📚 Year 11 Edexcel Maths: Case Study Practical Exercises | Year 11 Edexcel 数学：案例分析实战演练
In this article, we explore a series of real-world case studies that require you to apply the mathematical skills covered in the Year 11 Edexcel GCSE syllabus. Each scenario is designed to mirror exam-style problem-solving questions, helping you to build confidence in using numbers, algebra, geometry, statistics, and probability in practical contexts. Follow each worked example carefully and practise the methods on your own.
本文通过一系列真实案例，带您演练 Year 11 Edexcel GCSE 大纲中的数学技能。每个场景都模拟了考试风格的应用题，帮助您在实践情境中熟练使用数字、代数、几何、统计和概率。请仔细学习每个示例，并独立练习这些方法。
1. Home Budget Planning | 家庭预算规划
A family has a monthly income of £3200. Their fixed expenses are: rent £950, food £480, utility bills £210, transport £170, and entertainment £300. They aim to save at least £800 each month. We need to determine whether they can meet this savings goal and find what percentage of income remains after spending.
一个家庭月收入为 £3200。固定支出为：房租 £950，伙食 £480，水电燃气 £210，交通 £170，娱乐 £300。他们希望每月至少储蓄 £800。我们需要判断能否达成储蓄目标，并计算支出后剩余收入占原收入的百分比。
First, calculate the total monthly expenditure by adding all costs: £950 + £480 + £210 + £170 + £300 = £2110.
首先，计算月度总支出：£950 + £480 + £210 + £170 + £300 = £2110。
Subtract this total from the income to find the remaining amount: £3200 − £2110 = £1090.
从收入中减去总支出，得到剩余金额：£3200 − £2110 = £1090。
The remaining £1090 is greater than the savings target of £800, so the goal is achievable.
剩余 £1090 大于储蓄目标 £800，因此可以达成目标。
To express the savings as a percentage of income: (1090 ÷ 3200) × 100 ≈ 34.1% (1 d.p.).
将储蓄表示为收入的百分比：(1090 ÷ 3200) × 100 ≈ 34.1%（保留一位小数）。
This case tests addition, subtraction, and percentage calculations — all core number skills for the Edexcel exam.
这个案例检验了加法、减法和百分比计算——这些都是 Edexcel 考试的核心数字技能。
2. Kitchen Renovation: Area and Cost | 厨房翻新：面积与成本
A rectangular kitchen floor measures 4.2 m by 3.5 m. The owner plans to cover it with square tiles of side length 0.3 m. The tiles are sold in boxes of 10, costing £22.50 per box. To allow for cutting and breakage, an extra 5% of tiles is added. Find the total cost of the tiles.
一个矩形厨房地板长 4.2 m，宽 3.5 m。业主计划用边长 0.3 m 的方形瓷砖铺满地面。瓷砖每盒 10 块，每盒 £22.50。考虑到切割和破损，要增加 5% 的备用瓷砖。求购买瓷砖的总成本。
Calculate the floor area: 4.2 m × 3.5 m = 14.7 m².
计算地板面积：4.2 m × 3.5 m = 14.7 m²。
Calculate the area of one tile: 0.3 m × 0.3 m = 0.09 m².
计算单块瓷砖面积：0.3 m × 0.3 m = 0.09 m²。
Number of tiles needed without wastage: 14.7 ÷ 0.09 ≈ 163.33…, so at least 164 tiles. With 5% extra: 163.33… × 1.05 ≈ 171.5, so round up to 172 tiles.
不考虑损耗时需要瓷砖数：14.7 ÷ 0.09 ≈ 163.33…，至少 164 块。增加 5% 备用：163.33… × 1.05 ≈ 171.5，向上取整为 172 块。
Number of boxes required: 172 ÷ 10 = 17.2, so 18 boxes must be purchased.
需要购买的盒数：172 ÷ 10 = 17.2，因此必须购买 18 盒。
Total cost: 18 boxes × £22.50 = £405.00.
