📚 KS3 Maths: Essential Maths Book 9i Compressed Question Types Analysis | KS3 数学：Essential Maths Book 9i 压缩题型解析
The Essential Maths Book 9i Compressed edition is widely used in UK schools to reinforce and extend KS3 mathematics skills. This article breaks down the most common question types found in the book, providing clear strategies and examples that will help you tackle assessments with confidence. By understanding the structure and demands of each topic area, you can focus your revision effectively and avoid typical pitfalls.
Essential Maths Book 9i 压缩版是英国学校广泛使用的 KS3 数学巩固与拓展教材。本文拆解了书中常见的题型，提供了清晰的解题策略和范例，帮助你从容应对各类测验。通过了解每个专题的结构与要求，你可以更有针对性地复习，避开常见陷阱。
1. Number and Place Value | 数与位值
Question types here focus on reading and writing large numbers, understanding place value up to millions, and ordering positive and negative integers. You may be asked to write a number in words, identify the value of a digit, or round numbers to the nearest 10, 100, or 1000.
本部分题型重点是大数字的读写、百万以内位值理解，以及正负整数排序。题目可能要求用文字写出数字、指出某一位数字的值，或将数字四舍五入到最近的10、100或1000。
A typical task: 'Round 23,456 to the nearest thousand.' The key is to look at the hundreds digit: 4 (less than 5), so round down to 23,000. Another example involves ordering temperatures: -5°C, 3°C, -1°C, 0°C. The correct order from coldest to warmest is -5°C < -1°C < 0°C < 3°C.
典型题：'将23,456四舍五入到千位。' 关键是看百位数字：4（小于5），所以舍去，得到23,000。另一个例子是对温度排序： -5°C, 3°C, -1°C, 0°C。从冷到暖的正确顺序是 -5°C < -1°C < 0°C < 3°C。
Place value questions often ask: 'What is the value of the 5 in 3,509,420?' The answer is 500,000 (five hundred thousand), because the 5 occupies the hundred-thousands place. Always read the number carefully and label the columns (millions, hundred-thousands, ten-thousands, thousands, hundreds, tens, ones).
位值题目常问：'3,509,420中的5表示多少？' 答案是500,000（五十万），因为5位于十万位。务必仔细读数，标注列名（百万、十万、万、千、百、十、个）。
2. Addition, Subtraction, Multiplication and Division | 加减乘除运算
n (e.g., 935 ÷ 5), and multi-step problems combining addition and subtraction with money or measures.
这些基本运算不仅出现在直接计算题中，也隐藏在应用题里。典型题型包括长乘法（如234 × 56）、短除法（如935 ÷ 5），以及结合钱币或测量单位的多步骤加减问题。
s to get 13,104. For division, 935 ÷ 5, use the short division algorithm: 5 into 9 goes 1 remainder 4, carry to 3 making 43, 5 into 43 goes 8 remainder 3, carry to 5 making 35, 5 into 35 goes 7 exactly, so answer 187.
计算234 × 56时，使用网格法或竖式：234 × 50 = 11,700，234 × 6 = 1,404；相加得13,104。除法935 ÷ 5，用短除算法：5除9得1余4，移至3成43，5除43得8余3，移至5成35，5除35得7，答案187。
Word problems often involve real-life contexts: 'A box holds 24 pencils. How many pencils are there in 15 boxes?' Multiply 24 by 15 to get 360 pencils. Always show your method clearly, as marks are awarded for working even if the final answer is incorrect.
应用题常嵌入生活场景：'一个盒子装24支铅笔，15盒共有多少支？' 24乘15得360支。务必清晰地展示解题步骤，因为即使最终答案错误，过程也有步骤分。
3. Fractions, Decimals and Percentages | 分数、小数与百分数
This topic tests conversion between the three forms, finding fractions of amounts, adding/subtracting fractions, and percentage increase/decrease. A common question: 'Convert 3/8 to a decimal and a percentage.' Divide 3 by 8 to get 0.375, then multiply by 100 for 37.5%.
本专题考查三者之间的转换、求一个量的几分之几、分数加减法，以及百分比增减。常见题：'将3/8转换为小数与百分数。' 3除以8得0.375，乘以100得37.5%。
When finding a fraction of an amount, such as 'Calculate 2/5 of 60,' tors, like 1/4 + 2/5, find a common denominator (20): 1/4 = 5/20, 2/5 = 8/20, sum = 13/20.
