📚 Case Study Mastery: Practical Problem Solving for Year 10 Eduqas Maths | 案例分析实战演练：Eduqas Year 10 数学应用解题
Eduqas GCSE Mathematics places a strong emphasis on applying mathematical knowledge to real-world contexts. Case studies are not just textbook exercises; they are carefully constructed scenarios that require you to interpret data, choose appropriate techniques, and communicate your reasoning clearly. In this guide, we will walk through a series of practical case studies, mirroring the style of exam questions, covering number, algebra, geometry, statistics, and ratio. You will learn how to break down complex problems, extract relevant information, and justify your methods — skills essential for top marks.
Eduqas GCSE 数学非常重视将数学知识应用于现实情境。案例分析不仅仅是教科书上的练习；它们是精心构建的场景，要求你解读数据、选择合适的技巧，并清晰地表达你的推理。在本指南中，我们将通过一系列实践案例，模拟考试题的风格，涵盖数字、代数、几何、统计和比例。你将学会如何拆解复杂问题、提取相关信息并论证你的方法——这些是获得高分的必备技能。
1. Interpreting the Brief | 解读题目要求
Before rushing into calculations, read the case study brief at least twice. Highlight key numbers, units, constraints, and what the question is actually asking. Many marks are lost because students solve a different problem from the one set. Ask yourself: What information is given? What am I trying to find? Are there any hidden assumptions, such as VAT at 20% or rounding to the nearest penny?
在匆忙计算之前，至少把案例简介读两遍。标出关键数据、单位、限制条件以及题目真正在问什么。很多分数是因为学生解错了题而丢掉的。问问自己：给出了哪些信息？我要找的是什么？有没有隐藏的假设，比如20%的增值税或者要四舍五入到最接近的便士？
A typical Eduqas case study might involve planning a school event. For example: 'A school is organising a prom for 180 students. The venue costs £450, a DJ charges £120 per hour for 4 hours, and decorations cost £35.60 per pack (each pack decorates 3 tables of 8 students). Calculate the cost per student.' Note the layers: you must work out the number of tables, then the packs required, then total cost, then divide by 180.
一个典型的Eduqas案例可能涉及策划学校活动。例如：“学校要为180名学生组织毕业舞会。场地费450英镑，DJ每小时收费120英镑共4小时，装饰品每包35.60英镑（每包可以装饰3张桌子，每桌坐8名学生）。计算每名学生的费用。”注意层次：你必须先算出桌子数量，再算所需包数，然后总费用，最后除以180。
2. Budgeting a School Trip | 学校旅行的预算规划
This case study blends arithmetic, ratio, and percentage. A school plans to take 196 students and 14 staff to a theme park. The entry price is £22.50 per person, but groups of 15 or more get a 12% discount. The coach hire is £640 per coach, and each coach carries 53 passengers. The school wants to allocate one free staff place for every 15 paying students. Work out the total cost and the cost per paying student.
这个案例融合了算术、比例和百分比。学校计划带196名学生和14名员工去主题公园。门票每人22.50英镑，但15人或以上的团体可享受12%折扣。每辆大巴租金640英镑，每辆可载53人。学校希望每15名付费学生分配一个免费员工名额。计算总费用和每位付费学生的费用。
f go free. For every 15 paying students, 1 staff goes free. Number of paying students: 196, so free staff places = floor(196 / 15) = 13. Therefore, 13 staff go free, and 1 member of staff must pay. Total paying people = 196 + 1 = 197. Now calculate ticket cost: 197 × £22.50 = £4,432.50. Group discount: 12% off, so you pay 88%. Discounted ticket total = £4,432.50 × 0.88 = £3,900.60.
