📚 Year 7 AQA Computer Science: Cross-Curricular Integrated Questions | Year 7 AQA 计算机：跨学科综合题型训练
Computing is not an isolated subject. In Year 7, you will encounter questions that blend concepts from mathematics, science, geography, design, and even music. Mastering these cross-curricular problems will help you see how computational thinking applies everywhere and prepare you for realistic AQA-style assessments. This guide provides integrated question practice, showing how to connect computer science with other subjects.
计算机并不是一门孤立的学科。在七年级，你会遇到把数学、科学、地理、设计甚至音乐等概念融合在一起的题目。掌握这些跨学科问题将帮助你看清计算思维如何应用到各个地方，并为真实的 AQA 风格评估做好准备。本指南提供综合题型训练，展示如何将计算机科学与其他学科联系起来。
1. Binary and Mathematics | 二进制与数学
Computer data is built on binary digits – a direct link to place value and powers of two that you study in mathematics. Typical AQA questions ask you to convert between binary and decimal or to perform simple addition in both systems.
计算机数据建立在二进制数字之上——这与你数学中学到的位值和 2 的幂直接相关。典型的 AQA 题目会要求你在二进制与十进制之间进行转换，或在两种系统中进行简单的加法运算。
Example: Convert the binary number 1101₂ to decimal, add 3 to the result, then convert back to binary.
例题：将二进制数 1101₂ 转换为十进制，把结果加 3，然后再转换回二进制。
1101₂ = 1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8 + 4 + 0 + 1 = 13₁₀
16 is written as 10000₂ because 16 = 2⁴. This trains both arithmetic and logical thinking.
然后 13 + 3 = 16。十进制数 16 写作 10000₂，因为 16 = 2⁴。这既训练了算术能力，又锻炼了逻辑思维。
2. Flowcharts and Scientific Experiments | 流程图与科学实验
When you plan a science investigation, you follow a sequence of steps just like an algorithm. Drawing a flowchart for a lab experiment helps you spot decision points and repetitions, a key skill in computing.
当你规划一项科学调查时，你会像算法一样遵循一系列步骤。为实验室实验绘制流程图可以帮助你发现决策点和重复环节，这是计算机中的一项关键技能。
Consider an experiment to test if water freezes at 0°C. Design a flowchart that includes a temperature check, a loop for waiting, and an output when ice forms.
考虑一个测试水是否在 0°C 结冰的实验。设计一个流程图，包含温度检查、等待循环以及当冰形成时的输出。
Your flowchart might have a diamond-shaped decision box: 'Is temperature ≤ 0°C?' with arrows for 'Yes' and 'No'. If 'No', loop back to measure again after 1 minute. If 'Yes', display 'Ice detected'. This mirrors the if-else and loop blocks you use in Scratch or Python.
你的流程图中可能有一个菱形决策框：‘温度 ≤ 0°C 吗？’，并用箭头标出‘是’和‘否’。如果‘否’，则返回并在 1 分钟后再次测量。如果‘是’，则显示‘检测到冰’。这反映了你在 Scratch 或 Python 中使用的 if-else 和循环积木。
3. Data Representation in Geography | 地理中的数据表示
Digital maps use pixels and colours to represent real-world locations. In geography, you learn about grid references and satellite images – concepts that directly relate to bitmap graphics and resolution.
数字地图使用像素和颜色来表示真实世界的地点。在地理课中，你学习网格参考和卫星图像——这些概念直接关系到位图图形和分辨率。
culate the ground distance each pixel represents and explain why higher resolution helps identify buildings.
问题：一张城市的卫星图像覆盖 8 km × 8 km 区域，用一个 400 × 400 像素的位图表示。计算每个像素代表的地面距离，并解释为什么更高分辨率有助于识别建筑物。
creasing resolution to 800 × 800 pixels would make each pixel 10 m, capturing more details – just as you need more bits to store colour depth.
