📚 Year 11 Edexcel Maths: Case Study Practical Exercises | Year 11 Edexcel 数学：案例分析实战演练
Case studies are a powerful way to bridge the gap between textbook techniques and real-world problem solving. In Edexcel Year 11 Maths, exam questions often embed calculations within a practical context — from planning a budget to analysing scientific data. This article walks you through a variety of fully worked case studies, each designed to strengthen your ability to choose the right method, interpret results, and communicate your reasoning clearly.
案例分析是连接课本技巧与现实问题解决的强大桥梁。在 Edexcel Year 11 数学考试中，题目常把计算嵌入实际情境——从预算规划到科学数据分析。本文通过一系列详细解答的案例，帮助你强化选择正确方法、解读结果和清晰表达推理的能力。
1. Getting Started with Contextual Number Problems | 情境中的数字问题入门
A family is planning a holiday. The total budget is £2400. Flights cost £780 for four people. Accommodation is £95 per night for 7 nights. They plan to spend £45 per day on food and activities. Use estimation and exact calculation to check whether they stay within budget.
一个家庭正在计划假期，总预算为2400英镑。四人机票花费780英镑。住宿费用为每晚95英镑，共7晚。他们计划每天在饮食和活动上花费45英镑。使用估算和精确计算来检查他们是否在预算之内。
Step 1: Estimate costs. Round flights to £800, accommodation to £100 × 7 = £700, and daily spending to £50 × 7 = £350. Estimated total ≈ £800 + £700 + £350 = £1850, which looks comfortable.
步骤1：估算费用。将机票四舍五入为800英镑，住宿为100×7=700英镑，每日开支为50×7=350英镑。估算总费用约为800+700+350=1850英镑，看起来还算宽裕。
Step 2: Exact calculation. Accommodation: 95 × 7 = £665. Food and activities: 45 × 7 = £315. Total = 780 + 665 + 315 = £1760. Remaining = 2400 − 1760 = £640.
步骤2：精确计算。住宿：95×7=665英镑。饮食与活动：45×7=315英镑。总计：780+665+315=1760英镑。剩余：2400−1760=640英镑。
Conclusion: The family is well within budget, with enough left for extras. This case shows how rounding helps with quick checks before precise working.
结论：该家庭远在预算之内，并有足够的余额用于额外开销。本案例展示了在精确计算之前，四舍五入如何帮助快速检查。
2. Modelling with Algebra | 代数建模
A mobile phone contract charges a fixed monthly fee of £10 plus 5p per minute of calls. Another provider offers a deal with no fixed fee but charges 8p per minute. Determine the number of minutes where both contracts cost the same, and advise a customer who talks for 400 minutes per month.
一份手机合同每月收取10英镑固定费用，外加每分钟5便士的通话费。另一家运营商不收取固定费用，但每分钟通话收费8便士。求出两份合同费用相同时的通话分钟数，并为一位每月通话400分钟的客户提供建议。
Let the number of minutes be m. Contract A cost: 1000 + 5m (pence, converting £10 to 1000p). Contract B cost: 8m (pence). Set equal: 1000 + 5m = 8m.
设通话分钟数为 m。合同A费用：1000+5m（便士，将10英镑转换为1000便士）。合同B费用：8m（便士）。令两者相等：1000+5m=8m。
Solve: 1000 = 3m → m = 333 ⅓ minutes. For m = 400, A: 1000 + 5×400 = 3000p = £30; B: 8×400 = 3200p = £32. So contract A is cheaper.
解方程：1000=3m → m=333⅓分钟。当 m=400 时，A：1000+5×400=3000便士=30英镑；B：8×400=3200便士=32英镑。因此合同A更便宜。
This algebraic model allows easy comparison for any usage level. The break‑even point helps the customer see where the decision changes.
这个代数模型可以轻松比较任意使用量。盈亏平衡点帮助客户看清决策变化的临界点。
3. Ratio and Proportion in Recipes | 食谱中的比率与比例
A cake recipe for 6 people requires 200 g flour, 150 g sugar, 4 eggs, and 120 ml milk. Adapt the recipe for 10 people, keeping the same consistency. Then, if you have only 5 eggs, how many whole cakes (for 6 people) can you make?
