📚 Year 10 CAIE Computer Science Mock Exam Walkthrough | 10年级CAIE计算机单元测试模拟卷解析
This walkthrough covers a typical Year 10 CAIE IGCSE Computer Science unit test paper, focusing on core topics such as data representation, logic gates, hardware, software, networks, pseudocode, and cybersecurity. Each question is broken down with clear steps, key terms, and revision tips to help you secure top marks in your assessment.
这份模拟卷解析涵盖了典型的10年级CAIE IGCSE计算机科学单元测试内容，重点包括数据表示、逻辑门、硬件、软件、网络、伪代码及网络安全。每道题都配有清晰的解题步骤、关键术语和复习提示，助你在考试中稳稳拿高分。
1. Binary and Denary Conversion | 二进制与十进制转换
The question asks: Convert denary 75 into an 8-bit binary number. To solve this, start by placing column headings for 8-bit binary: 128, 64, 32, 16, 8, 4, 2, 1. Then work from left to right, seeing if the column value fits into the remaining number. 128 does not fit into 75 -> 0. 64 fits into 75 -> 1, remainder 11. 32 does not fit into 11 -> 0. 16 does not fit into 11 -> 0. 8 fits into 11 -> 1, remainder 3. 4 does not fit into 3 -> 0. 2 fits into 3 -> 1, remainder 1. 1 fits into 1 -> 1. Hence the 8-bit binary representation is 01001011₂. Always show the leading zero for 8-bit format.
题目要求将十进制数75转换为8位二进制数。解题时先写出8位列标题：128, 64, 32, 16, 8, 4, 2, 1。从左向右，如果列值小于或等于剩余数值则填1，否则填0。128>75 -> 0；64≤75 -> 1，余11；32>11 -> 0；16>11 -> 0；8≤11 -> 1，余3；4>3 -> 0；2≤3 -> 1，余1；1≤1 -> 1。得到结果为01001011₂。8位二进制一定要保留前导零。
2. Binary to Hexadecimal Conversion | 二进制转十六进制
Convert 10101110₂ into hexadecimal. Split the binary number into groups of four bits starting from the right: 1010 1110. The left nibble 1010 equals denary 10, which is A in hex. The right nibble 1110 equals denary 14, which is E in hex. Combine them to get AE₁₆ full binary to denary first: 128+32+8+4+2 = 174, then 174 ÷ 16 = 10 remainder 14, so AE₁₆.
将二进制数10101110₂转换为十六进制。从右向左每4位一组划分：1010 1110。左边半字节1010对应十进制10，即十六进制A；右边半字节1110对应十进制14，即十六进制E。合并得到AE₁₆。也可以先把整个二进制转十进制检验：128+32+8+4+2=174，174÷16得10余14，因此也是AE₁₆。
3. Logic Gates and Truth Tables | 逻辑门与真值表
valuate each row for the three inputs. For example, when A=0, B=0, C=0: A AND B = 0, NOT C = 1, so Q = 0 OR 1 = 1. Work through all eight combinations systematically. The completed table is shown below. Being careful with operator precedence (AND before OR) is essential.
题目给出布尔表达式 Q = (A AND B) OR (NOT C)，要求填写真值表。设置列A、B、C、A AND B、NOT C、Q。逐行评估输入。例如当A=0, B=0, C=0时，A AND B = 0，NOT C = 1，所以Q = 0 OR 1 = 1。按此逻辑处理全部八种组合。下方为完整的真值表。必须注意运算优先级：先括号内AND，再OR。
| A | B | C | A AND B | NOT C | Q |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 |
4. RAM vs ROM | 随机存取存储器与只读存储器
State two differences between RAM and ROM. First, RAM is volatile – its contents are lost when power is turned off – while ROM is non-volatile and retains data permanently. Second, RAM stores data and programs currently in use by the CPU, whereas ROM holds firmware or boot instructions that never change during normal operation. You can also mention that RAM is typically larger in capacity and is read/write, while ROM is often read-only in practice.
