📚 Mock Paper Walkthrough: Unit Test for Year 11 Cambridge Computer Science | Year 11 剑桥计算机：单元测试模拟卷解析
This article walks you through a typical Cambridge IGCSE Computer Science unit test, offering detailed solutions and explanations for each question. Use it to check your understanding, identify weak areas, and sharpen your exam technique. Each section focuses on a core topic from the syllabus, with English explanations followed by Chinese translations.
本文将以一份典型的剑桥 IGCSE 计算机科学单元测试卷为例，逐题提供详细解析。你可以通过本文检验自己的知识掌握情况，发现薄弱环节，并提升应试技巧。每个小节聚焦一个考纲核心主题，所有解释均采用英文在前、中文在后的形式呈现。
1. Data Representation: Binary and Hexadecimal Conversion | 数据表示：二进制与十六进制转换
A common question asks you to convert the decimal number 202 into an 8‑bit binary number and then into hexadecimal. Start by drawing an 8‑bit place‑ , leaving 10. Then 8 + 2 = 10, so bits for 8 and 2 are set. The binary result is 11001010.
一道常见题目要求你将十进制数 202 转换为 8 位二进制数，再转换为十六进制。首先列出 8 位位权表（128、64、32、16、8、4、2、1）。在能够组成 202 的位权下方写 1，其余写 0。128 + 64 = 192，剩余 10；再由 8 + 2 = 10，因此 8 和 2 对应的位为 1。最终二进制结果为 11001010。
₁₆, and 1010₂ = 10 = A₁₆. Hence 202₁₀ = CA₁₆. Always show your working to gain full marks.
将二进制 11001010 转为十六进制时，先把字节分成两个半字节：1100₂ 和 1010₂。1100₂ = 12（十进制）= C₁₆，1010₂ = 10 = A₁₆。因此 202₁₀ = CA₁₆。务必写出计算过程以获得满分。
2. Logic Gates: AND, OR, NOT and Truth Tables | 逻辑门：与门、或门、非门和真值表
You might be given a simple logic circuit and asked to complete its truth table. Consider an AND gate followed by a NOT gate. For two inputs A and B, the AND output is A AND B, then the NOT inverts it. The final column shows the NAND operation. Let’s construct the truth table:
题目可能给出一个简单逻辑电路，要求补全其真值表。例如，一个与门后接一个非门。对于两个输入 A 和 B，与门的输出为 A AND B，再经非门取反。最终输出列即为 NAND 运算。下面列出真值表：
| A | B | A AND B | NOT (A AND B) |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 0 |
ble to recognise gate symbols and write Boolean expressions (e.g. Q = NOT (A AND B) or Q = ¬(A∧B)) is essential.
记住，或门在至少一个输入为 1 时输出 1，非门则将输入取反。能够识别门电路符号并写出布尔表达式（如 Q = NOT (A AND B) 或 Q = ¬(A∧B)）至关重要。
3. CPU Architecture: The Fetch‑Decode‑Execute Cycle | CPU 架构：取指−解码−执行周期
Describe what happens during the fetch stage of the machine cycle. The Program Counter (PC) holds the address of the next instruction; this address is copied to the Memory Address Register (MAR). The control unit then sends a read signal, and the instruction is fetched from RAM into the Memory Data Register (MDR). Finally, the instruction is copied to the Current Instruction Register (CIR) and the PC is incremented.
描述机器周期的取指阶段。程序计数器 (PC) 存放下一条指令的地址；该地址被复制到内存地址寄存器 (MAR)。控制单元随后发出读信号，指令从 RAM 中被取出并放入内存数据寄存器 (MDR)。最后，指令复制到当前指令寄存器 (CIR)，PC 自增 1。
In the decode stage, the control unit interprets the opcode in the CIR to determine what operation is required. During execute, the CPU carries out the operation, which may involve the ALU for calculations or further data accesses. Clearly explaining each register’s role earns high marks.
