📚 Year 8 CAIE Mathematics: Case Study Practice | Year 8 CAIE 数学：案例分析实战演练
In Year 8 Mathematics, applying concepts to real-life situations is a key skill. The CAIE syllabus encourages students to use numbers, algebra, geometry, and statistics in practical contexts. This article presents a series of case-study exercises that will help you practise these topics step by step. Each problem is designed like a mini-investigation, giving you a chance to think mathematically and communicate your reasoning clearly.
在八年级数学中，将概念应用到现实生活场景是一项关键技能。CAIE 大纲鼓励学生在实际情境中使用数字、代数、几何和统计。本文提供了一系列案例分析练习，帮助你逐步练习这些主题。每个问题都像一个小型探究，让你有机会进行数学思考并清晰地表达推理过程。
1. Smart Shopping and Discounts | 精明购物与打折
Case: A jacket has an original price of £80. The shop offers a 20% discount to everyone, and members receive an additional 5% off the discounted price. How much does a member pay for the jacket?
案例：一件夹克原价 £80。商店对所有顾客提供 20% 折扣，会员还可以在折扣价基础上再享 5% 优惠。一位会员购买这件夹克需要支付多少钱？
First, apply the 20% discount. A 20% discount means paying 80% of the original price, so the price after the first discount is £80 × 0.80 = £64.
首先，应用 20% 的折扣。20% 折扣意味着支付原价的 80%，因此首次折扣后的价格为 £80 × 0.80 = £64。
Next, the member gets an extra 5% off £64. This means paying 95% of £64, which is £64 × 0.95 = £60.80.
接着，会员再享受 £64 基础之上的 5% 折扣。这意味着支付 £64 的 95%，即 £64 × 0.95 = £60.80。
Final price = Original price × (1 – 0.20) × (1 – 0.05) = £80 × 0.80 × 0.95 = £60.80
最终价格 = 原价 × (1 – 0.20) × (1 – 0.05) = £80 × 0.80 × 0.95 = £60.80
Rounding to the nearest penny gives £60.80. The member saves £19.20 compared with the original price.
精确到分，结果就是 £60.80。与原价相比，会员节省了 £19.20。
2. Baking a Cake: Ratio and Proportion | 烘烤蛋糕：比例与配比
A recipe for 8 cupcakes requires 200 g flour, 150 g sugar and 100 g butter. You need to bake 20 cupcakes for a party. How much of each ingredient will you need?
一份制作 8 个纸杯蛋糕的食谱需要 200 克面粉、150 克糖和 100 克黄油。你需要为聚会烘烤 20 个纸杯蛋糕，每种材料需要多少？
multiply each amount by the scale factor: (new number) ÷ (original number) = 20 ÷ 8 = 2.5.
由于数量必须保持相同配比，你可以将每种材料的用量乘以比例系数：（新数量）÷（原数量）= 20 ÷ 8 = 2.5。
00 g × 2.5 = 500 g. Sugar needed: 150 g × 2.5 = 375 g. Butter needed: 100 g × 2.5 = 250 g.
面粉需要：200 × 2.5 = 500 克。糖需要：150 × 2.5 = 375 克。黄油需要：100 × 2.5 = 250 克。
Always check that the sum of the scaled amounts still reflects the original ratio. In this case, 500 : 375 : 250 simplifies to 4 : 3 : 2, which matches the original ratio of flour, sugar and butter.
始终检查放大后的用量总和是否仍然体现原始配比。此处 500 : 375 : 250 化简后为 4 : 3 : 2，与原来面粉、糖和黄油的配比一致。
3. Planning a Road Trip: Distance, Speed and Time | 规划自驾游：距离、速度与时间
A family plans a road trip of 240 km. They expect an average speed of 60 km/h over the journey. How long, in hours, will the driving take?
一家人计划一次 240 公里的自驾游。他们预计全程平均速度为 60 公里/小时。驾驶时间是多少小时？
Use the formula: Time = Distance ÷ Speed. So driving time = 240 km ÷ 60 km/h = 4 hours.
使用公式：时间 = 距离 ÷ 速度。因此驾驶时间 = 240 ÷ 60 = 4 小时。
They plan a 30-minute lunch break at the halfway point. What is the total journey time from departure to arrival?
他们计划在半途休息 30 分钟吃午餐。从出发到到达的总旅行时间是多少？
= driving time + break time = 4 hours + 0.5 hours = 4.5 hours, or 4 hours and 30 minutes.
