📚 Essential Maths Book 8 Answers: Compressed Question Types Explained | Essential Maths Book 8 答案压缩版：题型解析
Welcome to our comprehensive breakdown of the question types found in Essential Maths Book 8, a popular Key Stage 3 resource. This compressed guide distils the most common problem formats into clear, manageable explanations, helping students master the skills tested at Year 8 level. Instead of providing a simple answer key, we focus on the underlying methods, step-by-step reasoning, and common pitfalls, so you can tackle any similar question with confidence.
欢迎阅读我们对 Essential Maths Book 8 中常见题型的全面解析，这是一本广受欢迎的 Key Stage 3 学习资料。这份压缩指南将最常见的题目格式提炼成了清晰易懂的解说，帮助学生掌握 Year 8 所考查的技能。我们不只提供答案，而是重点讲解潜藏的方法、分步推理以及常见易错点，让你能够自信地应对任何类似问题。
1. Integers and Order of Operations | 整数与运算顺序
Book 8 exercises on integers regularly combine negative numbers with BODMAS/BIDMAS rules. Students are required to evaluate expressions such as 15 − 2 × (−3) + 6 ÷ 2, following the correct hierarchy: brackets first, then orders (powers), then division and multiplication from left to right, and finally addition and subtraction.
Book 8 中关于整数的练习经常将负数与 BODMAS/BIDMAS 规则结合在一起。学生需要计算如 15 − 2 × (−3) + 6 ÷ 2 的表达式，遵循正确的层次：先括号，再指数（幂），然后从左到右计算乘除，最后是加减。
A typical compressed-answer puzzle asks you to fill in a missing number or sign. For example: ‘Copy and complete: 20 − __ × 4 = −4.’ Let the blank be n; the equation becomes 20 − 4n = −4. Subtracting 20 from both sides gives −4n = −24, so n = 6. Always verify by substituting back: 20 − 6 × 4 = 20 − 24 = −4, which is correct.
一道典型的压缩答案谜题要求填入缺失的数字或符号。例如：‘抄写并完成：20 − __ × 4 = −4。’ 设空白处为 n，等式变为 20 − 4n = −4。两边减去 20 得 −4n = −24，所以 n = 6。务必代回检验：20 − 6 × 4 = 20 − 24 = −4，正确无误。
Another favourite format is spotting and explaining a mistake. For instance, ‘Mia claims −5² = 25. Explain why she is wrong.’ The error arises because the exponent applies only to the 5, not the minus sign. According to the order of operations, −5² means −(5 × 5) = −25. The correct interpretation is that squaring comes before applying the negative, unless brackets are used: (−5)² = 25.
另一种常见的题型是找出并解释错误。例如，‘Mia 声称 −5² = 25。解释她为什么错了。’ 错误的原因在于指数只作用于 5，而不是负号。根据运算顺序，−5² 表示 −(5 × 5) = −25。正确的理解是平方运算先于负号的应用，除非使用括号：(−5)² = 25。
2. Fractions, Decimals and Percentages | 分数、小数和百分比
Book 8 consolidates fluency in converting between fractions, decimals and percentages, as well as performing all four operations with fractions. Compressed questions often take the form of ‘Write the missing number’ in a chain of equivalent forms, or ‘Arrange these fractions in ascending order’ after converting to a common denominator.
Book 8 巩固了分数、小数和百分比之间的转换，以及分数的四则运算。压缩题目通常以 ‘写出缺失的数字’ 的形式出现在等价形式链中，或者在转换成同分母之后 ‘按升序排列这些分数’。
To add or subtract fractions, find the lowest common multiple of the denominators. For example, 2/3 + 1/4 = 8/12 + 3/12 = 11/12. Always simplify the final answer where possible. In multiplication, simply multiply numerators and denominators: 3/5 × 2/7 = 6/35. For division, flip the second fraction and multiply: 5/8 ÷ 3/4 = 5/8 × 4/3 = 20/24 = 5/6 after simplifying.
