📚 Essential Maths 7S Homework Book Compressed: Key Question Types Explained | KS3 数学：Essential Maths 7S 练习册压缩题型解析
The KS3 Maths curriculum builds on primary skills and introduces more formal algebraic and geometric thinking. The Essential Maths 7S Homework Book provides a compressed set of exercises designed to reinforce key concepts. In this article, we analyse the most common question types found in the compressed homework tasks, offering step-by-step methods to tackle them effectively.
KS3 数学课程在小学技能的基础上，引入了更正式的代数和几何思维。《Essential Maths 7S Homework Book》提供了一套压缩练习，旨在巩固关键概念。本文将分析压缩家庭作业中最常见的题型，并提供逐步方法以有效解决它们。
1. Integer Operations | 整数运算
IDMAS (brackets, indices, division/multiplication, addition/subtraction). A typical question: Calculate –20 + 12 ÷ (–3) × 2.
整数运算混合了正负数的加、减、乘、除。这本练习册常结合 BIDMAS 规则（括号、指数、乘除、加减）。典型题目：计算 –20 + 12 ÷ (–3) × 2。
Step 1: Division first: 12 ÷ (–3) = –4. The expression becomes –20 + (–4) × 2.
步骤1：先算除法：12 ÷ (–3) = –4。表达式变为 –20 + (–4) × 2。
Step 2: Multiplication next: (–4) × 2 = –8. Now we have –20 + (–8).
步骤2：接着乘法：(–4) × 2 = –8。现在剩下 –20 + (–8)。
Step 3: Addition of two negatives: –20 + (–8) = –28. The answer is –28.
步骤3：两个负数相加：–20 + (–8) = –28。答案是 –28。
2. Fractions and Decimals | 分数与小数
Questions test converting between fractions, decimals and mixed numbers, and performing operations. For instance, work out 2 ½ ÷ 1 ¼ and give the answer as a decimal.
这类题目考查分数、小数和带分数之间的转换及运算。例如，计算 2 ½ ÷ 1 ¼，结果用小数表示。
First, convert mixed numbers to improper fractions: 2 ½ = 5/2, 1 ¼ = 5/4.
首先，把带分数化为假分数：2 ½ = 5/2，1 ¼ = 5/4。
Divide fractions: 5/2 ÷ 5/4 = 5/2 × 4/5 = 20/10 = 2.
分数相除：5/2 ÷ 5/4 = 5/2 × 4/5 = 20/10 = 2。
0. The homework often requires giving the answer to 1 decimal place, so 2.0 is acceptable.
写成小数：2 = 2.0。练习中常要求保留一位小数，所以 2.0 可以接受。
3. Percentages and Increase/Decrease | 百分比与增减
Typical problems involve finding a percentage of an amount, or percentage increase and decrease. For example: A coat costing £85 is reduced by 15% in a sale. Find the sale price.
典型问题包括求一个数的百分之几，或者求百分比增减。例如：一件原价 85 英镑的外套打 15% 折扣，求售价。
Find 15% of £85: 85 × 0.15 = £12.75. Subtract from original: £85 – £12.75 = £72.25.
先求 £85 的 15%：85 × 0.15 = £12.75。然后从原价减去：£85 – £12.75 = £72.25。
Alternatively, the multiplier method: 100% – 15% = 85%, so new price = 85 × 0.85 = £72.25. Both ways appear in the homework book.
另一种方法是用乘数：100% – 15% = 85%，因此新价格 = 85 × 0.85 = £72.25。两种方法在练习册中都会出现。
4. Ratio and Proportion | 比率与比例
Ratio questions often describe sharing in a given ratio or working out missing values. Example: Share £120 between Ali, Ben and Cath in the ratio 2 : 3 : 5. How much does each receive?
比率题通常要求按给定比例分配，或计算缺失值。例题：将 120 英镑按 2：3：5 分给 Ali、Ben 和 Cath。每人分得多少？
Total parts = 2 + 3 + 5 = 10. One part = £120 ÷ 10 = £12.
总份数 = 2 + 3 + 5 = 10。一份 = £120 ÷ 10 = £12。
Ali gets 2 × £12 = £24, Ben gets 3 × £12 = £36, Cath gets 5 × £12 = £60.
Ali 得 2 × £12 = £24，Ben 得 3 × £12 = £36，Cath 得 5 × £12 = £60。
5. Simplifying Algebraic Expressions | 代数表达式化简
These tasks require collecting like terms and using index laws. A common compressed question: Simplify 4x² + 3y – 2x + 7y + x² – 5x.
此类任务要求合并同类项并运用指数律。常见的压缩题：化简 4x² + 3y – 2x + 7y + x² – 5x。
Group like terms: x² terms: 4x² + x² = 5x². x terms: –2x – 5x = –7x. y terms: 3y + 7y = 10y.
合并同类项：x² 项：4x² + x² = 5x²。x 项：–2x – 5x = –7x。y 项：3y + 7y = 10y。
Simplified expression: 5x² – 7x + 10y. Always write terms in descending order of powers.
