📚 Year 9 Edexcel Maths: Case Study Practice | Year 9 Edexcel 数学：案例分析实战演练
Case study practice combines real-life scenarios with mathematical concepts, helping you see how abstract ideas become useful tools. This approach is essential for Year 9 Edexcel Maths because it prepares you for the problem-solving demands of GCSE and beyond. Rather than just memorising formulas, you will learn to interpret information, select the right methods, and check your answers in meaningful contexts.
案例分析实战演练将现实生活场景与数学概念相结合，帮助你看到抽象概念如何变成有用的工具。这种方法对 Year 9 Edexcel 数学至关重要，因为它为你应对 GCSE 及更高层次的解题要求做好准备。你不仅是在记忆公式，更要学会解读信息、选择恰当的方法，并在有意义的背景中检验答案。
1. Understanding the Case Study Approach | 理解案例分析方法
In a case study, you are given a situation – it might be planning a trip, running a school fair, or calculating building costs. Your job is to read the context carefully, identify the mathematical elements, and break the problem into smaller steps. Always start by underlining key numbers, units, and relationships in the text.
在案例分析中，你会得到一个情境——可能是规划一次旅行、举办学校义卖，或者计算建筑成本。你的任务是仔细阅读背景，找出其中的数学要素，并把问题分解成更小的步骤。务必先从文中划出关键数字、单位和关系。
You might see a mixture of topics in one case. For example, planning a party involves budgeting (arithmetic), scaling recipes (ratio and proportion), and measuring space (area). Practice telling yourself what each chunk of the problem is asking before you dive into calculations.
你可能会在一个案例中看到多个主题的组合。例如，策划一场派对涉及预算（算术）、按比例调整食谱（比和比例）以及测量空间（面积）。在开始计算之前，练习告诉自己每个部分在问什么。
2. Number Skills in Real Contexts | 真实情境中的数字运算
A youth club orders 240 cans of drink for an event. 35% are cola, ¼ are lemonade, and the rest are orange. How many orange cans are there? se ? This case tests your ability to work with percentages, fractions, and subtraction. First, find 35% × 240 = 84. Then ¼ of 240: 240 ÷ 4 = 60. Add these: 84 + 60 = 144. Subtract from 240: 240 – 144 = 96 orange cans.
一家青年俱乐部为活动订购了 240 罐饮料。35% 是可乐，¼ 是柠檬水，其余是橙汁。有多少罐橙汁？这个案例测试你处理百分数、分数和减法的能力。首先求出 240 的 35%：0.35 × 240 = 84。然后 240 的 ¼：240 ÷ 4 = 60。相加：84 + 60 = 144。从 240 中减去：240 – 144 = 96 罐橙汁。
Always express your working clearly. Write '35% of 240 = 84' nal answer. In an exam, method marks can be earned even if a small arithmetic slip occurs.
务必清晰地写出运算过程。写出“240 的 35% = 84”，而不仅仅是最终答案。在考试中，即使出现小的计算错误，方法分仍然可以拿到。
3. Algebraic Cost Analysis | 代数成本分析
Imagine you run a stall selling handmade bracelets. Each bracelet costs £2.50 to make, and venue hire is £15. The profit P (in pounds) when selling n bracelets at £5 each is given by: P = 5n – 2.5n – 15. Simplify this expression and find the number of bracelets needed to break even.
假设你经营一个摊位，售卖手工手链。每条手链的制作成本是 2.50 英镑，场地租金为 15 英镑。如果按每条 5 英镑出售 n 条手链，利润 P（英镑）的计算式为：P = 5n – 2.5n – 15。化简该表达式，并求出达到盈亏平衡时需要卖出的手链数量。
Simplify: 5n – 2.5n = 2.5n, so P = 2.5n – 15. Break even means profit = 0: 2.5n – 15 = 0 → 2.5n = 15 → n bracelets to cover costs. This uses simplifying expressions and solving linear equations.
化简：5n – 2.5n = 2.5n，所以 P = 2.5n – 15。盈亏平衡意味着利润为 0：2.5n – 15 = 0 → 2.5n = 15 → n = 15 ÷ 2.5 = 6。你需要卖出 6 条手链才能收回成本。这里用到了表达式化简和解一元一次方程。
4. Inequalities in Mobile Phone Plans | 手机话费套餐中的不等式
Company A charges £10 per month plus £0.05 per text. Company B charges £12 per month plus £0.03 per text. For what number of monthly texts (t) is Company A cheaper? Set up an inequality: 10 + 0.05t < 12 + 0.03t.
A 公司每月收取 10 英镑，外加每条短信 0.05 英镑。B 公司每月收取 12 英镑，外加每条短信 0.03 英镑。当月短信量为多少 (t) 时，A 公司更便宜？建立不等式：10 + 0.05t < 12 + 0.03t。
Subtract 10 from both sides: 0.05t < 2 + 0.03t. Subtract 0.03t: 0.02t < 2. Divide by 0.02: t < 100. So if you send fewer than 100 texts, Company A is cheaper. Make sure you interpret the inequality correctly; the question might also ask for the point where costs are equal, which is t = 100.
