📚 Essential Maths 7H Answers | 重要数学7H解答知识点精讲
Essential Maths 7H is widely used in Key Stage 3 to strengthen higher-tier mathematical foundations. This guide breaks down key topics by revisiting typical questions and their answers, explaining not just the 'how' but the 'why' behind each step. Mastering these answers means mastering the concepts – a vital step towards GCSE success.
《重要数学7H》是 KS3 阶段巩固高阶数学基础的常用教材。本指南通过回顾典型题目及其解答，既讲“怎么做”，也讲“为什么”这么做。掌握这些解答背后的思路，就是掌握核心概念——这是通向 GCSE 成功的关键一步。
1. Number Operations and BIDMAS | 整数运算与运算顺序
In the answer to any calculation, the order of operations matters. We use BIDMAS: Brackets, Indices, Division, Multiplication, Addition, Subtraction. Without it, the same expression can yield different results.
在任何计算的解答中，运算顺序至关重要。我们遵循 BIDMAS 法则：括号、指数、除法、乘法、加法、减法。缺失这一顺序会让同一个算式得出不同的结果。
Example problem: Calculate 24 ÷ (2 + 4) × 3 - 2²
示例题目：计算 24 ÷ (2 + 4) × 3 - 2²
work from left to right, so 24 ÷ 6 = 4, then 4 × 3 = 12. Finally subtraction: 12 - 4 = 8.
分步解答：先算括号：2 + 4 = 6。再算指数：2² = 4。此时式子变成 24 ÷ 6 × 3 - 4。除法和乘法从左往右运算，24 ÷ 6 = 4，然后 4 × 3 = 12。最后做减法：12 - 4 = 8。
Key to the answer: Always check for hidden brackets implied by fraction lines or roots. Incorrect use of BIDMAS is a common slip, even in 7H.
解答关键：时刻注意分数线或根号隐含的括号。即便在 7H 级别，BIDMAS 用法错误也是常见失分点。
2. Fractions: Simplifying and Operations | 分数：化简与运算
Fraction answers in 7H often require simplifying to the lowest terms or writing as a mixed number. Adding, subtracting, multiplying and dividing fractions rely on a solid understanding of equivalent fractions.
7H 中，分数解答经常要求化简到最简形式或写为带分数。分数的加减乘除都依赖于对等值分数的扎实理解。
Example: Work out 3/4 + 2/5 and write the answer as a mixed number.
示例：计算 3/4 + 2/5，并将答案写成带分数形式。
= 8/20. Add: 15/20 + 8/20 = 23/20. As a mixed number, 23 ÷ 20 = 1 remainder 3, so 1 3/20.
解答方法：找出 4 和 5 的最小公倍数 20。转换单位：3/4 = 15/20，2/5 = 8/20。相加：15/20 + 8/20 = 23/20。化为带分数，23 ÷ 20 = 1 余 3，所以答案是 1 3/20。
2/1 × 3/10 = 6/10, then simplify to 3/5. Always present the final answer in simplest form.
很多 7H 题目在做分数乘法时需要先约分以简化计算。例如 2/3 × 9/10，可约掉 3 和 9 得到 2/1 × 3/10 = 6/10，再化简为 3/5。最终答案务必保持最简形式。
3. Decimals and Percentages | 小数与百分数
Questions in Essential Maths 7H regularly link decimals, fractions and percentages. A typical answer involves converting between these forms or finding a percentage of an amount without a calculator.
《重要数学7H》中的题目经常联系小数、分数和百分数。典型解答往往涉及它们之间的相互转化，或不用计算器求一个数的百分含量。
Example: Find 15% of £240 and express your answer in pounds.
示例：求 £240 的 15%，并以英镑表示答案。
Answer steps: 10% of 240 is £24, 5% is half of that, so £12. Total 15% = £24 + £12 = £36. Alternatively, 15% = 0.15, so 0.15 × 240 = 36.
解答步骤：240 的 10% 是 £24，5% 是 10% 的一半即 £12。所以 15% = £24 + £12 = £36。也可以用 0.15 × 240 = 36 计算。
proportions, students must justify their answers. For instance, which is larger: 3/8 or 37%? Convert 3/8 to a decimal: 3 ÷ 8 = 0.375, and 37% ce 0.375 > 0.37, the fraction is larger. Such justifications are standard in mark schemes.
