📚 KS3 Maths: Essential Maths 8H Homework Book Compressed Question Types Explained | KS3 数学：Essential Maths 8H 作业本压缩题型解析
The Essential Maths 8H Homework Book is a popular resource for Year 8 students following the higher tier of the KS3 curriculum. Its ‘compressed’ sections bring together the most representative question types, enabling focused revision on the trickiest topics. This article breaks down each key question type, explains the common pitfalls, and illustrates effective strategies to help you master the content.
《Essential Maths 8H 作业本》是面向 KS3 高阶八年级学生的常用练习册。其中的“压缩”题型集中了最有代表性的问题类型，帮助学生有针对性地攻克难点。本文逐一解析这些核心题型，指出常见易错点，并给出高效的解题策略，助你彻底掌握相关内容。
1. Simplifying Algebraic Expressions | 化简代数表达式
In these questions, you are given an expression such as 5a + 3b − 2a + 7b and asked to collect like terms. The key is to identify terms with exactly the same variable and power. For example, 5a and −2a are like terms, while 3b and 7b can be combined.
这类题目通常会给出如 5a + 3b − 2a + 7b 的表达式，要求合并同类项。关键在于识别具有完全相同变量和指数的项。例如 5a 和 −2a 是同类项，3b 和 7b 可以合并。
A common mistake is to mishandle negative signs. Remember that the sign in front of a term belongs to it. So 5a − 2a gives 3a, and 3b + 7b gives 10b. The final simplified answer is 3a + 10b.
常见的错误是处理负号不当。记住，项前面的符号属于该项。因此 5a − 2a 得到 3a，3b + 7b 得到 10b。最终化简结果为 3a + 10b。
Always write your answer in alphabetical order if the variables differ, and never combine terms like a² and a, as they represent different degrees.
如果变量不同，最终答案应按字母顺序书写，并且绝不能合并 a² 和 a 这样的项，因为它们代表不同的次数。
2. Expanding Brackets | 展开括号
Expanding a single bracket like 4(2x + 3) means multiplying each term inside by the number or term outside. So 4 × 2x = 8x and 4 × 3 = 12, giving 8x + 12.
展开如 4(2x + 3) 这样的单个括号，需要将括号外的数或项乘入括号内每一项。因此 4 × 2x = 8x，4 × 3 = 12，得到 8x + 12。
When the bracket has a minus sign in front, such as −3(y − 5), be especially careful: −3 × y = −3y and −3 × (−5) = +15, so the result is −3y + 15.
当括号前是减号时，如 −3(y − 5)，要特别小心：−3 × y = −3y，而 −3 × (−5) = +15，因此结果是 −3y + 15。
For double brackets like (x + 2)(x + 5), use the FOIL method or a grid. Multiply First, Outer, Inner, Last and then collect like terms: x² + 5x + 2x + 10 = x² + 7x + 10.
对于 (x + 2)(x + 5) 这样的双括号，可使用 FOIL 法或表格法。依次乘出首项、外项、内项和末项，然后合并同类项：x² + 5x + 2x + 10 = x² + 7x + 10。
3. Factorising Expressions | 因式分解
Factorising is the reverse of expanding. For a simple expression like 6x + 9, look for the highest common factor (HCF) of the coefficients. The HCF of 6 and 9 is 3, so write 3(2x + 3). Check by expanding.
因式分解是展开的逆过程。对于 6x + 9 这样的简单表达式，寻找系数的最大公因数 (HCF)。6 和 9 的最大公因数是 3，因此写作 3(2x + 3)。通过展开可以验证。
When factorising a quadratic like x² + 6x + 8, find two numbers that multiply to the constant term (+8) and add to the coefficient of x (+6). These are +2 and +4, so the factorised form is (x + 2)(x + 4).
如因式分解 x² + 6x + 8 这样的二次式，需要找到两个数，其乘积等于常数项 (+8)，且其和等于 x 的系数 (+6)。这两个数是 +2 和 +4，因此因式分解形式为 (x + 2)(x + 4)。
A pitfall is forgetting to take out all common factors first. For 3x² + 6x, factor out 3x to get 3x(x + 2).
一个常见的陷阱是忘记先提取所有公因式。例如对于 3x² + 6x，应先提取 3x，得到 3x(x + 2)。
4. Solving Linear Equations | 解一元一次方程
A typical compressed question might be: Solve 2x + 7 = 19. The aim is to isolate x by performing inverse operations. Subtract 7 from both sides: 2x = 12, then divide by 2: x = 6.
典型的压缩题型可能是：解方程 2x + 7 = 19。目标是通过逆运算分离 x。两边同时减 7：2x = 12，然后除以 2：x = 6。
When the unknown appears on both sides, such as 5x − 4 = 3x + 8, collect x terms on one side and numbers on the other. Subtract 3x: 2x − 4 = 8, add 4: 2x = 12, so x = 6.
当未知数出现在两边时，如 5x − 4 = 3x + 8，把含 x 的项移到一边，常数项移到另一边。减去 3x：2x − 4 = 8，加 4：2x = 12，因此 x = 6。
Always verify your answer by substituting it back into the original equation. This catches sign errors and arithmetic mistakes.
