Essential Maths Book 9i Answers Explained | KS3 数学核心练习答案精讲

📚 Essential Maths Book 9i Answers Explained | 九年级数学核心练习答案精讲

In this revision guide, we walk through selected answers from Essential Maths Book 9i, providing step-by-step explanations for key topics like equations, inequalities, sequences, graphs, Pythagoras’ theorem, area, volume, ratio, statistics and probability. Each section pairs an English explanation with a matching Chinese version, helping you understand not just the ‘what’ but the ‘why’ behind every solution.

在本复习指南中，我们精选了 Essential Maths Book 9i 中的典型练习并给出分步解析，涵盖方程、不等式、数列、图像、毕达哥拉斯定理、面积、体积、比例、统计和概率等核心章节。每个例题都配有中英文双语讲解，帮助你不仅知道“答案是什么”，更理解“为什么这样算”。

1. Solving Linear Equations | 解一元一次方程

Example: Solve 3x + 7 = 22. Subtract 7 from both sides to get 3x = 15. Then divide both sides by 3, giving x = 5.

例题：解方程 3x + 7 = 22。两边同时减去 7 得 3x = 15。再将两边同时除以 3，得到 x = 5。

Always check your answer by substituting it back into the original equation. Here, 3(5) + 7 = 15 + 7 = 22, which is correct.

记得将解代回原方程进行验算。这里 3 × 5 + 7 = 15 + 7 = 22，结果正确。

For equations containing brackets, expand them first. Example: 2(x + 4) = 18 → 2x + 8 = 18 → 2x = 10 → x = 5.

对于有括号的方程，先去括号。例如 2(x + 4) = 18，展开得 2x + 8 = 18，移项得 2x = 10，所以 x = 5。

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2. Inequalities on a Number Line | 一元一次不等式及其数轴表示

Solve 5y − 3 > 12. Add 3 to both sides → 5y > 15 → y > 3. On a number line, draw an open circle at 3 and shade to the right.

解不等式 5y − 3 > 12。两边同时加 3 得 5y > 15，再除以 5 得 y > 3。在数轴上，用空心圆圈标在 3 上，并向右侧涂阴影。

When the inequality sign includes equality (e.g., ≤ or ≥), use a filled circle. For −2 ≤ x < 4, put a filled circle at −2, an open circle at 4, and shade the region between them.

若不等号带有等号（如 ≤ 或 ≥），则使用实心圆点。例如 −2 ≤ x < 4，在 −2 处画实心圆，在 4 处画空心圆，中间区域涂阴影。

Multiplying or dividing both sides by a negative number reverses the inequality sign. Solve −2p ≤ 10 → p ≥ −5.

当两边同乘或同除以负数时，不等号方向要改变。解 −2p ≤ 10，两边同除以 −2 得到 p ≥ −5。

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3. nth Term of a Linear Sequence | 等差数列的第 n 项

Find the nth term of the sequence: 7, 11, 15, 19, … The difference between terms is +4. The zeroth term (term before first) is 7 − 4 = 3. So the nth term is 4n + 3.

求数列 7, 11, 15, 19, … 的第 n 项。相邻两项差为 +4。第零项（第一项之前的那一项）为 7 − 4 = 3。因此第 n 项公式为 4n + 3。

To check, substitute n = 1: 4(1) + 3 = 7; n = 2: 4(2) + 3 = 11. Correct. Use the formula to find the 50th term: 4(50) + 3 = 203.

检验：n = 1 时 4 × 1 + 3 = 7；n = 2 时 4 × 2 + 3 = 11。正确。用公式求第 50 项：4 × 50 + 3 = 203。

For decreasing sequences, the common difference is negative. Sequence 20, 17, 14, 11, … has difference −3. The nth term is −3n + 23 (since 20 − (−3) = 23).

对于递减数列，公差为负数。数列 20, 17, 14, 11, … 的公差为 −3。第 n 项公式为 −3n + 23（因为第一项减公差：20 − (−3) = 23）。

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4. Straight Line Graphs (y = mx + c) | 直线图像（y = mx + c）

Plot the graph of y = 2x − 1. The y-intercept c = −1, so the line crosses the y-axis at (0, −1). The gradient m = 2, meaning for every 1 unit right, go up 2 units.

画出 y = 2x − 1 的图像。y 轴截距 c = −1，所以直线与 y 轴交于点 (0, −1)。斜率 m = 2，表示每向右移动 1 个单位，向上移动 2 个单位。

From (0, −1), move right 1 and up 2 to (1, 1), then to (2, 3). Draw a straight line through these points. A table of values can also be used.