一份6人份的蛋糕食谱需要200克面粉、150克糖、4个鸡蛋和120毫升牛奶。按同样的稠度调整为10人份。然后，如果你只有5个鸡蛋，最多能做多少个完整的6人份蛋糕？
× 5/3 = 250 g, eggs = 4 × 5/3 = 6 ⅔ eggs (practically 7 eggs), milk = 120 × 5/3 = 200 ml.
缩放因子 = 10÷6=5/3。新面粉用量 = 200×5/3=333⅓克，糖 = 150×5/3=250克，鸡蛋 = 4×5/3=6⅔个（实际需7个），牛奶 = 120×5/3=200毫升。
With 5 eggs, one 6‑ complete cake can be made. The remaining egg does not meet the proportion for a full cake.
有5个鸡蛋，每个6人份蛋糕需要4个鸡蛋。可做蛋糕数 = 5÷4=1个余1，故只能制作1个完整蛋糕。剩余的鸡蛋无法满足全比例的要求。
Ratio problems like this highlight the importance of maintaining ingredient balance, a key skill for scaling up or down in practical situations.
此类比例问题凸显了在实际操作中保持配料平衡的重要性，无论放大还是缩小都需要这一关键技能。
4. Geometry in Design: Area and Volume | 设计中的几何：面积与体积
An architect designs a cylindrical water tank with a diameter of 3 m and a height of 2.5 m. Calculate the capacity in litres. The tank needs painting on the inside curved surface with a paint that covers 8 m² per litre. How many litres of paint are required?
一位建筑师设计了一个圆柱形水箱，直径3米，高2.5米。计算其容量（以升为单位）。水箱内曲面需要涂漆，油漆的覆盖率为每升8平方米。需要多少升油漆？
5 × 2.5 = 5.625π m³. In litres (1 m³ = 1000 L): capacity ≈ 5.625 × 3.142 × 1000 ≈ 17677 L.
半径 r=1.5 m。体积 = πr²h = π×(1.5)²×2.5 = π×2.25×2.5 = 5.625π m³。以升为单位（1 m³ = 1000 L）：容量 ≈ 5.625×3.142×1000 ≈ 17677升。
× 1.5 × 2.5 = 7.5π m² ≈ 23.56 m². Paint needed = 23.56 ÷ 8 ≈ 2.95 litres, so buy 3 litres.
曲面面积 = 2πrh = 2×π×1.5×2.5 = 7.5π m² ≈ 23.56 m²。所需油漆 = 23.56÷8 ≈ 2.95升，因此需购买3升。
These calculations combine circle geometry and unit conversion — common exam themes. Always consider practical rounding up for materials.
这些计算结合了圆几何与单位转换——常见的考试主题。务必考虑实际材料需求的向上取整。
5. Interpreting Distance-Time Graphs | 解释距离-时间图
A delivery van travels from a depot to a shop 30 km away. The graph shows: 0–0.5 h: steady speed; 0.5–1.0 h: stopped; 1.0–1.5 h: continues journey; 1.5–2.0 h: unloads. The total distance is 30 km reached at 1.5 h. Sketch and interpret the graph, calculating speeds and discussing what each section represents.
一辆送货车从仓库开到30公里外的一家商店。图表显示：0–0.5小时：匀速行驶；0.5–1.0小时：停车；1.0–1.5小时：继续行驶；1.5–2.0小时：卸货。总距离30公里在1.5小时时达到。画出并解释该图表，计算速度并讨论各段含义。
h. Second leg stationary: horizontal line at 15 km for 0.5 h. Third leg: from 15 km to 30 km in 0.5 h, speed again 30 km/h. After reaching 30 km, the line stays flat for 0.5 h, indicating unloading with no distance change.
第一段：在0.5小时内距离从0增加到15公里，速度=15÷0.5=30 km/h。第二段静止：在15公里处水平线持续0.5小时。第三段：从15公里到30公里，用时0.5小时，速度仍为30 km/h。到达30公里后，直线保持水平0.5小时，表示卸货而无距离变化。
Gradients reveal speed; flat sections indicate stops. These graphs are frequently used to test understanding of motion and are perfect case studies for interpreting real journeys.