求一个数的几分之几，如'计算60的2/5'，除以分母乘分子：60 ÷ 5 = 12，12 × 2 = 24。异分母分数加减，如1/4 + 2/5，找公分母20：1/4 = 5/20，2/5 = 8/20，和为13/20。
Percentage change questions: 'A jacket costing £80 is reduced by 15%. What is the new price?' Find 10% (£8) and 5% (£4), so discount is £12, new price £68. Alternatively, multiply £80 by 0.85 directly.
百分比变化题：'一件夹克原价80英镑，打八五折后的新价格？' 10%是8英镑，5%是4英镑，折扣共12英镑，新价68英镑。也可以直接80 × 0.85。
4. Ratio and Proportion | 比和比例
Ratio questions involve simplifying ratios, sharing an amount in a given ratio, and solving proportion problems. Simplify ratios like a fraction: e.g., 12:18 divide both by 6 to get 2:3. Always write ratios in their simplest whole-number form.
比的问题包括化简比、按给定比分摊数量，以及比例应用。化简比类似分数：如12:18，两边同除以6得2:3。务必用最简整数的形式表示比。
A sharing question: 'Share £72 between Ali and Ben in the ratio 3:5.' Total parts = 3 + 5 = 8, one part = £72 ÷ 8 = £9. Ali gets 3 × £9 = £27, Ben gets 5 × £9 = £45. Double-check the sum: £27 + £45 = £72.
分配问题：'将72英镑按3:5分给Ali和Ben。' 总份数 = 3+5=8，每份=72÷8=9英镑。Ali得3×9=27英镑，Ben得5×9=45英镑。检验总和27+45=72。
In proportion problems, students might meet recipes adapted for more people: 'A recipe uses 300g of flour for 4 people. How much flour for 10 people?' find for 1 person: 300 ÷ 4 = 75g, then for 10: 75 × 10 = 750g. This is the unitary method.
比例应用题中，学生常遇食谱适配问题：'一份食谱供4人食用需要300克面粉，10人需要多少？' 先求1人量：300÷4=75克，再求10人：75×10=750克。这就是单比法。
5. Algebra: Expressions and Equations | 代数：表达式与方程
Algebraic question types in Book 9i include simplifying expressions, expanding single brackets, substituting values into formulas, and solving linear equations. Simplifying an expression like 3a + 2b - a + 5b yields 2a + 7b. Always combine like terms carefully.
Book 9i 中的代数题型包括化简表达式、展开单个括号、代入公式求值，以及解一元一次方程。化简如 3a + 2b - a + 5b 得到 2a + 7b。一定要仔细合并同类项。
sides: 2x = 8, then divide by 2: x = 4. Always check by substituting back: 2(4) + 3 = 11.
展开括号：4(x + 3) 得 4x + 12。记得括号内每一项都要乘以外面的数。解方程：2x + 3 = 11。两边减3：2x = 8，再除以2：x = 4。通过代回检验：2(4) + 3 = 11。
2x + 3 = 11 → x = 4
Problems may also ask you to write an expression for a situation: 'n apples cost p pence each. Write an expression for the total cost.' Answer: np pence. Being fluent in using letters to represent unknown quantities is essential.
题目还可能要求根据情境列式：'n个苹果，每个p便士，写出总成本表达式。' 答案：np 便士。熟练用字母表示未知量至关重要。
6. Sequences and Functions | 数列与函数
Question types cover finding the next terms of a sequence, identifying the term-to-term rule, and working out the nth term of an arithmetic sequence. A linear sequence such as 4, 7, 10, 13,... has a common difference of 3. The nth term is 3n + 1.
题型涵盖找出数列后几项、确定项与项之间的规则，以及求等差数列的第n项公式。线性数列如4, 7, 10, 13,... 的公差是3，第n项公式为3n + 1。
..., common difference = -4, zero term = 9 - (-4) = 13, so nth term = -4n + 13 or 13 - 4n.
求第n项公式，先求零项：首项减公差，4 - 3 = 1。第n项 = 公差 × n + 零项。数列9, 5, 1, -3,...，公差 = -4，零项 = 9 - (-4) = 13，所以第n项 = -4n + 13 或 13 - 4n。
ons may ask for the output when the input is 5, or to find the input given the output (inverse operations).
