📚 Year 10 Cambridge IGCSE Computer Science: Unit Test Mock Paper Analysis | 剑桥IGCSE计算机科学十年级：单元模拟试卷解析
This article provides a detailed walkthrough of a mock unit test paper for Year 10 Cambridge IGCSE Computer Science. Designed to mirror the style and content of the 0478 syllabus, the paper tests fundamental topics covered in the first year – from data representation and hardware to networks, security and programming logic. Alongside each question, you will find clear solutions, key explanations and common pitfalls to avoid.
本文详细解析一份针对十年级剑桥IGCSE计算机科学课程的单元模拟试卷。这份试卷贴合0478教学大纲的风格和内容，考查第一学年所学的核心主题——从数据表示、硬件到网络、安全以及编程逻辑。每道题目的旁边都附有清晰的答案、关键解释以及需要避免的常见错误。
1. Overview of the Mock Paper | 模拟试卷概览
The mock paper is structured into two sections to replicate the real IGCSE assessment. Section A contains 20 multiple-choice questions (20 marks) that quickly test recall and understanding of key concepts. Section B consists of 5 short-answer questions (30 marks) that require more detailed responses, calculations and structured reasoning. The total of 50 marks should be tackled within 60 minutes, making time management essential.
这份模拟卷分为两个部分，以模拟真实的IGCSE测试。A部分包含20道选择题（20分），快速考查对关键概念的回忆和理解。B部分则包含5道简答题（30分），要求给出更详细的回答、计算以及有条理的推理。试卷满分50分，需在60分钟内完成，因此时间管理至关重要。
A typical Section A question might ask you to identify the correct binary representation of a denary number or to spot the main function of the CPU. Section B often includes a file size calculation, a comparison of storage devices, a short trace table and an ethical scenario. Every topic from the first year of study can appear, so you must be comfortable with both theory and simple numerical work.
典型的A部分题目可能会要求你找出一个十进制数的正确二进制表示，或者辨认CPU的主要功能。B部分通常包含文件大小计算、存储设备比较、简单的跟踪表以及一个伦理情景题。第一学年学过的每个主题都可能出现，因此你必须同时熟练掌握理论和简单的数值计算。
2. Binary and Hexadecimal Conversions | 二进制与十六进制转换
Sample Question 1: Convert the denary number 173 into hexadecimal. Show your working.
示例题1： 将十进制数173转换为十六进制。请写出计算过程。
= 10 remainder 13. The denary number 10 is A in hex, and 13 is D. Therefore, 173₁₀ = AD₁₆.
答案：173 ÷ 16 = 10 余 13。十进制数10在十六进制中为A，13为D。因此，173₁₀ = AD₁₆。
Sample Question 2: Convert the hexadecimal value 2F₁₆ into an 8‑bit binary number.
示例题2： 将十六进制值2F₁₆转换成一个8位二进制数。
Answer: Each hex digit becomes a 4‑bit nibble: 2 = 0010, F = 1111. Combined, we get 0010 1111. As an 8‑bit binary number, this is 00101111₂ (the leading zeros are retained to show 8 bits).
答案：每个十六进制位转换为4位二进制：2 = 0010，F = 1111。合并后得到0010 1111。作为一个8位二进制数，即为00101111₂（保留前导零以显示8位）。
Common mistake: forgetting to pad each hex digit to exactly four bits. Always expand fully, then remove or keep leading zeros as the question requires.
常见错误：忘记将每个十六进制位补足到四位。务必完全展开，然后根据题目要求去掉或保留前导零。
3. Data Storage: File Size Calculation | 数据存储：文件大小计算
Sample Question: our depth of 24 bits. Calculate its file size in megabytes (MB), giving your answer to two decimal places.
示例题： 一幅未压缩的彩色图像分辨率为1024×768像素，色深为24位。计算其文件大小（以兆字节MB为单位），结果保留两位小数。
o kilobytes: 2,359,296 ÷ 1024 = 2304 KB. Then convert to megabytes: 2304 ÷ 1024 = 2.25 MB.
逐步解答：总位数 = 1024 × 768 × 24 = 18,874,368 位。除以8得到字节数：18,874,368 ÷ 8 = 2,359,296 字节。转换为千字节：2,359,296 ÷ 1024 = 2304 KB。再转换为兆字节：2304 ÷ 1024 = 2.25 MB。
stions often require you to round to a given number of decimal places, so read the instructions carefully. Leaving the answer as a fraction or forgetting the conversion step loses marks.
