📚 Year 7 CIE Maths: Case Study Practice | 七年级CIE数学案例分析实战演练
Welcome to this case study practice revision guide for Year 7 CIE Maths. Here we will explore real-life scenarios where you apply your knowledge of number operations, percentages, geometry, statistics, and algebra. Each case study presents a problem, breaks it down step by step, and shows how mathematical thinking helps solve everyday challenges. Practise these examples to build confidence and fluency in tackling word problems and multi-step tasks.
欢迎使用这份针对七年级CIE数学的案例分析实战演练复习指南。我们将一起探索真实生活场景，应用你在数字运算、百分数、几何、统计和代数方面的知识。每个案例都展示一个问题，逐步分解，展示数学思维如何帮助解决日常挑战。通过练习这些例子，你将建立解决文字题和多步骤任务的信心与熟练度。
1. Supermarket Discount and Multibuy | 超市折扣与多买优惠
A supermarket has a 25% discount on all boxes of cereal. Each box normally costs £2.80. Sarah wants to buy three boxes. She also has a voucher that gives an additional 10% off the total already reduced amount. How much does she finally pay?
一家超市所有盒装麦片打七五折（25%折扣）。每盒原价2.80英镑。萨拉想买三盒。她还拥有一张优惠券，可在已折扣总金额上再打九折。她最终支付多少？
First, find the price of one box after the 25% discount. 25% of £2.80 is 0.25 × 2.80 = £0.70. So the discounted price per box is £2.80 – £0.70 = £2.10.
首先，计算一盒在25%折扣后的价格。25% × 2.80英镑 = 0.25 × 2.80 = 0.70英镑。因此每盒折扣价为2.80 – 0.70 = 2.10英镑。
Discount per box = 25% × £2.80 = 0.25 × 2.80 = £0.70
Discounted price per box = £2.80 – £0.70 = £2.10
For three boxes, the total before the voucher is 3 × £2.10 = £6.30. The voucher gives an extra 10% off this total. So the final discount amount is 10% × £6.30 = £0.63. Final amount = £6.30 – £0.63 = £5.67.
三盒的金额在使用优惠券前为3 × 2.10英镑 = 6.30英镑。优惠券再减免总金额的10%。所以额外折扣为10% × 6.30 = 0.63英镑。最终应付6.30 – 0.63 = 5.67英镑。
Total for 3 boxes = 3 × £2.10 = £6.30
Final amount = £6.30 – (10% × £6.30) = £6.30 – £0.63 = £5.67
Sarah pays £5.67. Always check the order of discounts: here the voucher applies after the percentage reduction, so the discounts compound.
萨拉支付5.67英镑。始终注意折扣顺序：这里优惠券是在百分比折扣之后使用，所以折扣累加。
2. Planning a School Trip Budget | 规划学校旅行预算
A school organises a trip to a museum for 45 students and 5 teachers. The coach costs £320 for the day. Museum entry is £6.50 per student and £9.00 per teacher. The school wants to include a packed lunch costing £3.80 per person. Find the total cost of the trip and the amount each student needs to contribute if the teachers’ costs are shared equally among the students.
一所学校组织45名学生和5名教师参观博物馆。大巴车一天费用为320英镑。博物馆门票学生每人6.50英镑，教师每人9.00英镑。学校希望包含每人3.80英镑的打包午餐。计算旅行的总费用，以及如果将教师费用均摊到学生身上，每名学生需贡献多少。
Calculate costs separately. Coach: £320. Student entry: 45 × £6.50 = £292.50. Teacher entry: 5 × £9.00 = £45.00. Packed lunches: total people = 45 + 5 = 50; 50 × £3.80 = £190.00. Total cost = £320 + £292.50 + £45.00 + £190 = £847.50.