从 (0, −1) 出发，右移 1、上移 2 到 (1, 1)，再到 (2, 3)。通过这几点画直线。也可以使用数值表格法画图。

To find where two lines intersect, solve their equations simultaneously. Lines y = 2x − 1 and y = −x + 5 intersect when 2x − 1 = −x + 5 → 3x = 6 → x = 2, then y = 3. Intersection is (2, 3).

求两直线交点需联立方程。直线 y = 2x − 1 与 y = −x + 5 相交时，2x − 1 = −x + 5，解得 3x = 6，x = 2，代入得 y = 3。交点为 (2, 3)。

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5. Pythagoras’ Theorem | 毕达哥拉斯定理

In a right‑angled triangle, a² + b² = c², where c is the longest side (hypotenuse). Find the hypotenuse when the shorter sides are 6 cm and 8 cm: 6² + 8² = 36 + 64 = 100, so c = √100 = 10 cm.

在直角三角形中，a² + b² = c²，其中 c 为斜边（最长边）。已知两条直角边分别为 6 cm 和 8 cm，求斜边：6² + 8² = 36 + 64 = 100，所以 c = √100 = 10 cm。

To find a shorter side, rearrange the formula. A triangle has hypotenuse 13 cm and one side 5 cm. Find the other side: b = √(13² − 5²) = √(169 − 25) = √144 = 12 cm.

求一条直角边时需变形公式。已知斜边 13 cm，一直角边 5 cm，求另一直角边：b = √(13² − 5²) = √(169 − 25) = √144 = 12 cm。

Always identify the hypotenuse first. It is always opposite the right angle. The theorem only applies to right‑angled triangles.

解题时先确认斜边，斜边总是直角所对的边。这一定理仅适用于直角三角形。

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6. Area and Volume of Prisms | 棱柱的面积与体积

Volume of a prism = area of cross‑section × length. A triangular prism has a cross‑sectional triangle of base 6 cm, height 4 cm, and length 10 cm. Area of triangle = ½ × base × height = ½ × 6 × 4 = 12 cm². Volume = 12 × 10 = 120 cm³.

棱柱体积 = 横截面积 × 长度。一个三棱柱的横截面是底 6 cm、高 4 cm 的三角形，棱柱长 10 cm。三角形面积 = ½ × 底 × 高 = ½ × 6 × 4 = 12 cm²。体积 = 12 × 10 = 120 cm³。

Surface area of a prism: add the area of all faces. For a cuboid 3 cm × 4 cm × 5 cm: two faces 3×4, two faces 3×5, two faces 4×5. Total = 2(12 + 15 + 20) = 2 × 47 = 94 cm².

棱柱表面积：将所有面的面积相加。一个 3 cm × 4 cm × 5 cm 的长方体：包括两个 3×4 面、两个 3×5 面、两个 4×5 面。总面积 = 2(12 + 15 + 20) = 2 × 47 = 94 cm²。

For cylinders, the volume is πr²h and the surface area is 2πrh + 2πr². Use π ≈ 3.14 in KS3 exercises unless told otherwise.

圆柱体的体积公式为 πr²h，表面积为 2πrh + 2πr²。在 KS3 练习中，没有特别说明时 π 取 3.14。

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7. Ratio and Proportion | 比与比例

Simplify the ratio 24:36. Find the highest common factor (HCF) of 24 and 36, which is 12. Divide both sides by 12 → 2:3.

化简比 24:36。找出 24 和 36 的最大公因数（HCF）为 12，两边同时除以 12 得 2:3。

Divide £72 in the ratio 3:5. Total parts = 3 + 5 = 8. Value of one part = £72 ÷ 8 = £9. First share = 3 × £9 = £27, second share = 5 × £9 = £45.

将 72 英镑按 3:5 的比例分配。总份数 = 3 + 5 = 8。每份金额 = 72 ÷ 8 = 9 英镑。第一份 = 3 × 9 = 27 英镑，第二份 = 5 × 9 = 45 英镑。

Direct proportion: if 5 pens cost £3.50, find the cost of 8 pens. Cost per pen = £3.50 ÷ 5 = £0.70. For 8 pens: 8 × £0.70 = £5.60. Alternatively use the unitary method or a multiplier.

正比例问题：5 支笔价值 3.50 英镑，求 8 支笔的价格。每支笔价 = 3.50 ÷ 5 = 0.70 英镑。8 支笔 = 8 × 0.70 = 5.60 英镑。也可用单位法或倍数法求解。

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8. Mean, Median, Mode and Range | 平均数、中位数、众数和极差

For the data set: 4, 8, 6, 8, 5, 9, 2. Mean = (4+8+6+8+5+9+2) ÷ 7 = 42 ÷ 7 = 6. Mode = 8 (appears most). Range = 9 − 2 = 7.

