Year 9 Edexcel Computer Science: Unit Test Mock Paper Analysis | 九年级爱德思计算机：单元测试模拟卷解析

📚 Year 9 Edexcel Computer Science: Unit Test Mock Paper Analysis | 九年级爱德思计算机：单元测试模拟卷解析

As Year 9 students approach their Edexcel Computer Science unit test, a thorough review of key concepts and typical exam-style questions becomes essential. This mock paper analysis breaks down common question types, model answers, and frequent mistakes, guiding you through the core topics that appear time and again. By working through these examples, you will sharpen your understanding of hardware, data representation, networks, programming, and computational thinking — all vital for success.

随着九年级学生迎来爱德思计算机科学单元测试，系统复习核心概念与典型试题显得至关重要。本文通过模拟卷解析，拆解常见题型、标准答案与易错点，带领你逐个击破反复出现的重要主题。研读这些例题，你将加深对硬件、数据表示、网络、编程和计算思维等关键知识的理解，为考试成功夯实基础。

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1. Computer Hardware and Storage | 计算机硬件与存储

Question: Describe the roles of the CPU, RAM, and a hard disk drive (HDD) in a computer system. (4 marks)

Model Answer: The CPU (Central Processing Unit) executes program instructions and processes data, acting as the 'brain' of the computer. RAM (Random Access Memory) temporarily holds the operating system, applications, and data that are currently in use, enabling fast access by the CPU; it is volatile, meaning its contents are lost when power is turned off. The hard disk drive provides long-term, non-volatile storage for the operating system, software, and user files, retaining data even when the computer is switched off. (1 mark per component plus 1 mark for correctly distinguishing between volatile and non-volatile memory.)

Common Pitfall: Many students mistakenly believe that RAM permanently stores their files. Always remember that RAM clears when the computer shuts down — saving work to the hard disk is necessary for it to persist.

题目：阐述 CPU、RAM 和硬盘驱动器（HDD）在计算机系统中的作用。（4 分）

标准答案：CPU（中央处理器）执行程序指令并处理数据，相当于计算机的“大脑”。RAM（随机存取存储器）临时存放当前正在使用的操作系统、应用程序和数据，供 CPU 快速访问；它是易失性存储器，断电后内容丢失。硬盘驱动器提供长期、非易失性存储，用于保存操作系统、软件和用户文件，即使关机后数据依然保留。（每正确描述一个组件得 1 分，若能准确区分易失性与非易失性存储再得 1 分。）

常见错误：许多学生误以为 RAM 永久保存文件。务必牢记，计算机关机时 RAM 清空——只有保存至硬盘的工作才能在下次开机后继续使用。

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2. Binary and Denary Conversion | 二进制与十进制转换

Question: Convert the denary number 157 into an 8-bit binary number. Show your working clearly. (3 marks)

Model Answer: We repeatedly divide by 2 and record the remainders:

Division	Quotient	Remainder
157 ÷ 2	78	1
78 ÷ 2	39	0
39 ÷ 2	19	1
19 ÷ 2	9	1
9 ÷ 2	4	1
4 ÷ 2	2	0
2 ÷ 2	1	0
1 ÷ 2	0	1

Reading the remainders from bottom to top gives 10011101. As an 8-bit representation, we can pad with a leading zero if needed, but 10011101 already uses 8 bits. So 157 in denary equals 10011101 in binary. (1 mark for method, 1 mark for correct remainders, 1 mark for correct final answer.)

Common Pitfall: Reading the remainders in the wrong order — remember to take the last remainder as the most significant bit.

