📚 Year 10 CIE Computer Science: Unit Test Mock Exam Walkthrough | Year 10 CIE 计算机：单元测试模拟卷解析
Mock exams are a crucial part of preparing for CIE IGCSE Computer Science. This walkthrough covers a typical Year 10 unit test paper, breaking down key topics and showing you how to approach each type of question with confidence. We will revise number systems, logic gates, hardware, programming fundamentals, and more, explaining model answers step by step.
模拟考试是备战 CIE IGCSE 计算机科学的关键一环。本文解析一份典型的 Year 10 单元测试卷，拆解重要知识点，并展示如何自信地应对每一类题目。我们将复习数制、逻辑门、硬件、编程基础等内容，逐步讲解标准答案。
1. Number Systems and Conversions | 数制与转换
A typical question asks you to convert between binary, denary and hexadecimal. The key is remembering place values and grouping bits. For example, a question might give you 1101 1010₂ and require the denary and hex equivalents.
常见题目要求你在二进制、十进制和十六进制之间转换。关键是记住位权值以及位的分组。例如，题目可能会给出 1101 1010₂，要求写出对应的十进制和十六进制值。
To find the denary value, multiply each binary digit by its place value (128, 64, 32, 16, 8, 4, 2, 1). So 1101 1010₂ = (1×128) + (1×64) + (0×32) + (1×16) + (1×8) + (0×4) + (1×2) + (0×1) = 128 + 64 + 16 + 8 + 2 = 218 in denary.
求十进制值时，将每个二进制位乘以其位权值（128、64、32、16、8、4、2、1）。因此 1101 1010₂ = (1×128) + (1×64) + (0×32) + (1×16) + (1×8) + (0×4) + (1×2) + (0×1) = 128 + 64 + 16 + 8 + 2 = 218（十进制）。
For hexadecimal, split the binary into nibbles (4 bits): 1101 and 1010. 1101₂ = D (denary 13), 1010₂ = A (denary 10). Thus the hex is DA₁₆. Always show the base to avoid losing marks.
转换为十六进制时，将二进制按 4 位一组（半字节）分组：1101 和 1010。1101₂ = D（十进制 13），1010₂ = A（十进制 10）。因此十六进制是 DA₁₆。务必标出进制，以免失分。
2. Data Storage Units and File Size Calculations | 数据存储单位与文件大小计算
You need to be comfortable with bits, bytes, KiB, MiB, and how to calculate file sizes. A question might state “A colour image is 600 × 400 pixels, using 16-bit colour depth. Calculate the file size in KiB.”
你必须熟悉位、字节、KiB、MiB，并会计算文件大小。例如，题目可能给出“一幅彩色图像为 600 × 400 像素，使用 16 位色深。计算文件大小，以 KiB 为单位”。
Total bits = pixels × colour depth = 600 × 400 × 16 = 3 840 000 bits. Divide by 8 to get bytes: 480 000 bytes. Then divide by 1024 to get KiB: 480 000 ÷ 1024 = 468.75 KiB. Round appropriately if asked, or leave as a decimal to one place.
总位数 = 像素数 × 色深 = 600 × 400 × 16 = 3 840 000 位。除以 8 得到字节数：480 000 字节。再除以 1024 得到 KiB：480 000 ÷ 1024 = 468.75 KiB。如有要求可适当四舍五入，或保留一位小数。
Remember the difference between kB (1000 bytes) and KiB (1024 bytes). CIE normally uses powers of 2 for memory, so always clarify your divisor. Showing steps earns method marks even if the final answer slips.
记住 kB（1000 字节）和 KiB（1024 字节）的区别。CIE 在内存相关计算中通常使用 2 的幂次，因此务必注明除数。展示计算步骤即使最终答案有误也能获得方法分。
3. Logic Gates and Truth Tables | 逻辑门与真值表
A common task is to complete a truth table for a given logic circuit. Suppose the circuit is Q = (A AND B) OR (NOT A). You must identify every combination of A and B (00, 01, 10, 11) and work through the expression.
