📚 Case Study Practical Exercises | 案例分析实战演练
Year 7 AQA Maths often includes case study questions that test your ability to apply number skills, algebra, geometry and statistics to real-life situations. This revision guide provides a series of practical exercises designed to build your confidence. Each scenario requires you to read carefully, extract key information and choose the right mathematical tools. By working through these examples, you will strengthen your problem-solving skills and be better prepared for any assessment that asks you to think beyond simple calculations.
AQA Year 7 数学常常包含案例分析题，考查你将数字技巧、代数、几何和统计知识应用到现实生活中的能力。这份复习指南提供了一系列精心设计的实战演练，帮助树立信心。每个情境都要求你仔细阅读、提取关键信息并选择合适的数学工具。通过完成这些例子，你将提升解决问题的能力，为那些要求超越简单计算的评估做好更充分的准备。
1. Understanding Case Study Questions | 理解案例分析题
Case study questions describe a realistic scenario and then ask a series of connected problems. The first step is to read the entire description without doing any maths. Highlight the numbers, units and keywords such as "total", "each", "per person", "discount" and "change". Then identify what the question actually wants you to find. Finally, decide which operations you need: addition, subtraction, multiplication, division or a combination. Always show your working clearly and check that your answer makes sense in the context.
案例分析题会描述一个真实的情境，然后提出一系列关联的问题。第一步是不做任何运算，通读全部描述。标出数字、单位以及如 "总共"、"每个"、"每人"、"折扣" 和 "找零" 这样的关键词。然后确定题目究竟让你求什么。最后决定需要哪种运算：加法、减法、乘法、除法或组合。始终清晰地展示你的计算过程，并检查答案在背景中是否合理。
A good habit is to use estimation before calculating. For instance, if a trip costs £4.80 per student for 28 students, round to £5 and 30 to get an estimate of £150. Later, compare your exact answer with the estimate to spot careless errors. With practice, you will find that case study questions become enjoyable puzzles rather than scary obstacles.
一个好习惯是在计算前先估算。例如，如果一次旅行每名学生费用为 4.80 英镑，共 28 名学生，四舍五入为 5 英镑和 30 人，估算值约为 150 英镑。之后，将精确答案与估算值进行比较，以发现粗心导致的错误。经过练习，你会发现案例分析题会变成有趣的谜题，而不是可怕的障碍。
2. Budgeting a School Trip | 学校旅行预算规划
A class of 30 students is planning a visit to a science museum. The ticket costs £4.50 per student and a group booking fee of £15 is applied once per class. The school has provided a budget of £200. The trip organiser also needs to buy a packed snack for each student at £1.35 per snack. Will the budget be enough? If so, how much money will remain? If not, how much extra money is needed?
一个 30 人的班级计划参观科学博物馆。每名学生门票 4.50 英镑，每个班级一次性收取 15 英镑的团体预订费。学校提供了 200 英镑的预算。活动组织者还需要为每名学生购买一份打包零食，每份 1.35 英镑。预算够用吗？如果够，会剩下多少钱？如果不够，还需要额外多少资金？
Total ticket cost = 30 × £4.50 + £15 = £135 + £15 = £150
门票总花费 = 30 × 4.50 英镑 + 15 英镑 = 135 英镑 + 15 英镑 = 150 英镑
Next, find the cost of snacks: 30 × £1.35 = £40.50. Add this to the ticket total to get the overall expense: £150 + £40.50 = £190.50. Because £190.50 is less than the £200 budget, the trip can go ahead. Subtract to find the leftover amount: £200 − £190.50 = £9.50. This remaining £9.50 could be used for an emergency fund or an extra class treat.
接着计算零食花费：30 × 1.35 英镑 = 40.50 英镑。将这一数额加到门票总费用上得到总支出：150 英镑 + 40.50 英镑 = 190.50 英镑。由于 190.50 英镑小于 200 英镑的预算，旅行可以成行。做减法得出剩余金额：200 英镑 − 190.50 英镑 = 9.50 英镑。这省下的 9.50 英镑可用作应急基金或额外的班级庆祝。
3. Party Planning with Fractions | 用分数规划聚会
Tom is organising a birthday party for himself and 11 friends. He orders 4 identical rectangular cakes. Each cake is cut into 6 equal slices. Tom expects every child, including himself, to eat 3 slices of cake. He wonders whether he has ordered enough cake and wants to describe any leftovers as a fraction of a whole cake.