请说明RAM和ROM的两个区别。第一，RAM是易失性存储器，断电后数据丢失；ROM则是非易失性存储器，数据永久保留。第二，RAM存放CPU当前使用的程序和数据，而ROM存储固件或启动指令，正常运行时内容不变。你还可补充RAM容量通常更大且可读写，ROM在实际使用中通常只读。
5. Input, Output and Storage Devices | 输入、输出与存储设备
The question gives a list of devices: microphone, projector, SSD, barcode scanner, speakers, hard disk drive. Classify each as input, output, or storage. A microphone captures sound — input. A projector displays images — output. An SSD (solid-state drive) stores data — storage. A barcode scanner reads barcodes — input. Speakers produce sound — output. A hard disk drive is storage. Be prepared for devices that can belong to more than one category, for instance a touchscreen is both input and output, but here every device fits a single primary category.
题目给出一组设备：麦克风、投影仪、SSD、条码扫描器、扬声器、硬盘驱动器。请分类为输入、输出或存储设备。麦克风拾取声音——输入设备；投影仪显示影像——输出设备；SSD（固态硬盘）保存数据——存储设备；条码扫描器读取条码——输入设备；扬声器发声——输出设备；硬盘驱动器是存储设备。注意有些设备可能兼具多个类别，例如触摸屏既是输入也是输出，但本题中每个设备都仅属于一个主要类别。
6. File Size Calculation | 文件大小计算
The scenario: A 2-minute stereo audio file is recorded with a sample rate of 44.1 kHz and a bit depth of 16 bits. Calculate the approximate file size in megabytes. Use the formula:
File size (bits) = Sample rate (Hz) × Bit depth × Number of channels × Duration (s)
. First, 44,100 × 16 = 705,600; ×2 = 1,411,200; ×120 = 169,344,000 bits. Convert to bytes: 169,344,000 ÷ 8 = 21,168,000 bytes. Assuming 1 MB = 1,000,000 bytes, the size is 21.168 MB, approximately 21.2 MB. In CAIE exams, always read the question for which conversion factor (1000 or 1024) to use; when unspecified, 1,000,000 bytes per MB is common for storage.
场景：一段2分钟的立体声音频，采样率44.1 kHz，位深度16 bits。计算近似文件大小（以兆字节为单位）。使用公式：
文件大小（比特）= 采样率 (Hz) × 位深度 × 声道数 × 时长 (秒)
逐步计算：采样率44,100 Hz；位深度16；声道数2（立体声）；时长2×60=120秒。比特总数=44,100×16×2×120。先算44,100×16=705,600；×2=1,411,200；×120=169,344,000比特。转换为字节：169,344,000÷8=21,168,000 字节。假设1 MB=1,000,000 字节，文件大小约为21.168 MB，约21.2 MB。CAIE考试中需仔细审题确认采用1000还是1024换算；未明确时，存储容量常取1 MB=1,000,000字节。
7. Operating System Functions | 操作系统功能
Describe two functions of an operating system. One key function is memory management – the OS allocates RAM to running programs and ensures they do not interfere with each other. Another is providing a user interface, such as a graphical user interface (GUI) or command-line interface, which allows users to interact with the computer. Other valid answers include file management, managing peripheral devices (device drivers), and user account control for security.
请描述操作系统的两个功能。其一，内存管理——操作系统为正在运行的程序分配RAM，并保证它们互不干扰。其二，提供用户界面，如图形用户界面（GUI）或命令行界面，让用户与计算机交互。其他可接受的回答还包括文件管理、外设管理（设备驱动程序）以及用于安全管理的用户账户控制。
8. Network Topologies: Star vs Bus | 网络拓扑结构：星型与总线型
Explain what happens in a star network if one cable fails. In a star topology, every device connects to a central switch or hub via its own dedicated cable. If one cable fails, only the device connected by that cable loses network access; the rest of the network remains fully functional. This is a major advantage over a bus topology, where a single cable break can bring down the whole network segment. The question may also ask you to identify the topology from a diagram or suggest the most suitable topology for a school computer lab – star is often the preferred choice.