记住，对于文件存储，1 KB = 1024 字节，1 MB = 1024 KB。IGCSE题目经常要求四舍五入到指定小数位，因此要仔细阅读要求。若答案写成分数形式或遗漏转换步骤，均会被扣分。
4. CPU Components and Functions | CPU组件与功能
Sample Question: Name two components of the central processing unit (CPU) and describe the role of each.
示例题： 说出中央处理器（CPU）的两个组件，并描述各自的作用。
Answer: The Control Unit (CU) – it fetches instructions from memory, decodes them and then controls the flow of data between the CPU and other components. The Arithmetic Logic Unit (ALU) – it carries out all arithmetic operations (addition, subtraction) and logical operations (AND, OR, NOT) as directed by the CU.
答案：控制单元（CU）——它从内存中取出指令，进行译码，然后控制数据在CPU与其他组件之间的流动。算术逻辑单元（ALU）——按控制单元的指示执行所有算术运算（加法、减法）和逻辑运算（与、或、非）。
Many students confuse the CU with the ALU. The CU is the 'manager' that organises the fetch‑decode‑execute cycle; the ALU is the 'worker' that does the actual calculations and comparisons. Both are essential for processing data.
许多学生会混淆控制单元和算术逻辑单元。CU是组织“取指‑译码‑执行”周期的“管理者”；ALU则是执行实际计算和比较的“工作者”。两者对数据处理都至关重要。
5. Memory Types: RAM vs ROM | 存储器类型：RAM与ROM
Sample Question: Explain two key differences between RAM and ROM.
示例题： 解释RAM和ROM之间的两个主要区别。
First difference: RAM is volatile memory, meaning all stored data is lost when the computer is switched off. ROM is non‑volatile, so its contents are retained even without power.
区别一：RAM是易失性存储器，即计算机关机后所有存储的数据都会丢失。ROM是非易失性的，因此即使断电，其内容也会被保留。
Second difference: RAM can be both read from and written to during normal operation, making it suitable for storing programs and data currently in use. ROM is read‑only and cannot be changed by the user; it typically holds firmware or startup instructions (BIOS).
区别二：RAM在正常运行时既可读又可写，因此适合存放当前正在使用的程序和数据。ROM是只读的，用户无法修改；它通常存放固件或启动指令（BIOS）。
In the mock paper, you might also be asked to give an example of where each type of memory is used. RAM is used as main memory, while ROM is found in embedded systems and for the boot loader.
在模拟卷中，你可能还会被要求举例说明每种存储器的用途。RAM用作主存储器，而ROM则用于嵌入式系统和引导加载程序。
6. Input and Output Devices in Context | 输入输出设备在具体情境中的应用
Sample Question: A farmer wants to monitor conditions inside a greenhouse automatically. Suggest two suitable sensors and explain what each sensor measures.
示例题： 一位农民想自动监控温室内的环境状况。请推荐两种合适的传感器，并说明每种传感器测量的是什么。
Sensor 1: Temperature sensor – it measures the air temperature inside the greenhouse, allowing the system to activate heaters or vents when needed.
传感器1：温度传感器——测量温室内的空气温度，使系统在需要时启动加热器或通风口。
Sensor 2: Humidity sensor – it detects the amount of moisture in the air. The data helps control misting systems to maintain ideal growing conditions.
传感器2：湿度传感器——检测空气中的水分含量。这些数据有助于控制喷雾系统，以维持理想的生长条件。
Other acceptable answers include a light sensor (to control artificial lighting) or a soil moisture sensor (to manage irrigation). Marks are awarded for linking the sensor clearly to the measurement and the real‑world outcome.
其他可接受的答案包括光传感器（用于控制人工照明）或土壤湿度传感器（用于管理灌溉）。给分的关键在于，要清晰地建立传感器与测量量以及实际效果之间的关联。
7. Network Hardware and Topologies | 网络硬件与拓扑结构
Sample Question: A school office uses a star topology for its LAN. Describe one advantage and one disadvantage of this topology.
示例题： 某学校办公室的局域网采用星形拓扑。请描述该拓扑的一个优点和一个缺点。
Advantage: In a star network, each device is connected by its own cable to a central switch. If one cable fails, only that device is affected; the rest of the network continues to function normally. This makes fault finding straightforward.