常见任务是填写给定逻辑电路的真值表。假设电路为 Q = (A AND B) OR (NOT A)。你需要列出 A 和 B 的所有组合（00、01、10、11）并逐步计算表达式。
For A=0, B=0: (0 AND 0) = 0, NOT A = 1, so Q = 0 OR 1 = 1. For A=0, B=1: (0 AND 1)=0, NOT A=1, Q=1. For A=1, B=0: (1 AND 0)=0, NOT A=0, Q=0. For A=1, B=1: (1 AND 1)=1, NOT A=0, Q=1. So the output is 1,1,0,1.
当 A=0, B=0：(0 AND 0)=0，NOT A=1，因此 Q=0 OR 1=1。A=0, B=1：(0 AND 1)=0，NOT A=1，Q=1。A=1, B=0：(1 AND 0)=0，NOT A=0，Q=0。A=1, B=1：(1 AND 1)=1，NOT A=0，Q=1。因此输出为 1、1、0、1。
You may also be asked to draw the logic gate diagram. Use standard symbols: AND is a D-shape, OR a curved shape, NOT a triangle with a bubble. Always label inputs and outputs clearly.
你可能也需要画出逻辑门示意图。使用标准符号：AND 呈 D 形，OR 为弧形，NOT 为带小圆圈的三角形。务必清晰标注输入和输出。
4. Computer Architecture and the Fetch-Execute Cycle | 计算机体系结构与取指-执行周期
Questions often ask you to describe the role of the CPU components: Control Unit (CU), Arithmetic Logic Unit (ALU), and registers like the Program Counter (PC) and Memory Address Register (MAR).
题目经常会让你描述 CPU 各部件的作用：控制单元（CU）、算术逻辑单元（ALU），以及程序计数器（PC）和内存地址寄存器（MAR）等寄存器。
The CU directs operations by decoding instructions and sending control signals. The ALU performs calculations and logical operations. The PC holds the address of the next instruction, while the MAR holds the address being read from or written to.
CU 通过解码指令并发送控制信号来指挥操作。ALU 执行计算和逻辑运算。PC 存放下一条指令的地址，而 MAR 存放当前正在读取或写入的地址。
Explain the fetch-execute cycle step by step: (1) PC value copied to MAR, (2) PC incremented by 1, (3) instruction fetched from RAM into MDR, (4) instruction copied to CIR and decoded by CU, (5) CU executes the instruction, possibly using ALU. This sequence repeats millions of times per second.
分步解释取指-执行周期：(1) PC 的值复制到 MAR，(2) PC 加 1，(3) 指令从 RAM 取入 MDR，(4) 指令复制到 CIR 并由 CU 解码，(5) CU 执行指令，可能用到 ALU。该序列每秒重复数百万次。
5. Input and Output Devices | 输入与输出设备
You need to know how common devices work and where they are used. For instance, a barcode scanner uses a laser or camera to read black and white bars of varying widths, converting the reflected light into digital data.
你需要知道常见设备的工作原理及其应用场景。例如，条形码扫描器使用激光或摄像头读取宽度不同的黑白条纹，将反射光转换为数字数据。
Output devices like inkjet printers spray tiny droplets of ink onto paper, while laser printers use static charge and toner. Be prepared to compare their speed, cost per page, and typical use cases.
输出设备如喷墨打印机将微小墨滴喷到纸张上，而激光打印机使用静电和碳粉。准备好比较它们的速度、每页成本以及典型应用场景。
A typical exam question might ask: “Explain why a solid-state drive is more suitable than a hard disk drive in a laptop.” You would mention no moving parts, lower power consumption, faster read/write, and greater physical durability.
典型的考题可能会问：“解释为什么固态硬盘比机械硬盘更适合笔记本电脑。”你应该提到无移动部件、功耗更低、读写更快以及物理耐用性更强。
6. Networks and the Internet | 网络与互联网
Be ready to define LAN, WAN, and the hardware needed to set up a network: routers, switches, network interface cards (NICs), and transmission media. You might be asked to compare peer-to-peer and client-server networks.
