📚 Analysis of Question Types in Cambridge Primary Mathematics Learner’s Book 6, 2nd Edition | 剑桥小学数学学习者用书第六册第二版题型解析
The Cambridge Primary Mathematics Learner’s Book 6, 2nd Edition builds a strong foundation in key mathematical concepts for Stage 6 learners. This article breaks down the main question types found in the book, offering clear explanations and step-by-step strategies. Understanding these patterns will help students tackle assessments with confidence and deepen their reasoning skills.
《剑桥小学数学学习者用书第六册第二版》为六年级学生打下坚实的数学基础。本文梳理了教材中的主要题型，提供清晰的讲解和分步解题策略。掌握这些题型特征，能帮助学生自信应对测评，并深化推理能力。
1. Place Value and Number Patterns | 位值与数字模式
Place value questions test the understanding of digits in numbers up to millions and decimal fractions to thousandths. A typical task asks: ‘What is the value of the digit 8 in 482 509?’ Since 8 is in the ten-thousands place, its value is 80 000. For decimals, ‘Write 3.057 in expanded form’ would be 3 + 0.05 + 0.007.
位值类题目考查学生是否理解数字中每一位的含义，范围涵盖百万位及千分位小数。常见问法如“数字 482 509 中的 8 表示什么？”因为 8 在万位，所以它的值是 80 000。对于小数，“将 3.057 写成展开式”应为 3 + 0.05 + 0.007。
Number pattern exercises require identifying the rule and extending sequences, e.g. 0.25, 0.5, 0.75, 1.0, … Learners must recognise the step of +0.25. Another type involves completing missing terms in a pattern like 72, 36, 18, __, 4.5, where the rule is dividing by 2.
数字规律题要求学生找出规则并延续序列，例如 0.25, 0.5, 0.75, 1.0,… 学生须识别出每次增加 0.25 的规律。另一种题型是补全缺失项，如 72, 36, 18, __, 4.5，规律为每次除以 2。
2. Addition and Subtraction Strategies | 加减法策略
Mental and written methods are assessed through multi-step problems and estimation. A classic compensation example is 4567 + 2998. By adding 3000 and subtracting 2, we get 4567 + 3000 = 7567, then 7567 − 2 = 7565. Checking with the inverse operation (7565 − 2998 = 4567) confirms the answer.
心算与笔算策略通过多步问题和估算题来考查。典型的补偿法例子是 4567 + 2998。通过加上 3000 再减去 2，得到 4567 + 3000 = 7567，然后 7567 − 2 = 7565。用逆运算检验（7565 − 2998 = 4567）即可确认答案正确。
Questions often present missing digit puzzles, such as 3 __ 5 2 + 1 4 __ = 4 9 1 2. Students must use column addition and place value reasoning to find that the missing digits are 4 and 6. Subtraction tasks may include finding the difference between two timings or adjusting recipes.
经常出现的还有缺数谜题，例如 3 __ 5 2 + 1 4 __ = 4 9 1 2。学生需要运用竖式加法和位值推理，得出缺失数字分别为 4 和 6。减法任务则可能包含计算两个时刻之间的时长差或调整食谱分量。
3. Multiplication and Division | 乘除法
Learners multiply up to four-digit numbers by two-digit numbers and divide with remainders expressed as fractions or decimals. For 234 × 56, the area model or long multiplication yields: 234 × 50 = 11 700, 234 × 6 = 1404, total = 13 104. Another method is the grid, which reinforces place value.
学生需要掌握四位数乘两位数，以及将余数表示为分数或小数的除法。以 234 × 56 为例，采用面积模型或长乘法：234 × 50 = 11 700，234 × 6 = 1404，总和为 13 104。另一种格子法能强化位值概念。
Division questions often ask ‘How many 15s are there in 375?’ or ‘Divide 489 by 7, giving the remainder as a fraction.’ Using short division, 489 ÷ 7 = 69 with a remainder of 6, so 69 6/7. Real-life contexts include sharing money equally or finding numbers of packets.
除法题常问“375 里面有多少个 15？”或“计算 489 ÷ 7，将余数写成分数”。用短除法，489 ÷ 7 = 69 余 6，因此答案为 69 6/7。真实情境包括平均分钱或计算包装数量。
4. Fractions, Decimals and Percentages | 分数、小数和百分数
Conversions and comparisons form a major part of the curriculum. A typical ordering question: Arrange 0.6, ⅔, 65% from least to greatest. Converting all to decimals gives 0.6, 0.666…, 0.65, so the order is 0.6, 65%, ⅔. Learners must also simplify fractions like 18/24 to ¾.