一定要将答案代入原方程进行验证，这能帮助发现符号错误和计算失误。
5. Working with Fractions | 分数运算
Questions on adding and subtracting fractions require a common denominator. For ⅓ + ⅖, the lowest common multiple of 3 and 5 is 15. Rewrite as 5/15 + 6/15 = 11/15.
分数的加减法题型需要通分。对于 ⅓ + ⅖，3 和 5 的最小公倍数是 15。改写为 5/15 + 6/15 = 11/15。
Multiplying fractions is simpler: multiply numerators together and denominators together. ¾ × ⅔ = (3×2)/(4×3) = 6/12, which simplifies to ½.
分数乘法更简单：分子相乘，分母相乘。¾ × ⅔ = (3×2)/(4×3) = 6/12，约分为 ½。
To divide by a fraction, multiply by its reciprocal. So ⅘ ÷ ⅗ = ⅘ × 5/3 = 20/15 = 1⅓. Many errors occur when students forget to flip the second fraction.
除以一个分数等于乘以它的倒数。因此 ⅘ ÷ ⅗ = ⅘ × 5/3 = 20/15 = 1⅓。许多错误发生在学生忘记将第二个分数翻转。
Mixed numbers must be converted to improper fractions before multiplying or dividing. For instance, 2½ × 1⅓ becomes 5/2 × 4/3 = 20/6 = 3⅓.
带分数在乘除前必须先化成假分数。例如，2½ × 1⅓ 变为 5/2 × 4/3 = 20/6 = 3⅓。
6. Percentages and Percentage Change | 百分比与百分比变化
Basic percentage questions ask for a percentage of an amount. To find 15% of £240, find 10% (£24) and 5% (£12) and add them: £36. Or multiply £240 by 0.15.
基础百分比题要求找出一个数的百分之几。例如求 £240 的 15%，可以先算出 10% (£24) 和 5% (£12)，相加得到 £36。或者直接用 £240 × 0.15。
Percentage increase and decrease are tested heavily. A £60 jacket with a 20% increase: increase = 20% of £60 = £12, so new price = £72. For a decrease, subtract instead.
百分比增减是重点考查内容。一件 £60 的外套增加 20%：增加额 = £60 的 20% = £12，新价格为 £72。如果是减少，则相减。
Reverse percentage problems ask, ‘After a 25% increase, the price is £80. What was the original price?’ Here, £80 represents 125% of the original, so 1% is £80 ÷ 125 = £0.64, and 100% = £64. Or divide by 1.25.
逆推百分比的题目问：“一件商品涨价 25% 后售价 £80，原价是多少？” 此时 £80 代表原价的 125%，因此 1% 为 £80 ÷ 125 = £0.64，100% 为 £64。或者直接除以 1.25。
7. Ratio and Proportion | 比例与比率
A typical ratio question might state that the ratio of flour to sugar is 5 : 3. If 400 g of flour is used, how much sugar is needed? The scale factor is 400 ÷ 5 = 80, so sugar = 3 × 80 = 240 g.
典型的比例题可能描述面粉与糖的比例为 5 : 3。如果用了 400 g 面粉，需要多少糖？缩放因子为 400 ÷ 5 = 80，因此糖的量为 3 × 80 = 240 g。
When sharing an amount in a given ratio, like £60 in the ratio 3 : 2, find the total number of parts (5), one part = £60 ÷ 5 = £12, so the shares are 3 × £12 = £36 and 2 × £12 = £24.
当按给定比例分配一个总量时，例如把 £60 按 3 : 2 分配，先计算总份数 (5)，一份为 £60 ÷ 5 = £12，因此各部分分别为 3 × £12 = £36 和 2 × £12 = £24。
Direct proportion problems often involve converting between units or currencies. If 3 kg of apples cost £4.50, then 1 kg costs £1.50, and 7 kg cost £10.50. Use the unitary method to build confidence.
正比例问题常涉及单位换算或货币兑换。如果 3 kg 苹果售价 £4.50，那么每公斤 £1.50，7 kg 则需 £10.50。使用归一法能有效增强信心。
8. Angles in Polygons | 多边形内角
Questions frequently ask you to find missing angles in triangles and quadrilaterals, using the fact that angles in a triangle sum to 180°. In a quadrilateral, the sum is 360°.
题目经常要求利用三角形内角和为 180° 的性质求缺失的角。四边形内角和为 360°。
For a regular polygon, each interior angle can be found by dividing the total sum. A regular pentagon has sum (5−2) × 180° = 540°, so each interior angle is 540° ÷ 5 = 108°.
对于正多边形，每个内角可用总和除以边数求出。正五边形内角和为 (5−2) × 180° = 540°，因此每个内角为 540° ÷ 5 = 108°。
Parallel line angle rules are often combined. Look for alternate angles (Z shape), corresponding angles (F shape), and co-interior angles (C shape) which sum to 180°. Being able to spot these quickly saves time.
