📚 Year 7 AQA Further Mathematics: Case Study Practical Exercises | Year 7 AQA 进阶数学：案例分析实战演练
Mathematics comes alive when it is applied to real-world situations. This article presents a series of case study practical exercises designed for Year 7 students following the AQA Further Mathematics pathway. Each case study challenges you to think logically, apply multiple skills, and communicate your reasoning clearly. You will explore topics including number operations, algebra, ratio, geometry, statistics, and logic – all woven into practical scenarios that mirror everyday problem‑solving. Work through each exercise step by step, and use the detailed bilingual explanations to deepen your understanding.
当数学应用于真实场景时，它变得生动起来。本文为学习 AQA 进阶数学的 Year 7 学生提供一系列案例分析实战演练。每个案例都要求你进行逻辑思考，综合运用多种技能，并清晰地表达推理过程。你将探索运算、代数、比、几何、统计和逻辑等主题——全都融入日常生活中常见的实际问题场景。请逐步完成每个练习，并利用详细的中英双语讲解来加深理解。
1. Understanding Case Studies | 认识案例分析
A case study in mathematics is a detailed examination of a particular situation or problem that requires you to extract relevant data, choose suitable strategies, and justify your solutions. Unlike routine exercises, case studies often contain extra information or ambiguous details – you must decide what is important. Developing this skill helps you become a confident and independent problem‑solver, ready for the challenges of Further Mathematics at AQA level.
数学中的案例分析是对某个特定情境或问题的详细探究，你需要提取相关数据、选择合适的策略并论证你的解答。与常规练习题不同，案例分析常包含多余信息或模棱两可的细节——你必须判断哪些才是重要的。培养这项技能能让你成为自信、独立的问题解决者，为 AQA 进阶数学的挑战做好准备。
In each case study that follows, you will see a scenario, a set of questions, and a fully worked solution with explanations. Try to solve the problem on your own first, then compare your approach with the model answer.
在下面的每个案例中，你都将看到一个情境、一组问题，以及带有讲解的完整解答。请先尝试自己解决问题，然后将你的方法跟参考答案进行对比。
2. Case 1: The Sponsored Walk | 案例一：慈善步行
A group of Year 7 students organises a sponsored walk to raise money for charity. Each student finds sponsors who agree to pay a fixed amount for every kilometre walked. Amira walks 9 km and is sponsored by 4 people: two pay £0.50 per km, one pays £0.80 per km, and the fourth pays £1.20 per km. Liam walks 7.5 km and has 3 sponsors who all pay £0.90 per km. Question: Who raises more money, and by how much? Also, if the charity receives 85% of the total raised (the rest covers costs), how much does the charity actually receive?
一群 Year 7 学生组织慈善步行募捐。每个学生找到赞助人，赞助人同意按每公里支付固定金额。Amira 走了 9 km，有 4 位赞助人：两位付 £0.50/km，一位付 £0.80/km，第四位付 £1.20/km。Liam 走了 7.5 km，有 3 位赞助人，均付 £0.90/km。问题：谁筹到的钱更多？多多少？此外，如果慈善机构收到总金额的 85%（其余用作成本），那么慈善机构实际收到多少？
Step 1 – Amira's total: Amount from first two sponsors = 2 × (9 × 0.50) = 2 × 4.50 = £9.00. Third sponsor = 9 × 0.80 = £7.20. Fourth sponsor = 9 × 1.20 = £10.80. Total = 9.00 + 7.20 + 10.80 = £27.00.
步骤一 – Amira 的总金额：前两位赞助人 = 2 × (9 × 0.50) = 2 × 4.50 = £9.00。第三位 = 9 × 0.80 = £7.20。第四位 = 9 × 1.20 = £10.80。总计 = 9.00 + 7.20 + 10.80 = £27.00。
Step 2 – Liam's total: 3 sponsors × (7.5 × 0.90) = 3 × 6.75 = £20.25.
步骤二 – Liam 的总金额：3 位赞助人 × (7.5 × 0.90) = 3 × 6.75 = £20.25。
Step 3 – Comparison: Amira raises £27.00 − £20.25 = £6.75 more than Liam.
步骤三 – 比较：Amira 比 Liam 多筹集 £27.00 − £20.25 = £6.75。
Step 4 – Charity amount: Total raised = 27.00 + 20.25 = £47.25. 85% of £47.25 = 0.85 × 47.25 = £40.1625, which rounds to £40.16. The charity receives £40.16.