分别计算各项费用。大巴车：320英镑。学生门票：45 × 6.50 = 292.50英镑。教师门票：5 × 9.00 = 45.00英镑。打包午餐：总人数45 + 5 = 50；50 × 3.80 = 190.00英镑。总费用 = 320 + 292.50 + 45 + 190 = 847.50英镑。
Now, if the students cover the teachers’ entry (£45) and teachers’ lunches (5 × £3.80 = £19), total extra = £45 + £19 = £64. This amount divided among 45 students is £64 ÷ 45 ≈ £1.42 per student. Total each student pays = their own trip costs + this share = (£6.50 + £3.80) + £1.42 = £11.72. Alternatively, the total cost £847.50 divided by 45 gives £18.83 per student, but careful: this includes the coach cost which already all share. A simpler method: remove coach cost from student share? The problem states teachers’ costs shared equally. So we could say total without coach: £847.50 – £320 = £527.50; then added coach: each student pays £320/45 ≈ £7.11; plus their own entry & lunch (£10.30) = £17.41; plus teachers’ share (£64/45 ≈ £1.42) gives £17.41 + £1.42 = £18.83, which matches £847.50/45. So each student contributes £18.83.
现在，如果学生分摊教师门票（45英镑）和教师午餐（5 × 3.80 = 19英镑），额外总金额为45 + 19 = 64英镑。此金额由45名学生分摊为64 ÷ 45 ≈ 1.42英镑每人。每名学生总支付 = 自身旅行成本 + 此分摊 = (6.50 + 3.80) + 1.42 = 11.72英镑。但注意这未包含大巴车费。另一种计算：总费用847.50英镑除以45得18.83英镑每生。这更直接，因为包含了所有分摊。确实，大巴车费也由学生承担，所以每名学生应付847.50 ÷ 45 = 18.83英镑。因此答案是£18.83。
Total trip cost = £847.50
Amount per student = £847.50 ÷ 45 = £18.83
The school can then collect £18.83 from each student to cover all expenses.
学校即可向每名学生收取18.83英镑以覆盖全部开支。
3. Garden Design: Area and Perimeter | 花园设计：面积与周长
Alex wants to lay turf in a rectangular garden measuring 8.5 m by 5 m. The grass turf costs £4.20 per square metre. He also needs to put a fence around the garden. The fencing panels are 1.8 m wide and cost £12.50 each. He will have a 1 m wide gate on one of the shorter sides, which costs £45. Calculate the total cost for turf and fencing.
艾利克斯想在一个长8.5米、宽5米的矩形花园里铺设草皮。草皮每平方米售价4.20英镑。他还需要在花园四周围上围栏。围栏板每块宽1.8米，单价12.50英镑。他将在一条短边安装一扇1米宽的门，门价格为45英镑。计算草皮和围栏的总花费。
First, find the area of the garden: length × width = 8.5 m × 5 m = 42.5 m². Turf cost = 42.5 × £4.20 = £178.50. Next, perimeter of the rectangle: 2 × (8.5 + 5) = 2 × 13.5 = 27 m. The gate replaces 1 m of fence, so total fence length needed = 27 m – 1 m = 26 m. Number of panels: each panel covers 1.8 m, so divide 26 by 1.8. 26 ÷ 1.8 = 14.44… You cannot buy a fraction of a panel, so Alex must buy 15 panels. Cost of panels = 15 × £12.50 = £187.50. Add the gate: £45. Total fencing cost = £187.50 + £45 = £232.50. Overall total = £178.50 + £232.50 = £411.
首先，求花园面积：长 × 宽 = 8.5米 × 5米 = 42.5平方米。草皮费用 = 42.5 × 4.20 = 178.50英镑。接着，矩形周长 = 2 × (8.5 + 5) = 2 × 13.5 = 27米。门代替1米围栏，因此实际所需围栏长度为27 – 1 = 26米。围栏板数量：每块宽1.8米，所以26 ÷ 1.8 = 14.44… 不可购买零头，艾利克斯须购买15块。围栏板费用 = 15 × 12.50 = 187.50英镑。加上门45英镑，围栏总费用 = 187.50 + 45 = 232.50英镑。总花费 = 178.50 + 232.50 = 411英镑。
Area = 8.5 × 5 = 42.5 m²
Turf cost = 42.5 × £4.20 = £178.50
Fence length = 27 m – 1 m = 26 m
Panels needed: 26 ÷ 1.8 = 14.44 → 15 panels
Total cost = £178.50 + £232.50 = £411
Always round up when buying materials like fencing panels.