在解码阶段，控制单元对 CIR 中的操作码进行译码，确定需要执行的操作。执行阶段，CPU 完成实际操作，可能涉及算术逻辑单元 (ALU) 进行计算或进一步的数据访问。清晰说明各寄存器的角色可以赢得高分。
4. Memory: RAM vs ROM | 存储器：RAM 与 ROM 的比较
A typical comparison question asks you to distinguish between RAM and ROM. RAM (Random Access Memory) is volatile – it loses its contents when power is turned off – and is used to hold the operating system, programs and data currently in use. It is read‑write memory and generally has a larger capacity than ROM in a modern computer.
典型的比较题要求区分 RAM 和 ROM。RAM（随机存取存储器）是易失性的——断电后内容丢失——用于存放当前正在使用的操作系统、程序和数据。它是可读写的存储器，在现代计算机中容量通常比 ROM 大。
ROM (Read Only Memory) is non‑volatile and retains its contents without power. It stores the boot program (BIOS / firmware) needed to start the computer. ROM is usually read‑only under normal operation and has a much smaller capacity. Remember to mention volatility, purpose, read/write capability and typical size.
ROM（只读存储器）是非易失性的，断电后内容仍然保留。它存储启动计算机所需的引导程序（BIOS / 固件）。ROM 一般只读，容量远小于 RAM。回答时务必提及易失性、用途、读写能力和典型容量。
5. Networks: Star vs Bus Topology | 网络：星型拓扑与总线拓扑
In a star topology, all devices connect to a central switch or hub. If one cable fails, only that device is disconnected, making the network reliable. However, if the central device fails, the whole network goes down. Star networks are easy to expand and troubleshoot.
在星型拓扑中，所有设备都连接到一台中央交换机或集线器。一条线缆出现故障只会影响所连的那台设备，因此网络可靠性较高。然而，中央设备一旦故障，整个网络就会瘫痪。星型网络易于扩展和故障排查。
A bus topology uses a single backbone cable to which all devices are connected. It is cheaper to install but less reliable: a break in the backbone can bring down the entire network, and data collisions are more frequent. Performance degrades as more devices are added. These differences are common exam targets.
总线拓扑使用一条主干线缆连接所有设备。安装成本较低，但可靠性较差：主干线缆断裂可能导致整个网络瘫痪，且数据冲突更频繁。随着设备增加，网络性能会下降。这些差异是常见的考试目标。
6. Network Protocols: The TCP/IP Model | 网络协议：TCP/IP 模型
Exam questions may ask you to name the four layers of the TCP/IP model and describe their purposes. The Application layer provides network services to user applications (HTTP, FTP, SMTP). The Transport layer ensures reliable data delivery using TCP (connection‑oriented) or UDP (connectionless), handling acknowledgments and retransmissions.
考题可能要求你列出 TCP/IP 模型的四层并描述其用途。应用层为最终用户应用程序提供网络服务（HTTP、FTP、SMTP）。传输层使用 TCP（面向连接）或 UDP（无连接）确保可靠的数据传输，负责确认和重传。
The Internet layer handles logical addressing and routing, using IP addresses to get packets from source to destination across multiple networks. The Link layer (or Network Access layer) deals with the physical hardware and MAC addresses, defining how data is placed onto the physical medium. Relating protocols to layers impresses examiners.
网络层（Internet 层）处理逻辑寻址和路由，利用 IP 地址将数据包从源端跨多个网络送达目的端。链路层（网络接入层）负责物理硬件和 MAC 地址，定义数据如何放到物理媒介上。将协议与层次对应起来能给考官留下深刻印象。
7. Encryption: Symmetric vs Asymmetric | 加密：对称加密与非对称加密
Symmetric encryption uses a single shared key to both encrypt and decrypt data. It is fast and efficient, suitable for encrypting large volumes of data, but the key must be exchanged securely between sender and receiver – a significant challenge in open networks.
对称加密使用同一个共享密钥进行数据加密和解密。它速度快、效率高，适合加密大量数据，但密钥必须在发送方和接收方之间安全交换，这在开放网络中是一个重大挑战。
Asymmetric encryption uses a key pair: a public key (shared openly) and a private key (kept secret). Anything encrypted with the public key can only be decrypted with the corresponding private key. This solves the key‑exchange problem but is slower and computationally more intensive. Many systems combine both: use asymmetric encryption to exchange a symmetric session key.