每个像素覆盖 8000 m ÷ 400 = 20 m × 20 m。一栋小房子可能被忽略，因为 20 m 比典型的屋顶大。将分辨率提高到 800 × 800 像素将使每个像素变为 10 m，捕捉到更多细节——正如你需要更多比特来存储颜色深度一样。
4. Programming Loops in Music | 音乐中的循环编程
Music composition involves repetition and patterns, much like loops in programming. Using a block-based language you can compose a beat that repeats a drum sound for a fixed number of times or until a condition changes.
音乐创作涉及重复和模式，很像编程中的循环。使用基于积木的语言，你可以创作一段节拍，让鼓声重复固定的次数或直到条件改变为止。
Write a Scratch script that plays a snare drum note 8 times, then a bass drum note 4 times. Use a nested loop to play the whole pattern twice. Explain how this relates to musical structure, such as verses and choruses.
编写一个 Scratch 脚本，播放 8 次军鼓音符，然后播放 4 次大鼓音符。使用嵌套循环将整个模式重复两遍。解释这如何与音乐结构（如主歌和副歌）相关联。
| Musical concept | Programming concept |
|---|---|
| Repeat a riff 4 times | For loop (repeat 4) |
| Chorus after two verses | Call a function or broadcast a message |
| Fade out volume gradually | Variable decreasing in a loop |
5. Networks and Communication in History | 历史中的网络与通信
Before the internet, people used physical networks like postal systems and telegraph wires. Exploring how messages travelled in history helps you understand modern computer networks and packet switching.
在互联网出现之前，人们使用如邮政系统和电报线这样的物理网络。探索历史上信息如何传递有助于你理解现代计算机网络和数据包交换。
Compare sending a letter in the Roman Empire with sending an email today. Describe three similarities and three differences, considering routing, delivery time, and error correction.
比较罗马帝国时期寄信与今天发送电子邮件。描述三点相似和三点不同，从路由、送达时间和纠错方面考虑。
Similarities: both need an address (recipient), both rely on a chain of relay points (post stations vs routers), and both can be delayed or lost. Differences: email is near-instant; postal systems used physical transport; digital messages can carry error-checking codes like checksums, while letters relied on wax seals to detect tampering.
相似点：两者都需要地址（收件人），都依赖于一系列中继点（驿站 vs 路由器），且都可能被延迟或丢失。区别：电子邮件几乎瞬时到达；邮政系统使用实体运输；数字信息可以携带校验和之类的查错码，而信件则依靠蜡封来检测是否被篡改。
6. Computational Thinking in Art & Design | 艺术与设计中的计算思维
Decomposition and pattern recognition are as important in art as they are in computing. When you create a mosaic or a geometric pattern, you break the image into small parts, much like a programmer decomposes a problem.
分解与模式识别在艺术中和在计算机中一样重要。当你创作一幅马赛克或几何图案时，你会把图像分解成小部分，就像程序员分解问题一样。
n algorithm that places a star tile at every intersection where the row and column numbers are both even. Then shade the result.
任务：使用 10×10 网格设计一个伊斯兰几何星形图案。编写一个算法，在行号和列号都为偶数的每个交叉点放置一个星形瓦片。然后给结果上色。
The algorithm in pseudocode might read: FOR col = 1 to 10 DO; IF (row mod 2 = 0) AND (col mod 2 = 0) THEN drawStar at (row, col). This combines creativity with logical operators and modular arithmetic – skills that also appear in mathematics.
伪代码中的算法可能是：FOR row = 1 to 10 DO; FOR col = 1 to 10 DO; IF (row mod 2 = 0) AND (col mod 2 = 0) THEN 在 (row, col) 处画星形。 这结合了创造力与逻辑运算符以及模运算——这些技能也会在数学中出现。
7. E-Safety and PSHE | 电子安全与个人、社会、健康和经济教育
Keeping safe online is a cross-curricular topic shared between computing and PSHE. AQA-style questions often present a scenario about passwords, social media settings, or cyberbullying and ask you to apply both technical and ethical reasoning.
安全上网是计算机与个人、社会、健康和经济教育共有的跨学科主题。AQA 式题目常常呈现一个关于密码、社交媒体设置或网络欺凌的场景，要求你同时应用技术和伦理推理。
Scenario: Maya receives a friend request from a stranger who claims to go to the same school. The stranger asks for her phone number and a photo. Describe three security risks and suggest three technical or behavioural strategies Maya should use.