总时间 = 驾驶时间 + 休息时间 = 4 小时 + 0.5 小时 = 4.5 小时，即 4 小时 30 分钟。
If they aim to arrive by 2:00 pm, they must set off by 9:30 am at the latest.
如果他们希望下午 2:00 到达，最迟必须在上午 9:30 出发。
4. Designing a Rectangular Garden: Area and Perimeter | 设计矩形花园：面积与周长
A rectangular garden plot measures 15 m in length and 10 m in width. The owners want to lay new turf over the entire area and install a fence around the perimeter.
一个矩形花园地块长 15 米、宽 10 米。主人想在整个区域铺设新草皮，并在四周安装围栏。
Area of the garden = length × width = 15 m × 10 m = 150 m².
花园面积 = 长 × 宽 = 15 × 10 = 150 平方米。
Perimeter of the garden = 2 × (length + width) = 2 × (15 + 10) = 2 × 25 = 50 m.
花园周长 = 2 × (长 + 宽) = 2 × (15 + 10) = 2 × 25 = 50 米。
If turf costs £4 per square metre, the cost for turf is 150 × £4 = £600.
如果草皮每平方米 £4，草皮总费用为 150 × £4 = £600。
If fencing costs £12 per metre, the fence will cost 50 × £12 = £600.
如果围栏每米 £12，建造围栏的费用为 50 × £12 = £600。
The total cost for the project is £600 + £600 = £1200. Always check whether the budget allows for both materials.
项目总费用为 £600 + £600 = £1200。务必检查预算是否能够承担这两种材料的开销。
5. Analysing a Class Survey: Bar Charts | 分析班级调查：条形图
Thirty students were asked to name their favourite fruit. The results are shown in the table below.
30 名学生被问及他们最喜欢的水果。结果如下表所示。
| Fruit / 水果 | Frequency / 频数 |
|---|---|
| Apple / 苹果 | 8 |
| Banana / 香蕉 | 12 |
| Orange / 橙子 | 6 |
| Grapes / 葡萄 | 4 |
Draw a bar chart to represent the data, and calculate the percentage of students who chose apples.
绘制一个条形图来呈现数据，并计算选择苹果的学生所占的百分比。
To draw the bar chart, label the horizontal axis with the fruit names and the vertical axis with frequency (from 0 to at least 12). Draw bars of equal width whose heights match each frequency.
绘制条形图时，在水平轴上标注水果名称，垂直轴标注频数（从 0 到至少 12）。绘制等宽的条形，其高度与各频数相对应。
Percentage of students choosing apples = (number choosing apples ÷ total number) × 100% = (8 ÷ 30) × 100% ≈ 26.7%.
选择苹果的学生百分比 = (选择苹果的人数 ÷ 总人数) × 100% = (8 ÷ 30) × 100% ≈ 26.7%。
Bananas are the most popular choice, with 40% of the votes. Always check that the percentages add up to 100%.
香蕉是最受欢迎的选择，获得了 40% 的选票。务必检查各类百分比之和是否等于 100%。
6. Fair Game? : Probability | 公平游戏？：概率
A spinner is divided into 5 equal sections: 2 are coloured red, 1 is blue, and 2 are green. players win a prize if the spinner lands on red.
一个转盘被分为 5 个相等的部分：2 个红色，1 个蓝色，2 个绿色。如果转盘停在红色，玩家即可获奖。
gle spin. Probability (red) = number of red sections ÷ total sections = 2 ÷ 5 = 2/5 = 0.4.
计算单次旋转获胜的概率。获胜的概率（红） = 红色区域数量 ÷ 总区域数 = 2 ÷ 5 = 2/5 = 0.4。
Is this game fair? A game is unfair if a player’s chance of winning is less than 50% (or 0.5). Since 0.4 < 0.5, the player is more likely to lose than to win.
这个游戏公平吗？如果玩家获胜的概率低于 50%（即 0.5），游戏就是不公平的。因为 0.4 < 0.5，玩家更可能输而不是赢。
the player a greater chance. To balance exactly 0.5, you could redesign with 10 sections, making 5 red, or simply have a 1/2 chance by using a coin toss instead.
为了使游戏公平，你可以把另一个区域也涂成红色，使得 5 个区域中有 3 个红色。此时概率变为 3/5 = 0.6，有利于玩家。如果要刚好实现 0.5 的平衡，可以重新设计成 10 个区域，其中 5 个红色，或者干脆用抛硬币的方式获得 1/2 的机会。
7. Mean, Median and Mode from Rainfall Data | 降雨数据的平均数、中位数和众数
Daily rainfall (in mm) for one week was recorded as: 12, 15, 10, 12, 8, 18, 12. Find the mean, median and mode of this data set.
