📚 Year 7 CAIE Further Mathematics: Case Study Practical Exercises | Year 7 CAIE 进阶数学：案例分析实战演练
In Year 7 Further Mathematics, case studies transform abstract ideas into real investigations. You learn to choose the right tools—fractions, equations, graphs, or averages—to solve problems that mirror life outside the classroom.
在七年级进阶数学中，案例研究将抽象的概念转化为真实的探究。你将学会选择合适的工具——分数、方程、图表或平均数——来解决那些反映课外实际情境的问题。
1. What is a Case Study in Mathematics? | 数学中的案例分析是什么？
A case study presents a real‑world situation—such as budgeting a trip or designing a space—where mathematics is the key to making decisions. You need to read carefully, identify patterns, and apply multiple topics together.
案例分析呈现一个真实情境——例如为旅行做预算或设计空间——其中数学是做决策的关键。你需要仔细阅读、找出规律，并综合运用多个数学主题。
Unlike a textbook exercise that tells you which chapter to use, a case study may combine fractions, percentages, algebra, and geometry in one task. This mirrors how problems appear in daily life and develops confident problem‑solving skills.
与课本中告诉你使用哪一章知识的习题不同，一个案例研究可能将分数、百分比、代数和几何融合在一个任务中。这模拟了生活中出现的问题的方式，并培养你自信解决问题的能力。
2. Case Study 1: Planning a School Trip | 案例1：规划学校旅行
A school hires a coach for £120. Each student pays a subsidised fee of £5. The trip coordinator, Ms Li, needs to work out the total cost and the amount every student actually pays once the coach fee is shared equally. She also wants to explore what happens if the number of students changes.
一所学校以120英镑租用了一辆大巴。每位学生需支付5英镑的补贴费用。行程协调员李老师需要计算总费用以及当大巴费用均摊后，每位学生实际需要支付的金额。她还希望探究如果学生人数发生变化，情况会怎样。
Let n represent the number of students. The total bill is the fixed coach hire plus the student contributions: Total = 120 + 5n. The amount each student pays is (120 + 5n) ÷ n, which can also be written as 120/n + 5.
设 n 表示学生人数。总费用是固定的大巴租金加上学生个人费用：合计 = 120 + 5n。每位学生实际支付的金额为 (120 + 5n) ÷ n，也可写作 120/n + 5。
3. Mathematical Tools: Ratios, Proportions, and Simple Equations | 数学工具：比例、比率和简单方程
The trip case uses division, proportion, and equation rearrangement. When n = 40, the total is 120 + 5×40 = 320, so each student pays 320 ÷ 40 = 8. The coach cost of £120 is shared among 40 pupils, adding £3 to the base £5.
旅行案例用到了除法、比例以及方程的重组。当 n = 40 时，总额为 120 + 5×40 = 320，所以每位学生支付 320 ÷ 40 = 8。120英镑的大巴费用由40名同学均摊，在基本的5英镑上增加了3英镑。
If the number of students doubles to 80, the shared-coach part halves: 120/80 = 1.5, so each pays £6.50. This inverse relationship is important in resource sharing and scaling.
如果学生人数加倍至80人，均摊的大巴费用减半：120/80 = 1.5，因此每人支付6.50英镑。这种反比关系在资源共享和按比例缩放问题中很重要。
4. Solving the Trip Problem Step by Step | 逐步解决旅行问题
Step 1 – Define the variables: n = number of students, C = total cost, p = cost per student. Write the formulas: C = 120 + 5n and p = (120 + 5n)/n.
第一步 – 定义变量：n = 学生人数，C = 总费用，p = 人均费用。写出公式：C = 120 + 5n，p = (120 + 5n)/n。
Step 2 – Substitute values and compute. For n = 30, C = 120 + 150 = 270, p = 270 ÷ 30 = 9. For n = 60, C = 120 + 300 = 420, p = 420 ÷ 60 = 7. Organise results in a table to see the trend.
第二步 – 代入并计算。当 n = 30，C = 120 + 150 = 270，p = 270 ÷ 30 = 9。当 n = 60，C = 120 + 300 = 420，p = 420 ÷ 60 = 7。将结果整理成表格以观察趋势。
Step 3 – Check reasonableness. The per‑student cost should always be above £5 and decrease as more students join. If n is very small, say 10, p = (120 + 50)/10 = 17, which is high but logically correct.
第三步 – 检查合理性。人均费用应始终高于5英镑，并随着学生增多而降低。如果 n 非常小，比如10，p = (120 + 50)/10 = 17，虽然高昂但逻辑上是正确的。
5. Case Study 2: Designing a Rectangular Garden | 案例2：设计矩形花园
A gardener wants a rectangular flower bed where the length is 3 metres longer than the width. The area must be exactly 54 square metres. Find the dimensions and calculate the amount of edging needed for the perimeter.
一位园丁想要一个长方形的花坛，长度比宽度多3米。面积必须正好是54平方米。求花坛的尺寸，并计算围边所需的材料长度（周长）。
Let the width be w metres, then the length is w + 3. Area = length × width = (w + 3)w = 54. This leads to a quadratic equation: w² + 3w − 54 = 0.
设宽度为 w 米，则长度为 w + 3。面积 = 长 × 宽 = (w + 3)w = 54。于是得到一个二次方程：w² + 3w − 54 = 0。
6. Using Area and Perimeter Formulas | 使用面积和周长公式
To solve w² + 3w − 54 = 0, we look for two numbers that multiply to −54 and add to 3. These are 9 and −6, giving (w + 9)(w − 6) = 0. The positive solution is w = 6 metres, so length = 9 metres.
