📚 Year 7 CIE Advanced Mathematics: Case Study Practical Exercises | 七年级CIE进阶数学：案例分析实战演练
In the Cambridge International Lower Secondary Checkpoint and beyond, Advanced Mathematics for Year 7 challenges students to apply their knowledge of number theory, algebra, geometry, statistics and measurement to realistic scenarios. This article presents a series of case studies that show how foundational mathematical concepts – from prime factorisation to volume calculations – can be used to solve everyday problems. Each case is broken down into clear steps, encouraging you to reason logically and communicate your thinking.
在剑桥国际初中Checkpoint及更高层次的学习中，七年级进阶数学要求学生将数论、代数、几何、统计和测量等知识应用于实际情境。本文通过一系列案例分析，展示如何运用基础数学概念（从质因数分解到体积计算）解决日常问题。每个案例都被拆解为清晰的步骤，帮助你进行逻辑推理并清晰地表达思考过程。
1. Introduction to Case Studies | 案例分析简介
Case studies bridge the gap between abstract theory and practical application. They train you to identify relevant data, select appropriate methods, perform calculations accurately and interpret results. In Year 7 Advanced Mathematics, common themes include working with integers, fractions, decimals, percentages, algebraic expressions, equations, angle rules, area and volume, and statistical measures. Through these examples, you will build confidence in tackling multi-step problems under timed conditions such as those found in the Checkpoint test.
案例分析弥合了抽象理论与实际应用之间的鸿沟。它训练你识别相关数据、选择合适的方法、准确计算并解释结果。在七年级进阶数学中，常见主题包括整数、分数、小数、百分数、代数式、方程、角度规则、面积与体积及统计量。通过这些示例，你将增强信心，解决像Checkpoint测试中那样的多步骤限时问题。
2. Case 1: Prime Factorisation in Cryptography | 案例一：密码学中的质因数分解
Imagine you are designing a simple code that uses the product of two prime numbers as a public key. A friend gives you the number 91 and tells you it was made by multiplying two primes. Find the two prime factors and explain why prime factorisation makes breaking the code more difficult for someone who only knows the product.
假设你正在设计一种简单密码，使用两个质数的乘积作为公钥。一位朋友给你数字91，并告诉你它是由两个质数相乘得到的。找出这两个质因数，并解释为什么只知乘积时，质因数分解会使破解密码更困难。
9+1=10, not divisible by 3. Does not end in 0 or 5, so not divisible by 5. Divide by 7: 7×13 = 91 exactly. So the factors are 7 and 13. Both are prime. If an attacker only knows 91, they must trial divide by all possible primes up to the square root of 91 (≈9.5). That takes effort. Large numbers like 300-digit products are extremely difficult to factorise, which is why modern encryption relies on this property.
要找出质因数，列出小于91的质数：2, 3, 5, 7, 11, 13, 17, 19...检查整除性：91是奇数，不能被2整除。数字和9+1=10，不能被3整除。不以0或5结尾，不能被5整除。除以7：7×13恰好等于91。因此因数是7和13，两者都是质数。如果攻击者只知道91，他们必须尝试除以所有小于等于√91(≈9.5)的质数，这需要工作量。像300位数字的乘积极难分解，这就是现代加密依赖此特性的原因。
3. Case 2: Using Algebra to Model Savings | 案例二：用代数建立储蓄模型
Liam receives pocket money each week. He starts with £5 already saved. Each week he earns another £3. Write an algebraic expression for his total savings after w weeks. Then calculate how many weeks it will take for him to save enough to buy a £50 video game.
利亚姆每周都得到零花钱。他开始时已存了5英镑，每周再赚3英镑。写出w周后他总储蓄额的代数表达式。然后计算他需要多少周才能攒够钱买一款50英镑的电子游戏。
de by 3: w = 15. Liam needs 15 weeks. This linear model shows how algebra can represent a real-life pattern and answer ‘what if’ questions efficiently.
设总储蓄为T，起始金额为5，每周增加3。因此 T = 3w + 5。要找出T=50时的周数，解方程 3w + 5 = 50。两边减去5：3w = 45。除以3：w = 15。利亚姆需要15周。这个线性模型展示了代数如何表示现实生活模式并高效回答“假设”类问题。
4. Case 3: Area and Perimeter of Composite Shapes | 案例三：复合图形的面积与周长
und the entire garden. Find the total perimeter and the area of the garden. (Use π ≈ 3.14)
学校花园设计为一个12米乘5米的矩形，其一短边附着一个直径5米的半圆。整个花园周围要安装围栏。求花园的总周长和总面积。（使用π≈3.14）
2 = (3.14 × 5)/2 = 7.85 m. Total perimeter = 12 + 12 + 5 + 7.85 = 36.85 m. Area: Rectangle area = 12 × 5 = 60 m². Semicircle area = (π × r²)/2 = (3.14 × 2.5²)/2 = (3.14 × 6.25)/2 = 19.625/2 = 9.8125 m². Total area = 60 + 9.8125 = 69.8125 m². When painting or turfing, you would use area; for fencing, perimeter. Understanding composite shapes helps in real-world landscaping and DIY projects.