平行线角度规则经常综合出现。要识别内错角 (Z 形)、同位角 (F 形) 和同旁内角 (C 形)，后者互补 180°。快速发现这些关系可以节省时间。
9. Area and Circumference of Circles | 圆的面积与周长
These problems test your recall of the formulas. The circumference C = πd or C = 2πr. The area A = πr². Use the π button or 3.14 as instructed, and round answers correctly.
这类题目考查对公式的记忆。周长 C = πd 或 C = 2πr。面积 A = πr²。根据要求使用 π 键或 3.14，并正确四舍五入结果。
A compressed question might give the circumference and ask for the area. For example, if C = 31.4 cm, find r = 31.4 ÷ (2 × 3.14) = 5 cm, then area = 3.14 × 5² = 78.5 cm².
压缩题型可能给出周长要求面积。例如，若 C = 31.4 cm，求出 r = 31.4 ÷ (2 × 3.14) = 5 cm，那么面积 = 3.14 × 5² = 78.5 cm².
Be careful with half circles and quarter circles. The perimeter of a semicircle includes the diameter: πr + d. The area is half of the full circle: ½πr².
注意半圆和四分之一圆。半圆的周长包含直径：πr + d。面积是整圆的一半：½πr²。
10. Probability and Tree Diagrams | 概率与树状图
Basic probability is written as a fraction: P(event) = number of favourable outcomes / total number of outcomes. All probabilities sum to 1. So the probability of not rolling a 6 on a die is ⅚.
基础概率用分数表示：P(事件) = 有利结果数 / 总结果数。所有概率之和为 1。因此不掷出 6 的概率是 ⅚。
Tree diagrams help with combined events. When drawing a tree for flipping a coin twice, label branches with probabilities (½ each). Multiply along branches to find the probability of two heads: ½ × ½ = ¼.
树状图有助于解决复合事件。画掷两次硬币的树状图时，在分支上标出概率 (各 ½)。沿分支相乘可求出两次正面的概率：½ × ½ = ¼。
When events are ‘without replacement’, the probabilities change. For example, drawing two red sweets from a bag of 5 red and 3 green changes the denominator from 8 to 7 for the second pick. Update fractions carefully.
当事件是“不放回”时，概率会变化。例如，从 5 红 3 绿的袋中取两颗红色糖果，第二次抽取时分母由 8 变为 7。务必小心更新分数。
11. Mean, Median, Mode and Range | 平均数、中位数、众数和极差
The mode is the most frequent value. The median is the middle value when data is ordered. The mean is the sum of all values divided by how many there are. The range is the largest minus the smallest.
众数是出现频率最高的值。中位数是数据排序后居中的值。平均数是所有数据总和除以数据个数。极差是最大值减最小值。
A typical question gives a set of numbers, e.g., 4, 7, 2, 9, 7, 11, and asks for all four measures. Order them: 2, 4, 7, 7, 9, 11. Mode = 7, median = (7+7)/2 = 7, mean = (2+4+7+7+9+11) ÷ 6 = 6.67, range = 11 − 2 = 9.
典型题目会提供一组数，如 4, 7, 2, 9, 7, 11，要求计算全部四项统计量。排序：2, 4, 7, 7, 9, 11。众数 = 7，中位数 = (7+7)/2 = 7，平均数 = (2+4+7+7+9+11) ÷ 6 = 6.67，极差 = 11 − 2 = 9。
Be aware that the mean is sensitive to outliers, while the median and mode are more robust. An outlier can pull the mean up or down significantly.
注意平均数对极端值很敏感，而中位数和众数较稳健。一个极端值可能显著拉高或拉低平均数。
12. Sequences and nth Term | 数列与第 n 项
Linear sequences increase or decrease by a constant difference. The nth term for a sequence like 5, 8, 11, 14, … has difference 3, so the formula begins with 3n. Then adjust to match the first term: 3 × 1 = 3, we need 5, so +2. Thus the nth term is 3n + 2.
线性数列以固定的差递增或递减。数列 5, 8, 11, 14, … 的公差为 3，因此公式以 3n 开头。然后调整使首项匹配：3 × 1 = 3，需要 5，因此 +2。所以第 n 项公式为 3n + 2。
You may be asked to find the 10th term or to use the nth term to check if a number belongs to the sequence. Substitute n = 10: 3 × 10 + 2 = 32. To check if 50 is in the sequence, solve 3n + 2 = 50 → 3n = 48 → n = 16, so yes.
你可能被要求找出第 10 项，或者用第 n 项公式判断某个数是否属于该数列。代入 n = 10：3 × 10 + 2 = 32。要判断 50 是否在数列中，解方程 3n + 2 = 50 → 3n = 48 → n = 16，因此是。
For more complex patterns, like square or triangle numbers, recognise the pattern and describe it in words and symbols. Practice writing the nth term for sequences like 1, 4, 9, 16, … (n²) or 1, 3, 6, 10, … (n(n+1)/2).
对于更复杂的规律，如平方数或三角数，要能识别规律并用文字和符号描述。练习写出 1, 4, 9, 16, … 的 n² 以及 1, 3, 6, 10, … 的 n(n+1)/2 这类第 n 项公式。
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