优点：在星形网络中，每台设备通过专属线缆连接到中央交换机。如果某条线缆出故障，只有该设备受影响；网络其余部分仍可正常工作。这使得故障排查简单直接。
Disadvantage: The central switch represents a single point of failure. If the switch stops working, the entire network goes down, and no device can communicate. Additionally, the cost of cabling can be higher than with a bus topology.
缺点：中央交换机是一个单点故障源。一旦交换机停止工作，整个网络就会瘫痪，所有设备都无法通信。此外，布线成本可能高于总线拓扑。
When answering such questions, avoid vague terms. State precisely what the central device is (switch or hub) and explain the impact on network performance and reliability.
回答这类问题时，应避免使用模糊的术语。要明确说出中央设备是什么（交换机或集线器），并解释其对网络性能和可靠性的影响。
8. Software Classification and the Operating System | 软件分类与操作系统
Sample Question: Distinguish between system software and application software, giving one named example of each.
示例题： 区分系统软件和应用软件，并各举一个具体的例子。
System software is designed to manage the computer hardware and provide a platform for running application software. It operates in the background. An example is the operating system, such as Microsoft Windows or Linux.
系统软件旨在管理计算机硬件，并为运行应用软件提供平台。它在后台运行。例子包括操作系统，如微软Windows或Linux。
Application software performs specific tasks useful to the user. It requires the system software to function. A common example is a word processor like Microsoft Word, which allows the user to create and edit text documents.
应用软件执行对用户有用的具体任务，其运行依赖系统软件。常见的例子是文字处理软件，如微软Word，它让用户能够创建和编辑文本文档。
Some mock questions may ask you to name a utility program – these are a type of system software that performs maintenance tasks, such as disk defragmenters or antivirus tools.
有些模拟题可能会要求你说出一种工具软件的名称——这类软件属于系统软件，用于执行维护任务，如磁盘碎片整理程序或杀毒工具。
9. Cybersecurity and Data Protection | 网络安全与数据保护
Sample Question: Define the term 'phishing' and recommend one method a user can employ to avoid becoming a victim.
示例题： 定义“网络钓鱼”这一术语，并建议一种用户可以用于避免上当受骗的方法。
Phishing is a type of social engineering attack where cybercriminals send fraudulent emails or text messages that appear to come from a trusted source. The goal is to trick the recipient into revealing sensitive information such as passwords, bank details or personal identification numbers.
网络钓鱼是一种社会工程攻击，网络犯罪分子发送看似来自可靠来源的欺诈性电子邮件或短信，目的就是诱骗收件人泄露密码、银行信息或个人身份证号码等敏感信息。
To avoid phishing, users should never click on links or download attachments in unsolicited messages. It is safer to type the organisation's web address directly into the browser. Also, checking for poor grammar and urgent demands for action can help identify phishing attempts.
为避免网络钓鱼，用户绝不应点击来路不明信息中的链接或下载附件。更安全的方式是直接在浏览器地址栏中输入该机构的网址。此外，留意拼写错误和紧迫的行动要求也有助于识别钓鱼企图。
A full‑mark response might also mention spam filters, two‑factor authentication or reporting the email to the IT department. The key is to show understanding that phishing relies on human error rather than purely technical weaknesses.
一个能得满分的答案还可能提到垃圾邮件过滤器、双因素认证或将邮件报告给IT部门。关键在于要体现出理解——网络钓鱼依赖的是人的疏忽，而不仅仅是技术漏洞。
10. Ethics and Legislation | 伦理与立法
Sample Question: Name one law that governs the use of computers and personal data, and briefly describe one provision it makes.
示例题： 说出一个监管计算机使用和个人数据的法律，并简要描述其一项规定。
One relevant law is the Data Protection Act (or General Data Protection Regulation depending on jurisdiction). A key provision is that personal data must be processed fairly and lawfully, and must only be kept for as long as necessary. This means organisations cannot collect data without a clear purpose or hold it indefinitely.
一个相关的法律是《数据保护法》（根据司法管辖区不同也可能是《通用数据保护条例》）。其一项关键规定是，个人数据必须公平、合法地处理，并且只能保存必要长的时间。这意味着组织不能在没有明确目的的情况下收集数据，也不能无限期地持有数据。
Another commonly cited law is the Computer Misuse Act.
Published by TutorHao | Year 10 Computer Science Revision Series | aleveler.com
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