📚 Year 10 Eduqas Computer Science: Interdisciplinary Mixed-Topic Practice | Year 10 Eduqas 计算机科学：跨学科综合题型训练
Eduqas GCSE Computer Science is not an isolated subject; it thrives when connected to mathematics, physics, geography, biology, business, music and art. This article provides a collection of interdisciplinary mixed-topic practice questions, designed to strengthen your computational thinking while demonstrating how computer science concepts underpin real-world applications. Each section presents a themed problem, linking key Year 10 topics—algorithms, data representation, programming, networks, databases, cybersecurity, logic gates, spreadsheets, web technologies and recursion—to other academic disciplines. Use these challenges to prepare for exam-style questions that require you to transfer knowledge across boundaries.
Eduqas GCSE 计算机科学并不是一门孤立的学科；当它与数学、物理、地理、生物、商业、音乐和艺术联系起来时，会变得格外有生命力。本文提供了一系列跨学科综合题型训练，旨在强化你的计算思维，同时展示计算机科学概念如何支撑现实世界的应用。每个小节设计了一个主题问题，将算法、数据表示、编程、网络、数据库、网络安全、逻辑门、电子表格、网页技术和递归等 Year 10 核心知识点与其他学科领域关联起来。利用这些挑战来备考那些需要你跨领域迁移知识的考试题型。
1. Algorithmic Thinking in Mathematics: Prime Number Sieve | 数学中的算法思维：质数筛选
p to 100, you start by listing integers from 2 to 100 and repeatedly marking multiples of each unmarked number as composite. The efficiency relies on only iterating up to the square root of the upper bound, because any composite number n must have a factor ≤ √n.
厄拉多塞筛法是一个将数学理论与分步逻辑执行融合的经典算法。若要找出 100 以内的所有质数，首先列出 2 到 100 的整数，然后反复将每个未标记数的倍数标记为合数。其高效性在于只需迭代到上限的平方根即可，因为任何合数 n 必定有一个 ≤ √n 的因子。
d a loop structure. For N = 100, trace the first three iterations and state how many numbers remain unmarked after the process.
考试题型：用伪代码编写一个算法，输出直到给定输入 N 的所有质数。你的解答须使用一个布尔数组和循环结构。当 N = 100 时，追踪前三次迭代，并说明处理结束后还有多少个数字保持未标记状态。
p TO N STEP p: SET isPrime[multiple] TO false. OUTPUT all indices where isPrime is true. The mathematical grounding helps you understand why the inner loop starts at p².
伪代码片段：将 isPrime[2..N] 全部设置为 true。对于 p 从 2 到 sqrt(N)：如果 isPrime[p] 为真，则将 multiple 从 p×p 到 N 步长 p 的 isPrime[multiple] 设为 false。输出所有 isPrime 为真的下标。数学基础能帮助你理解为什么内层循环从 p² 开始。
2. Binary Data Representation: Measuring and Storing Scientific Data | 二进制数据表示：测量与存储科学数据
nding the relationship between binary word length and resolution is crucial. An 8-bit ADC (analogue-to-digital converter) divides a voltage range into 2⁸ = 256 levels. If the range is 0 V to 5 V, the smallest detectable change is 5 ÷ 255 ≈ 0.0196 V.
在科学实验中，温度、pH 或光强度等测量值常由传感器捕捉并以二进制值存储。理解二进制字长与分辨率之间的关系至关重要。一个 8 位模数转换器（ADC）将电压范围划分为 2⁸ = 256 个等级。若量程为 0 V 至 5 V，则最小可检测变化为 5 ÷ 255 ≈ 0.0196 V。
Convert the following analogue readings into 8-bit binary, assuming a 0–5 V range: 2.35 V, 4.10 V, 0.78 V. Also, calculate how many bits are needed to achieve a resolution of at least 0.005 V over the same range. Show your working and explain the trade-off with storage space.
假设量程为 0–5 V，将以下模拟读数转换为 8 位二进制：2.35 V、4.10 V、0.78 V。此外，计算要在此量程下实现至少 0.005 V 的分辨率需要多少位。展示计算过程，并解释与存储空间之间的权衡。
ts (1023 steps). Each sample stored as a 10-bit integer increases file size compared to 8-bit, which matters when collecting thousands of readings per second in environmental monitoring.
