📚 KS3 Maths: Essential Maths 8H Homework Answers – Compressed Question Type Analysis | KS3 数学：Essential Maths 8H 作业答案压缩题型解析
Essential Maths 8H is a widely used textbook for Year 8 students following a higher-tier scheme of work. The homework exercises often contain a mix of fluency, reasoning and problem‑solving questions that are deliberately compressed – meaning several skills are tested within a single task. This article unpacks common question types, shows how to arrive at full‑mark answers, and highlights the reasoning steps that examiners expect. Whether you are checking your own work or helping someone else, this breakdown will build confidence and accuracy.
《Essential Maths 8H》是 Year 8 高阶课程广泛使用的教材。课后作业常常包含混合了流畅性、推理与解决问题能力的“压缩式”题目，即一道题考查多种技能。本文拆解常见题型，展示如何得出满分答案，并突出阅卷者期望的推理步骤。无论你是在自查作业还是在辅导他人，这份解析都能帮你建立信心、提升准确率。
1. Operations with Negative Numbers | 负数运算
Many 8H homework questions combine addition, subtraction, multiplication and division of directed numbers in a single calculation.
很多 8H 作业题把有向数的加、减、乘、除合并在一个算式中考查。
Example question: Evaluate (-3)² + (-8) ÷ 2 – (-5).
例题：计算 (-3)² + (-8) ÷ 2 – (-5)。
Step 1: Handle the exponent first. (-3)² means (-3) × (-3) = 9.
第一步：先算指数。(-3)² 即 (-3) × (-3) = 9。
Step 2: Perform the division. (-8) ÷ 2 = -4.
第二步：进行除法。(-8) ÷ 2 = -4。
Step 3: Rewrite the expression: 9 + (-4) – (-5).
第三步：重写算式：9 + (-4) – (-5)。
Step 4: Subtracting a negative is equivalent to adding a positive, so -(-5) becomes +5.
第四步：减去一个负数等同于加上正数，-(-5) 变成 +5。
Now we have 9 + (-4) + 5. 9 + (-4) = 5, then 5 + 5 = 10.
现在得到 9 + (-4) + 5。9 + (-4) = 5，再加 5 得 10。
Final answer: 10.
最终答案：10。
2. Fractions, Decimals and Percentages Interchange | 分数、小数与百分比互换
Compressed tasks often ask you to convert between forms and then use the converted value in a comparison or a remaining‑amount problem.
压缩题型经常要求先进行形式转换，再运用到比较或剩余量问题中。
Example: Write 3/8 as a decimal and a percentage, then find 3/8 of £320.
例题：将 3/8 写为小数和百分数，然后求 £320 的 3/8 是多少。
To convert 3/8 to a decimal, divide 3 by 8: 3 ÷ 8 = 0.375.
将 3/8 化为小数：3 ÷ 8 = 0.375。
To change 0.375 to a percentage, multiply by 100: 0.375 × 100 = 37.5%.
把 0.375 化为百分数，乘以 100：0.375 × 100 = 37.5%。
Finding 3/8 of £320 means calculating (3/8) × 320. You can divide 320 by 8 to get 40, then multiply by 3: 40 × 3 = £120.
求 £320 的 3/8，即 (3/8) × 320。可以先用 320 除以 8 得 40，再乘以 3：40 × 3 = £120。
Alternatively, 0.375 × 320 also equals 120.
或者，0.375 × 320 同样等于 120。
Answer: 0.375, 37.5%, £120.
答案：0.375，37.5%，£120。
3. Simplifying Algebraic Expressions | 代数式化简
Homework in 8H frequently requires collecting like terms, expanding brackets and simplifying in one continuous expression.
8H 作业频繁要求合并同类项、展开括号并在一个连式中完成化简。
Example: Simplify 4a + 3b – 2a + 5b + 2(a – b).
例题：化简 4a + 3b – 2a + 5b + 2(a – b)。
Step 1: Expand the bracket: 2(a – b) = 2a – 2b.
第一步：展开括号：2(a – b) = 2a – 2b。
The expression becomes 4a + 3b – 2a + 5b + 2a – 2b.
式子变为 4a + 3b – 2a + 5b + 2a – 2b。
Step 2: Group a terms: 4a – 2a + 2a = 4a (because -2a + 2a cancels).