周长：图形由矩形的三条边（12 + 12 + 5）加上半圆的弧长组成。半圆弧长 = (π × d) / 2 = (3.14 × 5) / 2 = 7.85 米。总周长 = 12 + 12 + 5 + 7.85 = 36.85 米。面积：矩形面积 = 12 × 5 = 60 平方米。半圆面积 = (π × r²)/2 = (3.14 × 2.5²)/2 = (3.14 × 6.25)/2 = 19.625/2 = 9.8125 平方米。总面积 = 60 + 9.8125 = 69.8125 平方米。在刷漆或铺草坪时，你会用到面积；围栏则用到周长。理解复合图形有助于实际的景观设计和DIY项目。
5. Case 4: Data Analysis – Mean, Median, Mode | 案例四：数据分析——平均数、中位数、众数
A teacher recorded the scores of ten students in a mental maths test: 8, 9, 7, 10, 6, 9, 8, 9, 7, 8. Determine the mean, median and mode. Which measure best represents the typical score? Why might the mean be misleading if there were an extreme value?
一位老师记录了十名学生在速算测验中的分数：8, 9, 7, 10, 6, 9, 8, 9, 7, 8。计算平均数、中位数和众数。哪个统计量最能代表典型分数？如果有一个极端值，为什么平均数可能产生误导？
ode = most frequent score: 8 and 9 both appear three times, so bimodal: 8 and 9. In this symmetric distribution, mean and median are close. Mode gives the most common score. If a student scored 0, the mean would drop to 7.36, making it less representative. Data analysis lets you summarise and interpret information, a key skill in scientific research and business reports.
将分数排序：6, 7, 7, 8, 8, 8, 9, 9, 9, 10。平均数 = 总和 ÷ 10 = (6+7+7+8+8+8+9+9+9+10) ÷ 10 = 81 ÷ 10 = 8.1。中位数 = 中间两个值（第5和第6个）的平均数 = (8+8)/2 = 8。众数 = 出现最频繁的分数：8和9都出现三次，因此是双众数：8和9。在这个对称分布中，平均数和中位数接近。众数给出了最常见的分数。如果一名学生得0分，平均数将降至7.36，代表性降低。数据分析让你能够总结并解读信息，这是科学研究和商业报告中的关键技能。
6. Case 5: Sequences and Patterns in Nature | 案例五：自然界中的数列与模式
The Fibonacci sequence begins 1, 1, 2, 3, 5, 8, 13, ... . Each term after the first two is the sum of the two preceding terms. Count the spirals on a pineapple or the petals of a daisy – they often match Fibonacci numbers. Find the 10th term of the Fibonacci sequence and calculate the ratio of consecutive terms 8/5, 13/8, 21/13. What number does this ratio approach? (Use a calculator if needed)
斐波那契数列以 1, 1, 2, 3, 5, 8, 13... 开头，从第三项起每一项是前两项之和。数一数菠萝上的螺旋或雏菊的花瓣数——它们常常与斐波那契数吻合。求斐波那契数列的第10项，并计算连续两项的比值：8/5, 13/8, 21/13。这个比值接近什么数？（如需可用计算器）
pproach the golden ratio, approximately 1.618. This pattern appears in art, architecture and nature, showing how mathematics describes the world around us. Number sequences provide a foundation for later study of functions and series.
续写数列：1, 1, 2, 3, 5, 8, 13, 21, 34, 55。第10项是55。比值：8/5 = 1.6，13/8 = 1.625，21/13 ≈ 1.615。它们接近黄金比例，约1.618。这种模式出现在艺术、建筑和自然界中，展示了数学如何描述我们周围的世界。数列为日后学习函数和级数打下基础。
7. Case 6: Fraction and Ratio in Recipes | 案例六：食谱中的分数与比例
A recipe for 4 people requires 2½ cups of flour, ¾ cup of sugar and 1¼ cups of milk. You need to adapt it for 6 people. Find the new quantities. Then write the ratio of flour to sugar in simplest form for the original recipe.
一份供4人食用的食谱需要2½杯面粉、¾杯糖和1¼杯牛奶。你需要调整为6人份。计算新的用量。然后写出原食谱中面粉与糖的最简整数比。
2) = 15/8 = 1⅞ cups. Original ratio flour:sugar = 2½ : ¾ = (5/2) / (3/4) = (5/2) × (4/3) = 20/6 = 10/3, so 10:3. Working with fractions and ratios is essential for scaling recipes, mixing concrete or diluting chemicals accurately.
从4人到6人的缩放因子 = 6/4 = 1.5。新面粉量 = 2½ × 1.5 = (5/2) × (3/2) = 15/4 = 3¾ 杯。糖 = ¾ × 1.5 = ¾ × 3/2 = 9/8 = 1⅛ 杯。牛奶 = 1¼ × 1.5 = (5/4) × (3/2) = 15/8 = 1⅞ 杯。原面粉与糖的比例 = 2½ : ¾ = (5/2) / (3/4) = (5/2) × (4/3) = 20/6 = 10/3，即10:3。分数和比例的运用对于调整食谱分量、搅拌混凝土或精确稀释化学品至关重要。
8. Case 7: Speed, Distance, Time for Journey Planning | 案例七：行程规划中的速度、距离与时间
A family drives from town A to town B, a distance of 180 km, at an average speed of 60 km/h. On the return journey, traffic reduces their average speed to 45 km/h. Calculate the total time taken for the round trip and the overall average speed. Why is the overall average speed not simply (60+45)/2?
