Tag: KS3

📚 Cambridge KS3 Maths: Percentages, Ratios and Proportion | 剑桥KS3数学：百分数、比例与比

Percentages, ratios and proportion appear throughout the Cambridge KS3 mathematics curriculum, from shopping discounts and recipe changes to scale drawings and speed calculations. This revision guide brings the key ideas together so you can move from simple conversions to more demanding problem solving with confidence.

百分数、比和比例贯穿剑桥KS3数学课程，从购物折扣、配方调整到比例尺作图和速度计算都会用到。这份复习指南把核心概念集中起来，帮助你从简单转换逐步过渡到更有挑战性的问题求解。

1. Understanding Percentages | 理解百分数

A percentage is simply a number out of 100. The word ‘per cent’ means ‘per hundred’, so 37% means 37 out of every 100, and 100% means the whole amount.

百分数就是一个以100为分母的数。“per cent”表示“每一百”，所以37%表示每一百中有37，100%表示整体。

Percentages are useful because they give a standard way of comparing parts of different wholes. For example, 18 out of 25 and 70 out of 100 are hard to compare directly, but both can be written as percentages.

百分数很有用，因为它提供了比较不同整体部分的标准方法。例如，25人中的18人和100人中的70人直接比较比较困难，但两者都可以写成百分数。

percentage = (part ÷ whole) × 100

2. Converting Between Fractions, Decimals and Percentages | 分数、小数与百分数之间的转换

You need to be able to switch quickly between fractions, decimals and percentages. To change a fraction to a percentage, divide the numerator by the denominator and then multiply by 100.

你需要能够在分数、小数和百分数之间快速转换。要把分数转换为百分数，先用分子除以分母，再乘以100。

To change a percentage to a decimal, divide by 100. To change a decimal to a percentage, multiply by 100. For example, 0.45 = 45% and 8% = 0.08.

要把百分数转换为小数，除以100；要把小数转换为百分数，乘以100。例如，0.45 = 45%，8% = 0.08。

Common equivalents such as 1/2 = 0.5 = 50%, 1/4 = 0.25 = 25% and 3/4 = 0.75 = 75% should be memorised because they speed up calculations in exams.

常见的等值关系如 1/2 = 0.5 = 50%、1/4 = 0.25 = 25% 和 3/4 = 0.75 = 75% 应该记住，因为它们在考试中能加快计算速度。

3. Finding a Percentage of a Quantity | 求一个数量的百分数

Without a calculator, build up the answer using 10%, 5% and 1%. For example, to find 35% of 80, first find 10% = 8, then 5% = 4, so 30% = 24 and 5% = 4, giving 28.

在没有计算器的情况下，可以利用10%、5%和1%来逐步求出答案。例如，求80的35%，先求10% = 8，再求5% = 4，所以30% = 24，5% = 4，得到28。

With a calculator, convert the percentage to a decimal and multiply. For 35% of 80, calculate 0.35 × 80 = 28.

使用计算器时，把百分数转换为小数再相乘。求80的35%，计算 0.35 × 80 = 28。

When the percentage is greater than 100, the answer is larger than the original quantity. For instance, 120% of 50 = 1.2 × 50 = 60.

当百分数大于100时，结果比原数量大。例如，50的120%等于 1.2 × 50 = 60。

4. Percentage Increase and Decrease | 百分数增减

To increase a quantity by a percentage, add the percentage to 100% and then use the new percentage. For example, increasing 60 by 15% means finding 115% of 60, which is 1.15 × 60 = 69.

要把一个数量增加某个百分数，先把该百分数加到100%上，再求新的百分数。例如，把60增加15%就是求60的115%，即 1.15 × 60 = 69。

To decrease a quantity, subtract the percentage from 100%. Decreasing 80 by 20% means finding 80% of 80, which is 0.8 × 80 = 64.

要减少某个百分数，从100%中减去该百分数。把80减少20%就是求80的80%，即 0.8 × 80 = 64。

Always pay attention to whether the question asks for the increase only or the new amount. The increase on 60 at 15% is 9, but the new amount is 69.

始终注意题目要求的是只求增加量还是新数量。60增加15%，增加量是9，但新数量是69。

5. Reverse Percentages | 逆向百分数

Reverse percentage problems give the value after a percentage change and ask for the original amount. Write the new amount as a percentage of the original, then divide to find 1% and multiply to find 100%.

逆向百分数问题给出百分数变化后的值，要求原数量。把新数量写成原数量的百分数，然后除以该百分数求1%，再乘以100求100%。

For example, if a jacket costs £84 after a 30% reduction, £84 represents 70% of the original price. Divide by 70 to find 1% = £1.20, then multiply by 100 to get the original price £120.

例如，一件夹克降价30%后售价为84英镑，那么84英镑代表原价的70%。除以70得1% = 1.20英镑，再乘以100得原价120英镑。

Do not simply increase £84 by 30%, because 30% of £84 is not the same as 30% of the original price. The original amount is always the base for the percentage change.

不要简单地把84英镑增加30%，因为84英镑的30%与原价的30%并不相同。原数量才是百分数变化的基础。

6. Introduction to Ratio | 比的基础

A ratio compares quantities of the same kind. For example, if a fruit bowl has 3 apples and 5 oranges, the ratio of apples to oranges is 3 : 5.

比用来比较同类数量。例如，一个水果碗里有3个苹果和5个橙子，苹果与橙子的比就是 3 : 5。

The order of a ratio matters. A ratio of 2 : 1 is not the same as 1 : 2. Always write ratios in the order requested by the question.

比的顺序很重要。2 : 1 和 1 : 2 并不相同。一定要按照题目要求的顺序写出比。

Ratios can be written in different forms but still represent the same relationship. Multiplying or dividing each part by the same non-zero number gives an equivalent ratio, such as 2 : 3 = 4 : 6.

比可以写成不同形式但仍表示相同关系。把每一部分同时乘以或除以同一个非零数，会得到相等的比，例如 2 : 3 = 4 : 6。

7. Simplifying and Sharing in a Ratio | 简化比与按比分配

To simplify a ratio, divide all parts by their highest common factor. The ratio 12 : 18 : 24 simplifies by dividing each part by 6 to give 2 : 3 : 4.

要简化比，把所有部分同时除以它们的最大公因数。比 12 : 18 : 24 的每部分除以6，得到 2 : 3 : 4。

To share an amount in a ratio, add the parts to find the total number of shares, divide the amount by this total, then multiply by each part. For example, £90 shared in the ratio 2 : 3 gives 5 parts, so one part is £18; the shares are £36 and £54.

