📚 KS3 Mathematics: Essential Maths Book 9F Compressed – Question Type Analysis | KS3 数学：Essential Maths 9F 压缩版题型解析
Essential Maths Book 9F provides a comprehensive coverage of the Key Stage 3 curriculum for Year 9 students following a foundation tier pathway. This article analyses the compressed question types found in the book, helping learners recognise common patterns, master problem-solving techniques, and build confidence for assessments. Each section focuses on a core topic with worked examples and practical tips.
《Essential Maths 9F》全面覆盖了九年级基础阶段的关键阶段3课程。本文对该书压缩版中的题型进行解析，帮助学生识别常见模式、掌握解题技巧并增强考试信心。每个部分聚焦一个核心主题，配有示例和实用建议。
1. Number Operations and BIDMAS | 数字运算与运算顺序
Questions in this topic test your ability to perform mixed operations with whole numbers, decimals and negative numbers. The key rule is BIDMAS (Brackets, Indices, Division/Multiplication, Addition/Subtraction). Many mistakes happen when students ignore the order or mishandle negative signs.
本主题的题目考查混合运算能力，涉及整数、小数和负数。关键规则是 BIDMAS（括号、指数、乘除、加减）。很多错误源于忽略运算顺序或错误处理负号。
A typical compressed question asks you to evaluate an expression like: ( −3 ) + 8 × 2 − 4² ÷ 2. Always start inside brackets, then powers, then multiply/divide from left to right, and finally add/subtract from left to right.
典型的压缩题型要求计算如 ( −3 ) + 8 × 2 − 4² ÷ 2 的表达式。始终先算括号，再算幂，然后从左到右乘除，最后从左到右加减。
Worked example: Evaluate 5 − ( −2 ) × 3 + 12 ÷ ( −4 )² × 5
First, calculate the bracket: ( −2 ) stays as −2. The squared bracket: ( −4 )² = 16. Now rewrite: 5 − ( −2 ) × 3 + 12 ÷ 16 × 5. Multiply and divide left to right: ( −2 ) × 3 = −6, so now 5 − ( −6 ) + 12 ÷ 16 × 5. 12 ÷ 16 = 0.75, then 0.75 × 5 = 3.75. Expression becomes 5 − ( −6 ) + 3.75 = 5 + 6 + 3.75 = 14.75.
首先计算括号：( −2 ) 保持为 −2。平方括号：( −4 )² = 16。重写式子：5 − ( −2 ) × 3 + 12 ÷ 16 × 5。从左到右乘除：( −2 ) × 3 = −6，式变为 5 − ( −6 ) + 12 ÷ 16 × 5。12 ÷ 16 = 0.75，再 0.75 × 5 = 3.75。得到 5 − ( −6 ) + 3.75 = 5 + 6 + 3.75 = 14.75。
Be careful with ‘of’ word problems. For example, ‘What is 1/4 of 36?’ means (1/4) × 36. Distinguish ‘of’ from order of operations: it acts like multiplication but at the same level as division.
注意带有 ‘of’ 文字题。例如 ‘What is 1/4 of 36?’ 意思是 (1/4) × 36。区分 ‘of’ 和运算顺序：它如同乘法，与乘除同级。
2. Fractions, Decimals and Percentages (FDP) | 分数、小数与百分数
These three forms of representing parts of a whole are tested together. You need to convert fluently between them and compare values. A common question type presents a mix of fractions, decimals and percentages and asks you to place them in ascending order.
这三种表示部分与整体的形式经常一起考查。你需要熟练互化并能比较大小。常见的题型是给出分数、小数和百分数的混合，要求按升序排列。
To compare 3/8, 0.4, 35% and 2/5, convert all to decimals: 3/8 = 0.375, 0.4 stays 0.4, 35% = 0.35, 2/5 = 0.4. So ascending: 0.35 (35%), 0.375 (3/8), 0.4 (0.4 or 2/5). Always write the original form in the answer.
要比较 3/8、0.4、35% 和 2/5，全部转换为小数：3/8 = 0.375，0.4 保持 0.4，35% = 0.35，2/5 = 0.4。所以升序：0.35（35%），0.375（3/8），0.4（0.4 或 2/5）。答案中必须写原形式。
Another frequent type involves finding a percentage of a quantity without a calculator, e.g. 15% of £240. Use 10% and 5%: 10% = £24, 5% = £12, so 15% = £24 + £12 = £36. For calculator questions, convert percentage to a decimal and multiply.
