IGCSE AQA Computer Science: Typical Exam Questions Explained | IGCSE AQA 计算机：典型例题详解

📚 IGCSE AQA Computer Science: Typical Exam Questions Explained | IGCSE AQA 计算机：典型例题详解

This article walks you through a selection of typical IGCSE AQA Computer Science exam questions, covering key topics such as data representation, logic gates, programming concepts, networking, and databases. Each question is paired with a detailed solution and explanation, giving you a clear sense of the depth and style expected in the actual assessment. Use these worked examples to strengthen your understanding and boost your confidence.

本文精选 IGCSE AQA 计算机课程的典型考题，涵盖数据表示、逻辑门、程序设计、网络和数据库等核心主题。每道题目都配有详细的解答和解析，帮助你清晰把握考试的实际深度与风格。通过这些精讲例题，你可以巩固理解、提升备考信心。

1. Binary Addition and Overflow | 二进制加法与溢出

Question: Perform binary addition of 1101₂ and 0111₂, and explain whether overflow occurs if the result is stored in a 4-bit register.

问题：计算 1101₂ 与 0111₂ 的二进制加法，并说明如果结果存放在 4 位寄存器中是否发生溢出。

Solution: Align the bits and add column by column, carrying as needed. 1101₂ (13 in decimal) plus 0111₂ (7 in decimal) gives 10100₂ (20 in decimal). In a 4-bit register, only the lower 4 bits ‘0100’ can be stored, and the carry-out from the most significant bit is lost. This loss means the stored value (4) is incorrect relative to the actual sum, so overflow has occurred. Overflow in signed arithmetic here would also be indicated by a carry into the sign bit that does not match the carry-out.

解答：按位对齐，逐列相加并进位。1101₂（十进制13）加 0111₂（十进制7）得到 10100₂（十进制20）。在 4 位寄存器中，只能存储低 4 位 ‘0100’，最高位的进位丢失。由于实际和与存储值（4）不符，说明发生了溢出。在带符号运算中，若进入符号位的进位与出符号位的进位不一致，也标志溢出。

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2. Hexadecimal Conversion | 十六进制转换

Question: Convert the binary number 10111101₂ into hexadecimal. Show your working.

问题：将二进制数 10111101₂ 转换为十六进制。请展示转换过程。

Solution: Group the binary digits into nibbles (groups of four) from the right: 1011 1101. Convert each nibble: 1011₂ = B₁₆, 1101₂ = D₁₆. Therefore, the hexadecimal equivalent is BD₁₆. To verify, 1011₂ is 8+2+1=11 (B), 1101₂ is 8+4+1=13 (D).

解答：从右向左将二进制数每 4 位分为一组（nibble）：1011 1101。分别转换：1011₂ = B₁₆，1101₂ = D₁₆。因此十六进制结果为 BD₁₆。验证：1011₂ = 8+2+1=11（即 B），1101₂ = 8+4+1=13（即 D）。

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3. Logic Gates and Truth Tables | 逻辑门与真值表

Question: A circuit consists of an AND gate followed by a NOT gate. Inputs are A and B. Draw the truth table for the output Q and name the equivalent single logic gate.

问题：某电路由一个与门（AND）后接一个非门（NOT）组成。输入为 A 和 B。画出输出 Q 的真值表，并说出该组合等效的单一逻辑门名称。

Solution: First, compute X = A AND B, then Q = NOT X. The truth table is shown below. The output Q is 1 only when both inputs are 0, otherwise 0. This matches the behaviour of a NAND gate. Thus, the combination is equivalent to a NAND gate.

解答：首先计算 X = A AND B，然后 Q = NOT X。真值表如下。输出 Q 仅在两个输入都为 0 时为 1，其他情况下为 0。这正符合与非门（NAND）的行为，因此该组合等效于一个与非门。

A	B	X = A AND B	Q = NOT X
0	0	0	1
0	1	0	1
1	0	0	1
1	1	1	0

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4. Algorithm Design with Pseudocode | 算法设计：伪代码

Question: Write pseudocode to input 10 numbers and output the largest one. Use a loop and a variable to track the maximum.