分数加减时，先找出分母的最小公倍数。例如，2/3 + 1/4 = 8/12 + 3/12 = 11/12。若可能，最后要化简答案。在乘法中，分子乘分子、分母乘分母：3/5 × 2/7 = 6/35。除法则是将第二个分数倒置后相乘：5/8 ÷ 3/4 = 5/8 × 4/3 = 20/24 = 5/6（化简后）。
When dealing with percentages of amounts or percentage increase and decrease, a compressed-answer task may ask you to fill gaps in a calculation like ‘A jacket costs £40. In a sale it is reduced by 15%. New price = £40 × 0.85 = £___.’ The multiplier 0.85 comes from 100% − 15% = 85%. If the question gave the reduced price and asked for the original, you would divide by 0.85.
在处理百分比求值、百分比增减时，压缩答案题目可能要求你填补计算中的空缺，例如 ‘一件夹克售价 40 英镑。促销降价 15%。新价格 = 40 × 0.85 = ___ 英镑。’ 乘数 0.85 来源于 100% − 15% = 85%。如果题目给出降价后的价格，让你求原价，则需除以 0.85。
3. Ratio and Proportion | 比和比例
Ratio questions in Book 8 move beyond simple sharing into scaling recipes, maps and exchange rates. A compressed answer version may display a partially completed ratio table or ask you to complete the missing term in a proportion such as 3 : 7 = __ : 28.
Book 8 中的比和比例题从简单的分配扩展到食谱配比、地图比例尺和汇率换算。压缩答案形式可能展示一个部分完成的比表，或者要求你补全比例中的缺失项，如 3 : 7 = __ : 28。
To solve a proportion, identify the relationship between the known terms. Here, 7 has been multiplied by 4 to give 28, so the corresponding term must also be multiplied by 4: 3 × 4 = 12. The completed statement is 3 : 7 = 12 : 28. For sharing questions, add the parts of the ratio to find the total number of shares, then divide the given amount by that total and multiply by each part.
解比例时，先找出已知项之间的关系。此题中 7 乘以 4 得到 28，因此对应项也必须乘以 4：3 × 4 = 12。完整的式子是 3 : 7 = 12 : 28。对于分配问题，将比的各项相加得到总份数，再用给定总数除以总份数，然后乘以各项的份数。
A typical recipe problem: ‘A fruit salad for 4 people uses 300 g of strawberries. How many grams are needed for 10 people?’ The unitary method works well: for 1 person you need 300 ÷ 4 = 75 g, so for 10 people you need 75 × 10 = 750 g. Alternatively, use the multiplier 10/4 = 2.5, so 300 × 2.5 = 750 g. Compressed answers often just show the final number, but the working is what secures the marks.
一道典型的食谱问题：‘4 人份的水果沙拉需要 300 克草莓。10 人份需要多少克？’ 归一法很有效：1 人份需要 300 ÷ 4 = 75 克，所以 10 人份需要 75 × 10 = 750 克。或者，使用乘数 10/4 = 2.5，那么 300 × 2.5 = 750 克。压缩答案往往只显示最终数字，但解题过程才能确保得分。
4. Algebraic Simplification | 代数化简
In the algebra sections, students learn to collect like terms, expand single brackets, and factorise simple expressions. Compressed answer tasks might present several expressions and ask you to match them with their simplified forms, or to fill in the missing coefficient in an expansion such as 5(2x − 3) = 10x − __ .
在代数部分，学生学习合并同类项、展开单项括号以及简单的因式分解。压缩答案任务可能给出若干表达式，要求你与它们的简化形式配对，或者在展开式中填入缺失的系数，如 5(2x − 3) = 10x − __ 。
To collect like terms, combine terms with exactly the same variable part. For 3a + 4b − a + 2b, group the a-terms: 3a − a = 2a, and the b-terms: 4b + 2b = 6b. The simplified expression is 2a + 6b. Remember that constants (numbers without letters) are like terms only with each other.
合并同类项时，将变量部分完全相同的项组合在一起。对于 3a + 4b − a + 2b，把含 a 的项分为一组：3a − a = 2a，把含 b 的项分为一组：4b + 2b = 6b。化简后的表达式为 2a + 6b。请记住，常数（没有字母的数字）只能彼此作为同类项。
When expanding brackets, multiply the term outside by every term inside. For −4(x − 2y), calculate −4 × x = −4x, and −4 × (−2y) = +8y, giving −4x + 8y. A frequent error is mishandling the negative sign, so double-check the signs. Factorising is the reverse process: find the highest common factor and write it outside the bracket, e.g., 15x + 20 = 5(3x + 4).