首先确定总人数：196 + 14 = 210。由于免费员工名额基于付费学生，你需要算出多少员工免费。每15名付费学生，1名员工免费。付费学生数：196，所以免费员工名额 = 取整(196 / 15) = 13。因此，13名员工免费，1名员工需付费。总付费人数 = 196 + 1 = 197。现在计算门票费用：197 × £22.50 = £4,432.50。团体折扣：12% off，所以支付88%。折扣后门票总额 = £4,432.50 × 0.88 = £3,900.60。
210. Each coach holds 53. Number of coaches = ceil(210 / 53) = 4 coaches. Coach cost = 4 × £640 = £2,560. Total trip cost = £3,900.60 + £2,560 = £6,460.60. Cost per paying student: total cost / 196 = £6,460.60 / 196 = £ the nearest penny). Notice the careful rounding and interpretation of the free place rule — Eduqas examiners expect you to show this step clearly.
接下来，大巴：总人数210。每辆大巴载53人。大巴数量 = 向上取整(210 / 53) = 4辆。大巴费用 = 4 × £640 = £2,560。旅行总费用 = £3,900.60 + £2,560 = £6,460.60。每位付费学生的费用：总费用 / 196 = £6,460.60 / 196 = £32.96（精确到便士）。注意仔细的取整和对免费名额规则的解释——Eduqas考官希望看到你清晰地展示这个步骤。
3. Designing a Garden | 花园设计中的面积与成本
Geometry case studies often involve composite shapes and materials. Consider a rectangular garden measuring 12 m by 8 m, with a circular pond of diameter 3 m in the centre. The rest is to be turfed. Turf costs £4.25 per square metre, plus a delivery charge of £25. Calculate the total turf cost. You must also find the perimeter fencing cost if fencing is £15 per metre with a gate of width 1.2 m that costs £45.
几何案例通常涉及复合图形和材料。设想一个长12米、宽8米的矩形花园，中央有一个直径3米的圆形池塘。其余部分铺设草皮。草皮每平方米4.25英镑，外加25英镑配送费。计算草皮总费用。你还需要计算围栏成本，如果围栏每米15英镑，并有一个宽1.2米、售价45英镑的门。
a = 96 − 7.07 = 88.93 m² (accept 88.94 or similar reasonable rounding). Turf cost: 88.93 × £4.25 = £377.95. Add delivery: £377.95 + £25 = £402.95.
花园面积：12 × 8 = 96 平方米。池塘面积：π × (1.5)² = π × 2.25 ≈ 7.0686 平方米（使用π=3.14得7.065平方米）。草皮面积 = 96 − 7.07 = 88.93 平方米（接受88.94或类似合理取整）。草皮费用：88.93 × £4.25 = £377.95。加上配送费：£377.95 + £25 = £402.95。
(12 + 8) = 40 m. Subtract gate width: 40 − 1.2 = 38.8 m of fencing. Fencing cost = 38.8 × £15 = £582. Plus gate: £45. Total fencing cost = £627. Therefore, the combined project cost is £402.95 + £627 = £1,029.95. This demonstrates how to handle compound shapes and cost additions — typical in Eduqas problem-solving.
花园周长 = 2 × (12 + 8) = 40 米。减去门宽：40 − 1.2 = 38.8 米围栏。围栏费用 = 38.8 × £15 = £582。加上门：£45。围栏总费用 = £627。因此，项目总成本为 £402.95 + £627 = £1,029.95。这展示了如何处理复合形状和费用叠加——Eduqas解决问题中的典型题型。
4. Mobile Phone Tariff Comparison | 手机套餐比较
Linear functions and graphs are common in case studies. A mobile phone company offers two tariffs: Plan A: £10 per month plus 8p per minute. Plan B: £18 per month with 500 free minutes, then 6p per minute thereafter. Determine which is cheaper for 450 minutes and for 700 minutes. Write down the cost equations and find the break-even point.