步骤四 – 慈善金额：总筹集金额 = 27.00 + 20.25 = £47.25。£47.25 的 85% = 0.85 × 47.25 = £40.1625，四舍五入为 £40.16。慈善机构收到 £40.16。
3. Case 2: Designing a Rectangular Garden | 案例二：设计矩形花园
A rectangular garden is to be fenced on three sides, with the house wall forming the fourth side. The total length of fencing available is 14 metres. The garden must be at least 2 metres wide. The length perpendicular to the house is called the depth (d metres) and the width along the house is w metres. Find all possible integer dimensions (d, w) that use exactly 14 m of fencing. Then decide which pair gives the largest area for planting.
一个矩形花园三面围栏，屋墙作为第四边。可用围栏总长为 14 米。花园宽度不得小于 2 米。与房屋垂直的长度称为深度（d 米），沿房屋的宽度为 w 米。找出所有恰好用尽 14 米围栏的整数尺寸对 (d, w)。然后决定哪一对提供最大的种植面积。
width: 2d + w = 14, with w ≥ 2. Rearranging: w = 14 − 2d. Since w ≥ 2, we need 14 − 2d ≥ 2 ⇒ 12 ≥ 2d ⇒ ≤ 6. Also d must be positive. Integer values: d = 1, 2, 3, 4, 5, 6. Calculate w for each:
围栏覆盖两个深度和一个宽度：2d + w = 14，且 w ≥ 2。整理得：w = 14 − 2d。因为 w ≥ 2，所以 14 − 2d ≥ 2 ⇒ 12 ≥ 2d ⇒ d ≤ 6。d 还需为正数。整数值：d = 1, 2, 3, 4, 5, 6。计算对应的 w：
| d (m) | w = 14 − 2d (m) | Valid? |
| 1 | 12 | Yes |
| 2 | 10 | Yes |
| 3 | 8 | Yes |
| 4 | 6 | Yes |
| 5 | 4 | Yes |
| 6 | 2 | Yes |
All six pairs are valid. Now area A = d × w = d(14 − 2d) = 14d − 2d². Calculate area for each:
所有六对尺寸均有效。现在面积 A = d × w = d(14 − 2d) = 14d − 2d²。逐一计算面积：
- d=1: A = 1 × 12 = 12 m²
- d=2: A = 2 × 10 = 20 m²
- d=3: A = 3 × 8 = 24 m²
- d=4: A = 4 × 6 = 24 m²
- d=5: A = 5 × 4 = 20 m²
- d=6: A = 6 × 2 = 12 m²
sions 3 m × 8 m and 4 m × 6 m. Both give the same area, so the gardener can choose either.
最大面积为 24 m²，由 3 m × 8 m 和 4 m × 6 m 两种尺寸实现。两者面积相同，因此花园设计可任选其一。
4. Case 3: Party Planning and Ratio | 案例三：派对策划与比例
A party mix consists of nuts, pretzels, and dried fruit in the ratio 5 : 3 : 2 by mass. The organiser needs to prepare 2.5 kg of the mix. Furthermore, the cost prices are £8 per kg for nuts, £5 per kg for pretzels, and £6 per kg for dried fruit. What is the total cost of the ingredients? If the mix is sold in 150 g bags and each bag must yield a profit of 40% on the cost of ingredients, what should be the selling price per bag?
派对什锦按质量由坚果、椒盐脆饼和干果以 5 : 3 : 2 的比例组成。组织者需准备 2.5 kg 该什锦。此外，坚果成本 £8/kg，椒盐脆饼 £5/kg，干果 £6/kg。原料总成本是多少？若什锦以 150 g 一袋出售，且每袋需在原料成本基础上盈利 40%，每袋售价应为多少？
kg; Pretzels: 3 × 0.25 = 0.75 kg; Dried fruit: 2 × 0.25 = 0.5 kg. Cost: Nuts = 1.25 × 8 = £10.00; Pretzels = 0.75 × 5 = £3.75; Fruit = 0.5 × 6 = £3.00. Total ingredient cost = £10.00 + £3.75 + £3.00 = £16.75 for 2.5 kg.