斜率揭示速度；水平段表示停顿。这类图表常用于考查对运动的理解，是解释真实行程的绝佳案例。
6. Statistical Investigation: Comparing Data Sets | 统计调查：比较数据集
Two classes, 11A and 11B, took the same maths test. Summary statistics: 11A: mean 62, median 64, range 40, interquartile range 12. 11B: mean 58, median 57, range 55, IQR 22. Draw box plots and write a comparison, considering average and spread.
11A和11B两个班级参加了同一次数学测试。统计摘要：11A：均值62，中位数64，极差40，四分位距12。11B：均值58，中位数57，极差55，IQR 22。绘制箱线图并写出比较，考虑平均水平和离散程度。
mate LQ = 46, UQ = 68. (Exact values depend on full data, but these illustrate the shape.)
箱线图绘制：对于11A，下四分位数 ≈ 64−0.5×12=58，上四分位数=70。对于11B，中位数57，IQR 22，可估计LQ=46，UQ=68。（精确值取决于完整数据，但此处用来说明形状。）
Comparison: 11A has a higher median and mean, indicating better overall performance. Moreover, 11A’s smaller range and IQR show more consistent marks, while 11B’s wider spread suggests a greater mix of high and low achievers.
比较：11A具有更高的中位数和均值，说明整体表现更好。此外，11A较小的极差和IQR表明成绩更稳定，而11B较宽的分布暗示了高低分学生之间差异更大。
Statistical case studies develop the ability to analyse data critically — a skill tested through exam questions requiring comparisons and interpretations.
统计案例研究培养批判性分析数据的能力——这是考试中要求比较和解释的题目所考查的技能。
7. Probability and Risk Assessment | 概率与风险评估
A game involves spinning two fair spinners. Spinner 1 has sections numbered 1, 2, 3. Spinner 2 has sections 1, 2, 2, 4. You win if the sum is 5 or more. Calculate the probability of winning, and discuss whether it is worth paying £1 to play if a win pays £2.50.
一个游戏需要转动两个均匀的转盘。转盘1的区域标有1、2、3。转盘2的区域标有1、2、2、4。若两数之和为5或以上则获胜。计算获胜的概率，并讨论如果获胜可得2.50英镑，支付1英镑玩一次是否值得。
; (3,1)=4, (3,2)=5, (3,2)=5, (3,4)=7. Winning sums (≥5): (1,4), (2,4), (3,2), (3,2), (3,4) — 5 outcomes. So P(win) = 5/12.
列出所有结果（3×4=12种等可能组合）。可能和值： , (1,2)=3, (1,4)=5; (2,1)=3, (2,2)=4, (2,2)=4, (2,4)=6; (3,1)=4, (3,2)=5, (3,2)=5, (3,4)=7。获胜和值(≥5)：(1,4), (2,4), (3,2), (3,2), (3,4) — 共5个结果。因此 P(获胜)=5/12。
Expected winnings per game = (5/12) × 2.50 + (7/12) × 0 ≈ £1.04. Since £1.04 > £ in the long run. In reality, very few trials may show a loss, so risk must be considered.
每局期望赢得金额 = (5/12)×2.50 + (7/12)×0 ≈ 1.04英镑。由于1.04 > 1，长期来看游戏对玩家略为有利。实际上，试验次数很少时可能出现亏损，因此必须考虑风险。
This case demonstrates how probability models inform decision‑making under uncertainty, linking directly to Edexcel’s expectation and risk topics.
本案例展示了概率模型如何为不确定性下的决策提供信息，直接关联到Edexcel的期望与风险主题。
8. Compound Measures: Speed and Density | 复合单位：速度与密度
A solid metal sphere has a mass of 2.4 kg and a radius of 4 cm. Calculate its density in g/cm³. If the sphere is melted into a wire of cross‑sectional area 0.2 cm², find the length of wire produced.
