Essential Maths Book 8C: Question Type Analysis | KS3 数学：Essential Maths Book 8C 题型解析

📚 Essential Maths Book 8C: Question Type Analysis | KS3 数学：Essential Maths Book 8C 题型解析

Essential Maths Book 8C is a core resource for Key Stage 3 students, covering a wide range of mathematical topics aligned with the Year 8 curriculum in England. This article analyses the most common question types found in the book, providing detailed step-by-step solutions. By working through these examples, students can strengthen their problem-solving skills and build confidence for assessments.

Essential Maths Book 8C 是英格兰 KS3 阶段的核心教材，覆盖了 Year 8 课程中的广泛数学主题。本文分析了书中常见的题型，并提供了逐步解析。通过这些示例，学生可以加强解题技巧，为评估建立信心。

1. Understanding Place Value and Standard Form | 理解位值与标准形式

Book 8C introduces numbers up to millions and standard form conversions. A common question asks: Write 7.2 × 10⁴ as an ordinary number.

8C 教材介绍了百万级数字和标准形式的转换。常见题型：将 7.2 × 10⁴ 写成普通数字。

Solution: 7.2 × 10000 = 72000. This reinforces multiplying by powers of ten.

解答：7.2 × 10000 = 72000。这强化了乘以10的幂的运算。

Another question type: Round 0.0658 to 2 significant figures.

另一题型：将 0.0658 四舍五入到两位有效数字。

Answer: First two significant digits are 6 and 5, next digit is 8 (≥5), so the rounded value is 0.066.

答案：前两位有效数字是 6 和 5，下一位是 8（大于等于5），所以四舍五入后为 0.066。

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2. Operations with Fractions | 分数运算

Students practice adding and subtracting fractions with different denominators, often applied in real contexts.

学生练习不同分母分数的加法和减法，常应用于实际情境。

Example: Calculate 2/3 + 1/4, simplifying your answer.

例题：计算 2/3 + 1/4，并化简结果。

Solution: LCM of 3 and 4 is 12. 2/3 = 8/12, 1/4 = 3/12, so the sum is 11/12.

解答：3 和 4 的最小公倍数是 12。2/3 = 8/12，1/4 = 3/12，和为 11/12。

Questions also combine mixed numbers, e.g., 1 1/2 + 2 1/3.

题目还涉及带分数，例如 1 1/2 + 2 1/3。

Tip: Convert to improper fractions first: 1 1/2 = 3/2, 2 1/3 = 7/3. Common denominator 6 gives 9/6 + 14/6 = 23/6 = 3 5/6.

技巧：先化为假分数：1 1/2 = 3/2，2 1/3 = 7/3。通分分母 6 得到 9/6 + 14/6 = 23/6 = 3 5/6。

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3. Percentages and Real-Life Applications | 百分数与实际应用

Percentage increase and decrease questions are key. For example: A jacket costs £80 before a 15% discount. Find the sale price.

百分数增减题型很重要。例如：一件夹克原价 80 英镑，打 15% 折扣，求售价。

Method: 15% of £80 = 0.15 × 80 = £12, so sale price = £80 – £12 = £68.

方法：80 的 15% = 0.15 × 80 = 12 英镑，因此售价 = 80 – 12 = 68 英镑。

Alternatively, multiply by 0.85 directly: £80 × 0.85 = £68.

另一种方法：直接乘以 0.85：80 × 0.85 = 68 英镑。

Reverse percentage problems also appear: after a 20% increase, a quantity is 240. Find the original.

逆向百分数问题也出现：增加 20% 后值为 240，求原数。

Solution: If increased by 20%, the new value is 120% of the original, so original = 240 ÷ 1.2 = 200.

解答：如果增加20%，新值是原值的120%，因此原数 = 240 ÷ 1.2 = 200。

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4. Ratio and Proportion Problem Solving | 比与比例问题解决

Sharing in a ratio: Divide £56 in the ratio 3:5.

按比例分配：将 56 英镑按 3:5 分配。

Total parts = 3 + 5 = 8. One part = £56 ÷ 8 = £7. The amounts are: 3 parts = £21, 5 parts = £35.

