Essential Maths 9C Homework Book: Question Types Explained | KS3 数学 Essential Maths 9C 作业题型解析

📚 Essential Maths 9C Homework Book: Question Types Explained | KS3 数学 Essential Maths 9C 作业题型解析

The Essential Maths 9C Homework Book is a key resource for Year 9 students in Key Stage 3, covering a wide range of topics from number and algebra to geometry and statistics. This article analyses the typical question types found in the book, providing step-by-step approaches in both English and Chinese to help students master essential KS3 maths skills and tackle homework with confidence.

《Essential Maths 9C 作业本》是 KS3 阶段九年级学生的重要练习资料，全面覆盖数、代数、几何与统计等主题。本文深入解析书中常见题型，以中英双语提供分步解题方法，帮助学生掌握核心数学技能，自信应对作业挑战。

1. Arithmetic with Negative Numbers | 负数运算

This question type tests your ability to use the four operations with positive and negative integers. You must remember the rules: adding a negative is the same as subtracting its positive value, and when multiplying or dividing, two negatives make a positive.

这类题型考查正负整数的四则运算。必须牢记规则：加上一个负数等于减去其绝对值；在乘法和除法中，同号得正、异号得负。

Example: Work out -3 + 4 × (-2) – 6 ÷ (-3).

例题：计算 -3 + 4 × (-2) – 6 ÷ (-3)。

Solution steps:

解题步骤：

1. Follow the order of operations (BIDMAS). Perform multiplication and division first, from left to right: 4 × (-2) = -8, and 6 ÷ (-3) = -2.

1. 遵循运算顺序（括号、指数、乘除、加减）。先算乘除，从左到右：4 × (-2) = -8，6 ÷ (-3) = -2。

2. Rewrite the expression: -3 + (-8) – (-2).

2. 重写表达式：-3 + (-8) – (-2)。

3. Simplify the double signs: -3 – 8 + 2 = -9.

3. 去括号化简：-3 – 8 + 2 = -9。

Final answer: -9

最终答案：-9

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

Questions often require converting between fractions, decimals and percentages, and performing calculations with them, especially in real-life contexts like discounts or data. Knowing common equivalences (e.g. 1/8 = 0.125 = 12.5%) speeds up problem solving.

题目常要求分数、小数和百分数之间的相互转换，并完成混合运算，尤其涉及折扣或数据分析等实际情境。熟记常见等值关系（如 1/8 = 0.125 = 12.5%）能提高解题速度。

Example: Convert 0.375 to a fraction in its simplest form, and then find 37.5% of 240.

例题：将 0.375 化为最简分数，并计算 240 的 37.5%。

Solution:

解答：

1. 0.375 = 375/1000. Simplify by dividing numerator and denominator by 125: 375÷125 = 3, 1000÷125 = 8, giving 3/8.

1. 0.375 = 375/1000。分子和分母同时除以 125：375÷125 = 3，1000÷125 = 8，得到 3/8。

2. 37.5% is equivalent to 0.375, so 37.5% of 240 = 0.375 × 240 = 90.

2. 37.5% 即 0.375，因此 240 的 37.5% = 0.375 × 240 = 90。

Answer: 90

答案：90

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3. Percentage Increase and Decrease | 百分比增减

These problems involve increasing or decreasing an amount by a given percentage. A reliable method uses a multiplier: for an increase of x%, multiply by 1 + x/100; for a decrease, multiply by 1 – x/100. This is essential for compound interest and repeated percentage change.

这类问题要求按照给定的百分比对数量进行增减。可靠的乘数法是：增加 x% 则乘以 1 + x/100；减少 x% 则乘以 1 – x/100。这为复利和多次百分比变化问题打下基础。

Example: A jacket originally costs £80. In a sale, the price is reduced by 15%. Find the sale price.

例题：一件夹克原价 80 英镑，打八五折（降价 15%）。折后价是多少？

Solution:

解答：

Decrease multiplier = 1 – 15/100 = 0.85. Sale price = 80 × 0.85 = £68.

减少乘数 = 1 – 15/100 = 0.85。售价 = 80 × 0.85 = 68 英镑。

Alternatively, you can first find 15% of 80 (= 12) and subtract from 80: 80 – 12 = 68.

另一种方法是先算出 80 的 15%（即 12），再从 80 中减去：80 – 12 = 68。

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4. Ratio and Proportion | 比与比例

Ratio questions ask you to share a quantity in a given ratio, scale recipes, or interpret maps. The key strategy is to find the value of one part by dividing the total quantity by the total number of parts. Once you have one part, you can find any amount.

比的问题通常需要按给定比例分配数量、缩放配方或解读地图。核心策略是用总量除以总份数，求出每份的具体数值。已知一份的量，即可求出任意数量。

Example: The ratio of boys to girls in a school is 3:5. If there are 720 students altogether, how many are girls?

例题：学校中男生与女生的人数比为 3:5。如果总共有 720 名学生，女生有多少人？

Solution:

解答：

Total number of parts = 3 + 5 = 8. One part = 720 ÷ 8 = 90.

总份数 = 3 + 5 = 8。一份 = 720 ÷ 8 = 90。

Number of girls = 5 parts × 90 = 450.

女生人数 = 5 份 × 90 = 450。

There are 450 girls.

女生有 450 人。

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5. Algebraic Expressions and Simplification | 代数表达式与化简

This topic covers collecting like terms, using the distributive law to expand brackets, and simplifying expressions with powers. Remember that terms must have exactly the same variable and exponent to be combined.

该主题涉及合并同类项、运用分配律展开括号以及化简含有幂的表达式。只有变量和指数完全相同的项才能合并。

Example: Simplify 3x² + 2x – 5 + 4x² – 3x + 1.

例题：化简 3x² + 2x – 5 + 4x² – 3x + 1。

Solution:

解答：

Group like terms: (3x² + 4x²) gives 7x²; (2x – 3x) gives -x; and (-5 + 1) gives -4.

合并同类项：(3x² + 4x²) 得 7x²；(2x – 3x) 得 -x；(-5 + 1) 得 -4。

Simplified expression: 7x² – x – 4

化简结果：7x² – x – 4

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6. Solving Linear Equations | 解一元一次方程

Linear equations have the unknown raised to the power of 1. Use inverse operations on both sides to isolate the variable. Always keep the equation balanced by doing the same thing to both sides, and check

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