汤姆为自己和 11 个朋友组织生日聚会。他订了 4 个相同的长方形蛋糕。每个蛋糕切成相等的 6 片。汤姆预计每个孩子（包括他自己）吃 3 片蛋糕。他想知道蛋糕是否够，并希望用几分之一个整蛋糕来表示剩余量。
If each child eats 3 slices, the total slices needed = 12 × 3 = 36 slices. Since 36 > 24, there are not enough slices. Tom is short of 36 − 24 = 12 slices. In terms of whole cakes, 12 slices out of 6 slices per cake equals 12/6 = 2 whole cakes. So Tom needed 2 more whole cakes, making a total of 6 cakes.
首先计算可用的总片数：4 块蛋糕 × 每块 6 片 = 24 片。儿童总数为 1（汤姆）+ 11 = 12 人。如果每个孩子吃 3 片，所需总片数 = 12 × 3 = 36 片。由于 36 > 24，蛋糕片数不够。汤姆缺少 36 − 24 = 12 片。以整块蛋糕为单位，12 片除以每块 6 片等于 12/6 = 2 块整蛋糕。因此汤姆还需要 2 块整蛋糕，总共需要 6 块。
4. Measuring for a DIY Project | DIY 项目测量
Amira wants to cover one rectangular wall of her bedroom with sticky photo tiles. The wall is 2.5 metres high and 3.8 metres wide. Each photo tile is a square with side length 0.5 metres and costs £1.20. The tiles are sold in packs of 8. Calculate the wall area, the number of tiles needed, the number of packs required and the total cost. Amira has a budget of £50. Will she have any money left?
阿米拉想用可粘贴的照片瓷砖覆盖卧室的一面长方形墙壁。墙高 2.5 米、宽 3.8 米。每块照片瓷砖是边长为 0.5 米的正方形，价格为 1.20 英镑。瓷砖每 8 块为一包出售。请计算墙壁面积、所需瓷砖数量、需要的包数以及总花费。阿米拉的预算为 50 英镑，她会有剩余的钱吗？
Wall area = 2.5 m × 3.8 m = 9.5 m²
墙壁面积 = 2.5 米 × 3.8 米 = 9.5 平方米
ks: 38 ÷ 8 = 4.75, so she must buy 5 packs, giving her 5 × 8 = 40 tiles. Total cost = 40 × £1.20 = £48. Her budget is £50, so she will have £50 − £48 = £2 left. The extra 2 tiles can be kept as spares.
一块瓷砖的面积 = 0.5 × 0.5 = 0.25 平方米。所需瓷砖数量 = 9.5 ÷ 0.25 = 38 块。因为瓷砖每包 8 块，计算包数：38 ÷ 8 = 4.75，所以她必须购买 5 包，得到 5 × 8 = 40 块瓷砖。总花费 = 40 × 1.20 英镑 = 48 英镑。她的预算是 50 英镑，因此会剩余 50 − 48 = 2 英镑。多余的 2 块瓷砖可留作备用。
5. Interpreting a Bar Chart | 解读条形图
A school library recorded the number of books borrowed each day from Monday to Friday. The bar chart data shows: Monday 45 books, Tuesday 62 books, Wednesday 55 books, Thursday 70 books, Friday 83 books. The librarian needs to report the total weekly borrowing, the mean number of books per day, and the difference between the busiest and quietest day.
学校图书馆记录了周一到周五每天借出的图书数量。条形图数据为：周一 45 本，周二 62 本，周三 55 本，周四 70 本，周五 83 本。图书馆员需要报告一周的总借阅量、日均借阅量以及借出量最多和最少的日子的差值。
iest day is Friday with 83 books and the quietest is Monday with 45 books. The difference = 83 − 45 = 38 books. If the librarian wants to display this as a bar chart, the vertical axis should be labelled "Number of books" and the horizontal axis "Day", with each bar properly scaled.