题目：将十进制数 157 转换为 8 位二进制数。请清晰展示计算过程。（3 分）

标准答案：反复除以 2 并记录余数：

除法	商	余数
157 ÷ 2	78	1
78 ÷ 2	39	0
39 ÷ 2	19	1
19 ÷ 2	9	1
9 ÷ 2	4	1
4 ÷ 2	2	0
2 ÷ 2	1	0
1 ÷ 2	0	1

从下往上读取余数得到 10011101。作为 8 位表示，该数值已占满 8 位，无需补零。因此十进制 157 等于二进制 10011101。（方法 1 分，正确余数 1 分，最终答案正确 1 分。）

常见错误：以错误顺序读取余数——请记住，应将最后得到的余数作为最高有效位。

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

Question: Draw the truth table for a two-input AND gate. Then explain how an AND gate could be used in a car's safety system. (3 marks)

Model Answer: Truth table for AND gate:

Input A	Input B	Output Q (A AND B)
0	0	0
0	1	0
1	0	0
1	1	1

eatbelt buckle sensor (1 = fastened). The output of the AND gate connects to a warning light. The light only illuminates when the driver is seated AND the seatbelt is NOT fastened. By using a NOT gate after the buckle sensor, the AND gate receives a 1 only when the belt is unfastened. Thus the warning activates when both conditions (driver present AND belt unfastened) are true. (1 mark for truth table, 2 marks for plausible application with correct logic.)

题目：绘制双输入与门（AND gate）的真值表。然后解释如何在汽车安全系统中使用与门。（3 分）

标准答案：与门真值表：

输入 A	输入 B	输出 Q (A AND B)
0	0	0
0	1	0
1	0	0
1	1	1

在汽车中，与门可用于安全带警告系统。一个输入来自座椅传感器（1 = 驾驶员入座），另一个来自安全带卡扣传感器（1 = 已系好）。与门的输出连接警示灯。警示灯仅在驾驶员入座且安全带未系时亮起：将卡扣传感器的信号先通过非门（NOT gate）处理，只有未系时向与门送入 1，这样当驾驶员在座且未系安全带这两个条件同时满足时，警告才会激活。（真值表 1 分，逻辑正确的应用 2 分。）

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4. Data Representation: Text, Images, Sound | 数据表示：文本、图像与声音

Question: Explain how a black-and-white bitmap image is represented digitally. Your answer should refer to pixels, resolution, and file size. (4 marks)

Model Answer: for white, 1 for black. The resolution describes the number of pixels in each row and column, e.g., 800 × 600. Higher resolution means more detail but also a larger file size. The file size in bits can be estimated as width × height (in pixels), because each pixel uses 1 bit. Metadata such as the image dimensions and colour depth must also be stored, adding a small overhead. (1 mark for pixel/bit relationship, 1 mark for resolution, 1 mark for file size calculation, 1 mark for mentioning metadata.)

题目：解释黑白位图图像是如何用数字方式表示的。请提及像素、分辨率和文件大小。（4 分）

标准答案：位图图像由称为像素的小方格网格构成。在黑白（单色）图像中，每个像素可用一个比特表示：0 代表白色，1 代表黑色。分辨率描述每行和每列的像素数量，如 800 × 600。分辨率越高细节越丰富，但文件也更大。文件大小（比特）可估算为宽度 × 高度（像素数），因为每个像素占用 1 比特。此外还需存储元数据，如图像尺寸和色深，这会增加少量额外开销。（像素/比特关系 1 分，分辨率 1 分，文件大小计算 1 分，提及元数据 1 分。）

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5. Networks and Topologies | 网络与拓扑结构

Question: Compare a star network topology with a mesh topology. Give one advantage and one disadvantage of each. (4 marks)

Model Answer: In a star topology, all devices connect to a central switch or hub. Advantage: If one cable fails, only that device is affected, and the rest of the network continues working. Disadvantage: The central device is a single point of failure — if it fails, the entire network goes down. In a mesh topology, every device (or many devices) connects to several others, often providing multiple paths. Advantage: High reliability because if one connection breaks, data can take an alternative route. Disadvantage: Expensive to install and maintain due to the large number of cables and connections required. (1 mark per valid advantage/disadvantage, max 4.)