按比分配时，先把比的各部分相加得到总份数，用总量除以总份数，再分别乘以各部分。例如，90英镑按 2 : 3 分配，总份数为5，一份是18英镑，因此两份为36英镑，三份为54英镑。

When a ratio includes more than two parts, the same method applies. Adding all the parts gives the total number of shares used in the division.

当比包含两个以上的部分时，方法相同。把所有部分相加得到用于分配的总份数。

8. Proportion and Unitary Method | 比例与单位法

Two quantities are in direct proportion when they increase or decrease at the same rate. For example, if 5 pens cost £3, then 15 pens cost three times as much, £9.

当两个数量以相同速率增加或减少时，它们成正比。例如，5支笔3英镑，那么15支笔就是三倍价格，9英镑。

The unitary method finds the value of one unit first. If 8 kg of flour costs £6, then 1 kg costs £6 ÷ 8 = £0.75, so 12 kg costs £0.75 × 12 = £9.

单位法先求出1个单位的值。如果8千克面粉6英镑，那么1千克就是 6 ÷ 8 = 0.75英镑，所以12千克为 0.75 × 12 = 9英镑。

Use the unitary method for recipe problems, currency conversions, speeds and maps. It helps you show clear working and reduces mistakes when quantities are not multiples of each other.

单位法可用于配方问题、货币换算、速度和地图问题。它能展示清晰的计算过程，并在数量不是倍数关系时减少错误。

9. Problem Solving with Percentages and Ratios | 百分数与比的问题求解

Many KS3 exam questions combine several skills in one problem. Read the question carefully and underline key information such as ‘of’, ‘increase’, ‘decrease’, ‘share’, and ‘in the ratio’.

许多KS3考试题把多项技能结合在一个问题中。仔细读题，并在“的”“增加”“减少”“分配”“按比例”等关键信息下划线。

For example, a bike worth £240 is reduced by 25% in a sale, then a further 10% is taken off the sale price. The first reduction gives £240 × 0.75 = £180, and the second gives £180 × 0.90 = £162. The final price is not 65% of £240 because the second reduction applies to a smaller amount.

例如，一辆价值240英镑的自行车在促销中降价25%，然后在促销价上再减10%。第一次降价后为 240 × 0.75 = 180英镑，第二次为 180 × 0.90 = 162英镑。最终价格不是240英镑的65%，因为第二次降价是在较小金额上计算的。

When a ratio changes after a quantity is added or removed, set up the before and after ratios using a common multiplier. For example, if the ratio of red to blue counters is 3 : 5 and 12 blue counters are removed so the ratio becomes 3 : 2, let the original numbers be 3x and 5x, then solve 5x − 12 = 2x to find x = 4.

当添加或移除数量后比值发生变化时，使用共同乘数建立变化前后的比。例如，红蓝计数器的比为 3 : 5，拿走12个蓝色计数器后比变为 3 : 2，设原来的数量分别为 3x 和 5x，解方程 5x − 12 = 2x，得 x = 4。

10. Common Misconceptions and Exam Tips | 常见误区与应试技巧

A common mistake is confusing percentages with absolute amounts. A 50% increase on a small number can be a much smaller actual change than a 10% increase on a large number.

一个常见误区是把百分数和绝对数量混淆。对较小数字增加50%，其实际变化可能远小于对较大数字增加10%。

Another mistake is adding ratios incorrectly. When a ratio is 2 : 3 : 5, the total number of shares is 10, not 3. Always add the parts before dividing the quantity.

另一个错误是错误地相加比。当比为 2 : 3 : 5 时，总份数是10，不是3。在分配数量之前，一定要先相加各部分。

In exams, show all working even when using a calculator. Marks are given for method, and writing the multiplier or division step helps you avoid hidden errors.

考试中，即使使用计算器也要写出所有步骤。方法可以得分，写出乘数或除法步骤有助于避免隐藏错误。

Finally, check that your answer makes sense in the context of the question. A percentage decrease should not give a negative amount unless the question allows it, and a shared amount should always add back to the original total.

最后，检查答案在题目情境中是否合理。百分数减少不应得到负值，除非题目允许；按比分配的各个部分相加后应等于原来的总量。

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📚 Cambridge KS3 Mathematics: Solving Linear Equations Step by Step | 剑桥 KS3 数学：逐步解线性方程

Linear equations are one of the most important topics in the Cambridge KS3 mathematics curriculum. They appear in number, algebra and problem-solving questions, and they provide the foundation for more advanced work in graphs, sequences and formulae. In this article, you will learn how to solve linear equations step by step, avoid common errors and apply your skills to word problems.

线性方程是剑桥 KS3 数学课程中最重要的主题之一。它们出现在数、代数与问题解决题中，并且为图像、数列和公式等更高级的内容打下基础。在本文中，你将逐步学习如何解线性方程、避免常见错误，并把技能应用到文字题中。

1. What Is a Linear Equation? | 什么是线性方程？

A linear equation is an equation in which the unknown appears only to the first power. It usually contains one variable, such as x, and it can be written in the form ax + b = c, where a, b and c are numbers.

线性方程是指未知数只出现一次幂的方程。它通常包含一个变量，例如 x，并且可以写成 ax + b = c 的形式，其中 a、b 和 c 是数。

For example, 3x + 2 = 14, 5y − 7 = 18 and x/4 + 1 = 6 are all linear equations. The variable is not squared, cubed or placed in a denominator.

例如，3x + 2 = 14、5y − 7 = 18 和 x/4 + 1 = 6 都是线性方程。变量没有被平方、立方，也没有出现在分母中。

The main goal is to find the value of the unknown that makes the equation true. This value is called the solution or root.

主要目标是求出使等式成立的未知数的值。这个值称为方程的解或根。

2. Key Vocabulary | 关键术语

Before you start solving, make sure you understand these key words. Knowing the correct vocabulary will help you follow each step and explain your method clearly in class.

在开始解题之前，请确保你理解以下关键词。掌握正确的术语将帮助你跟上每一个步骤，并在课堂上清晰地解释你的方法。

Understanding these terms will help you follow each step without confusion and answer short-answer questions accurately.

理解这些术语将帮助你清晰地跟上每一个步骤，并准确回答简答题。

3. The Balancing Method | 天平法（等式平衡法）

An equation is like a balance scale. Whatever you do to one side, you must do to the other side. This is the golden rule of solving equations.

方程就像一架天平。你对一边所做的任何操作，都必须对另一边做同样的操作。这是解方程的黄金法则。

If you add, subtract, multiply or divide one side by a number, you must do exactly the same to the other side to keep the equation balanced. Think of the two pans of a scale staying level.

