📚 Year 8 CIE Maths: Case Study Practical Drills | 案例分析实战演练
In Year 8 CIE Maths, applying mathematical concepts to real-life scenarios helps deepen understanding and prepares you for higher-level problem solving. This article presents ten case study drills covering ratios, percentages, algebra, statistics, measurement, and more. Follow each step-by-step solution to strengthen your skills.
在八年级CIE数学中，将数学概念应用于现实情境有助于加深理解，为更高层次的问题解决做好准备。本文提供十个案例实战练习，涵盖比率、百分数、代数、统计、测量等。跟随逐步求解，提升你的能力。
1. Planning a School Trip | 规划学校旅行
of coaches required is 240 ÷ 48 = 5. Checking: 5 × 48 = 240, exactly correct.
某学校为240名学生组织博物馆之旅。每辆巴士可舒适乘坐48人，因此所需巴士数为 240 ÷ 48 = 5 辆。验证：5 × 48 = 240，完全正确。
t would pay £650 ÷ 240 = £2.7083..., which we can round to £2.71.
每辆巴士租金为130英镑。总费用 = 5 × 130 = 650英镑。每位学生需支付 650 ÷ 240 ≈ 2.71英镑。
p 2 gets (4/10) × 240 = 96 students; Group 3 gets (5/10) × 240 = 120 students.
学生按1:4:5的比例分为三个活动小组。总份数 = 1+4+5 = 10。第一组 240 × 1/10 = 24人；第二组 240 × 4/10 = 96人；第三组 240 × 5/10 = 120人。
tal cost = 6 × £130 = £780, and the cost per student rises to £3.25.
若因行李限制每辆巴士只能载45人，所需数量为 240 ÷ 45 ≈ 5.33，因此需要6辆巴士。新总费用 = 6 × 130 = 780英镑，人均费用升至3.25英镑。
minutes). In kilometres, using 1 mile ≈ 1.6 km, the distance is 45 × 1.6 = 72 km.
博物馆距离45英里，巴士平均速度为30英里/小时。行驶时间 = 距离 ÷ 速度 = 45 ÷ 30 = 1.5小时（1小时30分钟）。转化为千米：1英里 ≈ 1.6千米，距离为 45 × 1.6 = 72千米。
2. Designing a Rectangular Garden | 设计矩形花园
th is (w + 3) m. Perimeter formula: 2(length + width) = 2(w + 3 + w) = 2(2w + 3) = 4w + 6.
园丁想设计一个长比宽多3米的矩形花圃，周长为26米。设宽为 w 米，则长为 (w + 3) 米。周长公式：2(长+宽) = 2(w + 3 + w) = 2(2w + 3) = 4w + 6。
h is 5 m and the length is 5 + 3 = 8 m. The area of the garden = 8 × 5 = 40 m²;.
建立方程 4w + 6 = 26 → 4w = 20 → w = 5。因此宽为5米，长为 5+3 = 8米。花园面积 = 8 × 5 = 40平方米。
.5 = 6 m. The area including the path = 9 × 6 = 54 m². The path area alone = 54 - 40 = 14 m².
花园四周修建一条0.5米宽的小路。包含小路的总尺寸：长 = 8 + 2 × 0.5 = 9米，宽 = 5 + 2 × 0.5 = 6米。包含小路的总面积 = 9 × 6 = 54平方米。小路本身的面积 = 54 - 40 = 14平方米。
ab = 0.5 × 0.5 = 0.25 m²;. Number of slabs needed = 14 ÷ 0.25 = 56 slabs.
铺设小路使用边长为0.5米的正方形地砖。一块地砖的面积 = 0.5 × 0.5 = 0.25平方米。所需地砖数量 = 14 ÷ 0.25 = 56块。
3. Analysing Sports Team Statistics | 分析运动队统计数据
mean, add all values: 12+18+10+20+15 = 75. Divide by 5: mean = 75 ÷ 5 = 15 points.
一支篮球队记录了五名球员在一场比赛中的得分：12, 18, 10, 20, 15。计算平均数：总和 = 12+18+10+20+15 = 75。除以5：平均数 = 75 ÷ 5 = 15分。
no mode because all values appear only once. The range = maximum - minimum = 20 - 10 = 10.
将数据从小到大排列：10, 12, 15, 18, 20。中位数为15。由于每个数值只出现一次，没有众数。极差 = 最大值 - 最小值 = 20 - 10 = 10。
These statistics help the coach understand consistency and performance. For a bar chart, label the x-axis with player names and y-axis with points; draw bars of heights 12, 18, 10, 20, 15. A bar chart makes comparisons visually immediate.
这些统计量帮助教练了解球员表现的稳定性和水平。绘制条形图时，x轴标球员姓名，y轴示得分，绘制高度为12, 18, 10, 20, 15的长条。条形图可直观比较。
4. Shopping Discounts and Tax | 购物折扣与税收
A jacket has an original price of £80. The shop offers a 20% discount. The discount amount = 20% und;16. The discounted price = £80 - £16 = £64. Alternatively, paying 80% of original: 0.80 × 80 = £64.