一个实心金属球的质量为2.4千克，半径为4厘米。计算其密度，单位为 g/cm³。若将该球熔化制成横截面积为0.2 cm²的金属丝，求金属丝的长度。
(4/3)π×64 = (256/3)π cm³ ≈ 268.08 cm³. Density = mass/volume = 2400 ÷ 268.08 ≈ 8.95 g/cm³.
将质量转换为克：2.4 kg = 2400 g。球体积 = (4/3)πr³ = (4/3)π(4)³ = (4/3)π×64 = (256/3)π cm³ ≈ 268.08 cm³。密度 = 质量/体积 = 2400÷268.08 ≈ 8.95 g/cm³。
When reshaped, volume stays the same. Wire length L satisfies: cross‑sectional area × L = volume. So 0.2 L = 268.08 → L = 268.08 ÷ 0.2 = 1340.4 cm ≈ 13.4 m.
重塑时体积保持不变。金属丝长度L满足：横截面积×L=体积。因此0.2L=268.08 → L=268.08÷0.2=1340.4 cm ≈ 13.4 m。
Compound measures like density and volume conservation are standard GCSE challenges, requiring careful unit management and formula rearrangement.
密度和体积守恒等复合单位问题是GCSE常见的挑战，要求仔细处理单位并灵活变换公式。
9. Sequences in Practical Patterns | 实际模式中的数列
Square tiles are laid around a rectangular flower bed. The first design has a 1 m by 2 m bed with 10 tiles around it. Each subsequent design increases the length and width by 1 m. Find a formula for the number of tiles T in the n-th design, and use it to predict the tiles needed for the 20th design.
方形瓷砖围绕一个矩形花坛铺设。第一种设计为1米×2米花坛，周围用了10块瓷砖。随后每种设计的长宽均增加1米。求出第n种设计所需瓷砖数T的公式，并用它预测第20种设计所需的瓷砖数量。
Assume the bed dimensions for design n are (n) by (n+1) metres? e, number of tiles equals length in metres (if tile side 1 m) plus one extra at the corner? side perimeter and corners counted once correctly). For L=1, W=2: 2+4+4=10, which matches.
假设设计n的花坛尺寸为n 米乘 (n+1) 米？检查：n=1：1米×2米，周长=2×(1+2)=6，但瓷砖围绕铺设：每边的瓷砖数量等于长度（米），角上需加一。围绕L×W矩形铺设的已知公式为 2L+2W+4（瓷砖铺在外周长，角块正确计数一次）。当L=1, W=2：2+4+4=10，相符。
2 + 4 = 4n + 6. Check n=2 (2×3): 4×2+6=14, and manual count: 2×2+2×3+4=4+6+4=14. Correct.
因此对于设计n，L=n, W=n+1。T=2n+2(n+1)+4=4n+2+4=4n+6。验证n=2（2×3）：4×2+6=14，手工计数：2×2+2×3+4=4+6+4=14，正确。
For n=20: T = 4×20+6 = 86 tiles. This linear sequence is typical in GCSE pattern‑spotting questions that link to algebraic modelling.
当n=20时：T=4×20+6=86块瓷砖。这种线性数列是GCSE中典型的模式识别题目，与代数建模相连。
10. Trigonometry for Heights and Distances | 用于高度和距离的三角学
From a point 50 m away from the base of a tower, the angle of elevation to the top is 28°. A flagpole on top of the tower creates an additional angle of 5° to its top. Find the height of the tower and the length of the flagpole.
从塔底外50米处的一点测得塔顶的仰角为28°。塔顶上方有一旗杆，到旗杆顶端的仰角再增加5°。求塔的高度和旗杆的长度。
t tower height = h. tan 28° = h / 50 → h = 50 × tan 28° ≈ 50 × 0.5317 = 26.585 m ≈ 26.6 m.
设塔高 = h。tan 28° = h/50 → h = 50 × tan 28° ≈ 50 × 0.5317 = 26.585 m ≈ 26.6 m。
n 33° ≈ 50 × 0.6494 = 32.47 m. Flagpole length = H − h = 32.47 − 26.585 = 5.885 m ≈ 5.9 m.