总比例份数 = 5 + 3 + 2 = 10。每份质量 = 2.5 kg ÷ 10 = 0.25 kg。坚果：5 × 0.25 = 1.25 kg；椒盐脆饼：3 × 0.25 = 0.75 kg；干果：2 × 0.25 = 0.5 kg。成本：坚果 = 1.25 × 8 = £10.00；脆饼 = 0.75 × 5 = £3.75；干果 = 0.5 × 6 = £3.00。原料总成本 = £10.00 + £3.75 + £3.00 = £16.75（对应 2.5 kg）。
bag pricing based on the mix, so consider 16 bags from this batch). Total cost per batch = £16.75, so cost per bag = £16.75 ÷ 16 = £1.046875 ≈ £1.05. For a 40% profit on cost, selling price = cost × 1.40 = £1.046875 × 1.40 = £1.465625 ≈ £1.47 per bag.
2.5 kg 可制成 150 g 的袋数 = 2500 g ÷ 150 g = 16.666...，即可做 16 整袋。整批成本 £16.75，因此每袋成本 = £16.75 ÷ 16 ≈ £1.05。按成本盈利 40%，售价 = 成本 × 1.40 = £1.046875 × 1.40 ≈ £1.47/袋。
5. Case 4: School Book Fair Statistics | 案例四：学校书展统计
During a book fair, the number of books sold each day from Monday to Friday was recorded: Mon 24, Tue 31, Wed 28, Thu 35, Fri 42. The librarian also noted the types: fiction and non-fiction. On Friday, 45% of sales were non‑fiction. The total number of non‑fiction books sold over the five days was 73. Find the number of non‑fiction books sold on Friday and then determine the percentage of non‑fiction sales for the whole week. Finally, calculate the mean number of fiction books sold per day.
书展期间，周一至周五每天售书量记录如下：周一 24，周二 31，周三 28，周四 35，周五 42。图书管理员还记录了书籍类型：虚构类和非虚构类。周五的销售中 45% 为非虚构类。五天售出的非虚构类图书总数为 73。求周五售出的非虚构类图书数量，然后确定整周非虚构类销售占比。最后计算平均每天售出的虚构类图书数量。
Friday total sales = 42. Non‑fiction on Friday = 45% le; typical interpretation: 18.9 means 19 books, or use exact 18.9 for further calculation? Let's assume 45% of 42 = 18.9, so 19 if we round, but it's safer to keep 18.9 and note that real books must be integers. Here we' for percentage we can use exact. For realistic case, suppose the librarian counted 19 non‑fiction on Friday. Then total non‑fiction = 73, so Mon–Thu non‑fiction = 73 − 19 = 54 books.
周五总销量 = 42。周五非虚构类 = 42 的 45% = 0.45 × 42 = 18.9，按四舍五入为 19 本。若非虚构总数 = 73，则周一至周四非虚构 = 73 − 19 = 54 本。
Total weekly sales = 24 + 31 + 28 + 35 + 42 = 160 books. Total non‑fiction = 73. Percentage non‑fiction = (73 ÷ 160) × 100% = 45.625% ≈ 45.6%.
周总销量 = 24 + 31 + 28 + 35 + 42 = 160 本。非虚构总数为 73。非虚构占比 = (73 ÷ 160) × 100% = 45.625% ≈ 45.6%。
87 books. Mean fiction per day = 87 ÷ 5 = 17.4, so on average 17.4 fiction books per day.
虚构类总数 = 160 − 73 = 87 本。平均每天虚构类 = 87 ÷ 5 = 17.4，即平均约 17.4 本/天。
6. Case 5: Mobile Phone Tariffs | 案例五：手机资费对比
A student wants to choose between two monthly phone plans. Plan X charges £15 per month plus 5p per minute of calls. Plan Y charges £22 per month but includes 200 minutes free, then charges 8p per additional minute. In a typical month, the student uses between 180 and 300 minutes. Analyse which plan is cheaper for 180, 250, and 300 minutes. At what usage level (in minutes) do the two plans cost the same?
某学生要在两个月话费套餐中选择。套餐 X 月租 £15，通话每分钟 5p。套餐 Y 月租 £22，含 200 分钟免费通话，超出后每分钟 8p。典型月份内，该生通话时长为 180 至 300 分钟。分析通话 180 分钟、250 分钟和 300 分钟时哪个套餐更便宜。当通话时长为多少分钟时，两种套餐花费相等？
Convert pence to pounds: 5p = £0.05, 8p = £0.08.
将便士转换为英镑：5p = £0.05，8p = £0.08。
- For 180 min: Plan X cost = 15 + 180×0.05 = 15 + 9 = £24.00. Plan Y: 180 ≤ 200, so cost = £22.00. Plan Y is cheaper by £2.
