Cambridge Primary Mathematics Workbook 6, 2nd Edition Question Type Analysis | Cambridge 小学数学练习册6 第二版题型解析

📚 Cambridge Primary Mathematics Workbook 6, 2nd Edition Question Type Analysis | Cambridge 小学数学练习册6 第二版题型解析

Stepping into the final year of primary mathematics, Workbook 6 of the Cambridge Primary Mathematics series (2nd Edition) bridges foundational skills and the more abstract thinking required for lower secondary. This article distills the core question types found within each unit, offering learners and educators a clear roadmap to tackle everything from multi-step word problems to nascent algebraic reasoning. By understanding the structure and intention behind each exercise, students can transform workbook practice from rote completion into genuine mathematical confidence.

进入小学阶段最后一年的数学学习，Cambridge Primary Mathematics Workbook 6（第二版）连接了基础技能与初中所需更抽象的思维。本文提炼了每个单元中的核心题型，为学习者和教育者提供清晰路线图，以应对从多步骤应用题到初步代数推理的各类问题。通过理解每个练习背后的结构与意图，学生可以将练习册从机械完成转化为真正的数学信心。

1. Place Value and Ordering Whole Numbers up to Ten Million | 数位与千万以内整数的排序

Exercises on place value consistently ask learners to dissect multi-digit numbers by identifying the value of a specific digit. For example, ‘What is the value of the digit 7 in the number 4,728,105?’ The answer is 700,000, not just 7. Questions also demand ordering a set of six or seven large numbers from smallest to largest, requiring careful comparison of digits in the millions, hundred-thousands, and thousands places. A secondary skill tested here is rounding large numbers to the nearest 10, 100, 1000, or even 100,000, where attention to the digit immediately to the right of the target place is paramount.

关于数位的练习持续要求学习者剖析多位数，识别特定位数的数值。例如：’数字 4,728,105 中，7 这个数字的价值是什么？’ 答案是 700,000，而不仅仅是 7。题目还要求将六个或七个大数按从小到大的顺序排列，这需要仔细比较百万位、十万位和千位上的数字。这里测试的另一项技能是将大数四舍五入到最近的 10、100、1000 甚至 100,000，此时关注目标数位紧右边的数字至关重要。

2. Multiplying and Dividing by Powers of 10 with Decimals | 利用10的幂进行小数乘除

This question type cements the decimal point’s movement rather than simply appending or removing zeros. When multiplying a decimal such as 0.458 by 100, the digits shift two places to the left, yielding 45.8. Corresponding division questions ask for 34.7 divided by 100, resulting in 0.347. Workbook 6 frequently embeds these calculations within problem-solving contexts, such as converting between centimetres and metres (1.2 m = 120 cm) or grams and kilograms (45 g = 0.045 kg). Some problems challenge learners to complete a statement like 0.56 x _____ = 560, which requires recognising a shift of 1000, or three places.

这一题型强化小数点的移动规律，而非简单地增加或移除零。当将 0.458 这样的小数乘以 100 时，数字向左移动两位，得到 45.8。相应的除法题要求将 34.7 除以 100，结果是 0.347。练习册 6 经常将这些计算嵌入到问题解决的背景中，例如在厘米和米之间进行转换（1.2 米 = 120 厘米），或在克与千克之间进行转换（45 克 = 0.045 千克）。有些问题要求学习者完成诸如 0.56 × _____ = 560 这样的等式，这需要识别出移动三位，即乘以 1000。

3. Comparing, Adding, and Subtracting Fractions with Unlike Denominators | 异分母分数的比较和加减

Learners frequently encounter questions requiring them to find the lowest common denominator (LCD) for fractions like 2/3 and 3/5 to order them or calculate their sum. A typical workbook exercise might ask: ‘Which is greater, 5/8 or 7/12? Show your working.’ The solution requires converting both fractions to a common denominator of 24 — giving 15/24 and 14/24 — thus determining 5/8 is greater. Addition and subtraction problems push mastery further by mixing three fractions, often with one improper fraction or a mixed number. Diagrams of fraction bars sometimes accompany these tasks to support visual understanding.

学习者经常会遇到需要为 2/3 和 3/5 这样的分数寻找最小公分母（LCD）来排序或计算总和的问题。一道典型的练习册题目可能会问：’5/8 和 7/12 哪个更大？展示你的运算步骤。’ 解答时需要将两个分数转换为公分母 24 — 分别得到 15/24 和 14/24 — 从而确定 5/8 更大。加减法问题通过混合三个分数进一步突破掌握程度，通常包含一个假分数或带分数。这些任务有时会附带分数条图示，以支持视觉理解。

4. Long Multiplication with Two Digits and Area Models | 两位数长乘法与面积模型

Workbook 6 systematically develops multi-digit multiplication using the formal column method and the area model. A standard question presents 324 x 17 and expects a full written method, emphasizing the placeholder zero when multiplying by the tens digit. The area model (or grid method) appears as a strategy question: partition 324 into 300, 20, and 4, and 17 into 10 and 7; then calculate partial products such as 300 x 10 = 3000, 300 x 7 = 2100, 20 x 10 = 200, 20 x 7 = 140, 4 x 10 = 40, and 4 x 7 = 28; finally sum all parts to get 5508. Word problems contextualized these skills in real-life scenarios — for instance, calculating the total number of paper sheets in 256 packs of 24.

