📚 Year 7 Cambridge Computing: Interdisciplinary Integrated Question Practice | 七年级剑桥计算机：跨学科综合题型训练
Welcome to this interdisciplinary revision session designed for Year 7 Cambridge Computing students. The real world rarely presents challenges neatly labelled ‘computing’ – instead, you will need to combine computational thinking with knowledge from mathematics, science, geography, history, and even art. This article brings together a collection of integrated question types and worked examples to help you build confidence in applying algorithms, data representation, programming concepts, and digital literacy across different subjects. By working through these scenarios, you will see how binary numbers describe colours, how flowcharts solve maths problems, how spreadsheets analyse science experiments, and much more. Let us begin this cross-curricular journey.
欢迎参加本次为七年级剑桥计算机学生设计的跨学科复习训练。真实世界中的挑战很少贴着“计算机”标签——你需要将计算思维与数学、科学、地理、历史甚至艺术知识结合起来。本文汇集了一系列综合题型和示例，帮助你自信地将算法、数据表示、编程概念和数字素养应用于不同学科。通过探索这些情景，你将看到二进制数如何描述颜色，流程图如何解决数学问题，电子表格如何分析科学实验，以及更多精彩内容。让我们一起开启这趟跨学科之旅。
1. Mathematics Meets Binary: Numbers and Colours | 数学与二进制：数字与颜色
Computing often uses binary to represent numbers and colours. In a simple 24-bit colour model, each pixel is made of three 8-bit components: red, green, and blue. Each component can be expressed as an 8-bit binary number, which can be converted to a decimal value between 0 and 255. This directly links to place value and powers of two, which you study in maths.
计算机常使用二进制表示数字和颜色。在简单的24位颜色模型中，每个像素由三个8位分量组成：红色、绿色和蓝色。每个分量都可以表示为一个8位二进制数，并可转换为0到255之间的十进制值。这与你在数学中学到的位值和2的幂直接相关。
Example problem: A pixel in an image has the following binary components: Red = 01100100, Green = 10101010, Blue = 00001111. Convert each binary number to decimal. Then describe the likely colour of this pixel.
示例问题：图像中的一个像素具有以下二进制分量：红色 = 01100100，绿色 = 10101010，蓝色 = 00001111。将每个二进制数转换为十进制。然后描述该像素的可能颜色。
To convert, we use the place value method. For red 01100100₂: columns are 2⁷ (128), 2⁶ (64), 2⁵ (32), 2⁴ (16), 2³ (8), 2² (4), 2¹ (2), 2⁰ (1). Add the values where a 1 appears: 64 + 32 + 4 = 100. For green 10101010₂: 128 + 32 + 8 + 2 = 170. For blue 00001111₂: 8 + 4 + 2 + 1 = 15. So the RGB triplet is (100, 170, 15). This combination produces a leaf-green shade because the green component is strong, red is moderate, and blue is very low.
转换时，我们使用位值法。对于红色 01100100₂：数位依次为 2⁷ (128), 2⁶ (64), 2⁵ (32), 2⁴ (16), 2³ (8), 2² (4), 2¹ (2), 2⁰ (1)。将1对应的值相加：64 + 32 + 4 = 100。对于绿色 10101010₂：128 + 32 + 8 + 2 = 170。对于蓝色 00001111₂：8 + 4 + 2 + 1 = 15。因此RGB三元组为 (100, 170, 15)。这种组合产生一种叶绿色调，因为绿色分量很强，红色适中，蓝色极低。
This exercise demonstrates how binary numbers form the foundation of digital images, linking computing with mathematics and creativity in art.
这个练习展示了二进制数如何构成数字图像的基础，将计算机与数学以及艺术创意联系起来。
2. Flowcharts and Mathematical Algorithms: Finding the GCD | 流程图与数学算法：寻找最大公约数
Flowcharts are a visual way to design algorithms. When solving a mathematical problem such as finding the greatest common divisor (GCD) of two numbers, we can use the Euclidean algorithm and represent it with a flowchart. This connects computational thinking to the maths topic of number theory.
流程图是设计算法的可视化方式。当解决如求两数最大公约数 (GCD) 的数学问题时，我们可以使用欧几里得算法并用流程图表示。这将计算思维与数论数学主题联系起来。
Problem: Use a flowchart to find the GCD of 48 and 18. Trace through the steps and write the final answer.
