📚 Case Study Practical Exercises | 案例分析实战演练
In Year 9 CAIE Mathematics, students often encounter case study questions that combine multiple mathematical skills into a single real-world scenario. These questions test not only your ability to perform calculations but also your skill in interpreting data, identifying relevant information, and communicating your reasoning clearly. This article presents a series of practical case studies designed to mirror the style and complexity of CAIE examination questions, covering algebra, geometry, statistics, ratio, and more. Each case study breaks down the problem-solving process step by step, helping you build confidence for your assessments.
在九年级 CAIE 数学课程中，学生经常会遇到将多种数学技能融合在单一真实场景中的案例分析题。这些问题不仅考察你的计算能力，还考察你解读数据、识别相关信息以及清晰表达推理过程的能力。本文呈现了一系列实践性案例研究，旨在反映 CAIE 考试题目的风格和复杂度，涵盖代数、几何、统计、比例等内容。每个案例逐步分解解题过程，帮助你为评估建立信心。
1. The Garden Fence Problem | 花园围栏问题
A homeowner wants to build a rectangular garden against an existing wall, so only three sides need fencing. The total length of fencing available is 24 metres. The garden’s length (parallel to the wall) is twice its width. Calculate the dimensions of the garden and the total area that can be planted.
一位房主想靠着现有墙壁建一个矩形花园，因此只需要围三面。可用的围栏总长度为 24 米。花园的长度（平行于墙壁）是其宽度的两倍。请计算花园的尺寸和可种植的总面积。
Let the width of the garden be w metres. The length is 2w metres. The fencing covers two widths and one length, so we write the equation: 2w + 2w = 24, which simplifies to 4w = 24, giving w = 6. The width is 6 metres and the length is 12 metres. The area is length × width = 12 × 6 = 72 m². This case demonstrates how algebraic equations can model physical constraints.
设花园的宽度为 w 米，则长度为 2w 米。围栏覆盖两个宽度和一个长度，因此我们写出方程：2w + 2w = 24，化简得 4w = 24，得 w = 6。宽度为 6 米，长度为 12 米。面积为长 × 宽 = 12 × 6 = 72 m²。这个案例展示了代数方程如何模拟物理约束。
Extension question: If the homeowner instead used all 24 metres for four sides (no existing wall), what would be the maximum possible area? Investigate how the area changes with different length-to-width ratios.
扩展问题：如果房主将全部 24 米用于四面围栏（没有现有墙壁），最大可能面积是多少？探究面积如何随不同的长宽比变化。
2. Designing a Pizza Box | 设计披萨盒
A pizza company wants to design a square box for a circular pizza of diameter 30 cm. The box must have a 2 cm clearance on all sides of the pizza for packaging material. Calculate the side length of the square base and the area of cardboard needed for the base. Then find the percentage of the base area that is not covered by the pizza.
一家披萨公司想为一个直径 30 厘米的圆形披萨设计一个方形盒子。盒子必须在披萨四周留有 2 厘米的间隙用于包装材料。计算方形底面的边长和所需的纸板面积。然后求底面未被披萨覆盖的面积百分比。
The pizza diameter is 30 cm, so the minimum internal width of the box is 30 + 2 + 2 = 34 cm. The base is square, so the side length is 34 cm. The area of the square base is 34² = 1156 cm². The pizza area is π × (15)² = π × 225 ≈ 706.86 cm². The uncovered area is 1156 − 706.86 = 449.14 cm². The percentage uncovered is (449.14 ÷ 1156) × 100% ≈ 38.85%. This problem integrates geometry with percentage calculations.
披萨直径为 30 厘米，因此盒子的最小内部宽度为 30 + 2 + 2 = 34 厘米。底面为正方形，因此边长为 34 厘米。方形底面的面积为 34² = 1156 平方厘米。披萨面积为 π × (15)² = π × 225 ≈ 706.86 平方厘米。未覆盖面积为 1156 − 706.86 = 449.14 平方厘米。未覆盖百分比为 (449.14 ÷ 1156) × 100% ≈ 38.85%。此问题将几何与百分比计算相结合。
Further exploration: If the box height is 5 cm, calculate the total surface area of cardboard required, including a lid. Assume no overlapping flaps for simplicity.