另一种常见题型是无计算器求一个数的百分之几，如 15% of £240。使用 10% 和 5%：10% = £24，5% = £12，所以 15% = £24 + £12 = £36。对于可使用计算器的题目，将百分数转换为小数后相乘。
| Fraction | Decimal | Percentage |
|---|---|---|
| 1/2 | 0.5 | 50% |
| 1/3 | 0.3̅ | 33 1/3% |
| 2/3 | 0.6̅ | 66 2/3% |
| 1/4 | 0.25 | 25% |
3. Simplifying Algebraic Expressions | 代数式的化简
These questions check your ability to collect like terms, expand brackets and take out common factors. You must recognise terms with the same variable and power, such as 3a and 5a being like terms, while 3a and 3a² are not.
这类题目检测你合并同类项、展开括号和提取公因式的能力。你必须能识别相同变量和幂次的项，例如 3a 和 5a 是同类项，而 3a 和 3a² 不是。
To simplify 4x + 2y − x + 5y − 3, group x terms: 4x − x = 3x; y terms: 2y + 5y = 7y; constant remains −3. Result: 3x + 7y − 3. When expanding, multiply each term inside the bracket by the factor outside: 5(2a − 3) = 10a − 15.
化简 4x + 2y − x + 5y − 3，分组 x 项：4x − x = 3x；y 项：2y + 5y = 7y；常数项保持 −3。结果是 3x + 7y − 3。展开括号时，用外面的因式乘以括号内每一项：5(2a − 3) = 10a − 15。
Factorising is the reverse process: 12mn + 8n = 4n(3m + 2). Look for the highest common factor. Be precise with signs; for −6b + 9, factor out −3 to get −3(2b − 3).
因式分解是逆过程：12mn + 8n = 4n(3m + 2)。寻找最大公因式。注意符号；对于 −6b + 9，提取 −3 得到 −3(2b − 3)。
Substitution questions provide a formula and values: if x = −2 and y = 5, find 3x² − y. First, x² = 4, then 3 × 4 = 12, minus y gives 12 − 5 = 7.
代入题型给出公式和数值：若 x = −2，y = 5，求 3x² − y。先算 x² = 4，然后 3 × 4 = 12，减去 y 得 12 − 5 = 7。
4. Solving Linear Equations | 解一次方程
Linear equations appear frequently, ranging from one-step to multi-step varieties involving brackets and fractions. The golden rule is: whatever you do to one side, you must do to the other to maintain balance.
一次方程经常出现，从一步到多步，涉及括号和分数。黄金法则是：对等式一边所做的任何操作，必须对另一边同样执行以保持平衡。
For 3x + 5 = 20, subtract 5 from both sides → 3x = 15, then divide by 3 → x = 5. When the unknown appears on both sides, eliminate one side first: 7y − 4 = 3y + 8. Subtract 3y: 4y − 4 = 8, add 4: 4y = 12, y = 3.
对于 3x + 5 = 20，两边减 5 → 3x = 15，然后除以 3 → x = 5。当未知数出现在两边，先消去一边：7y − 4 = 3y + 8。减去 3y：4y − 4 = 8，加 4：4y = 12，y = 3。
Equations with brackets require expansion first: 2(x + 3) = 4x − 2 → 2x + 6 = 4x − 2 → 6 + 2 = 4x − 2x → 8 = 2x → x = 4. Always check your solution by substituting back into the original equation.
带括号的方程需要先展开：2(x + 3) = 4x − 2 → 2x + 6 = 4x − 2 → 6 + 2 = 4x − 2x → 8 = 2x → x = 4。务必代回原方程检验。
Fraction example: Solve 3n/4 − 1 = 2n/3 + 5
Multiply every term by the LCM of 4 and 3, which is 12: 12×(3n/4) − 12×1 = 12×(2n/3) + 12×5 → 9n − 12 = 8n + 60 → n = 72.
：3n/4 − 1 = 2n/3 + 5，每一项乘以 4 和 3 的最小公倍数 12：12×(3n/4) − 12×1 = 12×(2n/3) + 12×5 → 9n − 12 = 8n + 60 → n = 72。
5. Sequences and the nth Term | 数列与第n项
Sequences assess your ability to spot patterns and generate rules. The most common type asks for the nth term of an arithmetic sequence, where the difference between consecutive terms is constant.
数列考查你发现规律并生成通项的能力。最常见的题型是求等差数列的第 n 项，其中相邻两项的差为常数。
For the sequence 7, 12, 17, 22, 27, … the difference is +5. The nth term is 5n + 2. The ‘a’ value (5) is the difference, and the ‘+ 2’ adjusts for the starting term: when n = 1, 5(1) + 2 = 7. To find the 20th term, substitute n = 20: 5(20) + 2 = 102.
