KS3 Maths: Essential Maths Book 9i Answers – Question Type Analysis | KS3 数学：Essential Maths Book 9i Answers 题型解析

📚 KS3 Maths: Essential Maths Book 9i Answers – Question Type Analysis | KS3 数学：Essential Maths Book 9i Answers 题型解析

Welcome to our in-depth analysis of the most common question types found in Essential Maths Book 9i, a widely used resource for Year 9 students. Mastering these exercise patterns is essential for building a strong foundation in KS3 mathematics and bridging smoothly to GCSE. In this article, we break down typical problem formats, provide worked examples, and share step-by-step solutions that match the style of Book 9i answer keys. Each section focuses on a specific topic area, from number operations to geometry and probability, helping you understand not only the final answer but also the logical process behind it.

欢迎阅读我们针对 Essential Maths Book 9i 中最常见题型所做的深入解析，这本题集在 Year 9 学生中使用广泛。掌握这些练习的题型对于打好 KS3 数学基础并顺利衔接到 GCSE 至关重要。本文拆解了典型的题目形式，提供了与 Book 9i 答案解析风格一致的范例和逐步求解过程。每一节都聚焦一个特定主题领域，从数的运算到几何与概率，帮助你不仅理解最终答案，更掌握其背后的逻辑步骤。

1. Operations with Fractions and Decimals | 分数和小数运算

Students frequently encounter questions that mix fractions, decimals, and whole numbers, requiring careful application of the order of operations (BIDMAS). A classic Book 9i example is: Evaluate 1 1/2 + 2.5 × 3/4. First, convert mixed numbers and decimals to improper fractions or a single form: 1 1/2 = 3/2, 2.5 = 5/2. The multiplication takes priority: 5/2 × 3/4 = 15/8. Then add: 3/2 + 15/8 = 12/8 + 15/8 = 27/8 = 3 3/8.

学生经常会遇到分数、小数和整数混合的题目，需要认真使用运算顺序（括号、指数、乘除、加减）。Book 9i 中一个典型的例子是：计算 1 1/2 + 2.5 × 3/4。首先将带分数和小数都转换成假分数或统一的形式：1 1/2 = 3/2，2.5 = 5/2。乘法优先：5/2 × 3/4 = 15/8。然后做加法：3/2 + 15/8 = 12/8 + 15/8 = 27/8 = 3 3/8。

When subtracting fractions, always find a common denominator first. For instance, 5/6 – 2/9: the least common multiple of 6 and 9 is 18, so rewrite as 15/18 – 4/18 = 11/18. With decimals, align the decimal points and use column addition or subtraction. A common pitfall is forgetting to borrow correctly, so writing the numbers in a clear column helps.

在做分数减法时，一定要先通分。例如 5/6 – 2/9：6 和 9 的最小公倍数是 18，改写为 15/18 – 4/18 = 11/18。处理小数时，将小数点对齐并采用竖式加减。学生常见的错误是借位出错，因此将数字清晰地排成竖式会很有帮助。

2. Solving Linear Equations | 解线性方程

Linear equations in one variable appear throughout Book 9i, often with brackets and unknowns on both sides. Consider the equation 3(2x – 1) = 4x + 7. Begin by expanding the bracket: 6x – 3 = 4x + 7. Next, collect the x terms on one side by subtracting 4x from both sides: 2x – 3 = 7. Add 3 to both sides to give 2x = 10, and finally divide by 2 to obtain x = 5.

一元一次线性方程在 Book 9i 中贯穿始终，常常带有括号且两边都有未知数。考虑方程 3(2x – 1) = 4x + 7。先展开括号：6x – 3 = 4x + 7。然后将含 x 的项移到一边，两边减去 4x 得 2x – 3 = 7。两边加上 3 得到 2x = 10，最后除以 2，解得 x = 5。

If the equation contains fractions, eliminate denominators early by multiplying every term by the lowest common multiple. For example, in x/2 + 3 = x/3 + 5, multiply through by 6: 3x + 18 = 2x + 30. Then 3x – 2x = 30 – 18, leading to x = 12. Always verify your solution by substituting it back into the original equation.

如果方程中含有分数，可以一开始就用最小公倍数去分母。例如在 x/2 + 3 = x/3 + 5 中，两边乘以 6 得：3x + 18 = 2x + 30。然后 3x – 2x = 30 – 18，得到 x = 12。求出解后一定要代回原方程检验。

3. Simplifying Expressions and Expanding Brackets | 化简代数式和展开括号

Collecting like terms is a fundamental skill tested regularly. An expression such as 7a – 3b + 2a + 8b – b simplifies to (7a + 2a) + (–3b + 8b – b) = 9a + 4b. Pay close attention to signs, especially when subtracting a term with a negative coefficient.

