📚 Year 7 Cambridge Maths: Deep Dive into Past Papers | 剑桥 Year 7 数学历年真题深度解析
Welcome to this in-depth analysis of Cambridge Year 7 Mathematics past paper questions. This article carefully unpacks the most common question types, step-by-step solutions, and key strategies to help you master every topic. Each section focuses on a core concept, shows a typical exam-style question, and explains the solution in clear detail, with both English and Chinese commentary.
欢迎来到本次剑桥 Year 7 数学历年真题深度解析。本文精心拆解最常见的题型、逐步解题过程和关键策略，助你掌握每一个主题。每个部分聚焦一个核心概念，展示典型真题风格的题目，并用清晰的中英双语解析。
1. Understanding Place Value and Rounding | 理解位值与四舍五入
Place value questions often ask you to identify the value of a specific digit or to round a number to a given decimal place or significant figure. In Cambridge Year 7 past papers, you might see: “Write down the value of the 5 in 3.752.”
位值题常见于要求你指出某一位数字的值，或将数字四舍五入到指定的小数位或有效数字。在剑桥 Year 7 真题中，会出现：“写出 3.752 中数字 5 的值。”
The digit 5 is in the hundredths place, so its value is 5/100 or 0.05. Remember the place value names: tenths, hundredths, thousandths, and so on. When rounding, look at the next digit: if it is 5 or more, round up.
数字 5 在百分位，因此它的值是 5/100 或 0.05。记住位值名称：十分位、百分位、千分位等。四舍五入时，看下一位数字：若大于等于 5，则进位。
Example from a typical paper: Round 4.768 to 1 decimal place. Step 1: Identify the first decimal digit (7). Step 2: Look at the next digit (6). Since 6 ≥ 5, we increase 7 to 8. Answer: 4.8.
典型真题示例：将 4.768 四舍五入到一位小数。步骤 1：找到第一位小数（7）。步骤 2：看下一位数字（6）。由于 6 ≥ 5，把 7 加到 8。答案：4.8。
A common mistake is not checking the next digit carefully. Also, remember that if you are rounding to the nearest whole number, you look at the first decimal digit only.
常见错误是没有仔细检查下一位数字。还要记住，如果四舍五入到整数，只需看第一位小数。
2. Addition, Subtraction, Multiplication, and Division | 加减乘除运算
Calculations with whole numbers and decimals often involve the correct order of operations (BIDMAS/BODMAS). Past papers might test this with questions like: “Work out (64 – 16) ÷ 8 + 3 × 2.”
整数和小数的计算常常涉及正确的运算顺序（BIDMAS/BODMAS）。真题中可能出现：“计算 (64 – 16) ÷ 8 + 3 × 2。”
First, brackets: 64 – 16 = 48. Then division: 48 ÷ 8 = 6. Then multiplication: 3 × 2 = 6. Finally, addition: 6 + 6 = 12. The answer is 12.
首先计算括号：64 – 16 = 48。然后除法：48 ÷ 8 = 6。接着乘法：3 × 2 = 6。最后加法：6 + 6 = 12。答案是 12。
When multiplying large numbers, such as 234 × 17, use column method or the grid method. Always double-check your carrying. For division, a common exam question is dividing a 3-digit number by a 1-digit number, for example 672 ÷ 6 = 112.
当计算大数乘法时，如 234 × 17，使用竖式或网格法。务必检查进位。对于除法，常见考题是三位数除以一位数，例如 672 ÷ 6 = 112。
In word problems, read the question carefully to decide which operation to use. A past paper might say: “There are 135 apples in a box. They are packed into bags of 9. How many bags are needed?” This is a division problem: 135 ÷ 9 = 15 bags.
在应用题中，仔细读题确定用哪种运算。真题可能说：“一箱有 135 个苹果，装入每袋 9 个的袋子。需要多少个袋子？”这是除法问题：135 ÷ 9 = 15 袋。
3. Fractions, Decimals, and Percentages | 分数、小数与百分数
Converting between fractions, decimals, and percentages is a key skill. A common past paper question asks: “Write 0.35 as a fraction in its simplest form.”