非对称加密使用密钥对：公钥（公开共享）和私钥（保密）。用公钥加密的内容只有对应的私钥才能解密。这解决了密钥交换难题，但速度较慢且计算成本更高。许多系统将两者结合：用非对称加密交换对称会话密钥。
8. Algorithm Design: Pseudocode for Finding the Maximum | 算法设计：求最大值的伪代码
You might be asked to write a pseudocode algorithm that finds the largest number in a list of ten numbers input by the user. A solid approach: initialise a variable max to the first entered value. Then loop from the second item to the tenth, comparing each input with max and updating max if the input is greater.
题目可能要求你编写伪代码算法，找出用户输入的十个数字中的最大值。一种可靠的做法是：将变量 max 初始化为第一个输入值，然后从第二项循环至第十项，将每次输入与 max 比较，若大于 max 则更新 max。
max ← INPUT
FOR i ← 2 TO 10
num ← INPUT
IF num > max THEN
max ← num
ENDIF
NEXT i
OUTPUT max
Mark schemes reward clear variable names, correct loops, and efficient logic. Avoid unnecessary complexity; Cambridge IGCSE pseudocode is straightforward and similar to the above. Practise tracing your algorithm with sample data to be sure it works.
评分标准奖励清晰的变量命名、正确的循环和高效的逻辑。无需过度复杂；剑桥 IGCSE 的伪代码简洁明了，与上例相似。建议用样本数据手动追踪你的算法，确保其可行。
9. Databases: Writing SQL Queries | 数据库：编写 SQL 查询
Given a table STUDENTS(StudentID, Name, Grade, Tutor), a question might ask: "Write an SQL statement to display the names of all students in Grade 11, ordered alphabetically by Name." The answer uses SELECT, FROM, WHERE and ORDER BY:
假设有一个表 STUDENTS(StudentID, Name, Grade, Tutor)，题目可能问：“写一条 SQL 语句，显示所有 11 年级学生的姓名，按姓名字母顺序排序。”答案会用到 SELECT、FROM、WHERE 和 ORDER BY：
SELECT Name FROM STUDENTS WHERE Grade = 11 ORDER BY Name ASC;
Another common task is to find summary information, e.g. "how many students each tutor looks after." Use GROUP BY and COUNT:
另一种常见任务是查找汇总信息，例如“每位导师负责多少学生”。使用 GROUP BY 和 COUNT：
SELECT Tutor, COUNT(*) FROM STUDENTS GROUP BY Tutor;
Remember that SQL keywords are not case‑sensitive, but writing them in uppercase makes your answer clearer. Always ensure your query returns exactly the required fields and nothing more.
注意 SQL 关键字不区分大小写，但大写书写能使答案更清晰。务必确保查询只返回所要求的字段，不多不少。
10. Ethics and Legal: Data Protection Act | 伦理与法律：数据保护法案
The Data Protection Act sets rules for how personal data can be collected, stored, processed and shared. Organisations must keep data accurate and up‑to‑date, hold it only for as long as necessary, and protect it from unauthorised access. Individuals have the right to see the data held about them and to correct errors.
数据保护法案为个人数据的收集、存储、处理和共享制定了规则。机构必须确保数据准确、保持更新，仅在必要时保留数据，并防止未经授权的访问。个人有权查看自己被持有的数据并更正错误。
Exam questions often provide a scenario, such as a school keeping student medical records, and ask you to discuss implications. You need to apply the principles: consent, purpose limitation, security, and the right of access. Mentioning the consequences of non‑compliance (fines, reputation damage) demonstrates deeper understanding.
考试常给出一段情景，如学校保存学生的医疗记录，要求你讨论其影响。你需要运用相关原则：知情同意、目的限制、安全性和访问权。提及不合规的后果（罚款、声誉受损）可以展示更深的理解。
Published by TutorHao | Computer Science Revision Series | aleveler.com
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