准备好定义 LAN、WAN，以及组建网络所需的硬件：路由器、交换机、网卡和传输介质。你可能会被要求比较对等网络和客户-服务器网络。
In a peer-to-peer network, each device acts as both client and server, sharing resources directly. This suits small offices with no dedicated server. Client-server models have a central server that manages security, files, and backups; better for large organisations.
在对等网络中，每台设备同时作为客户和服务器，直接共享资源。这适合没有专用服务器的小型办公室。客户-服务器模式有中央服务器管理安全、文件和备份；更适合大型机构。
When explaining IP and MAC addresses, remember: MAC is a 48-bit hardware address burned into the NIC, while IP is a logical address assigned manually or via DHCP, used for routing between networks.
解释 IP 和 MAC 地址时记住：MAC 是烧录在网卡上的 48 位硬件地址，而 IP 是逻辑地址，手动或通过 DHCP 分配，用于网络间的路由。
7. Problem-Solving with Pseudocode | 使用伪代码解决问题
CIE papers feature pseudocode that follows a specific syntax: INPUT, OUTPUT, IF … THEN … ELSE … ENDIF, FOR … TO … NEXT, WHILE … DO … ENDWHILE. Practice writing algorithms for tasks like finding the maximum of three numbers.
CIE 试卷包含遵循特定语法的伪代码：INPUT、OUTPUT、IF … THEN … ELSE … ENDIF、FOR … TO … NEXT、WHILE … DO … ENDWHILE。练习编写解决如求三个数最大值等任务的算法。
For example: INPUT a, INPUT b, INPUT c. max ← a. IF b > max THEN max ← b ENDIF. IF c > max THEN max ← c ENDIF. OUTPUT max. Using the left arrow for assignment is standard.
例如：INPUT a, INPUT b, INPUT c。max ← a。IF b > max THEN max ← b ENDIF。IF c > max THEN max ← c ENDIF。OUTPUT max。使用左箭头进行赋值是标准做法。
A common mistake is forgetting to initialise variables. Always assign a starting value before using a variable in a loop or condition. Your solution should also account for edge cases, such as all values being equal.
常见的错误是忘记初始化变量。在循环或条件中使用变量前，务必先赋初值。你的解法还应考虑边界情况，例如所有值都相等。
8. Trace Tables and Dry Runs | 跟踪表与手动执行
Trace tables help you follow the flow of an algorithm and verify its correctness. A question will provide pseudocode and ask you to complete a table showing variable values after each step.
跟踪表帮助你追踪算法流程并验证其正确性。题目会提供伪代码，要求你完成表格，显示每一步后变量的值。
Suppose the code is: total ← 0. FOR i ← 1 TO 3 total ← total + i NEXT. The trace table would have columns for i and total. Step 1: i=1, total=1. Step 2: i=2, total=3. Step 3: i=3, total=6. The final total is 6.
假设代码为：total ← 0。FOR i ← 1 TO 3 total ← total + i NEXT。跟踪表包含 i 和 total 两列。第一步：i=1，total=1。第二步：i=2，total=3。第三步：i=3，total=6。最终 total 为 6。
When loops contain conditions, record every iteration carefully. Pay attention to when the loop terminates and whether the condition is checked before or after the loop body.
当循环包含条件时，仔细记录每次迭代。注意循环何时终止，以及条件是在循环体之前还是之后检查。
9. Error Detection Methods | 错误检测方法
Understand parity bits, checksums, and check digits. For parity, an extra bit is added to make the total number of 1s even (even parity) or odd (odd parity). For a byte 0101 1001, count the 1s: there are four, so even parity bit would be 0.
理解奇偶校验位、校验和和校验位。对于奇偶校验，增加一个额外位使所有 1 的总数为偶数（偶校验）或奇数（奇校验）。对于字节 0101 1001，1 的个数为 4，因此偶校验位为 0。
Checksums add up data blocks and transmit the sum. The receiver recalculates and compares. Check digits, such as the last digit of an ISBN, use modulus algorithms to detect single-digit errors.