在比较大小时，学生需给出判断依据。例如，3/8 和 37% 哪个更大？将 3/8 化为小数：3 ÷ 8 = 0.375，37% = 0.37。因为 0.375 > 0.37，所以分数更大。这类解释是评分标准里常见的得分点。
4. Negative Numbers | 负数
Handling directed numbers correctly is tested throughout 7H. Answers involving addition, subtraction, multiplication and division of negatives must account for the signs.
在有向数运算上不丢分，是 7H 数学的通关要求。涉及负数加减乘除的解答必须正确处理好符号。
Example: Calculate –3 + (–5) – (–2) + 6.
示例：计算 –3 + (–5) – (–2) + 6。
Answer: Think of –3 + (–5) as –3 – 5 = –8. Then –(–2) becomes +2, so –8 + 2 = –6. Finally –6 + 6 = 0.
解答：将 –3 + (–5) 看作 –3 – 5 = –8。然后 –(–2) 变为 +2，因此 –8 + 2 = –6。最后 –6 + 6 = 0。
d division rules: positive × negative = negative; negative × negative = positive. Example: –12 ÷ (–3) × (–2) = 4 × (–2) = –8. Lay out the stages clearly, because the answer may require the working steps.
乘除法规则：正 × 负 = 负；负 × 负 = 正。例如 –12 ÷ (–3) × (–2) = 4 × (–2) = –8。清晰展示每一步骤，因为答案可能要求给出完整推导。
5. Ratio and Proportion | 比例与比率
Ratio problems in 7H often appear in the context of sharing amounts or scaling recipes. The answer must include clear simplification and, where needed, finding the value of one part.
7H 中的比率题常出现在分摊金额或调整配方的情景下。解答必须包含清晰的化简过程，并在需要时先求出一份量的值。
Example: Share £96 in the ratio 3 : 5. How much does each person receive?
示例：按 3 : 5 的比例分享 £96。每人各得多少？
Answer: Total parts = 3 + 5 = 8. One part = £96 ÷ 8 = £12. First person gets 3 × £12 = £36. Second person gets 5 × £12 = £60.
解答：总份数 = 3 + 5 = 8。一份量 = £96 ÷ 8 = £12。第一个人得 3 × £12 = £36，第二个人得 5 × £12 = £60。
For scaling problems, e.g. 'A recipe for 6 people uses 4 eggs. How many eggs for 9 people?' ften expected as a simplified fraction in the working, even if the final number is an integer.
对于缩放问题，如“6 人份食谱用 4 个鸡蛋，9 人份需要几个蛋？”，解答运用单位法。每人需 4 ÷ 6 = 2/3 个蛋。9 人需 9 × 2/3 = 6 个蛋。注意推算过程中常要用最简分数书写，哪怕最终结果是整数。
6. Algebraic Expressions | 代数表达式
Simplifying expressions like 4a + 3b – 2a + 5b is a core 7H skill. The answer must group like terms correctly and respect the signs.
化简诸如 4a + 3b – 2a + 5b 的表达式是 7H 核心技能。解答须正确合并同类项并注意符号。
Example: Simplify 5x – (2x – 4) + 3(1 – x).
示例：化简 5x – (2x – 4) + 3(1 – x)。
Answer step-by-step: Distribute the minus: –(2x – 4) = –2x + 4. Expand 3(1 – x) = 3 – 3x. Now the expression is 5x – 2x + 4 + 3 – 3x. Collect like terms: 5x – 2x – 3x = 0x, and constant: 4 + 3 = 7. Final answer: 7.
分步解答：去括号注意符号：–(2x – 4) = –2x + 4。展开 3(1 – x) = 3 – 3x。现在表达式为 5x – 2x + 4 + 3 – 3x。合并同类项：5x – 2x – 3x = 0x，常数项：4 + 3 = 7。最终答案：7。
In mark schemes, the answer alone is not enough; the expansion and collection steps carry marks. So always show each stage clearly, even for simple-looking questions.
在评分标准中，光有答案是不够的；展开和合并同类项的步骤同样得分。所以即使是看似简单的题，也要清晰展示每一步。
7. Solving Linear Equations | 解一元一次方程
Essential Maths 7H introduces two-step and multi-step equations. A correct answer involves isolating the unknown by performing inverse operations on both sides.