线性函数和图像在案例分析中很常见。一家手机公司提供两种套餐：计划A：每月10英镑加每分钟8便士。计划B：每月18英镑含500分钟免费通话，超出部分每分钟6便士。确定在450分钟和700分钟时哪个更便宜。写出费用方程，并求出盈亏平衡点。
8; if m > 500, C = 18 + 0.06(m − 500). For 450 minutes: Plan A = 10 + 0.08×450 = 10 + 36 = £46. Plan B = £ (since 450 ≤ 500). Plan B is cheaper. For 700 minutes: Plan A = 10 + 0.08×700 = 10 + 56 = £66. Plan B = 18 + 0.06×(200) = 18 + 12 = £ int where A becomes cheaper. Set 10 + 0.08m = 18 + 0.06(m − 500) for m > 500. Solve: 10 + 0.08m = 18 + 0.06m − 30 → 10 + 0.08m = 0.06m − 12 → 0.02m = −22 → m = −1100, not possible. Therefore for all m > 500, B is cheaper. Below 500, compare 10 + 0.08m = 18 → 0.08m = 8 → m = 100 minutes. So for m < 100, A is cheaper; at m = 100, equal; between 100 and 500, B cheaper. This algebraic analysis is precisely what an Eduqas case study expects.
设C为费用（英镑），m为分钟数。计划A：C = 10 + 0.08m。计划B：若m ≤ 500，C = 18；若m > 500，C = 18 + 0.06(m − 500)。450分钟时：计划A = 10 + 0.08×450 = 10 + 36 = £46。计划B = £18（因为450 ≤ 500）。计划B更便宜。700分钟时：计划A = 10 + 0.08×700 = 10 + 56 = £66。计划B = 18 + 0.06×(200) = 18 + 12 = £30。再次B更便宜。盈亏平衡发生在计划A费用等于计划B费用时。鉴于这些值B总是更便宜，我们需要检查是否存在A变得更便宜的点。对于m > 500，设10 + 0.08m = 18 + 0.06(m − 500)。求解：10 + 0.08m = 18 + 0.06m − 30 → 10 + 0.08m = 0.06m − 12 → 0.02m = −22 → m = −1100，不可能。因此对所有m > 500，B更便宜。低于500，比较10 + 0.08m = 18 → 0.08m = 8 → m = 100分钟。因此当m < 100时，A更便宜；m = 100时相等；在100到500之间，B更便宜。这个代数分析正是Eduqas案例分析所期待的。
5. Population Growth and Exponential Change | 人口增长与指数变化
Case studies may feature exponential growth or decay. A city's population is 240,000 and is expected to grow by 2.3% each year. Write an expression for the population after t years. Calculate the population after 8 years and the year when it first exceeds 300,000.
案例分析可能涉及指数增长或衰减。某城市人口为240,000，预计每年增长2.3%。写出t年后的人口表达式。计算8年后的人口，并求出人口首次超过300,000的年份。
The growth model is P = 240000 × (1.023)t. After 8 years: P = 240000 × (1.023)8. You can use a calculator: 1.0238 ears, population ≈ 287,950 (3 s.f.). To find when it exceeds 300,000, set 240000 × (1.023)t > 300000 → (1.023)t (rounding up, since year 10 is the first integer when it exceeds). At t=9: 240000 × 1.0239 ≈ 281,400; t=10: 287,952? Wait, that was t=8. Recalculate: 1.0239 = 1.0238 × 1.023 ≈ 1.1998×1.023≈1.2274, P≈294,576. t=10: 1.02310 000 in the 10th year. This models typical Eduqas exponential questions where you must interpret the inequality and use trial or logs.