在分数、小数和百分数之间转换是一项关键技能。真题常考：“将 0.35 写成最简分数。”
0.35 means 35/100. Simplify by dividing both numerator and denominator by 5 to get 7/20. Always remember to simplify fractions to their lowest terms.
0.35 即 35/100。分子分母同时除以 5 约分得 7/20。始终记住把分数化到最简形式。
Here is a quick reference table for common equivalences:
下面是一个常见等值快速参考表格：
| Fraction | Decimal | Percentage |
|---|---|---|
| 1/2 | 0.5 | 50% |
| 1/4 | 0.25 | 25% |
| 3/4 | 0.75 | 75% |
| 1/5 | 0.2 | 20% |
| 1/10 | 0.1 | 10% |
Ordering fractions can be tricky. When you see “Arrange 2/5, 0.45, 35% from smallest to largest,” convert all to the same form. 2/5 = 0.4, 35% = 0.35, so the order is 35%, 2/5, 0.45.
比较分数大小可能较难。当看到“将 2/5、0.45、35% 从小到大排列”时，把所有数化为同一种形式。2/5 = 0.4，35% = 0.35，因此顺序是 35%、2/5、0.45。
To find a percentage of an amount, like 15% of 200, multiply the amount by the decimal equivalent: 200 × 0.15 = 30.
求一个数的百分比，如 200 的 15%，用该数乘对应的小数：200 × 0.15 = 30。
4. Simplifying Ratios and Proportions | 简化比例与比率
Ratio questions often appear in context, such as sharing money or mixing ingredients. A Cambridge Year 7 question may state: “Simplify the ratio 24:36.” Divide both sides by their highest common factor (12) to get 2:3.
比例题常出现在具体情境中，如分钱或混合配料。剑桥 Year 7 的题目可能说：“简化比例 24:36。”两边同时除以它们的最大公因数（12）得到 2:3。
Another common type: “A recipe uses 300 g of flour and 100 g of sugar. Write the ratio of flour to sugar in its simplest form.” The ratio is 300:100, which simplifies to 3:1.
另一种常见类型：“一个食谱用 300 克面粉和 100 克糖。写出面粉与糖的最简比。”比例为 300:100，化简为 3:1。
When combining ratios with actual amounts, use the unitary method. For example, “The ratio of boys to girls in a class is 3:4. If there are 12 boys, how many girls are there?” Since 3 parts = 12 boys, 1 part = 4, so 4 parts = 16 girls.
当比例与实际数量结合时，使用单位法。例如，“一个班级男生与女生的比例是 3:4。如果有 12 名男生，有多少女生？”3 份 = 12 名男生，1 份 = 4，所以 4 份 = 16 名女生。
Always check if the question requires simplifying a ratio or using a total amount. Misreading can lead to the wrong answer.
务必检查题目是要求化简比例，还是使用总量。看错题意会导致错误答案。
5. Introduction to Algebra: Expressions and Equations | 代数初步：表达式与方程
Algebra in Year 7 includes forming expressions and solving simple linear equations. A typical past paper question: “Ellie thinks of a number, triples it, and subtracts 5. If the result is 16, what is the number?”
Year 7 的代数包括列出表达式和求解简单的一元一次方程。典型真题：“Ellie 想一个数，乘以 3 再减去 5。如果结果是 16，这个数是多少？”
Let the number be n. The equation is 3n – 5 = 16. Add 5 to both sides: 3n = 21. Divide by 3: n = 7. You can check: 3 × 7 – 5 = 21 – 5 = 16.
设这个数为 n。方程为 3n – 5 = 16。两边加 5：3n = 21。除以 3：n = 7。检验：3 × 7 – 5 = 21 – 5 = 16。
Forming expressions from words is another common exam task. “Write an expression for 5 less than 2x” would be 2x – 5. Pay attention to the order: “less than” reverses the terms.
