📚 Essential Maths Book 7 Answers: Compressed Question-Type Analysis | 基础数学第7册答案：压缩题型解析
Mastering KS3 Mathematics requires a clear understanding of the common question types that appear in textbooks like Essential Maths Book 7. This article breaks down the most frequently tested topics into compact, easy‑to‑follow revision notes. Each section explains the key idea in English, immediately followed by a Chinese translation, and provides step‑by‑step approaches to typical questions. By studying these compressed question‑type analyses, students can quickly recall methods and avoid common mistakes when working through their own Book 7 exercises.
掌握 KS3 数学需要对《Essential Maths Book 7》这类教材中常见的题型有清晰的理解。本文将最常见的考点浓缩为简洁、易读的复习笔记。每一节先用英语解释核心概念，紧接着给出中文翻译，并提供典型题目的分步解法。通过学习这些压缩题型解析，同学们在完成自己的第7册练习时，可以快速回顾方法并避免常见错误。
1. Integer Arithmetic and Order of Operations | 整数运算与运算顺序
When solving questions with positive and negative integers, always identify the signs first. For example, to calculate (-7) + (+4), start at -7 on the number line and move 4 units to the right, reaching -3. If the question involves multiplication or division, remember that two matching signs give a positive result, while two different signs give a negative result.
解答含正负整数的题目时，一定要先确定符号。例如计算 (-7) + (+4)，从数轴上的 -7 出发，向右移动 4 个单位，到达 -3。如果题目涉及乘除，记住同号得正，异号得负。
= -6, then 5 + (-6) = -1. Brackets take highest priority: (5 + 3) × (-2) = 8 × (-2) = -16.
对于 5 + 3 × (-2) 这样的混合运算，必须遵守运算顺序。乘法优先于加法，因此 3 × (-2) = -6，然后 5 + (-6) = -1。括号的优先级最高：(5 + 3) × (-2) = 8 × (-2) = -16。
Tip: Use the BIDMAS (or PEMDAS) rule to stay safe – Brackets, Indices, Division/Multiplication, Addition/Subtraction.
技巧：使用 BIDMAS（或 PEMDAS）规则保证正确——括号、指数、乘除、加减。
2. Fractions: Addition and Subtraction | 分数的加法与减法
= 6/15 and 1/3 = 5/15, so the sum is 11/15. Always simplify the final answer if possible.
分数加减首先要统一分母。计算 2/5 + 1/3 时，最小公分母是 15。转换分数：2/5 = 6/15，1/3 = 5/15，相加得 11/15。最后结果能约分的一定要约分。
t: 1½ = 3/2, 2⅓ = 7/3. Common denominator 6 gives 9/6 + 14/6 = 23/6, which simplifies to 3⅚. This avoids mistakes with whole number parts.
带分数如 1½ + 2⅓ 可先化为假分数：1½ = 3/2，2⅓ = 7/3。公分母 6 使算式变为 9/6 + 14/6 = 23/6，化简为 3⅚。这样做可以避免在整数部分出错。
3. Decimals, Fractions and Percentages Conversion | 小数、分数与百分比的互化
A typical question asks: 'Write 0.35 as a fraction in its simplest form.' 35/100, then divide numerator and denominator by 5 to get 7/20. For percentages, 0.35 = 35% because you multiply by 100. Conversely, 40% = 40/100 = 2/5 = 0.4.
典型题目如“将 0.35 化为最简分数”。先写成 35/100，分子分母同时除以 5 得 7/20。对于百分数，0.35 = 35%，因为乘以 100。反过来，40% = 40/100 = 2/5 = 0.4。
When ordering a mixture of 0.6, 58%, and ⅗, convert all to the same form. 0.6 = 60%, ⅗ = 60%, so 0.6 and ⅗ are equal and larger than 58%. This method avoids confusion and allows easy comparison.
比较 0.6、58% 和 ⅗ 大小时，统一为同一种形式。0.6 = 60%，⅗ = 60%，因此 0.6 与 ⅗ 相等，且都大于 58%。这个方法避免混淆，使比较一目了然。
4. Simplifying Algebraic Expressions | 化简代数式
Collect like terms to simplify expressions. For 3a + 5b - a + 2b, group a‑terms (3a - a = 2a) and b‑ 5b + 2b = 7b). The result is 2a + 7b. The sign in front of each term belongs to that term.
通过合并同类项化简式子。对于 3a + 5b - a + 2b，将 a 项合并（3a - a = 2a），b 项合并（5b + 2b = 7b），得到 2a + 7b。每项前面的符号都归该项所有。
= 8x, 4 × (-3) = -12, so the expansion is 8x - 12. Then combine with any other like terms.
展开括号如 4(2x - 3)，将括号内每一项乘以外面的数：4 × 2x = 8x，4 × (-3) = -12，展开得到 8x - 12。如有其他同类项，再进一步合并。
5. Solving Linear Equations | 解一元一次方程
des by 2: x = 4. Always check your answer by substituting it back: 2(4) + 3 = 11, correct.
