📚 Year 7 Cambridge Advanced Mathematics: Mock Unit Test Analysis | Year 7 Cambridge 进阶数学：单元测试模拟卷解析
Welcome to this detailed walkthrough of a Year 7 Cambridge Advanced Mathematics mock unit test. This analysis is designed to help students identify key topics, understand typical question formats, and master problem-solving techniques. Each section below presents a sample exam-style question, followed by step-by-step reasoning, common pitfalls, and essential revision notes.
欢迎阅读这份 Year 7 Cambridge 进阶数学单元测试模拟卷的详细解析。本分析旨在帮助学生识别核心主题、理解典型题目形式并掌握问题解决技巧。以下各节均提供了一道模拟试题，并附有逐步推理过程、常见错误提示和关键复习要点。
1. Number and Place Value | 数字与位值
Question: Write the number 4 027 316 in words and state the value of the digit '2'. Explain how you would round the number to the nearest hundred thousand.
问题：用文字写出数字 4 027 316，并说明数字 '2' 的值。解释如何将该数四舍五入到最近的十万位。
Solution: The number in words is 'four million, twenty-seven thousand, three hundred and sixteen'. The digit '2' is in the ten-thousands place, so its value is 20 000. To round to the nearest hundred thousand, we look at the ten-thousands digit (2). Since 2 < 5, we round down, giving 4 000 000.
解析：该数字的中文读法是 '四百零二万七千三百一十六'。数字 '2' 位于万位，因此它的值是 20 000。四舍五入到最近的十万位时，看万位数字（2）。因为 2 小于 5，所以舍去，得到 4 000 000。
- Key skill: Recognising place value up to millions. | 关键技能：识别百万以内的位值。
- Common mistake: Confusing 'ten thousands' with 'hundred thousands'. | 常见错误：混淆 '万位' 与 '十万位'。
2. Fractions, Decimals and Percentages | 分数、小数和百分数
Question: Convert 3/8 into a decimal and a percentage. Then find 3/8 of £240.
问题：将 3/8 转换为小数和百分数。然后求 £240 的 3/8。
Solution: 3/8 = 3 ÷ 8 = 0.375. As a percentage, multiply by 100: 0.375 × 100 = 37.5%. To find 3/8 of £240, calculate (240 ÷ 8) × 3 = 30 × 3 = £90.
解析：3/8 = 3 ÷ 8 = 0.375。转换为百分数，乘以 100：0.375 × 100 = 37.5%。求 £240 的 3/8，计算 (240 ÷ 8) × 3 = 30 × 3 = £90。
- Equivalent forms: 1/8 = 0.125 = 12.5%. | 等值形式：1/8 = 0.125 = 12.5%。
- Use fraction-of-amount method: divide by denominator, multiply by numerator. | 使用求一个量的几分之几的方法：除以分母，乘以分子。
3. Algebraic Expressions | 代数表达式
Question: Simplify the expression 5a + 3b – 2a + 7b. Then evaluate it when a = 4 and b = –1.
问题：化简表达式 5a + 3b – 2a + 7b。当 a = 4 且 b = –1 时，求其值。
Solution: Collect like terms: 5a – 2a = 3a, and 3b + 7b = 10b. The simplified expression is 3a + 10b. Substituting a = 4, b = –1 gives 3(4) + 10(–1) = 12 – 10 = 2.
解析：合并同类项：5a – 2a = 3a，3b + 7b = 10b。化简后的表达式为 3a + 10b。代入 a = 4，b = –1 得 3(4) + 10(–1) = 12 – 10 = 2。
- Only combine terms with exactly the same letter part. | 只能合并字母部分完全相同的项。
- Carefully handle negative substitutions using brackets. | 使用括号小心处理负数代入。
4. Solving Equations | 解方程
Question: Solve the equation 2(x – 3) = 8. Then check your answer.
问题：解方程 2(x – 3) = 8。然后验证你的答案。
Solution: Expand the bracket: 2x – 6 = 8. Add 6 to both sides: 2x = 14. Divide by 2: x = 7. Check: 2(7 – 3) = 2 × 4 = 8, which matches the right-hand side.
解析：展开括号：2x – 6 = 8。两边加 6：2x = 14。除以 2：x = 7。验证：2(7 – 3) = 2 × 4 = 8，与右边一致。
- Always check your solution by substituting it back. | 一定要通过代回原方程来检验解。
- Alternative method: divide both sides by 2 first, then x – 3 = 4, so x = 7. | 另一种方法：先两边除以 2，得 x – 3 = 4，所以 x = 7。
5. Sequences and Patterns | 序列与模式
Question: The first three terms of a sequence are 6, 11, 16. Find the term-to-term rule and the 10th term.
问题：一个序列的前三项是 6、11、16。求逐项规则和第 10 项。
Solution: The sequence increases by 5 each time (11 – 6 = 5, 16 – 11 = 5). The term-to-term rule is 'add 5' st term = 6, common difference = 5. nth term = 5n + 1. So the 10th term = 5 × 10 + 1 = 51.
解析：序列每次增加 5（11 – 6 = 5，16 – 11 = 5）。逐项规则是 '加 5'。为求第 10 项，使用位置与项的关系式：首项 = 6，公差 = 5。第 n 项 = 5n + 1。因此第 10 项 = 5 × 10 + 1 = 51。
- Recognise arithmetic sequences: constant difference. | 识别等差数列：差恒定。
- Derive nth term: difference × n + (first term – difference). | 推导第 n 项：公差 × n + （首项 – 公差）。
6. Angles and Lines | 角与直线
Question: Two angles are on a straight line. One angle is 132°. Find the other angle. Classify both angles.