根据文字列出表达式是另一常见考题。“比 2x 少 5 的表达式”应写为 2x – 5。注意顺序：“少”会调换各项位置。
Simplifying expressions by collecting like terms also appears, such as 4a + 2b – a + 3b = 3a + 5b. Only add or subtract the coefficients of the same variable.
通过合并同类项简化表达式，如 4a + 2b – a + 3b = 3a + 5b。只有相同变量的系数才能相加减。
6. Sequences and Patterns | 数列与规律
Sequence questions ask you to find the next terms or find the nth term. A classic past paper problem: “Here is a sequence: 5, 8, 11, 14, … Find the 10th term.”
数列题会要求找出后几项或找到第 n 项。经典真题问题：“数列：5、8、11、14……求第 10 项。”
Identify the difference: 8 – 5 = 3, so the sequence increases by 3 each time. The first term is 5. The term-to-term rule is “add 3.” The position-to-term rule (nth term) is 3n + 2 (because 3×1 + 2 = 5). For n = 10, term = 3×10 + 2 = 32.
找出公差：8 – 5 = 3，因此数列每次加 3。首项为 5。相邻两项的规则是“加 3”。位置到项的规则（第 n 项）是 3n + 2（因为 3×1 + 2 = 5）。当 n = 10 时，项 = 3×10 + 2 = 32。
Sometimes sequences are not linear; they could be based on a pattern of shapes or numbers. Always write the sequence carefully and test your nth term formula.
有时数列并非线性，可能基于图形或数字模式。务必仔细写出数列，并检验你的第 n 项公式。
To find the nth term, subtract the common difference from the first term to determine the constant: first term – common difference = 5 – 3 = 2, so nth term = 3n + 2.
求第 n 项时，用首项减去公差得到常数：首项 – 公差 = 5 – 3 = 2，因此第 n 项 = 3n + 2。
7. Angles, Lines, and Shapes | 角、线与图形
Angle facts are tested regularly. A question might show a triangle with two angles given, like 45° and 75°, and ask for the third angle. The sum of angles in a triangle is 180°, so the missing angle is 180 – (45 + 75) = 60°.
角的性质常考。题目可能给出一个三角形已知两个角，如 45° 和 75°，求第三个角。三角形内角和为 180°，因此缺失角为 180 – (45 + 75) = 60°。
For straight lines and around a point: angles on a straight line add up to 180°, angles around a point sum to 360°. These are used to find missing angles in diagrams.
对于直线和一点周角：直线上的角合计 180°，一点周围的角合计 360°。这些性质用于求图形中的未知角。
Parallel lines create equal corresponding and alternate angles. In a typical past paper diagram, if two parallel lines are crossed by a transversal, you can use these rules to calculate unknown angles.
平行线产生相等的同位角和内错角。在典型真题图形中，若两条平行线被一条截线穿过，可用这些规则计算未知角。
Remember to give a brief reason for each angle you calculate, such as “angles in a triangle sum to 180°” or “corresponding angles are equal.” This is often required for full marks.
记得为你计算的每个角给出简要理由，例如“三角形内角和 180°”或“同位角相等”。这常常是拿满分的必要条件。
8. Perimeter, Area, and Volume | 周长、面积与体积
Calculating perimeter and area of rectangles, triangles, and compound shapes appears frequently. The area of a rectangle is length × width. A question may give a rectangle with length 8 cm and width 5 cm and ask for its perimeter and area.
计算矩形、三角形及复合图形的周长和面积是常考题。矩形面积为长 × 宽。题目可能给出长为 8 厘米、宽为 5 厘米的矩形，要求求周长和面积。
Perimeter = 2 × (8 + 5) = 26 cm. Area = 8 × 5 = 40 cm². Remember to write the correct units (cm, cm²).
