📚 Case Study Practice: Mastering Real-Life Maths | 案例分析实战演练
Mathematics is not just about abstract equations — it is a powerful tool for solving everyday problems. In this article, you will work through ten carefully designed case studies that mirror the kind of real-world scenarios found in CCEA Year 11 examinations. Each case challenges you to apply number, algebra, geometry, statistics, and probability skills in context. Use these examples to sharpen your problem-solving approach and build the confidence to tackle any applied question that comes your way.
数学不只是抽象的方程式，它是解决日常问题的强大工具。本文将通过十个精心设计的案例分析，带你演练CCEA十一年级考试中常见的现实场景。每个案例都将挑战你在实际情境中运用算术、代数、几何、统计和概率的技能。请利用这些例题磨炼解题思路，积累信心，从容面对各类应用题。
1. Budgeting for a School Trip | 学校旅行预算
A school is planning a day trip to a science museum. The total cost includes a fixed coach hire of £240 and an entry fee of £7.50 per student. The school wants to keep the total cost per student below £15. If the trip is only viable when at least 30 students sign up, determine the range of student numbers that makes the trip possible while staying within budget.
一所学校计划组织去科技馆的一日游。总费用包括240英镑的客车固定租赁费和每名学生7.50英镑的门票。学校希望将每名学生的人均总费用控制在15英镑以下。如果只有至少30名学生报名时行程才可行，请确定既能成行又不超出预算的学生人数范围。
Let the number of students be n 5n. The cost per student is (240 + 7.5n) ÷ n. We need this to be less than 15, and n ≥ 30.
设学生人数为 n。总费用为 240 + 7.5n。人均费用为 (240 + 7.5n) ÷ n。要求该值低于15，且 n ≥ 30。
n 32. Combined with n ≥ 30, the trip is possible for n > 32, meaning at least 33 students are required to keep the cost per student below £15.
列出不等式：(240 + 7.5n) / n < 15。两边同乘 n（正数）：240 + 7.5n < 15n → 240 < 7.5n → n > 32。因此 n 必须大于32。结合 n ≥ 30，当 n > 32 时旅行可行，即至少需要33名学生才能使人均费用低于15英镑。
2. Mobile Phone Plan Comparison | 手机套餐比较
Rachel wants a new mobile contract. Plan A charges a £20 monthly fee plus 5p per minute of calls. Plan B charges £12 per month plus 8p per minute. Over a month, she expects to use 350 minutes. Which plan is cheaper, and by how much? At what usage are both plans equal in cost?
雷切尔想签一份新手机合约。套餐A月租20英镑，通话每分钟5便士。套餐B月租12英镑，通话每分钟8便士。一个月里她预计使用350分钟。哪个套餐更便宜，便宜多少？通话时长达到多少时两个套餐的费用相同？
Cost for Plan A: 2000p + 5 × 350 = 2000 + 1750 = 3750p = £37.50. Plan B: 1200p + 8 × 350 = 1200 + 2800 = 4000p = £40.00. Plan A is cheaper by £2.50.
套餐A费用：2000 + 5 × 350 = 2000 + 1750 = 3750便士 = 37.50英镑。套餐B：1200 + 8 × 350 = 1200 + 2800 = 4000便士 = 40.00英镑。套餐A便宜2.50英镑。
nutes. 2000 + 5m = 1200 + 8m → 800 = 3m → m ≈ 266.67 minutes. At 267 minutes, Plan A costs £33.35 and Plan B costs £ if Rachel talks less than 267 minutes, Plan B is cheaper; above 267 minutes, Plan A wins.
为求平衡点，设通话时长为 m 分钟。2000 + 5m = 1200 + 8m → 800 = 3m → m ≈ 266.67分钟。在267分钟时，套餐A费用33.35英镑，套餐B费用33.36英镑，几乎相等。因此如果雷切尔通话少于267分钟，套餐B更便宜；超过267分钟，套餐A更划算。
3. Designing a Rectangular Garden | 矩形花园设计
A gardener has 30 metres of fencing to enclose a rectangular vegetable patch against an existing wall, meaning only three sides need fencing. What dimensions will give the maximum possible area, and what is that area?
一位园丁有30米长的围栏，要做成一个靠墙的矩形菜地，因此只需围三面。什么样的尺寸能得到最大面积？最大面积是多少？
The fencing used is 2x + y = 30, so y = 30 − 2x. Area A = x × y = x(30 − 2x) = 30x − 2x².