问题：使用流程图求48和18的最大公约数。逐步跟踪并写出最终答案。
A typical flowchart for the Euclidean algorithm starts with two numbers A and B. Check if B = 0. If yes, output A as the GCD. If not, set Temp = B, B = A mod B, A = Temp, and loop back. Let us trace: A=48, B=18. B is not 0, so Temp=18, B=48 mod 18 = 12, A=18. Now A=18, B=12. B not 0, Temp=12, B=18 mod 12 = 6, A=12. Now A=12, B=6. B not 0, Temp=6, B=12 mod 6 = 0, A=6. Now B=0, output A=6. The GCD of 48 and 18 is 6. You can draw this using standard flowchart symbols (ovals for start/end, parallelograms for input/output, rectangles for processes, diamonds for decisions). This method links mathematical reasoning with structured algorithm design.
欧几里得算法的典型流程图从两个数A和B开始。检查B是否等于0。若是，输出A作为GCD。若否，设置 Temp = B, B = A mod B, A = Temp，并返回循环。我们跟踪：A=48, B=18。B不为0，因此 Temp=18, B=48 mod 18 = 12, A=18。现在 A=18, B=12。B不为0，Temp=12, B=18 mod 12 = 6, A=12。现在 A=12, B=6。B不为0，Temp=6, B=12 mod 6 = 0, A=6。现在 B=0，输出A=6。48和18的GCD为6。你可以使用标准流程图符号（椭圆形表示开始/结束，平行四边形表示输入/输出，矩形表示处理，菱形表示判断）来绘制。这一方法将数学推理与结构化算法设计联系起来。
3. Spreadsheets and Science Experiments: Analysing Hooke’s Law | 电子表格与科学实验：分析胡克定律
Spreadsheet software is a powerful tool for handling scientific data. In a physics investigation of Hooke’s Law, students measure the extension of a spring when different forces are applied. Computing skills help to calculate averages, plot graphs, and identify patterns – exactly what scientists do.
电子表格软件是处理科学数据的强大工具。在胡克定律的物理探究中，学生测量弹簧在不同拉力下的伸长量。计算机技能有助于计算平均值、绘制图表并识别模式——这正是科学家所做的。
Problem: The table below shows the results of an experiment. Use a spreadsheet to calculate the average extension for each force. Then determine if the spring obeys Hooke’s Law (Force is proportional to extension).
问题：下表显示了实验结果。使用电子表格计算每个力对应的平均伸长量。然后判断弹簧是否遵守胡克定律（力与伸长量成正比）。
| Force (N) | Extension trial 1 (mm) | Extension trial 2 (mm) | Extension trial 3 (mm) |
|---|---|---|---|
| 1.0 | 15 | 17 | 16 |
| 2.0 | 30 | 32 | 31 |
| 3.0 | 45 | 46 | 47 |
| 4.0 | 60 | 59 | 61 |
In a spreadsheet, you would add a column ‘Average extension’ and use a formula like =AVERAGE(B2:D2) for the first row, then copy down. The averages become: 1.0 N → 16 mm; 2.0 N → 31 mm; 3.0 N → 46 mm; 4.0 N → 60 mm. Plotting a scatter graph of Force vs Average extension shows points lying almost exactly on a straight line through the origin. The ratios Force/Extension are approximately 0.0625, 0.0645, 0.0652, 0.0667 – the small variations are due to experimental error but generally consistent, confirming the proportional relationship. This task integrates data handling from computing with scientific data analysis.
在电子表格中，你会添加一列“平均伸长量”，并对第一行使用类似 =AVERAGE(B2:D2) 的公式，然后向下填充。各平均值变为：1.0 N → 16 mm；2.0 N → 31 mm；3.0 N → 46 mm；4.0 N → 60 mm。绘制力-平均伸长量的散点图，显示这些点几乎恰好落在一条穿过原点的直线上。力/伸长量的比值约为0.0625、0.0645、0.0652、0.0667——微小的差异来自实验误差，但总体一致，证实了比例关系。此任务将计算机的数据处理与科学数据分析相结合。
4. Network Delay and Geography: Calculating Data Transmission Time | 网络延迟与地理：计算数据传输时间
When you load a website, data travels through cables across continents and under oceans. Network delay (latency) can be estimated using the speed of light in fibre optics, combined with distances between cities. This topic blends computer networking with geography and physics.
