Year 11 AQA Maths: Case Study Practice Walkthrough | Year 11 AQA 数学：案例分析实战演练

📚 Year 11 AQA Maths: Case Study Practice Walkthrough | Year 11 AQA 数学：案例分析实战演练

Case study questions form a vital part of the AQA GCSE Mathematics exam. They present a real-world situation where you need to select and apply the right mathematical tools. This walkthrough offers eight complete case studies, each broken down into clear steps, so you can see exactly how to extract data, choose a strategy, and avoid common errors. Practising these scenarios will build your confidence for any problem-solving task in the exam.

案例分析题在 AQA GCSE 数学考试中至关重要。它们给出一个真实情境，要求你选择并运用正确的数学工具。本实战演练提供了八个完整的案例，逐一分解为清晰步骤，让你看到如何提取数据、选择策略并避开常见错误。练习这些情境，将帮助你从容应对考试中的任何问题解决任务。

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1. Understanding Case Study Questions | 理解案例分析题

In a case study, you are given a paragraph or table of information that describes a practical situation. Your first job is to read the whole description and identify the quantities, units, and the specific question being asked. Often, several topics are mixed together, such as percentages, area, and speed.

在案例分析中，你会看到一段文字或表格，描述一个实际情况。你的首要任务是通读全部描述，确定数量、单位和要解决的具体问题。题目常常混合多个主题，例如百分比、面积和速度。

Break the problem into smaller parts. For example, if you need to find the total cost of laying turf, you might first calculate the area, then subtract the area of any obstacles, and finally multiply by the unit price. Always check whether units need converting and whether your final answer is reasonable.

把问题分解成更小的部分。例如，如果需要计算铺草皮的总费用，你可以先计算总面积，再减去障碍物的面积，最后乘以单价。始终检查是否需要转换单位，以及最终答案是否合理。

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2. Case 1: Family Budget Planning | 案例一：家庭预算规划

The Johnson family have a monthly income of £2500. Their regular expenses are: rent £800, food £400, utilities £200, transport £150, entertainment £250, savings £300 and other items £400. The family want to know what percentage of their income goes to each category and how a 10% cut in entertainment would boost their savings.

约翰逊一家的月收入为 2500 英镑。他们的固定支出如下：房租 800 英镑，食品 400 英镑，水电 200 英镑，交通 150 英镑，娱乐 250 英镑，储蓄 300 英镑，其他 400 英镑。他们想知道各项支出占收入的百分比，以及若娱乐支出减少 10%，储蓄会增加多少。

Category	Amount (£)
Rent	800
Food	400
Utilities	200
Transport	150
Entertainment	250
Savings	300
Other	400

To find each percentage, divide the category amount by the total income and multiply by 100. For example, the rent percentage is (800 ÷ 2500) × 100 = 32%.

计算各项百分比时，用各类别金额除以总收入再乘以 100。例如，房租百分比为 (800 ÷ 2500) × 100 = 32%。

Percentage = (Part ÷ Whole) × 100

If entertainment is cut by 10%, the saving is 10% of £250 = £25. Add that £25 to the original savings: £300 + £25 = £325. Always recalculate the new percentages to make sure they still sum to 100%.

如果娱乐支出削减 10%，节省的金额为 250 英镑的 10%，即 25 英镑。把 25 英镑加到原本的储蓄中：300 + 25 = 325 英镑。务必重新计算新的百分比，确保总和仍为 100%。

Common trap: using the original total to find the new percentage after a change. The whole amount remains £2500, so the new savings percentage becomes (325 ÷ 2500) × 100 = 13%.

常见陷阱：在数值发生变化后仍用原总数计算新百分比。总收入仍为 2500 英镑，因此新的储蓄百分比为 (325 ÷ 2500) × 100 = 13%。

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3. Case 2: Garden Design | 案例二：花园设计

A rectangular garden measures 12 m by 8 m. The plan includes a square patio of side 3 m and a circular pond of radius 1.2 m. The remaining ground will be covered with turf costing £4.80 per square metre, plus a fixed labour charge of £50. Find the total cost.