设垂直于墙的边长为 x 米（有两边），平行于墙的边长为 y 米。围栏总长 2x + y = 30，因此 y = 30 − 2x。面积 A = x × y = x(30 − 2x) = 30x − 2x²。
= −30 / (2 × −2) = 7.5 m. Then y = 30 − 2(7.5) = 15 m. Maximum area = 7.5 × 15 = 112.5 m².
这是一个二次方程。为求最大面积，可用配方法或对称轴公式 x = −b / (2a)。此处 a = −2，b = 30，得 x = −30 / (2 × −2) = 7.5 米。于是 y = 30 − 2(7.5) = 15 米。最大面积 = 7.5 × 15 = 112.5 平方米。
4. Survey Analysis and Pie Charts | 调查分析与饼图
200 students were asked about their favourite way to travel to school: walk, bus, cycle, or car. The results were: walk 60, bus 80, cycle 30, car 30. Draw a pie chart to represent this data. Calculate the angle for each sector and determine the proportion of students who use active travel (walk or cycle).
200名学生被问及他们最喜欢的上学方式：步行、公交、自行车或汽车。结果为：步行60人，公交80人，自行车30人，汽车30人。请绘制饼图表示数据。计算每个扇形的角度，并确定采用主动出行（步行或自行车）的学生比例。
(30 ÷ 200) × 360° = 54°. Car: (30 ÷ 200) × 360° = 54°. Check: 108 + 144 + 54 + 54 = 360°.
学生总数200。步行角度：(60 ÷ 200) × 360° = 108°。公交：(80 ÷ 200) × 360° = 144°。自行车：(30 ÷ 200) × 360° = 54°。汽车：(30 ÷ 200) × 360° = 54°。验证：108 + 144 + 54 + 54 = 360°。
ve travel (walk + cycle) = 60 + 30 = 90 students. Proportion = 90 / 200 = 9/20 = 0.45 = 45%. This means nearly half the students use a physically active method to get to school.
主动出行（步行+自行车）= 60 + 30 = 90人。比例 = 90 / 200 = 9/20 = 0.45 = 45%。这意味着接近一半的学生采用体力活动方式上学。
5. Probability in Board Games | 棋盘游戏中的概率
In a board game, a player rolls two fair six-sided dice and sums the scores. They need a total of 7 to land on a special square. What is the probability of rolling exactly 7? If the game also requires rolling a double (both dice the same) to unlock a bonus, what is the probability of rolling a double? Are the events 'sum of 7' and 'double' mutually exclusive?
在一款棋盘游戏中，玩家掷两个公平的六面骰子并求和。他们需要掷出7点才能落在一个特殊格子上。掷出恰好7点的概率是多少？如果游戏还要求掷出一对（两个骰子点数相同）来解锁奖励，掷出一对的概率是多少？事件“和为7”和“掷出一对”是否互斥？
likely outcomes. Combinations giving a sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 outcomes. So P(sum of 7) = 6/36 = 1/6.
共有 6 × 6 = 36 种等可能结果。和为7的组合：(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) — 共6种。因此 P(和为7) = 6/36 = 1/6。
Doubles outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) — um of 7 and a double cannot happen at the same time because all doubles give an even sum (2,4,6,8,10,12) and 7 is odd. Thus the events are mutually exclusive.
一对的结果：(1,1),(2,2),(3,3),(4,4),(5,5),(6,6) — 共6种。P(一对) = 6/36 = 1/6。和为7与一对不可能同时发生，因为所有一对的点数和都是偶数（2,4,6,8,10,12），而7是奇数。因此这两个事件互斥。
6. Speed, Distance and Time in a Journey | 旅程中的速度、距离与时间
A delivery van travels from Belfast to Derry, a distance of 112 km, at an average speed of 64 km/h. It then returns by the same route but encounters traffic, averaging only 48 km/h. Calculate the total time taken for the round trip and the overall average speed. Explain why the average speed is not simply the mean of 64 and 48.
一辆送货车从贝尔法斯特开往德里，距离112公里，平均时速64公里。随后沿原路返回，但遇到交通拥堵，平均时速只有48公里。计算往返总耗时和全程平均速度。解释为什么平均速度不是64和48的简单算术平均数。
). Return journey time: 112 ÷ 48 = 7/3 ≈ 2.333 hours (2h 20min). Total time = 1.75 + 2.333… = 4.0833 hours (4h 5min).
去程时间：距离 ÷ 速度 = 112 ÷ 64 = 1.75小时（1小时45分钟）。回程时间：112 ÷ 48 = 7/3 ≈ 2.333小时（2小时20分钟）。总时间 = 1.75 + 2.333… = 4.0833小时（4小时5分钟）。
1/48) = 112 × (3/192 + 4/192) = 112 × 7/192 = 784/192 = 49/12 hours. Average speed = 224 ÷ (49/12) = 224 × 12/49 = (224/49) × 12 = 32/7 × 12 = 384/7 ≈ 54.86 km/h. This is less than the arithmetic mean (56 km/h) because the slower return trip takes longer, giving it greater weight in the overall average.