为了解 w² + 3w − 54 = 0，我们寻找两个数，乘积为 −54，总和为 3。这两个数是 9 和 −6，因此得到 (w + 9)(w − 6) = 0。正数解为 w = 6 米，所以长度 = 9 米。
Perimeter = 2 × (length + width) = 2 × (9 + 6) = 30 metres. The gardener must buy 30 m of edging. Checking: 9 × 6 = 54 m², and 9 is indeed 3 more than 6.
周长 = 2 × (长 + 宽) = 2 × (9 + 6) = 30 米。园丁需要购买30米的围边。验证：9 × 6 = 54 m²，且9确实比6多3。
If the area were different, for instance 72 m², the equation would be w² + 3w − 72 = 0, which factorises to (w + 12)(w − 6)[check? Actually −6 × 12 = −72, sum 6, no]. We would need to try integer factors. The method is the same.
如果面积不同，例如72 m²，方程变为 w² + 3w − 72 = 0，可分解为？我们需要尝试整数因数分解。方法是相同的。
7. Case Study 3: Analysing Sports Scores | 案例3：分析运动比赛得分
A Year 7 football team recorded their goals in 8 matches: 2, 5, 1, 3, 0, 4, 7, 2. As a team analyst, you need to summarise performance using averages and charts.
一支七年级足球队记录了8场比赛的进球数：2, 5, 1, 3, 0, 4, 7, 2。作为球队分析师，你需要使用平均数和图表来总结球队表现。
First, order the data: 0, 1, 2, 2, 3, 4, 5, 7. The range = highest − lowest = 7 − 0 = 7, showing large variability. The mode is 2 goals (appears twice).
首先，将数据排序：0, 1, 2, 2, 3, 4, 5, 7。极差 = 最大值 − 最小值 = 7 − 0 = 7，显示出较大的波动性。众数为2球（出现了两次）。
8. Handling Data: Charts, Averages, and Range | 数据处理：图表、平均数和极差
Mean = (0+1+2+2+3+4+5+7) ÷ 8 = 24 ÷ 8 = 3 goals per game. Median is the average of the 4th and 5th values: (2+3)/2 = 2.5. The mean is slightly higher than the median, suggesting a couple of high‑scoring games pull the average up.
平均数 = (0+1+2+2+3+4+5+7) ÷ 8 = 24 ÷ 8 = 每场3球。中位数是第4和第5个值的平均数：(2+3)/2 = 2.5。平均数略高于中位数，表明有几场高分比赛拉高了平均值。
A bar chart can display the frequency of each score. Draw a horizontal axis with goals 0–7 and vertical axis with frequency. Each bar height shows how many matches had that many goals, helping visualise the common results.
柱状图可以展示每个得分的频数。用水平轴表示进球数0–7，垂直轴表示频数。每个柱形的高度表示有多少场比赛打进了那么多球，有助于可视化常见的结果。
9. Common Mistakes When Working with Case Studies | 案例分析中的常见错误
Mistake 1: Rushing into calculations without understanding the context. Always read the scenario at least twice and highlight key numbers and units.
错误1：尚未理解情境就仓促计算。务必至少阅读两遍情境叙述，并标出关键数字和单位。
Mistake 2: Forgetting to check if an answer is sensible. A negative width or a cost of £0 per student signals an error. Use estimation to verify.
错误2：忘记检查答案是否合理。负数的宽度或学生人均0英镑的费用都表明有错误。使用估算来验证。
Mistake 3: Mixing units—using metres for area but centimetres for perimeter without conversion. Keep the same units throughout the calculation.
错误3：单位混用——例如面积用米而周长用厘米却未换算。整个计算过程中应保持单位一致。
Mistake 4: Overcomplicating. Often a case study tests basic operations combined creatively. Try a simpler version of the problem first.
错误4：过于复杂化。案例分析往往是通过创造性组合来考查基本运算。可以先尝试问题的一个简化版。
10. Practice Challenges and Tips | 实战练习与提示
Challenge 1: A cinema hire costs £150. Each ticket is sold for £7. If 50 tickets are sold, what is the profit? What is the minimum number of tickets to cover costs?
挑战1：一家电影院租金为150英镑。每张票售价7英镑。若售出50张票，利润是多少？至少需要售出多少张票才能覆盖成本？
Challenge 2: A rectangular lawn is twice as long as it is wide. Its perimeter is 36 m. Find the area. Hint: let width = w, length = 2w, and set up 2(w + 2w) = 36.
挑战2：一片矩形草坪的长是宽的两倍。其周长为36米。求面积。提示：设宽为 w，长为 2w，列出方程 2(w + 2w) = 36。
Tip: Always show your working clearly. In a case study, layout matters as much as the answer. Use tables and organised steps to make your reasoning easy to follow.
提示：始终清晰展示解题过程。在案例分析中，呈现方式与答案同样重要。使用表格和有组织的步骤，让你的推理过程易于理解。
Tip: Practise reverse‑checking. After finding garden dimensions, plug them back into the area formula; after computing an average, re‑add the numbers to ensure no slip. This builds exam confidence.
提示：练习反向检验。在求出花园尺寸后，回代入面积公式检验；计算出平均数后，重新加总数值确保无误。这能提高考试信心。
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