展开括号时，用括号外的项乘以括号内的每一项。对于 −4(x − 2y)，计算 −4 × x = −4x，−4 × (−2y) = +8y，得到 −4x + 8y。一个常见错误是处理负号失误，因此要仔细检查符号。因式分解是逆过程：找出最大公因数并将其写在括号外，例如，15x + 20 = 5(3x + 4)。
5. Solving Linear Equations | 解一元一次方程
Linear equations in Book 8 range from one-step through two-step to those with unknowns on both sides. Compressed answer sheets often expect you to fill in a missing step in a solution sequence or to correct a deliberately mistaken solution. Understanding the balance method is essential.
Book 8 中的一元一次方程从一步方程到两步方程，再到未知数在两边的情况。压缩答题纸通常要求你在解题序列中填补缺失的步骤，或者纠正故意写错的解法。理解平衡法是关键。
For a two-step equation like 2x + 7 = 19, subtract 7 from both sides: 2x = 12, then divide both sides by 2: x = 6. Always perform the same operation to both sides to maintain equality. When the equation has unknowns on both sides, such as 5x − 3 = 3x + 7, first collect x-terms on one side by subtracting 3x from both sides, giving 2x − 3 = 7. Then add 3: 2x = 10, so x = 5.
对于像 2x + 7 = 19 这样的两步方程，两边减去 7：2x = 12，然后两边除以 2：x = 6。始终在方程两边执行相同的运算以保持等式。当方程两边都有未知数时，例如 5x − 3 = 3x + 7，首先通过两边减去 3x 将含 x 的项集中到一侧，得到 2x − 3 = 7。然后加 3：2x = 10，因此 x = 5。
A compressed-type question might show: ‘Solve 4(x + 2) − 3 = 3x + 6. Step 1: 4x + 8 − 3 = 3x + 6 → 4x + 5 = 3x + 6. Step 2: 4x − 3x = 6 − 5 → x = 1. What is the next step to check?’ Checking involves substituting x = 1 back into the original equation: left side = 4(1+2)−3 = 4×3−3 = 9; right side = 3×1+6 = 9; both equal, so correct.
一道压缩型题目可能展示：‘解方程 4(x + 2) − 3 = 3x + 6。步骤 1：4x + 8 − 3 = 3x + 6 → 4x + 5 = 3x + 6。步骤 2：4x − 3x = 6 − 5 → x = 1。下一步该如何检验？’ 检验需要将 x = 1 代回原方程：左边 = 4(1+2)−3 = 4×3−3 = 9；右边 = 3×1+6 = 9；两者相等，所以正确。
6. Linear Sequences | 线性数列
Sequences work in Book 8 focuses on finding the nth term rule and using it to generate terms or decide whether a number belongs to the sequence. Compressed answers often present a table with missing terms or ask you to write the rule in the form an + b.
Book 8 中数列的内容侧重于寻找第 n 项通项公式，并利用它来生成项或判断某个数是否属于该数列。压缩答案常常给出一个缺项的表格，或要求你写出 an + b 形式的通项公式。
For the sequence 5, 9, 13, 17, …, find the common difference (4). This is the coefficient a. So the rule begins 4n. When n = 1, 4n = 4, but the first term is 5, so b = 1. The nth term is 4n + 1. To find the 50th term, substitute n = 50: 4 × 50 + 1 = 201. To check if 150 is in the sequence, solve 4n + 1 = 150 → 4n = 149 → n = 37.25, not an integer, so 150 is not a term.
对于数列 5, 9, 13, 17, …，找出公差 (4)。这就是系数 a。所以通项以 4n 开始。当 n = 1 时，4n = 4，但首项是 5，因此 b = 1。第 n 项为 4n + 1。要求第 50 项，代入 n = 50：4 × 50 + 1 = 201。要检验 150 是否在数列中，解方程 4n + 1 = 150 → 4n = 149 → n = 37.25，不是整数，所以 150 不是数列的项。
Some questions provide the nth term and ask you to complete the first three terms. If the rule is 2n − 3, then for n=1: −1; n=2: 1; n=3: 3. Compressed answer tasks may also require drawing the next pattern in a sequence of shapes and linking the number of matchsticks or dots to a linear rule.
