AQA Year 9 Computer Science: Past Paper Deep Dive | AQA 九年级计算机：历年真题深度解析

📚 AQA Year 9 Computer Science: Past Paper Deep Dive | AQA 九年级计算机：历年真题深度解析

Past papers are the gold standard for AQA Computer Science preparation. They reveal recurring question styles, common pitfalls, and the exact depth of knowledge expected. This in-depth analysis breaks down typical Year 9 / GCSE-level questions from previous exam series, guiding you through core topics such as data representation, programming, hardware, and networking. By studying these annotated solutions, you will sharpen your exam technique and build confidence.

历年真题是备考 AQA 计算机科学的最佳利器。它们揭示了反复出现的题型、常见陷阱以及考纲要求的深度。本文深度解析九年级/GCSE 水平真题中的典型题目，带你逐一攻克数据表示、编程、硬件和网络等核心主题。通过研读这些带注解的解法，你将提升应试技巧、增强信心。

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1. Binary and Hexadecimal Number Systems | 二进制与十六进制数制

AQA papers frequently include questions on converting between binary, denary, and hexadecimal. A typical task: 'Convert the 8-bit binary number 10101100 into hex.' The quick method is to split the binary into two nibbles: 1010 (A) and 1100 (C), giving AC. Understanding place values (128, 64, 32, 16, 8, 4, 2, 1) is essential.

AQA 真题经常要求进行二进制、十进制和十六进制转换。一个典型题目：“将八位二进制数 10101100 转换为十六进制。” 快捷方法是把二进制拆成两个半字节：1010 (A) 和 1100 (C)，得到 AC。掌握位值（128、64、32、16、8、4、2、1）至关重要。

Another common trap is binary addition and overflow. For example, 11001010 + 01101100 results in a 9-bit number, but only 8 bits are stored, causing an overflow error. The exam may ask you to identify the error and explain its consequence.

另一个常见陷阱是二进制加法与溢出。例如 11001010 + 01101100 产生 9 位结果，但只存储 8 位，导致溢出错误。试卷可能要求你识别该错误并解释后果。

emainder 8. 12 is C in hex, so the answer is C8. Always check using binary as a bridge if unsure.

将十进制转十六进制时，连续除以 16 是首选方法。转换 200₁₀：200 ÷ 16 = 12 余 8。12 对应十六进制的 C，因此答案为 C8。拿不准时可用二进制作为桥梁验算。

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

Logic gate questions appear almost every year. You might be given a circuit like (A AND B) OR (NOT C) and asked to complete a truth table. First, recall the symbols: AND outputs 1 only when both inputs are 1; OR outputs 1 when at least one input is 1; NOT inverts the input.

逻辑门题目几乎每年都考。你可能会遇到类似 (A AND B) OR (NOT C) 的电路，并需要完成真值表。首先回顾符号：AND 仅在两个输入都为 1 时输出 1；OR 在至少一个输入为 1 时输出 1；NOT 将输入反转。

truth table for the expression Q = (A AND B) OR (NOT C), using 1 for TRUE and 0 for FALSE.

下面是针对表达式 Q = (A AND B) OR (NOT C) 的示例真值表，用 1 表示 TRUE，0 表示 FALSE。

A	B	C	Q
0	0	0	1
0	0	1	0
0	1	0	1
0	1	1	0
1	0	0	1
1	0	1	0
1	1	0	1
1	1	1	1

Always build intermediate columns in your working (e.g., A AND B, NOT C) to avoid mistakes. Exam marks are often awarded for showing these steps.

务必在草稿中建立中间列（例如 A AND B、NOT C），以避免错误。试卷通常会给展示这些步骤的过程分。

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3. Programming Errors and Debugging | 编程错误与调试

Debugging snippets are classic Year 9 questions. Consider this faulty Python code:
print("Enter a number")
num = input()
if num > 10
    print("big")
else
    print("small")
There are two errors: the if and else statements are missing colons, and input() returns a string, so the comparison num > 10 will cause a type error.

调试代码片段是九年级经典题目。考虑以下有问题的 Python 代码：
print("Enter a number")
num = input()
if num > 10
    print("big")
else
    print("small")
存在两处错误：if 和 else 语句缺少冒号，并且 input() 返回字符串，因此比较 num > 10 会引发类型错误。

To fix it, add colons after if num > 10 and else, and convert the input to an integer using int(input()). In the exam you might be asked to circle errors or write the corrected code. Look carefully for indentation mistakes too, as Python relies on consistent indentation.

修正方法：在 if num > 10 和 else 后添加冒号，并使用 int(input()) 将输入转换为整数。考试中可能要求你圈出错误或写出改正后的代码。同时要仔细检查缩进，因为 Python 依赖一致的缩进。

Logic errors are trickier. For instance, a loop that intends to sum even numbers might mistakenly include all numbers if the condition is set as if num % 2 == 1 instead of 0. Always trace the code with a test value to catch such mistakes.

逻辑错误更隐蔽。例如，一个循环本意是求偶数和，但如果条件写成 if num % 2 == 1 而不是 0，就会错误包含奇数。务必用测试值跟踪代码来发现这类错误。

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4. Sorting Algorithms: Bubble Sort | 排序算法：冒泡排序

Bubble sort is a favourite for examining algorithmic thinking. Given the list [6, 2, 5, 3, 9], show the list after one full pass. Compare adjacent pairs and swap if they are in the wrong order.

