📚 IGCSE WJEC Computer Science: Unit Test Paper | IGCSE WJEC 计算机：单元测试卷
This unit test paper provides a comprehensive set of exam‑style questions covering key topics from the WJEC IGCSE Computer Science specification. Each question is followed by a detailed answer and explanation to reinforce your understanding. Use this resource to test your knowledge and identify areas for further revision.
本单元测试卷提供了一套全面的考试风格题目，覆盖 WJEC IGCSE 计算机科学大纲的核心知识点。每道题后都附有详细的答案与解析，以巩固你的理解。请利用此资源检验所学，并找出需要进一步复习的内容。
1. Binary to Hexadecimal Conversion | 二进制与十六进制转换
Convert the 8‑bit binary number 11010110 into its hexadecimal equivalent. Show your working.
将 8 位二进制数 11010110 转换为等值的十六进制数。请写出计算过程。
To convert, split the binary number into two 4‑bit nibbles: 1101 and 0110. 1101 in binary equals 13 in decimal, which is D in hexadecimal. 0110 equals 6 in decimal, which is 6 in hexadecimal. Therefore the result is D6.
转换时，先将二进制数拆分为两个 4 位半字节：1101 和 0110。1101 二进制相当于十进制 13，即十六进制的 D。0110 相当于十进制 6，即十六进制的 6。因此结果为 D6。
2. Logic Gates and Truth Tables | 逻辑门与真值表
Complete the truth table for the logic circuit: Q = (A AND B) OR (NOT C). The inputs A, B, C are each either 0 or 1. Write the output Q for every possible combination of inputs.
完成以下逻辑电路的真值表：Q = (A AND B) OR (NOT C)。输入 A、B、C 的取值均为 0 或 1。写出所有输入组合对应的输出 Q。
For the 8 combinations (000 to 111), evaluate: A AND B gives 1 only when A=1 and B=1; NOT C inverts C. The OR combines them. The outputs are: 000→1, 001→0, 010→1, 011→0, 100→1, 101→0, 110→1, 111→1.
对于 8 种组合（000 到 111），计算过程为：A AND B 仅在 A=1 且 B=1 时得 1；NOT C 将 C 取反；最后 OR 运算合并二者。输出结果依次为：000→1，001→0，010→1，011→0，100→1，101→0，110→1，111→1。
3. Components of the CPU | CPU 的组成部件
Name three key components inside the central processing unit (CPU) and briefly explain the role of each.
说出中央处理器（CPU）内部的三个关键部件，并简要说明各自的作用。
The control unit (CU) directs the operation of the processor by fetching, decoding and coordinating instructions. The arithmetic logic unit (ALU) performs calculations and logical operations. Registers are very fast storage locations that hold data, addresses or instructions temporarily during execution. Together they form the fetch–decode–execute cycle.
控制单元（CU）通过取指、译码和协调指令来指挥处理器的工作。算术逻辑单元（ALU）执行算术与逻辑运算。寄存器是速度极快的暂存位置，在执行过程中临时保存数据、地址或指令。三者共同完成取指–译码–执行周期。
4. Primary and Secondary Storage | 主存储与辅助存储
Distinguish between RAM and ROM in terms of volatility and typical usage. Then give one example of a secondary storage device and explain why secondary storage is needed.
从易失性和典型用途两个方面区分 RAM 和 ROM。然后举一个辅助存储设备的例子，并解释为什么需要辅助存储。
RAM is volatile, meaning its contents are lost when power is turned off; it is used to hold the operating system, programs and data currently in use. ROM is non‑volatile, retains data without power, and typically stores firmware such as BIOS or bootstrap code. A solid‑state drive (SSD) is an example of secondary storage. Secondary storage is needed because it provides permanent, large‑capacity storage for software and user data, and retains data when the computer is off.
RAM 是易失性的，断电后内容丢失，用于存放当前运行的操作系统、程序和数据。ROM 为非易失性，掉电后数据保留，通常存储固件，如 BIOS 或引导代码。固态硬盘（SSD）是辅助存储的一例。需要辅助存储是因为它能提供永久的大容量存储，保存软件和用户数据，并且在关机后仍可保留信息。
5. Network Topologies | 网络拓扑
Compare a star topology with a mesh topology in terms of fault tolerance and cabling requirements. State one advantage and one disadvantage of each.
从容错性和布线要求方面比较星形拓扑和网状拓扑。指出每种拓扑的一个优点和一个缺点。
In a star topology, all devices connect to a central switch. An advantage is easy fault isolation — if one cable fails only that device is affected. A disadvantage is the central switch is a single point of failure. Cabling is moderate. In a full mesh topology, each device connects to every other device. It offers excellent fault tolerance because multiple redundant paths exist. However, it requires extensive cabling, making it expensive and difficult to scale. Star is therefore cheaper and simpler, while mesh provides higher reliability.
在星形拓扑中，所有设备连接到中央交换机。优点是故障隔离容易——某条线路故障只影响该设备。缺点是中央交换机是单点故障点。布线量适中。在全网状拓扑中，每台设备都与其他所有设备直连。它提供极佳的容错性，因为存在多条冗余路径。但需要大量布线，成本高且不易扩展。因此星形拓扑较便宜、简单，而网状拓扑可靠性更高。
6. Network Security | 网络安全
Describe what a brute‑force attack is and explain two measures that can help protect an online account from this type of threat.