第二步：合并 a 项：4a – 2a + 2a = 4a（因为 -2a + 2a 抵消）。
Step 3: Group b terms: 3b + 5b – 2b = 6b.
第三步：合并 b 项：3b + 5b – 2b = 6b。
Simplified answer: 4a + 6b.
化简结果：4a + 6b。
4. Solving Linear Equations with Unknowns on Both Sides | 解含未知数在两侧的线性方程
Typical 8H problems demand solving equations such as 5x + 2 = 3x + 10, often followed by a substitution check.
典型的 8H 题目要求解如 5x + 2 = 3x + 10 的方程，通常还要求代入检验。
Example: Solve 7y – 4 = 3y + 12 and verify your answer.
例题：解方程 7y – 4 = 3y + 12 并检验答案。
Step 1: Subtract 3y from both sides: 7y – 4 – 3y = 3y + 12 – 3y → 4y – 4 = 12.
第一步：两边减去 3y：7y – 4 – 3y = 3y + 12 – 3y → 4y – 4 = 12。
Step 2: Add 4 to both sides: 4y = 16.
第二步：两边加 4：4y = 16。
Step 3: Divide by 4: y = 4.
第三步：除以 4：y = 4。
Check: Left side = 7(4) – 4 = 28 – 4 = 24; Right side = 3(4) + 12 = 12 + 12 = 24. They match.
检验：左边 = 7(4) – 4 = 28 – 4 = 24；右边 = 3(4) + 12 = 12 + 12 = 24。两边相等。
Final answer: y = 4.
最终答案：y = 4。
5. Generating Terms of a Sequence from the nth Term | 由第 n 项公式求数列各项
These questions give the nth term rule and ask for the first few terms, or ask you to find whether a certain number is in the sequence.
这种题给出第 n 项公式，要求写出前几项，或判断某个数是否在数列中。
Example: The nth term of a sequence is 3n + 5. Write down the first three terms. Is 50 a term in this sequence?
例题：某数列的第 n 项公式为 3n + 5。写出前三项。50 是这个数列的项吗？
First term (n = 1): 3(1) + 5 = 8.
第一项 (n = 1)：3(1) + 5 = 8。
Second term (n = 2): 3(2) + 5 = 11.
第二项 (n = 2)：3(2) + 5 = 11。
Third term (n = 3): 3(3) + 5 = 14. So the sequence begins 8, 11, 14.
第三项 (n = 3)：3(3) + 5 = 14。数列开头为 8, 11, 14。
To check if 50 is a term, set 3n + 5 = 50 → 3n = 45 → n = 15. Since n is an integer, 50 is the 15th term.
要判断 50 是否在数列中，设 3n + 5 = 50 → 3n = 45 → n = 15。因为 n 是整数，所以 50 是第 15 项。
Answer: First three terms: 8, 11, 14. Yes, 50 is a term.
答案：前三项：8, 11, 14。是的，50 是其中一项。
6. Angles on Parallel Lines and in Triangles | 平行线上的角与三角形内角
8H geometry questions mix alternate, corresponding and co‑interior angles with triangle angle sums, often in a single multi‑step diagram.
8H 几何题常将同位角、内错角、同旁内角与三角形内角和混合在一个多步图形问题中。
Example: In a diagram, two parallel lines are cut by a transversal. One given angle is 72°. Find the sizes of the other seven angles marked, explaining your reasoning.
例题：图中两条平行线被一条截线所截。已知一个角为 72°。求另外七个标记角的大小并解释理由。
When a transversal crosses parallel lines, alternate angles are equal, corresponding angles are equal, and co‑interior angles sum to 180°.
当截线穿过平行线时，内错角相等，同位角相等，同旁内角之和为 180°。
If the given angle is 72° and is, say, an acute alternate angle, its alternate is also 72°. The adjacent angle on a straight line is 180° – 72° = 108°. Then use corresponding facts to label all others.
如果已知角 72° 是一个锐角内错角，那么其内错角也是 72°。同一直线上的邻角为 180° – 72° = 108°。再用同位角等关系标出所有角。
In a combined triangle inside the diagram, interior angles sum to 180°, so you can find a missing third angle once two are known.