有些题目给出第 n 项公式，要求你写出前三项。如果公式是 2n − 3，那么 n=1 时：−1；n=2 时：1；n=3 时：3。压缩答案任务还可能要求画出图形序列中的下一个图案，并将火柴棍或圆点的数量与线性公式联系起来。
7. Angles and Parallel Lines | 角与平行线
Geometry questions in Book 8 rely on angle facts: angles on a straight line sum to 180°, angles around a point total 360°, vertically opposite angles are equal, and the special relationships when a transversal crosses parallel lines. Compressed answer formats often show a diagram with some angles labelled and others left blank, expecting you to fill them in using reasoning.
Book 8 中的几何题依赖于角的基本事实：直线上的角之和为 180°，绕一点的角之和为 360°，对顶角相等，以及截线穿过平行线时的特殊关系。压缩答案形式通常展示一个标有部分角度、其余留空的图形，期望你通过推理填出空白处。
When a transversal intersects two parallel lines, alternate angles are equal, corresponding angles are equal, and co-interior (allied) angles sum to 180°. For example, if a corresponding angle to a given 65° angle is requested, the answer is simply 65°. If a co-interior angle paired with 65° is needed, it is 180° − 65° = 115°.
当一条截线与两条平行线相交时，内错角相等，同位角相等，同旁内角（互补角）之和为 180°。例如，要求一个已知 65° 角的同位角，答案就是 65°。若要求与 65° 成同旁内角的角度，则为 180° − 65° = 115°。
Triangles bring in the rule that interior angles sum to 180°. A typical multi-step problem: in a triangle, one angle is 50°, an exterior angle is 130°. Find the other interior angles. Since the exterior angle equals the sum of the two opposite interior angles, one of those is 50°, so the other is 130° − 50° = 80°. Then the third angle is 180° − 50° − 80° = 50°. Identifying the correct angle facts is the key to efficient solutions.
三角形的内角和为 180°。一道典型的多步问题：三角形中一个角为 50°，一个外角为 130°。求其他内角。外角等于与它不相邻的两个内角之和，其中一个内角为 50°，因此另一个为 130° − 50° = 80°。那么第三个角为 180° − 50° − 80° = 50°。识别正确的角关系是高效解题的关键。
8. Area and Perimeter of Compound Shapes | 组合图形的面积与周长
Book 8 extends earlier work to compound shapes made from rectangles, triangles, and sometimes circles. Compressed questions often present a shape with some side lengths hidden, requiring you to deduce them before calculating area or perimeter. Using known properties of shapes and subtraction of lengths is essential.
Book 8 将之前学到的内容扩展到由矩形、三角形，有时还有圆形构成的组合图形。压缩题目通常给出一个部分边长隐藏的图形，要求你在计算面积或周长之前先推导出它们。利用已知的图形性质和长度的减法至关重要。
For a compound rectangle, split the shape into simpler parts. If an L-shape is formed from two rectangles, calculate the area of each and add them. The formula for the area of a rectangle is A = l × w, and for a triangle it is A = ½ × base × height. For a circle, A = πr² and circumference C = 2πr (or C = πd). In KS3, π is often taken as 3.14 or the exact answer is left in terms of π.
对于组合矩形，将图形分割成更简单的部分。如果一个 L 形由两个矩形组成，分别计算各部分的面积再相加。矩形面积公式为 A = l × w，三角形面积为 A = ½ × 底 × 高。对于圆，A = πr²，周长 C = 2πr（或 C = πd）。在 KS3 阶段，π 通常取 3.14，或答案保留为 π 表示。
Perimeter requires adding all the outer side lengths. A common pitfall is forgetting to include the ‘inner’ edges that become outer edges after subtracting a shape. For a rectangle with a smaller rectangle cut out from a corner, carefully list every outer segment. Compressed answer drills may ask: ‘The perimeter of this shape is 42 cm. Find the missing side length.’ Set up an equation by summing all known sides plus the unknown and equating to the total perimeter.
周长需要将所有外侧边长相加。一个常见误区是忘记将减掉一个图形后变成外侧边的 ‘内部’ 边计入。对于从一个
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