and division before addition, and adding a negative is the same as subtracting its positive.
记住运算顺序：先乘除后加减，并且加上一个负数等于减去它的正数。
3. Fractions, Decimals and Percentages | 分数、小数与百分数
Question: Shade 3/8 of a strip divided into 8 equal parts and express 3/8 as a decimal and a percentage.
题目：将一条均匀分成 8 份的条带涂满 3/8，并把 3/8 写成小数和百分数。
Solution: Three parts shaded out of eight. As a decimal: 3 ÷ 8 = 0.375. As a percentage: 0.375 × 100% = 37.5%.
解答：涂满 8 份中的 3 份。小数：3 ÷ 8 = 0.375。百分数：0.375 × 100% = 37.5%。
Use short division or known facts (1/8 = 0.125) to convert fractions quickly. Always simplify percentages to one decimal place if needed.
使用短除法或已知事实 (1/8 = 0.125) 快速转换分数。百分数通常按需要保留一位小数。
4. Factors, Multiples and Primes | 因数、倍数与质数
Question: Find all the factors of 36. Write 60 as a product of prime factors in index form.
题目：找出 36 的所有因数，并用质因数指数形式表示 60。
Solution: Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. Prime factorisation of 60: 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5.
解答：36 的因数：1, 2, 3, 4, 6, 9, 12, 18, 36。60 的质因数分解：60 = 2 × 2 × 3 × 5 = 2² × 3 × 5。
A common error is missing factor pairs like (3,12). Work systematically from 1 upwards. For index form, group identical primes and use exponents.
常见错误是遗漏因数对如 (3,12)。从 1 开始系统地找出所有因数。指数形式需将相同质数分组并写成幂。
5. Algebraic Expressions and Equations | 代数表达式与方程
Question: Simplify 5a + 3b − 2a + 4b. Solve the equation 4x − 7 = 13.
题目：化简 5a + 3b − 2a + 4b，并解方程 4x − 7 = 13。
Solution: 5a − 2a = 3a, 3b + 4b = 7b, so simplified expression is 3a + 7b. For the equation: add 7 to both sides: 4x = 20; then divide by 4: x = 5.
解答：5a − 2a = 3a，3b + 4b = 7b，所以化简为 3a + 7b。方程：两边加 7 得 4x = 20；再除以 4 得 x = 5。
Only like terms can be combined. When solving equations, always do the same operation to both sides to maintain balance.
只有同类项可以合并。解方程时，始终在等式两边执行相同运算以保持平衡。
6. Sequences and Patterns | 数列与规律
Question: The first three terms of a sequence are 7, 12, 17. Write down the term-to-term rule and find the 10th term.
题目：一个数列的前三项是 7, 12, 17。写出项与项的递推规则，并求第 10 项。
Solution: Term-to-term rule: add 5 each time. This is an arithmetic sequence with first term 7 and common difference 5. nth term formula: 5n + 2. For n = 10: 5×10 + 2 = 52.
解答：递推规则：每次加 5。这是一个等差数列，首项 7，公差 5。第 n 项公式：5n + 2。当 n = 10 时：5×10 + 2 = 52。
Avoid simply adding 5 nine times without checking. Learn to derive the position-to-term rule (nth term) to find any term efficiently.
避免不检查而简单连加 9 次 5。学习推导位置-项公式（第 n 项），可以更有效地求出任意项。
7. Geometry: Angles and Shapes | 几何：角与图形
Question: In triangle ABC, angle A = 48° and angle B = 72°. Calculate angle C. Identify the type of triangle.
题目：在三角形 ABC 中，角 A = 48°，角 B = 72°。计算角 C，并指出三角形的类型。
Solution: Sum of angles in a triangle = 180°. Angle C = 180 − (48 + 72) = 180 − 120 = 60°. All angles are less than 90°, so it is an acute triangle.
解答：三角形内角和为 180°。角 C = 180 − (48 + 72) = 180 − 120 = 60°。所有角均小于 90°，所以是锐角三角形。
Check that the sum of your calculated angles equals 180°. An acute triangle has three acute angles; an obtuse triangle has one angle > 90°.
检查所算角度之和是否等于 180°。锐角三角形有三个锐角；钝角三角形有一个角大于 90°。
8. Perimeter, Area and Volume | 周长、面积与体积
Question: A rectangle has length 9 cm and width 5 cm. Find its perimeter and area. A cube has side length 4 cm; find its volume.
题目：一个长方形的长是 9 cm，宽是 5 cm。求其周长和面积。一个正方体的棱长是 4 cm，求其体积。
Solution: Perimeter = 2 × (9 + 5) = 28 cm. Area = 9 × 5 = 45 cm². Volume of cube = 4 × 4 × 4 = 64 cm³.