校验和对数据块求和并传输该和数。接收端重新计算并比较。校验位，如 ISBN 的最后一位，使用模算法检测单个数字错误。
You might be asked to identify which bit has been corrupted given a received frame with incorrect parity. This tests your ability to count 1s and decide whether the error is in the data or parity bit.
你可能会被要求根据收到的具有错误奇偶校验的帧，找出哪一位被破坏。这考验你数 1 的能力，并判断错误发生在数据位还是校验位。
10. Cybersecurity Basics | 网络安全基础
Know about malware (viruses, worms, Trojans), phishing, brute-force attacks, and denial of service. Questions often ask for prevention methods: firewalls, anti-malware software, strong passwords, and user education.
了解恶意软件（病毒、蠕虫、木马）、网络钓鱼、暴力攻击和拒绝服务。题目常要求给出预防方法：防火墙、反恶意软件、强密码和用户教育。
Explain how a firewall works: it monitors incoming and outgoing traffic, allowing or blocking data based on predefined rules. This creates a barrier between a trusted internal network and untrusted external networks.
解释防火墙如何工作：它监控进出流量，根据预定义规则允许或阻止数据。这在可信的内部网络和不可信的外部网络之间建立一道屏障。
When discussing encryption, describe the process of scrambling plaintext into ciphertext using a key. Symmetric encryption uses a single shared key, whereas asymmetric uses a public/private key pair. Even if intercepted, the data cannot be read without the key.
讨论加密时，描述使用密钥将明文加扰成密文的过程。对称加密使用单一共享密钥，而非对称加密使用公/私钥对。即使被截获，没有密钥也无法读取数据。
Remember to link your answers to the scenario given. If a question mentions a school network, suggest proxy servers to filter web content and user authentication to prevent unauthorised access.
记住将答案与给定场景关联。如果问题提到校园网，建议使用代理服务器过滤网页内容，并使用用户认证防止未授权访问。
11. High- and Low-Level Languages | 高级与低级语言
Differentiate between machine code, assembly, and high-level languages like Python or Java. A question could ask: “Why do programmers prefer high-level languages?” The answer points to easier debugging, portability, and readability.
区分机器码、汇编和高级语言（如 Python 或 Java）。题目可能会问：“为什么程序员更喜欢高级语言？”答案指向更易调试、可移植性和可读性。
Assemblers and compilers translate code. An assembler converts assembly mnemonics into machine code one-to-one. A compiler translates an entire high-level source file into machine code before execution; an interpreter does it line by line.
汇编器和编译器翻译代码。汇编器将汇编助记符一对一转换成机器码。编译器在执行前将整个高级源文件翻译成机器码；解释器则逐行翻译。
When you need to explain the advantage of a compiler over an interpreter, mention that compiled code runs faster, but interpreted code can be tested immediately without a lengthy compilation step.
当你需要解释编译器相比解释器的优势时，提到编译后的代码运行更快，但解释型代码可立即测试，无需漫长的编译步骤。
12. Exam Technique and Common Pitfalls | 考试技巧与常见陷阱
Always read the command words: “State” requires a brief answer, “Describe” needs a step-by-step account, “Explain” asks for reasons and how/why. Many students lose marks by not expanding enough for higher-mark questions.
始终注意指令词：“State”要求简短回答，“Describe”需要逐步说明，“Explain”要求给出原因和方式/原因。许多学生因为未充分扩展高分值题目而失分。
Manage your time: allocate 1 minute per mark. If you are stuck, move on and come back. For program-writing questions, even incomplete pseudocode with correct structure can earn marks. Never leave a table blank — attempt every part.
管理好时间：每分题分配 1 分钟。卡住时先跳过，稍后再回做。对于编写程序的题目，即使伪代码不完整，结构正确也能得分。绝不留任何表格空白——每题都要尝试。
Finally, revise by practising past papers under timed conditions. The more you expose yourself to the style of CIE questions, the more confident you will become on the real test day.
最后，通过在限时条件下练习历年真题来复习。你越熟悉 CIE 的出题风格，真正考试那天就会越自信。
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