一个矩形花园长 12 米，宽 8 米。设计包括一个边长为 3 米的正方形露台，以及一个半径为 1.2 米的圆形池塘。其余地面将铺上草皮，每平方米 4.80 英镑，另加固定人工费 50 英镑。求总费用。

First, calculate the total garden area: 12 × 8 = 96 m². Then subtract the patio area: 3 × 3 = 9 m². For the pond, use the circle area formula A = π × r². Using π ≈ 3.14, the pond area is 3.14 × 1.2² = 3.14 × 1.44 ≈ 4.52 m².

首先，计算花园总面积：12 × 8 = 96 平方米。然后减去露台面积：3 × 3 = 9 平方米。池塘面积用圆面积公式 A = π × r²。取 π ≈ 3.14，池塘面积约为 3.14 × 1.2² = 3.14 × 1.44 ≈ 4.52 平方米。

Area of a circle = π × r²

Turf area = 96 − 9 − 4.52 = 82.48 m². Turf cost = 82.48 × 4.80 ≈ 395.90. Total cost = 395.90 + 50 = £445.90. If the question asks for an answer to the nearest penny, write 445.90.

铺草皮区域 = 96 − 9 − 4.52 = 82.48 平方米。草皮费用 = 82.48 × 4.80 ≈ 395.90。总费用 = 395.90 + 50 = 445.90 英镑。若题目要求精确到便士，则答案写作 445.90。

Watch out: when subtracting areas, keep as much precision as possible. Rounding too early can give an inaccurate final total.

注意：做面积减法时，尽可能保留更多小数位数。过早四舍五入可能导致最终结果不准确。

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4. Case 3: School Trip | 案例三：学校旅行

A coach travels a distance of 184 km at an average speed of 56 km/h. There are two breaks of 20 minutes each. If the coach leaves at 09:15, at what time does it arrive?

一辆巴士以 56 公里/小时的平均速度行驶 184 公里。途中有两次休息，每次 20 分钟。如果巴士 09:15 出发，到达时间是几点？

First, find the driving time: travel time = distance ÷ speed = 184 ÷ 56 ≈ 3.2857 hours. Convert the decimal to minutes: 0.2857 × 60 ≈ 17 minutes, so driving time is about 3 hours 17 minutes. Add the two breaks: 2 × 20 = 40 minutes. Total journey time = 3 hours 17 min + 40 min = 3 hours 57 min.

首先，求出行驶时间：行驶时间 = 距离 ÷ 速度 = 184 ÷ 56 ≈ 3.2857 小时。将小数部分转换为分钟：0.2857 × 60 ≈ 17 分钟，因此行驶时间约为 3 小时 17 分钟。加上两次休息：2 × 20 = 40 分钟。总行程时间 = 3 小时 17 分 + 40 分 = 3 小时 57 分。

Time = Distance ÷ Speed

Add 3 hours 57 minutes to 09:15. 09:15 + 3 hours = 12:15, then + 57 minutes = 13:12. So arrival time is 13:12 (or 1:12 pm).

将 3 小时 57 分钟加到 09:15 上。09:15 + 3 小时 = 12:15，再加 57 分钟 = 13:12。因此到达时间为 13:12（或下午 1:12）。

A typical mistake is to forget to add the break time or to convert the minutes incorrectly. Always check whether the question expects the answer in 24‑hour format.

典型的错误是忘记加休息时间或分钟换算错误。务必确认题目是否要求用 24 小时制作答。

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5. Case 4: Survey Data Analysis | 案例四：调查数据分析

Thirty students rated a school canteen from 1 (poor) to 5 (excellent). The results: four students gave 1, six gave 2, ten gave 3, seven gave 4, and three gave 5. Calculate the mean, median and mode.

