Essential Maths 7C Homework Book: Question Types Explained | 《Essential Maths 7C 作业本》题型解析

📚 Essential Maths 7C Homework Book: Question Types Explained | 《Essential Maths 7C 作业本》题型解析

The Essential Maths 7C Homework Book is a key resource for Year 7 students following the KS3 curriculum. It offers a wide range of practice questions that consolidate classroom learning and build fluency in core topics. This article breaks down the most common question types found in the book, providing step-by-step strategies and examples to help students tackle homework with confidence.

《Essential Maths 7C 作业本》是 KS3 阶段七年级学生的重要练习资料，涵盖大量巩固课堂知识、提升核心技能的训练题目。本文深度解析书中常见题型，结合例题和解题策略，帮助学生从容应对作业，提高数学自信与成绩。

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1. Number Operations and Order of Operations | 数的运算与运算顺序

Many questions test the BIDMAS rule (Brackets, Indices, Division/Multiplication, Addition/Subtraction). For example, evaluate 24 – 6 × 2 + 4² ÷ 2. The correct order: first handle the index: 4² = 16, then division and multiplication: 16 ÷ 2 = 8 and 6 × 2 = 12. Finally, work left to right: 24 – 12 + 8 = 12 + 8 = 20. Forgetting the rule leads to common errors, such as subtracting before multiplying.

大量题目检验 BIDMAS 规则（括号、指数、除/乘、加/减）。例如计算 24 – 6 × 2 + 4² ÷ 2。正确顺序：先算指数 4²=16，再同时处理除法和乘法：16÷2=8 及 6×2=12，最后从左到右加减：24–12+8=12+8=20。忽视规则会导致典型错误，如先减后乘。

Worded problems often embed multi-step operations. For instance: ‘Sarah buys 3 packs of pens at £2.50 each and a notebook for £1.75. She pays with a £10 note. How much change does she get?’ Solution: total spent = 3 × 2.50 + 1.75 = 7.50 + 1.75 = 9.25. Change = 10.00 – 9.25 = £0.75. This encourages students to write clear workings.

应用题常嵌入多步运算。如：“莎拉买了3包笔，每包2.50英镑，再加一本1.75英镑的笔记本，付10英镑，应找零多少？”解法：总花费=3×2.50+1.75=7.50+1.75=9.25，找零=10.00–9.25=0.75英镑。这要求学生写出清晰步骤。

The book also includes negative numbers alongside BIDMAS, e.g., –12 ÷ 3 + 5 × (–2). Apply division and multiplication first: –4 + (–10) = –14. Use number line logic or the ‘two negatives make a positive’ rule for multiplication/division.

书中还将负数与 BIDMAS 结合，如 –12÷3+5×(–2)。先算除法和乘法：–4+(–10)=–14。借用数轴逻辑或负负得正规则处理乘除。

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2. Fractions, Decimals and Percentages | 分数、小数与百分数

Adding and subtracting fractions with different denominators is a key skill. For 2/5 + 3/4, find a common denominator (20): 2/5 = 8/20, 3/4 = 15/20, sum = 23/20 = 1 3/20. Encourage students to always simplify answers.

异分母分数加减法是必会技能。例如 2/5+3/4，找公分母 20：2/5=8/20，3/4=15/20，和为 23/20 = 1 3/20。提醒学生最后化为最简形式或带分数。

Converting between forms: a typical question asks to write 0.35 as a simplified fraction and as a percentage. 0.35 = 35/100 = 7/20; 0.35 × 100% = 35%. Knowing common equivalents (e.g., 0.5 = 1/2 = 50%) speeds up work.

形式互化：典型题要求将 0.35 写为最简分数和百分数。0.35=35/100=7/20；0.35×100%=35%。熟记常见等价关系（如 0.5=1/2=50%）可提速。

Finding a percentage of an amount: 15% of 240. Method 1: 10% = 24, 5% = 12, so 15% = 36. Method 2: 0.15 × 240 = 36. Discount problems (‘25% off an £80 jacket’) are frequent: sale price = 80 × 0.75 = £60.