总距离 = 112 × 2 = 224公里。全程平均速度 = 总距离 ÷ 总时间 = 224 ÷ (112/64 + 112/48)。精确计算：总时间 = 112(1/64 + 1/48) = 112 × (3/192 + 4/192) = 112 × 7/192 = 784/192 = 49/12小时。平均速度 = 224 ÷ (49/12) = 224 × 12/49 = (224/49) × 12 = 32/7 × 12 = 384/7 ≈ 54.86公里/小时。这低于算术平均数（56公里/小时），因为较慢的回程耗时更长，在整体平均中权重更大。
7. Best Buy in Supermarket Offers | 超市优惠中的最划算购买
Three supermarkets sell the same brand of pasta. Store X: 500 g pack for £1.10. Store Y: 750 g pack for £1.59 with a 'buy one get one half price' offer. Store Z: 1 kg pack for £2.00, but currently with 20% off. Which store offers the lowest price per 100 g for a shopper who wants to buy 1.5 kg of pasta? Remember to consider the actual amount you need to buy.
三家超市销售同品牌意大利面。X店：500克装1.10英镑。Y店：750克装1.59英镑，有“买一送一半价”优惠。Z店：1公斤装2.00英镑，目前打八折。对于想购买1.5公斤意大利面的顾客，哪家店每100克价格最低？记得要考虑实际需要购买的数量。
Calculate for each store. Store X: need 1.5 kg = 1500 g = three 500 g packs. Cost = 3 × £1.10 = £3.30. Price per 100g = £3.30 ÷ 15 = £0.22.
逐店计算。X店：需1.5千克 = 1500克 = 三包500克。费用 = 3 × 1.10 = 3.30英镑。每100克价格 = 3.30 ÷ 15 = 0.22英镑。
Store Y: 750 g pack. To get 1.5 kg, buy two packs. With 'buy one get one half price', cost = £1.59 + £0.795 = £2.385. Total mass = 1.5 kg. Price per 100g = 2.385 ÷ 15 ≈ £0.159.
Y店：750克装。要买1.5千克，需两包。有“买一送一半价”优惠，费用 = 1.59 + 0.795 = 2.385英镑。总质量 = 1.5千克。每100克价格 = 2.385 ÷ 15 ≈ 0.159英镑。
Store Z: 1 kg pack with 20% off. Discounted price = £2.00 × 0.8 = £ 500 g, but Store Z only sells 1 kg packs. So you must buy two 1 kg packs, total cost = 2 × £1.60 = £ If we consider cost for exactly needed mass, price per 100g for the used pasta = 3.20 ÷ 15? No, you used 1.5 kg out of 2 kg, cost per 100g based on purchase = 3.20 ÷ 20 = £0.16 for 2000g, but effective cost for 1500g = £3.20. This is £0.213 per 100g if you only consider what you eat. It's better to calculate the cost per 100g of the pasta you actually buy and use: £3.20 for 2 kg is £ implies you must buy enough to cover at least 1.5 kg. So Store Z: you buy 2 kg, total cost £3.20, unit cost per 100g = £0.16. So Store Y at £0.159/100g is cheapest.
Z店：1公斤装打八折。折扣价 = 2.00 × 0.8 = 1.60英镑/公斤。要买1.5千克，可能需要一包1公斤和另一包500克，但Z店只卖1公斤装。因此必须买两包1公斤，总费用 = 2 × 1.60 = 3.20英镑，得到2千克。如果你只考虑所需1.5千克，会有500克浪费。按购买量计算每100克价格 = 3.20 ÷ 20 = 0.16英镑；如果只考虑使用部分，有效成本是3.20英镑/1.5千克 ≈ 0.213英镑/100克。通常题目要求按实际购买量计算单价，所以Z店为0.16英镑/100克。比较：X店0.22英镑，Y店0.159英镑，Z店0.16英镑。因此Y店最便宜。
8. Compound Interest on Savings | 储蓄的复利计算
Aoife places £800 in a savings account that pays 3.5% compound interest per annum, added at the end of each year. She leaves the money untouched for 4 years. How much will she have at the end? What is the total interest earned? If she had instead chosen a simple interest account at 4% per year, which would have given a better return after 4 years?
