📚 Cambridge Primary Mathematics Workbook 5 Key Concepts Explained | 剑桥小学数学练习册5（第二版）知识点精讲
The Cambridge Primary Mathematics Workbook 5 (2nd Edition) is designed to strengthen foundational maths skills for learners typically aged 9-10. It aligns with the Cambridge Primary curriculum and provides extensive practice in number, geometry, measurement, data handling and problem solving. In this article we , we will walk through the essential topics covered in the workbook, explaining key concepts and the standard methods behind them, so that students can build confidence and achieve mastery.
剑桥小学数学练习册5（第二版）专为培养 9 至 10 岁学生的基础数学能力而编写，内容紧密贴合剑桥小学课程大纲。该练习册涵盖数、几何、测量、数据处理和解决问题等广泛内容，并提供大量练习。本文将精讲练习册涉及的核心主题，解析关键概念与标准方法，帮助学生树立信心、学深悟透。
1. Place Value and Number Sense up to 1,000,000 | 百万以内的位值与数感
In Year 5, learners extend their understanding of place value to numbers up to one million. They are expected to read, write, compare and order these numbers confidently. The key concept is that each digit has a value ten times greater as it moves one place to the left, so a digit's worth depends on its position, not just on the digit itself.
在五年级，学生需要将位值概念拓展到百万以内的数，应能熟练地读、写、比较和排序这些数。核心概念是：每向左移动一位，数值就变为原来的十倍，因此一个数字的值取决于它所在的位置，而不只是数字本身。
In the number 506,432, the digit 5 represents five hundred thousands, 0 holds the ten thousands place, 6 stands for six thousands, 4 for four tens and 2 for two ones. Partitioning a number into its place-value parts supports both mental arithmetic and written calculation.
在 506,432 中，数字 5 代表五十万，0 占住万位，6 代表六千，4 代表四个十，2 代表两个一。把一个数按位值拆开，有助于心算和笔算。
392,147 = 300,000 + 90,000 + 2,000 + 100 + 40 + 7
Comparing numbers up to 1,000,000 involves checking the highest place value first. For instance, 745,230 > 745,023 because both have seven hundred thousands and four ten thousands, but the thousands place differs: 5 thousands against 0 thousands. Rounding uses the same logic - to round to the nearest 1,000, look at the hundreds digit and round up when it is 5 or more.
比较百万以内的数时，首先看最高位。例如 745,230 > 745,023，因为两者都有 7 个十万和 4 个万，但千位不同：5 个千和 0 个千。四舍五入遵循同样的位值逻辑——四舍五入到最接近的 1,000 时，看百位数字，如果是 5 或更大就向前进一。
0), C (100), D (500) and M (1,000). For example, MCMLXXXIV = 1,000 + 900 + 80 + 4 = 1,984.
本阶段还会学习 1,000 以内的罗马数字，符号包括 I（1）、V（5）、X（10）、L（50）、C（100）、D（500）和 M（1,000）。例如 MCMLXXXIV = 1,000 + 900 + 80 + 4 = 1,984。
2. Negative Numbers in Real-Life Contexts | 负数的实际应用
Negative numbers are introduced through temperature readings, floors below ground, bank balances and positions on a number line. Learners need to count forwards and backwards through zero, which is the trickiest step: the sequence goes 3, 2, 1, 0, -1, -2, -3.
负数通过温度读数、地下楼层、银行余额以及数轴上的位置引入。学生需要能跨过 0 正向和反向数数，这是最容易出错的一步：数列依次为 3、2、1、0、-1、-2、-3。
On a number line, numbers to the right are greater than numbers to the left, so -2 is greater than -7. A common mistake is to think that -7 is larger because 7 > 2; the minus sign reverses the order.
在数轴上，右边的数总比左边的数大，所以 -2 大于 -7。常见错误是认为 -7 更大，因为 7 > 2；负号会颠倒大小顺序。
Temperature questions are the most frequent exam-style task. The rise from -3 °C to 5 °C is found by counting up: 3 steps to reach 0, then 5 more steps, giving a rise of 8 °C. Finding the difference between a positive and a negative value always means adding the two distances from zero.
温度类问题是练习册中出现最多的题型。从 -3 °C 升到 5 °C，用数数的方式求解：先数 3 步到 0，再数 5 步，共升高 8 °C。求一个正数与一个负数之间的差，总是把它们到 0 的距离相加。
Difference = 5 − (−3) = 5 + 3 = 8 °C
3. The Four Operations and Written Methods | 四则运算与笔算方法
oduces a carry of 1 into the tens. Subtraction works the same way in reverse: when the top digit is too small, exchange one unit from the column on its left.
加法和减法的笔算方法，关键在于把相同位值的数字对齐。3,476 + 2,589 = 6,065 中，个位列 6 + 9 = 15，向十位进 1。减法原理相同，只是方向相反：当被减数的某一位不够减时，从左边相邻的位借 1。
ial products: 1,652 + 9,440 = 11,092. Long division reverses the process, and learners must be ready for a zero in the quotient, as in 4,368 ÷ 6 = 728.
长乘法把两位数乘数拆成十位和个位。计算 236 × 47 时，先乘 7 得 1,652，再乘 40 得 9,440，最后把部分积相加：1,652 + 9,440 = 11,092。长除法是相反的过程，学生还要会处理商中间出现 0 的情况，例如 4,368 ÷ 6 = 728。
Estimation is used to check answers: rounding 236 to 200 and 47 to 50 gives about 10,000, which confirms that 11,092 is reasonable. The order of operations follows BODMAS - brackets, orders (powers and roots), division and multiplication, then addition and subtraction.
估算用来检验答案：把 236 看成 200、47 看成 50，大约得 10,000，说明 11,092 是合理的。运算顺序遵循 BODMAS——先括号，再乘方（幂与根），然后乘除，最后加减。
6 + 4 × 5 = 6 + 20 = 26 (not 50)
4. Factors, Multiples, Primes, Squares and Cubes | 因数、倍数、质数、平方数与
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