解释星型网络中一根电缆发生故障时会出现什么情况。星型拓扑中，每台设备都通过专用电缆连接到中央交换机或集线器。如果某根电缆断开，只有该电缆连接的设备失去网络访问，其余设备仍然正常工作。这是星型拓扑相比总线拓扑的一大优势——总线拓扑中单根电缆断裂可能导致整个网段瘫痪。考题还可能要求根据示意图识别拓扑结构，或为学校机房推荐最合适的拓扑——星型通常是首选。
9. Pseudocode: Find Maximum | 伪代码：查找最大值
The problem: Write pseudocode to input 10 numbers and output the largest one. A typical solution initialises a variable max to the first input, then uses a loop to compare each subsequent number with max, updating max if a larger value is found. Pseudocode should be clear and language-agnostic.
题目要求：编写伪代码，输入10个数字并输出其中的最大值。常见解法是先将第一个输入赋值给变量max，然后利用循环逐一比较后续输入与max，若找到更大的值则更新max。伪代码需清晰且不依赖特定编程语言。
Pseudocode / 伪代码：
INPUT num
max ← num
FOR i ← 2 TO 10
INPUT num
IF num > max THEN
max ← num
ENDIF
END FOR
OUTPUT max
The above first reads the initial number and sets it as the maximum, then repeats for the remaining nine inputs. This logic works for any set of numbers and demonstrates correct use of iteration and selection. In CAIE exams, minor syntax variations (e.g., using 'while' instead of 'for') are acceptable as long as the algorithm is correct.
上面先读取第一个数并将其设为最大值，然后对剩余九个输入重复判断。这个逻辑适用于任何一组数字，并正确展示了循环与选择结构的用法。在CAIE考试中，只要算法正确，语法上的细微差异（如使用WHILE代替FOR）也是可以接受的。
10. Cybersecurity Threats and Prevention | 网络安全威胁与防范
Describe phishing and give one way to prevent it. Phishing is a social engineering attack where a malicious party sends a fraudulent message, often an email, that appears to come from a trusted source. The goal is to trick the recipient into revealing sensitive information such as passwords, credit card details, or personal data. A common prevention method is to educate users to check the sender's email address carefully, avoid clicking suspicious links, and verify requests through official channels. Other correct answers include using email spam filters, multi-factor authentication, or reporting suspected phishing.
请描述网络钓鱼并给出一种预防方法。网络钓鱼是一种社会工程攻击，攻击者发送看似来自可信来源的欺诈性信息（通常是电子邮件），目的是诱骗收件人泄露密码、信用卡信息或个人数据等敏感内容。常见的预防措施是教育用户仔细核对发件人邮箱地址，避免点击可疑链接，并通过官方渠道核实请求。其他正确回答还包括使用垃圾邮件过滤器、多因素身份认证或举报可疑钓鱼行为。
11. Top Revision Tips for the Mock Exam | 模拟考试高分复习建议
Before the test, practise conversions (binary, denary, hex) until they feel automatic. Memorise truth tables for the basic logic gates (AND, OR, NOT, NAND, NOR). Create simple flashcards for key hardware and software terms. When tackling calculation questions, always show your working – marks are awarded for method steps even if the final answer has a minor slip. Finally, try past paper questions under timed conditions to build confidence and speed.
考前多练习进制转换（二进制、十进制、十六进制），做到熟练掌握。熟记基本逻辑门（AND, OR, NOT, NAND, NOR）的真值表。制作简易抽认卡记忆关键硬件和软件术语。解答计算题时务必写出推导过程——即使最终答案有小错，步骤正确仍可得分。最后，限时完成历年真题，提高信心和答题速度。
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