奥伊芙将800英镑存入一个年利率3.5%的复利储蓄账户，每年年底计息。她不动用这笔钱，存期4年。到期时她将拥有多少钱？总利息是多少？如果她选择年利率4%的单利账户，4年后哪种方式回报更高？
Compound amount = Principal × (1 + rate)n se: 1.035² = 1.071225, 1.035⁴ = (1.071225)² = 1.147523 (approx). Amount = 800 × 1.147523 = £918.02. Interest = 918.02 − 800 = £118.02.
复利总额 = 本金 × (1 + 利率)n = 800 × (1 + 0.035)⁴。逐步计算：1.035² = 1.071225，1.035⁴ = (1.071225)² ≈ 1.147523。总额 = 800 × 1.147523 = 918.02英镑。利息 = 918.02 − 800 = 118.02英镑。
Simple interest: total interest = Principal × rate × time = 800 × 0.04 × 4 = £128. Final amount = £928. The simple interest account gives £928, which is better than the compound account (£ s case because the simple rate is higher. Always compare both rate and compounding method.
单利：总利息 = 本金 × 利率 × 时间 = 800 × 0.04 × 4 = 128英镑。最终金额 = 928英镑。该单利账户给出928英镑，比复利账户的918.02英镑更好，因为单利利率更高。始终要同时比较利率和计息方式。
9. VAT and Discounts in a Sale | 销售中的增值税与折扣
A laptop is priced at £540 excluding VAT. VAT is charged at 20%. During a sale, the store offers a 15% discount on the VAT-inclusive price. What is the final sale price? If the store instead applied the discount first and then added VAT, would the final price be the same? Show calculations.
一台笔记本电脑不含增值税的标价为540英镑，增值税率为20%。在一场促销中，商店对含税价提供15%的折扣。最终售价是多少？如果商店先打折再加增值税，最终价格会一样吗？请计算说明。
First method: Price including VAT = 540 × 1.20 = £648. Discount 15%: sale price = 648 × (1 − 0.15) = 648 × 0.85 = £550.80.
方法一：含税价 = 540 × 1.20 = 648英镑。折扣15%：售价 = 648 × 0.85 = 550.80英镑。
Second method: Discount applied to pre-VAT price: discounted price excl VAT = 540 × 0.85 = £459. Then add VAT: 459 × 1.20 = £ se multiplication is commutative, (540 × 1.20) × 0.85 = 540 × (1.20 × 0.85) = 540 × 1.02 = £ . However, in real life, stores usually discount the VAT-inclusive price, and the VAT fraction is applied to the discounted price, so the customer pays VAT on the discounted amount only, which is exactly the same as in method two.
方法二：先打折再加增值税：不含税折扣价 = 540 × 0.85 = 459英镑；含税 = 459 × 1.20 = 550.80英镑。有趣的是，因为乘法满足交换律，(540 × 1.20) × 0.85 = 540 × (1.20 × 0.85) = 540 × 1.02 = 550.80英镑。顺序不影响结果。前提是折扣百分比作用于同一税基。但在实际操作中，商店通常对含税价打折，相当于顾客在折后金额上支付增值税，结果相同。
10. Interpreting Graphs from a Science Experiment | 科学实验图表解读
nds. The data is plotted on a graph with time (minutes) on the x-axis and temperature (°C) on the y-axis. The graph shows a straight line from (0, 20) to (5, 85). What is the rate of temperature increase per minute? quation to predict the temperature after 7.5 minutes, assuming the linear trend continues.
一名学生加热烧杯中的水，每30秒记录一次温度。数据绘制成图，横轴为时间（分钟），纵轴为温度（摄氏度）。图中显示一条从 (0, 20) 到 (5, 85) 的直线。每分钟的升温速率是多少？以 T = a × t + b 的形式写出直线方程。用方程预测7.5分钟时的温度，假设线性趋势持续。
ient a = (85 − 20) ÷ (5 − 0) = 65 ÷ 5 = 13 °C per minute. So the equation is T = 13t + 20.
直线过点 (0,20) 和 (5,85)。y截距 b = 20。斜率 a = (85 − 20) ÷ (5 − 0) = 65 ÷ 5 = 13 °C/分钟。因此方程为 T = 13t + 20。
at 100 °C under normal pressure, so the linear model would break down beyond the boiling point. This reminds us that mathematical models have their limits — even perfect lines must be interpreted within a real-life context.
当 t = 7.5 分钟时，T = 13 × 7.5 + 20 = 97.5 + 20 = 117.5 °C。但常压下水的沸点为100 °C，因此线性模型在超过沸点后不再适用。这提醒我们数学模型有其局限性——即便是完美的直线，也必须结合现实背景加以解读。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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