- 通话 180 分钟：套餐 X 费用 = 15 + 180×0.05 = 15 + 9 = £24.00。套餐 Y：180 ≤ 200，费用 = £22.00。套餐 Y 便宜 £2。
- For 250 min: Plan X = 15 + 250×0.05 = 15 + 12.50 = £27.50. Plan Y = 22 + (250 − 200)×0.08 = 22 + 4 = £26.00. Plan Y is cheaper.
- 通话 250 分钟：套餐 X = 15 + 250×0.05 = 15 + 12.50 = £27.50。套餐 Y = 22 + (250 − 200)×0.08 = 22 + 4 = £26.00。套餐 Y 更便宜。
- For 300 min: Plan X = 15 + 300×0.05 = £30.00. Plan Y = 22 + (100)×0.08 = 22 + 8 = £30.00. Both cost £30.
- 通话 300 分钟：套餐 X = 15 + 300×0.05 = £30.00。套餐 Y = 22 + 100×0.08 = 22 + 8 = £30.00。两者相同。
m > 200: 15 + 0.05m = 22 + 0.08(m − 200). Solve: 15 + 0.05m = 22 + 0.08m − 16 = 6 + 0.08m ⇒ 15 − 6 = 0.08m − 0.05m ⇒ 9 = 0.03m ⇒ m = 300. For m ≤ 200, Plan Y is always cheaper because £22 < £15 + 0.05×200 = £25. So the plans cost the same at 300 minutes.
当 m > 200 时令花费相等：15 + 0.05m = 22 + 0.08(m − 200)。解得：15 + 0.05m = 22 + 0.08m − 16 = 6 + 0.08m ⇒ 15 − 6 = 0.08m − 0.05m ⇒ 9 = 0.03m ⇒ m = 300。当 m ≤ 200 时，套餐 Y 总是更便宜（£22 < £15 + 0.05×200 = £25）。因此两套餐在 300 分钟时花费相同。
7. Case 6: Constructing a Triangular Flower Bed | 案例六：三角形花坛的构建
A triangular flower bed has sides in the ratio 3 : 4 : 5. The perimeter is 36 metres. Find the length of each side. Then calculate the area of the triangle. If soil costs £2.50 per square metre, what is the total cost of soil needed to cover the bed? (Hint: a 3‑4‑5 triangle is right‑angled.)
一个三角形花坛三边之比为 3 : 4 : 5，周长为 36 米。求每条边的长度。然后计算三角形的面积。若土壤每平方米 £2.50，覆盖整个花坛需要多少土壤费用？（提示：3‑4‑5 三角形为直角三角形。）
t side (15 m) is the hypotenuse; the perpendicular sides are 9 m and 12 m. Area of a right‑ ½ × base × height. Using 9 m and 12 m as the legs: Area = (9 × 12) ÷ 2 = 108 ÷ 2 = 54 m².
边长比例份数：3 + 4 + 5 = 12。每份 = 36 ÷ 12 = 3 m。边长：3×3 = 9 m，4×3 = 12 m，5×3 = 15 m。最长边（15 m）为斜边；两垂直边分别为 9 m 和 12 m。直角三角形面积 = ½ × 底 × 高。以 9 m 和 12 m 为直角边：面积 = (9 × 12) ÷ 2 = 108 ÷ 2 = 54 m²。
Cost of soil = 54 × £2.50 = £135.00.
土壤费用 = 54 × £2.50 = £135.00。
8. Case 7: Number Patterns and the Bouncing Ball | 案例七：数字模式与弹跳球
A ball is dropped from a height of 128 cm. Each time it bounces, it reaches ¾ of its previous height. What height does it reach after the 3rd bounce? After how many bounces is the height first less than 20 cm? Write the sequence of heights for the first four bounces.
一个球从 128 cm 高度落下。每次弹跳都达到前一次高度的 ¾。第 3 次弹跳后高度是多少？经过多少次弹跳高度首次低于 20 cm？写出前四次弹跳的高度序列。
= 72 cm. 3rd: 72 × ¾ = 54 cm. 4th: 54 × ¾ = 40.5 cm. Sequence: 128, 96, 72, 54, 40.5, ...