总份数 = 3+5=8，一份 = 56÷8=7 英镑，分配为：3份=21英镑，5份=35英镑。

Map scale problems: a distance of 5 cm on a map represents 2 km. How far apart are two towns if the map distance is 8 cm?

地图比例尺问题：图上 5 cm 代表实际 2 km。若两镇图上距离 8 cm，实际距离多远？

Solution: 1 cm represents 2 ÷ 5 = 0.4 km. So 8 cm represents 8 × 0.4 = 3.2 km.

解答：1 cm 代表 2÷5=0.4 km，所以 8 cm 代表 8×0.4=3.2 km。

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5. Simplifying Algebraic Expressions | 代数式化简

A typical simplification: 5a + 3b – 2a + 7b.

典型化简题：5a + 3b – 2a + 7b。

Collect like terms: 5a – 2a = 3a, and 3b + 7b = 10b, giving 3a + 10b.

合并同类项：5a – 2a = 3a，3b + 7b = 10b，得到 3a + 10b。

Expanding single brackets: 4(2x + 3).

单项括号展开：4(2x + 3)。

Multiply each term inside by 4: 4 × 2x = 8x, 4 × 3 = 12, result is 8x + 12.

括号内每项乘以4：4×2x = 8x，4×3 = 12，结果为 8x + 12。

Book 8C also introduces simple factorising, e.g., 6y – 9 = 3(2y – 3).

8C 教材还引入简单因式分解，如 6y – 9 = 3(2y – 3)。

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6. Solving Equations with Brackets and Unknowns on Both Sides | 含括号与未知数在两侧的方程求解

Two-step equation: 2x + 5 = 17.

两步方程：2x + 5 = 17。

Subtract 5 from both sides: 2x = 12, then divide by 2: x = 6.

两边减5：2x = 12，然后除以2：x = 6。

Equations with brackets: 3(x – 4) = 2x + 5.

含括号的方程：3(x – 4) = 2x + 5。

Expand LHS: 3x – 12 = 2x + 5. Then 3x – 2x = 5 + 12, so x = 17.

展开左边：3x – 12 = 2x + 5。然后 3x – 2x = 5 + 12，所以 x = 17。

Equations with unknowns on both sides: 5y + 7 = 2y + 22.

未知数在两边：5y + 7 = 2y + 22。

Subtract 2y from both sides: 3y + 7 = 22. Then 3y = 15, y = 5.

两边减 2y：3y + 7 = 22，然后 3y = 15，y = 5。

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7. Angles in Polygons and Parallel Lines | 多边形与平行线中的角度

Angles on a straight line sum to 180°. If one angle is 72°, find the other.

平角和为180°。若一角为72°，求另一角。

Answer: 180° – 72° = 108°.

答案：180° – 72° = 108°。

Angles in a triangle: In a triangle, two angles are 45° and 70°. Find the third.

三角形内角：三角形两角为45°和70°，求第三角。

Sum = 180°, so third angle = 180° – (45° + 70°) = 65°.

内角和180°，第三角 = 180° – (45°+70°) = 65°。

Parallel lines: Corresponding angles are equal. If a transversal creates an angle of 55° with a parallel line, find all missing angles and state reasons.

平行线：同位角相等。若一横截线与平行线夹角55°，求所有未知角并说明理由。

Typically, corresponding angle = 55°, alternate angle = 55°, interior angle = 125° (supplementary).

典型解答：同位角 = 55°，内错角 = 55°，同旁内角 = 125°（互补）。

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8. Area and Circumference of Circles | 圆的面积与周长

Circumference of a circle: C = πd or 2πr.

圆的周长：C = πd 或 2πr。

Example: A circle has radius 7 cm. Find its circumference, giving your answer in terms of π.

例题：圆半径为7 cm，求周长，用 π 表示。

Solution: C = 2 × π × 7 = 14π cm.

解答：C = 2 × π × 7 = 14π cm。

Area: A = πr².

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