📚 7S Homework Contents List: Complete 12-Section Revision Checklist | 7S 数学作业内容清单：十二模块完整复习清单
This revision guide presents a complete contents list for the 7S mathematics homework programme. It organises the key topics into twelve manageable sections, each with the essential skills, common question types and useful study tips. Use the list as a checklist to track your progress and identify areas for further practice.
本复习指南呈现 7S 数学作业计划的完整内容清单。它将关键主题分为十二个易于掌握的模块，每个模块都列明核心技能、常见题型和实用学习提示。请将本清单用作进度检查表，以跟踪学习进展并找出需要进一步练习的薄弱环节。
Every section states what the homework covers, shows worked examples in the same style as the actual tasks, and lists the items you should be able to do without help before you tick that section off. Read each English paragraph together with the Chinese paragraph that follows it: the English sentences carry the mathematical vocabulary your teacher uses in class and in tests, while the Chinese sentences explain the same idea in the language you think in.
每个模块都会说明作业的覆盖范围，给出与真实题目同风格的示例，并列出在打勾确认完成之前你应能独立完成的条目。请把每一段英文与其后的中文对照阅读：英文句子承载课堂上和考试中老师会使用的数学词汇，中文句子则用你思考时所用的语言解释同一概念。
1. Number and Place Value | 数与位值
Work in this section covers reading, writing, ordering and comparing whole numbers, decimals and negative numbers. Students learn to identify place value, round to any specified accuracy, and use estimation to check answers.
本部分的作业涵盖整数、小数和负数的读法、写法、排序与比较。学生将学习识别位值、按要求精度进行四舍五入，并利用估算检查答案。
A typical homework set asks learners to round 5,643 to the nearest 100, write the value of the digit 7 in 4,782,309, and arrange -7, -2, 0, 3, 5 on a number line. These tasks build fluency with large numbers and signed numbers.
一套典型的作业会要求学生将 5,643 四舍五入到最近的百位，写出 4,782,309 中数字 7 的值，并在数轴上排列 -7、-2、0、3、5。这些任务帮助建立大数和带符号数的运算流利度。
Place value questions are marked quickly but lose marks easily, because a single misplaced digit changes the value by a factor of ten. Always write the digits in columns first, label the column headings (millions, hundred thousands, ten thousands, thousands, hundreds, tens, ones) and only then answer.
位值题的批改速度很快，却极易丢分，因为一个放错位置的数字就会让数值相差十倍。请先把数字按列写出，标出列标题（百万、十万、万、千、百、十、个），然后才作答。
- Identify place value up to 1,000,000,000 | 识别最大到 1,000,000,000 的位值
- Round 5,643 to the nearest 100 | 将 5,643 四舍五入到最近的百位
- Order negative and positive numbers | 排列负数和正数
- Use estimation to check a calculation | 用估算检验计算结果
4,782,309 = 4 × 10⁶ + 7 × 10⁵ + 8 × 10⁴ + 2 × 10³ + 3 × 10² + 0 × 10¹ + 9 × 10⁰
5,643 ≈ 5,600 (to the nearest 100)
When rounding, look only at the digit immediately to the right of the required place: 4 or less rounds down, 5 or more rounds up. When ordering signed numbers, remember that negative numbers get smaller as they move away from zero, so -7 < -2 < 0 < 3 < 5.
四舍五入时只看目标数位右边紧邻的那一位：4 及以下舍去，5 及以上进位。排列带符号数时记住，负数离零越远越小，因此 -7 < -2 < 0 < 3 < 5。
2. Four Operations | 四则运算
This section consolidates mental and written methods for addition, subtraction, multiplication and division. Homework includes long multiplication, short and long division, and word problems that require choosing the correct operation.
本节巩固加减乘除的心算与笔算方法。作业包括长乘法、短除法和长除法，以及需要选择正确运算的文字题。
MAS. Questions such as 3 + 4 × 5 must be solved as 3 + (4 × 5) = 23, not (3 + 4) × 5 = 35.
学生还必须正确运用运算顺序，通常记忆为 BIDMAS 或 BODMAS。例如 3 + 4 × 5 应按 3 + (4 × 5) = 23 计算，而不是 (3 + 4) × 5 = 35。
The five letters stand for Brackets, Indices (or Orders), Division and Multiplication, Addition and Subtraction. Division and multiplication have the same priority, and so do addition and subtraction: work from left to right within the same level of priority.
