KS3 Maths: Worked Examples Explained | KS3 数学：典型例题详解

📚 KS3 Maths: Worked Examples Explained | KS3 数学：典型例题详解

In KS3 Mathematics, building a solid foundation means working through a wide range of typical problems. This article presents twelve carefully chosen examples with step-by-step solutions, covering number, algebra, geometry, data and probability. Each worked example shows not only how to get the right answer, but also why each step is necessary, helping you develop problem-solving skills you can rely on.

在 KS3 数学中，打下扎实的基础需要练习大量的典型题目。本文精选了十二道例题并提供逐步详解，涵盖数、代数、几何、数据处理与概率。每一道题不仅展示如何求得正确答案，也说明每一步的原因，帮助你建立可靠的问题解决能力。

1. Order of Operations (BIDMAS) | 整数混合运算

Example: Evaluate 15 + 3 × (8 − 2)² ÷ 9.

例题：计算 15 + 3 × (8 − 2)² ÷ 9。

Step 1: Work inside the brackets first: 8 − 2 = 6.

第一步：先计算括号内部：8 − 2 = 6。

Step 2: Apply the exponent (power): 6² = 36.

第二步：计算指数（幂）：6² = 36。

Step 3: Now multiplication and division from left to right. 3 × 36 = 108, then 108 ÷ 9 = 12.

第三步：然后从左到右计算乘除。3 × 36 = 108，接着 108 ÷ 9 = 12。

Step 4: Finally, addition: 15 + 12 = 27.

第四步：最后计算加法：15 + 12 = 27。

Answer: 27

答案：27

2. Adding and Subtracting Fractions | 分数加减法

Example: Work out 2/3 + 1/4 − 5/12.

例题：计算 2/3 + 1/4 − 5/12。

Step 1: Find a common denominator for 3, 4 and 12. The lowest common multiple is 12.

第一步：找到 3、4 和 12 的公分母。最小公倍数是 12。

Step 2: Convert each fraction: 2/3 = 8/12, 1/4 = 3/12, and 5/12 stays the same.

第二步：转换每个分数：2/3 = 8/12，1/4 = 3/12，5/12 保持不变。

Step 3: Add and subtract the numerators: 8 + 3 − 5 = 6. So the result is 6/12.

第三步：加减分子：8 + 3 − 5 = 6。得到 6/12。

Step 4: Simplify the fraction by dividing numerator and denominator by 6: 6/12 = 1/2.

第四步：约分，分子分母同时除以 6：6/12 = 1/2。

Answer: ½

答案：½

3. Decimals and Percentages | 小数与百分数

Example: (a) Write 0.625 as a percentage. (b) Find 15% of 80.

例题：(a) 把 0.625 写成百分数。(b) 求 80 的 15%。

Part (a): Multiply the decimal by 100%: 0.625 × 100% = 62.5%.

第 (a) 部分：小数乘以 100%：0.625 × 100% = 62.5%。

Part (b): ‘Percent’ means ‘out of 100’, so 15% = 15/100 = 0.15. Multiply: 80 × 0.15 = 12.

第 (b) 部分：百分数表示“百分之”，15% = 15/100 = 0.15。相乘：80 × 0.15 = 12。

Always remember to convert percentages to decimals before multiplying.

始终记住，相乘前要把百分数转换成小数。

4. Simplifying Algebraic Expressions | 代数表达式化简

Example: Simplify 3a + 5b − a + 2b + 4a.

例题：化简 3a + 5b − a + 2b + 4a。

Step 1: Identify and group like terms. Terms with ‘a’: 3a, −a, 4a. Terms with ‘b’: 5b, 2b.

第一步：找出并合并同类项。含 ‘a’ 的项：3a、−a、4a。含 ‘b’ 的项：5b、2b。

Step 2: Combine the ‘a’ terms: 3a − a + 4a = (3 − 1 + 4)a = 6a.

第二步：合并 ‘a’ 项：3a − a + 4a = (3 − 1 + 4)a = 6a。

Step 3: Combine the ‘b’ terms: 5b + 2b = 7b.

第三步：合并 ‘b’ 项：5b + 2b = 7b。

Step 4: Write the simplified expression: 6a + 7b.

第四步：写出化简后的表达式：6a + 7b。

Simplified: 6a + 7b

化简结果：6a + 7b

5. Solving Linear Equations | 解一次方程

Example: Solve 5x − 7 = 2x + 8.

例题：解方程 5x − 7 = 2x + 8。

Step 1: Get all variable terms on one side. Subtract 2x from both sides: 5x − 2x − 7 = 8, which simplifies to 3x − 7 = 8.

第一步：把所有含未知数的项移到一边。两边同减 2x：5x − 2x − 7 = 8，化简得 3x − 7 = 8。

Step 2: Isolate the variable term by adding 7 to both sides: 3x = 15.

第二步：两边同时加 7，把含未知数的项单独留在左边：3x = 15。

Step 3: Divide both sides by 3 to find x: x = 5.

第三步：两边同除以 3，得到 x：x = 5。

x = 5

6. Ratio and Proportion | 比与比例

Example: A map has a scale of 1 : 25 000. How many centimetres on the map represent a real distance of 5 km?

例题：一幅地图的比例尺是 1 : 25 000。实际距离 5 km 在地图上表示多少厘米？

Step 1: Convert the real distance to centimetres: 5 km = 5 × 1000 m = 5000 m, and 5000 m = 5000 × 100 cm = 500 000 cm.

第一步：把实际距离换算成厘米：5 km = 5 × 1000 m = 5000 m，5000 m = 5000 × 100 cm = 500 000 cm。

Step 2: The scale ratio 1 : 25 000 means 1 cm on the map stands for 25 000 cm on the ground. Divide the real distance by 25 000: 500 000 ÷ 25 000 = 20.