在图内嵌套的三角形中，内角和为 180°，因此知道两个角就可以求出第三个未知角。
Always write a brief reasoning beside each answer, such as “angles on a straight line” or “corresponding angles are equal”.
每个答案旁边务必写出简要理由，例如“平角上的角”或“同位角相等”。
7. Area and Perimeter of Composite Shapes | 组合图形的面积与周长
Composite shapes made of rectangles and triangles appear frequently. Students must split the shape, find missing side lengths, then calculate area and perimeter separately.
由矩形和三角形组合的图形经常出现。学生需要分解图形，找出缺失的边长，然后分别计算面积和周长。
Example: An L‑shaped figure is formed from two rectangles. The outer dimensions are 10 cm by 8 cm, with a cut‑out rectangle of 4 cm by 6 cm. Find the area and perimeter.
例题：一个 L 形由两个矩形构成。外部尺寸为 10 cm × 8 cm，切去了一个 4 cm × 6 cm 的矩形。求面积和周长。
Area method 1: Calculate the area of the large rectangle (10 × 8 = 80 cm²) and subtract the cut‑out (4 × 6 = 24 cm²). Area = 80 – 24 = 56 cm².
面积方法一：计算大矩形面积 (10 × 8 = 80 cm²)，减去切去的面积 (4 × 6 = 24 cm²)。面积 = 80 – 24 = 56 cm²。
Perimeter: Walk around the outside. The outline involves the 10 cm base, 8 cm height, a step inwards of 6 cm, 4 cm, etc. Add all outward edges: 10 + 8 + 6 + 4 + 4 + 4 = 36 cm (check your step lengths).
周长：沿着外轮廓走一圈。轮廓包括底边 10 cm、高 8 cm、内凹进 6 cm、4 cm 等。将所有外边长相加：10 + 8 + 6 + 4 + 4 + 4 = 36 cm（仔细核对步长）。
Always show a labelled sketch with missing sides calculated using subtraction.
务必画出带标注的草图，并用减法算出缺失边。
8. Interpreting Pie Charts and Bar Charts | 解读饼图与条形图
Data handling questions often present a pie chart with a frequency table or ask you to construct a pie chart from given data.
数据处理题经常给出带频数表的饼图，或要求根据数据绘制饼图。
Example: In a survey of 60 students, 20 prefer football, 15 prefer tennis, 10 prefer basketball and the rest prefer hockey. Calculate the angle for each sector and draw a pie chart.
例题：在一项 60 名学生的调查中，20 人喜欢足球，15 人喜欢网球，10 人喜欢篮球，其余喜欢曲棍球。计算每个扇形的角度并绘制饼图。
Total frequency = 60. Angle per student = 360° ÷ 60 = 6°.
总频数为 60。每个学生对应的角度 = 360° ÷ 60 = 6°。
Football angle = 20 × 6° = 120°; Tennis = 15 × 6° = 90°; Basketball = 10 × 6° = 60°; Hockey has 60 – 20 – 15 – 10 = 15 students, so angle = 15 × 6° = 90°.
足球角度 = 20 × 6° = 120°；网球 = 15 × 6° = 90°；篮球 = 10 × 6° = 60°；曲棍球人数 = 60 – 20 – 15 – 10 = 15，角度 = 15 × 6° = 90°。
Check: 120° + 90° + 60° + 90° = 360°. Then draw the circle and measure angles accurately.
检验：120° + 90° + 60° + 90° = 360°。接着画圆并精确度量角度。
9. Ratio and Proportion Word Problems | 比与比例文字题
Compressed ratio problems involve sharing in a ratio, then using the parts to find a total or a difference.
压缩式比例题涉及按比例分配，然后使用各份来求总数或差。
Example: The ratio of boys to girls in a class is 3:5. There are 12 more girls than boys. How many students are in the class?
例题：某班级男孩与女孩的比是 3:5。女孩比男孩多 12 人。班级共有多少名学生？
The difference in parts is 5 – 3 = 2 parts. These 2 parts represent 12 students. So 1 part = 12 ÷ 2 = 6 students.