总借阅量 = 45 + 62 + 55 + 70 + 83 = 315 本。平均值通过总数除以天数得出：315 ÷ 5 = 63 本/天。借阅量最多的一天是周五（83 本），最少的是周一（45 本）。差值 = 83 − 45 = 38 本。如果图书馆员想用条形图展示这些数据，纵轴应标注 "图书数量"，横轴标 "星期"，每个条形应适当确定刻度。
The library charges a fine of £0.05 per day for each overdue book. If the total fine collected over the week was £12.60, how many overdue days were paid for in total? Divide £12.60 by £ 60 ÷ 0.05 = 252 overdue days. This figure can be used to discuss average overdue patterns.
图书馆对每本逾期图书每天收取 0.05 英镑的罚款。如果本周收到的罚款总额为 12.60 英镑，一共为多少逾期天数缴纳了罚款？用 12.60 英镑除以 0.05 英镑：12.60 ÷ 0.05 = 252 个逾期天数。该数字可用于讨论平均逾期模式。
6. Using Ratios in Recipes | 在食谱中使用比率
A basic biscuit recipe uses 150 g of butter, 100 g of caster sugar, 250 g of self-raising flour and 1 medium egg to make 20 biscuits. A student needs to produce 70 biscuits for a charity bake sale. Write the recipe ratio for butter to sugar to flour in its simplest form and then scale up the ingredients to make 70 biscuits.
一份基础饼干食谱使用 150 克黄油、100 克细砂糖、250 克自发面粉和 1 个中等大小的鸡蛋，可制作 20 块饼干。一名学生需为慈善烘焙义卖制作 70 块饼干。请写出黄油、砂糖和面粉的食谱比率的最简形式，然后按比例增加配料以制作 70 块饼干。
lour. To find the scale factor for 70 biscuits, divide 70 by 20: factor = 3.5. Multiply each original amount by 3.5: butter = 150 × 3.5 = 525 g, sugar = 100 × 3.5 = 350 g, flour = 250 × 3.5 = 875 g, and eggs = 1 × 3.5 = 3.5 eggs. Since half an egg is impractical, the baker might use 4 eggs and adjust slightly, but 3.5 is acceptable for a maths exercise.
黄油 : 砂糖 : 面粉的比率 = 150 : 100 : 250。所有部分除以 50 化简得：3 : 2 : 5。因此，每 3 份黄油对应 2 份砂糖和 5 份面粉。为求出制作 70 块饼干的倍数因子，用 70 除以 20：因子 = 3.5。将每种原用量乘以 3.5：黄油 = 150 × 3.5 = 525 克，砂糖 = 100 × 3.5 = 350 克，面粉 = 250 × 3.5 = 875 克，鸡蛋 = 1 × 3.5 = 3.5 个。由于半个鸡蛋不太实际，烘焙者可能使用 4 个鸡蛋并稍作调整，但 3.5 在数学练习中是可接受的。
7. Solving Time Problems | 解决时间问题
Liam catches a bus from his home to the leisure centre. The timetable shows departure at 14:25 and arrival at 15:07. How long does the journey take? After swimming, he needs to catch a bus back that leaves at 16:40. If he stays exactly 1 hour and 20 minutes at the leisure centre, will he make the return bus?
利亚姆从家乘公交车去休闲中心。时刻表显示出发时间为 14:25，到达时间为 15:07。行程需要多长时间？游泳后，他需要赶回程公交车，发车时间为 16:40。如果他在休闲中心呆刚好 1 小时 20 分钟，他能赶上回程公交吗？
Journey time = 15:07 − 14:25 = 42 minutes
行程时长 = 15:07 − 14:25 = 42 分钟
so yes, he easily makes the bus. If he had wanted to leave exactly at 16:40, he would have needed to start his activity by 16:40 − 1 h 20 min = 15:20, which is 13 minutes after arrival. Always use a timeline to visualise adding and subtracting times.
他于 15:07 到达，并呆 1 小时 20 分钟。相加：15:07 + 1 小时 20 分 = 16:27。回程车 16:40 出发，因此他有 13 分钟的等待时间，所以是的，他完全能赶上。如果他想要刚好在 16:40 离开，他需在 16:40 − 1 小时 20 分 = 15:20 就开始活动，即到达后 13 分钟。始终使用时间线来直观地做时间加减。
8. Algebra in Real Life: Mobile Phone Plans | 现实中的代数：手机套餐
Two mobile phone companies offer different deals. Company X charges a flat monthly fee of £8 plus £0.04 per minute of calls. Company Y charges £12 per month and £0.02 per minute. Let m represent the number of minutes used in a month. Write an algebraic expression for the monthly cost of each company. Then find the number of minutes for which both plans cost the same.