这五个字母分别代表括号（Brackets）、指数或幂（Indices / Orders）、除法与乘法、加法与减法。除法与乘法优先级相同，加法与减法也相同：同一优先级内从左到右依次计算。
- Multiply 234 × 36 using a formal written method | 用规范笔算方法计算 234 × 36
- Divide 4,536 ÷ 8 using short division | 用短除法计算 4,536 ÷ 8
- Apply BIDMAS to a mixed calculation | 对混合运算运用 BIDMAS
- Choose the correct operation in a word problem | 在文字题中选择正确的运算
- Check every answer with an inverse operation or an estimate | 用逆运算或估算检验每个答案
3 + 4 × 5 = 3 + 20 = 23
(3 + 4) × 5 = 7 × 5 = 35
234 × 36 = 8,424 4,536 ÷ 8 = 567
In word problems, highlight the key phrase before calculating. Words such as "altogether", "increase by" and "sum" signal addition; "difference", "decrease by" and "how many more" signal subtraction; "product", "each" and "times as many" signal multiplication; "share equally", "per" and "grouped into" signal division.
做文字题时先划出关键短语再计算。"总共""增加""之和"提示加法；"差""减少""多多少"提示减法；"积""每个""是几倍"提示乘法；"平均分成""每""分成几组"提示除法。
3. Factors, Multiples, Primes and Index Notation | 因数、倍数、质数与指数记法
Homework here covers listing factors and multiples, identifying prime numbers up to 100, finding the highest common factor (HCF) and lowest common multiple (LCM), writing a number as a product of its prime factors, and using square and cube numbers with their roots.
本部分作业涵盖列举因数与倍数、找出 100 以内的质数、求最大公因数（HCF）与最小公倍数（LCM）、把数写成质因数乘积，以及平方数、立方数及其方根。
2, and the three index laws you need are aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ and (aᵐ)ⁿ = aᵐ⁽ⁿ⁾.
指数记法同时引入：2⁵ 表示 2 × 2 × 2 × 2 × 2 = 32。你需要掌握的三条指数法则是 aᵐ × aⁿ = aᵐ⁺ⁿ、aᵐ ÷ aⁿ = aᵐ⁻ⁿ 以及 (aᵐ)ⁿ = aᵐⁿ。
- List all factors of 36 | 列举 36 的所有因数
- Find HCF(24, 36) and LCM(4, 6) | 求 HCF(24, 36) 与 LCM(4, 6)
- Write 360 as a product of prime factors | 把 360 写成质因数乘积
- Evaluate √144 and 5³ | 计算 √144 与 5³
- Use the index laws to simplify algebraic powers | 用指数法则化简代数幂
360 = 2³ × 3² × 5
HCF(24, 36) = 12 LCM(4, 6) = 12
√144 = 12 5³ = 125 2⁵ = 32
Use factor trees or repeated division for prime factorisation, then write the answer with the primes in ascending order and in index form. For the HCF, take the product of the shared prime factors with the lowest powers; for the LCM, take the product of all prime factors with the highest powers.
用因数树或连续除法进行质因数分解，然后把质数按从小到大、用指数形式写出答案。求 HCF 时取公共质因数的最低次幂之积；求 LCM 时取所有质因数的最高次幂之积。
4. Fractions, Decimals and Percentages | 分数、小数与百分数
This section links the three ways of writing the same number. Homework includes simplifying fractions, converting between fractions, decimals and percentages, calculating a fraction or percentage of an amount, and applying percentage increase and decrease.
本节把同一个数的三种写法联系起来。作业包括约分、在分数、小数与百分数之间互化、计算某量的几分之几或百分之几，以及应用百分比的增减。
s, multiply and divide fractions, and work with common equivalences such as 1/2 = 0.5 = 50%, 1/4 = 0.25 = 25%, 3/4 = 0.75 = 75%, 1/5 = 0.2 = 20% and 1/8 = 0.125 = 12.5%.
你应能对分母不同的分数做加减、对分数做乘除，并熟练使用常见等价关系，如 1/2 = 0.5 = 50%、1/4 = 0.25 = 25%、3/4 = 0.75 = 75%、1/5 = 0.2 = 20% 以及 1/8 = 0.125 = 12.5%。
- Simplify 18/24 to its lowest terms | 把 18/24 约到最简
- Convert 3/4 to a decimal and a percentage | 把 3/4 化成小数和百分数
- Add 2/5 + 1/4 | 计算 2/5 + 1/4
- Find 15% of 240 | 求 240 的 15%
- Increase 80 by 12% and decrease 150 by 20% | 把 80 增加 12%，把 150 减少 20%
2/5 + 1/4 = 8/20 + 5/20 = 13/20
15% of 240 = 0.15 × 240 = 36
80 × 1.12 = 89.6 150 × 0.8 = 120
For percentage increase use the multiplier 1 + r, and for percentage decrease use 1 - r, where r is the rate written as a decimal. A 20% decrease means multiplying by 0.8, and a 12% increase means multiplying by 1.12. Always check whether the question asks for the change or the final amount.