如果你对一边加、减、乘或除以一个数，你必须对另一边做完全相同的操作，以保持等式平衡。想象天平的两个托盘始终保持水平。

x + 3 = 10 → x + 3 − 3 = 10 − 3 → x = 7

This method allows you to isolate the unknown on one side of the equation. Always write the operation you are performing on both sides on the same line.

这种方法可以让你把未知数单独分离到等式的一边。请始终在同一行中写出你对两边所执行的运算。

4. Solving One-Step Equations | 解一步方程

One-step equations require only one operation to isolate the variable. For addition or subtraction equations, use the inverse operation.

一步方程只需要一步运算即可分离变量。对于加法或减法方程，使用逆运算即可。

Example 1: Solve x + 9 = 20.

例 1：解方程 x + 9 = 20。

x + 9 − 9 = 20 − 9 → x = 11

Example 2: Solve y − 5 = 12.

例 2：解方程 y − 5 = 12。

y − 5 + 5 = 12 + 5 → y = 17

For multiplication or division equations, divide or multiply both sides by the same non-zero number.

对于乘法或除法方程，两边同时除以或乘以同一个非零数。

Example 3: Solve 4x = 28.

例 3：解方程 4x = 28。

4x ÷ 4 = 28 ÷ 4 → x = 7

Example 4: Solve x/5 = 3.

例 4：解方程 x/5 = 3。

x/5 × 5 = 3 × 5 → x = 15

Always check your answer by substituting it back into the original equation. For Example 3, 4 × 7 = 28, so the answer is correct.

始终通过将答案代回原方程来检验。例如例 3 中，4 × 7 = 28，因此答案正确。

5. Solving Two-Step Equations | 解两步方程

Two-step equations involve two operations, such as multiplication and addition. To solve them, undo the operations in the reverse order: first the constant, then the coefficient.

两步方程涉及两种运算，例如乘法和加法。解这类方程时，按相反顺序逆运算：先处理常数，再处理系数。

Example 1: Solve 3x + 2 = 14.

例 1：解方程 3x + 2 = 14。

3x + 2 − 2 = 14 − 2 → 3x = 12

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📚 KS3 Cambridge Mathematics: Mastering Linear Equations | KS3 剑桥数学：掌握一元一次方程

Linear equations are one of the most important building blocks of KS3 mathematics. At Cambridge Lower Secondary level, you need to be able to solve equations fluently, use the balancing method, and apply equations to real-world problems. This article will guide you through every key skill and help you avoid common mistakes.

一元一次方程是 KS3 数学中最重要的基础知识之一。在剑桥初中阶段，你需要能够熟练解方程，使用等式平衡法，并运用方程解决实际问题。本文将带你掌握每一项核心技能，并帮助你规避常见错误。

1. What Is a Linear Equation? | 什么是一元一次方程？

A linear equation in one variable is any equation that can be written in the form ax + b = c, where a, b and c are numbers and x is the unknown. The word “linear” means that the variable x only appears to the power 1. There are no x² or x³ terms in this type of equation.

一元一次方程是可以写成 ax + b = c 形式的方程，其中 a、b、c 是数，x 是未知数。“线性”一词表示变量 x 只以一次幂出现。在这类方程中没有 x² 或 x³ 项。

For example, all of the following are linear equations:

例如，以下方程都是一元一次方程：

x + 3 = 7, 2x − 5 = 9, 4(x + 1) = 20, 3x = 15

The solution of an equation is the value of x that makes the equation true. In the first example, x = 4 because 4 + 3 = 7.

方程的解是使等式成立的 x 值。在第一个例子中，x = 4，因为 4 + 3 = 7。

2. The Balancing Method | 等式平衡法

The balancing method is the safest way to solve linear equations. Think of the equation as a balance scale: both sides must stay equal at all times. If you add, subtract, multiply or divide one side, you must do exactly the same to the other side.

等式平衡法是解一元一次方程最稳妥的方法。把方程想象成一个天平：两边必须始终保持相等。如果你对一边进行加、减、乘或除，就必须对另一边做完全相同的运算。

This method helps you isolate x step by step. The key is to use inverse operations. The table below shows the most common inverse pairs.

这个方法能帮助你一步步把 x 分离出来。关键是使用逆运算。下表展示了最常见的逆运算对应关系。

Always write the operation you are doing on both sides. This helps you avoid making mistakes and shows clear working in your exercise book.

每次都要在两边写出你正在进行的运算。这能帮助你避免出错，并且在练习本中展示清晰的解题过程。

3. Solving Simple Equations | 解简单方程

Let us solve x + 5 = 12. To isolate x, subtract 5 from both sides. The equation becomes x + 5 − 5 = 12 − 5, which simplifies to x = 7.

让我们来解 x + 5 = 12。为了分离 x，两边同时减去 5。方程

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📚 Mastering Linear Equations for Cambridge KS3 Mathematics | 掌握剑桥 KS3 数学中的线性方程

Linear equations are one of the most important building blocks in the Cambridge KS3 mathematics curriculum. They appear in nearly every topic, from solving word problems to understanding graphs and formulas. If you can solve a linear equation confidently, you already have a powerful tool for later work in algebra, geometry and even science. This article takes you through the key ideas step by step, with clear explanations, worked examples and helpful practice tips. It is designed to support classroom learning and revision for tasks such as the exercise on page 183, question 2.

线性方程是剑桥 KS3 数学课程中最重要的基础模块之一。它们几乎出现在每个主题中，从解文字题到理解图像和公式都离不开它。如果你能自信地解出一个线性方程，就已经掌握了代数、几何甚至科学后续学习的强大工具。本文将带你一步步梳理核心概念，提供清晰的解释、例题和实用的练习建议，旨在帮助你巩固课堂所学并复习类似第 183 页第 2 题这样的练习。

1. What is a Linear Equation? | 什么是线性方程？

A linear equation is an equation in which the highest power of the variable is 1. This means the variable, often written as x, is not squared, cubed or raised to any higher power. The general form of a linear equation in one variable can be written as ax + b = c, where a, b and c are numbers and a is not zero.

线性方程是指未知数的最高次数为 1 的方程。这意味着未知数（通常写作 x）不会被平方、立方或进行更高次运算。一元一次方程的一般形式可以写成 ax + b = c，其中 a、b、c 是数字，且 a 不等于 0。

For example, 3x + 5 = 17 and 2(x – 4) = 10 are linear equations. The word ‘linear’ suggests a straight line: when these equations are plotted on a graph, they produce a straight line. Understanding this link between algebra and graphs will help you later with coordinate geometry.

