📚 Year 10 CCEA Computer Science: Unit Test Mock Paper Walkthrough | Year 10 CCEA 计算机：单元测试模拟卷解析
This article provides a detailed walkthrough of a typical Year 10 CCEA Computer Science unit test mock paper. Each section covers a key topic, presents a representative question, explains the correct answer, and highlights common pitfalls. Use this guide to boost your exam confidence and deepen your understanding of core concepts.
本文详细解析了一套典型的 Year 10 CCEA 计算机科学单元测试模拟卷。每个小节涵盖一个重要主题，呈现一道典型题目，讲解正确答案并指出常见错误。善用本指南，可以增强备考信心，加深对核心概念的理解。
1. Binary to Denary Conversion | 二进制转十进制
Question: Convert the 8‑ e a 1 appears. 128 + 32 + 16 + 4 + 2 = 182. Many students forget to align the bits from right to left; always double‑check the column headings.
题目：将八位二进制数 10110110 转换为十进制，需写出计算过程。正确做法是写出从 128 到 1 的位权，将出现 1 的对应权值相加：128 + 32 + 16 + 4 + 2 = 182。很多同学忘记从右向左对齐位权；务必再次核对列标题。
- 位权：128，64，32，16，8，4，2，1
- 二进制：1，0，1，1，0，1，1，0
- 结果：128 + 32 + 16 + 4 + 2 = 182
2. Hexadecimal Representation | 十六进制表示法
: convert to binary first (200 = 11001000₂), then split into nibbles: 1100 (C) and 1000 (8). Avoid mixing up the letters: A=10, B=11, C=12, D=13, E=14, F=15.
题目：十进制数 200 用十六进制表示为 C8。请解释原因。200 ÷ 16 = 12 余 8；12 用十六进制表示为 C，因此组合成 C8。另一种方法是先转为二进制（200 = 11001000₂），然后拆分为半字节：1100（C）和 1000（8）。注意不要混淆字母：A=10，B=11，C=12，D=13，E=14，F=15。
200₁₀ → 11001000₂ → C8₁₆
3. Logic Gates and Truth Tables | 逻辑门与真值表
Question: Complete the truth table for a two‑ member that an OR gate would give 1 when at least one input is 1; a NOT gate inverts a single input.
题目：完成双输入与门的真值表。与门只有在两个输入都为 1 时才输出 1。对于输入 A、B：0,0→0；0,1→0；1,0→0；1,1→1。记住，或门会在至少一个输入为 1 时输出 1；非门则将单个输入取反。
| A | B | Output (AND) |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
4. CPU Components | 中央处理器组件
Question: Name the three main parts of the CPU and briefly describe the function of the Control Unit. The three parts are the Control Unit (CU), the Arithmetic Logic Unit (ALU), and the registers. The CU sends control signals to direct the flow of data, fetches instructions from memory, and decodes them. Do not confuse the CU with the ALU, which performs calculations and logical operations.
题目：说出 CPU 的三个主要部件，并简述控制单元的功能。三个部件是控制单元（CU）、算术逻辑单元（ALU）和寄存器。控制单元发送控制信号以指挥数据流动，从内存中取指、译码。不要混淆控制单元和算术逻辑单元，后者负责运算和逻辑操作。
- 控制单元（CU）：取指、译码、产生控制信号
- 算术逻辑单元（ALU）：算术运算和逻辑运算
- 寄存器：高速暂存数据
5. Input and Output Devices | 输入与输出设备
Question: For a supermarket checkout, which input device is used to read product barcodes, and why is a thermal printer sometimes chosen for receipts? The barcode scanner (or barcode reader) uses a laser or LED to read the pattern of bars. Thermal printers are fast, quiet, and have no ink cartridges to replace, which suits high‑volume retail environments. However, thermal prints can fade over time.
题目：在超市收银处，哪种输入设备用于读取商品条码？为什么有时选择热敏打印机打印收据？条码扫描器（条码阅读器）使用激光或 LED 读取条码图案。热敏打印机速度快、安静、无需更换墨盒，适合大批量零售场景。但热敏打印会随时间褪色。
6. Networks: LAN vs WAN | 网络：局域网与广域网
Question: Compare a LAN and a WAN in terms of geographical area and ownership. A LAN (Local Area Network) covers a small geographical area such as a school or office, and the hardware is typically owned by the organisation. A WAN (Wide Area Network) spans a large area, often across cities or countries, and uses external infrastructure rented from telecom providers. The Internet is the largest WAN.
