Year 9 AQA Computer Science: Unit Test Mock Paper Walkthrough | Year 9 AQA 计算机：单元测试模拟卷解析

📚 Year 9 AQA Computer Science: Unit Test Mock Paper Walkthrough | Year 9 AQA 计算机：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test designed for Year 9 students following the AQA Computer Science specification. It covers key topics such as data representation, logic gates, programming fundamentals, and networks. Each question is explained step by step to help you consolidate your understanding and improve exam technique.

本文详细解析一份针对AQA计算机科学课程设计的九年级单元测试模拟卷。内容涵盖数据表示、逻辑门、编程基础以及网络等核心主题。每道题目都进行了逐步讲解，帮助你巩固知识、提升应试技巧。

---

1. Mock Paper Overview | 模拟试卷概览

The mock paper consists of 10 questions, testing your ability to convert between number systems, analyse logic circuits, understand data types, trace algorithms, write simple code, and recall networking concepts. Total marks: 40. Time allowed: 45 minutes.

模拟卷共10题，考查数制转换、逻辑电路分析、数据类型理解、算法追踪、简单代码编写以及网络概念记忆。总分40分，时间45分钟。

All answers are expected to be clear and show working where required. The following sections will guide you through each question with model solutions and key revision points.

所有答案都要求清晰，并在必要时展示计算过程。以下各节将逐一解析每道题目，提供标准解答和关键复习点。

---

2. Question 1: Binary to Denary Conversion | 问题1：二进制转十进制

Question: Convert the 8-bit binary number 01101010 to denary. Show your working. (4 marks)

题目：将8位二进制数01101010转换为十进制。请写出计算过程。（4分）

For 01101010: 0×128 + 1×64 + 1×32 + 0×16 + 1×8 + 0×4 + 1×2 + 0×1 = 64 + 32 + 8 + 2 = 106.

将二进制各位写在对应权值（128, 64, 32, 16, 8, 4, 2, 1）下。对于01101010：0×128 + 1×64 + 1×32 + 0×16 + 1×8 + 0×4 + 1×2 + 0×1 = 64 + 32 + 8 + 2 = 106。

Answer: 106

答案：106

---

3. Question 2: Denary to Binary Conversion | 问题2：十进制转二进制

Question: Convert denary 78 into an 8-bit binary number. (3 marks)

题目：将十进制78转换为8位二进制数。（3分）

der 0; 1 ÷ 2 = 0 remainder 1. Reading remainders from bottom to top gives 1001110. Pad to 8 bits by adding a leading 0: 01001110.

方法1：连续除以2取余。78 ÷ 2 = 39 余 0；39 ÷ 2 = 19 余 1；19 ÷ 2 = 9 余 1；9 ÷ 2 = 4 余 1；4 ÷ 2 = 2 余 0；2 ÷ 2 = 1 余 0；1 ÷ 2 = 0 余 1。从下往上读余数得1001110。补足8位，前加0得01001110。

14 fits 8 (2³), leaving 6; 6 fits 4 (2²), leaving 2; 2 fits 2 (2¹). So active bits: 64, 32? Wait, only 64, 8, 4, 2. So binary: 0 1 0 0 1 1 1 0? Check: 64+8+4+2=78. Yes, 01001110.

方法2：权值法。78中有64（2⁶），剩14；14中有8（2³），剩6；6中有4（2²），剩2；2中有2（2¹）。因此有效位为64, 8, 4, 2，对应二进制01001110。

Answer: 01001110

答案：01001110

---

4. Question 3: Hexadecimal Conversion | 问题3：十六进制转换

Question: Convert denary 200 to hexadecimal. Show your working. (4 marks)

题目：将十进制200转换为十六进制。写出计算过程。（4分）

C (A=10, B=11, C=12). The remainder 8 stays as 8. Therefore the hexadecimal number is C8.

除以16。200 ÷ 16 = 12 余 8。商12对应十六进制数字C（A=10，B=11，C=12）。余数8不变。因此十六进制数为C8。

You can verify by converting back: C×16 + 8 = 12×16 + 8 = 192 + 8 = 200.