化简结果为：5x² – 7x + 10y。始终按次数降序书写各项。
6. Solving Linear Equations | 解一元一次方程
known appears frequently. The homework book compresses multi-step problems like: Solve 5(x – 3) + 2 = 2x + 7.
解含一个未知数的方程频繁出现。练习册中会压缩多步骤问题，如：解方程 5(x – 3) + 2 = 2x + 7。
Expand brackets: 5x – 15 + 2 = 2x + 7 → 5x – 13 = 2x + 7.
去括号：5x – 15 + 2 = 2x + 7 → 5x – 13 = 2x + 7。
Collect x on one side: 5x – 2x = 7 + 13 → 3x = 20.
移项使 x 在一边：5x – 2x = 7 + 13 → 3x = 20。
both sides by 3: x = 20/3 or 6 ⅔. Often the answer is left as a mixed number or fraction.
两边除以 3：x = 20/3 或 6 ⅔。答案常保留为带分数或分数形式。
7. Sequences and the nth Term | 序列与第 n 项
nth term). A typical task: Write down the first 4 terms of the sequence with nth term = 4n – 3. Then find the 20th term.
KS3 序列侧重寻找项与项之间的关系以及位置与项的关系（第 n 项公式）。典型任务：写出通项公式为 4n – 3 的序列的前 4 项，然后求第 20 项。
For n = 1: 4(1) – 3 = 1; n = 2: 4(2) – 3 = 5; n = 3: 4(3) – 3 = 9; n = 4: 4(4) – 3 = 13. The sequence begins 1, 5, 9, 13.
n = 1：4(1) – 3 = 1；n = 2：4(2) – 3 = 5；n = 3：4(3) – 3 = 9；n = 4：4(4) – 3 = 13。序列开始为 1, 5, 9, 13。
20th term: 4(20) – 3 = 80 – 3 = 77. The difference between terms is constant; this is an arithmetic sequence.
第 20 项：4(20) – 3 = 80 – 3 = 77。各项之差恒定，这是一个等差数列。
8. Angles and Straight Lines | 角与直线
Angle questions test rules around points, straight lines, triangles and parallel lines. For example: In a triangle, two angles are 47° and 68°. Find the third angle. Then state whether the triangle is acute, right or obtuse.
角度题考查点、直线、三角形和平行线的角度规则。例如：三角形中两个角分别是 47° 和 68°，求第三个角，并判断三角形是锐角、直角还是钝角三角形。
Sum of angles in a triangle is 180°. Third angle = 180° – (47° + 68°) = 180° – 115° = 65°.
三角形内角和为 180°。第三角 = 180° – (47° + 68°) = 180° – 115° = 65°。
All angles (47°, 68°, 65°) are less than 90°, so it is an acute triangle. The compressed book often mixes reasoning with calculation.
所有角 (47°, 68°, 65°) 都小于 90°，所以是锐角三角形。压缩练习册常将推理与计算结合。
9. Area, Perimeter and Volume | 面积、周长与体积
Pupils must calculate area and perimeter of compound shapes, and volume of cuboids. A standard question: A rectangle measures 8 cm by 5 cm. A square has the same perimeter. Find the side length of the square and compare their areas.
学生需要计算组合图形的面积与周长，以及长方体的体积。标准题：一个长方形长 8 cm，宽 5 cm。一个正方形周长与之相等。求正方形的边长，并比较它们的面积。
gle perimeter = 2 × (8 + 5) = 26 cm. Square perimeter = 4 × side → side = 26 ÷ 4 = 6.5 cm.
长方形周长 = 2 × (8 + 5) = 26 cm。正方形周长 = 4 × 边长 → 边长 = 26 ÷ 4 = 6.5 cm。
× 5 = 40 cm². Square area = 6.5 × 6.5 = 42.25 cm². The square has a slightly larger area.
长方形面积 = 8 × 5 = 40 cm²。正方形面积 = 6.5 × 6.5 = 42.25 cm²。正方形面积稍大。
10. Statistical Diagrams and Averages | 统计图表与平均数
Questions involve interpreting bar charts, pictograms, and calculating mean, median, mode and range. Example: The scores of 6 students are 5, 7, 6, 8, 5, 9. Find the mean score and the range.
这类题目要求解释条形图、象形图，并计算平均数、中位数、众数和极差。例题：6 名学生的分数为 5, 7, 6, 8, 5, 9。求平均分和极差。
es ÷ number of values. Sum = 5 + 7 + 6 + 8 + 5 + 9 = 40. Mean = 40 ÷ 6 = 6.67 (to 2 d.p.).
平均数 = 数据总和 ÷ 数据个数。总和 = 5 + 7 + 6 + 8 + 5 + 9 = 40。平均数 = 40 ÷ 6 ≈ 6.67（保留两位小数）。
Range = highest – lowest = 9 – 5 = 4. In the homework book, you often need to draw conclusions from the averages.
极差 = 最大值 – 最小值 = 9 – 5 = 4。在练习册中，经常需要根据平均数得出结论。
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