问题：两个角位于一条直线上。其中一个角是 132°。求另一个角。对这两个角进行分类。
Solution: Angles on a straight line sum to 180°. So the other angle = 180° – e 132° angle is obtuse (between 90° and 180°), and the 48° angle is acute (less than 90°).
解析：直线上的角之和为 180°。因此另一个角 = 180° – 132° = 48°。132° 的角是钝角（在 90° 到 180° 之间），48° 的角是锐角（小于 90°）。
- Recall angle facts: straight line = 180°, right angle = 90°, full turn = 360°. | 记住角度知识：直线 180°，直角 90°，周角 360°。
- Classification: acute (< 90°), right angle (90°), obtuse (90°–180°), reflex (> 180°). | 分类：锐角（< 90°），直角（90°），钝角（90°–180°），优角（> 180°）。
7. Perimeter and Area | 周长与面积
Question: A rectangle has length 8.5 cm and width 4 cm. Calculate its perimeter and area. Then find the side length of a square with the same area.
问题：一个矩形的长是 8.5 cm，宽是 4 cm。计算它的周长和面积。然后求面积相同的正方形的边长。
= 25 cm. Area = length × width = 8.5 × 4 = 34 cm². For the square: area = side², so side = √34 ≈ 5.83 cm (to 2 decimal places).
解析：周长 = 2 × (长 + 宽) = 2 × (8.5 + 4) = 2 × 12.5 = 25 cm。面积 = 长 × 宽 = 8.5 × 4 = 34 cm²。对于正方形：面积 = 边长²，所以边长 = √34 ≈ 5.83 cm（保留两位小数）。
- Always include units: cm for perimeter, cm² for area. | 务必写单位：周长用 cm，面积用 cm²。
- Understand that area remains the same when changing shape. | 理解形状改变时面积不变。
8. Data Handling | 数据处理
Question: The number of books read by 15 students is: 3, 5, 2, 4, 5, 1, 5, 3, 2, 6, 4, 3, 5, 2, 4. Construct a frequency table and find the mode, median, and range.
问题：15 名学生的阅读书籍数量为：3, 5, 2, 4, 5, 1, 5, 3, 2, 6, 4, 3, 5, 2, 4。制作一个频数表，并求众数、中位数和极差。
Frequency table:
频数表：
| Number of books | 书籍数量 | Frequency | 频数 |
| 1 | 1 |
| 2 | 3 |
| 3 | 3 |
| 4 | 3 |
| 5 | 4 |
| 6 | 1 |
lues in order: 1,2,2,2,3,3,3,4,4,4,5,5,5,5,6. The 8th value is 4, so median = 4. Range = 6 – 1 = 5.
解析：众数 = 5（出现次数最多）。求中位数时，按顺序排列所有数值：1,2,2,2,3,3,3,4,4,4,5,5,5,5,6。第 8 个数值是 4，所以中位数 = 4。极差 = 6 – 1 = 5。
- Mode can be read directly from a frequency table. | 众数可以直接从频数表读出。
- For the median with an odd count, find the middle position: (n+1)/2. | 对于奇数个数，中位数位置是 (n+1)/2。
9. Geometry: Symmetry and Transformations | 几何：对称与变换
Question: A triangle has coordinates A(1,2), B(4,2), C(3,5). Reflect it in the y-axis. Write the coordinates of the image. State whether the triangle has any lines of symmetry.
问题：一个三角形的顶点坐标为 A(1,2), B(4,2), C(3,5)。将其关于 y 轴反射。写出镜像三角形的坐标。说明该三角形是否有对称轴。
Solution: Reflecting in the y-axis changes the sign of the x-coordinate, so A'(–1,2), B'(–4,2), C'(– original triangle is isosceles (AB is horizontal, AC = BC), so it has one line of symmetry – the vertical line x = 2.5 passing through C.
解析：关于 y 轴反射会改变 x 坐标的符号，因此 A'(–1,2), B'(–4,2), C'(–3,5)。原三角形是等腰三角形（AB 水平，AC = BC），因此它有一条对称轴——通过点 C 的垂直线 x = 2.5。
- Reflection rules: x-axis → (x, –y); y-axis → (–x, y). | 反射规则：x 轴反射 → (x, –y)；y 轴反射 → (–x, y)。
- Check for symmetry by folding or comparing side lengths. | 通过折叠或比较边长来检查对称性。
10. Problem-Solving Strategies | 问题解决策略
Question: Maya thinks of a number. She multiplies it by 3, subtracts 7, and gets 20. Write an equation and find the number. Then explain how a bar model could represent this problem.
问题：Maya 想了一个数。她将数乘以 3，减去 7，得到 20。写出方程并求出该数。然后解释如何用条形图表示这个问题。
Solution: Let the number be n. The equation is 3n – bar model would show three equal bars for 3n, with a section of 7 removed, leaving 20. The total length of the three bars must be 27.
解析：设这个数为 n。方程为 3n – 7 = 20。加 7：3n = 27。除以 3：n = 9。条形图可展示三个表示 3n 的等长条形，去掉一段长度为 7 的部分，剩下 20。三个条形的总长度必定是 27。
- Define the unknown clearly. | 清晰地定义未知数。
- Bar models help visualise inverse operations. | 条形图有助于直观理解逆运算。
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