可通过逆向转换验证：C×16 + 8 = 12×16 + 8 = 192 + 8 = 200。

Answer: C8

答案：C8

---

5. Question 4: Logic Gate Truth Table | 问题4：逻辑门真值表

Question: Complete the truth table for the logic expression A AND (NOT B). (4 marks)

题目：补全逻辑表达式 A AND (NOT B) 的真值表。（4分）

The circuit takes inputs A and B, inverts B, then performs an AND operation. The truth table must cover all four combinations of A and B (00, 01, 10, 11).

该电路输入A和B，将B取反后再与A进行AND运算。真值表需覆盖A和B的四种组合（00, 01, 10, 11）。

A	B	NOT B	A AND (NOT B)
0	0	1	0
0	1	0	0
1	0	1	1
1	1	0	0

output 1. All other cases give 0. This is a common pattern used in conditional operations.

仅当A=1且B=0时输出为1。其余情况均为0。这是条件操作中常见的一种模式。

---

6. Question 5: Data Storage Units | 问题5：数据存储单位

Question: (a) How many bytes are there in 2 kilobytes? (b) How many bits are there in 2 KiB? (2 marks)

题目：(a) 2千字节中有多少字节？(b) 2 KiB中有多少比特？（2分）

Part (a) uses the decimal prefix: 1 kilobyte (KB) = 1000 bytes. So 2 KB = 2000 bytes.

第(a)小题使用十进制前缀：1千字节（KB）= 1000字节。因此2 KB = 2000字节。

KiB = 1024 bytes. 2 KiB = 2048 bytes. Since 1 byte = 8 bits, 2048 bytes × 8 = 16384 bits.

第(b)小题涉及千位二进制字节（KiB），为二进制前缀：1 KiB = 1024 字节。2 KiB = 2048 字节。因为1字节=8比特，2048 × 8 = 16384 比特。

Many exam questions deliberately test the difference between KB (1000) and KiB (1024). Always check which unit is given.

许多考试题故意考查KB（1000）与KiB（1024）的区别。务必留意给出的单位。

Answers: (a) 2000 bytes (b) 16384 bits

答案：(a) 2000 字节 (b) 16384 比特

---

广告 / Sponsored

7. Question 6: Variables and Data Types | 问题6：变量与数据类型

Question: State the most suitable data type (integer, real, Boolean, string, character) for storing each of the following: a person's name; age in years; temperature reading in Celsius; whether a door is open; a student's grade letter. (4 marks)

题目：为以下各项选择最合适的数据类型（整型、实型、布尔型、字符串、字符）：一个人的姓名；年龄（岁）；摄氏温度读数；门是否开启；学生等级字母。（4分）

Name: string (sequence of characters). Age: integer (whole number). Temperature: real (may contain decimal places). Door open: Boolean (true/false or 0/1). Grade letter: character (single symbol).

姓名：字符串（字符序列）。年龄：整型（整数）。温度：实型（可能含小数）。门开启：布尔型（真/假或0/1）。等级字母：字符（单个符号）。

It is important to distinguish between string and character: a name like 'Alice' requires a string, while a single grade like 'A' can be stored as a character.

注意区分字符串与字符：像'Alice'这样的姓名需要字符串，而单个等级如'A'可用字符存储。

---

8. Question 7: Algorithm Flowchart Trace | 问题7：算法流程图追踪

Question: ce the algorithm for (i) x = 7 and (ii) x = 12, stating the output in each case. (4 marks)

题目：某流程图算法先输入数字x。若x > 10，则y = x * 2；否则y = x + 5。然后输出y。分别对(i) x = 7 和 (ii) x = 12 追踪算法，说出每种情况下的输出。（4分）

x = 7: 7 is not greater than 10, so the else branch executes. y = 7 + 5 = 12. Output: 12.

对于x = 7：7不大于10，因此执行else分支。y = 7 + 5 = 12。输出：12。

For x = 12: 12 > 10 is true, so y = 12 * 2 = 24. Output: 24.

对于x = 12：12 > 10为真，因此y = 12 × 2 = 24。输出：24。

This flowchart demonstrates a simple selection construct (if-else). Knowing how to dry-run with test data is essential for debugging and exam success.