例如，3x + 5 = 17 和 2(x – 4) = 10 都是线性方程。“线性”一词意味着直线：当这些方程在坐标图上绘制时，它们会形成一条直线。理解代数与图像之间的这种联系，将有助于你后续学习坐标几何。

2. The Balancing Method | 平衡法

Think of an equation as a set of balance scales. The left-hand side and the right-hand side must always have the same value. Whatever you do to one side of the equation, you must do exactly the same to the other side. This is the golden rule of solving equations.

把方程想象成一台天平。左边和右边的值必须始终相等。你对方程的一边做了什么，就必须对另一边也做完全相同的操作。这是解方程的金科玉律。

• Add the same number to both sides. 两边同时加上同一个数。
• Subtract the same number from both sides. 两边同时减去同一个数。
• Multiply both sides by the same non-zero number. 两边同时乘以同一个不为 0 的数。
• Divide both sides by the same non-zero number. 两边同时除以同一个不为 0 的数。

This method keeps the equation balanced and helps you isolate the variable step by step.

这种方法能保持方程平衡，帮助你逐步分离出未知数。

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📚 Cambridge KS3 Maths: Pythagoras’ Theorem | 剑桥初中数学：勾股定理

This KS3 Cambridge revision guide covers the key facts you need for Pythagoras’ theorem, with worked examples and exam-style tips. It is designed for students working through page 181, exercise 2 and similar questions.

这本剑桥 KS3 复习指南涵盖勾股定理所需的关键知识点，配有例题和考试技巧。它适合正在学习第 181 页练习 2 及类似题目的学生。

1. What is Pythagoras’ Theorem? | 什么是勾股定理？

Pythagoras’ theorem is a rule that works only for right-angled triangles. It states that the square of the longest side, called the hypotenuse, is equal to the sum of the squares of the other two shorter sides.

勾股定理是一条只适用于直角三角形的规则。它指出，最长边（称为斜边）的平方等于另外两条较短边的平方和。

In symbols, for a right-angled triangle with legs a and b, and hypotenuse c:

用符号表示，对于直角边为 a 和 b、斜边为 c 的直角三角形：

a² + b² = c²

2. The Hypotenuse and the Legs | 斜边与直角边

The hypotenuse is always the side opposite the right angle. It is the longest side in the triangle. The other two sides are called the legs, and they form the right angle.

斜边始终是直角三角形中直角对面的那条边。它是三角形中最长的边。另外两条边称为直角边，它们形成直角。

Always label the hypotenuse as c before substituting into the formula. This avoids many common mistakes.

在代入公式之前，始终将斜边标为 c。这可以避免许多常见错误。

3. The Formula a² + b² = c² | 公式 a² + b² = c²

The formula says that if you square both shorter sides and

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📚 Solving Linear Equations for KS3 Cambridge | KS3剑桥数学：解一元一次方程

Linear equations are one of the most important skills in the Cambridge KS3 Mathematics curriculum. They appear in algebra, geometry, and word problems across Year 7 to Year 9. This article explains how to solve them step by step, check answers, and avoid common errors.

一元一次方程是剑桥 KS3 数学课程中最重要的技能之一。它们出现在代数、几何以及 Year 7 到 Year 9 的文字题中。本文将逐步讲解如何解方程、检验答案并避免常见错误。

1. What is a Linear Equation? | 什么是一元一次方程？

A linear equation is an equation where the unknown variable, often written as x, is only raised to the power of 1. For example, 2x + 3 = 11 is linear. If you see x² or x³, it is not a linear equation at KS3 level.

一元一次方程是指未知数（通常用 x 表示）的指数为 1 的方程。例如 2x + 3 = 11 就是一元一次方程。如果出现 x² 或 x³，就不是 KS3 阶段的一元一次方程。

Linear equations can look different, but they all mean the same thing: two expressions have the same value. Your job is to find the value of the unknown that makes the equation true.

一元一次方程可以有不同形式，但它们都表示同一个意思：两个表达式的值相等。你的任务是找出使方程成立的未知数的值。

2. The Balancing Method | 天平法（等式平衡）

Think of an equation as a balance scale. Whatever you do to one side, you must do to the other side. This keeps the equation true. If you add 5 to the left, add 5 to the right.

把方程想象成一台天平。你对一边做的任何运算，必须对另一边也做同样的运算，这样等式才能保持成立。如果左边加 5，右边也要加 5。

If a = b, then a + c = b + c, a – c = b – c, a × c = b × c, a ÷ c = b ÷ c (c ≠ 0)

This rule is the basis of every method shown below. Never add, subtract, multiply, or divide only one side of an equation.

这条规则是下面所有方法的基础。绝不能只对等式的一边进行加、减、乘或除。

3. Solving One-Step Equations | 解一步方程

Use inverse operations to isolate the unknown. For x + 4 = 9, subtract 4 from both sides to get x = 5. For 3x = 18, divide both sides by 3 to get x = 6.

使用逆运算将未知数单独留在一边。对于 x + 4 = 9，两边同时减 4，得 x = 5。对于 3x = 18，两边同时除以 3，得 x = 6。

If the equation is x ÷ 4 = 5, multiply both sides by 4 to get x = 20. The inverse of addition is subtraction, and the inverse of multiplication is division.

如果方程是 x ÷ 4 = 5，两边同时乘以 4，得 x = 20。加法的逆运算是减法，乘法的逆运算是除法。

• x + 7 = 15 → x = 8
• x – 3 = 9 → x = 12
• 5x = 35 → x = 7
• x ÷ 6 = 4 → x = 24

4. Solving Two-Step Equations | 解两步方程

For an equation such as 2x + 3 = 11, undo the addition or subtraction first, then undo the multiplication or division. Subtract 3 from both sides to get 2x = 8, then divide both sides by 2 to get x = 4.

对于 2x + 3 = 11 这样的方程，先处理加减法，再处理乘除法。两边先减 3，得 2x = 8，然后两边除以 2，得 x = 4。

2x + 3 = 11 → 2x = 8 → x = 4

This order is important because the term 2x is being grouped as one value before 3 is added. You must remove the 3 first to reveal 2x.

这个顺序很重要，因为 2x 在被加上 3 之前是作为一个整体存在的。你必须先去掉 3，才能得到 2x。

5. Equations with Brackets | 含括号的方程

Always expand brackets first unless the bracket is being divided or multiplied by a single number. For 3(x + 2) = 21, expand to 3x + 6 = 21, then subtract 6 and divide by 3 to get x = 5.