换算与比较是本阶段的重要部分。典型排序题：将 0.6、⅔ 和 65% 从小到大排列。全部转换为小数后得到 0.6、0.666…、0.65，因此正确顺序是 0.6、65%、⅔。学生还需给分数约分，例如将 18/24 化为 ¾。
| Fraction | Decimal | Percentage | 分数 | 小数 | 百分数 |
|---|---|---|---|---|---|
| ½ | 0.5 | 50% | ½ | 0.5 | 50% |
| ⅓ | 0.333… | 33 ⅓% | ⅓ | 0.333… | 33 ⅓% |
| ¾ | 0.75 | 75% | ¾ | 0.75 | 75% |
Problem-solving tasks often involve finding a fraction of an amount, e.g. ‘What is ⅗ of 200?’ The method: divide 200 by 5 to get 40, then multiply by 3 to obtain 120. Percentage increase and decrease problems also appear, such as ‘A £60 shirt is reduced by 15%. What is the sale price?’ 15% of £60 = £9, so sale price = £51.
解决问题类题型常涉及求一个数量的几分之几，例如“200 的 ⅗ 是多少？” 解法：200 ÷ 5 = 40，再乘以 3 得到 120。此外还有百分数增减问题，比如“一件 60 英镑的衬衫降价 15%，售价是多少？” 60 英镑的 15% 是 9 英镑，所以售价为 51 英镑。
5. Ratio and Proportion | 比和比例
Learners simplify ratios and use proportion to scale recipes or mixtures. A question might state: ‘A fruit salad uses 24 grapes and 16 blueberries. Write the ratio in simplest form.’ Dividing both by 8 gives 3 : 2. Another common task: ‘A model car is built to a scale of 1 : 50. If the real car is 4.5 m long, how long is the model in cm?’ Convert 4.5 m to 450 cm, then divide by 50 to obtain 9 cm.
学生需要化简比并利用比例调整食谱或混合物。例如：“一份水果沙拉用了 24 颗葡萄和 16 颗蓝莓，写出最简整数比。”将两者同除以 8 得 3 : 2。另一种常见题型：“一辆汽车模型按 1 : 50 的比例制作。真车长 4.5 米，模型长多少厘米？” 先将 4.5 米化为 450 厘米，再除以 50 得到 9 厘米。
Sharing in a given ratio is also assessed. ‘Share £120 in the ratio 2 : 3’ means there are 5 parts in total; one part = £120 ÷ 5 = £24, so the shares are 2 × £24 = £48 and 3 × £24 = £72. These problems connect division with ratio concepts.
按给定比例分配的题目同样会出现。“将 120 英镑按 2 : 3 分配”意味着总共 5 份；每份为 120 ÷ 5 = 24 英镑，因此两份为 48 英镑，三份为 72 英镑。这类问题将除法与比的概念联系起来。
6. Algebra and Simple Equations | 代数与简单方程
Simple equations and function machines build early algebraic thinking. A typical equation: 3x + 5 = 20. Subtract 5 from both sides to get 3x = 15, then divide by 3 to find x = 5. Learners also solve problems like ‘I think of a number, multiply it by 4, add 7, and get 35. What is the number?’ This leads to 4n + 7 = 35, so n = 7.
简单方程与函数机器培养初步的代数思维。典型方程如 3x + 5 = 20。两边同时减去 5 得 3x = 15，再除以 3 得出 x = 5。学生还需解答类似“我想一个数，乘以 4，再加 7，得到 35。我想的数是多少？”的问题，可列出 4n + 7 = 35，解得 n = 7。
Function machines show input → rule → output. Given the rule ‘×3 then –2’, and input 5, the output is 13. Reversing the machine to find the input from the output strengthens inverse operation skills. Completing number sequences with algebraic rules, like the nth term 3n+1 giving 4, 7, 10…, is another key area.
函数机器呈现“输入 → 规则 → 输出”的模式。已知规则为“×3 然后 –2”，输入 5，则输出为 13。反向运算以根据输出求出输入，能强化逆运算技能。用代数规律补全数列也是重点之一，例如第 n 项为 3n+1 的数列：4, 7, 10…
7. Measurement and Unit Conversions | 测量与单位换算
Questions involve converting between units of length, mass, volume and time. For instance, ‘Convert 3.6 km to metres’ requires multiplying by 1000 to get 3600 m. ‘A bottle holds 1.5 litres; how many 200 ml cups can be filled?’ 1.5 L = 1500 ml, 1500 ÷ 200 = 7.5, so 7 full cups with 100 ml left over.
测量类题目涉及长度、质量、容积和时间的单位换算。例如“将 3.6 千米换算为米”，需乘以 1000 得 3600 米。“一瓶装 1.5 升水，可以装满多少个 200 毫升的杯子？”1.5 升 = 1500 毫升，1500 ÷ 200 = 7.5，即装满 7 杯，余 100 毫升。
Time calculations may ask: ‘A film starts at 14:35 and runs for 1 hour 50 minutes. What time does it end?’ Adding gives 15:25 + 50 minutes = 16:25. Time zone problems and timetables also feature. Perimeter and area problems often appear in measurement sections (detailed later).
时间计算可能问：“一部电影 14:35 开始，片长 1 小时 50 分钟，什么时候结束？”相加得 15:25 + 50 分钟 = 16:25。时区问题和时间表也是常见题型。周长与面积问题经常出现在测量章节中（将在后文详述）。
8. Geometry: Angles, Triangles and Shapes | 几何：角、三角形与形状
Learners calculate missing angles using known facts: angles on a straight line sum to 180°, angles around a point sum to 360°, and the interior angles of a triangle total 180°. For example, if a triangle has angles of 55° and 70°, the third angle is 180° − (55° + 70°) = 55°. In a parallelogram, opposite angles are equal, and adjacent angles sum to 180°.