问题：编写伪代码，输入 10 个数并输出其中的最大值。使用循环和一个变量记录当前最大值。

Solution: Initialise ‘max’ to a very small value or the first input. Then loop 10 times: read number, if number > max then max = number. Finally, print max. Example pseudocode:
max ← -∞
FOR i ← 1 TO 10
   INPUT num
   IF num > max THEN
     max ← num
   ENDIF
ENDFOR
OUTPUT max

解答：将变量 max 初始化为极小值或第一个输入。循环 10 次：读入一个数，若该数大于 max，则更新 max。最后输出 max。示例伪代码如下：
max ← -∞
FOR i ← 1 TO 10
   INPUT num
   IF num > max THEN
     max ← num
   ENDIF
ENDFOR
OUTPUT max

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5. Arrays and Records | 数组与记录

Question: A school stores student data using a record structure with fields: StudentID (integer), Name (string), Grade (character). Define an array of 30 such records and write pseudocode to find and output the name of the student with the highest StudentID.

问题：学校用记录结构存储学生数据，字段包括：StudentID（整数）、Name（字符串）、Grade（字符）。定义一个包含 30 个此类记录的数组，并编写伪代码找出并输出具有最大 StudentID 的学生姓名。

Solution: Declare an array students[1:30] of record type. Then assume the first student has the highest ID. Iterate through the array; if a student’s ID is greater than the current highest, update the highest ID and store the name. Finally, output the stored name. Example pseudocode:
DECLARE students : ARRAY[1:30] OF RECORD
   id : INTEGER
   name : STRING
   grade : CHAR
ENDRECORD
highestID ← students[1].id
bestName ← students[1].name
FOR i ← 2 TO 30
   IF students[i].id > highestID THEN
     highestID ← students[i].id
     bestName ← students[i].name
   ENDIF
ENDFOR
OUTPUT bestName

解答：定义一个记录类型，包含所需字段，再定义数组 students[1:30]。先将第一个学生视为 ID 最大者，然后遍历数组，如果当前学生的 ID 更大，则更新最大 ID 和相应姓名。最后输出该姓名。伪代码如上所示。

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6. Conditional Statements and Loops | 条件语句与循环

Question: A program needs to repeatedly ask the user to enter a password until the correct word ‘TutorHao2025’ is entered. After a correct entry, display ‘Access granted’ and end. Write pseudocode using a WHILE loop.

问题：程序需要反复要求用户输入密码，直到输入正确密码 ‘TutorHao2025’。输入正确后显示 ‘Access granted’ 并结束。使用 WHILE 循环编写伪代码。

Solution: Use a variable ‘password’ initialised to an empty string. While password is not equal to ‘TutorHao2025’, prompt the user and input password. After the loop, output ‘Access granted’. Pseudocode:
password ← ”
WHILE password ≠ ‘TutorHao2025’ DO
   OUTPUT ‘Enter password: ‘
   INPUT password
ENDWHILE
OUTPUT ‘Access granted’

解答：将变量 password 初始化为空字符串。当 password 不等于 ‘TutorHao2025’ 时，循环提示并输入密码。循环结束后输出 ‘Access granted’。伪代码：
password ← ”
WHILE password ≠ ‘TutorHao2025’ DO
   OUTPUT ‘Enter password: ‘
   INPUT password
ENDWHILE
OUTPUT ‘Access granted’

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7. IP Addressing and MAC Addresses | IP 地址与 MAC 地址

Question: Explain the difference between an IP address and a MAC address. State which layer of the TCP/IP model each belongs to and why both are necessary in a network.

问题：解释 IP 地址与 MAC 地址之间的区别。说明它们分别属于 TCP/IP 模型的哪一层，并阐述为何网络中两者都必不可少。

Solution: An IP address (e.g., 192.168.1.10) is a logical address used for routing packets across different networks; it belongs to the Internet layer. A MAC address (e.g., 00:1A:2B:3C:4D:5E) is a physical hardware address embedded in a network interface card, used for local delivery within the same network segment; it operates at the Link layer. Both are necessary because IP addressing enables global end-to-end communication, while MAC addressing allows actual frame delivery on the local physical network.

解答：IP 地址（如 192.168.1.10）是一种逻辑地址，用于在不同网络之间路由数据包，属于互联网层。MAC 地址（如 00:1A:2B:3C:4D:5E）是烧录在网卡中的物理硬件地址，用于同一网段内的本地交付，工作在链路层。两者缺一不可：IP 寻址实现全局的端到端通信，而 MAC 寻址确保帧在本地物理网络中的实际传输。

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8. Network Security Threats | 网络安全威胁

Question: Describe two forms of malware and explain how a firewall helps protect a network from unauthorised access.

问题：描述两种恶意软件的形式，并解释防火墙如何帮助保护网络免受未授权访问。

Solution: A virus is a malicious program that attaches itself to legitimate files and spreads when those files are shared. A trojan horse disguises itself as useful software but performs harmful actions once executed. A firewall monitors incoming and outgoing network traffic based on predefined security rules. It can block packets from suspicious IP addresses, close specific ports, and prevent unauthorised external connections, thus reducing the risk of intrusion.