使用公式 步数 = 2ⁿ − 1，若需 0.005 V 分辨率，则要求 5 ÷ 0.005 = 1000 个步数，因此 2ⁿ − 1 ≥ 1000，得到 n = 10 位（1023 个步数）。与 8 位相比，每个样本存储为 10 位整数会增大文件体积，这在进行环境监测每秒采集数千个读数时不可忽视。
3. Programming a Physics Simulation: Projectile Motion | 编程物理模拟：抛体运动
d gravitational acceleration g = 9.81 m/s², the horizontal and vertical displacements at time t are: x = u × cosθ × t, y = u × sinθ × t − ½ g t². You can model the motion in a loop until y becomes negative.
模拟抛体轨迹将运动学方程与编程结构结合起来。给定初速度 u、发射角 θ 和重力加速度 g = 9.81 m/s²，时刻 t 的水平与垂直位移分别为：x = u × cosθ × t，y = u × sinθ × t − ½ g t²。你可以在循环中对运动建模，直到 y 变为负值。
rojectile lands. Assume no air resistance. Use a suitable loop and remember to convert the angle to radians. What is the maximum height reached if u = 20 m/s and θ = 45°?
编写一个 Python 风格的程序，要求用户输入速度（m/s）和角度（度），然后每 0.1 秒打印 (x, y) 坐标，直到抛体落地。假设无空气阻力。使用合适的循环，并记得将角度转换为弧度。若 u = 20 m/s、θ = 45°，最大高度是多少？
n track the y position and stop when y ≤ 0, appending coordinates to a list for later plotting, thus linking programming with data visualisation.
H = (u² × sin²θ) ÷ (2g)
最大高度公式为 H = (u² × sin²θ) ÷ (2g)。代入数值：H = (20² × (sin 45°)²) ÷ (2 × 9.81) ≈ (400 × 0.5) ÷ 19.62 ≈ 10.19 m。在代码中，你可以追踪 y 轴位置，当 y ≤ 0 时停止，并将坐标加入列表以便后续绘图，从而将编程与数据可视化联系起来。
4. Networks and Geography: Network Topologies for Smart Cities | 网络与地理：智慧城市的网络拓扑
Urban planners use network topology concepts to design communication infrastructures. A city-wide sensor network might adopt a mesh topology for resilience, where each node (traffic light, air quality monitor) relays data. Alternatively, a star topology with central hubs can reduce cable length but introduces a single point of failure. Geography influences the choice: a compact city centre may benefit from star, while a sprawling suburb needs mesh.
城市规划者运用网络拓扑概念设计通信基础设施。一个全市范围的传感器网络可能采用网状拓扑以增强韧性，其中每个节点（交通灯、空气质量监测仪）都会中继数据。或者，使用带有中央集线器的星形拓扑可以减少线缆长度，但引入了单点故障。地理环境影响着选择：紧凑的市中心适合星形，而蔓延的郊区则需要网状。
Question: A smart city covers an area of 50 km² with 500 sensor nodes distributed evenly. Compare the total cable length required for a star network with a single central hub at the geographic centre versus a full mesh network. Assume average distance from centre to node is 2.8 km. For mesh, each connection is a dedicated cable; calculate the approximate number of cables and comment on scalability and fault tolerance.
问题：某智慧城市占地 50 km²，均匀分布 500 个传感器节点。比较以地理中心设置单个中央集线器的星形网络与全连接网状网络所需的总线缆长度。假设从中心到节点的平均距离为 2.8 公里。对网状结构，每条连线为专用线缆；计算大约的线缆数量，并评论可扩展性和容错性。
0. Although mesh offers extreme redundancy, the cost and complexity are unmanageable, so a hybrid topology or wireless mesh is used in reality—demonstrating how geography and economics shape network design.
星形拓扑总线缆 ≈ 500 × 2.8 公里 = 1400 公里。全连接网状结构中，线缆数量 = n(n−1)/2 = 500×499/2 = 124,750。尽管网状提供了极高的冗余度，但成本与复杂性无法承受，因此现实中会采用混合拓扑或无线网状网络——这展示了地理和经济因素如何影响网络设计。
5. Databases and Biology: Designing a Species Classification Database | 数据库与生物学：设计物种分类数据库
Biological classification relies on hierarchical data perfectly suited for relational databases. A flat table with fields SpeciesID, CommonName, Genus, Family, Order, Class, Phylum, Kingdom would introduce redundancy. Normalisation leads to separate tables for each taxonomic rank, linked by foreign keys. You can then use SQL to query, for instance, all species belonging to the Carnivora order.