购买围栏板等材料时，务必向上取整。
4. Travel Timetable and Average Speed | 旅行时刻表与平均速度
A bus leaves the school at 09:15 and arrives at the activity centre at 11:00. The distance between the two places is 63 km. Calculate the average speed of the bus in km/h. The students have activities until 15:30, then the bus returns, taking the same time as the morning journey. At what time will they arrive back at school?
一辆巴士于09:15离开学校，11:00到达活动中心。两地距离63公里。计算巴士的平均速度（公里/小时）。学生们的活动持续到15:30，之后巴士返回，回程用时与去程相同。他们将于几点回到学校？
First, find the travel time in the morning. From 09:15 to 11:00 is 1 hour and 45 minutes. Convert to hours: 45 minutes = 45/60 = 0.75 hours, so total time = 1.75 hours. Average speed = distance ÷ time = 63 km ÷ 1.75 h = 36 km/h. The return journey will also take 1.75 hours = 1 hour 45 minutes. Departure from activity centre at 15:30, add 1 hour 45 min: 15:30 + 1:45 = 17:15. So arrival back at school is at 17:15.
首先，计算上午行程用时。09:15到11:00为1小时45分钟。转换为小时：45分钟 = 45/60 = 0.75小时，总时间 = 1.75小时。平均速度 = 距离 ÷ 时间 = 63公里 ÷ 1.75小时 = 36公里/小时。回程同样需要1小时45分钟。15:30离开活动中心，加上1小时45分：15:30 + 1:45 = 17:15。他们回到学校的时间为17:15。
Time = 1 h 45 min = 1.75 h
Average speed = 63 km ÷ 1.75 h = 36 km/h
Return arrival = 15:30 + 1 h 45 min = 17:15
Another check: 36 km/h means they travel 36 km in one hour; in 1.75 h they cover 36 × 1.75 = 63 km, correct.
验证：36公里/小时表示每小时行驶36公里，1.75小时行驶36 × 1.75 = 63公里，正确。
5. Analysing Sports Scores: Averages | 分析体育得分：平均数
A basketball player scores the following points in six games: 18, 22, 15, 30, 10, 25. Find the mean, median, and range of her scores. If she wants to raise her mean score to 22 points after the seventh game, what must she score in that game?
一位篮球运动员在六场比赛中的得分为：18、22、15、30、10、25。求她得分的平均数、中位数和极差。如果她想在第七场比赛后将平均得分提升至22分，她在该场必须得多少分？
First, order the scores: 10, 15, 18, 22, 25, 30. Mean = sum of scores ÷ number of games = (18+22+15+30+10+25) ÷ 6 = 120 ÷ 6 = 20 points. Median with six numbers: average of the 3rd and 4th values in order: (18+22)/2 = 20. Range = highest – lowest = 30 – 10 = 20. For the seventh game, total points needed for a mean of 22 after 7 games = 22 × 7 = 154. Current total = 120, so required score = 154 – 120 = 34 points.
首先，将得分排序：10、15、18、22、25、30。平均数 = 总分 ÷ 场次 = (18+22+15+30+10+25) ÷ 6 = 120 ÷ 6 = 20分。六个数据的中位数：第3和第4个值的平均数 (18+22)/2 = 20。极差 = 最大值减最小值 = 30 – 10 = 20。对于第七场比赛，7场比赛后平均22分所需总分 = 22 × 7 = 154。当前总分为120，因此需要得154 – 120 = 34分。
Mean after 6 games = 20
Median = 20
Range = 20
Required score in 7th game = 34
The player needs a very high score in the next game; this shows how one score can influence the mean.