《重要数学7H》引入两步及多步方程。正确解答需要通过逆运算在等式两边逐步隔离未知数。
Example: Solve 2(x – 3) + 5 = 11.
示例：解方程 2(x – 3) + 5 = 11。
Answer method: Expand: 2x – 6 + 5 = 11 → 2x – 1 = 11. Add 1 to both sides: 2x = 12. Divide both sides by 2: x = 6. Check: 2(6 – 3) + 5 = 2×3 + 5 = 11, correct.
解答方法：展开：2x – 6 + 5 = 11 → 2x – 1 = 11。两边加 1：2x = 12。两边除以 2：x = 6。检验：2(6 – 3) + 5 = 2×3 + 5 = 11，正确。
subtract 4, and divide by 2 to give x = 3. Always finish by substituting back as a check.
当未知数出现在两边时，先移项归并。如 3x + 4 = x + 10，两边减 x 得 2x + 4 = 10，再减 4、除以 2 得 x = 3。务必代回原方程加以验证。
8. Angles and Parallel Lines | 角与平行线
Geometry answers in 7H need reasons based on angle facts: angles on a straight line sum to 180°, vertically opposite angles are equal, and alternate or corresponding angles on parallel lines are equal.
7H 几何解答需要给出基于角度规律的依据：直线上的角之和为 180°，对顶角相等，平行线中的内错角或同位角相等等。
Example: Two parallel lines are cut by a transversal. One angle is 120°. Find the other three angles at that intersection and state reasons.
示例：两条平行线被一条截线所截，已知一个角为 120°。求该交点处另外三个角，并说明理由。
te to 120° is also 120° (vertically opposite angles). The angle on the straight line adjacent to 120° is 180° – . Its vertically opposite angle is 60°. The corresponding angle on the other parallel line is 120°, and so on.
解答：与 120° 对顶的角也是 120°（对顶角相等）。与 120° 共线的邻补角为 180° – 120° = 60°（直线上的角）。60° 的对顶角也是 60°。另一平行线上同位角为 120°，其余类推。
Marks are almost always split between finding the angle and giving the correct reason. So write both clearly in your answer. Do not just write '180 – 120', but name the rule.
分数通常分别给在求角数值和给出正确理由上。因此解答中务必两者都写清。不要只写‘180 – 120’，而要写出角的关系名称。
9. Area and Perimeter | 面积与周长
Calculating area and perimeter of rectangles, triangles and compound shapes is common. Answers must include units and often a step-by-step breakdown for composite figures.
计算矩形、三角形和组合图形的面积与周长是常考题。解答须包含单位，对组合图形通常需要分步拆解。
Example: Find the area and perimeter of a rectangle of length 8 cm and width 5 cm, and then a triangle with base 8 cm and height 5 cm.
示例：求一个长 8 cm、宽 5 cm 的矩形的面积和周长，以及一个底 8 cm、高 5 cm 的三角形的面积。
a = (8 × 5) ÷ 2 = 20 cm². For perimeter of triangle, we would need all three side lengths.
解答：矩形面积 = 8 × 5 = 40 cm²，周长 = 2 × (8 + 5) = 26 cm。三角形面积 = (8 × 5) ÷ 2 = 20 cm²。三角形周长需要知道三条边长。
For compound shapes, divide the figure into simple shapes, calculate each area separately, then add or subtract as needed. Always double-check that you didn't count any edge twice. State the area expression before substituting numbers, as many 7H mark schemes give credit for the correct formula.
对于组合图形，将其拆分成简单图形，分别计算面积后再加减。务必检查是否重复计算了某条边。写入数值前先写出面积公式，因为很多 7H 评分标准会给公式分。
10. Averages and Range | 平均数与范围
Data-handling answers require finding mean, median, mode and range. In 7H, questions often ask to compare two data sets using these statistics.
数据处理解答要求计算平均数、中位数、众数和范围。7H 常要求用这些统计量比较两组数据。
Example: Find the mean and range of 4, 7, 9, 6, 4, 10.