通常先展开括号，除非括号被一个数除或乘。对于 3(x + 2) = 21，展开为 3x + 6 = 21，然后两边减 6，再除以 3，得 x = 5。

3(x + 2) = 21 → 3x + 6 = 21 → 3x = 15 → x = 5

If you see a negative outside the bracket, be careful with signs. For -2(x – 4) = 10, expand to -2x + 8 = 10, then solve to find x = -1.

如果括号外是负数，要注意符号。对于 -2(x – 4) = 10，展开为 -2x + 8 = 10，然后解方程得 x = -1。

6. Variables on Both Sides | 两边含未知数的方程

When the unknown appears on both sides, collect the variable terms on one side first. Solve 5x – 7 = 2x + 5 by subtracting 2x from both sides to get 3x – 7 = 5. Then add 7 and divide by 3 to find x = 4.

当未知数出现在等式两边时，先把含未知数的项合并到同一边。解 5x – 7 = 2x + 5，两边先减 2x，得 3x – 7 = 5，然后加 7，再除以 3，得 x = 4。

5x – 7 = 2x + 5 → 3x – 7 = 5 → 3x = 12 → x = 4

A common choice is to subtract the smaller variable term from both sides. This keeps the coefficient of x positive, which is easier to work with.

一个常见做法是从两边减去较小的未知数项。这样可以使 x 的系数保持为正，计算起来更容易。

7. Equations with Fractions | 含分数的方程

To clear fractions, multiply every term on both sides by the denominator. For x/3 + 2 = 6, multiply all terms by 3 first: x + 6 = 18, then subtract 6 to get x = 12.

为了消除分数，将两边每一项都乘以分母。对于 x/3 + 2 = 6，先把所有项都乘以 3：x + 6 = 18，然后两边减 6，得 x = 12。

x/3 + 2 = 6 → x + 6 = 18 → x = 12

If there is more than one fraction, multiply by the lowest common denominator of all fractions. For x/2 + x/3 = 5, multiply by 6: 3x + 2x = 30, so 5x = 30 and x = 6.

如果有多个分数，要乘以所有分母的最小公分母。对于 x/2 + x/3 = 5，两边乘以 6：3x + 2x = 30，所以 5x = 30，x = 6。

8. Forming Equations from Word Problems | 从文字题建立方程

Translate words into algebra. The sentence “I think of a number, double it and add 7; the answer is 19” becomes 2n + 7 = 19. Define the variable clearly before writing the equation.

把文字翻译成代数。句子“我想一个数，把它乘 2 再加 7，结果是 19”可以写成 2n + 7 = 19。在写方程前要先清楚地定义变量。

2n + 7 = 19 → 2n = 12 → n = 6

Geometry also uses linear equations. In a triangle, if the angles are x°, 2x°, and 3x°, the sum is 180°, so x + 2x + 3x = 180. This gives 6x = 180 and x = 30.

几何中也会用到一元一次方程。在一个三角形中，如果三个角分别是 x°、2x° 和 3x°，则内角和为 180°，所以 x + 2x + 3x = 180。由此得 6x = 180，x = 30。

9. Checking Your Answer | 检验答案

Always substitute your value back into the original equation. If 2x + 3 = 11 and x = 4, then 2(4) + 3 = 8 + 3 = 11. Since both sides match, the answer is correct.

始终把你求得的值代回原方程进行检验。如果 2x + 3 = 11 且 x = 4，那么 2(4) + 3 = 8 + 3 = 11。两边相等，答案就是正确的。

The check step is especially useful in exams because it catches sign errors and arithmetic mistakes before you move on.

检验这一步在考试中特别有用，因为它能在你继续做下一题前发现符号错误和计算错误。

10. Common Mistakes and How to Avoid Them | 常见错误及如何避免

The first common mistake is operating on only one side of the equation. Remember that any operation must be applied to both sides to keep the balance.

第一个常见错误是只对等式的一边进行运算。记住，任何运算都必须同时作用于两边，才能保持等式平衡。

The second mistake is sign errors when moving terms. For example, in 3x – 4 = 2x + 6, subtracting 2x from both sides gives x – 4 = 6, not x + 4 = 6.

第二个常见错误是移项时出现符号错误。例如在 3x – 4 = 2x + 6 中，两边减 2x 得到 x – 4 = 6，而不是 x + 4 = 6。

Finally, remember to check whether the unknown is just x, not x². Linear equations only include x to the

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📚 Mastering Linear Equations for KS3 Cambridge Mathematics | KS3 剑桥数学：掌握线性方程

Linear equations are one of the most important building blocks in KS3 Cambridge Mathematics. They appear in every topic from geometry to statistics, and mastering them early will make Year 9 and IGCSE work much easier. This article walks you through the key ideas, methods and common mistakes step by step, with practice questions at the end.

线性方程是 KS3 剑桥数学中最重要的基础模块之一。从几何到统计，它们几乎出现在每个主题中；尽早掌握线性方程，会让 Year 9 和 IGCSE 的学习轻松很多。本文逐步讲解核心概念、解题方法和常见错误，并在最后配有练习题。

1. What is a Linear Equation? | 什么是线性方程？

A linear equation is a statement that two expressions are equal, and the unknown appears only to the power of 1. The word ‘linear’ means that if you draw its graph, you get a straight line. Examples include x + 5 = 12, 3x – 7 = 8 and 2(x + 4) = 18.

线性方程是表示两个表达式相等的一种陈述，未知数的次数仅为 1 次。”线性”这个词意味着如果你画出它的图像，会得到一条直线。例如 x + 5 = 12、3x – 7 = 8 和 2(x + 4) = 18。

linear equation: ax + b = c, where a ≠ 0

线性方程：ax + b = c，其中 a ≠ 0

In KS3 you will mostly solve equations where the unknown is called x, y or n. The goal is always the same: find the value of the unknown that makes the equation true.

在 KS3 阶段，你主要会解未知数用 x、y 或 n 表示的方程。目标始终相同：求出使方程成立的未知数的值。

2. The Balancing Method | 等式的平衡法

Think of an equation as a balance scale. Whatever you do to one side, you must do exactly the same to the other side, otherwise the scale tips. This is the golden rule of solving equations.

把方程想象成一架天平。无论你对一边做什么运算，另一边也必须做完全相同的运算，否则天平就会倾斜。这是解方程的黄金法则。

• Add the same number to both sides: both sides add the same number
• Subtract the same number from both sides: both sides subtract the same number
• Multiply or divide both sides by the same non-zero number: both sides multiply or divide by the same non-zero number

• 两边同时加上同一个数
• 两边同时减去同一个数
• 两边同时乘以或除以同一个非零的数

For example, if x + 3 = 10, we can subtract 3 from both sides to keep the equation balanced: x + 3 – 3 = 10 – 3, so x = 7.