练习册 6 使用正式的竖式方法和面积模型系统地培养多位数乘法技能。一个标准题目给出 324 × 17，期望呈现完整的书面方法，强调乘以十位数时使用占位零。面积模型（或网格法）作为一种策略题出现：将 324 分解为 300、20 和 4，将 17 分解为 10 和 7；然后计算部分积，如 300 × 10 = 3000、300 × 7 = 2100、20 × 10 = 200、20 × 7 = 140、4 × 10 = 40、4 × 7 = 28；最后将所有部分相加得到 5508。应用题将这些技能置于真实生活场景中——例如，计算 256 包每包 24 张的纸张总数。

5. Dividing by a Two-Digit Number with Remainders | 除以两位数的带余除法

Division questions in the workbook progress from simple short division to the formal long division layout when dividing by numbers between 11 and 25. A typical calculation is 987 divided by 15. The algorithm is presented stepwise: how many 15s in 98? 6 (that makes 90), remainder 8, bring down the 7 to make 87. How many 15s in 87? 5 (making 75), remainder 12. The answer is expressed as 65 remainder 12, or increasingly, as a fraction 65 12/15 simplified to 65 4/5, or a decimal 65.8. Many problems then ask learners to interpret the remainder contextually — for example, if 350 children need minibuses that hold 15, how many minibuses are needed? The answer is 24, because 23 full minibuses leave 5 children stranded.

练习册中的除法问题从简单的短除法过渡到除以 11 到 25 之间数字时的正式长除法格式。一个典型计算是 987 除以 15。算法逐步呈现：98 里面有多少个 15？有 6 个（得出 90），余数 8，把 7 放下来形成 87。87 里面有多少个 15？有 5 个（得出 75），余数 12。答案表示为 65 余 12，或者越来越多地表示为分数 65 12/15，化简为 65 4/5，或表示为小数 65.8。许多问题随后要求学习者根据上下文解释余数——例如，如果 350 名儿童需要每辆载客 15 人的小巴，需要多少辆小巴？答案是 24 辆，因为 23 辆满载小巴后会剩下 5 名儿童。

6. Operations with Decimals and Money | 小数与货币运算

Money calculations provide a motivating context for adding, subtracting, multiplying, and dividing decimals to two decimal places. A typical question presents an item priced at £3.49 with a 20% discount, asking for the new price. The learner first finds 10% (0.349, rounded practically to 0.35) and doubles it to get the discount of £0.70, then subtracts from £3.49 to get £2.79. Other problems involve finding the total cost for multiple items, such as 6 notebooks at £1.25 each. Multiplication of a decimal by a single digit — 1.25 x 6 — gets the answer 7.50. Rounding to the nearest pound or penny appears as an independent skill: rounding £9.08 to the nearest pound gives £9, while £9.58 rounds to £10.

货币计算为加减乘除两位小数提供了一个激励性的背景。一道典型题目给出的商品标价为 3.49 英镑，享受 20% 折扣，要求计算出新价格。学习者首先求出 10%（0.349，实际四舍五入为 0.35），然后翻倍得到 0.70 英镑的折扣，再从 3.49 英镑中减去得到 2.79 英镑。其他问题涉及求多件商品的总价，例如 6 本笔记本每本 1.25 英镑。小数乘以一位数——1.25 × 6 ——得到答案 7.50。四舍五入到最近的英镑或便士作为一项独立技能出现：将 9.08 英镑四舍五入到最近的英镑是 9 英镑，而 9.58 英镑四舍五入后为 10 英镑。

7. Converting and Calculating with Fractions, Decimals, and Percentages | 分数、小数和百分比的转换与计算

Fluency in converting between forms is tested intensively. Questions present a table with one column filled and the others blank: e.g., Fraction = 3/5, Decimal = ?, Percentage = ?. The learner fills in Decimal = 0.6 and Percentage = 60%. More complex problems ask for percentages of quantities, such as 15% of 240. Methods vary — some learners find 10% (24) and 5% (12) and sum them to reach 36, while others multiply 240 by 0.15. Workbook 6 extends this to finding the whole from a given percentage: ‘If 30% of a number is 45, what is the number?’ Setting up the equation 0.3 x n = 45 leads to n = 150, or reasoning that 1% is 45 / 30 = 1.5, so 100% is 150.