描述什么是暴力攻击，并解释两种可帮助保护在线账户免受此类威胁的措施。
A brute‑force attack involves systematically trying all possible passwords until the correct one is found. To defend against this, accounts can enforce strong password policies — requiring a mix of characters, length and complexity — which makes guessing computationally infeasible. Additionally, implementing account lockout after a certain number of failed attempts prevents an attacker from making unlimited guesses. Two‑factor authentication (2FA) adds another protection layer even if the password is cracked.
暴力攻击是指系统地试遍所有可能的密码，直到找出正确的一个。为防御此类攻击，账户可强制实施强密码策略——要求混合字符、长度和复杂性——使猜测在计算上变得不可行。此外，设定多次登录失败后锁定账户，可以阻止攻击者无限次试图猜测。双因素认证（2FA）即使密码被破解也能增加一层保护。
7. Programming Concepts – Loops and Variables | 编程概念 – 循环与变量
Look at the pseudocode below. It is meant to calculate the sum of all even numbers between 1 and 20 inclusive. Identify and correct the error in the algorithm.
阅读以下伪代码。其目的是计算 1 到 20（含）之间所有偶数的总和。找出并修正算法中的错误。
total ← 0
FOR i ← 1 TO 20
IF i MOD 2 = 0 THEN
total ← total + i
ENDIF
NEXT i
OUTPUT total
The error is that the loop bounds are correct but the statement total ← total + i appears only when the condition is true, which is correct. Actually there is no syntax error; logically the algorithm sums even numbers correctly. However, a common mistake could be counting from 0 to 20 or misplacing the IF. If a student writes FOR i ← 0 TO 20 by mistake, the loop would include 0 (which is even but does not affect sum) but also might include 20 correctly. The corrected version is as shown: total initialised to 0, loop from 1 to 20, add i only when i MOD 2 equals 0. The output is 110.
错误可能出现在边界或条件位置。若学生错误地写成 FOR i ← 0 TO 20，循环将包含 0（偶数但不影响总和）和 20。正确的版本是：total 初始化为 0，循环从 1 到 20，仅当 i MOD 2 = 0 时才将 i 加到 total。最终输出为 110。因此算法本身是正确的，但需注意循环起始值应为 1。
8. Data Compression | 数据压缩
Explain the difference between lossy and lossless compression. Give a scenario where lossy compression is more suitable and justify your choice.
解释有损压缩与无损压缩的区别。给出一个有损压缩更合适的场景，并说明理由。
Lossless compression reduces file size without losing any information — the original data can be perfectly reconstructed. It is essential for text documents and executable files. Lossy compression permanently removes some data considered less important, achieving much smaller file sizes but with some quality reduction. For instance, streaming music over a mobile network: lossy formats like MP3 greatly reduce data usage, and the human ear can tolerate slight quality loss in exchange for faster loading and lower bandwidth cost.
无损压缩在减少文件体积的同时不丢失任何信息——原始数据可以被完全复原。它对文本文档和可执行文件至关重要。有损压缩则永久性地移除某些被判定为不重要的数据，能获得更小的文件，但会有一定的质量下降。例如，在移动网络中流式播放音乐：MP3 等有损格式可大幅减少数据用量，人耳可以容忍轻微的音质损失，以换取更快的加载速度和更低的带宽成本。
9. Operating System Functions | 操作系统功能
List three main functions of an operating system and explain how they contribute to the efficient operation of a computer.
列出操作系统的三个主要功能，并解释它们如何帮助计算机高效运行。
First, memory management allocates RAM to programs and frees it when no longer needed, preventing conflicts and optimising multitasking. Second, process scheduling decides the order and time slice for each running program, ensuring the CPU is used fairly and responsively. Third, file management organises data on storage devices, providing a hierarchical directory structure and access controls, so users and applications can save and retrieve files reliably. These functions together create a stable environment for software execution.
首先，内存管理为程序分配 RAM 并在不再需要时释放，避免冲突并优化多任务处理。其次，进程调度决定每个运行程序的执行顺序和时间片，保证 CPU 公平使用并维持响应速度。第三，文件管理在存储设备上组织数据，提供层次化的目录结构和访问控制，使得用户和应用程序能够可靠地保存与提取文件。这些功能共同为软件执行创造稳定的环境。
10. Ethical and Environmental Issues | 道德与环境问题
Discuss one ethical issue and one environmental concern associated with the widespread use of computing devices. Suggest responsible practices that individuals or organisations can adopt.
讨论与计算机设备广泛使用相关的一个道德问题和一个环境问题。提出个人或组织可以采取的负责任做法。
An ethical issue is the collection and misuse of personal data without consent, leading to privacy violations and identity theft. Organisations should follow data protection regulations, obtain clear consent, and minimise data collection. An environmental concern is the growing volume of electronic waste (e‑waste), which contains toxic materials and pollutes ecosystems when improperly disposed. Responsible practices include recycling devices through certified schemes, donating functional equipment for reuse, and designing products with longer lifespans and easier repairability.
一个道德问题是未经同意收集和滥用个人数据，导致隐私侵犯和身份盗用。组织应遵循数据保护法规，获得明确同意并尽量减少数据收集。一个环境问题是日益增长的电子废弃物（电子垃圾），其中含有有毒物质，不当处置会污染生态系统。负责任的做法包括通过认证渠道回收设备、捐赠仍可使用的器材以促进再利用，以及设计寿命更长、更易维修的产品。
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