目标是分离变量。解 2x + 3 = 11，两边同时减去 3 得 2x = 8。然后两边除以 2 得 x = 4。务必把答案代回原方程检验：2(4) + 3 = 11，正确。
For equations with unknowns on both sides, like 5x - 2 = 2x + 7, bring x‑ the other. Subtract 2x from both sides: 3x - 2 = 7. Add 2 to both sides: 3x = 9, so x = 3.
对于诸如 5x - 2 = 2x + 7 这样未知数在等号两边的方程，将 x 项移到一边，常数移到另一边。两边同时减去 2x：3x - 2 = 7。两边加 2：3x = 9，因此 x = 3。
6. Number Sequences and the nth Term | 数列与第 n 项
rk out b: when n = 1, 3(1) = 3, but the first term is 4, so b = 1. The nth term is 3n + 1.
等差数列 4, 7, 10, 13... 的公差为 3。第 n 项的表达式可采用 an + b 的形式。公差（3）作为 n 的系数：3n。再求 b：当 n = 1 时，3(1) = 3，但首项为 4，因此 b = 1。第 n 项为 3n + 1。
ting out the whole list. Some questions ask whether a given number appears in the sequence – 1 equal to that number and solve for n; if n is an integer, the number is in the sequence.
利用第 n 项求任意项。第 20 项，将 n = 20 代入 3n + 1：3 × 20 + 1 = 61。这样无需列出整个数列。有些问题会问某个数是否在数列中，将 3n + 1 设为该数，解出 n；若 n 为整数，则该数属于数列。
7. Angles and Properties of Shapes | 角与图形性质
diagram with an angle of 120°, the adjacent angle on a straight line is 180° - 120° = 60°.
有关角的题目常涉及以下事实：直线上的角之和为 180°，周角为 360°，对顶角相等。若图上有一个 120° 的角，其邻补角即为 180° - 120° = 60°。
or quadrilaterals, the sum is 360°. Identifying isosceles triangles (two equal angles) or parallel lines (alternate and corresponding angles) is key to solving multi‑step problems.
在三角形中，内角和为 180°。已知两角 50° 和 70°，第三角为 180° - 50° - 70° = 60°。四边形的内角和为 360°。识别等腰三角形（两角相等）或平行线（内错角、同位角）是解决多步问题的关键。
8. Perimeter, Area and Volume | 周长、面积与体积
same rectangle is length × width = 5 × 3 = 15 cm². Remember to use square units for area.
周长是图形边界一周的长度。长 5 厘米、宽 3 厘米的长方形，周长为 2 × (5 + 3) = 16 厘米。同一长方形的面积是长 × 宽 = 5 × 3 = 15 平方厘米。面积单位切勿遗漏“平方”。
000 000 cm³. Always check whether the dimensions are in the same unit before calculating.
长方体体积 = 长 × 宽 × 高。一个 2 厘米 × 3 厘米 × 4 厘米的盒子，体积为 2 × 3 × 4 = 24 立方厘米。单位换算时，1 平方米 = 10 000 平方厘米，1 立方米 = 1 000 000 立方厘米。计算前务必确保所有尺寸单位一致。
9. Statistics: Mean, Median, Mode and Range | 统计：平均数、中位数、众数和极差
numbers, so mean = 35 ÷ 5 = 7. The median is the middle value when ordered: 4, 6, 8, 8, 9 – n = 8. If there is an even number of data points, take the mean of the two middle numbers.
平均数（均值）为所有数值之和除以数值个数。数据集 4, 8, 6, 8, 9 的总和为 35，共 5 个数，平均数 = 35 ÷ 5 = 7。中位数是按大小排序后居中的数：4, 6, 8, 8, 9，中位数为 8。若有偶数个数据，则取中间两个数的平均值。
The mode is the most frequent value: here 8 appears twice, so mode = 8. Range = maximum – scribe a data set, and questions often ask you to compare two sets using these statistics.
众数是出现次数最多的值：此处 8 出现两次，众数为 8。极差 = 最大值 - 最小值 = 9 - 4 = 5。这些统计量有助于描述一组数据，题目经常要求用它们比较两组数据。
10. Probability and Sample Spaces | 概率与样本空间
of favourable outcomes) / (total number of outcomes). Rolling an even number on a fair six‑ided dice gives 3 favourable outcomes (2, 4, 6) out of 6, so the probability is 3/6 = 1/2.
概率用 0 到 1 之间的分数表示：事件概率 =（成功结果数）/（总可能结果数）。掷一个均匀的六面骰子得到偶数的成功结果有 3 个（2, 4, 6），总结果 6 个，概率为 3/6 = 1/2。
robability of getting a head and an even number, count the favourable outcomes: H2, H4, H6 – 3 outcomes, so probability = 3/12 = 1/4.
对于两件事情，比如抛硬币与掷骰子，可以列出样本空间（H1, H2, … T6）。总结果数为 2 × 6 = 12。求得到正面且偶数的概率，数出成功结果：H2, H4, H6——3 种结果，概率 = 3/12 = 1/4。
Probability trees or listing outcomes systematically are essential tools for more complex problems, and they appear regularly in Book 7 assessments.
概率树或系统列举结果是解决更复杂问题的必备工具，这在第 7 册的评估中经常出现。
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