题目：从地理范围和所有权角度比较局域网和广域网。局域网覆盖的物理范围较小，如学校或办公室，硬件通常由组织自行拥有。广域网覆盖面积大，常跨越城市或国家，需租用电信供应商的外部基础设施。互联网是最大的广域网。
7. Network Topologies | 网络拓扑结构
Question: Draw a star topology and state one advantage and one disadvantage. In a star topology, all devices connect to a central switch or hub. Advantage: if one cable fails, only that node is affected. Disadvantage: the whole network depends on the central device; if it fails, the network goes down. Practice sketching neat diagrams with clear labels.
题目：画出星型拓扑并说明一个优点和一个缺点。在星型拓扑中，所有设备都连接到中央交换机或集线器。优点：如果一条电缆出现故障，只有该节点受影响。缺点：整个网络依赖中央设备；如果它出现故障，网络将瘫痪。练习画出整齐并带有清晰标注的简图。
8. Flowcharts and Pseudocode | 流程图与伪代码
Question: A program asks for a number and outputs “Even” if it is divisible by 2, otherwise “Odd” his logic. The flowchart should start with an input oval, then a decision diamond checking “number MOD 2 = 0?”. If true, output “Even”; if false, output “Odd” symbols: rounded rectangle for start/end, rectangle for processes, diamond for decisions.
题目：一个程序要求输入一个数字，如果它可以被 2 整除则输出“偶数”，否则输出“奇数”。请画出表示该逻辑的流程图。流程图应以起始椭圆形开始，接着是一个判断菱形，检查“数字 MOD 2 = 0？”如果为真，输出“偶数”；如果为假，输出“奇数”。使用正确的符号：圆角矩形表示开始/结束，矩形表示处理，菱形表示判断。
开始 → 输入 n → n % 2 = 0? → 是: 输出 "Even" → 结束; 否: 输出 "Odd" → 结束
9. Programming Errors | 编程错误类型
Question: Identify the type of error in this Python code: print("The answer is" + 42). This is a runtime error (specifically a TypeError) because you cannot concatenate a string and an integer without explicit conversion. Syntax errors would be detected before execution (e.g. missing colon). Logic errors produce wrong results without crashing.
题目：指出下列 Python 代码中的错误类型：print("The answer is" + 42)。这是一个运行时错误（具体是类型错误），因为无法在没有显式转换的情况下将字符串和整数拼接。语法错误会在执行前被检测到（例如遗漏冒号）。逻辑错误会导致错误结果，但不会使程序崩溃。
10. Database Terminology | 数据库术语
Question: Define the terms field, record, and primary key in a database table. A field is a single piece of data within a record (e.g. "Student Name"). A record is a complete set of fields about one entity (one row). A primary key is a field that uniquely identifies each record, such as Student ID. Foreign keys create relationships between tables.
题目：定义数据库表中的术语：字段、记录和主键。字段是记录中的单个数据项（例如“学生姓名”）。记录是关于某个实体的整组字段（一整行）。主键是唯一标识每条记录的字段，如学生编号。外键用于创建表之间的关系。
11. Cybersecurity Threats | 网络安全威胁
Question: Describe what a phishing attack is and how users can protect themselves. Phishing involves fraudulent emails or messages that mimic legitimate organisations to steal personal data. Protection measures: check the sender’s address, avoid clicking suspicious links, hover over links to preview URLs, and use multi‑factor authentication. Never share passwords via email.
题目：描述什么是网络钓鱼攻击，以及用户如何保护自己。钓鱼攻击是指仿冒合法机构的欺诈性邮件或消息，旨在窃取个人数据。保护措施：核实发件人地址，不要点击可疑链接，悬停鼠标预览 URL，并使用多因素认证。切勿通过邮件分享密码。
12. Sound and Image Representation | 声音与图像表示
Question: Explain how increasing the sampling rate affects the quality and file size of a digital sound recording. A higher sampling rate captures more samples per second, which improves the sound quality and fidelity but increases the file size because more data is stored. The resolution (bit depth) also affects quality.
题目：解释提高采样率会如何影响数字录音的质量和文件大小。更高的采样率每秒采集更多样本，从而提升音质和保真度，但因为存储更多数据，文件大小会随之增加。分辨率（位深）也会影响质量。
Sound quality ∝ sampling rate × bit depth; File size = rate × depth × duration (channels)
Published by TutorHao | Computer Science Revision Series | aleveler.com
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