增长模型为 P = 240000 × (1.023)t。8年后：P = 240000 × (1.023)8。可使用计算器：1.0238 ≈ 1.1998，所以P ≈ 240000 × 1.1998 = 287,952。因此8年后人口约287,950（3位有效数字）。要求出何时超过300,000，设240000 × (1.023)t > 300000 → (1.023)t > 1.25。取对数：t log(1.023) > log(1.25)。t > log(1.25)/log(1.023) ≈ 0.09691 / 0.00988 ≈ 9.81。所以10年后（向上取整，因为第10年是首个满足的整数）。在t=9时：240000 × 1.0239 ≈ 294,576；t=10: ≈301,368。故在第10年超过300,000。这符合Eduqas典型的指数问题，需要解读不等式并使用试值或对数。
6. Health and Fitness: BMI and Calorie Analysis | 健康与健身：BMI与卡路里分析
l. Classify their BMI according to standard ranges: underweight < 18.5, normal 18.5-24.9, overweight 25-29.9, obese ≥ 30. Next, if they want to reach a BMI of 24, what target weight should they aim for? e the weekly calorie deficit and the weight loss over 10 weeks (1 kg of fat ≈ 7,700 calories).
另一个常见主题是在情境中使用公式。身体质量指数公式为 BMI = 体重(kg) / (身高(m))²。某人体重85公斤，身高1.78米。根据标准范围对其BMI进行分类：偏瘦 < 18.5，正常18.5-24.9，超重25-29.9，肥胖 ≥ 30。接下来，如果他们想达到BMI为24，目标体重应是多少？此外，他们每骑30分钟自行车消耗300卡路里，每日摄入2,400卡路里。如果他们每周骑车5次，每次45分钟，计算每周的卡路里赤字以及10周内的减重量（1公斤脂肪 ≈ 7,700卡路里）。
16 kg ≈ 76.0 kg. So they need to lose about 9 kg. Calorie calculation: cycling per session: (45/30) × 300 = 1.5 × 300 = 450 calories. Weekly cycling calories = 5 × 450 = 2,250 calories. Daily intake = 2,400, weekly intake = 2,400 × 7 = 16,800. Basal metabolic needs not given, but if we assume maintenance calories are exactly their intake (so weight stays constant without exercise), then the cycling creates a deficit. Weekly deficit = 2,250 calories. Over 10 weeks: 22,500 calories. Weight loss = 22,500 / 7,700 ≈ 2.92 kg. This multi-step calculation is exactly the type of linked reasoning Eduqas case studies demand.
BMI = 85 / (1.78)² = 85 / 3.1684 ≈ 26.83。属于超重类别。求BMI为24的目标体重：24 = m / (1.78)² → m = 24 × 3.1684 = 76.0416 kg ≈ 76.0 kg。因此需要减掉约9 kg。卡路里计算：每次骑车：(45/30) × 300 = 1.5 × 300 = 450卡路里。每周骑车消耗 = 5 × 450 = 2,250卡路里。每日摄入2,400，每周摄入 = 2,400 × 7 = 16,800。未给出基础代谢需求，但如果我们假设维持体重所需热量恰好等于摄入（不运动时体重不变），那么骑车创造了赤字。每周赤字 = 2,250卡路里。10周总计22,500卡路里。减重量 = 22,500 / 7,700 ≈ 2.92 kg。这种多步骤计算正是Eduqas案例分析所要求的连锁推理。
7. Business Profit and Break-Even Analysis | 企业利润与盈亏平衡分析
Mathematical modelling for a small business: A company makes wooden toys. Fixed costs (rent, insurance) are £820 per month. Each toy costs £4.20 in materials and is sold for £9.99. Write equations for total cost C(x) and revenue R(x). Find the break-even number of toys per month. If they sell 200 toys, what is the profit? Then, consider a 15% discount for bulk orders of 50 or more: how does that change the break-even if 40% of sales are bulk orders?
小企业的数学建模：一家公司生产木质玩具。每月固定成本（租金、保险）为820英镑。每个玩具的材料成本为4.20英镑，售价为9.99英镑。写出总成本C(x)和收入R(x)的方程。求每月盈亏平衡的玩具数量。如果他们卖出200个玩具，利润是多少？然后考虑对50个及以上的批量订单给予15%折扣：如果40%的销售是批量订单，盈亏平衡点如何变化？
o sell 142 toys. Profit for 200 toys: R(200) = 9.99 × 200 = 1,998. C(200) = 820 + 4.20×200 = 820 + 840 = 1,660. Profit = 1,998 − 1,660 = £338.