周长 = 2 × (8 + 5) = 26 厘米。面积 = 8 × 5 = 40 平方厘米。记得标注正确单位（cm，cm²）。
For a triangle, area = ½ × base × height. If base is 10 cm and height is 6 cm, area = ½ × 10 × 6 = 30 cm².
三角形的面积 = ½ × 底 × 高。若底为 10 厘米，高为 6 厘米，面积 = ½ × 10 × 6 = 30 平方厘米。
Volume of a cuboid = length × width × height. A typical question: Find the volume of a box 4 cm by 3 cm by 2 cm. Volume = 4 × 3 × 2 = 24 cm³.
长方体的体积 = 长 × 宽 × 高。典型题目：求一个长 4 厘米、宽 3 厘米、高 2 厘米的盒子的体积。体积 = 4 × 3 × 2 = 24 立方厘米。
Watch out for compound shapes by splitting them into smaller rectangles. Add or subtract areas accordingly.
对于复合图形，可将其拆分为较小的矩形。相应加上或减去面积。
9. Data Handling: Averages and Charts | 数据处理：平均数与图表
Mean, median, mode, and range are the “four measures” you need to know. A past paper dataset: 6, 8, 5, 9, 5, 12. Find the mean, median, mode, and range.
均值、中位数、众数和范围是你需要掌握的“四大指标”。真题数据集：6、8、5、9、5、12。求均值、中位数、众数和范围。
Mode = 5 (most frequent). Range = 12 – 5 = 7. To find median, order the numbers: 5, 5, 6, 8, 9, 12. Since there are 6 numbers, median is the mean of the 3rd and 4th: (6 + 8) ÷ 2 = 7.
众数 = 5（出现最频繁）。范围 = 12 – 5 = 7。求中位数时，将数字排序：5、5、6、8、9、12。因为是 6 个数，中位数为第 3 和第 4 个数的平均：(6 + 8) ÷ 2 = 7。
Mean = sum of numbers ÷ number of data points. Sum = 5+5+6+8+9+12 = 45. Mean = 45 ÷ 6 = 7.5.
均值 = 数据总和 ÷ 数据个数。总和 = 5+5+6+8+9+12 = 45。均值 = 45 ÷ 6 = 7.5。
Interpreting bar charts and pictograms is also tested. You may need to read frequencies from a chart and then calculate the mean or total. Always check the scale and the key.
解读条形图和象形图也是考点。你可能需要从图表中读取频数，然后计算均值或总数。务必检查尺度和图例。
10. Probability Basics | 概率基础
Probability is expressed as a fraction, decimal, or percentage between 0 and 1. A typical question: “A bag contains 3 red, 2 blue, and 5 green balls. One ball is chosen at random. What is the probability it is red?”
概率用 0 到 1 之间的分数、小数或百分数表示。典型问题：“一个袋子中有 3 个红球、2 个蓝球和 5 个绿球。随机取一个球，取到红色的概率是多少？”
Total balls = 3+2+5 = 10. Number of red balls = 3. Probability = 3/10. Write it as a fraction in its simplest form. 3/10 is already simplified.
总球数 = 3+2+5 = 10。红球数 = 3。概率 = 3/10。用最简分数表示，3/10 已是最简。
You might be asked about the probability of “not red” or the complement. Probability of not red = 1 – 3/10 = 7/10. Also, probability can be placed on a scale from 0 (impossible) to 1 (certain).
可能问“不是红色”的概率或互补事件的概率。不是红色的概率 = 1 – 3/10 = 7/10。同时，概率可从 0（不可能）到 1（一定）的尺度上表示。
Sometimes you need to list outcomes for two events, like tossing a coin and rolling a dice. A sample space diagram helps. Probability then is the number of favorable outcomes over total outcomes.
有时需要列出两个事件的结局，如抛硬币和掷骰子。样本空间图可帮助分析。概率即为有利结局数除以总可能结局数。
Always write your final answer as a simplified fraction unless the question specifies otherwise.
除非题目特别说明，否则最终答案要写成最简分数。
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