两边同时减去 10：0.05t < 2 + 0.03t。减去 0.03t：0.02t < 2。除以 0.02：t < 100。因此，如果你发送的短信少于 100 条，A 公司更便宜。确保正确解读不等式；题目可能还会问费用相等的点，即 t = 100。
5. Sequences in Tile Patterns | 地砖图案中的数列
A pathway is made using grey and white tiles. Pattern 1 has 4 grey tiles, pattern 2 has 7, pattern 3 has 10. Find the rule for the number of grey tiles in pattern n, and predict the number in pattern 50.
一条小径用灰色和白色地砖铺设。图案 1 有 4 块灰砖，图案 2 有 7 块，图案 3 有 10 块。找出图案 n 中灰砖数量的规律，并预测图案 50 中的数量。
The sequence 4, 7, 10, … has a common difference of 3. Using the formula for the nth term of an arithmetic sequence: first term a = 4, difference d = 3, so term = a + (n – 1)d = 4 + 3(n – 1). Simplify: 4 + 3n – 3 = 3n + 1. For pattern 50, substitute n = 50: 3(50) + 1 = 151 grey tiles.
数列 4, 7, 10, … 的公差为 3。使用等差数列第 n 项的公式：首项 a = 4，公差 d = 3，因此第 n 项 = a + (n – 1)d = 4 + 3(n – 1)。化简：4 + 3n – 3 = 3n + 1。对于图案 50，代入 n = 50：3(50) + 1 = 151 块灰砖。
6. Geometry in Garden Design | 花园设计中的几何
A triangular flower bed has two equal sides of 5 m each and an included angle of 70°. The gardener wants to edge it with a border. What is the perimeter, and what type of triangle is it? This case mixes properties of triangles and perimeter calculations.
一个三角形花坛的两条边相等，各长 5 米，这两条边的夹角为 70°。园丁想用边框把它围起来。周长是多少？它是什么类型的三角形？这个案例结合了三角形性质和周长的计算。
angled; the problem would likely provide the length or use construction. Suppose the base is given as 6 m. Then perimeter = 5 + 5 + 6 = 16 m. Always check angle properties: the base angles are equal, each (180° – 70°) ÷ 2 = 55°.
由于有两条边相等，它是一个等腰三角形。要求第三边，Year 9 阶段可能不涉及三角学（除非是直角三角形）；题目可能会直接给出底边长度，或者用尺规作图。假设底边已知为 6 米，那么周长 = 5 + 5 + 6 = 16 米。务必检查角的性质：底角相等，每个底角 = (180° – 70°) ÷ 2 = 55°。
7. Volume and Capacity in Aquariums | 鱼缸的容积与容量
An aquarium is a cuboid measuring 80 cm by 40 cm by 50 cm. The owner fills it to ⅘ of its height to leave space. How many litres of water are needed? 1 litre = 1000 cm³.
一个鱼缸是长方体，尺寸为 80 cm × 40 cm × 50 cm。主人注水至高度的 ⅘ 处，以留出空间。需要多少升水？1 升 = 1000 cm³。
Full volume = 80 × 40 × 50 = 160,000 cm³. Water height = ⅘ m³. Convert to litres: 128,000 ÷ 1000 = 128 litres. Alternatively, multiply full volume by ⅘: 160,000 × ⅘ = 128,000 cm³. This shows that fraction skills and unit conversions are vital.
满容积 = 80 × 40 × 50 = 160,000 cm³。水的高度 = ⅘ × 50 = 40 cm，因此水的体积 = 80 × 40 × 40 = 128,000 cm³。换算成升：128,000 ÷ 1000 = 128 升。或者用满容积乘以 ⅘：160,000 × ⅘ = 128,000 cm³。这展示了分数运算和单位换算至关重要。
8. Ratio and Proportion in Recipes | 食谱中的比与比例
A cookie recipe for 16 cookies requires 200 g flour, 100 g butter, and 50 g sugar. You need to make 40 cookies for a bake sale. How much of each ingredient do you need? This is a classic proportion problem.
一份饼干食谱可制作 16 块饼干，需要 200 克面粉、100 克黄油和 50 克糖。你需要为糕点义卖制作 40 块饼干。每种配料需要多少？这是一个经典的比例问题。
n also set up a ratio table. The key is to maintain the same ratio of ingredients; any change must be proportional.
首先求出倍数因子：40 ÷ 16 = 2.5。每种配料乘以 2.5：面粉 = 200 × 2.5 = 500 克，黄油 = 100 × 2.5 = 250 克，糖 = 50 × 2.5 = 125 克。你也可以用比列表格。关键是保持配料比例不变；任何变动都必须是成比例的。
9. Statistical Averages from Survey Data | 调查数据中的统计平均数
Students record the number of texts sent in an hour: {8, 12, 15, 9, 14, 12, 10, 8, 13}. Calculate the mean, median, mode, and range. Write a sentence comparing these values.
