📚 Year 8 CIE Advanced Mathematics: Case Study Practice & Problem Solving | Year 8 CIE 进阶数学：案例分析实战演练
In advanced mathematics at Year 8, solving real-world problems through case studies sharpens your analytical skills and prepares you for more complex challenges. This article presents a series of practical scenarios where you can apply algebra, geometry, statistics, and logical reasoning. Each case is broken down step by step, so you can clearly see how mathematical methods lead to reliable solutions.
在八年级的进阶数学中，通过案例研究解决现实问题能够锻炼你的分析能力，为应对更复杂的挑战做好准备。本文提供了一系列实际场景，让你能够应用代数、几何、统计和逻辑推理。每个案例都逐步拆解，让你清晰看到数学方法如何导出可靠的解答。
1. Designing a School Garden: Area and Perimeter | 设计学校花园：面积与周长
A school plans to build a rectangular garden against an existing wall. The fencing for the other three sides is 20 metres long. The goal is to maximise the planting area. Let the width perpendicular to the wall be x metres and the side parallel to the wall be y metres. The total fencing gives the equation 2x + y = 20, so y = 20 – 2x.
一所学校打算靠着现有墙壁建一个矩形花园。另外三面的围栏总长度为 20 米。目标是最大化种植面积。设垂直于墙的宽度为 x 米，平行于墙的边长为 y 米。由围栏总长得到方程 2x + y = 20，因此 y = 20 – 2x。
The area A of the garden is A = x × y = x(20 – 2x) = 20x – 2x². Since this is a quadratic expression, we can test integer values for x to find the maximum area.
花园的面积 A = x × y = x(20 – 2x) = 20x – 2x²。因为这是一个二次式，我们可以通过检验 x 的整数值来找到最大面积。
| Width x (m) | Length y (m) | Area A (m²) |
|---|---|---|
| 3 | 14 | 42 |
| 4 | 12 | 48 |
| 5 | 10 | 50 |
| 6 | 8 | 48 |
| 7 | 6 | 42 |
The table shows that the maximum area of 50 m² occurs when x = 5 m and y = 10 m. The shape is not a square because one side is against the wall. If a semicircular flower bed with diameter 4 m is added inside the garden, its area is ½ × π × (2²) = 2π ≈ 6.28 m², leaving 43.72 m² for planting.
表格显示，当 x = 5 米、y = 10 米时出现最大面积 50 平方米。该形状不是正方形，因为有一侧靠墙。如果在花园内添加一个直径为 4 米的半圆形花坛，其面积为 ½ × π × (2²) = 2&pi； ≈ 6.28 平方米，剩余种植面积为 43.72 平方米。
2. Fundraising Ticket Sales: Linear Equations | 筹款门票销售：线性方程
A school concert sold 200 tickets in total. Adult tickets cost $8 each and child tickets cost $5 each. The total income from ticket sales was $1300. We need to find how many adult and child tickets were sold. Let a be the number of adult tickets and c the number of child tickets.
一场学校音乐会共售出 200 张票。成人票每张 8 美元，儿童票每张 5 美元。门票总收入为 1300 美元。我们需要求出卖掉多少张成人票和儿童票。设 a 为成人票数量，c 为儿童票数量。
From the total number of tickets: a + c = 200. From the total income: 8a + 5c = 1300. We can solve these simultaneous equations. Multiply the first equation by 5: 5a + 5c = 1000. Subtract this from the income equation: (8a + 5c) – (5a + 5c) = 1300 – 1000, giving 3a = 300, so a = 100.
由总票数得：a + c = 200。由总收入得：8a + 5c = 1300。我们可以解这两个联立方程。将第一个方程乘以 5：5a + 5c = 1000。从收入方程中减去该式：(8a + 5c) – (5a + 5c) = 1300 – 1000，得到 3a = 300，所以 a = 100。
Substitute a = 100 into a + c = 200: 100 + c = 200, so c = 100. Therefore, 100 adult tickets and 100 child tickets were sold. Checking the income: 8×100 + 5×100 = 800 + 500 = 1300, which matches.
将 a = 100 代入 a + c = 200：100 + c = 200，故 c = 100。因此，卖出了 100 张成人票和 100 张儿童票。核对收入：8×100 + 5×100 = 800 + 500 = 1300，一致。
3. Comparing Mobile Phone Plans: Graphs and Cost Analysis | 比较手机套餐：图表与费用分析
Two mobile phone plans are available. Plan A has a monthly fee of $10 and charges $0.05 per minute of calls. Plan B has a monthly fee of $15 and charges $0.03 per minute. We want to find the number of minutes for which both plans cost the same, and decide which plan is cheaper for different usage levels. Let m be the number of minutes used in a month.
有两种手机套餐可选。套餐 A 月租 10 美元，通话每分钟 0.05 美元。套餐 B 月租 15 美元，通话每分钟 0.03 美元。我们想找出使两个套餐费用相同的通话分钟数，并判断在不同使用量下哪个套餐更便宜。设 m 为每月使用的分钟数。
Cost for Plan A: CA = 10 + 0.05m. Cost for Plan B: CB = 15 + 0.03m. Set them equal: 10 + 0.05m = 15 + 0.03m. Subtract 0.03m from both sides: 10 + 0.02m = 15. Subtract 10: 0.02m = 5. Divide by 0.02: m = 250 minutes.