第二步：比例尺 1 : 25 000 表示地图上 1 cm 代表实际 25 000 cm。将实际距离除以 25 000：500 000 ÷ 25 000 = 20。

Map distance = 20 cm

图上距离 = 20 cm

7. Perimeter and Area | 周长与面积

Example: A rectangle has length 8 cm and width 5 cm. Find its perimeter and area. A right triangle inside has base 6 cm and height 4 cm; find its area.

例题：一个长方形长 8 cm，宽 5 cm。求它的周长和面积。一个直角三角形底 6 cm，高 4 cm；求它的面积。

Rectangle perimeter: P = 2(length + width) = 2(8 + 5) = 2 × 13 = 26 cm.

长方形周长：P = 2 × (长 + 宽) = 2 × (8 + 5) = 26 cm。

Rectangle area: A = length × width = 8 × 5 = 40 cm².

长方形面积：A = 长 × 宽 = 8 × 5 = 40 cm²。

Triangle area: A = ½ × base × height = ½ × 6 × 4 = 12 cm².

三角形面积：A = ½ × 底 × 高 = ½ × 6 × 4 = 12 cm²。

8. Volume of a Cuboid | 长方体体积

Example: A cuboid measures 10 cm in length, 4 cm in width and 3 cm in height. Calculate its volume.

例题：一个长方体长 10 cm，宽 4 cm，高 3 cm。计算它的体积。

Formula: Volume = length × width × height.

公式：体积 = 长 × 宽 × 高。

Substitute the values: V = 10 cm × 4 cm × 3 cm = 120 cm³.

代入数值：V = 10 cm × 4 cm × 3 cm = 120 cm³。

Volume = 120 cm³

体积 = 120 cm³

9. Mean, Median and Mode | 平均数、中位数与众数

Example: The data set is 4, 7, 2, 9, 5, 4. Find the mean, median and mode.

例题：数据集为 4, 7, 2, 9, 5, 4。求平均数、中位数和众数。

Mean: Sum the numbers: 4+7+2+9+5+4 = 31. Divide by the count (6): 31 ÷ 6 ≈ 5.17 (to 2 d.p.).

平均数：求和：4+7+2+9+5+4 = 31。除以个数 6：31 ÷ 6 ≈ 5.17（保留两位小数）。

Median: Put the numbers in order: 2, 4, 4, 5, 7, 9. With an even count, the median is the average of the middle two: (4 + 5) ÷ 2 = 4.5.

中位数：按顺序排列：2, 4, 4, 5, 7, 9。偶数个数据时，中位数是中间两个数的平均数：(4 + 5) ÷ 2 = 4.5。

Mode: The number that appears most often is 4.

众数：出现次数最多的数是 4。

10. Basic Probability | 基础概率

Example: A bag contains 5 red balls, 3 blue balls and 2 green balls. One ball is picked at random. Find P(red) and P(not blue).

例题：一个袋子中有 5 个红球，3 个蓝球和 2 个绿球。随机抽取一个球。求 P(红) 和 P(非蓝)。

Total number of balls = 5 + 3 + 2 = 10.

球的总数 = 5 + 3 + 2 = 10。

P(red) = number of red balls / total = 5/10 = 1/2 = 0.5.

P(红) = 红球数量 / 总数 = 5/10 = 1/2 = 0.5。

P(not blue) means picking red or green. Number of favourable outcomes = 5 + 2 = 7, so P(not blue) = 7/10 = 0.7.

P(非蓝) 表示抽到红或绿。有利结果数 = 5 + 2 = 7，所以 P(非蓝) = 7/10 = 0.7。

11. Negative Numbers | 负数运算

Example: Evaluate (−6) × (−4) + (−8) ÷ 2.

例题：计算 (−6) × (−4) + (−8) ÷ 2。

Step 1: Multiplication of two negatives gives a positive: (−6) × (−4) = 24.

第一步：两个负数相乘得正数：(−6) × (−4) = 24。

Step 2: Division with one negative gives a negative: (−8) ÷ 2 = −4.

第二步：一负一正相除得负数：(−8) ÷ 2 = −4。

Step 3: Substitute back: 24 + (−4) = 24 − 4 = 20.

第三步：代回表达式：24 + (−4) = 24 − 4 = 20。

Answer: 20

答案：20

12. Coordinates and Midpoints | 坐标与中点

Example: Point A is (2, 3) and point B is (8, 7). Find the midpoint of AB and the length of AB.

例题：点 A 为 (2, 3)，点 B 为 (8, 7)。求线段 AB 的中点坐标和长度。

Midpoint formula: ( (x₁+x₂)/2 , (y₁+y₂)/2 ). Substituting: ((2+8)/2, (3+7)/2) = (10/2, 10/2) = (5, 5).

中点公式：( (x₁+x₂)/2 , (y₁+y₂)/2 )。代入：((2+8)/2, (3+7)/2) = (10/2, 10/2) = (5, 5)。

Length (distance) formula: √[(x₂−x₁)² + (y₂−y₁)²]. Calculate differences: 8−2=6, 7−3=4. Then √(6²+4²) = √(36+16) = √52 = √(4×13) = 2√13 ≈ 7.21 (2 d.p.).

长度（距离）公式：√[(x₂−x₁)² + (y₂−y₁)²]。计算差值：8−2=6，7−3=4。然后 √(6²+4²) = √(36+16) = √52 = √(4×13) = 2√13 ≈ 7.21（保留两位小数）。

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KS3 Maths: Worked Examples Explained | KS3 数学：典型例题详解