例如，如果 x + 3 = 10，我们可以从两边同时减去 3 来保持等式平衡：x + 3 – 3 = 10 – 3，因此 x = 7。

3. Solving One-Step Equations | 解一步方程

One-step equations can be solved by performing one inverse operation. If x has been increased by a number, subtract that number from both sides. If x has been multiplied by a number, divide both sides by that number.

一步方程只需使用一次逆运算即可解出。如果 x 被加上了一个数，就在两边同时减去这个数。如果 x 被乘以了一个数，就在两边同时除以这个数。

Example 1: Solve x + 8 = 15.

示例 1：解方程 x + 8 = 15。

x + 8 – 8 = 15 – 8 => x = 7

x + 8 – 8 = 15 – 8 => x = 7

Example 2: Solve 5x = 35.

示例 2：解方程 5x = 35。

5x ÷ 5 = 35 ÷ 5 => x = 7

5x ÷ 5 = 35 ÷ 5 => x = 7

Always use the inverse operation: addition reverses subtraction, multiplication reverses division, and vice versa.

始终使用逆运算：加法抵消减法，乘法抵消除法，反之亦然。

4. Solving Two-Step Equations | 解两步方程

Two-step equations require two inverse operations, usually in reverse order of BIDMAS. For example, in 3x – 7 = 8, x is first multiplied by 3 and then 7 is subtracted. To solve, undo the subtraction first, then undo the multiplication.

两步方程需要使用两次逆运算，通常按 BIDMAS 的相反顺序进行。例如在 3x – 7 = 8 中，x 先乘以 3，再减去 7。解方程时，先消除减法，再消除乘法。

Solve 3x – 7 = 8.

解方程 3x – 7 = 8。

3x – 7 + 7 = 8 + 7 => 3x = 15

3x – 7 + 7 = 8 + 7 => 3x = 15

3x ÷ 3 = 15 ÷ 3 => x = 5

3x ÷ 3 = 15 ÷ 3 => x = 5

It helps to write each new line directly below the previous one. This makes your working clear and easy to check in an exam.

把每一个新的步骤直接写在前一行下方会很有帮助。这样在考试中你的解题过程清晰且易于检查。

5. Equations with Brackets | 带括号的方程

When an equation contains brackets, expand them first using the distributive law. For example, 2(x + 4) = 18 becomes 2x + 8 = 18, then solve as a two-step equation.

当方程中含有括号时，首先使用分配律展开括号。例如 2(x + 4) = 18 可以展开为 2x + 8 = 18，然后按两步方程来解。

Solve 2(x + 4) = 18.

解方程 2(x + 4) = 18。

2x + 8 = 18

2x + 8 = 18

2x + 8 – 8 = 18 – 8 => 2x = 10

2x + 8 – 8 = 18 – 8 => 2x = 10

x = 5

x = 5

Be careful with negative signs: 3(2x – 5) = 21 expands to 6x – 15 = 21, not 6x – 5 = 21.

注意负号：3(2x – 5) = 21 展开后是 6x – 15 = 21，而不是 6x – 5 = 21。

6. Equations with Variables on Both Sides | 两边都有未知数的方程

When the unknown appears on both sides, collect all variable terms on one side and all number terms on the other. Choose the side with the larger coefficient of x to keep things positive if possible.

当未知数出现在等式两边时，把所有含未知数的项移到一边，把所有常数项移到另一边。如果可能，选择 x 系数较大的一边，以保持未知数系数为正。

Solve 7x – 3 = 4x + 9.

解方程 7x – 3 = 4x + 9。

7x – 4x – 3 = 4x – 4x + 9 => 3x – 3 = 9

7x – 4x – 3 = 4x – 4x + 9 => 3x – 3 = 9

3x – 3 + 3 = 9 + 3 => 3x = 12

3x – 3 + 3 = 9 + 3 => 3x = 12

x = 4

x = 4

Always subtract the smaller variable term from both sides to avoid negative coefficients at the start.

总是从两边减去较小的含未知数项，以避免一开始就出现负系数。

7. Forming Equations from Word Problems | 根据应用题建立方程

Word problems ask you to translate a real-life situation into an equation. Start by letting the unknown be x, then write each part of the sentence as a mathematical expression.

应用题要求你把现实情境转化为方程。首先设未知数为 x，然后把句子中的每一部分写成数学表达式。

Example: Sam is 5 years older than Tom. The sum of their ages is 27. Find Tom’s age.

示例：Sam 比 Tom 大 5 岁。他们的年龄之和是 27。求 Tom 的年龄。

Let Tom’s age = x. Then Sam’s age = x + 5.

设 Tom 的年龄为 x。则 Sam 的年龄为 x + 5。

x + (x + 5) = 27 => 2x + 5 = 27

x + (x + 5) = 27 => 2x + 5 = 27

2x = 22 => x = 11

2x = 22 => x = 11

Tom is 11 years old and Sam is 16 years old. Always re-read the question to check your answer makes sense in the context.

Tom 11 岁，Sam 16 岁。始终重新读一遍题目，检查答案在情境中是否合理。

8. Checking Your Solution | 检验你的解

After solving an equation, substitute your answer back into the original equation. If the left side equals the right side, your solution is correct.

解完方程后，把答案代回原方程。如果左边等于右边，你的解就是正确的。

For 3x – 7 = 8, we found x = 5.

对于 3x – 7 = 8，我们求出 x = 5。

Left side: 3(5) – 7 = 15 – 7 = 8 => Right side: 8

左边：3(5) – 7 = 15 – 7 = 8 => 右边：8

Since both sides equal 8, x = 5 is correct. This habit is especially useful in exams where small arithmetic slips are common.

由于两边都等于 8，x = 5 是正确的。这个习惯在考试中特别有用，因为小的算术失误很常见。

9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Many marks are lost in KS3 equations through avoidable errors. The table below shows some frequent mistakes and the correct approach.

在 KS3 方程题中，许多丢分都来自可以避免的错误。下表列出了一些常见错误和正确做法。

10. Practice Questions and Quick Answers | 练习题与快速答案

Try these questions without looking at the answers. Then check your working step by step.

先尝试不看答案完成以下题目，然后逐步检查你的解题过程。

• Solve x + 9 = 20
• Solve 4x = 28
• Solve 2x + 3 = 17
• Solve 3(x – 2) = 12
• Solve 5x + 4 = 2x + 19
• Form and solve: Twice a number plus 6 equals 30. Find the number.