设x为玩具数量。C(x) = 820 + 4.20x。R(x) = 9.99x。盈亏平衡时R(x) = C(x)：9.99x = 820 + 4.20x → 5.79x = 820 → x = 820 / 5.79 ≈ 141.62，因此需要卖出142个玩具。200个玩具的利润：R(200) = 9.99 × 200 = 1,998。C(200) = 820 + 4.20×200 = 820 + 840 = 1,660。利润 = 1,998 − 1,660 = £338。
Now with mixed sales: 40% of toys sold at discounted price of £9.99 × 0.85 = £8.4915, say £8.49 (or keep exact). Let' uation: 9.3906x = 820 + 4.20x → 5.1906x = 820 → x = 820 / 5.1906 ≈ 157.98 → need 158 toys. This demonstrates sensitivity analysis, a higher-order skill.
现在考虑混合销售：40%的玩具以折扣价卖出，售价为£9.99 × 0.85 = £8.4915，取£8.49（或保留精确值）。我们保留精确值：折扣价 = 9.99 × 0.85 = 8.4915。加权平均售价 = 0.6 × 9.99 + 0.4 × 8.4915 = 5.994 + 3.3966 = 9.3906。盈亏平衡方程：9.3906x = 820 + 4.20x → 5.1906x = 820 → x = 820 / 5.1906 ≈ 157.98 → 需要158个玩具。这展示了敏感性分析，属于高阶技能。
8. Packaging Design: Volume and Surface Area | 包装设计：体积与表面积
², and for the ends costs 0.6 pence per cm². Find the radius r that minimises the total material cost, and calculate that minimum cost. Use π = 3.14.
一家公司想设计一个圆柱形罐子来装500 cm³的糖果。罐子有底和盖。侧面材料成本为每平方厘米0.4便士，两端材料成本为每平方厘米0.6便士。求出使总材料成本最小的半径r，并计算该最小成本。使用π = 3.14。
Volume V = π r² h = 500 → h = 500 / (π r²). Curved surface area Aside = 2π r h = 2π r × (500/(π r²)) = 1000 / r. Ends area total Aends alues or trial and improvement (since calculus is beyond the syllabus). Try r = 4: C = 400/4 + 3.768×16 = 100 + 60.288 = 160.288 pence. r = 5: C = 400/5 + 3.768×25 = 80 + 94.2 = 174.2 pence. r = 3.5: C = 400/3.5 + 3.768×12.25 ≈ 114.2857 + 46.158 = 160.44. r = 3.8: 400/3.8≈105.263 + 3.768×14.44≈54.41 total 159.67. r = 3.9: 400/3.9≈102.564 + 3.768×15.21≈57.32 total 159.88. r = 3.7: 400/3.7≈108.108 + 3.768×13.69≈51.57 total 159.68. Minimum around r = 3.75: 400/3.75=106.667 + 3.768×14.0625≈52.97 total 159.64. So minimum cost ≈ 159.6 pence, or £1.60. This trial method is perfectly acceptable in Eduqas and shows iterative thinking.
体积 V = π r² h = 500 → h = 500 / (π r²)。侧面积 Aside = 2π r h = 2π r × (500/(π r²)) = 1000 / r。两端总面积 Aends = 2π r²。成本 C = 0.4 × (1000/r) + 0.6 × (2π r²) = 400/r + 1.2π r²。代入π = 3.14：C = 400/r + 3.768 r²。为求最小值，10年级学生可使用表格试值或逐次逼近（微积分超纲）。试r = 4：C = 400/4 + 3.768×16 = 100 + 60.288 = 160.288便士。r = 5：C = 400/5 + 3.768×25 = 80 + 94.2 = 174.2便士。r = 3.5：C ≈ 160.44。r = 3.8：159.67。r = 3.9：159.88。r = 3.7：159.68。最小值大约在r = 3.75：106.667 + 52.97 = 159.64。所以最小成本约159.6便士，或£1.60。这种试值法在Eduqas中完全可以接受，并展示了迭代思维。
9. Interpreting Statistical Diagrams | 解读统计图表
zzle. Use the graph to estimate the median, interquartile range, and the number of students who took longer than 35 minutes. Without the actual diagram, imagine a smooth curve: at 50% cumulative frequency (60 students), the time is 28 minutes; at 25% (30) it's 22 min; at 75% (90) it' requency: say 100 students, so number taking longer than 35 min = 120 − 100 = 20 students.