进一步探索：如果盒子的高度为 5 厘米，计算所需纸板的总表面积，包括盖子。为简化起见，假设没有重叠的翻盖。
3. School Sports Day Results | 学校运动会成绩
During a school sports day, the times (in seconds) for the 100-metre sprint were recorded for eight students: 13.2, 14.1, 12.8, 15.3, 13.9, 12.5, 14.6, and 13.5. Calculate the mean, median, and range of these times. Determine which measure of central tendency best represents the typical performance and explain why.
在学校运动会上，记录了八名学生 100 米短跑的时间（以秒为单位）：13.2、14.1、12.8、15.3、13.9、12.5、14.6 和 13.5。计算这些时间的平均数、中位数和极差。判断哪种集中趋势度量最能代表典型成绩，并解释原因。
First, arrange the times in ascending order: 12.5, 12.8, 13.2, 13.5, 13.9, 14.1, 14.6, 15.3. The mean is the sum (109.9) divided by 8, which equals 13.7375 seconds. The median is the average of the 4th and 5th values: (13.5 + 13.9) ÷ 2 = 13.7 seconds. The range is 15.3 − 12.5 = 2.8 seconds. The median is a better measure here because the time 15.3 is somewhat of an outlier and pulls the mean upward. The median gives a more robust indication of typical performance.
首先，将成绩按升序排列：12.5、12.8、13.2、13.5、13.9、14.1、14.6、15.3。平均数为总和 (109.9) 除以 8，等于 13.7375 秒。中位数为第 4 和第 5 个值的平均数：(13.5 + 13.9) ÷ 2 = 13.7 秒。极差为 15.3 − 12.5 = 2.8 秒。在这里中位数是更好的度量，因为 15.3 秒在某种程度上是一个异常值，拉高了平均数。中位数更能稳健地反映典型成绩。
Data visualisation: Construct a box-and-whisker plot for this dataset. Identify the lower quartile, median, and upper quartile, and discuss what the interquartile range tells you about the spread of the central 50% of the data.
数据可视化：为这组数据绘制箱线图。识别下四分位数、中位数和上四分位数，并讨论四分位距告诉你关于中间 50% 数据分布的信息。
4. Recipe Scaling Scenario | 食谱缩放场景
A recipe for 6 people requires 300 g of flour, 200 g of sugar, 150 g of butter, and 3 eggs. You need to adapt this recipe for 10 people. Calculate the required quantities of each ingredient. If butter is sold in 250 g blocks and eggs are sold in boxes of 6, how many blocks of butter and boxes of eggs must you purchase?
一份供 6 人食用的食谱需要 300 克面粉、200 克糖、150 克黄油和 3 个鸡蛋。你需要将此食谱调整为 10 人份。计算每种原料所需的量。如果黄油以 250 克块装出售，鸡蛋以 6 个盒装出售，你必须购买多少块黄油和多少盒鸡蛋？
The scaling factor is 10 ÷ 6 = 5/3 ≈ 1.667. Multiply each quantity: flour = 300 × (5/3) = 500 g; sugar = 200 × (5/3) ≈ 333.3 g; butter = 150 × (5/3) = 250 g; eggs = 3 × (5/3) = 5 eggs. You need exactly one 250 g block of butter. Since eggs come in boxes of 6, you must buy one full box, leaving one spare egg. This case study involves direct proportion and practical rounding considerations.
缩放因子为 10 ÷ 6 = 5/3 ≈ 1.667。将每种量相乘：面粉 = 300 × (5/3) = 500 克；糖 = 200 × (5/3) ≈ 333.3 克；黄油 = 150 × (5/3) = 250 克；鸡蛋 = 3 × (5/3) = 5 个鸡蛋。你恰好需要一块 250 克的黄油。由于鸡蛋按 6 个盒装出售，你必须购买一整盒，剩余一个鸡蛋。本案例涉及正比例和实际取整的考虑。
Cost analysis: If flour costs £0.80 per 500 g, sugar costs £1.20 per kg, butter costs £1.90 per 250 g block, and eggs cost £2.40 per box, calculate the total cost of ingredients for the adapted recipe. Determine the cost per person.
成本分析：如果面粉每 500 克 0.80 英镑，糖每公斤 1.20 英镑，黄油每 250 克块 1.90 英镑，鸡蛋每盒 2.40 英镑，计算调整后食谱的原料总成本。并确定每人的成本。
5. The Carnival Game | 嘉年华游戏
At a school carnival, a game involves spinning a fair spinner divided into 5 equal sectors numbered 1 to 5, and simultaneously drawing a marble from a bag containing 3 red marbles and 2 blue marbles. A player wins a prize if the spinner lands on an even number AND a red marble is drawn. Calculate the probability of winning. Determine whether the game favours the organisers if a prize costs £1.50 and the entry fee is £0.50 per play.
