📚 Cambridge Lower Secondary Mathematics WB 8 Answers: Question Type Breakdown | 剑桥初中数学练习册8答案题型全解析
Cambridge Lower Secondary Mathematics Workbook 8 is carefully structured to reinforce and extend the Stage 8 curriculum. The answer pages reveal a wide variety of question types that test both procedural fluency and conceptual understanding. In this analysis, we unpack the common formats, explain the thinking behind each solution, and highlight the skills students need to master. By studying how the answers are constructed, learners can refine their problem‑solving strategies and avoid typical mistakes.
剑桥初中数学练习册第8册经过精心设计，旨在巩固和拓展阶段8课程内容。答案页展示了丰富多样的题型，既考查运算熟练度，又检验概念理解。本文逐一剖析常见题型，解释每类解答背后的思路，并指出学生需要掌握的核心技能。通过研究答案的构造方式，学习者可以优化解题策略，避开常见陷阱。
1. Integers and Order of Operations | 整数与运算顺序
Questions on integers often combine addition, subtraction, multiplication and division with negative numbers. A typical task asks students to evaluate an expression like −6 + (−4) × 3. The answer must follow the order of operations (BIDMAS/BODMAS): multiplication first, then addition. The solution shows that −4 × 3 = −12, and then −6 + (−12) = −18. Common errors include forgetting that adding a negative is equivalent to subtraction or performing the addition before the multiplication.
整数运算题通常将加、减、乘、除与负数结合。常见题型要求计算如−6 + (−4) × 3 的表达式。答案必须遵循运算顺序（括号、指数、乘除、加减）：先乘法，后加法。解答显示 −4 × 3 = −12，然后 −6 + (−12) = −18。典型错误包括忘记加上一个负数等同于减法，或在乘法之前先做加法。
Another frequent format is word problems involving temperature changes, bank balances or elevator floors. The answer demonstrates the use of a number line or the rule ‘two negatives make a positive’ for products. For instance, (−2) × (−5) = 10. The reasoning explains that a negative times a negative yields a positive. Workbook answers often include short justifications to reinforce the rule.
另一种常见题型是涉及温度变化、银行余额或电梯楼层的文字题。答案中展示了数轴的使用方法，或“负负得正”的乘法法则。例如 (−2) × (−5) = 10。推理过程说明负数乘以负数得正数。练习册答案常附有简短的理由说明，以强化这一规则。
2. Fractions, Decimals and Percentages | 分数、小数与百分比
Workbook 8 frequently asks students to convert between fractions, decimals and percentages, and to order them. An answer might require converting ⅗ to a decimal (0.6) and then to a percentage (60%). The working shows division or equivalent fraction strategies. Special attention is given to recurring decimals: ⅓ = 0.333…, which is written with a dot notation in the answer, but the explanation highlights that it is acceptable to write 0.3̇.
练习册8频繁要求学生进行分数、小数和百分比的互化并排序。一个答案可能需要将 ⅗ 化为小数 0.6，再化为百分比 60%。解答过程展示了除法或等值分数策略。循环小数得到特别关注：⅓ = 0.333…，答案中用点标记表示，但解释强调写作 0.3̇ 亦可。
Addition and subtraction of fractions with different denominators appear regularly. The answer first finds the least common denominator (LCD). For ⅖ + ½ , the LCD is 10, giving 4/10 + 5/10 = 9/10. Many solutions include a step‑by‑step breakdown to illustrate how equivalent fractions are built. Misconceptions like adding numerators and denominators separately are corrected through clear modelling.
异分母分数的加减法也频繁出现。答案首先找出最小公分母（ LCD）。例如 ⅖ + ½ ，LCD 为10，得出 4/10 + 5/10 = 9/10。许多解答包含逐步拆分过程，以说明如何构造等值分数。诸如分子分母分别相加的错误观念，通过清晰的模型得到纠正。
3. Algebraic Expressions and Substitution | 代数表达式与代入
This section tests the ability to substitute values into expressions. A typical item provides an expression like 3a + 2b and asks for its value when a = 4 and b = −1. The answer plugs the numbers in: 3 × 4 + 2 × (−1) = 12 − 2 = 10. Brackets are often used around negative numbers in the working to avoid sign errors. The solution also sometimes includes a check with a different set of values.