总成本：18 盒 × £22.50 = £405.00。
This problem combines area, unit conversions, and percentage increases — common in Edexcel compound measure questions.
这个问题结合了面积、单位换算和百分比增加——常见于 Edexcel 复合测量题。
3. Travel Planning: Distance, Speed and Time | 旅行计划：距离、速度和时间
trip from London to Edinburgh, a distance of 400 miles. She expects an average speed of 50 he leaves at 08:00. Also, her car uses fuel at a rate of 40 miles per gallon. Petrol costs £1.65 per litre. Given that 1 gallon = 4.546 litres, calculate the arrival time and the fuel cost.
一位司机计划从伦敦开车到爱丁堡，距离 400 英里。她预计平均时速 50 mph，中途休息 45 分钟。她早上 8:00 出发。她的汽车油耗为 40 英里/加仑，汽油价格为 £1.65 每升。已知 1 加仑 = 4.546 升，计算到达时间和油费。
Driving time = distance ÷ speed = 400 ÷ 50 = 8 hours. Total time including break = 8 h + 0.75 h = 8.75 hours.
驾驶时间 = 距离 ÷ 速度 = 400 ÷ 50 = 8 小时。总时长（含休息）= 8 h + 0.75 h = 8.75 小时。
Arrival time: 08:00 + 8 hours 45 minutes = 16:45.
到达时间：08:00 + 8 小时 45 分 = 16:45。
Fuel needed: 400 miles ÷ 40 miles per gallon = 10 gallons. In litres: 10 × 4.546 = 45.46 litres.
所需燃油：400 英里 ÷ 40 英里/加仑 = 10 加仑。换算为升：10 × 4.546 = 45.46 升。
Petrol cost: 45.46 litres × £1.65 = £75.01 (to the nearest penny).
油费：45.46 升 × £1.65 = £75.01（精确到便士）。
This case practises speed-distance-time relationships and unit conversions, which are fundamental in the Edexcel exam.
这个案例练习了速度-距离-时间关系及单位换算，是 Edexcel 考试的基础内容。
4. Savings and Compound Interest | 储蓄与复利
A student invests £2000 in a savings account that pays 3.5% compound interest per year. She leaves the money untouched for 5 years. Compare the final amount with what she would have earned under simple interest.
一名学生将 £2000 投入年复利 3.5% 的储蓄账户，5 年不动用。比较最终金额与单利下的收益。
The compound interest formula is:
复利公式为：
A = P(1 + r/100)n
where P = £2000, r = 3.5, n = 5.
其中 P = £2000, r = 3.5, n = 5。
pute stepwise: 1.035² = 1.071225; 1.035³ ≈ 1.108718; 1.035⁴ ≈ 1.147523; 1.035⁵ ≈ 1.187686.
计算括号内：1 + 3.5/100 = 1.035。五次方：1.035⁵。逐步计算：1.035² = 1.071225；1.035³ ≈ 1.108718；1.035⁴ ≈ 1.147523；1.035⁵ ≈ 1.187686。
Final amount A = £2000 × 1.187686 = £2375.37 (to 2 d.p.).
最终金额 A = £2000 × 1.187686 = £2375.37（保留两位小数）。
With simple interest, total interest = £2000 × 0.035 × 5 = £350, giving a total of £2350. Compound interest yields £25.37 more.
单利计算，总利息 = £2000 × 0.035 × 5 = £350，总额为 £2350。复利多出 £25.37。
This illustrates the power of compound growth and is a typical Edexcel percentage/proportion problem.
这展示了复利的增长力量，是 Edexcel 典型的百分比/比例问题。
5. Probability in Games | 游戏中的概率
A spinner at a funfair has 8 equal sections: 2 red, 3 blue and 3 green. A prize is won if the spinner lands on red. Calculate the probability of winning on one spin, and the probability of winning exactly once in two independent spins.