两家手机公司提供不同的套餐。公司 X 收取 8 英镑的月固定费外加每分钟 0.04 英镑的通话费。公司 Y 每月收 12 英镑，每分钟 0.02 英镑。设 m 代表一个月使用的分钟数。写出每家公司的月费用代数表达式。然后求出使得两个套餐花费相同的分钟数。
Cost of X = 8 + 0.04m
公司 X 的费用 = 8 + 0.04m
Cost of Y = 12 + 0.02m
公司 Y 的费用 = 12 + 0.02m
ivide by 0.02: m = 4 ÷ 0.02 = 200 minutes. If you use fewer than 200 minutes, Company X is cheaper; if more, Company Y is the better deal. This demonstrates how algebra helps you make informed decisions.
为求费用何时相等，令两式相等：8 + 0.04m = 12 + 0.02m。两边同时减去 0.02m：8 + 0.02m = 12。再减 8：0.02m = 4。除以 0.02：m = 4 ÷ 0.02 = 200 分钟。如果使用少于 200 分钟，公司 X 更便宜；如果多于 200 分钟，公司 Y 更划算。这展示了代数如何帮助你做出明智的决定。
9. Area and Perimeter for a Garden | 花园的面积与周长
A rectangular garden measures 15 metres in length and 10 metres in width. In the centre, there is a circular flower bed with a radius of 2 metres. The rest of the garden is covered with lawn. The owner wants to build a fence exactly along the four outer edges of the garden. How much lawn area is there? How many metres of fencing are needed? Use 3.14 as an approximation for π.
一个长方形花园长 15 米、宽 10 米。正中央有一个半径为 2 米的圆形花坛。花园其余部分覆盖着草坪。园主想沿着花园的四条外边沿修建围栏。草坪的面积是多少？需要多少米围栏？使用 3.14 作为 π 的近似值。
Total garden area = 15 × 10 = 150 m²
花园总面积 = 15 × 10 = 150 平方米
Flower bed area = π × r² = 3.14 × 2² = 3.14 × 4 = 12.56 m²
花坛面积 = π × r² = 3.14 × 2² = 3.14 × 4 = 12.56 平方米
eter = 2 × (length + width) = 2 × (15 + 10) = 2 × 25 = 50 metres. The fence goes around the outside, so the flower bed does not affect the perimeter measurement. Therefore, 50 metres of fencing is required.
草坪面积 = 总面积 − 花坛面积 = 150 − 12.56 = 137.44 平方米。长方形花园的周长是外边缘一圈的距离。周长 = 2 × (长 + 宽) = 2 × (15 + 10) = 2 × 25 = 50 米。围栏建在外部，因此花坛不影响周长计算。所以需要 50 米围栏。
10. Mixed Practice: School Fundraising | 综合练习：学校筹款
The student council organises a bake sale to raise money for new sports equipment. They buy ingredients for cupcakes at a total cost of £18.60 and make 120 cupcakes. They sell each cupcake for 35 pence. In addition, they run a raffle that collects £27. The target is to raise at least £75. How much profit do they make from cupcakes alone? Do they meet the overall target?
学生会组织烘焙义卖为购置新体育器材筹款。他们花 18.60 英镑购买纸杯蛋糕原料，制作了 120 个纸杯蛋糕。每个蛋糕售价 35 便士。此外，他们还组织了一次抽奖，筹集了 27 英镑。目标是至少筹集 75 英镑。单凭纸杯蛋糕他们能获利多少？他们是否达到了总目标？
Revenue from cupcakes = 120 × £0.35 = £42
纸杯蛋糕销售收入 = 120 × 0.35 英镑 = 42 英镑
Cupcake profit = revenue − costs = £42 − £18.60 = £23.40. Total money raised = cupcake profit + raffle money = £23.40 + £27 = £50.40. Since the target is £75, they fall short by £75 − £50.40 = £ example combines decimals, percentages and financial reasoning in one realistic scenario.
纸杯蛋糕利润 = 销售收入 − 成本 = 42 英镑 − 18.60 英镑 = 23.40 英镑。总筹集金额
Published by TutorHao | Year 7 Mathematics Revision Series | aleveler.com
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