函数常用数字机器表示：输入 → ×2 → +3 → 输出，可写成 y = 2x + 3。考试题可能问输入为5时的输出，或已知输出求输入（逆运算）。
7. Coordinates and Graphs | 坐标与图像
Typical tasks include plotting points on a coordinate grid, drawing straight-line graphs from a table of values, and interpreting distance-time or conversion graphs. Points are written as (x, y). Remember 'along the corridor, up the stairs': x first, then y.
典型任务有在坐标网格上描点、由数值表绘制直线图，以及解读距离-时间图或换算图。点写为(x, y)。记住“先横后竖”：先沿走廊走（x），再上楼（y）。
1, 1, 3; plot those points and draw a straight line through them. Label the axes clearly and use a ruler.
画 y = 2x - 1 的图像：选 x = -1, 0, 1, 2；计算 y 值分别为 -3, -1, 1, 3；描点并用直尺过点画直线。清晰标注坐标轴。
e ready to calculate speed from the gradient: speed = vertical change / horizontal change.
解读距离-时间图：水平线段表示物体静止；斜直线表示匀速；越陡的线表示速度越快。需会从斜率计算速度：速度 = 纵向变化 ÷ 横向变化。
8. Geometry: Angles and Shapes | 几何：角与图形
Geometry question types involve calculating missing angles on a straight line (sum 180°), around a point (360°), and in triangles (180°). Parallel lines angles (alternate, corresponding, vertically opposite) are also tested. Use these facts to set up equations.
几何题型包括计算直线上缺失的角（和180°）、围绕点的角（360°）、三角形内角和（180°）。平行线中的角（内错角、同位角、对顶角）也会考查。利用这些事实列方程。
s one angle of 110°, the adjacent angle is 70° (co-interior), and the opposite angle is 110°.
示例：三角形中两个角分别为45°和70°，第三个角 = 180° - (45°+70°) = 65°。平行四边形中若有一个角是110°，邻角为70°（同旁内角互补），对角为110°。
on, octagon) and calculate the sum of interior angles of an n-sided polygon: (n-2) × 180°.
学生还会接触四边形性质、对称性以及简单尺规作图。要能命名多边形（五边形、六边形、八边形），并计算n边形内角和：(n-2)×180°。
9. Perimeter, Area and Volume | 周长、面积与体积
angle = ½ × base × height. Remember that perimeter is the total distance around the shape.
矩形、三角形及组合图形的面积与周长是本专题的核心。使用公式：矩形面积 = 长 × 宽；三角形面积 = ½ × 底 × 高。记住周长是围绕形状一周的总长度。
For compound shapes, split the shape into simpler rectangles, calculate their areas, then add or subtract as needed. Example: an L-shape can be divided into two rectangles, find each area and sum them.
对于组合图形，将其拆分为更简单的矩形，分别计算面积，然后按要求相加或相减。示例：L形可分割为两个长方形，分别求面积再求和。
Area of a triangle = ½ × b × h
pe is made of unit cubes, you can count them. Know that 1 cm³ = 1 mL, a useful conversion.
立方体和长方体体积：体积 = 长 × 宽 × 高。若图形由单位立方块组成，可直接计数。须知 1 cm³ = 1 mL，这是个有用的换算。
10. Statistics and Probability | 统计与概率
Statistical question types include reading and drawing bar charts, pie charts, and line graphs; calculating mean, median, mode, and range; and interpreting data from tables. The mean is the sum of values divided by the number of values; the median is the middle value when ordered; the mode is the most frequent.
统计题型包括读、画条形图、饼图和折线图；计算平均数、中位数、众数和极差；以及解读表格数据。平均数是总和除以数据个数；中位数是排序后中间的值；众数是出现次数最多的数。
A typical question: 'Find the mean of 8, 12, 5, 7, 8.' is 8, and putting them in order (5, 7, 8, 8, 12) the median is 8. The range is 12 - 5 = 7.
典型题：'求8, 12, 5, 7, 8的平均数。' 总和=40，除以5得平均数为8。众数是8，排序后(5,7,8,8,12)中位数是8，极差为12-5=7。
s / total number of possible outcomes. The probability scale goes from 0 (impossible) to 1 (certain).
概率题要求用分数、小数或百分数表示事件发生的可能性。概率 = 有利结果数 / 所有可能结果数。概率标度从0（不可能）到1（必然）。
blue marble = 2/(3+2+5) = 2/10 = 1/5. Always simplify fractions and show clear reasoning.
示例：袋中有3个红球、2个蓝球和5个绿球。抽到蓝球的概率 = 2/(3+2+5) = 2/10 = 1/5。记得化简分数并展示清晰的推理过程。
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