学生利用已知几何事实计算未知角：直线上的角之和为 180°，绕一个点的所有角之和为 360°，三角形内角和为 180°。例如，已知三角形两角为 55° 和 70°，则第三角为 180° − (55° + 70°) = 55°。平行四边形中，对角相等，邻角互补。
Questions on properties of shapes ask learners to classify triangles as equilateral, isosceles or scalene based on side lengths and angles. Recognising nets of 3D shapes, identifying lines of symmetry and drawing reflections are included. For instance, ‘A regular pentagon has how many lines of symmetry?’ Answer: 5.
图形性质类题目要求学生根据边长和角度将三角形分类为等边、等腰或不等边三角形。此外还包括辨认立体图形的展开图、找出对称轴以及绘制反射图形。例如：“正五边形有几条对称轴？”答案是 5 条。
9. Perimeter, Area and Volume | 周长、面积和体积
Formulas for rectangles are applied: area = length × width, perimeter = 2 × (length + width). A typical problem: ‘A rectangular field is 28 m long and 15 m wide. Find the area and the perimeter.’ Area = 28 × 15 = 420 m²; perimeter = 2 × (28 + 15) = 86 m. Learners often must work backwards from area or perimeter to find missing side lengths.
矩形公式的应用：面积 = 长 × 宽，周长 = 2 × (长 + 宽)。典型题目：“一块长方形田地长 28 米、宽 15 米。求它的面积和周长。”面积 = 28 × 15 = 420 m²；周长 = 2 × (28 + 15) = 86 m。学生经常需要从面积或周长反推缺失的边长。
Compound shapes are broken into rectangles. For an L-shaped figure, students divide it into two rectangles, compute each area and sum them. Volume of cubes and cuboids is calculated as length × width × height, with answers in cm³ or m³. ‘A box is 6 cm by 5 cm by 4 cm. What is its volume?’ Volume = 6 × 5 × 4 = 120 cm³.
组合图形的面积通过分割为矩形来计算。对L形图，学生将其分成两个矩形，分别求面积再相加。正方体和长方体的体积使用长 × 宽 × 高计算，单位为 cm³ 或 m³。“一个盒子长 6 厘米、宽 5 厘米、高 4 厘米，体积是多少？”体积 = 6 × 5 × 4 = 120 cm³。
10. Statistics: Graphs and Averages | 统计：图表与平均数
Interpreting bar charts, line graphs and pie charts is a core skill. A bar chart question might ask: ‘How many more pupils chose blue than red?’ Students read the scale and subtract. Line graphs often show temperature or distance over time; learners describe trends and answer values at specific points.
解读柱状图、折线图和饼图是一项核心技能。柱状图题可能问：“选择蓝色的学生比选择红色的多多少人？”学生读取刻度并做减法。折线图常展示温度或距离随时间的变化；学生需要描述趋势并回答特定时刻的数值。
The mean average is calculated by dividing the sum by the number of values. For instance, ‘Find the mean of 14, 18, 22, 26.’ Sum = 80, mean = 20. Some exercises require finding the total from the mean and number of items. Learners also compare two sets of data using range as a measure of spread.
平均数通过总和除以数据个数得出。例如“求 14、18、22、26 的平均数”。总和为 80，平均数为 20。有些练习要求根据平均数和个数反推总和。学生还会用极差作为衡量离散程度的指标，对比两组数据。
11. Probability | 概率
Probability is expressed as a fraction of favourable outcomes over total possible outcomes. A bag contains 3 red, 2 blue and 5 green marbles. The probability of picking a blue marble is 2/10, simplified to ⅕. The probability of not picking red is 1 − 3/10 = 7/10.
概率用有利结果数目除以可能结果总数所得的分数表示。一个袋子里有 3 个红球、2 个蓝球和 5 个绿球。抽到蓝球的概率是 2/10，约简为 ⅕。抽不到红球的概率是 1 − 3/10 = 7/10。
Questions on the probability scale place events from 0 (impossible) to 1 (certain). For example, ‘Rolling a standard die and getting a 7’ is 0, while ‘getting a number less than 7’ is 1. Spinner problems with equally sized sectors test addition of fractions for multiple event outcomes.
概率标尺题目将事件从 0（不可能）到 1（一定）进行排列。例如，“掷一枚标准骰子得 7”的概率是 0，而“得到小于 7 的数”的概率是 1。转盘问题中各扇区大小相等，考查多个事件结果对应分数的加和运算。
12. Word Problems and Logical Reasoning | 文字题与逻辑推理
Multi-step word problems integrate various skills. An example: ‘Tom has twice as many stickers as Jane. Together they have 36 stickers. How many does each have?’ Let Jane’s stickers be x, Tom’s 2x, so 3x = 36
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