对于数列 7, 12, 17, 22, 27, …，差值为 +5。第 n 项为 5n + 2。’5′ 是公差，’+ 2′ 调整首项：当 n = 1 时，5(1) + 2 = 7。求第 20 项，代入 n = 20：5(20) + 2 = 102。
If the difference is negative, e.g. −3, the coefficient of n is negative: 20, 17, 14, 11, … gives nth term −3n + 23. Check n = 2 → −3(2) + 23 = 17.
若公差为负，如 −3，则 n 的系数为负：20, 17, 14, 11, … nth term = −3n + 23。检验 n = 2 → −3(2) + 23 = 17。
Non-linear sequences also appear: 1, 4, 9, 16, 25, … (square numbers) nth term = n². Another pattern might be multiplying by a constant, like 3, 6, 12, 24 (×2), which is geometric and not typically asked for nth term at KS3 foundation, but you recognise the rule.
非线性数列也会出现：1, 4, 9, 16, 25, …（平方数）nth term = n²。另一种模式可能是乘以常数，如 3, 6, 12, 24（×2），这是等比数列，KS3 基础阶段通常不要求写出通项，但需识别规则。
6. Ratio and Proportion | 比与比例
Ratio questions involve sharing quantities, simplifying ratios, and solving word problems. Always write ratios in their simplest integer form, similar to fractions.
比的问题涉及分配量、简化比和解决文字题。始终将比写成最简整数形式，类似于分数。
To share £60 in the ratio 3 : 5, find the total parts: 3 + 5 = 8 parts. One part = £60 ÷ 8 = £7.50. First share = 3 × £7.50 = £22.50, second share = 5 × £7.50 = £37.50.
按 3 : 5 分配 £60，先求总份数：3 + 5 = 8 份。一份 = £60 ÷ 8 = £7.50。第一份 = 3 × £7.50 = £22.50，第二份 = 5 × £7.50 = £37.50。
Ratio problems often include changing quantities or mixing. For example, a drink is made from cordial and water in the ratio 1 : 4. How much cordial is needed for 2.5 litres? Total parts = 5, one part = 2.5 ÷ 5 = 0.5 litres, so cordial = 0.5 litres.
比的问题常涉及量的变化或混合。例如，一种饮料由浓缩液和水按 1 : 4 勾兑。制作 2.5 升需要多少浓缩液？总份数 = 5，一份 = 2.5 ÷ 5 = 0.5 升，所以浓缩液 = 0.5 升。
Direct proportion questions ask you to find missing values in tables where y = kx. If 5 pens cost £2.80, the cost for 11 pens = (2.80/5) × 11 = £6.16. Use the unitary method: find the value for one unit first.
正比例问题要求根据 y = kx 求表格中的缺失值。若 5 支笔 £2.80，11 支笔的费用 = (2.80/5) × 11 = £6.16。用归一法：先求一个单位的值。
7. Perimeter, Area and Volume | 周长、面积与体积
Foundation questions cover rectangles, triangles, parallelograms and compound shapes. Remember the key formulae and use correct units: perimeter is a length (cm, m), area is in square units (cm², m²), volume in cubic units (cm³, m³).
基础题型涵盖长方形、三角形、平行四边形和组合图形。牢记关键公式并使用正确单位：周长是长度（cm, m），面积是平方单位（cm², m²），体积是立方单位（cm³, m³）。
For a rectangle 8 cm by 5 cm, perimeter = 2(8+5) = 26 cm, area = 8 × 5 = 40 cm². Triangle area = base × height / 2. Parallelogram area = base × perpendicular height.
一个 8 cm × 5 cm 的长方形，周长 = 2(8+5) = 26 cm，面积 = 8 × 5 = 40 cm²。三角形面积 = 底 × 高 ÷ 2。平行四边形面积 = 底 × 垂直高。
Volume of a cuboid: length × width × height. For a cube of side 4 cm, volume = 4³ = 64 cm³. Surface area questions ask for the total area of all faces: a cuboid 6 cm × 3 cm × 2 cm has surface area = 2(6×3 + 6×2 + 3×2) = 2(18+12+6) = 72 cm².
长方体体积：长 × 宽 × 高。边长 4 cm 的立方体，体积 = 4³ = 64 cm³。表面积问题求所有面的总面积：一个 6 cm × 3 cm × 2 cm 的长方体，表面积 = 2(6×3 + 6×2 + 3×2) = 2(18+12+6) = 72 cm²。
Compound shapes require splitting into standard figures. Remember to identify missing side lengths by using opposite sides in rectilinear figures.