一件夹克原价80英镑，商店打八折（20%折扣）。折扣金额 = 80 × 0.20 = 16英镑。折后价格 = 80 - 16 = 64英镑。也可直接计算原价的80%：0.80 × 80 = 64英镑。
VAT at 20% is then added to the discounted price. VAT amount = 20% und;64 + £12.80 = £76.80. Or multiply by 1.20: 64 × 1.20 = £76.80.
然后在折后价上加收20%的增值税。增值税金额 = 64 × 0.20 = 12.80英镑。最终价格 = 64 + 12.80 = 76.80英镑。或者直接乘以1.20：64 × 1.20 = 76.80英镑。
discount, the calculation would be: price after VAT = 80 × 1.20 = £96, then 20% nd;76.80. The order does not change the final price because multiplication is commutative.
若先加增值税再打折：含税价 = 80 × 1.20 = 96英镑，然后打八折：96 × 0.80 = 76.80英镑。由于乘法具有交换性，顺序不影响最终价格。
5. Adjusting a Baking Recipe | 调整烘焙配方
each ingredient by 2.5: flour = 200 × 2.5 = 500 g, sugar = 100 × 2.5 = 250 g, butter = 50 × 2.5 = 125 g.
一份制作12块饼干的配方需面粉200克、糖100克、黄油50克。要制作30块饼干，比例因子为 30 ÷ 12 = 2.5。每种材料乘以2.5：面粉 = 200 × 2.5 = 500克，糖 = 100 × 2.5 = 250克，黄油 = 50 × 2.5 = 125克。
12 ≈ 4.167 g. With 80 g, maximum cookies = 80 ÷ (50/12) = 80 × 12/50 = 960/50 = 19.2, so 19 whole cookies can be made (the limiting ingredient is butter).
如果只有80克黄油，求最多能做多少块饼干。原配方50克黄油做12块，所以每块饼干用黄油量约为 50 ÷ 12 ≈ 4.167克。用80克黄油，最多数量 = 80 ÷ (50/12) = 80 × 12/50 = 19.2，因此最多可制作19块饼干（黄油是限制材料）。
12) × 19 ≈ 316.7 g, well below 500 g. So butter is indeed the limiting factor.
验证其他材料充足：19块饼干所需面粉 = (200/12) × 19 ≈ 316.7克，远少于500克。所以黄油确实是限制因素。
6. Train Timetables and Travel Time | 火车时刻表与旅行时间
A train departs from Station A at 10:15. It travels for 1 hour 45 minutes to Station B, stops for 15 minutes, then continues for 2 hours 20 minutes to Station C. Find the arrival time at Station C.
一列火车于10:15从A站出发，行驶1小时45分钟到达B站，停车15分钟，再行驶2小时20分钟到达C站。求到达C站的时间。
00. Add 15 min stop = 12:15. Add 2 h 20 min = 14:35. The train arrives at 14:35 (2:35 pm).
10:15 + 1小时45分 = 12:00。加15分钟停站 = 12:15。再加2小时20分 = 14:35。火车于14:35（下午2:35）到达。
stance = speed × time = 80 × 49/12 = 3920/12 = 326 ⅔ km (≈ 326.67 km).
总行驶时间 = 1小时45分 + 2小时20分 = 4小时5分 = 245分钟。若平均速度为80千米/时，将时间转换为小时：4 + 5/60 = 49/12小时。距离 = 速度 × 时间 = 80 × 49/12 = 3920/12 = 326⅔千米 (≈ 326.67 km)。
7. Surveying Student Favourites | 调查学生喜好
In a class of 30 students, favourite fruits are recorded: Apple 10, Banana 8, Orange 7, Grape 5. To show these as percentages: Apple (10/30)×100% ≈ 33.3%, Banana 26.7%, Orange 23.3%, Grape 16.7%.
一个班30名学生，最喜欢的水果记录如下：苹果10人，香蕉8人，橘子7人，葡萄5人。转换为百分比：苹果 (10/30)×100% ≈ 33.3%，香蕉26.7%，橘子23.3%，葡萄16.7%。
, Orange = 84°, Grape = 60°. The mode is Apple since it has the highest frequency.
绘制饼图时计算各扇形角度：苹果 = (10/30) × 360° = 120°，香蕉 = 96°，橘子 = 84°，葡萄 = 60°。众数为苹果，因其频数最高。
A bar chart plotting fruit categories on the x-axis and frequency on the y-axis with bar heights 10, 8, 7, 5 would also clearly display the distribution.
若绘制条形图，x轴为水果种类，y轴为频数，柱高分别为10、8、7、5，也可清晰显示分布。
8. Calculating Volume and Surface Area of a Box | 计算箱子体积与表面积
; 30 × 25 = 30 000 cm³;. Since 1000 cm³ = 1 litre, the volume is 30 litres.
一个长方体箱子长40厘米、宽30厘米、高25厘米。体积 = 长 × 宽 × 高 = 40 × 30 × 25 = 30000立方厘米。因1000立方厘米=1升，故体积为30升。
2950 = 5900 cm²;. This is the amount of wrapping paper needed to cover all six faces.
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