形式之间的转换流利度受到密集测试。题目会给出一个表格，其中一列已填好，其余空白：例如，分数 = 3/5，小数 = ?，百分比 = ?。学习者填入小数 = 0.6 和百分比 = 60%。更复杂的问题要求求出一个数量的百分比，例如 240 的 15%。方法各异——一些学习者求出 10%（24）和 5%（12）并相加得到 36，而其他人用 240 乘以 0.15。练习册 6 将此延伸至根据给定百分比求整体：’如果一个数的 30% 是 45，这个数是多少？’ 设立方程 0.3 × n = 45 得出 n = 150，或者推理 1% 是 45 ÷ 30 = 1.5，所以 100% 就是 150。

8. Geometric Reasoning: Angles, Triangles, and Quadrilaterals | 几何推理：角、三角形和四边形

Questions on angles frequently give a straight line with one angle known (e.g., 117°) and ask for the adjacent missing angle (63°). More advanced exercises embed angle-finding within intersecting lines or around a point, where the sum must be 360°. In triangles, learners use the fact that the sum of interior angles is 180° to find an unknown angle given the other two. Quadrilateral problems extend this to 360°. Some workbook pages involve drawing triangles with given side lengths using a ruler and compass, and then classifying them as scalene, isosceles, or equilateral. Others ask for the coordinates of vertices after a shape has been reflected or translated on a grid.

关于角的题目经常给出一条直线上的一个已知角（例如 117°），然后要求求出相邻的未知角（63°）。更高级的练习将求角问题嵌入相交线或围绕一点的情境中，此时总和必须为 360°。在三角形中，学习者利用内角和为 180° 这一事实，在已知另外两个角的情况下求出未知角。四边形问题将此延伸至 360°。一些练习册页面涉及使用直尺和圆规绘制给定边长的三角形，然后将其分类为不等边、等腰或等边三角形。其他问题要求在网格上对一个图形进行反射或平移后给出顶点的坐标。

9. Statistical Graphs and Interpreting Data | 统计图表与数据解读

The workbook presents a variety of data displays, including bar charts, dual bar charts, line graphs, pie charts, and Carroll diagrams. A typical double-bar chart question asks comparative questions: ‘On which day was the difference between the number of hot drinks and cold drinks sold the greatest?’ This demands reading the graph accurately and subtracting two values. Pie chart exercises require linking proportions to percentages and fractions — if one sector of a pie chart representing ‘walk’ occupies a quarter of the circle, that represents 25% of students, or 1/4. More subtle questions give the total number surveyed and ask learners to calculate exact frequencies from sector angles or percentages.

练习册展示了多种数据展示形式，包括条形图、复式条形图、折线图、饼图和卡罗尔图。一道典型的复式条形图题目会提出比较性问题：’哪一天热饮和冷饮的销售数量差异最大？’ 这需要准确读取图表并将两个数值相减。饼图练习要求将比例与百分比和分数联系起来——如果代表’步行’的饼图扇区占据了圆的四分之一，那代表 25% 的学生，即 1/4。更精细的问题会给出被调查的总人数，并要求学习者根据扇区角度或百分比计算确切频数。

10. Algebra, Sequences, and Simple Equations | 代数、数列与简单方程

Algebraic thinking emerges strongly in the second half of the workbook. Sequence questions move beyond simple linear patterns to those involving a rule applied to the previous term. A sequence like 3, 6, 12, 24, 48… follows the rule ‘multiply by 2’. Finding the nth term uses patterns such as ‘multiply by 3 and add 1’ leading to the expression 3n + 1. Equation-solving exercises start with single-step equations like x + 7 = 15, where x = 8. They quickly progress to two-step equations: 4y – 3 = 17. Here learners add 3 to both sides (4y = 20) and then divide by 4 (y = 5). Word problems form the culmination, translating phrases like ‘I think of a number, multiply it by 6, subtract 7, and get 35’ into 6n – 7 = 35 and solving for n = 7.

代数思维在练习册的后半部分强势出现。数列问题从简单的线性模式过渡到涉及对前一项应用运算规则的模式。像 3, 6, 12, 24, 48… 这样的数列遵循’乘以 2’的规则。求第 n 项用到的模式如’乘以 3 再加 1’导致表达式 3n + 1。解方程练习从单步方程开始，例如 x + 7 = 15，其中 x = 8。它们迅速进展到两步方程：4y – 3 = 17。这里学习者给两边加上 3（得到 4y = 20），然后除以 4（y = 5）。应用题是顶峰，将诸如’我想了一个数，将它乘以 6，减去 7，得到 35’这样的语句转化为 6n – 7 = 35 并解出 n = 7。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Cambridge Primary Mathematics Workbook 6, 2nd Edition Question Type Analysis | Cambridge 小学数学练习册6 第二版 题型解析