一周的日降雨量（单位：毫米）记录如下：12、15、10、12、8、18、12。求这组数据的平均数、中位数和众数。
s: 87 ÷ 7 = 12.43 (rounded to two decimal places). So the mean rainfall is about 12.43 mm.
要求平均数，先把所有数值加起来：12 + 15 + 10 + 12 + 8 + 18 + 12 = 87。再除以天数：87 ÷ 7 = 12.43（保留两位小数）。因此平均降雨量约为 12.43 毫米。
To find the median, put the values in order: 8, 10, 12, 12, 12, 15, 18. The middle value is the 4th one, which is 12. So the median rainfall is 12 mm.
要求中位数，将数值排序：8、10、12、12、12、15、18。中间的数是第 4 个，即 12。因此中位降雨量为 12 毫米。
The mode is the value that appears most often. Here, 12 appears three times, while others appear once, so the mode is 12 mm.
众数是出现次数最多的数值。这里 12 出现了三次，其他数值只出现一次，因此众数为 12 毫米。
In this case, median and mode are the same, but the mean is slightly higher because the value 18 pulls the average up.
在本例中，中位数和众数相同，但平均数略高，因为数值 18 拉高了平均值。
8. Budgeting Pocket Money: Simple Algebra | 零花钱预算：简单代数
Tom receives a fixed amount of pocket money each week, which we can call £x. This week, he spends £5 on a toy, and then exactly half of the remaining money on a book. After all purchases, he has £3 left.
Tom 每周收到固定数量的零花钱，我们称它为 £x。这个星期，他花 £5 买了一个玩具，然后把剩余钱的一半用于买书。购物后，他还剩下 £3。
alf on a book means he spends (x − 5) ÷ 2. The money left after the book is (x − 5) − (x − 5) ÷ 2. Since this equals £3, we write:
我们列一个方程。买玩具后剩余的钱 = x − 5。把剩余钱的一半用于买书，意味着他花了 (x − 5) ÷ 2。买书后剩下的钱是 (x − 5) − (x − 5) ÷ 2。由于这等于 £3，我们写出：
(x − 5) − (x − 5)/2 = 3
(x − 5) − (x − 5)/2 = 3
side. (x − 5) − (x − 5)/2 is the same as (x − 5)/2. So the equation becomes (x − 5)/2 = 3.
简化左边。(x − 5) − (x − 5)/2 等于 (x − 5)/2。因此方程变为 (x − 5)/2 = 3。
both sides by 2: x − 5 = 6. Then add 5 to both sides: x = 11. Tom’s weekly pocket money is £11.
两边同时乘以 2：x − 5 = 6。然后两边加 5：x = 11。Tom 每周的零花钱是 £11。
Check the answer: £11 − £5 = £6. Half of £6 is £3 spent on the book, leaving exactly £3. The solution works.
检验答案：£11 − £5 = £6。£6 的一半是 £3 用于买书，正好剩下 £3。答案正确。
9. Scale Drawing of a Bedroom | 卧室的比例图
A bedroom has a real length of 4 m and a real width of 3 m. An architect makes a scale drawing using a scale of 1:50. This means 1 cm on the drawing represents 50 cm in real life.
一间卧室的实际长度为 4 米，实际宽度为 3 米。建筑师使用 1:50 的比例尺绘制平面图，这意味着图纸上 1 厘米代表实际 50 厘米。
st, convert real measurements to centimetres: length = 4 m = 400 cm, width = 3 m = 300 cm.
首先，把实际尺寸转换为厘米：长度 = 4 米 = 400 厘米，宽度 = 3 米 = 300 厘米。
Using the scale, drawing length = 400 cm ÷ 50 = 8 cm. Drawing width = 300 cm ÷ 50 = 6 cm.
应用比例尺，图纸上的长度 = 400 ÷ 50 = 8 厘米，图纸上的宽度 = 300 ÷ 50 = 6 厘米。
the drawing is 48 cm², but the real area is 400 cm × 300 cm = 120 000 cm², which is 12 m².
在平面图上，卧室将用一个 8 厘米 × 6 厘米的矩形表示。图纸上的面积为 48 平方厘米，但实际面积是 400 × 300 = 120 000 平方厘米，即 12 平方米。
Scales help you visualise large objects or spaces in a manageable size. Always check the units and the scale factor when working with drawings.
比例尺帮助你在可管理的尺寸内呈现大型物体或空间。处理图纸时，务必检查单位和比例因子。
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