初始第 0 次弹跳高度 = 128 cm。第 1 次弹跳后：128 × ¾ = 96 cm。第 2 次：96 × ¾ = 72 cm。第 3 次：72 × ¾ = 54 cm。第 4 次：54 × ¾ = 40.5 cm。序列：128, 96, 72, 54, 40.5, ...
22.78125 × ¾ = 17.0859375 cm. So after the 7th bounce, height < 20 cm. Answer: 7 bounces.
继续：第 5 次弹跳：40.5 × ¾ = 30.375 cm。第 6 次：30.375 × ¾ ≈ 22.78 cm。第 7 次：22.78 × ¾ ≈ 17.09 cm。因此第 7 次弹跳后高度首次低于 20 cm。答案：7 次弹跳。
9. Case 8: The Birthday Logic Puzzle | 案例八：生日逻辑谜题
Three friends, Ali, Ben, and Chloe, have birthdays in different months. Use the clues to work out who was born in which month. Clues: (1) Ali’s birthday month has more letters than Ben’s. (2) Chloe’s birthday month is not in summer (June, July, August). (3) Ben’s birthday month has exactly 30 days. (4) The three months are February, June, and September. Determine each person’s birthday month.
三位朋友 Ali、Ben 和 Chloe 的生日在不同月份。利用线索推断每个人的生日月份。线索：(1) Ali 生日月份字母数比 Ben 的多。(2) Chloe 生日月份不在夏季（六月、七月、八月）。(3) Ben 的生日月份恰好有 30 天。(4) 三个月份分别为二月、六月和九月。确定每个人的生日月份。
Months given: February (letters: 8), June (4), September (9). 30‑day months among these: June has 30 days; September has 30 days; February has 28 or 29, not 30. So Ben must be either June or September. Clue 1: Ali’s month has more letters than Ben’s. Clue 2: Chloe not in summer, so Chloe cannot be June. So Chloe could be February or September.
给定月份：二月（字母数 8）、六月（4）、九月（9）。其中 30 天的月份：六月 30 天；九月 30 天；二月 28/29 天。因此 Ben 必为六月或九月。线索 1：Ali 的月份字母数多于 Ben。线索 2：Chloe 不在夏季，因此 Chloe 不能是六月。所以 Chloe 可能是二月或九月。
loe ≠ summer, so Chloe could be February or September. This is possible. Try Ben = September (9 letters). Then Ali must have >9 letters? None of the months has >9 letters. Impossible. Therefore Ben must be June.
尝试 Ben = 六月（4 字母）。则 Ali 字母数须多于 4，Ali 可为二月（8）或九月（9）。Chloe 取剩下的。若 Ben=六月，Chloe 不能是夏季，故 Chloe 可为二月或九月，可行。再试 Ben=九月（9 字母），则 Ali 须多于 9，无月份符合，不可能。因此 Ben 只能是六月。
oes not exclude February or September. So both are possible for Chloe. Need more deduction. We have no further clues? ave >4, so both work. Chloe would then be the other month. Is there any restriction on Ali? No. So two possibilities? Let's re‑read clues: "The three months are February, June, and September." So it' uary or September, we seem to have two solutions. But perhaps we missed that Chloe’s month "is not in summer" lutions: (Ali=Sep, Ben=Jun, Chloe=Feb) or (Ali=Feb, Ben=Jun, Chloe=Sep). Both satisfy clues. Sometimes a logic puzzle might have multiple or need an additional implicit assumption. Perhaps the wording "Ali's month has more letters than Ben's" – ution 1: Ali Sep (9), Ben Jun (4). In solution 2: Ali Feb (8), Ben Jun (4). Both valid. We'll present both and note that both satisfy all clues. Alternatively, check if "Chloe's birthday month is not in summer" could restrict September? September? er is autumn, not summer. So indeed both work. So the puzzle has two possible assignments.
Ben 确定为六月后，剩余二月和九月。Chloe 不能是夏季（六月已属于 Ben），因此 Chloe 可为二月或九月，均非夏季。Ali 则取另一月。得到两种可能：(Ali 九月, Ben 六月, Chloe 二月) 或 (Ali 二月, Ben 六月, Chloe 九月)。两解均满足所有线索。因此存在两种可能分配。
10. Case 9: Fitness Tracker and Averages | 案例九：健身追踪器与平均值
Anika wears a fitness tracker that records the number of steps she takes each day. Over the first 9 days of a month, her mean steps per day was 8430. On the 10th day, she walked only 3200 steps due to heavy
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