• 解方程 x + 9 = 20
• 解方程 4x = 28
• 解方程 2x + 3 = 17
• 解方程 3(x – 2) = 12
• 解方程 5x + 4 = 2x + 19
• 列方程求解：一个数的两倍加上 6 等于 30。求这个数。

Quick answers: x = 11; x = 7; x = 7; x = 6; x = 5; x = 12.

快速答案：x = 11；x = 7；x = 7；x = 6；x = 5；x = 12。

If you got any wrong, identify the step where the mistake happened and re-read the relevant section above.

如果你有任何一题做错，找出错误发生在哪一步，并重新阅读上文相关小节。

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📚 Cambridge KS3 Maths: Scale Drawings and Map Scales (p166 Q2 Style) | 剑桥 KS3 数学：比例图与地图比例尺（p166 第2题类型）

Scale drawings and map scales are essential tools in Cambridge KS3 Mathematics. They connect classroom ratio work to real-world maps, architectural plans and scale models. A question such as p166 Q2 typically asks you to interpret a scale, convert a measurement, or draw a scaled version of a shape. This article explains the key methods step by step.

比例图和地图比例尺是剑桥 KS3 数学的重要工具。它们把课堂上的比例知识与真实地图、建筑平面图和比例模型联系起来。像 p166 第2题这类题目通常会要求你解读比例尺、换算测量值，或绘制图形的缩放版本。本文一步步讲解关键方法。

1. What Is a Scale Drawing? | 什么是比例图？

A scale drawing is a copy of a real object or distance in which all lengths are multiplied or divided by the same number. The drawing looks the same in shape, but its size is changed. Maps, model cars, floor plans and technical diagrams are all examples of scale drawings.

比例图是真实物体或距离的复制图，其中所有长度都乘以或除以同一个数。图形的形状不变，但大小改变。地图、汽车模型、房屋平面图和技术图表都是比例图的例子。

For example, a plan might use the scale 1 : 100. This means that every 1 cm on the plan represents 100 cm in real life. The ratio 1 : 100 is the scale factor written as a ratio.

例如，一张平面图可能使用 1 : 100 的比例尺。这表示图上的每 1 cm 代表实际生活中的 100 cm。比值 1 : 100 就是以比值形式书写的比例因子。

scale = drawing length : real length

比例尺 = 图上长度 : 实际长度

2. Reading and Writing Map Scales | 读取和书写地图比例尺

A map scale can be written in three main ways: as a ratio, as a statement, or as a scale bar. In Cambridge KS3 questions you will usually see the ratio form, such as 1 : 25 000 or 1 : 50 000.

地图比例尺可以写成三种主要形式：比值形式、文字说明形式或比例条。在剑桥 KS3 题目中，你通常会看到比值形式，例如 1 : 25 000 或 1 : 50 000。

The ratio 1 : 25 000 means that 1 unit on the map represents 25 000 of the same units on the ground. If you measure in centimetres, then 1 cm on the map represents 25 000 cm in real life. Since 100 cm = 1 m and 1000 m = 1 km, this is the same as 250 m or 0.25 km.

比值 1 : 25 000 表示地图上的 1 个单位代表地面上 25 000 个相同单位。如果以厘米为单位测量，那么地图上的 1 cm 代表实际 25 000 cm。由于 100 cm = 1 m，1000 m = 1 km，这等于 250 m 或 0.25 km。

Always check whether the question wants the answer in centimetres, metres or kilometres. Converting units carefully is part of the skill being tested.

始终检查题目要求答案使用厘米、米还是千米。仔细换算单位正是考查能力的一部分。

3. Converting Map Distances to Real Distances | 将地图距离转换为实际距离

To find a real distance from a map, multiply the measured map distance by the scale factor. This works because the map has shrunk the real length by that factor.

要根据地图求实际距离，将测量的地图距离乘以比例因子即可。这是因为地图已经按该因子缩小了实际长度。

real distance = map distance × scale factor

实际距离 = 地图距离 × 比例因子

Suppose a map has scale 1 : 50 000 and two villages are 4 cm apart on the map. The real distance is 4 × 50 000 = 200 000 cm. Then convert to kilometres: 200 000 cm = 2000 m = 2 km.

假设一张地图的比例尺是 1 : 50 000，两个村庄在地图上相距 4 cm。实际距离为 4 × 50 000 = 200 000 cm。然后换算为千米：200 000 cm = 2000 m = 2 km。

You can also set up an equivalent ratio. Write 1 : 50 000 = 4 : x, then solve by multiplying 4 × 50 000. Both methods give the same result, and choosing one consistent method helps you avoid errors.

你也可以列出等价比。写为 1 : 50 000 = 4 : x，然后通过 4 × 50 000 求解。两种方法结果相同，选择一种一致的方法有助于避免错误。

4. Converting Real Distances to Map Distances | 将实际距离转换为地图距离

To find how far apart two places should be on a map, divide the real distance by the scale factor. You must first put both quantities in the same unit before dividing.

要求两个地点在地图上应相距多远，将实际距离除以比例因子。在相除之前，必须先把两个量化为相同单位。

map distance = real distance ÷ scale factor

图上距离 = 实际距离 ÷ 比例因子

For example, a road is 3 km long in real life and the map scale is 1 : 25 000. Write 3 km as 300 000 cm. Then map distance = 300 000 ÷ 25 000 = 12 cm.

例如，一条道路实际长 3 km，地图比例尺为 1 : 25 000。将 3 km 写为 300 000 cm。那么图上距离 = 300 000 ÷ 25 000 = 12 cm。

Some questions ask you to draw a line segment of this length, so measure it carefully with a ruler from 0, not from the edge of the ruler.

有些题目会要求你画出这个长度的线段，所以用尺子从 0 刻度开始仔细测量，而不是从尺子边缘开始。

5. Changing Units Correctly | 正确换算单位

Unit conversion is one of the most important sub-skills in scale problems. In Cambridge KS3 Mathematics you need to move confidently between centimetres, metres and kilometres.

单位换算是比例尺问题中最重要的子技能之一。在剑桥 KS3 数学中，你需要熟练地在厘米、米和千米之间转换。

1 cm = 10 mm
1 m = 100 cm
1 km = 1000 m = 100 000 cm

1 cm = 10 mm，1 m = 100 cm，1 km = 1000 m = 100 000 cm。

A useful shortcut for map scales is to remember that 100 000 cm = 1 km. So if you have 200 000 cm, divide by 100 000 to get 2 km. If you have 3 km, multiply by 100 000 to get 300 000 cm.