案例分析经常提供图表。例如，一个累积频率图显示了120名学生完成一个拼图的时间。使用图表估计中位数、四分位距以及完成时间超过35分钟的学生人数。虽然没有实际图表，但想象一条平滑曲线：在50%累积频率（60名学生）处，时间为28分钟；25%（30名）处为22分钟；75%（90名）处为34分钟。所以中位数=28，IQR=34−22=12分钟。对于35分钟，读取累积频率：假设为100名学生，则超过35分钟的人数 = 120−100 = 20名学生。
You might also encounter box plots or histograms. Practice describing comparisons: 'The median time for boys was 5 minutes shorter than for girls, and the interquartile range was wider, indicating more variation in boys' times.' Always quote values from the graph and relate them to the context.
你可能还会遇到箱线图或直方图。练习描述比较：“男孩的中位数时间比女孩短5分钟，且四分位距更宽，表明男孩时间的离散程度更大。”始终引用图表中的数值，并将其与情境联系起来。
10. Money and Best Buys | 货币与最佳购买方案
Ratio and proportion case studies are very common. A supermarket sells rice in three sizes: 500 g for £0.95, 1.5 kg for £2.65, and 2 kg for £3.50. Which is the best value for money? Show your working by calculating the price per 100 g. Also, if a recipe requires 2.8 kg of rice, what is the cheapest combination of packs to buy without wastage?
比例和比率案例非常普遍。一家超市出售三种规格的大米：500克0.95英镑，1.5公斤2.65英镑，2公斤3.50英镑。哪种最划算？通过计算每100克的价格来展示你的计算过程。另外，如果一个食谱需要2.8公斤大米，怎样组合购买最便宜且没有浪费？
Price per 100 g: 500 g pack: (0.95 / 5) = £0.19 per 100 g. 1.5 kg = 1500 g: (2.65 / 15) = £0.1767≈17.7p. 2 kg = 2000 g: (3.50 / 20) = £ value. To buy 2.8 kg without wastage, try one 2 kg pack and one 500 g pack and one 1.5 kg? That' 1.5 kg (1500 g) = 3500 g (too much). 2 kg + 500 g + 1.5 kg = 4000 g. 2 kg + 500 g + 300 g? . The cheapest combination that meets or exceeds 2.8 kg: two 1.5 kg packs = 3 kg, cost 2 × £2.65 = £5.30. Or one 2 kg plus one 1.5 kg = 3.5 kg, cost £3.50 + £2.65 = £6.15. Or 2 kg + 500 g + 500 g = 3 kg, cost £3.50 + 2×0.95 = £3.50 + £1.90 = £5.40. The 2×1.5 kg is cheapest at £5.30. But could we use 2 kg + two 500 g = 3 kg at £5.40, more expensive. So £5.30 is best. This tests flexibility with ratio and real-world constraints.
每100克价格：500克装：(0.95 / 5) = £0.19 per 100 g。1.5 kg = 1500 g：(2.65 / 15) = £0.1767≈17.7p。2 kg = 2000 g：(3.50 / 20) = £0.175 = 17.5p每100 g。所以2公斤装最划算。要购买2.8 kg且无浪费，试试一包2 kg加一包500 g再加一包1.5 kg？那是4 kg，太多了。实际上2
Published by TutorHao | Year 10 Mathematics Revision Series | aleveler.com
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