百分比增加用乘数 1 + r，百分比减少用乘数 1 - r，其中 r 是写成小数的比率。减少 20% 就是乘以 0.8，增加 12% 就是乘以 1.12。请务必看清题目问的是"变化量"还是"最终量"。
5. Ratio, Proportion and Rates of Change | 比、比例与变化率
Homework in this section covers simplifying ratios, dividing a quantity in a given ratio, solving direct proportion problems with the unitary method, using scale drawings and maps, and applying the speed, distance and time formulae.
本模块作业涵盖比的化简、按给定比分配数量、用单位法解正比例问题、使用比例尺图与地图，以及应用速度、路程与时间的公式。
A typical question asks for a quantity to be shared in a given ratio, or for a recipe, map or model to be scaled up or down. In every case, work out the value of one part first, then multiply by the number of parts you need.
典型题目要求按给定比分配某个量，或把配方、地图、模型按比例放大或缩小。无论哪种情况，都先求出一份的值，再乘以你需要的份数。
- Simplify 24 : 36 to 2 : 3 | 把 24 : 36 化简为 2 : 3
- Share £240 in the ratio 3 : 5 | 按 3 : 5 分配 £240
- Use the unitary method for a recipe scaled for 12 people | 用单位法为 12 人调整配方
- Use a scale of 1 : 50,000 on a map | 使用 1 : 50,000 的地图比例尺
- Calculate speed, distance or time | 计算速度、路程或时间
240 ÷ (3 + 5) = 30 3 × 30 = £90 5 × 30 = £150
speed = distance ÷ time 150 km ÷ 2.5 h = 60 km/h
Rates appear in many forms: miles per hour, price per kilogram, litres per minute, and pay per hour. Write the units into the formula as you substitute, then cancel them; if the units work out, the calculation is almost certainly set up correctly.
变化率有多种形式：每小时英里数、每千克价格、每分钟升数、每小时工资。代入时把单位写进公式中再约去；如果单位对得上，计算几乎可以肯定列式正确。
6. Algebraic Expressions, Formulae and Sequences | 代数式、公式与数列
This section introduces algebra as generalised arithmetic. Homework includes collecting like terms, multiplying out single brackets, substituting numbers into formulae, writing formulae from words, and generating and describing sequences.
本节把代数介绍为"一般化的算术"。作业包括合并同类项、展开单括号、把数代入公式、由文字描述写出公式，以及生成并描述数列。
For sequences you must be able to find the next term, spot the rule, and derive the nth term of a linear sequence. A linear sequence changes by the same amount each time, so the nth term takes the form an + b where a is the common difference.
对于数列，你必须能找到下一项、看出规律并推导线性数列的第 n 项。线性数列每次变化的量相同，因此第 n 项形如 an + b，其中 a 是公差。
- Simplify 3a + 5a - 2a | 化简 3a + 5a - 2a
- Expand 4(x + 3) | 展开 4(x + 3)
- Substitute x = 4 into 3x² - 5 | 把 x = 4 代入 3x² - 5
- Find the nth term of 5, 8, 11, 14, ... | 求 5, 8, 11, 14, ... 的第 n 项
- Explain whether a term appears in a given sequence | 判断某个数是否为给定数列的项
3a + 5a - 2a = 6a 4(x + 3) = 4x + 12
3x² - 5 with x = 4 gives 3 × 16 - 5 = 43
5, 8, 11, 14, ... nth term = 3n + 2
, which is 2 less than the sequence each time, so the answer is 3n + 2. Always test your rule on the first three terms before writing it down.
求 5、8、11、14、... 的第 n 项时，先看出公差为 3，再试 3n：当 n = 1、2、3 时得到 3、6、9，每次都比原数列少 2，所以答案是 3n + 2。写下规则前一定要用前三项检验。
7. Equations and Inequalities | 方程与不等式
Homework here covers solving linear equations with one unknown, including equations with brackets and with the unknown on both sides, forming equations from word problems, and solving and representing simple inequalities.
本部分作业
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