在学校的嘉年华上，有一个游戏需要旋转一个被分为 5 个相等扇区（标有数字 1 至 5）的公平转盘，同时从装有 3 个红色弹珠和 2 个蓝色弹珠的袋子中抽出一个弹珠。如果转盘停在偶数上并且抽到红色弹珠，玩家就赢得奖品。计算获胜的概率。如果奖品成本为 1.50 英镑，每次游戏入场费为 0.50 英镑，判断游戏是否对组织者有利。
The probability of spinning an even number (2 or 4) is 2/5. The probability of drawing a red marble is 3/5. The events are independent, so the probability of both occurring (winning) is (2/5) × (3/5) = 6/25 = 0.24. The expected payout per play is 0.24 × £1.50 = £0.36. Since the entry fee is £0.50, the organisers expect to gain £0.50 − £0.36 = £0.14 per play on average. The game indeed favours the organisers over many trials.
转出偶数（2 或 4）的概率为 2/5。抽出红色弹珠的概率为 3/5。这两个事件是独立的，因此两者都发生（获胜）的概率为 (2/5) × (3/5) = 6/25 = 0.24。每次游戏的预期支出为 0.24 × 1.50 英镑 = 0.36 英镑。由于入场费为 0.50 英镑，组织者平均每次游戏预期获利 0.50 − 0.36 = 0.14 英镑。经过多次试验，游戏确实对组织者有利。
Extension: If the organisers want the expected profit per play to be exactly £0.20, what should they charge as the entry fee, keeping the prize value the same? Set up and solve an equation.
扩展：如果组织者希望每次游戏的预期利润恰好为 0.20 英镑，在奖金价值不变的情况下，他们应该收取多少入场费？建立并求解方程。
6. Mobile Phone Plan Comparison | 手机套餐比较
Two mobile phone companies offer the following monthly plans. Plan A: a fixed charge of £15 plus £0.08 per minute of calls. Plan B: a fixed charge of £10 plus £0.12 per minute of calls. Determine the number of minutes for which the total monthly cost is the same for both plans. For a customer who makes 200 minutes of calls per month, which plan is cheaper, and by how much?
两家手机公司提供以下月度套餐。A 套餐：固定费用 15 英镑加上每分钟通话费 0.08 英镑。B 套餐：固定费用 10 英镑加上每分钟通话费 0.12 英镑。确定使得两种套餐月度总费用相同的通话分钟数。对于每月通话 200 分钟的客户，哪种套餐更便宜，便宜多少？
Let m be the number of minutes. Cost A = 15 + 0.08m; Cost B = 10 + 0.12m. Set them equal: 15 + 0.08m = 10 + 0.12m, subtract 10 from both sides: 5 + 0.08m = 0.12m, then 5 = 0.04m, so m = 125 minutes. For 200 minutes: Plan A costs 15 + 0.08(200) = £31; Plan B costs 10 + 0.12(200) = £34. Plan A is cheaper by £3. This problem uses linear equations to model real-world choices.
设通话分钟数为 m。A 套餐费用 = 15 + 0.08m；B 套餐费用 = 10 + 0.12m。令两者相等：15 + 0.08m = 10 + 0.12m，两边减去 10：5 + 0.08m = 0.12m，然后 5 = 0.04m，因此 m = 125 分钟。对于 200 分钟：A 套餐费用为 15 + 0.08(200) = 31 英镑；B 套餐费用为 10 + 0.12(200) = 34 英镑。A 套餐便宜 3 英镑。此问题使用线性方程模拟现实世界中的选择。
Graphical interpretation: Plot both cost equations on a graph with minutes on the horizontal axis and cost on the vertical axis. Identify the break-even point and shade the regions where each plan is more economical.
图形解释：在一张图上绘制两个费用方程，横轴为分钟数，纵轴为费用。识别盈亏平衡点，并标出每种套餐更经济的区域。
7. The Family Road Trip | 家庭公路旅行
A family departs from home at 09:00 and drives at an average speed of 80 km/h for 1.5 hours. They then stop for a 30-minute rest. After the break, they continue at an average speed of 60 km/h for another 2 hours. Draw a distance–time graph representing this journey. Calculate the total distance travelled and the average speed for the entire trip (including the rest period).