场景：Maya 收到一个自称是同校学生的陌生人的好友请求。这个陌生人索要她的电话号码和照片。描述三个安全风险，并建议 Maya 应采用的三个技术或行为策略。
- Risks: identity theft, grooming, location tracking.
- Strategies: set profile to private, block and report the user, never share personal information without checking with a trusted adult.
- Technical tips: use a strong password with symbols and numbers, enable two-factor authentication, review privacy settings monthly.
- 风险：身份盗窃、诱导、位置跟踪。
- 策略：将个人资料设为私密、屏蔽并举报该用户、未经与信任的成人确认绝不要分享个人信息。
- 技术提示：使用包含符号和数字的强密码、启用双因素认证、每月检查隐私设置。
8. Logic Gates and Electronics | 逻辑门与电子学
Logic gates are the building blocks of digital circuits and link computing directly to physics lessons about electricity. In Year 7 you may meet AND, OR and NOT gates through truth tables and simple combinations.
逻辑门是数字电路的基本构件，并将计算机直接与物理课中的电学联系起来。在七年级，你可能会通过真值表和简单的组合接触到与门、或门和非门。
Draw a circuit diagram with two switches (inputs A and B) controlling a lamp. The lamp should light up only when both A is ON and B is OFF. Identify the gate combination and write its truth table.
画一个电路图，两个开关（输入 A 和 B）控制一盏灯。只有当 A 打开且 B 关闭时灯才亮。识别门组合并写出其真值表。
| A | B | NOT B | A AND (NOT B) | Lamp (Output) |
|---|---|---|---|---|
| 0 | 0 | 1 | 0 | OFF |
| 0 | 1 | 0 | 0 | OFF |
| 1 | 0 | 1 | 1 | ON |
| 1 | 1 | 0 | 0 | OFF |
tem that permits access only when a door is closed (B=OFF) and a key card is valid (A=ON).
组合需要一个非门用于输入 B，以及一个与门用于 (A AND NOT B)。这与安全系统中的逻辑相同，只有当门关闭（B=OFF）且钥匙卡有效（A=ON）时才允许进入。
9. Data Analysis in Sport | 体育数据分析
Sports statistics are a rich source of data for analysis. You can collect, filter and visualise numbers using spreadsheets, linking computing to physical education and maths.
体育统计数据是数据分析的丰富来源。你可以使用电子表格收集、筛选和可视化数字，将计算机与体育教育和数学联系起来。
Collect the number of goals scored by your five favourite football players over four matches. Create a formula in a spreadsheet to calculate their average goals, and use conditional formatting to highlight averages above 1.5.
收集你最喜欢的五名足球运动员在四场比赛中的进球数。在电子表格中创建一个公式计算他们的平均进球数，并使用条件格式突出显示平均值大于 1.5 的单元格。
| Player | Match 1 | Match 2 | Match 3 | Match 4 | Average |
|---|---|---|---|---|---|
| Ali | 2 | 1 | 0 | 3 | 1.5 |
| Ben | 0 | 0 | 1 | 0 | 0.25 |
The average formula in cell G2 would be =AVERAGE(B2:E2). Conditional formatting with a rule 'Cell value > 1.5' can automatically turn a cell green. This integrates data handling with statistical measures from maths.
单元格 G2 中的平均值公式将是 =AVERAGE(B2:E2)。使用规则‘单元格值 > 1.5’的条件格式可以自动将单元格变为绿色。这将数据处理与数学中的统计量度结合了起来。
10. Cryptography and English | 密码学与英语
Secret codes have been used throughout history and literature. In computing, cryptography protects data. By writing and breaking simple ciphers, you improve both pattern recognition and your understanding of letter frequency.
历史上和文学作品中一直使用秘密代码。在计算机中，密码学保护数据安全。通过编写和破解简单密码，你可以提高模式识别能力并理解字母频率。
Use a Caesar cipher with a shift of +3 to encode the message 'MEET ME AT DAWN'. Then decode the response 'WKH EULGJH'.