当你加载一个网站时，数据通过电缆穿越大陆和海洋。网络延迟（时延）可以利用光在光纤中的速度和城市间的距离来估算。这一主题融合了计算机网络、地理和物理。
Problem: A submarine fibre-optic cable connects London to New York. The straight-line distance is approximately 5,600 km. Light travels in fibre at about 2.0 × 10⁸ m/s. If routing and processing add an extra 10 ms delay at each end, compute the minimum round-trip time (RTT) for a data packet to travel from London to New York and back.
问题：一条海底光缆连接伦敦和纽约。直线距离约为 5,600 km。光在光纤中的传播速度约为 2.0 × 10⁸ m/s。如果路由和处理在每端增加额外10 ms延迟，计算一个数据包从伦敦到纽约并返回的最短往返时间 (RTT)。
First, convert distance to metres: 5,600 km = 5,600,000 m = 5.6 × 10⁶ m. The one-way propagation time is distance / speed = 5.6×10⁶ m ÷ 2.0×10⁸ m/s = 0.028 s = 28 ms. A round trip involves going there and back, so two-way propagation is 2 × 28 ms = 56 ms. Processing adds 10 ms at London and 10 ms at New York, totalling 20 ms. Minimum RTT = 56 ms + 20 ms = 76 ms. This theoretical minimum can be checked against real-world ping times, which often are slightly higher due to additional network devices. This exercise reinforces your understanding of units, speed formula, and the global nature of the internet, while practising geography knowledge of transatlantic routes.
首先将距离转换为米：5,600 km = 5,600,000 m = 5.6 × 10⁶ m。单程传播时间为距离/速度 = 5.6×10⁶ m ÷ 2.0×10⁸ m/s = 0.028 s = 28 ms。往返需要一来一回，因此双向传播时间为 2 × 28 ms = 56 ms。处理过程在伦敦和纽约各增加10 ms，合计20 ms。最短 RTT = 56 ms + 20 ms = 76 ms。这个理论最小值可对照真实世界的 ping 时间，后者往往由于额外的网络设备而略高。此练习强化了你对单位、速度公式以及互联网全球化特性的理解，同时练习了跨大西洋路径的地理知识。
5. Sorting Algorithms and History: Ordering Historical Events | 排序算法与历史：排列历史事件
Sorting is a fundamental concept in computing. Many sorting algorithms exist, such as bubble sort and insertion sort. We can practise these algorithms by ordering historical events by year, linking computing to chronology skills learnt in history.
排序是计算中的一个基本概念。存在许多排序算法，例如冒泡排序和插入排序。我们可以通过按年份排列历史事件来练习这些算法，将计算机与历史课上学到的年代学技能联系起来。
Problem: Given the following events and their years: A – Norman Conquest (1066), B – Signing of Magna Carta (1215), C – The Black Death peaks in England (1348), D – Columbus reaches the Americas (1492), E – The Great Fire of London (1666). Use bubble sort to arrange these events in chronological order. Show the list after each pass.
问题：给定以下事件及其年份：A – 诺曼征服 (1066)、B – 签署《大宪章》(1215)、C – 黑死病在英格兰达到高峰 (1348)、D – 哥伦布到达美洲 (1492)、E – 伦敦大火 (1666)。使用冒泡排序按时间顺序排列这些事件。显示每次遍历后的列表。
We treat the list as [A, B, C, D, E] with years [1066, 1215, 1348, 1492, 1666] and aim to sort from earliest to latest. Bubble sort repeatedly steps through the list, compares adjacent elements, and swaps them if they are out of order. Pass 1: Compare 1066 and 1215 – no swap. Compare 1215 and 1348 – no swap. Compare 1348 and 1492 – no swap. Compare 1492 and 1666 – no swap. List remains [A, B, C, D, E]. Since no swaps were made, the algorithm would detect the list is already sorted and terminate. If we intentionally set an unsorted order, say [E, C, B, D, A], students must perform swaps: Pass 1 would compare 1666 and 1348 → swap, resulting [C, E, B, D, A]; next compare 1666 and 1215 → swap, [C, B, E, D, A]; and so on. This task perfectly merges algorithmic thinking with historical knowledge, demonstrating how computers process chronological data.