本节考查将数值代入表达式的能力。一个典型题目给出如 3a + 2b 的表达式，要求当 a = 4、b = −1 时求值。答案代入数字：3 × 4 + 2 × (−1) = 12 − 2 = 10。计算过程中常在负数外加括号以避免符号错误。解答有时还会用另一组数值进行验算。
Writing algebraic expressions from word statements is another core question type. For example, ‘Think of a number, multiply it by 5 and subtract 3’ translates to 5x − 3. The answer clarifies the variable choice and the order of operations. Reverse problems, where the expression is given and students write a scenario, also appear, encouraging flexible thinking.
根据文字叙述写出代数表达式是另一类核心题型。例如，“想一个数，乘以5再减去3”转换为 5x − 3。答案明确了变量的选择与运算顺序。反向问题——给出表达式让学生编写情景——也时有出现，促进思维的灵活性。
4. Simplifying and Expanding Brackets | 化简与去括号
Collecting like terms is a fundamental skill. Answers show how to simplify expressions such as 7x + 2y − 3x + 5y by grouping x terms and y terms separately, yielding 4x + 7y. The working often underlines like terms or uses different colours in the textbook, but the answer script simply rearranges them clearly. Misplacement of signs when moving terms is a highlighted trap.
合并同类项是一项基础技能。答案展示了如何简化如 7x + 2y − 3x + 5y 的表达式，分别将 x 项和 y 项分组，得出 4x + 7y。解题过程常在教科书上用下划线或不同颜色标注同类项，但答案文本只是将其清晰地重新排列。移项时符号错位是一个被特别强调的易错点。
Expanding brackets involves the distributive law. A question may ask to expand 4(2a − 3b). The solution writes 4 × 2a = 8a and 4 × (−3b) = −12b, then combines to 8a − 12b. Double brackets like (x + 2)(x + 5) are expanded using the FOIL method, resulting in x² + 5x + 2x + 10 = x² + 7x + 10. Answers sometimes include area model diagrams to illustrate the process.
去括号运用分配律。题目可能要求展开 4(2a − 3b)。解答先写 4 × 2a = 8a 和 4 × (−3b) = −12b，然后合并为 8a − 12b。像 (x + 2)(x + 5) 这样的双括号使用FOIL方法展开，得到 x² + 5x + 2x + 10 = x² + 7x + 10。答案有时会附带面积模型图来说明这一过程。
5. Solving Linear Equations | 解线性方程
One‑step equations such as x + 7 = 15 are solved by inverse operations: subtract 7 from both sides to get x = 8. The answer usually notes the balancing method. Two‑step equations, e.g. 3y − 4 = 11, require first adding 4 and then dividing by 3. The working clearly shows each step on a new line, maintaining equality.
形如 x + 7 = 15 的一步方程通过逆运算求解：两边同时减去7得 x = 8。答案通常注明平衡法。两步方程如 3y − 4 = 11 则需要先加4，再除以3。解答清晰地在每一行显示每一步骤，保持等式成立。
Equations with variables on both sides, like 5x + 2 = 3x + 8, appear frequently. The answer brings variable terms to one side and constants to the other: subtract 3x → 2x + 2 = 8, then subtract 2 → 2x = 6, so x = 3. Some solutions include a check by substituting the value back. Word problems that form equations (e.g. age or perimeter problems) are modelled with a ‘let’ statement before solving.
带有两边变量的方程，如 5x + 2 = 3x + 8，出现频率很高。答案将含变量项移到一边，常数移到另一边：减 3x → 2x + 2 = 8，然后减 2 → 2x = 6，得 x = 3。一些解答还包含代回验算。需要建立方程的应用题（如年龄或周长问题）在求解前会先用“设”语句建模。
6. Ratio, Proportion and Rates | 比、比例与速率
Ratio questions ask to simplify ratios like 24:36 to 2:3 by dividing by the highest common factor (12). The answer may also show the ratio in unit form or as a fraction. Sharing in a given ratio, for example dividing £56 in the ratio 3:5, is solved by finding the total parts (8), calculating £56 ÷ 8 = £7 per part, and then giving 3 × £7 = £21 and 5 × £7 = £35. The method is explained stepwise.