使用位移为 +3 的凯撒密码将消息 'MEET ME AT DAWN' 编码。然后解码回复 'WKH EULGJH'。
M → P, E → H, T → W, so 'MEET' becomes 'PHHW'. Full encryption: 'PHHW PH DW GDZQ'.
For decoding, shift each letter back by 3: W → T, K → H, H → E, so 'WKH' = 'THE'; E → B, U → R, L → I, J → G, H → E → 'BRIDGE'. The decrypted message is 'THE BRIDGE'. This is a fun way to combine English spelling with algorithmic transformation.
解码时，每个字母向回移 3 位：W → T, K → H, H → E，所以 'WKH' = 'THE'；E → B, U → R, L → I, J → G, H → E → 'BRIDGE'。解密后的消息是 'THE BRIDGE'。这是将英语拼写与算法转换相结合的一种有趣方式。
11. Robotics and Physics | 机器人与物理
Programming a robot involves sensors, motors and control structures, closely tied to physics concepts such as speed, distance and force. A typical integrated question asks you to calculate how many wheel rotations are needed to travel a certain distance.
给机器人编程涉及传感器、电机和控制结构，与速度、距离和力等物理概念密切相关。一个典型的综合问题会要求你计算机器人移动特定距离需要多少轮子旋转。
A robot has wheels with a diameter of 7 cm. How many complete rotations must the wheels turn to travel 1 metre? Write a program loop that drives the robot forward until an ultrasonic sensor detects an obstacle less than 15 cm away.
一个机器人的轮子直径为 7 cm。要行进 1 metre，轮子必须转多少圈？编写一个程序循环，使机器人向前行驶，直到超声波传感器检测到障碍物小于 15 cm 为止。
Circumference = π × d = 3.14 × 7 ≈ 21.98 cm.
÷ 21.98 cm ≈ 4.55, so 5 complete rotations (round up for practical movement). Pseudocode: WHILE ultrasonic distance > 15 cm DO; forward(50); END WHILE; stop(); This combines geometry, unit conversion and programming logic.
100 cm ÷ 21.98 cm ≈ 4.55，所以需要 5 圈完整的旋转（实际运动取整）。伪代码：WHILE 超声波距离 > 15 cm DO; forward(50); END WHILE; stop(); 这结合了几何、单位转换和程序逻辑。
12. Integrated Practice Test | 综合练习测试
Work through the following multi-step problem that draws together binary, flowcharts, and e-safety. This is typical of the extended questions you may see in an end-of-topic assessment.
完成以下结合了二进制、流程图和电子安全的多步骤问题。这是你在单元末评估中可能遇到的典型扩展题。
Scenario: A school library uses a 4-bit binary code to label bookshelves. Shelf '1001₂' holds computing books. Draw a flowchart that checks if a student's age is 12 or above and whether they have a permission slip. If both conditions are true, output 'Access Granted' along with the shelf location in decimal. If not, output 'Access Denied' and report to the teacher. Also explain why storing ages digitally requires encryption under e-safety rules.
场景：一所学校图书馆使用 4 位二进制代码来标记书架。书架 '1001₂' 存放计算机书籍。绘制一个流程图，检查学生的年龄是否在 12 岁及以上以及他们是否有家长许可单。如果两个条件都为真，则输出‘Access Granted’以及十进制书架位置。如果不符合，则输出‘Access Denied’并向老师报告。同时解释为什么根据电子安全规则，以数字方式存储年龄需要加密。
owchart must include two decision diamonds: one for age ≥ 12 and one for permission slip = 'Yes' so that if unauthorised users access the database, they cannot read sensitive information – protecting student identity.
转换：1001₂ = 9₁₀。流程图必须包含两个决策菱形：一个用于 age ≥ 12，另一个用于 permission slip = 'Yes'。用 AND 逻辑条件组合。对于电子安全，像年龄这样的个人数据必须加密，这样即使未授权用户访问了数据库，他们也无法读取敏感信息——从而保护学生身份。
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