生物分类依赖于非常适合于关系数据库的层次化数据。若使用一张包含字段 SpeciesID、通用名、属、科、目、纲、门、界的平面表会引入冗余。规范化会产生针对各个分类等级的独立表，并通过外键关联。接着你便能用 SQL 查询，例如，所有属于食肉目的物种。
Task: Create a 3NF database design for a wildlife catalogue. Define tables: Kingdom (KingdomID, KingdomName), Phylum (PhylumID, PhylumName, KingdomID), …, Species (SpeciesID, ScientificName, GenusID). Write an SQL statement to count how many species are in the Mammalia class. Explain why this design reduces update anomalies when a family name changes.
任务：为野生动物目录创建一个满足第三范式的数据库设计。定义如下表：Kingdom (KingdomID, KingdomName)，Phylum (PhylumID, PhylumName, KingdomID)，......，Species (SpeciesID, ScientificName, GenusID)。编写一条 SQL 语句来统计哺乳纲内的物种数量。解释当科名更改时，该设计为何能减少更新异常。
amily.OrderID = OrderC.OrderID JOIN Class ON OrderC.ClassID = Class.ClassID WHERE Class.ClassName = 'Mammalia' amily name is stored only once in the Family table, a single UPDATE avoids inconsistencies—a direct application of database theory to biological data management.
SQL 语句： y.OrderID = OrderC.OrderID JOIN Class ON OrderC.ClassID = Class.ClassID WHERE Class.ClassName = 'Mammalia'; 由于科名仅存储在 Family 表中一次，一条 UPDATE 即可避免不一致——这是数据库理论在生物数据管理中的直接应用。
6. Cybersecurity and Ethics in Healthcare: Patient Data Protection | 医疗保健中的网络安全与伦理：患者数据保护
Electronic Health Records (EHRs) store sensitive personal information, making cybersecurity a critical concern. Encryption, access controls and penetration testing are technical measures, but they intersect with ethical principles such as confidentiality, integrity and availability (CIA triad). Year 10 students must recognise how the Computer Misuse Act and GDPR apply to healthcare systems.
电子健康记录（EHR）存储着敏感的个人信息，使网络安全成为关键问题。加密、访问控制和渗透测试属于技术措施，但它们与保密性、完整性和可用性（CIA 三元组）等伦理原则相交织。Year 10 学生必须认识到《计算机滥用法》和《通用数据保护条例》（GDPR）如何适用于医疗保健系统。
Scenario: A hospital app uses a four-digit PIN for patient login. Describe two technical vulnerabilities and suggest stronger authentication. Then discuss the ethical implications if an unauthorised staff member accesses a celebrity’s medical records out of curiosity. Reference relevant legislation.
情景：某医院应用程序使用四位数字 PIN 进行患者登录。描述两个技术漏洞并提出更强的身份验证方案。然后讨论若一名未经授权的员工出于好奇访问了某位名人的医疗记录，会带来哪些伦理影响。援引相关法律。
Four-digit PIN has only 10,000 combinations; brute-force attacks are feasible. Also, shoulder surfing is a risk. Stronger authentication could use multi-factor authentication (MFA) with a one-time code and biometrics. Ethically, accessing records without a legitimate reason violates GDPR principles and patient trust; the staff member could face prosecution under the Computer Misuse Act for unauthorised access, highlighting the real-world weight of cybersecurity ethics.
四位 PIN 仅有一万种组合，暴力破解是可行的。此外，肩窥攻击也存在风险。更强的身份验证可采用多因素认证（MFA），结合一次性验证码和生物特征。伦理上，无正当理由访问病历违反了 GDPR 原则与患者信任；涉事员工可能因未经授权的访问而根据《计算机滥用法》面临起诉，这凸显了网络安全伦理的现实分量。
7. Logic Gates and Electrical Circuits: Building a Half Adder | 逻辑门与电路：构建半加器
Digital electronics underpins computing hardware. A half adder, built from an XOR gate (for the sum) and an AND gate (for the carry), demonstrates how binary addition is implemented physically. This topic links Boolean algebra and truth tables from computer science with circuit design in physics.
数字电子学是计算硬件的基石。由异或门（用于求和）和与门（用于进位）构成的半加器，展示了如何物理上实现二进制加法。这一主题将计算机科学中的布尔代数与真值表同物理学中的电路设计联系起来。
Draw the logic circuit for a half adder and produce its truth table with inputs A, B and outputs SUM, CARRY. Then, explain how two half adders and an OR gate can be combined to form a full adder capable of adding three single-bit binary numbers.
绘制半加器的逻辑电路图，并给出输入为 A、B、输出为 SUM、CARRY 的真值表。然后，解释如何将两个半加器和一个或门组合成一个全加器，以相加三个单比特二进制数。
| A | B | SUM (A XOR B) | CARRY (A AND B) |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
Half adder truth table (above). For a full adder: the first half adder processes inputs A and B, producing intermediate sum S1 and carry C1. The second half adder takes S1 and input Cin ARRY_OUT is C1 OR C2. This cascading mirrors the way processors handle multi-bit addition.