组合图形需要拆分成标准图形。记得利用直线图形中对边相等的性质找出缺失边长。
8. Angles and Properties of Shapes | 角与图形的性质
Angle facts are crucial: angles on a straight line sum to 180°, around a point sum to 360°, vertically opposite angles are equal. In triangles, interior angles sum to 180°; in quadrilaterals, 360°.
角的性质至关重要：直线上的角之和为 180°，绕一点一周为 360°，对顶角相等。三角形内角和为 180°；四边形内角和为 360°。
Parallel line questions use alternate angles (Z-shape), corresponding angles (F-shape) and co-interior angles (C-shape). Alternate angles are equal, corresponding angles are equal, co-interior sum to 180°.
平行线题目涉及内错角（Z 形）、同位角（F 形）和同旁内角（C 形）。内错角相等，同位角相等，同旁内角互补（和为 180°）。
Example: Find angle x when two parallel lines are cut by a transversal and a given angle is 75°.
If the 75° angle and x are alternate, then x = 75°. If co-interior, then x = 180° − 75° = 105°. Always state the reason.
若两条平行线被截线所截，已知角为 75°，求 x。若 75° 与 x 为内错角，则 x = 75°。若为同旁内角，则 x = 180° − 75° = 105°。务必写明理由。
Polygon interior angle sum = (n − 2) × 180°. For a regular pentagon, one interior angle = (5−2)×180°/5 = 108°. Exterior angles of any polygon always sum to 360°.
多边形内角和 = (n − 2) × 180°。正五边形每个内角 = (5−2)×180°/5 = 108°。任何多边形的外角和总是 360°。
9. Statistics: Averages and Charts | 统计：平均数与图表
You need to calculate mean, median, mode and range from a list or frequency table. Be precise with the formulas and steps: mean = sum of values ÷ number of values; median = middle value when ordered; mode = most frequent; range = largest − smallest.
你需要从列表或频数表中计算平均数、中位数、众数和极差。公式和步骤要精确：平均数 = 数值总和 ÷ 数据个数；中位数 = 排序后中间的数；众数 = 出现频率最高的值；极差 = 最大值 − 最小值。
For grouped data, estimate the mean using midpoints. From a bar chart, read frequencies correctly. Pie charts require converting angles to fractions of 360°.
对于分组数据，用组中值估算平均数。从条形图中正确读取频数。饼图需要将角度转换为 360° 的分数。
When comparing two data sets, state both averages and the spread. A higher mean with a smaller range often indicates more consistent better performance.
比较两组数据时，要说明平均数和离散程度。较高的平均数加上较小的极差，通常表明表现更好且更稳定。
Probability basics: probability scale from 0 to 1. Probability of an event = number of favourable outcomes / total outcomes. Events can be certain (1), likely, even chance (1/2), unlikely or impossible (0).
概率基础：概率范围从 0 到 1。事件的概率 = 有利结果数 / 所有可能结果数。事件可分为确定 (1)、很可能、等可能 (1/2)、不太可能或不可能 (0)。
10. Transformations and Symmetry | 变换与对称
Four types of transformation are tested: reflection, rotation, translation and enlargement. You must describe each fully or perform them on a grid.
考查四种变换：反射、旋转、平移和放大。你必须完整描述或在网格上执行变换。
Reflection: state the mirror line (e.g. x = 2, y = -1). Rotation: give centre of rotation, angle and direction (clockwise/anticlockwise). Translation: use a column vector, e.g. 3 right and -2 down.
反射：说明镜面线（如 x = 2, y = -1）。旋转：给出旋转中心、角度和方向（顺时针/逆时针）。平移：用列向量表示，如向右 3、向下 −2。
Enlargement: specify scale factor and centre. A scale factor of 2 doubles all lengths; a fractional scale factor (e.g. 1/3) reduces the shape. If the centre is not the origin, draw rays from the centre through vertices and measure distances.
放大：给出比例因子和中心。比例因子 2 将所有长度加倍；分数比例因子（如 1/3）将图形缩小。如果中心不是原点，从中心通过顶点画射线并测量距离。
Symmetry: line symmetry is a mirror line where the shape folds onto itself; rotational symmetry is how many times the shape matches itself in a full turn (order). A square has 4 lines and order 4.
对称性：线对称是图形可沿一条线对折重合，称为对称轴；旋转对称是指图形绕一点旋转一周能与自身重合的次数（旋转对称阶数）。正方形有 4 条对称轴，旋转对称阶数为 4。
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