一家人从A镇驾车到B镇，距离180公里，平均速度60公里/小时。返程时交通导致平均速度降至45公里/小时。计算往返的总时间和整体平均速度。为什么整体平均速度不能简单取(60+45)/2？
total distance ÷ total time = 360 ÷ 7 ≈ 51.4 km/h. That is less than (60+45)/2 = 52.5 km/h because more time was spent at the slower speed. The formula v = d/t and the concept of weighted averages are crucial in physics and logistics.
去程时间 = 距离 ÷ 速度 = 180 ÷ 60 = 3 小时。返程时间 = 180 ÷ 45 = 4 小时。总时间 = 7 小时。总距离 = 360 公里。整体平均速度 = 总距离 ÷ 总时间 = 360 ÷ 7 ≈ 51.4 公里/小时。这比(60+45)/2 = 52.5公里/小时要小，因为更多时间花在了较慢的速度上。公式 v = d/t 以及加权平均的概念在物理和物流中至关重要。
9. Case 8: Angles and Bearings in Navigation | 案例八：航海中的角度与方位角
A ship leaves a harbour and sails 15 km on a bearing of 060°, then turns and sails 10 km on a bearing of 150°. Draw a sketch and find the angle between the two paths. Hence, determine how far apart the ship is from its starting point, using construction or Pythagoras’ theorem if you form a right-angled triangle. (Assume straight lines and a flat Earth for this exercise)
一艘船离开港口，以方位角060°航行15公里，然后转向以方位角150°航行10公里。画出示意图，求两段路径之间的夹角。然后，如果能构成直角三角形，通过作图或勾股定理求出船只与起点的距离。（本练习假设直线和平坦地表）
ond leg: 150°. The angle between the two paths is the difference in their directions: 150° – the hypotenuse of a right triangle with legs 15 km and 10 km. Using Pythagoras: distance = √15² + 10² = √225 + 100 km. Bearings and angle calculations are used in aviation and marine navigation every day.
方位角从正北按顺时针测量。第一段：060°。第二段：150°。两条路径之间的夹角是方向之差：150° – 60° = 90°。因此两段路径互相垂直。起点到终点的距离构成直角三形的斜边，直角边分别为15公里和10公里。运用勾股定理：距离 = √15² + 10² = √225 + 100 = √325 ≈ 18.03 公里。方位角和角度计算每天都在航空和航海中应用。
10. Case 9: Volume of Prisms in Packaging | 案例十：包装中的棱柱体积
A company designs a gift box shaped as a triangular prism. The triangular cross-section has a base of 8 cm and a height of 6 cm. The length of the prism is 20 cm. Calculate the volume of the box. If the box is to hold 1 litre (1000 cm³) of popcorn, is this box large enough? If not, what length would be needed to reach exactly 1000 cm³, keeping the triangle the same?
一家公司设计了一款三角柱形礼盒。三角形截面的底为8厘米，高为6厘米，棱柱长度20厘米。计算盒子的体积。如果盒子要盛装1升（1000立方厘米）的爆米花，这个盒子够大吗？如果不够，保持三角形不变，需要多长的棱柱才能刚好达到1000立方厘米？
cm³, so it is not large enough. To find required length L: 24 × L = 1000 → L = 1000 ÷ 24 ≈ 41.7 cm. Prism volume calculations apply to packaging, swimming pools and industrial containers.
棱柱体积 = 横截面积 × 长度。三角形面积 = ½ × 底 × 高 = ½ × 8 × 6 = 24 平方厘米。体积 = 24 × 20 = 480 立方厘米。远小于1000立方厘米，因此不够大。求所需长度 L：24 × L = 1000 → L = 1000 ÷ 24 ≈ 41.7 厘米。棱柱体积计算可应用于包装、游泳池和工业容器。
11. Summary and Key Takeaways | 总结与关键要点
These case studies demonstrate that Year 7 Advanced Mathematics is not just a collection of rules but a toolkit for solving authentic problems. Whether you factorise numbers, use algebraic formulas, calculate areas and volumes, or interpret data, the key is to break down the problem into manageable steps, check your working and always relate your answer back to the context. Practise similar examples and you will develop the flexible thinking required for success in the Checkpoint exam and further maths challenges.
这些案例分析表明，七年级进阶数学不仅是一系列规则，更是解决实际问题的工具箱。无论是分解因数、使用代数公式、计算面积和体积，还是解读数据，关键在于将问题分解为可管理的步骤、检查计算过程，并将答案与情境相联系。练习类似的题目，你将培养灵活思维，在Checkpoint考试和更高级的数学挑战中取得成功。
Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com
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