30 名学生为学校食堂打分，从 1（差）到 5（优）。结果如下：4 人给 1 分，6 人给 2 分，10 人给 3 分，7 人给 4 分，3 人给 5 分。计算平均数、中位数和众数。

For the mean, multiply each score by its frequency, sum the products, then divide by the total number of students. (1×4) + (2×6) + (3×10) + (4×7) + (5×3) = 4 + 12 + 30 + 28 + 15 = 89. Mean = 89 ÷ 30 ≈ 2.97 (to 2 decimal places).

计算平均数时，将每个分数乘以频数，求和后再除以学生总人数。(1×4) + (2×6) + (3×10) + (4×7) + (5×3) = 4 + 12 + 30 + 28 + 15 = 89。平均数 = 89 ÷ 30 ≈ 2.97（保留两位小数）。

The median is the middle value when all 30 scores are listed in order. The 15th and 16th values both fall in the score 3 group, so median = 3. The mode is the score with the highest frequency, which is 3.

中位数是将全部 30 个分数按顺序排列后位于中间的值。第 15 和第 16 个值都在分数 3 那一组，因此中位数 = 3。众数是出现频数最高的分数，即 3。

Mean = Σ(f×x) ÷ Σf

When working with grouped data, always use the exact boundaries. In this case the data is discrete, so finding the median is straightforward.

处理分组数据时，务必使用精确的边界。本例为离散数据，因此求中位数较简单。

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6. Case 5: Recipe Adjustment | 案例五：食谱调整

A recipe for 6 people requires 450 g of flour, 200 g of sugar and 3 eggs. You need to make a portion for 10 people and also reduce the amount of sugar per person by 10%. Find the new quantities.

一份供 6 人食用的食谱需要 450 克面粉、200 克糖和 3 个鸡蛋。你需要做出供 10 人食用的分量，并且每人糖的用量要减少 10%。求新的用量。

First, find the scaling factor: 10 ÷ 6 = 10/6 = 5/3. Multiply the original flour by this factor: 450 × 5/3 = 750 g. For sugar: 200 × 5/3 = 1000/3 ≈ 333.3 g. But then apply the 10% reduction: new sugar = 333.3 × 0.9 = 300 g (rounded to the nearest gram). Eggs: 3 × 5/3 = 5 eggs.

首先求出比例因子：10 ÷ 6 = 10/6 = 5/3。将原面粉用量乘以该因子：450 × 5/3 = 750 克。糖：200 × 5/3 = 1000/3 ≈ 333.3 克。但接着要减少 10%，新糖量 = 333.3 × 0.9 = 300 克（四舍五入到克）。鸡蛋：3 × 5/3 = 5 个。

New quantity = Original × (new servings ÷ original servings)

Always check whether ingredients like eggs can be given as decimals. In reality you would round to the nearest whole egg, but the calculation gives a whole number here.

始终检查类似鸡蛋这样的配料能否用小数表示。实际中你可能要四舍五入到最接近的整数，但本题计算结果正好是整数。

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7. Case 6: Savings and Interest | 案例六：储蓄与利息

You deposit £1500 in a savings account that offers 2.5% annual compound interest. How much will the account be worth after 3 years? Also calculate the final amount if simple interest were used instead.

你将 1500 英镑存入一个年利率为 2.5% 的复利储蓄账户。3 年后账户价值是多少？如果使用单利，最终金额又是多少？

Compound interest formula: A = P × (1 + r)ᵗ, where P = 1500, r = 0.025, t = 3. So A = 1500 × (1.025)³. First calculate (1.025)³ = 1.025 × 1.025 × 1.025 ≈ 1.07689. Then A = 1500 × 1.07689 = 1615.34 (to the nearest penny).

复利公式：A = P × (1 + r)ᵗ，其中 P = 1500，r = 0.025，t = 3。因此 A = 1500 × (1.025)³。先计算 (1.025)³ = 1.025 × 1.025 × 1.025 ≈ 1.07689。然后 A = 1500 × 1.07689 = 1615.34（精确到便士）。

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