该球员需要在下一场比赛中得到很高的分数；这展示了一个数据如何影响平均数。
6. Scaling a Recipe: Ratio and Proportion | 调整食谱分量：比和比例
A recipe for 6 people requires 250 g of flour, 150 g of sugar, 4 eggs, and 100 ml of milk. How much of each ingredient is needed for 10 people? Express the new amounts.
一份供6人食用的食谱需要250克面粉、150克糖、4个鸡蛋和100毫升牛奶。供10人食用需要每种原料多少？请写出新的用量。
The scaling factor from 6 to 10 people is 10/6 = 5/3 ≈ 1.6667. Multiply each ingredient by this factor. Flour: 250 g × (10/6) = 250 × 10 ÷ 6 = 2500 ÷ 6 = 416.67 g (or 416 2/3 g). Sugar: 150 g × (10/6) = 1500 ÷ 6 = 250 g. Eggs: 4 × (10/6) = 40 ÷ 6 ≈ 6.67, but you can’t use a fraction of an egg practically, so you might round to 7 eggs, or use 6 eggs and adjust slightly; mathematically, 6.67 eggs. Milk: 100 ml × (10/6) = 1000 ÷ 6 ≈ 166.67 ml. For practical cooking, we might say 7 eggs and slightly more milk.
从6人量放大到10人量的比例因子为10/6 = 5/3 ≈ 1.6667。每样原料乘以该因子。面粉：250 g × (10/6) = 250 × 10 ÷ 6 = 2500 ÷ 6 = 416.67克（或416 2/3克）。糖：150 g × (10/6) = 1500 ÷ 6 = 250克。鸡蛋：4 × (10/6) = 40 ÷ 6 ≈ 6.67，实际中不能使用零头鸡蛋，因此可取7个或微调用6个；数学上为6.67个。牛奶：100 ml × (10/6) = 1000 ÷ 6 ≈ 166.67毫升。实际烹饪中，我们可能说7个鸡蛋并略增牛奶。
Scaling factor = 10/6 = 5/3
Flour: 250 g × 10/6 ≈ 416.7 g
Sugar: 150 g × 10/6 = 250 g
Eggs: 4 × 10/6 ≈ 6.67 → 7 eggs in practice
Milk: 100 ml × 10/6 ≈ 166.7 ml
Using ratios ensures the taste remains consistent when scaling recipes up or down.
使用比例可确保食谱在增减人数时味道一致。
7. Mobile Phone Plans: Algebraic Comparison | 手机套餐：代数比较
Two mobile phone plans: Plan A charges a fixed monthly fee of £8 plus 5p per minute of calls. Plan B charges £12 fixed fee but only 4p per minute. For how many minutes of calls per month will the total cost be the same for both plans? Write an equation and solve it.
两个手机套餐：A套餐每月固定费用8英镑，通话每分钟5便士。B套餐固定费用12英镑，通话每分钟仅4便士。每月通话多少分钟时，两种套餐总费用相同？列出方程并求解。
Let the number of minutes be m. Cost for Plan A = 800p + 5m pence (since £8 = 800p). Cost for Plan B = 1200p + 4m pence. Set them equal: 800 + 5m = 1200 + 4m. Subtract 4m from both sides: 800 + m = 1200. Then subtract 800: m = 400 minutes. So at 400 minutes, both plans cost the same. Check: Plan A = 800 + 5×400 = 800+2000 = 2800p = £28. Plan B = 1200 + 4×400 = 1200+1600 = 2800p = £28.