求一个数的百分之几：240 的 15%。方法一：10%=24，5%=12，合计 15%=36。方法二：0.15×240=36。打折题常见（如“80 英镑外套打 75 折”）：售价=80×0.75=60 英镑。

Fraction of a quantity: 3/8 of 32 kg = (32 ÷ 8) × 3 = 4 × 3 = 12 kg. The homework often mixes fractions of shapes and quantities.

求一个数的几分之几：32 kg 的 3/8 = (32÷8)×3=4×3=12 kg。作业本常将图形的分数与数量分数结合考查。

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3. Introduction to Algebra | 代数基础

Translating word descriptions into algebraic expressions is a core task. ‘I think of a number, multiply it by 5, then add 3’ becomes 5n + 3. Using function machines and writing expressions for perimeters (e.g., rectangle length x, width x + 2) are common.

将文字描述转为代数式是核心任务。“我想一个数，乘以5，再加3”写作 5n+3。书中常通过函数机器和写周长表达式（如长 x、宽 x+2 的矩形）来练习。

Simplifying expressions by collecting like terms: 4a + 3b – 2a + 5b = 2a + 8b. Students must learn not to combine unlike terms – a mistake like 3x + 2y = 5xy reveals a misunderstanding of terms.

合并同类项化简：4a+3b–2a+5b=2a+8b。学生须明白不同类项不能合并——像 3x+2y=5xy 的错误表明对“项”的理解不足。

Substitution exercises: if x = 3 and y = –2, evaluate 2x² – y. Work carefully: 2(3)² – (–2) = 2×9 + 2 = 18 + 2 = 20. Brackets help avoid sign errors with negative numbers.

代入求值：若 x=3, y=–2，求 2x²–y。仔细计算：2(3)²–(–2)=2×9+2=20。使用括号能避免负数符号错误。

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4. Solving Linear Equations | 解线性方程

One-step equations lay the groundwork: x + 7 = 12 → x = 5; 4x = 20 → x = 5. Students must understand that the same operation must be applied to both sides to maintain balance.

一步方程奠定基础：x+7=12 → x=5；4x=20 → x=5。学生必须理解等式两边需同时进行相同运算以保持平衡。

Two-step equations: 2x + 3 = 11. Reverse BIDMAS: subtract 3 first – 2x = 8, then divide by 2 – x = 4. The homework book often presents these in real-life contexts, such as ‘2x + 3 = total cost’.

两步方程：2x+3=11。逆向使用运算顺序：先减3得 2x=8，再除以2得 x=4。作业本常将其置于生活场景，如“2x+3=总花费”。

Equations with unknowns on both sides: 5y – 4 = 3y + 6. Begin by moving the smaller y-term: subtract 3y → 2y – 4 = 6, then add 4 → 2y = 10, y = 5. Encourage checking by substituting back.

两边含未知数的方程：5y–4=3y+6。先把较小的 y 项移走：两边减 3y 得 2y–4=6，再加 4 得 2y=10，y=5。提倡代入验算。

Forming equations from word problems: ‘The perimeter of a rectangle is 26 cm; its length is 8 cm. Find the width.’ 2(8 + w) = 26 → 16 + 2w = 26 → 2w = 10 → w = 5 cm. Drawing a diagram helps.

应用题列方程：“矩形周长26 cm，长8 cm，求宽。” 2(8+w)=26 → 16+2w=26 → 2w=10 → w=5 cm。画图辅助理解。

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5. Angles and Triangle Properties | 角与三角形性质

Angles on a straight line sum to 180°. If one angle is 47°, its adjacent supplementary angle is 180° – 47° = 133°. Around a point, angles total 360°. These facts are used to find missing angles without a protractor.

直线上的角之和为 180°。若一角为 47°，其邻补角=180°–47°=133°。一点周角之和为 360°。这些事实用于无用量角器求未知角。

In any triangle, angles sum to

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