比的问题要求化简如 24:36 的比，通过除以最大公因数12得到 2:3。答案可能还以单位形式或分数形式表示比。按给定比例分配，例如将56英镑按 3:5 分配，先算出总份数8，计算 £56 ÷ 8 = £7 每份，然后给出 3 × £7 = £21 和 5 × £7 = £35。解题方法逐步说明。
Direct proportion and rate problems include speed, exchange rates or recipes. A typical answer sets up a proportion: if 5 pens cost £2.75, then 1 pen costs £0.55, so 8 pens cost £4.40. The unitary method is clearly laid out. Graphs of proportional relationships are interpreted by reading coordinates; answers emphasize that a straight line through the origin indicates direct proportion.
正比例和速率问题涵盖速度、汇率或食谱。典型答案建立比例关系：若5支笔花费£2.75，则1支笔 £0.55，所以8支笔 £4.40。归一法被清晰地展示。正比例关系的图像通过读取坐标来解读；答案强调经过原点的直线表示正比例。
7. Angles and Parallel Lines | 角与平行线
This section focuses on calculating missing angles around a point, on a straight line, and within intersecting lines. Answers use the facts that angles on a straight line sum to 180°, and angles around a point sum to 360°. For vertically opposite angles, the solution states they are equal. Diagrams are referenced, but the text explanation walks through the deduction, e.g. ‘angle a = 180° − 65° = 115° (angles on a straight line)’.
本节重点计算点周围、直线上以及相交线中的未知角。答案运用直线上角之和为180°、点周围角之和为360°的事实。对于对顶角，解答指出它们相等。虽然参照了图示，但文字解释仍逐步演绎，例如“角 a = 180° − 65° = 115°（直线上角）”。
With parallel lines, questions ask to identify alternate, corresponding and co‑interior angles. The answer names the angle relationship and then writes an equation. For instance, if corresponding angles are equal, 2x = 110°, so x = 55°. Some multi‑step problems require finding several angles, linking parallel line properties with basic angle sums. The answer chain shows each logical step.
在涉及平行线的问题中，要求识别内错角、同位角和同旁内角。答案先标明角的关系，再写出等式。例如，若同位角相等，则 2x = 110°，得 x = 55°。一些多步问题需要求多个角，将平行线性质与基本角和相结合。答案链展示每一个逻辑步骤。
8. Properties of Triangles and Quadrilaterals | 三角形与四边形的性质
Triangles are classified by both sides and angles. Answers often begin by identifying the triangle type, then applying angle sum (180°). An isosceles triangle problem gives one base angle of 42°, so the other base angle is also 42°, and the vertex angle is 180° − 2 × 42° = 96°. For scalene triangles, all three angles must be calculated from given data. The explanation stresses that the longest side is opposite the largest angle.
三角形按边和角分类。答案通常先识别三角形类型，然后应用内角和180°。一道等腰三角形问题给出底角42°，则另一底角也是42°，顶角为 180° − 2 × 42° = 96°。对于不等边三角形，必须根据已知数据求出三个角。解释强调最长边对大角。
Quadrilateral questions require using angle sums (360°) and properties of special quadrilaterals such as parallelograms, rhombuses and trapeziums. A parallelogram answer might state that opposite angles are equal, so if one angle is 70°, the opposite is 70°, and adjacent angles sum to 180°, giving 110°. The working is systematic, with reasons like ‘co‑interior angles’ or ‘opposite angles of a parallelogram’.
四边形问题需用到内角和360°以及特殊四边形（平行四边形、菱形、梯形）的性质。平行四边形的答案可能说对角相等，因此若一个角为70°，其对角落为70°，邻角之和为180°，得110°。求解过程系统性强，理由如“同旁内角”或“平行四边形对角”等。
9. Perimeter, Area and Volume | 周长、面积与体积
Perimeter of rectilinear shapes is found by summing all side lengths. Workbook answers sometimes involve finding missing lengths before adding. For area, squares, rectangles, triangles and compound shapes are common. The area of a triangle is ½ × base × height; the answer shows careful identification of the perpendicular height. For a compound shape, the solution splits it into rectangles, calculates each area and sums them.
直线型图形周长通过将所有边长相加求得。练习册答案有时需要先找出未知长度再求和。面积方面，常见的有正方形、长方形、三角形及组合图形。三角形面积 = ½ × 底 × 高；答案展示如何仔细确定垂直高度。对于组合图形，解答将其分割为几个长方形，分别计算面积再相加。
Volume of cuboids is calculated as length × width × height, with units cubed. Some questions present nets or require working backwards from volume to a missing dimension. The answer writes the formula, substitutes and solves. Multi‑step problems combine surface area and volume, or convert between cm³ and ml. The reasoning is set out methodically, highlighting unit consistency.