解答：病毒是一种恶意程序，它附着在合法文件上，当这些文件被共享时传播。特洛伊木马则伪装成有用的软件，但一旦执行就会进行有害操作。防火墙根据预设的安全规则监控传入和传出的网络流量，可以阻断来自可疑 IP 地址的数据包、关闭特定端口并阻止未授权的外部连接，从而降低入侵风险。

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9. System Software – Operating System Functions | 系统软件 – 操作系统功能

Question: Identify three key functions of an operating system and briefly explain how each contributes to the operation of a computer.

问题：指出操作系统的三个关键功能，并简要解释每个功能如何促进计算机的运行。

Solution: 1. Memory management – the OS allocates RAM to processes and frees it when no longer needed, preventing conflicts and enabling multitasking. 2. File management – the OS organises data into files and directories, controlling read/write access and storage. 3. Processor scheduling – the OS decides which process gets CPU time, ensuring fair and efficient use of the processor. All three functions work together to provide a stable environment for software and users.

解答：1. 内存管理 – 操作系统为进程分配 RAM，并在不再需要时释放，防止冲突并支持多任务处理。2. 文件管理 – 操作系统将数据组织成文件和目录，控制读写访问和存储。3. 处理器调度 – 操作系统决定哪个进程获得 CPU 时间，确保公平且高效地使用处理器。三个功能协同工作，为软件和用户提供稳定的操作环境。

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10. Database Queries (SQL) | 数据库查询 (SQL)

Question: A table ‘Products’ contains fields: ProductID, Name, Price, Category. Write an SQL statement to retrieve the names and prices of all products in the ‘Electronics’ category with a price less than 500, ordered by price descending.

问题：有一个表 ‘Products’，包含字段：ProductID、Name、Price、Category。请写出一条 SQL 语句，查询所有属于 ‘Electronics’ 类别且价格低于 500 的产品的名称与价格，并按价格降序排列。

Solution:
SELECT Name, Price
FROM Products
WHERE Category = ‘Electronics’ AND Price < 500
ORDER BY Price DESC;

解答：
SELECT Name, Price
FROM Products
WHERE Category = ‘Electronics’ AND Price < 500
ORDER BY Price DESC;

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11. Binary Shifts and Logical Operations | 二进制移位与逻辑运算

Question: Perform a logical left shift of two places on the binary number 00001110₂ (8-bit). State the result in both binary and decimal, and explain the effect of this shift.

问题：对 8 位二进制数 00001110₂ 进行两次逻辑左移。用二进制和十进制表示结果，并说明这种移位的作用。

Solution: Left shift by two: move all bits two positions left, fill rightmost bits with zeros. Original: 00001110₂ (14 in decimal). After shift: 00111000₂ , which is 56 in decimal. A logical left shift by n places multiplies the unsigned number by 2ⁿ (here ×4). Bits shifted beyond the leftmost position are discarded; if they were 1, an overflow flag might be set.

解答：左移两位：将所有位向左移动两个位置，右侧空位补零。原数 00001110₂（十进制 14）。移位后为 00111000₂，即十进制 56。对无符号数进行一次逻辑左移 n 位相当于乘以 2ⁿ（本例 ×4）。移出最左侧的位被丢弃；若那些位为 1，则可能设置溢出标志。

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12. Ethical, Legal and Environmental Issues | 道德、法律与环境问题

Question: A software company collects user location data to improve services. Discuss one ethical and one legal concern related to this practice, and suggest how the company could address them.

问题：某软件公司收集用户位置数据以改善服务。讨论与此做法相关的一个道德问题和一个法律问题，并建议公司如何应对。

Solution: Ethically, collecting location data without explicit informed consent violates user privacy and autonomy. Legally, under regulations such as GDPR, companies must obtain clear consent and provide transparency about data usage. To address these, the company should implement an opt-in mechanism where users actively agree, publish a clear privacy policy explaining purpose and retention, and allow users to access or delete their data.

解答：道德层面，未经明确知情同意收集位置数据侵犯了用户隐私和自主权。法律层面，根据 GDPR 等法规，公司必须获得清晰同意并透明化数据用途。应对措施包括：实施选入机制，让用户主动同意；发布清晰的隐私政策，说明目的和保存期限；并允许用户访问或删除其数据。

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Published by TutorHao | Computer Science Revision Series | aleveler.com

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