示例：求数据 4, 7, 9, 6, 4, 10 的平均数和范围。
ues, so mean = 40 ÷ 6 = 6.666... which rounds to 6.7 or we leave as 6 2/3. Range = largest – smallest = 10 – 4 (most frequent). For median, order data: 4, 4, 6, 7, 9, 10. Median = (6 + 7) ÷ 2 = 6.5.
解答：总和 = 4 + 7 + 9 + 6 + 4 + 10 = 40。共 6 个数值，平均数 = 40 ÷ 6 = 6.666...，可四舍五入为 6.7 或保留 6 2/3。范围 = 最大值 – 最小值 = 10 – 4 = 6。众数是 4（出现最频繁）。中位数需排序：4, 4, 6, 7, 9, 10。中位数 = (6 + 7) ÷ 2 = 6.5。
When comparing, we might say the first set has a higher mean but a similar range, indicating greater central tendency but similar spread. This type of comment, when appropriate, demonstrates a deeper understanding that satisfies the 'explain' part of a question.
进行比较时，我们可能会说第一组数据平均数更高但范围相近，说明中心趋势更大而离散度相似。这类评价，在合适时使用，可以展现出对问题的深层理解，满足题目中的“解释”要求。
11. Sequences and Patterns | 序列与规律
7H sequences go beyond simple linear patterns to find the nth term. Answers require expressing the rule in algebra and using it to find, for example, the 100th term.
7H 的数列超越简单的线性模式，要求找出第 n 项的通项。解答需要用代数表达规律，并用以计算例如第 100 项。
Example: For the sequence 5, 9, 13, 17, …, find the nth term and the 50th term.
示例：对于序列 5, 9, 13, 17, …，找出第 n 项和第 50 项。
ch time, so the nth term relates to 4n. When n = 1, 4n = 4, but the first term is 5, so it's 4n + 1. Check: n = 2 gives 9, correct. 50th term = 4 × 50 + 1 = 201.
解答：每次增加 4，所以第 n 项与 4n 有关。n = 1 时，4n = 4，而首项为 5，所以通项是 4n + 1。检验 n = 2 得到 9，正确。第 50 项 = 4 × 50 + 1 = 201。
Some questions present pattern growth: 'Draw the next shape' or 'How many tiles in Pattern 10?' Answers link the picture to the algebraic rule. Write the sequence of numbers from the diagram, find the nth term, then apply it. This is a skill that bridges concrete patterns to abstract algebra.
有些题目以图形生长方式呈现：“画出下一个图形”或“图案 10 有多少块砖？”解答需将图形与代数规则联系起来。从图中写出数列，找出第 n 项，然后代入使用。这项技能正是从具体模式通向抽象代数的桥梁。
12. Introduction to Probability | 概率入门
Probability answers in 7H are expressed as fractions, decimals or percentages between 0 and 1. The answer to 'What is the probability of rolling a prime number on a fair six-sided die?' must be justified.
7H 概率解答以分数、小数或百分数表示，范围在 0 到 1 之间。对于“掷一个均匀六面骰，掷出素数的概率是多少？”这类问题，解答必须给出依据。
Example: A bag contains 3 red, 2 blue and 5 green counters. One counter is taken at random. Find the probability it is blue, and the probability it is not red.
示例：一个袋子里有 3 个红色、2 个蓝色和 5 个绿色筹码，随机取出一个。求抽到蓝色的概率，以及抽到不是红色的概率。
= 2/10 = 1/5. Not red means blue or green, total 7, so P(not red) = 7/10. Alternatively, 1 – P(red) = 1 – 3/10 = 7/10.
解答：总数 = 3 + 2 + 5 = 10。P(蓝) = 蓝色个数 / 总数 = 2/10 = 1/5。不是红色即蓝色或绿色，共 7 个，所以 P(非红) = 7/10。或用 1 – P(红) = 1 – 3/10 = 7/10。
sked, state the probability in words. In 7H, the concept of expected frequency may appear: 'If you pick 50 times, how many times do you expect blue?' Multiply probability by number of trials: 1/5 × 50 = 10 times.
务必化简分数，并根据题目要求用文字表述概率。7H 有时会涉及期望次数：“如果抽取 50 次，你期望抽到蓝色几次？”用概率乘以试验次数：1/5 × 50 = 10 次。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图