份数差为 5 – 3 = 2 份。这 2 份代表 12 名学生。因此 1 份 = 12 ÷ 2 = 6 名学生。
Boys = 3 parts = 18, girls = 5 parts = 30. Total students = 18 + 30 = 48.
男孩 = 3 份 = 18 人，女孩 = 5 份 = 30 人。总人数 = 18 + 30 = 48。
The answer can be verified: 30 – 18 = 12, matching the condition.
可验证：30 – 18 = 12，符合条件。
10. Substitution into Formulae and Using BIDMAS | 公式代入与运算法则
Essential Maths 8H includes substituting negative values into expressions like v = u + at, or evaluating algebraic expressions with powers.
《Essential Maths 8H》包含将负值代入公式，如 v = u + at，或计算含幂次的代数式。
Example: Given a = 3, b = -2, c = -4, evaluate 2a² – 3b + c.
例题：已知 a = 3，b = -2，c = -4，求 2a² – 3b + c 的值。
Step 1: Substitute carefully: 2(3)² – 3(-2) + (-4).
第一步：仔细代入：2(3)² – 3(-2) + (-4)。
Step 2: Apply index first: 3² = 9, so 2 × 9 = 18.
第二步：先算指数：3² = 9，2 × 9 = 18。
Step 3: Multiplication: -3 × (-2) = +6.
第三步：乘法：-3 × (-2) = +6。
Step 4: Now the expression is 18 + 6 – 4. 18 + 6 = 24, 24 – 4 = 20.
第四步：式子变为 18 + 6 – 4。18 + 6 = 24，24 – 4 = 20。
Final answer: 20.
最终答案：20。
Remember BIDMAS: Brackets, Indices, Division/Multiplication, Addition/Subtraction.
牢记运算法则：括号、指数、除法和乘法、加法和减法。
11. Using Metric and Imperial Conversions in Context | 情境中的公制与英制单位转换
Some 8H tasks embed conversion factors (e.g., 1 inch ≈ 2.54 cm, 1 kg ≈ 2.2 lb) into multi‑step problems involving best buys or travel.
有些 8H 任务将单位换算系数（例如 1 英寸 ≈ 2.54 厘米，1 千克 ≈ 2.2 磅）融入涉及最佳购买或出行的多步问题中。
Example: A road sign shows 50 miles. If 5 miles ≈ 8 km, convert 50 miles to kilometres.
例题：路标显示 50 英里。如果 5 英里 ≈ 8 公里，把 50 英里转换为公里。
Using the ratio: 5 miles : 8 km, so 1 mile ≈ 8 ÷ 5 = 1.6 km.
根据比例：5 英里 : 8 公里，所以 1 英里 ≈ 8 ÷ 5 = 1.6 公里。
Then 50 miles ≈ 50 × 1.6 = 80 km.
因此 50 英里 ≈ 50 × 1.6 = 80 公里。
Alternatively, 50 is 10 times 5 miles, so kilometres = 10 × 8 = 80 km.
或者，50 英里是 5 英里的 10 倍，所以公里数 = 10 × 8 = 80 公里。
Always include the approximate symbol ≈ when using conversion factors.
使用换算系数时务必加上约等号 ≈。
12. Expanding Double Brackets and Factorising | 展开双括号与因式分解
By the end of 8H, students begin quadratic expansions like (x + 3)(x – 7) and simple factorising into single brackets.
在 8H 后期，学生开始学习二次展开如 (x + 3)(x – 7)，以及简单的提公因式分解。
Example: Expand and simplify (x + 5)(x – 2).
例题：展开并化简 (x + 5)(x – 2)。
Use the FOIL method: First: x × x = x². Outer: x × (-2) = -2x. Inner: 5 × x = 5x. Last: 5 × (-2) = -10.
使用 FOIL 法则：首项相乘：x × x = x²；外项相乘：x × (-2) = -2x；内项相乘：5 × x = 5x；末项相乘：5 × (-2) = -10。
Combine like terms: x² + 3x – 10.
合并同类项：x² + 3x – 10。
For factorising, e.g., 6a + 8, find the highest common factor: 2. Write 2(3a + 4).
因式分解如 6a + 8，找出最大公因数：2。写成 2(3a + 4)。
Always expand your answer to check.
始终将答案展开以检验正确性。
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