学生们记录了一小时内发送的短信数量：{8, 12, 15, 9, 14, 12, 10, 8, 13}。计算平均数、中位数、众数和极差。写一句话比较这些值。
2 (5th value). Mean = sum ÷ 9 = (8+8+9+10+12+12+13+14+15) ÷ 9 = 101 ÷ 9 ≈ 11.2. Range = 15 – The mean is slightly lower than the median, suggesting a slight skew toward lower values.
排序后：8, 8, 9, 10, 12, 12, 13, 14, 15。众数 = 8 和 12（双众数）。中位数 = 中间值 = 12（第 5 个值）。平均数 = 总和 ÷ 9 = (8+8+9+10+12+12+13+14+15) ÷ 9 = 101 ÷ 9 ≈ 11.2。极差 = 15 – 8 = 7。平均数略低于中位数，说明数据略微偏向低值。
10. Probability in Games | 游戏中的概率
A spinner has 5 equal sections: red, blue, green, yellow, and pink. It is spun twice. What is the probability of landing on red both times? Draw a sample space or use the multiplication rule.
一个转盘有 5 个相等的扇区：红、蓝、绿、黄、粉。转两次。两次都落在红色的概率是多少？画出样本空间或使用乘法规则。
/5 = 1/25. The sample space has 25 outcomes; only (R,R) satisfies. Another case might ask: 'What is the probability of getting at least one yellow?' er to find the complement: P(no yellow) = 4/5 × 4/5 = 16/25, so P(at least one yellow) = 1 – 16/25 = 9/25.
单次转出红色的概率 = 1/5。由于两次转动独立，P(红且红) = 1/5 × 1/5 = 1/25。样本空间有 25 种结果；只有 (红,红) 符合。另一种情况可能问：“至少得一次黄色的概率是多少？”此时用补集会更简单：P(无黄) = 4/5 × 4/5 = 16/25，所以 P(至少一次黄) = 1 – 16/25 = 9/25。
11. Mixed Problem: School Prom Budget | 综合问题：学校毕业舞会预算
A prom committee has a budget of £1200. The venue charges £400 plus £15 per person. Catering costs 40% of the venue total. Decorations are £150 fixed. If p people attend, write a total cost expression and find the maximum possible p within budget. This blends algebra, percentages, and inequalities.
毕业舞会委员会有 1200 英镑预算。场地费用为 400 英镑外加每人 15 英镑。餐饮成本为场地总费用的 40%。装饰固定 150 英镑。如果有 p 人参加，写出总成本表达式，并求出预算内最多可容纳的人数。这混合了代数、百分数和不等式。
Venue cost = 400 + 15p. Catering = 40% of (400 + 15p) = 0.4(400 + 15p). Total cost = (400 + 15p) + 0.4(400 + 15p) + 150 = 1.4(400 + 15p) + 150. Simplify: 560 + 21p + 150 = 710 + 21p. Set ≤ 1200: 21p ≤ 490 → p ≤ 490 ÷ 21 ≈ 23.33. So maximum 23 people. Always round down in this context.
场地成本 = 400 + 15p。餐饮 = 场地总费用的 40% = 0.4(400 + 15p)。总成本 = (400 + 15p) + 0.4(400 + 15p) + 150 = 1.4(400 + 15p) + 150。化简：560 + 21p + 150 = 710 + 21p。令其 ≤ 1200：21p ≤ 490 → p ≤ 490 ÷ 21 ≈ 23.33。所以最多 23 人。在此情境下务必向下取整。
12. Review and Practice Tips | 复习与实战技巧
After working through case studies, reflect on what slowed you down. Did you misread a percentage? Forget to convert units? Keep a log of common mistakes. Reattempt similar cases with small variations: change the numbers, add a condition, or swap the question. The more you practise interpreting scenarios, the faster you become at selecting the right maths toolkit.
在完成案例分析后，反思是什么拖慢了你的速度。是误读了百分数？忘记转换单位？记录常见错误。用微小变化重新尝试类似案例：改变数字、增加条件，或调换问题。对情境解读练习得越多，你就能越快选出正确的数学工具箱。
Use the Edexcel specification list to tick off topics as you master them in context. Even a simple 'plan-a-trip' case tests a wide range: reading timetables (time intervals), fuel costs (unit conversion), and splitting bills (division and rounding). Aim to build confidence in multi-step reasoning – exactly what GCSE problem-solving rewards.
对照 Edexcel 考纲列表，在情境中掌握一个主题就勾掉一个。即便是简单的“规划出行”案例，也能测试众多内容：阅读时间表（时间间隔）、油费（单位转换）、分摊账单（除法和四舍五入）。目标是建立多步推理的信心——这正是 GCSE 解题给分所看重的。
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