长方体体积按 长 × 宽 × 高 计算，单位为立方。有些题目呈现展开图或需要从体积反求缺失尺寸。答案写出公式，代入并求解。多步骤问题结合了表面积与体积，或在 cm³ 与 ml 之间转换。推理过程有条不紊地展开，强调单位的一致性。
10. Statistics and Data Representation | 统计与数据表示
Data handling questions ask students to read and interpret bar charts, pie charts and line graphs. A typical answer extracts specific values from a bar chart, then calculates the total, mean, median or mode. The mean is shown as sum of values divided by the number of values. For the median, the data are ordered, and the middle value is identified. The explanation clarifies how the average is affected by outliers.
数据处理题要求学生阅读并解读条形图、饼图和折线图。典型答案从条形图中提取具体数值，然后计算总数、平均数、中位数或众数。平均数表示为数值总和除以数值个数。对于中位数，将数据排序后找到中间值。解释说明了极端值如何影响平均数。
Pie chart questions involve converting frequencies to angles. If a category represents 15 out of 60, its angle is 15/60 × 360° = 90°. Answers show the proportional calculation clearly. Two‑way tables and scatter graphs also appear; the solution identifies correlation types (positive, negative or none) and, where possible, draws a line of best fit. The focus is on using data to support conclusions.
饼图问题涉及将频数转化为角度。若某类别代表60中的15，则其角度为 15/60 × 360° = 90°。答案清晰地显示比例计算。双向表和散点图也出现；解答识别相关类型（正相关、负相关或无相关），并在可能时画出最佳拟合线。重点在于用数据支撑结论。
11. Probability Basics | 概率初步
Probability is expressed as a fraction, decimal or percentage between 0 and 1. Questions often list outcomes for a single event, like rolling a fair dice. The probability of rolling an even number is 3/6 = ½. Answers stress that a probability of 0 means impossible, and 1 means certain. Multi‑part items compare the likelihood of different events by calculating their probabilities.
概率用0到1之间的分数、小数或百分比表示。题目常列出单个事件的所有可能结果，如掷一枚公平骰子。掷出偶数的概率为 3/6 = ½。答案强调概率为0表示不可能，为1表示必然。多问答题通过计算概率来比较不同事件的可能性。
Sample space diagrams and systematic listing are used for combined events, such as flipping two coins. The answer may use a table to show all four outcomes: HH, HT, TH, TT. The probability of at least one tail is then 3/4. The solution methodically counts favourable outcomes and total outcomes. For mutually exclusive events, the addition rule is applied, with a note that the events cannot happen simultaneously.
样本空间图和系统列举法用于组合事件，如抛两枚硬币。答案可能用表格列出全部四种结果：HH, HT, TH, TT。至少出现一次反面的概率为 3/4。解答系统地数出有利结果和所有可能结果。对于互斥事件，使用加法法则，并注明事件不能同时发生。
12. Sequences and Patterns | 数列与规律
Number sequences often ask for the next terms and the nth term rule. A linear sequence like 5, 8, 11, 14… has a common difference of 3. The answer derives the nth term as 3n + 2. It shows how to test the formula with n = 1, 2, 3. Word‑based patterns, such as dots or matchsticks, are translated into numbers, and then the rule is found. The reasoning connects the visual pattern to the algebraic expression.
数列题常要求写出后续项及第 n 项规则。线性数列如 5, 8, 11, 14… 的公差为3。答案推导出第 n 项为 3n + 2。它演示了如何用 n = 1, 2, 3 检验公式。基于图形的规律（如圆点或火柴棍）先转化为数字，再找出规则。推理过程将视觉规律与代数表达式联系起来。
Non‑linear sequences, including square numbers and triangular numbers, also appear. Answers identify the pattern: square numbers 1, 4, 9, 16… have nth term n². Geometric sequences with a constant multiplier are introduced. The solution verifies the ratio between consecutive terms and gives the next few terms. Students are encouraged to describe the pattern in words before writing the algebraic rule.
非线性数列，包括平方数和三角数，亦会出现。答案识别规律：平方数 1, 4, 9, 16… 的第 n 项为 n²。公比恒定的等比数列也有引入。解答验证连续项之间的比值，并给出后续若干项。鼓励学生在写出代数规则前先用文字描述规律。
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