半加器真值表（如上）。对于全加器：第一个半加器处理输入 A 和 B，产生中间和 S1 与进位 C1；第二个半加器接收 S1 与输入进位 Cin，生成最终 SUM 与另一进位 C2；最终 CARRY_OUT 为 C1 OR C2。这种级联方式反映了处理器处理多位加法的方式。
8. Spreadsheet Modelling and Business Finance: Break-Even Analysis | 电子表格建模与商业金融：盈亏平衡分析
Spreadsheets are powerful tools for financial modelling. A break-even analysis determines the number of units a business must sell to cover its costs. Using formulas, absolute cell references and charting features, you can visualise the intersection of total revenue and total cost lines. This blends business studies with computational spreadsheet skills.
电子表格是金融建模的强大工具。盈亏平衡分析用于确定企业必须销售多少件产品才能覆盖成本。通过运用公式、绝对单元格引用和图表功能，你可以将总收入与总成本线的交点可视化。这融合了商科学习与电子表格计算技能。
Model: Fixed costs = £5000, variable cost per unit = £12, selling price per unit = £ reak-even point. Then produce a line chart showing both cost and revenue against quantity.
模型：固定成本 = 5000 英镑，每件可变成本 = 12 英镑，每件售价 = 20 英镑。创建一个电子表格，计算从 0 到 1000、步长为 100 的数量所对应的总成本、总收入和利润。使用公式计算精确的盈亏平衡点。然后制作一个折线图，显示成本与收入随数量变化的情况。
e cost) = 5000 ÷ (20 − 12) = 5000 ÷ 8 = 625 units. In a spreadsheet, cell formulas like =B$1+B2*B$3 (total cost) and =B2*B$4 (revenue) use absolute references ($ es understanding of spreadsheet modelling and the importance of accurate cell referencing.
Break-even = FC ÷ (SP − VC)
盈亏平衡点数量 = 固定成本 ÷ (售价 − 可变成本) = 5000 ÷ (20 − 12) = 5000 ÷ 8 = 625 件。在电子表格中，诸如 =B$1+B2*B$3（总成本）和 =B2*B$4（收入）的单元格公式使用绝对引用（$）来在向下拖拽时保持常量固定。这项任务强化了对电子表格建模及准确单元格引用重要性的理解。
9. Web Technologies and Art: Creating an Online Gallery with HTML/CSS | 网页技术与艺术：用 HTML/CSS 创建在线画廊
Web development offers a platform to blend creative design with technical coding. An online art gallery can be built using HTML for structure and CSS for styling. Images, headings, paragraphs and hyperlinks are basic elements, while CSS flexbox or grid can arrange thumbnails beautifully. This cross-curricular task engages visual arts and digital literacy.
网页开发提供了一个融合创意设计与技术编码的平台。在线艺术画廊可以用 HTML 搭建结构、用 CSS 进行样式设计。图像、标题、段落和超链接是基本元素，而 CSS 的 flexbox 或 grid 可以美观地排布缩略图。这项跨学科任务融合了视觉艺术与数字素养。
Build a simple HTML page for a gallery of four famous paintings. Each painting should have an image, a title in an alt text, and a caption. Use CSS to display the images in a row, centred, with a shadow effect on hover. Explain how you used the box model and the hover pseudo-class. How would you make the gallery responsive for mobile screens?
为四幅著名画作创建一个简单的 HTML 页面。每幅画应包含一个图像、一个替代文字（alt）标题和一个说明。使用 CSS 使图像排成一行、居中显示，并在悬停时呈现阴影效果。解释你是如何运用框模型和 hover 伪类的。同时，你会如何使画廊在手机屏幕上也能够响应式显示？
Sample CSS: .gallery { display: flex; justify-content: center; } .gallery img { width: 200px; margin: 10px; box-shadow: none; transition: box-shadow 0.3s; } .gallery img:hover { box-shadow: 5px 5px 10px rgba(0,0,0,0.5); }. For responsiveness, use @media queries to change flex-direction to column on smaller screens. This activity shows how artistic presentation is enabled by careful use of web technology, linking aesthetic principles with coding rules.
示例 CSS：.gallery { display: flex; justify-content: center; } .gallery img { width: 200px; margin: 10px; box-shadow: none; transition: box-shadow 0.3s; } .gallery img:hover { box-shadow: 5px 5px
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