一个家庭早上 9:00 从家出发，以 80 公里/小时的平均速度行驶了 1.5 小时。然后他们停下来休息 30 分钟。休息后，他们以 60 公里/小时的平均速度继续行驶了 2 小时。绘制表示这次旅程的距离–时间图。计算总行驶距离和整个行程（包括休息时间）的平均速度。
First leg: distance = 80 × 1.5 = 120 km, arriving at 10:30. Rest from 10:30 to 11:00 (horizontal line on the graph). Second leg: distance = 60 × 2 = 120 km, arriving at 13:00. Total distance = 240 km. Total time from 09:00 to 13:00 = 4 hours. Average speed = total distance ÷ total time = 240 ÷ 4 = 60 km/h. The graph should show two sloping segments with a flat section in between, clearly labelled with times and distances.
第一段：距离 = 80 × 1.5 = 120 公里，于 10:30 到达。休息时间为 10:30 至 11:00（图中为水平线段）。第二段：距离 = 60 × 2 = 120 公里，于 13:00 到达。总距离 = 240 公里。从 9:00 到 13:00 的总时间 = 4 小时。平均速度 = 总距离 ÷ 总时间 = 240 ÷ 4 = 60 公里/小时。图表应显示两条倾斜线段中间夹着一段平坦部分，并清晰标注时间和距离。
Interpretation question: How would the graph differ if the family encountered heavy traffic during the second leg, reducing their speed to 40 km/h? What would be the new arrival time and average speed?
解读问题：如果家庭在第二段行程中遇到交通拥堵，速度降至 40 公里/小时，图形会有什么不同？新的到达时间和平均速度是多少？
8. The Theatre Seating Plan | 剧院座位安排
A theatre has rows of seats arranged in an arithmetic sequence. The first row has 20 seats. Each subsequent row has 3 more seats than the previous row. There are 15 rows in total. Find the number of seats in the 10th row, the total number of seats in the theatre, and the row number that has 50 seats (if it exists).
一家剧院的座位排数按等差数列排列。第一排有 20 个座位。每一后续排比前一排多 3 个座位。总共有 15 排。求第 10 排的座位数、剧院的总座位数，以及拥有 50 个座位的排数（如果存在的话）。
The nth term of an arithmetic sequence is given by aₙ = a₁ + (n − 1)d, where a₁ = 20 and d = 3. For the 10th row: a₁₀ = 20 + (10 − 1) × 3 = 20 + 27 = 47 seats. The total number of seats Sₙ = n/2 × (a₁ + aₙ). First find the 15th term: a₁₅ = 20 + 14 × 3 = 62. S₁₅ = 15/2 × (20 + 62) = 7.5 × 82 = 615 seats. To find the row with 50 seats: solve 20 + (n − 1) × 3 = 50, which gives (n − 1) × 3 = 30, n − 1 = 10, n = 11. The 11th row has exactly 50 seats.
等差数列的第 n 项公式为 aₙ = a₁ + (n − 1)d，其中 a₁ = 20，d = 3。对于第 10 排：a₁₀ = 20 + (10 − 1) × 3 = 20 + 27 = 47 个座位。总座位数 Sₙ = n/2 × (a₁ + aₙ)。首先求第 15 项：a₁₅ = 20 + 14 × 3 = 62。S₁₅ = 15/2 × (20 + 62) = 7.5 × 82 = 615 个座位。求有 50 个座位的排数：解 20 + (n − 1) × 3 = 50，得 (n − 1) × 3 = 30，n − 1 = 10，n = 11。第 11 排恰好有 50 个座位。
Contextual analysis: Why might theatre designers use arithmetic sequences for seating? Discuss how this arrangement affects sightlines and the audience experience from different rows.
情境分析：为什么剧院设计师可能使用等差数列来安排座位？讨论这种安排如何影响不同排的视线和观众体验。
9. The Age Puzzle | 年龄谜题
Three years ago, a father was four times as old as his son. In five years’ time, the father will be three times as old as his son. Determine the current ages of the father and the son using simultaneous equations. Verify that your solution satisfies both conditions.