对于数据集：4, 8, 6, 8, 5, 9, 2。平均数 = (4+8+6+8+5+9+2) ÷ 7 = 42 ÷ 7 = 6。众数 = 8（出现次数最多）。极差 = 9 − 2 = 7。

To find the median, order the data: 2, 4, 5, 6, 8, 8, 9. The middle value is 6, so median = 6. With an even number of values, take the mean of the two middle numbers.

找中位数时先排序：2, 4, 5, 6, 8, 8, 9。中间值为 6，所以中位数 = 6。若数据个数为偶数，则取中间两数的平均数。

The range tells us about spread. A smaller range means the data is more consistent. Compare two sets: Set A range = 5, Set B range = 15 – Set A is less spread out.

极差反映数据的分散程度。极差越小，数据越集中。比较两组数据：A 组极差为 5，B 组极差为 15，A 组分布更紧凑。

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9. Probability Basics | 概率基础

Probability of an event = number of favourable outcomes ÷ total number of possible outcomes. A fair 6‑sided die: P(rolling a 4) = ⅙.

事件的概率 = 有利结果的数量 ÷ 所有可能结果的总数。抛掷一枚公平的六面骰子，掷出 4 的概率为 ⅙。

Probabilities always lie between 0 and 1, or can be expressed as fractions, decimals or percentages. An event that is certain has probability 1; impossible event has probability 0.

概率值始终在 0 到 1 之间，可以用分数、小数或百分数表示。必然事件的概率为 1，不可能事件的概率为 0。

From a bag of 3 red, 2 blue and 5 green counters, find P(blue) = 2/10 = ⅕. P(not red) = 7/10. The sum of probabilities of all outcomes is 1.

一个袋子中有 3 个红色、2 个蓝色和 5 个绿色筹码。抽到蓝色的概率 = 2/10 = ⅕。抽不到红色的概率 = 7/10。所有可能结果的概率之和为 1。

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10. Written Calculations and Estimation | 笔算与估算

Multiply 3.4 by 2.6 without a calculator: treat them as 34 × 26 = 884. Since each number had 1 decimal place, the product has 2 decimal places: 8.84.

不用计算器计算 3.4 × 2.6：先看作 34 × 26 = 884。因为每个因数有 1 位小数，积应有 2 位小数，结果为 8.84。

Estimate the value of 9.8 × 7.1 by rounding: 10 × 7 = 70. The exact answer is 69.58, so the estimate is reasonable. Estimation helps check calculator answers.

估算 9.8 × 7.1 的值：四舍五入为 10 × 7 = 70。精确答案为 69.58，估算值合理。估算有助于检验计算器答案。

Division with decimals: 4.5 ÷ 0.3. Multiply both by 10 to get 45 ÷ 3 = 15. So 4.5 ÷ 0.3 = 15. Converting to whole numbers makes division simpler.

包含小数的除法：4.5 ÷ 0.3。将被除数和除数同时乘以 10，变为 45 ÷ 3 = 15，所以 4.5 ÷ 0.3 = 15。化为整数可使除法变简单。

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11. Angles in Parallel Lines | 平行线中的角度关系

When a transversal crosses parallel lines, alternate angles are equal, corresponding angles are equal, and co‑interior (allied) angles sum to 180°.

当一条截线与平行线相交时，内错角相等，同位角相等，同旁内角互补（和为 180°）。

Example: Two parallel lines are cut by a transversal. One angle is given as 110°. Find the other angles. The alternate angle is 110°, the corresponding angle is 110°, and the co‑interior angle is 180° − 110° = 70°.

例题：两条平行线被一条截线所截，已知一个角为 110°。求其他角。其内错角为 110°，同位角为 110°，同旁内角为 180° − 110° = 70°。

Vertically opposite angles are also equal. The sum of angles on a straight line is 180°. Use these facts together to solve multi‑step angle problems.

对顶角也相等。平角为 180°。结合这些性质可以解决多步角度计算问题。

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12. Review and Tips for Exam Success | 复习与考试技巧

Re‑work all examples without looking at the solution first. Write down every step clearly – method marks matter. If stuck, try a simpler case or draw a diagram.

先不看答案，独立重做所有例题。清晰地写下每一步过程——步骤分同样重要。遇到困难时，可尝试简化问题或画图辅助。

In KS3 tests, always check units, show working fully, and use a pencil for graphs. When interpreting questions, underline key numbers and command words.

在 KS3 考试中，务必检查单位、完整展示运算过程、使用铅笔画图。审题时，在关键数字和指令词下划线标记。

For multi‑step problems, break them down into smaller parts. Label angles, write given ratios, and keep your work organised. Practising past questions is the best preparation.

对于多步问题，将其分解为更小的部分。给角度标上字母、写出已知比例、保持卷面整洁。多做历年真题是最好的备考方式。

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