套餐 A 的费用：CA = 10 + 0.05m。套餐 B 的费用：CB = 15 + 0.03m。令两者相等：10 + 0.05m = 15 + 0.03m。两边减去 0.03m：10 + 0.02m = 15。减去 10：0.02m = 5。除以 0.02：m = 250 分钟。
If you talk less than 250 minutes per month, Plan A is cheaper because its fixed fee is lower. If you talk more than 250 minutes, the lower per-minute rate of Plan B makes it the better choice. Plotting both linear functions would show the intersection point at m = 250.
如果每月通话少于 250 分钟，套餐 A 更便宜，因为其固定费用较低。如果通话多于 250 分钟，套餐 B 较低的每分钟费率使其更划算。绘制这两个线性函数的图像会在 m = 250 处显示交点。
4. Baking Proportion Problems: Ratios and Scaling | 烘焙比例问题：比率与缩放
A cake recipe requires 200 g of flour, 150 g of sugar, 100 g of butter, and 4 eggs to serve 8 people. You need to bake a larger cake for 20 people. Since the number of servings increases by a factor of 20 ÷ 8 = 2.5, every ingredient must be multiplied by 2.5.
一份蛋糕食谱需要 200 克面粉、150 克糖、100 克黄油和 4 个鸡蛋，可供应 8 人。你需要为 20 人烤一个更大的蛋糕。由于份数增加了 20 ÷ 8 = 2.5 倍，每种材料都必须乘以 2.5。
- Flour: 200 g × 2.5 = 500 g
- Sugar: 150 g × 2.5 = 375 g
- Butter: 100 g × 2.5 = 250 g
- Eggs: 4 × 2.5 = 10 eggs
- 面粉：200 克 × 2.5 = 500 克
- 糖：150 克 × 2.5 = 375 克
- 黄油：100 克 × 2.5 = 250 克
- 鸡蛋：4 × 2.5 = 10 个
This direct proportion method works because the relationship between servings and each ingredient is linear. If you only had 7 eggs, the limiting factor would be the eggs, requiring you to scale down the recipe to serve 7/4 × 8 = 14 people.
这种直接比例方法有效，因为份数与每种材料之间是线性关系。如果你只有 7 个鸡蛋，鸡蛋便成为限制因素，你需要将食谱缩减为供应 7/4 × 8 = 14 人。
5. Savings and Compound Interest: Percentage Increase | 储蓄与复利：百分比增长
Jamie deposits $1000 in a savings account that pays 4% interest per annum, compounded annually. The balance after n years is given by A = 1000 × (1.04)ⁿ. After 2 years, the amount is 1000 × 1.04² = 1000 × 1.0816 = $1081.60. After 5 years, it is 1000 × 1.04&sup5; ≈ 1000 × 1.21665 = $1216.65.
杰米将 1000 美元存入一个年利率为 4% 的储蓄账户，按年复利计算。n 年后的余额由公式 A = 1000 × (1.04)ⁿ 给出。2 年后，金额为 1000 × 1.04² = 1000 × 1.0816 = 1081.60 美元。5 年后，约为 1000 × 1.04&sup5; ≈ 1000 × 1.21665 = 1216.65 美元。
To calculate the interest earned, subtract the original principal. After 5 years, the interest is $216.65. If the interest were simple, the annual interest would be $40, totaling $200 over 5 years, so compounding gives an extra $16.65. The effect grows over longer periods.
要计算所得利息，从最终金额中减去本金。5 年后的利息为 216.65 美元。如果是单利，每年利息为 40 美元，5 年总计 200 美元，因此复利多出 16.65 美元。时间越长，复利效应越明显。
6. Surveying Students: Data Handling and Charts | 学生调查：数据处理与图表
A class of 30 students voted for their favourite sport. The results: Football 12, Basketball 8, Swimming 6, Other 4. We can organise the data in a frequency table and compute relative frequencies and pie chart angles.
一个 30 名学生的班级投票选出了最喜欢的运动。结果如下：足球 12 票，篮球 8 票，游泳 6 票，其他 4 票。我们可以用频数表整理数据，并计算相对频率和饼图角度。
| Sport | Frequency | Fraction | Angle (°) |
|---|---|---|---|
| Football | 12 | 12/30 = 2/5 | 144 |
| Basketball | 8 | 8/30 = 4/15 | 96 |
| Swimming | 6 | 6/30 = 1/5 | 72 |
| Other | 4 | 4/30 = 2/15 | 48 |
Pie chart angles are calculated by multiplying each fraction by 360°. For football, 12/30 × 360° = 144°. The mode is football because it has the highest frequency. The range of the data is 12 – 4 = 8. If we draw a bar chart, the height of each bar represents the frequency.
饼图角度由每个分数乘以 360° 计算得到。足球对应的角度为 12/30 × 360° = 144°。众数是足球，因为其频数最高。数据的极差为 12 – 4 = 8。如果绘制条形图，每个条形的高度代表频数。
7. Probability in a Game: Fair or Unfair? | 游戏中的概率：公平与否？
A spinner is divided into four coloured
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