三年前，一位父亲的年龄是他儿子的四倍。五年后，父亲的年龄将是儿子的三倍。使用联立方程确定父亲和儿子当前的年龄。验证你的解满足两个条件。
Let the father’s current age be F and the son’s current age be S. First condition: F − 3 = 4(S − 3), which simplifies to F − 3 = 4S − 12, giving F = 4S − 9. Second condition: F + 5 = 3(S + 5), which simplifies to F + 5 = 3S + 15, giving F = 3S + 10. Equate the two expressions for F: 4S − 9 = 3S + 10, subtract 3S from both sides: S − 9 = 10, so S = 19. Substitute back: F = 3(19) + 10 = 67. The father is currently 67 years old and the son is 19. Verification: three years ago, father was 64 and son was 16 (64 = 4 × 16). In five years, father will be 72 and son 24 (72 = 3 × 24).
设父亲当前的年龄为 F，儿子当前的年龄为 S。第一个条件：F − 3 = 4(S − 3)，化简为 F − 3 = 4S − 12，得 F = 4S − 9。第二个条件：F + 5 = 3(S + 5)，化简为 F + 5 = 3S + 15，得 F = 3S + 10。将两个 F 的表达式等同：4S − 9 = 3S + 10，两边减去 3S：S − 9 = 10，因此 S = 19。代回：F = 3(19) + 10 = 67。父亲目前 67 岁，儿子 19 岁。验证：三年前，父亲 64 岁，儿子 16 岁（64 = 4 × 16）。五年后，父亲 72 岁，儿子 24 岁（72 = 3 × 24）。
Alternative method: Solve this problem using a single variable by letting the son’s age three years ago be x. Compare this approach with the simultaneous equations method and discuss the efficiency of each.
替代方法：通过设儿子三年前的年龄为 x，使用单个变量解决此问题。将此方法与联立方程方法进行比较，并讨论每种方法的效率。
10. The School Fair Project | 学校义卖项目
Students plan to sell handmade badges at the school fair. The materials cost £0.35 per badge, and the stall rental fee is £12. They intend to sell each badge for £1.20. Calculate the number of badges they must sell to break even. If they sell 80 badges, determine the profit made. Also find the selling price per badge needed to break even at exactly 50 badges sold, assuming costs remain unchanged.
学生们计划在学校义卖上出售手工徽章。每个徽章的材料成本为 0.35 英镑，摊位租赁费为 12 英镑。他们打算每个徽章卖 1.20 英镑。计算他们必须卖出多少个徽章才能达到盈亏平衡。如果卖出 80 个徽章，求获得的利润。同时，假设成本不变，求出在恰好售出 50 个徽章时达到盈亏平衡所需的每个徽章售价。
Let n be the number of badges. Total cost = 12 + 0.35n. Total revenue = 1.20n. Break-even occurs when revenue equals cost: 1.20n = 12 + 0.35n, subtract 0.35n: 0.85n = 12, n = 12 ÷ 0.85 ≈ 14.12, so they must sell at least 15 badges (since badges are whole units). For 80 badges: revenue = 1.20 × 80 = £96; cost = 12 + 0.35 × 80 = £40; profit = 96 − 40 = £56. To break even at 50 badges, let the selling price be p: 50p = 12 + 0.35 × 50, 50p = 12 + 17.50 = 29.50, so p = 29.50 ÷ 50 = £0.59.
设徽章数量为 n。总成本 = 12 + 0.35n。总收入 = 1.20n。当收入等于成本时达到盈亏平衡：1.20n = 12 + 0.35n，减去 0.35n：0.85n = 12，n = 12 ÷ 0.85 ≈ 14.12，因此他们必须至少卖出 15 个徽章（因为徽章是整数单位）。对于 80 个徽章：收入 = 1.20 × 80 = 96 英镑；成本 = 12 + 0.35 × 80 = 40 英镑；利润 = 96 − 40 = 56 英镑。要在 50 个徽章时达到盈亏平衡，设售价为 p：50p = 12 + 0.35 × 50，50p = 12 + 17.50 = 29.50，因此 p = 29.50 ÷ 50 = 0.59 英镑。
Reflective discussion: Identify any assumptions made in this model. What factors in a real school fair might affect the accuracy of these calculations? How could the students use this analysis to make informed decisions about pricing and production?
反思讨论：识别此模型中所做的任何假设。在真实的学校义卖中，哪些因素可能影响这些计算的准确性？学生如何利用此分析做出关于定价和生产的明智决策？
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply