Year 10 WJEC Computer Science: Unit Test Mock Paper Analysis | 单元测试模拟卷解析

📚 Year 10 WJEC Computer Science: Unit Test Mock Paper Analysis | 单元测试模拟卷解析

This walkthrough analyses a complete Year 10 WJEC Computer Science unit test mock paper, covering typical questions from both Unit 1 and the computational thinking elements of Unit 2. Each section breaks down a real-style question, explains the expected answer, highlights common mistakes, and reinforces the theory behind the marks. Use this article to audit your understanding, sharpen your exam technique, and build confidence for the real assessment.

本文详细解析一份完整的 Year 10 WJEC 计算机科学单元测试模拟卷，内容覆盖 Unit 1 和 Unit 2 计算思维部分的典型题型。每一节会拆解一道真实风格的题目，解释参考答案与得分点，指出常见错误并巩固背后的理论。你可以用这篇文章来检查自己的知识掌握情况、磨练应试技巧、为正式测评树立信心。

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1. System Software and OS Concepts | 系统软件与操作系统概念

Question 1: “State two functions of an operating system.” (2 marks)

题目 1：”列出操作系统的两个功能。” (2 分)

A full-mark response could name memory management (allocating RAM to programs) and providing a user interface (such as a graphical user interface or command line). Many students lose marks by writing vague answers like “it manages the computer”. Examiners expect precise terms: file management, multitasking, peripheral management, and user management are all valid.

满分的答案可以是”内存管理”（为程序分配 RAM）和”提供用户界面”（如图形用户界面或命令行）。很多学生因为答案模糊而丢分，例如只写”管理计算机”。考官要求精确的术语：文件管理、多任务处理、外设管理、用户管理都是有效答案。

Another typical follow-up question asks: “Explain why an operating system needs to use interrupts.” An interrupt is a signal from hardware or software that tells the CPU to pause its current task and handle something urgent, such as a key press or a printer running out of paper. Without interrupts, the CPU would not be able to respond to real-time events efficiently.

另一道常见的后续问题是：”解释为什么操作系统需要使用中断。” 中断是硬件或软件发出的信号，告诉 CPU 暂停当前任务去处理更紧急的事情，比如按键按下或打印机缺纸。没有中断，CPU 就无法高效地响应实时事件。

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2. Binary and Hexadecimal Conversions | 二进制与十六进制转换

Question 2: “Convert the decimal number 206 into an 8-bit binary number and then into hexadecimal. Show your working.” (4 marks)

题目 2：”将十进制数 206 转换为 8 位二进制数，再转换为十六进制。请写出计算过程。” (4 分)

Start by building the binary value using place values 128, 64, 32, 16, 8, 4, 2, 1. 206 – 128 = 78 → 1; 78 – 64 = 14 → 1; 14 cannot subtract 32 → 0; 14 – 16 → cannot, 0; 14 – 8 = 6 → 1; 6 – 4 = 2 → 1; 2 – 2 = 0 → 1; 0 for 1 → 0. The binary result is 11001110₂.

首先根据位值 128, 64, 32, 16, 8, 4, 2, 1 构建二进制数。206 – 128 = 78 → 1；78 – 64 = 14 → 1；14 不够减 32 → 0；14 – 16 不够 → 0；14 – 8 = 6 → 1；6 – 4 = 2 → 1；2 – 2 = 0 → 1；个位 1 为 0。最终二进制结果为 11001110₂。

To convert to hexadecimal, split the binary into two nibbles: 1100₂ and 1110₂. 1100₂ = 12₁₀ = C, and 1110₂ = 14₁₀ = E. Therefore, the hexadecimal equivalent is CE₁₆. A common mistake is forgetting to group from the right, or miscounting the nibble values — practise quick nibble-to-hex lookups to save time.

转换为十六进制时，将二进制数拆成两个半字节：1100₂ 和 1110₂。1100₂ = 12₁₀ = C，1110₂ = 14₁₀ = E。因此十六进制等价为 CE₁₆。常见错误是忘记从右边开始分组，或计算半字节值时出错——多练习半字节与十六进制的速查可以节省时间。

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3. Logic Gates and Truth Tables | 逻辑门与真值表

Question 3: “Complete the truth table for a two-input AND gate, then draw a simple logic circuit diagram that uses an AND, an OR, and a NOT gate to express the condition: ‘Alarm sounds if door is open OR window is open AND system is armed’.” (6 marks)

题目 3：”完成一个两输入与门的真值表，然后画出一个简单的逻辑电路图，使用与门、或门和非门来表达以下条件：‘如果门开着 或 窗开着 且 系统已布防，则报警器响起。’” (6 分)

For the AND truth table, output is 1 only when both inputs are 1. The table:

与门真值表中，仅当两个输入均为 1 时输出为 1。表格如下：

A	B	Output
0	0	0
0	1	0
1	0	0
1	1	1

For the circuit, let D = door open (1 = open), W = window open, A = system armed. The Boolean expression is Alarm = (D OR W) AND A. In the diagram, D and W feed into an OR gate; the OR output and A feed into an AND gate; the AND output drives the alarm. No NOT gate is strictly needed unless inverting a signal, but you could show a NOT gate used to detect when the system is not armed. Precision with symbols and labels is essential; a hand-drawn diagram should use standard logic symbols and clear wiring.

对于电路，设 D = 门开 (1 表示开)，W = 窗开，A = 系统布防。布尔表达式为 Alarm = (D OR W) AND A。图中 D 和 W 输入一个或门；或门输出与 A 输入一个与门；与门输出驱动报警器。严格来说不需要非门，除非想表示信号取反，但你可以用一个非门来检测系统未布防的状态。画图时必须准确使用标准逻辑符号并清楚连线，手绘图也要符号规范、标注清晰。

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4. Network Topologies and Protocols | 网络拓扑与协议

Question 4: “Describe one advantage and one disadvantage of a star network topology. Then name the protocol used to securely transfer web pages.” (4 marks)

题目 4：”描述星形网络拓扑的一个优点和一个缺点。然后说出用于安全传输网页的协议名称。” (4 分)

Advantage: In a star topology, each node has its own dedicated cable to the central switch or hub. If one cable fails, only that node is affected — the rest of the network continues to function. This fault tolerance is a key benefit.

优点：在星形拓扑中，每个节点都有自己专用线缆连接到中央交换机或集线器。如果某根线缆出现故障，只有该节点受影响，网络其余部分仍可正常运行。这种容错能力是一个关键优点。

Disadvantage: The whole network depends on the central device; if the switch or hub fails, the entire network goes down. It is a single point of failure. Additionally, cabling costs can be higher than in a bus topology because every node needs its own cable back to the centre.

缺点：整个网络依赖于中央设备；如果交换机或集线器发生故障，整个网络就会瘫痪。这是一个单点故障。此外，布线成本可能比总线拓扑更高，因为每个节点都需要独立线缆连接至中心。

The protocol for secure web page transfer is HTTPS (Hypertext Transfer Protocol Secure). It encrypts data using TLS/SSL so that information such as passwords or payment details cannot be intercepted and read easily. Many students confuse it with HTTP, FTP or SMTP — remember that the ‘S’ stands for secure.

用于安全传输网页的协议是 HTTPS（超文本传输安全协议）。它利用 TLS/SSL 对数据进行加密，防止密码或支付信息等被轻易拦截和读取。很多学生会将它与 HTTP、FTP 或 SMTP 混淆——请记住 “S” 代表安全。

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5. Cybersecurity Threats and Prevention | 网络安全威胁与防护

Question 5: “Explain the difference between a virus and a worm. Then describe one way a firewall helps protect a network.” (4 marks)

题目 5：”解释病毒与蠕虫之间的区别。然后描述防火墙帮助保护网络的一种方式。” (4 分)

A virus is a piece of malicious code that attaches itself to a legitimate program or file and requires human action — such as opening an infected email attachment — to spread. Once activated, it can corrupt or delete data. A worm, on the other hand, is a standalone malware that self-replicates and spreads across networks without any user interaction, often exploiting security vulnerabilities.

病毒是一种恶意代码，它附着在合法程序或文件上，依赖人的操作（例如打开受感染的电子邮件附件）才能传播。一旦激活，病毒可以毁坏或删除数据。而蠕虫是一种独立运作的恶意软件，它能自我复制并自动通过网络传播，无需任何用户操作，通常利用安全漏洞扩散。

A firewall acts as a barrier between a trusted internal network and untrusted external networks. It inspects incoming and outgoing data packets according to a set of security rules, and blocks traffic that does not meet the criteria — for instance, traffic from suspicious IP addresses or requests to unapproved ports. This significantly reduces the attack surface.

防火墙充当可信内部网络与不可信外部网络之间的屏障。它根据一组安全规则检查进出数据包，并阻止不符合条件的流量，例如来自可疑 IP 地址的流量或未经批准的端口请求。这能显著减少网络的攻击面。

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6. Data Storage, Units and Compression | 数据存储、单位与压缩

Question 6: “A stereo audio track is sampled at 44.1 kHz with a 16-bit sample depth. Calculate the approximate file size in megabytes (MB) for a 10-second recording. Show your working. Then explain why lossy compression might be used for this file.” (5 marks)

题目 6：”一条立体声音轨以 44.1 kHz 采样率、16 位采样深度录制。计算一段 10 秒 录音的文件大小，结果用兆字节 (MB) 表示。请写出计算过程。然后解释为何对此文件可能使用有损压缩。” (5 分)

Step 1: 44.1 kHz = 44,100 samples per second per channel. Stereo means 2 channels. Bits per sample = 16. Duration = 10 s. Total bits = 44,100 × 16 × 2 × 10 = 14,112,000 bits. Step 2: Convert bits to bytes: 14,112,000 ÷ 8 = 1,764,000 bytes. Step 3: Convert bytes to megabytes: 1 MB = 2²⁰ bytes = 1,048,576 bytes. So 1,764,000 ÷ 1,048,576 ≈ 1.68 MB. Acceptable answers may round to 1.7 MB. Always show each unit conversion clearly to secure method marks.

步骤 1：44.1 kHz = 每秒每声道 44,100 次采样。立体声意味着 2 个声道。每采样 16 位，时长 10 秒。总位数 = 44,100 × 16 × 2 × 10 = 14,112,000 比特。步骤 2：比特转字节：14,112,000 ÷ 8 = 1,764,000 字节。步骤 3：字节转兆字节：1 MB = 2²⁰ 字节 = 1,048,576 字节，因此 1,764,000 ÷ 1,048,576 ≈ 1.68 MB。可接受的答案可四舍五入为 1.7 MB。一定要清晰展示每一步单位换算，以确保获得方法分。

Lossy compression permanently removes some audio data that is less likely to be noticed by human ears, often exploiting psychoacoustic masking. This drastically reduces file size — ideal for streaming or portable devices — but it sacrifices quality because the original data cannot be perfectly reconstructed. In contrast, lossless compression (e.g., FLAC) retains all information. For most consumer audio, lossy formats like MP3 offer an acceptable trade-off between size and quality.

有损压缩会永久移除人耳不易察觉的部分音频数据，通常利用心理声学掩蔽效应。这能大幅减小文件体积，非常适合流媒体或便携设备，但牺牲了音质，因为无法完全重建原始数据。相比之下，无损压缩（如 FLAC）保留全部信息。对大多数消费级音频，像 MP3 这类有损格式提供了大小与质量之间可接受的折中。

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7. Programming Concepts and Pseudocode | 编程概念与伪代码

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