从地面到旗杆顶的总高度：角度=33°。故 H = 50 × tan 33° ≈ 50 × 0.6494 = 32.47 m。旗杆长度 = H − h = 32.47 − 26.585 = 5.885 m ≈ 5.9 m。
This practical case uses right‑angled trigonometry (SOH CAH TOA) to solve a two‑stage problem. It mirrors questions where you must apply tan twice with care.
本实用案例运用直角三角形三角学（SOH CAH TOA）解决两步问题，仿照了需要仔细两次应用正切函数的考题。
11. Simultaneous Equations in Business | 商业中的联立方程
A coffee shop sells two blends: Classic and Premium. On Monday they sold 30 Classic and 20 Premium for total revenue of £190. On Tuesday they sold 15 Classic and 35 Premium for £185. Find the price per pack of each blend.
一家咖啡店销售两种混合咖啡：经典和特级。星期一售出30包经典和20包特级，总收入190英镑。星期二售出15包经典和35包特级，收入185英镑。求每种咖啡的单价。
Let Classic price = £c, Premium price = £p. Equations: 30c + 20p = 190 (1) ; 15c + 35p = 185 p = 370. Subtract (1) from this: (30c+70p) − (30c+20p) = 370 − 190 → 50p = 180 → p = 3.60.
设经典咖啡单价为 c 英镑，特级为 p 英镑。方程：30c + 20p = 190 (1)；15c + 35p = 185 (2)。将(2)乘以2：30c+70p=370。减去(1)：(30c+70p)−(30c+20p)=370−190 → 50p=180 → p=3.60。
= 3.60 into (1): 30c + 20×3.60 = 190 → 30c + 72 = 190 → 30c = 118 → c = 118/30 ≈ 3.933, so £3.93. Check in (2): 15×3.933 + 35×3.60 ≈ 59.00 + 126 = 185. Correct.
将p=3.60代入(1)：30c+20×3.60=190 → 30c+72=190 → 30c=118 → c=118/30 ≈ 3.933，即3.93英镑。验证(2)：15×3.933+35×3.60 ≈ 59.00+126=185。正确。
Business case studies like this demonstrate how simultaneous equations model revenue and cost, a key skill for the Edexcel higher tier.
此类商业案例展示了联立方程如何为收入与成本建立模型，是Edexcel高等级考试的关键技能。
12. Quadratic Applications in Projectile Motion | 二次方程在抛体运动中的应用
rmine the time when it returns to the ground, the maximum height, and the time interval during which the height exceeds 40 m.
一枚烟花火箭从地面发射。t秒后其高度h（米）由公式 h = 30t − 5t² 给出。求它落回地面的时间、最大高度，以及高度超过40米的时间区间。
und when h = 0: 30t − 5t² = 0 → 5t(6 − t) = 0 → t = 0 or t = 6. So flight lasts 6 seconds.
落回地面时 h=0：30t−5t²=0 → 5t(6−t)=0 → t=0 或 t=6。因此飞行持续6秒。
m height at vertex: t = −b/2a for h = −5t² + 30t; a = −5, b = 30 → t = −30/(2×−5) = 3 s. hmax = 30×3 − 5×9 = 90 − 45 = 45 m.
最大高度在顶点处：对于 h=−5t²+30t，t=−b/2a；a=−5, b=30 → t=−30/(2×−5)=3秒。hmax = 30×3−5×9=90−45=45米。
− 2)(t − 4) < 0. Solution: 2 < t < 4. So the rocket is above 40 m between 2 and 4 seconds.
高度>40：解不等式 −5t²+30t−40>0 → 乘以−1：5t²−30t+40<0 → 除以5：t²−6t+8<0 → (t−2)(t−4)<0。解为 2 Quadratic functions appear frequently in kinematic contexts. This case integrates factorising, completing the square, and inequality interpretation — all vital for Edexcel problem‑solving questions. 二次函数常出现在运动学情境中。本案例融合了因式分解、配方法和不等式解读——对Edexcel解题性问题均至关重要。
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