地图比例尺的一个有用速记是：100 000 cm = 1 km。因此如果有 200 000 cm，除以 100 000 得到 2 km。如果有 3 km，乘以 100 000 得到 300 000 cm。

When a question mixes units, convert to the smaller unit first. This keeps the arithmetic simple and reduces the chance of dividing in the wrong direction.

当题目混用单位时，先换算为较小单位。这样算式更简单，也能减少除错方向的可能。

6. Scale Drawings of Shapes | 图形的比例图

Scale drawing is not only about maps. You may need to enlarge or reduce a rectangle, triangle or irregular shape using a given scale. Every side must be multiplied by the same scale factor.

比例图不仅用于地图。你可能需要按给定比例放大或缩小矩形、三角形或不规则图形。每条边都必须乘以相同的比例因子。

For example, a rectangle measures 5 cm by 3 cm. It is enlarged by scale factor 3. The new dimensions are 15 cm by 9 cm. The shape remains a rectangle because both dimensions changed by the same factor.

例如，一个矩形长 5 cm、宽 3 cm。按比例因子 3 放大后，新尺寸为 15 cm 和 9 cm。由于两个尺寸都按相同因子变化，图形仍然是矩形。

If you are asked to draw a scale version, start by labelling the original lengths. Then multiply each length by the given factor, and measure the new lengths with a ruler. Check that angles stay the same: scaling changes size, not shape.

如果要求你画一个比例版本，先标注原始长度。然后将每个长度乘以给定因子，并用尺子测量新长度。注意角度保持不变：比例缩放改变大小，不改变形状。

7. Bearings on Scale Diagrams | 比例图上的方位角

Some Cambridge KS3 questions combine scale drawing with bearings. A bearing measures direction clockwise from north. Bearings are always written with three digits, such as 045° or 135°.

一些剑桥 KS3 题目将比例图与方位角结合考查。方位角是从北顺时针测量的方向。方位角始终写成三位数，例如 045° 或 135°。

To draw a bearing, place a protractor over the point, align 0° with north, and measure clockwise. Then use the scale to draw the correct distance along that bearing line.

要画方位角，将量角器放在点上，使 0° 对准北方，然后顺时针测量。接着使用比例尺沿着该方位线画出正确的距离。

For example, a ship sails from a harbour on a bearing of 060° for 5 km. If the scale is 1 : 50 000, then 5 km = 500 000 cm, and the map distance is 10 cm. Draw a north line, measure 60° clockwise, and draw a 10 cm line.

例如，一艘船从港口出发，沿 060° 方位航行 5 km。如果比例尺为 1 : 50 000，那么 5 km = 500 000 cm，图上距离为 10 cm。画出指北线，顺时针量出 60°，并画出 10 cm 长的线段。

Questions often ask for the return bearing. To find it, add 180° if the original bearing is less than 180°, or subtract 180° if it is 180° or more.

题目经常要求返航方位角。求法：如果原方位角小于 180°，则加 180°；如果大于或等于 180°，则减 180°。

8. Area Scale Factor | 面积比例因子

When a shape is enlarged by a linear scale factor k, its area is enlarged by k². This is a very common trap in scale questions, especially when converting map areas.

当一个图形按线性比例因子 k 放大时，其面积按 k² 放大。这是比例尺题目中非常常见的陷阱，尤其是在换算地图面积时。

area scale factor = k²

面积比例因子 = k²

Suppose a map scale is 1 : 50 000. The linear scale factor is 50 000. If a lake has area 2 cm² on the map, the real area is 2 × 50 000² = 2 × 2 500 000 000 = 5 000 000 000 cm². Converting to km² gives 0.5 km².

假设地图比例尺为 1 : 50 000。线性比例因子是 50 000。如果一座湖在地图上的面积是 2 cm²，那么实际面积为 2 × 50 000² = 2 × 2 500 000 000 = 5 000 000 000 cm²。换算为 km² 得到 0.5 km²。

Remember: 1 m² = 10 000 cm² and 1 km² = 1 000 000 m². Do not apply the linear factor to an area. If you find yourself multiplying an area by 50 000 instead of 50 000², stop and check.

记住：1 m² = 10 000 cm²，1 km² = 1 000 000 m²。不要将线性因子直接用于面积。如果你发现自己在用 50 000 而不是 50 000² 乘面积，请停下来检查。

9. Common Mistakes | 常见错误

One common mistake is forgetting to convert units before multiplying or dividing. For example, writing 3 km as 3 cm and multiplying by the scale factor gives a nonsense result.

一个常见错误是在乘法或除法之前忘记换算单位。例如把 3 km 写成 3 cm 再乘以比例因子，会得到荒谬的结果。

Another mistake is reading 1 : 25 000 as 1 cm = 25 km. It is 1 cm = 25 000 cm, which is only 0.25 km. Always move from cm to km by dividing by 100 000.

另一个错误是把 1 : 25 000 读成 1 cm = 25 km。实际上它是 1 cm = 25 000 cm，仅为 0.25 km。始终通过除以 100 000 将 cm 换算为 km。

Students also confuse area scale factor with linear scale factor. Use k for lengths and k² for areas. Finally, when drawing bearings, remember to measure clockwise from north, not from the nearest vertical grid line.

学生也容易混淆面积比例因子和线性比例因子。长度使用 k，面积使用 k²。最后，画方位角时记住从北顺时针测量，而不是从最近的竖直线开始测量。

10. Worked Exam-Style Example | 考试题型示例

A map has scale 1 : 25 000. Two towns are 6.4 cm apart on the map. Calculate the actual distance in kilometres.

一张地图的比例尺为 1 : 25 000。两个城镇在地图上相距 6.4 cm。计算实际距离，以 km 为单位。

Real distance = 6.4 × 25 000 = 160 000 cm. Since 100 000 cm = 1 km, 160 000 cm = 1.6 km. The actual distance is 1.6 km.

实际距离 = 6.4 × 25 000 = 160 000 cm。由于 100 000 cm = 1 km，160 000 cm = 1.6 km。实际距离为 1.6 km。

On the same map, a river is 3 km long in real life. Work out the length of the river on the map in centimetres.

在同一张地图上，一条河流实际长 3 km。计算该河流在地图上的长度，以 cm 为单位。

3 km = 300 000 cm. Map distance = 300 000 ÷ 25 000 = 12 cm. A 12 cm line should be drawn on the map.

3 km = 300 000 cm。图上距离 = 300 000 ÷ 25 000 = 12 cm

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