此流程图演示了简单的选择结构（if-else）。掌握用测试数据手工走查对于调试和考试成功至关重要。

---

9. Question 8: Simple Python Program | 问题8：简单Python程序

Question: Write a Python program that asks the user to enter a number and then prints 'Even' if the number is even, otherwise prints 'Odd'. (4 marks)

题目：编写一个Python程序，要求用户输入一个数字，若该数字为偶数则输出'Even'，否则输出'Odd'。（4分）

You need to use the input() function and the modulo operator %. Even numbers have a remainder of 0 when divided by 2.

你需要使用 input() 函数和取模运算符 %。偶数除以2余数为0。

Model answer:

参考答案：

num = int(input("Enter a number: "))
if num % 2 == 0:
    print("Even")
else:
    print("Odd")

Note the conversion to integer using int(), because input() returns a string. Indentation is required in Python to define the if and else blocks.

注意使用 int() 将输入转换为整数，因为 input() 返回字符串。Python中必须使用缩进来定义if和else代码块。

---

10. Question 9: Networking Basics | 问题9：网络基础

Question: What is the difference between a LAN and a WAN? Give one example of each. (3 marks)

题目：LAN和WAN之间有何区别？各举一例。（3分）

A LAN (Local Area Network) covers a small geographical area, such as a single building or school campus. It is typically owned and managed by a single organisation. Example: the school's computer network.

LAN（局域网）覆盖较小地理区域，如单个建筑或学校校园。通常由一个组织拥有和管理。例如：学校的计算机网络。

A WAN (Wide Area Network) spans a large geographic area, often between cities or countries. It usually uses leased telecommunications lines or the internet infrastructure. Example: the Internet itself, or a company's network connecting offices in London and Manchester.

WAN（广域网）跨越较大地理区域，常在城市或国家之间。通常使用租用的电信线路或互联网基础设施。例如：互联网本身，或连接伦敦和曼彻斯特办公室的公司网络。

---

11. Question 10: Cyber Security | 问题10：网络安全

Question: Explain what a phishing attack is and describe one way to protect against it. (3 marks)

题目：解释什么是网络钓鱼攻击，并描述一种防范方法。（3分）

Phishing is a social engineering attack where criminals send fraudulent emails or messages that appear to be from a trusted source. The goal is to trick recipients into revealing sensitive information such as passwords, bank details, or to install malware. For example, an email claiming to be from a bank asking you to 'verify your account' by clicking a link.

网络钓鱼是一种社会工程攻击，犯罪分子发送看似来自可信来源的欺诈邮件或消息，诱骗收件人泄露密码、银行信息等敏感数据或安装恶意软件。例如，一封自称来自银行的邮件要求你点击链接'验证账户'。

Protection methods include: checking the sender's email address carefully, never clicking suspicious links, hovering over links to see the actual URL, using spam filters, and not sharing personal information in response to unsolicited messages.

防范方法包括：仔细检查发件人邮箱地址、不点击可疑链接、将鼠标悬停在链接上查看实际URL、使用垃圾邮件过滤器、不回复主动发来的信息提供个人资料。

---

12. Summary and Revision Tips | 总结与备考建议

This mock test highlights the foundational topics for Year 9 AQA Computer Science: number conversions, logic, data types, basic programming, and networking. To prepare effectively, practise plenty of binary and hex conversions, learn truth tables for AND/OR/NOT, and write and test small Python programs. Always double-check whether units use decimal (KB) or binary (KiB) prefixes.

本次模拟测试突出了九年级AQA计算机科学的基础主题：数制转换、逻辑、数据类型、基础编程和网络。为有效备考，需大量练习二进制与十六进制转换，熟记AND/OR/NOT真值表，并编写测试小程序。务必反复核查单位使用的是十进制（KB）还是二进制（KiB）前缀。

When facing a flowchart or algorithm trace, work through it step by step with a pencil and paper. For programming questions, remember the syntax rules of Python and comment your code where helpful.

在面对流程图或算法追踪时，用纸笔逐步走查。编程题中牢记Python语法规则，并在有帮助处添加注释。

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

更多咨询请联系16621398022（同微信）