游乐园的转盘有 8 个等份：2 个红色，3 个蓝色，3 个绿色。转到红色即可获奖。计算转一次获奖的概率，以及独立转两次恰好获奖一次的概率。
Probability of red on one spin: 2 ÷ 8 = 1/4 = 0.25.
一次转到红色的概率：2 ÷ 8 = 1/4 = 0.25。
For two spins, use a probability tree or the binomial situation. Winning exactly once can happen as (win, lose) or (lose, win).
对于两次旋转，可使用概率树或二项情境。恰好一次获奖可以是（中奖，未中）或（未中，中奖）。
P(win then lose) = (1/4) × (3/4) = 3/16. P(lose then win) = (3/4) × (1/4) = 3/16.
P（中然后不中）= (1/4) × (3/4) = 3/16。P（不中然后中）= (3/4) × (1/4) = 3/16。
Total probability = 3/16 + 3/16 = 6/16 = 3/8 = 0.375.
总概率 = 3/16 + 3/16 = 6/16 = 3/8 = 0.375。
This problem reinforces the 'and' and 'or' rules in probability, essential for Edexcel Paper 2.
这道题巩固了概率中的“与”和“或”规则，是 Edexcel 卷二的重要内容。
6. Analysing Survey Data | 调查数据分析
Forty students were asked how many hours they spend on social media per day. The results are grouped:
40 名学生被问及每天花在社交媒体上的小时数，结果按组分类：
| Hours (h) | Frequency |
| 0 ≤ h < 1 | 6 |
| 1 ≤ h < 2 | 10 |
| 2 ≤ h < 3 | 12 |
| 3 ≤ h < 4 | 7 |
| 4 ≤ h ≤ 5 | 5 |
Calculate an estimate of the mean, the modal class, and the class containing the median.
计算平均值的估计值、众数所在组以及包含中位数的组。
d sum: (0.5×6) + (1.5×10) + (2.5×12) + (3.5×7) + (4.5×5) = 3 + 15 + 30 + 24.5 + 22.5 = 95.
取每个区间的中点：0.5, 1.5, 2.5, 3.5, 4.5。乘以频数并求和：(0.5×6) + (1.5×10) + (2.5×12) + (3.5×7) + (4.5×5) = 3 + 15 + 30 + 24.5 + 22.5 = 95。
Estimated mean = 95 ÷ 40 = 2.375 hours.
均值估计 = 95 ÷ 40 = 2.375 小时。
The modal class is 2 ≤ h < 3 with frequency 12.
众数所在组为 2 ≤ h < 3，频数为 12。
e median lies in the class where cumulative frequency first exceeds 20; that is 2 ≤ h < 3.
累计频数：6, 16, 28, 35, 40。中位数落在累计频数首次超过 20 的组，即 2 ≤ h < 3。
This task addresses grouped frequency statistics, a key GCSE skill.
这道题针对分组频率统计，是 GCSE 的关键技能。
7. Algebraic Modelling for Mobile Plans | 手机套餐的代数建模
Two mobile network providers offer monthly plans: Provider A charges a £10 fixed fee plus £0.05 per minute of calls. Provider B charges £15 fixed fee plus £0.03 per minute. Determine the number of minutes for which both plans cost the same, and advise which plan is cheaper for light users.
两家移动网络提供商的月度套餐为：A 收取 £10 固定月租加上每分钟 £0.05 的通话费。B 收取 £15 固定月租加上每分钟 £0.03。求两种套餐费用相同时的通话分钟数，并指出轻度用户选哪个更便宜。
thly cost: for A: C = 10 + 0.05m; for B: C = 15 + 0.03m, where m is the number of minutes.
写出月花费表达式：A：C = 10 + 0.05m；B：C = 15 + 0.03m，其中 m 为分钟数。
Set them equal: 10 + 0.05m = 15 + 0.03m.
使两者相等：10 + 0.05m = 15 + 0.03m。
Solve: subtract 0.03m from both sides: 10 + 0.02m = 15. Then 0.02m = 5
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