我们将列表视为 [A, B, C, D, E]，对应年份 [1066, 1215, 1348, 1492, 1666]，目标是按从早到晚排序。冒泡排序反复遍历列表，比较相邻元素，若顺序错误则交换。第一次遍历：比较1066和1215——不交换。比较1215和1348——不交换。比较1348和1492——不交换。比较1492和1666——不交换。列表保持 [A, B, C, D, E]。由于无交换，算法将检测到列表已排好序并终止。如果我们故意设置一个未排序列，如 [E, C, B, D, A]，学生必须进行交换：第一次遍历将比较1666和1348 → 交换得到 [C, E, B, D, A]；然后比较1666和1215 → 交换得 [C, B, E, D, A]；依此类推。这项任务完美地将算法思维与历史知识融合，展示了计算机如何处理年代数据。
6. Probability Simulation and Programming: Dice Roll Experiment | 概率模拟与编程：骰子投掷实验
Simulation is a powerful computing technique for exploring probability. By writing a program to roll two dice many times, we can experimentally verify the chances of different sums, such as the most likely outcome of 7. This links programming constructs like loops and random numbers to the mathematics of probability.
模拟是一种强大的计算机技术，用于探索概率。通过编写一个多次投掷两个骰子的程序，我们可以实验验证不同和值的概率，例如最可能出现的7。这将循环和随机数等编程结构与概率论数学联系起来。
Problem: Describe an algorithm (or Scratch script) that simulates rolling two six-sided dice 100 times and counts how many times the sum is exactly 7. What proportion would you expect, and how does the simulation approach this value?
问题：描述一个算法（或 Scratch 脚本），模拟投掷两个六面骰子100次，并统计和为7的次数。你期望的比例是多少？模拟如何接近该值？
A possible algorithm: set a counter count7 to 0. Repeat 100 times: generate a random integer between 1 and 6 for die1; generate another for die2; if die1 + die2 = 7, add 1 to count7. After the loop, calculate proportion = count7 / 100. In Scratch, you would use the ‘pick random 1 to 6’ block twice, add them, and use an ‘if’ block to change the counter variable. The theoretical probability of sum 7 is 6/36 = 1/6 ≈ 0.1667, because there are 6 favourable combinations (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) out of 36 possible outcomes. With only 100 trials, the simulation might yield a proportion between 0.1 and 0.23, but running more trials (e.g., 10,000) brings the experimental proportion closer to 0.1667. This highlights how computational approaches reinforce the law of large numbers taught in maths.
一种可能的算法：设置计数器 count7 为0。重复100次：生成一个1到6之间的随机整数作为 die1；生成另一个作为 die2；如果 die1 + die2 = 7，将 count7 加1。循环结束后，计算比例 = count7 / 100。在 Scratch 中，你将会两次使用“在1到6间选随机数”积木，将它们相加，并使用“如果”积木来改变计数器变量。和为7的理论概率是 6/36 = 1/6 ≈ 0.1667，因为在36种可能结果中有6种有利组合（1+6、2+5、3+4、4+3、5+2、6+1）。仅进行100次试验，模拟得到的比例可能在0.1到0.23之间，但进行更多次试验（如10,000次）会使实验比例更接近0.1667。这突出了计算方法如何强化数学课上讲授的大数定律。
7. Data Compression and Music: Storing a Simple Melody | 数据压缩与音乐：存储简单旋律
Computers need efficient ways to store and transmit media. In music, a melody can be represented as a sequence of notes, and run-length encoding (RLE) is a basic lossless compression technique that replaces repeated values with a count and the value. This brings together computing concepts with music literacy.
计算机需要高效的方式来存储和传输媒体。在音乐中，一段旋律可以表示为一串音符，而游程编码 (RLE) 是一种基本的无损压缩技术，它用计数值和值本身替换重复的值。这结合了计算概念与音乐素养。
Problem: The opening of ‘Twinkle Twinkle Little Star’ can be notated with numbers representing scale degrees: 1,1
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