合并同类项是一项经常考查的基本技能。例如表达式 7a – 3b + 2a + 8b – b 化简为 (7a + 2a) + (–3b + 8b – b) = 9a + 4b。特别注意符号，尤其是减去一个带有负系数的项时。

Expanding double brackets requires the systematic FOIL method. For (x + 4)(x – 3), multiply First terms: x·x = x²; Outer: x·(–3) = –3x; Inner: 4·x = 4x; Last: 4·(–3) = –12. Combine the middle terms: x² + x – 12. A common error is to forget that the product of a positive and a negative term is negative.

展开双括号需要系统地使用首外内尾（FOIL）法则。对于 (x + 4)(x – 3)，首项相乘 x·x = x²；外项 x·(–3) = –3x；内项 4·x = 4x；尾项 4·(–3) = –12。合并中间项得到 x² + x – 12。常见的错误是忘记正数与负数相乘结果为负。

4. Ratio and Proportion | 比例和正比例

Ratio problems are often presented in everyday contexts, such as sharing money or scaling recipes. If James and Sophie share £48 in the ratio 3:5, first find the total number of parts: 3 + 5 = 8. One part is worth £48 ÷ 8 = £6. James receives 3 × £6 = £18, Sophie receives 5 × £6 = £30.

比例问题常出现在日常生活情境中，如分钱或调整食谱比例。如果 James 和 Sophie 按 3:5 比分享 £48，先求出总份数：3+5=8，一份为 £48 ÷ 8 = £6。James 得到 3×£6=£18，Sophie 得到 5×£6=£30。

Direct proportion can be tackled using the unitary method. If 8 identical batteries cost £5.60, find the cost of 14 batteries. One battery costs £5.60 ÷ 8 = £0.70, so 14 batteries cost 14 × £0.70 = £9.80. Alternatively, set up a proportion: 8/5.60 = 14/x and cross-multiply to find x = (14 × 5.60) ÷ 8 = £9.80.

正比例问题可以用归一法求解。如果 8 节同样的电池售价 £5.60，求 14 节电池的价格。一节电池的价格是 £5.60 ÷ 8 = £0.70，因此 14 节的价格为 14 × £0.70 = £9.80。也可以列出比例式 8/5.60 = 14/x，交叉相乘求得 x = (14 × 5.60) ÷ 8 = £9.80。

5. Percentages, Increase and Decrease | 百分比增减

Percentage increase and decrease questions appear in shopping, salary, and population contexts. To increase £250 by 12%, multiply by 1.12: £250 × 1.12 = £280. For a decrease, such as a 20% discount on a £65 dress, multiply by 0.80: £65 × 0.80 = £52. Using the multiplier method is faster and reduces errors compared to finding the percentage amount and adding or subtracting.

百分比增减题常见于购物、薪资和人口场景。将 £250 增加 12%，乘以 1.12：£250 × 1.12 = £280。对于减少，例如一件 £65 的裙子打八折，应乘以 0.80：£65 × 0.80 = £52。使用乘数法比先算百分量再加减更快且减少错误。

Sometimes students need to work backwards to find the original amount before a percentage change. If the sale price of a computer after a 15% reduction is £680, then 85% of the original price equals £680. The original price = £680 ÷ 0.85 = £800. Checking: 15% of £800 is £120, and £800 – £120 = £680, confirming the result.

有时学生需要反推百分数变化前的原值。如果一台电脑减价 15% 后售价为 £680，那么原价的 85% 等于 £680。原价 = £680 ÷ 0.85 = £800。检验：£800 的 15% 为 £120，£800 – £120 = £680，结果正确。

6. Pythagoras’ Theorem | 勾股定理

Right-angled triangle problems test both finding the hypotenuse and a shorter side. The theorem states that a² + b² = c², where c is the hypotenuse. For a triangle with legs 5 cm and 12 cm, the hypotenuse c = √(5² + 12²) = √(25 + 144) = √169 = 13 cm. Always identify the hypotenuse as the side opposite the right angle, the longest side.

直角三角形的题目既考查求斜边，也考查求直角边。勾股定理表述为 a² + b² = c²，其中 c 为斜边。对于直角边为 5 cm 和 12 cm 的三角形，斜边 c = √(5² + 12²) = √(25 + 144) = √169 = 13 cm。务必将斜边识别为直角所对的最长边。

When a shorter side is missing, rearrange the formula: a = √(c² – b²). If the hypotenuse is 17 m and one leg is 8 m, the other leg = √(17² – 8²) = √(289 – 64) = √225 = 15 m. A common mistake is to add instead of subtract when solving for a leg; the diagram can help you decide.

当一条直角边未知

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