设通话分钟数为 m。A套餐费用 = 800便士 + 5m 便士（因8英镑 = 800便士）。B套餐费用 = 1200便士 + 4m 便士。令二者相等：800 + 5m = 1200 + 4m。两边同时减去4m：800 + m = 1200。再减去800：m = 400分钟。因此通话400分钟时，两套餐费用相同。验证：A = 800 + 5×400 = 800+2000 = 2800便士 = 28英镑。B = 1200 + 4×400 = 1200+1600 = 2800便士 = 28英镑。
800 + 5m = 1200 + 4m
m = 400 minutes
Equal cost = £28
If you talk less than 400 minutes per month, Plan A is cheaper; if more, Plan B is cheaper. Algebra helps make this decision.
如果每月通话少于400分钟，A套餐更便宜；若多于400分钟，B套餐更便宜。代数有助于做出决定。
8. Bearing and Distance: Finding the Shortcut | 方位与距离：寻找捷径
From her house, Anna walks 2 km on a bearing of 060° to reach a park. Then she walks 3 km on a bearing of 150° to a library. Calculate the straight-line distance from her house to the library. (Use scaled drawing or cosine rule? Year 7 may use a scale drawing approach. We will use a scaled drawing and measurement.)
安娜从家出发，沿060°方位行走2公里到达公园，然后沿150°方位行走3公里到达图书馆。计算从她家到图书馆的直线距离。（可以采用比例绘图或余弦定理，但七年级可借助比例绘图和测量。）
Sketch the route: From start (H), draw a line 2 cm (scale 1 cm = 1 km) at 60° from north. From that point, draw 3 cm at 150° (which is 150° clockwise from north, i.e. south-east direction). The angle between the two path segments? The first bearing 060°, second 150°, the difference is 90°? Let’s check: 150° – 60° = 90°, so the two paths are perpendicular. This makes the house-library distance the hypotenuse of a right-angled triangle with legs 2 km and 3 km. Use Pythagoras’ theorem: distance² = 2² + 3² = 4 + 9 = 13, so distance = √13 ≈ 3.61 km. Thus, straight-line distance is about 3.6 km.
画出路线：从出发点（H）以1厘米代表1公里，从正北顺时针60°画2厘米线段；从该点起以150°方向画3厘米（即从北顺时针150°，大致东南方向）。两段路径之间的夹角？第一条方位060°，第二条150°，差为90°，因此两段路径互相垂直。这样，家到图书馆的距离就是以2公里和3公里为直角边的直角三角形斜边。使用勾股定理：距离² = 2² + 3² = 4 + 9 = 13，因此距离 = √13 ≈ 3.61公里。所以直线距离约为3.6公里。
Angle between paths = 150° – 060° = 90°, so right-angled
Distance = √(2² + 3²) = √13 ≈ 3.61 km
Bearing calculations often create right triangles, making the shortcut easy to find with Pythagoras.
方位计算常形成直角三角形，从而利用勾股定理轻松求出捷径。
9. Saving Plan: Simple Interest and Sequences | 储蓄计划：单利与数列
Lucy saves £50 each month into an account that pays simple interest of 3% per year on the total amount saved at the end of each year. She starts on 1st January with £0 and saves £50 on the first day of each month. What will her balance be at the end of the first year (after interest is added)? How much interest does she earn in that year?
露西每月储蓄50英镑到一个账户，该账户每年末对当年存款总额支付3%的单利。她于1月1日开始，余额为零，每月第一天存入50英镑。第一年末（加入利息后）她的余额是多少？她当年获得多少利息？
First, total saved during the year = 12 × £50 = £600. The interest is 3% of this total, but simple interest typically is calculated on the amount deposited over time. However, since all deposits are made by December, we can treat the total £600 as the principal for the whole year if interest is added at the end. Actually, simple interest formula: Interest = Principal × rate × time. If she deposits monthly, the earlier deposits earn interest for longer. But if the account pays simple interest on the year-end balance of deposits only, it’s common to calculate on the average balance. Let’s assume the bank pays 3% on the total amount saved by the end of year, i.e., on £600, for the entire year (time = 1 year). Then interest = £600 × 0.03 × 1 = £18. So balance = £600 + £
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