冒泡排序是考察算法思维的常见题。给定列表 [6, 2, 5, 3, 9]，写出一次完整遍历后的列表。比较相邻对，若顺序错误则交换。

Pass 1 comparisons:

第一次遍历的比较过程：

• Compare 6 and 2 - swap => [2, 6, 5, 3, 9]
• Compare 6 and 5 - swap => [2, 5, 6, 3, 9]
• Compare 6 and 3 - swap => [2, 5, 3, 6, 9]
• Compare 6 and 9 - no swap => [2, 5, 3, 6, 9]

After the first pass, the largest element (9) has bubbled to the end. The list becomes [2, 5, 3, 6, 9]. The exam may ask how many passes are needed to fully sort n items (n-1 maximum) or to identify the state after a given pass.

第一次遍历后，最大元素 (9) 已冒泡到末尾。列表变为 [2, 5, 3, 6, 9]。试卷可能问对 n 个元素完全排序最多需要多少次遍历（n-1 次），或要求写出某次遍历后的状态。

Be prepared to write a simple bubble sort algorithm in pseudocode or a flowchart. Emphasise the use of a swap flag to improve efficiency by stopping early if no swaps occur in a pass.

请准备好用伪代码或流程图编写简单的冒泡排序算法。重点在于使用交换标志来提高效率，如果一次遍历中没有发生交换，即可提前终止。

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5. Network Protocols and the TCP/IP Model | 网络协议与TCP/IP模型

Questions on networking often ask you to complete missing layers or protocols in the TCP/IP model. The four layers are Application, Transport, Internet, and Link. Typical protocols: HTTP, HTTPS, FTP (Application); TCP, UDP (Transport); IP (Internet); Ethernet, Wi-Fi (Link).

网络相关题目常要求补全 TCP/IP 模型中的缺失层或协议。四层分别是应用层、传输层、互联网层和链路层。常见协议：HTTP、HTTPS、FTP（应用层）；TCP、UDP（传输层）；IP（互联网层）；以太网、Wi-Fi（链路层）。

You may be given a scenario, such as sending an email, and be asked which protocol at the Application layer is used (SMTP for sending, IMAP/POP3 for receiving). Remember that TCP provides error-free, ordered delivery, while UDP is faster but less reliable – suited to video streaming.

题目可能给出发送电子邮件的场景，问应用层使用哪个协议（发送用 SMTP，接收用 IMAP/POP3）。记住 TCP 提供无差错、有序的传输，而 UDP 更快但可靠性较低，适合视频流。

A frequently tested concept is the role of the Internet layer in routing packets using IP addresses, while the Link layer handles the physical transmission of frames. Diagrams showing data encapsulation are common: data -> segment -> packet -> frame.

一个常考概念是互联网层使用 IP 地址路由数据包，而链路层处理帧的物理传输。展示数据封装的示意图很常见：数据 -> 段 -> 包 -> 帧。

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6. Sound and Image Representation | 声音与图像表示

Calculating file sizes from specification is a core skill. For a bitmap image:

Image file size (bits) = width × height × colour depth

根据参数计算文件大小是一项核心技能。对于位图图像：

图像文件大小（比特）= 宽度 × 高度 × 色深

es by dividing by 8: 2,359,296 bytes. To get kilobytes, divide by 1024: approximately 2,304 KB, or about 2.25 MB (dividing by 1024 again). The exam often requires answers in appropriate units.

例题：一张 1024×768 图片，色深 24 位。文件大小 = 1024 × 768 × 24 = 18,874,368 比特。除以 8 转为字节：2,359,296 字节。再除以 1024 得到约 2,304 KB，或约 2.25 MB（再除以 1024）。考试通常要求以合适单位作答。

For sound:

Sound file size (bits) = sample rate × bit depth × duration (seconds) × channels

声音文件：

声音文件大小（比特）= 采样率 × 位深 × 时长（秒）× 声道数

ut 2.52 MB. If stereo, multiply by 2. Be careful with unit prefixes like k (kilo) meaning 1000 in sample rate but 1024 in storage; always use 1024 for bytes conversions unless directed otherwise.

一段 30 秒的单声道录音，采样率 44.1 kHz，位深 16 位，需要 44,100 × 16 × 30 = 21,168,000 比特，约 2.52 MB。立体声则需乘以 2。注意单位前缀：k 在采样率中表示 1000，而在存储量中又为 1024；除非另有说明，字节转换一律使用 1024。

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7. Pseudocode Tracing and Writing | 伪代码追踪与编写

Being able to trace pseudocode is tested through dry-run questions. Study this loop:

通过干运行题来测试伪代码追踪能力。分析以下循环：

total ← 0
FOR count ← 1 TO 5
    total ← total + count
ENDFOR
OUTPUT total

lates 1+2+3+4+5 = 15, so the output is 15. A typical trace table would list each iteration's values for count and total, gaining marks for systematic tracking.

该循环运行五次。变量 total 累加 1+2+3+4+5 = 15，因此输出 15。一个典型的追踪表会列出每次迭代的 count 和 total 值，系统化追踪能得分。

Writing pseudocode from a problem statement is also common, e.g., 'Input 10 numbers and output the largest.' Your solution might:

根据问题描述编写伪代码也很常见，如“输入 10 个数字，输出最大值”。你的解法可以是：

largest ← first input
FOR i ← 2 TO 10
    next ← input
    IF next > largest THEN
        largest ← next
    ENDIF
ENDFOR
OUTPUT largest

Marks are awarded for clear, unambiguous steps and correct use of loop structures. Do not use language-specific syntax;

Published by TutorHao | Year 9 Computer Science Revision Series | aleveler.com

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