📚 Case Study Practice Exercises in Year 8 CIE Maths | 八年级CIE数学案例分析实战演练
In Year 8 CIE Mathematics, applying concepts to real-world case studies strengthens problem-solving skills. This article presents a series of practical exercises covering key topics such as number operations, algebra, geometry, statistics, and probability. Each scenario is designed to mirror everyday situations, helping you build confidence in using maths beyond the classroom.
在八年级CIE数学中，将概念应用于现实案例研究可以增强解决问题的能力。本文提供一系列实战练习，涵盖数字运算、代数、几何、统计和概率等关键主题。每个场景都模拟日常生活情境，帮助你在课堂之外自信地运用数学。
1. Shopping Budget Challenge | 购物预算挑战
You have a budget of £60 to buy supplies for a school event. A pack of pens costs £4.50, a notebook £3.20, and a folder £2.80. You need to buy 5 packs of pens, 8 notebooks, and some folders. If you want to spend the whole budget exactly, how many folders can you buy?
你的预算是60英镑，为学校活动购买用品。一包笔4.50英镑，一个笔记本3.20英镑，一个文件夹2.80英镑。你需要购买5包笔、8个笔记本和一些文件夹。如果想正好花完预算，你能买多少个文件夹？
First, calculate the cost of pens: 5 × £4.50 = £22.50. Cost of notebooks: 8 × £3.20 = £25.60. Total spent on pens and notebooks: £22.50 + £25.60 = £48.10. Money left for folders: £60.00 - £48.10 = £11.90. Number of folders: £11.90 ÷ £2.80 = 4.25. Since you can only buy whole folders, you can buy 4 folders and will have £ ssible under these fixed purchases. This demonstrates real-world constraints in budgeting.
首先，计算笔的费用：5 × £4.50 = £22.50。笔记本费用：8 × £3.20 = £25.60。笔和笔记本合计：£22.50 + £25.60 = £48.10。剩余买文件夹的钱：£60.00 - £48.10 = £11.90。文件夹数量：£11.90 ÷ £2.80 = 4.25。由于只能购买整数个文件夹，你可以买4个，剩下£0.70。根据给定的固定购买量，无法正好花完预算。这展示了预算中的现实约束。
Key skills: multiplication, addition, subtraction, division, and interpreting remainders in context.
关键技能：乘法、加法、减法、除法以及在上下文中解释余数。
2. Travel Time and Distance | 旅行时间与距离
A coach travels from Manchester to London, a distance of 200 miles. The coach travels at an average speed of 50 miles per hour for the first 80 miles, then at 60 mph for the rest of the journey. How long does the journey take in hours and minutes?
一辆长途汽车从曼彻斯特开往伦敦，距离200英里。前80英里平均时速50英里，之后路程平均时速60英里。整个旅程需要多少小时和分钟？
cond part = 120 ÷ 60 = 2 hours. Total time = 1.6 + 2 = 3.6 hours = 3 hours and 36 minutes.
第一部分时间 = 距离÷速度 = 80 ÷ 50 = 1.6小时（即1小时36分钟）。剩余距离 = 200 - 80 = 120英里。第二部分时间 = 120 ÷ 60 = 2小时。总时间 = 1.6 + 2 = 3.6小时 = 3小时36分钟。
This case uses division, unit conversion (hours to minutes), and the relationship between speed, distance, time.
此案例使用了除法、单位换算（小时转分钟）以及速度、距离、时间的关系。
3. Designing a Rectangular Garden | 设计长方形花园
A gardener wants to create a rectangular flower bed with an area of 48 m². The length is twice the width. Find the dimensions and calculate the length of fencing needed to enclose it.
一位园丁想建造一个面积为48平方米的长方形花坛，长是宽的两倍。求尺寸并计算围栏所需的长度。
Let width = w, length = 2w. Area = length × width = 2w × w = 2w² = 48. So w² = 24, w = √24 = 2√6 m (approximately 4.90 m). Length = 4√6 m. Fencing perimeter = 2(l + w) = 2(4√6 + 2√6) = 12√6 m (approximately 29.39 m). If fencing costs £5.50 per metre, total cost ≈ 12√6 × 5.50 = 66√6 ≈ £161.64.
设宽为w，长为2w。面积 = 长×宽 = 2w×w = 2w² = 48。因此w² = 24，w = √24 = 2√6 米（约4.90米）。长度 = 4√6 米。围栏周长 = 2(4√6 + 2√6) = 12√6 米（约29.39米）。如果围栏每米5.50英镑，总成本 ≈ 12√6 × 5.50 = 66√6 ≈ £161.64。
Skills: forming equations, square roots, perimeter calculation, and cost analysis.
技能：建立方程、平方根、周长计算和成本分析。
4. Data Survey: Favourite Snacks | 数据调查：最喜欢的零食
A class of 30 students surveyed their favourite snack: crisps, fruit, biscuits, or chocolate. The results: 12 chose crisps, 8 fruit, 6 biscuits, and 4 chocolate. Represent this data in a bar chart and calculate the mean number of students per snack category.
一个30名学生的班级调查了他们最喜欢的零食：薯片、水果、饼干或巧克力。结果：12人选薯片，8人选水果，6人选饼干，4人选巧克力。用条形图表示数据，并计算每个零食类别的平均学生人数。
. For a bar chart, label axes: Snack type (horizontal), Frequency (vertical). Draw bars for each category. The mode is crisps. The range is 12 - 4 = 8.
总学生30人，4个类别，每类平均 = 30 ÷ 4 = 7.5人。即使数据分布不均，平均数仍是一个中心度量。条形图：x轴为零食种类，y轴为频数。众数是薯片，极差为12 - 4 = 8。
Key statistical concepts: frequency, mode, range, mean, and data representation.
关键统计概念：频数、众数、极差、平均数和数据表示。
5. Probability: Rolling Two Dice | 概率：掷两个骰子
You roll two fair six-sided dice and add the scores. What is the probability that the sum is 7? What about a sum less than 5?
你掷两个公平的六面骰子，并将点数相加。总和为7的概率是多少？总和小于5的概率呢？
includes sums 2, 3, 4. Sum 2: (1,1) 1 way; sum 3: (1,2),(2,1) 2 ways; sum 4: (1,3),(2,2),(3,1) 3 ways. Total favourable = 1+2+3 = 6. Probability = 6/36 = 1/6. So both probabilities are 1/6.
总可能结果 = 6×6 = 36。和为7的组合：(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)共6种。概率 = 6/36 = 1/6。和小于5包括2,3,4。和为2: (1,1)1种；和3: (1,2),(2,1)2种；和4: (1,3),(2,2),(3,1)3种。有利结果共6种，概率 = 6/36 = 1/6。因此两者概率均为1/6。
This exercise covers sample spaces, listing outcomes, and simplifying fractions.
本练习涵盖样本空间、列出结果和化简分数。
6. Fraction Recipes | 分数与食谱
A recipe for 4 people requires 1 ½ cups of flour, ¾ cup of sugar, and ⅓ cup of oil. Adapt the quantities for 6 people, expressing answers as mixed numbers where appropriate.
一份4人份的食谱需1½杯面粉、¾杯糖和⅓杯油。调整为6人份的量，用带分数表示结果。
1.5. Flour: 1½ = 3/2, multiplied by 3/2 gives (3/2) × (3/2) = 9/4 = 2¼ cups. Sugar: ¾ × 3
Published by TutorHao | Year 8 Mathematics Revision Series | aleveler.com
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