解答：周长 = 2 × (9 + 5) = 28 cm。面积 = 9 × 5 = 45 cm²。正方体体积 = 4 × 4 × 4 = 64 cm³。
Don’t confuse perimeter (distance around) with area (surface covered). Volume units are always cubed, area units are squared.
不要混淆周长（绕图形的长度）与面积（覆盖的表面）。体积单位始终是立方，面积单位是平方。
9. Coordinates and Graphs | 坐标与图表
Question: Plot the points A(2,5), B(6,5), C(6,1) and D(2,1) on a coordinate grid. Join them and name the shape. Give its area.
题目：在坐标网格上标出点 A(2,5), B(6,5), C(6,1) 和 D(2,1)。连接它们并说出形状，求其面积。
Solution: The shape is a rectangle. Length AB = 4 units (from x=2 to x=6), width AD = 4 units (from y=5 down to y=1). Area = 4 × 4 = 16 square units.
解答：该形状是一个长方形。长 AB = 4 单位（x 从 2 到 6），宽 AD = 4 单位（y 从 5 到 1）。面积 = 4 × 4 = 16 平方单位。
Count squares carefully: horizontal distance is difference in x-coordinates, vertical distance is difference in y-coordinates.
仔细数格子：水平距离是 x 坐标之差，垂直距离是 y 坐标之差。
10. Data Handling and Averages | 数据处理与平均数
Question: The test scores of 10 pupils are: 14, 18, 12, 14, 20, 14, 16, 18, 12, 14. Find the mean, median, mode and range.
题目：10 名学生的测验分数为：14, 18, 12, 14, 20, 14, 16, 18, 12, 14。求平均数（均值）、中位数、众数和极差。
Solution: First order the data: 12,12,14,14,14,14,16,18,18,20. Mean = (12+12+14+14+14+14+16+18+18+20) ÷ 10 = 152 ÷ 10 = 15.2. Median = (5th + 6th)/2 = (14+14)/2 = 14. Mode = 14 (appears most often). Range = 20 − 12 = 8.
解答：先排序：12,12,14,14,14,14,16,18,18,20。平均数 = (12+12+14+14+14+14+16+18+18+20) ÷ 10 = 152 ÷ 10 = 15.2。中位数 = (第5项+第6项)/2 = (14+14)/2 = 14。众数 = 14（出现最频繁）。极差 = 20 − 12 = 8。
Always order the data before finding median and range. The mode is the value, not its frequency.
在求中位数和极差之前务必将数据排序。众数是数值本身，不是它的频数。
11. Word Problems and Reasoning | 应用题与推理
Question: Three friends share £120 in the ratio 1:2:3. How much does each receive? A bus carries 54 pupils. If 2/3 are Year 7, how many Year 7 pupils are on the bus?
题目：三个朋友按 1:2:3 的比例分享 £120。每人各得多少？一辆巴士载有 54 名学生，如果其中 2/3 是 Year 7 学生，那么巴士上有多少名 Year 7 学生？
Solution: Total parts = 1+2+3 = 6. One part = £120 ÷ 6 = £20. Amounts: £20, £40, £60. For the bus: 2/3 of 54 = (54 ÷ 3) × 2 = 18 × 2 = 36 pupils.
解答：总份数 = 1+2+3 = 6。每份 = £120 ÷ 6 = £20。金额分别为 £20, £40, £60。巴士问题：54 的 2/3 = (54 ÷ 3) × 2 = 18 × 2 = 36 名学生。
Ratio questions often require adding the parts. For fractions of an amount, divide by the denominator then multiply by the numerator.
比例问题通常需要先将份数相加。求一个数的几分之几，先除以分母再乘以分子。
12. Common Mistakes and Exam Tips | 常见错误与考试技巧
Key pitfalls: forgetting place value when reading large numbers, mixing up area and perimeter formulas, ignoring BIDMAS/BODMAS in calculations, and not showing working steps. Always show your method — method marks can be awarded even if the final answer is wrong.
关键陷阱：读大数时忘记位值，混淆面积与周长公式，计算中忽略运算顺序，以及不展示计算步骤。始终写出解题步骤——即使最终答案错误，也可能得到方法分。
Keep an eye on units: convert all measurements to the same unit before calculating. Check your answers by working backwards or estimating. Read each question carefully — underlined keywords help.
注意单位：计算前将所有量度转换为相同单位。通过逆运算或估算来检查答案。仔细读题——划出关键词很有帮助。
Use this mock paper analysis to identify your weaker topics and revise them with targeted practice. Good luck!
用这份模拟卷解析找出你的薄弱知识点，并进行有针对性的练习。祝你好运！
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