题目：比较星型拓扑和网状拓扑。各给出一个优点和一个缺点。（4 分）

标准答案：星型拓扑中，所有设备连接到中央交换机或集线器。优点：某根线缆出现故障只影响该设备，网络其余部分仍可正常工作。缺点：中央设备是单点故障源——一旦它失效，整个网络瘫痪。网状拓扑中，每个（或多数）设备与多个其他设备直接相连，通常提供冗余路径。优点：高可靠性，因为一条连接断掉时数据可走备用路由。缺点：因需要大量线缆和连接，安装维护成本高昂。（每个有效优点或缺点 1 分，满分 4 分。）

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6. Internet and Web Technologies | 互联网与网络技术

Question: State the function of HTTP, HTTPS, and DNS. Explain why HTTPS is particularly important for online shopping. (4 marks)

Model Answer: HTTP (Hypertext Transfer Protocol) is used to transfer web pages and resources between a web server and a browser. HTTPS is the secure version of HTTP; it encrypts data using SSL/TLS to protect against eavesdropping. DNS (Domain Name System) translates human-friendly domain names (e.g., www.example.com) into IP addresses that computers use to locate each other. HTTPS is crucial for online shopping because customers enter sensitive information such as credit card numbers and personal details. Encryption ensures that this data cannot be intercepted and read by attackers during transmission, protecting both the customer's privacy and the integrity of the transaction. (1 mark per function, 1 mark for applying HTTPS significance to e-commerce.)

题目：说明 HTTP、HTTPS 和 DNS 的功能。解释为什么 HTTPS 对网上购物尤为重要。（4 分）

标准答案：HTTP（超文本传输协议）用于在 Web 服务器与浏览器之间传输网页和资源。HTTPS 是 HTTP 的安全版本，利用 SSL/TLS 加密数据以防止窃听。DNS（域名系统）将人类易记的域名（如 www.example.com）转换为计算机用于相互定位的 IP 地址。HTTPS 对网上购物至关重要，因为顾客在此过程中输入信用卡号、个人资料等敏感信息。加密确保这些数据在传输途中不会被攻击者截获与读取，既保护了顾客隐私，也维护了交易的完整性。（每个功能 1 分，将 HTTPS 的重要性与电子商务结合 1 分。）

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7. Introduction to Programming (Python) | 编程入门（Python）

Question: Write a Python program that asks the user for their name and age, then prints a message: 'Hello [name], next year you will be [age+1].' Ensure you handle the age as an integer. (3 marks)

Model Answer:

name = input('Enter your name: ')
age = int(input('Enter your age: '))
next_age = age + 1
print('Hello ' + name + ', next year you will be ' + str(next_age) + '.')

The code uses input() to read strings, int() to convert the age string into an integer so arithmetic can be performed, and then str() to concatenate the integer back into the output message. (1 mark for correct input and variable assignment, 1 mark for integer casting and addition, 1 mark for correct print statement with concatenation.)

题目：编写一个 Python 程序，提示用户输入姓名和年龄，然后输出消息：“Hello [姓名], next year you will be [年龄+1]。”请确保将年龄作为整数处理。（3 分）

标准答案：

name = input('Enter your name: ')
age = int(input('Enter your age: '))
next_age = age + 1
print('Hello ' + name + ', next year you will be ' + str(next_age) + '.')

代码使用 input() 读取字符串，用 int() 将年龄字符串转换为整数以进行算术运算，然后用 str() 把整数转换回字符串以便与输出信息拼接。（正确输入与变量赋值 1 分，整数类型转换与加法运算 1 分，正确使用拼接的打印语句 1 分。）

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8. Algorithm and Pseudocode | 算法与伪代码

Question: Write pseudocode to find the largest number in a list of integers, e.g., [23, 45, 12, 67, 34]. (3 marks)

Model Answer:

FUNCTION findMax(list)
  max ← list[0]
  FOR i ← 1 TO LENGTH(list) - 1
    IF list[i] > max THEN
      max ← list[i]
    ENDIF
  ENDFOR
  RETURN max
ENDFUNCTION

The algorithm initialises max with the first element, then iterates through the list. Whenever an element larger than the current max is found, it updates max. This is a linear search approach. (1 mark for initialisation, 1 mark for correct loop structure, 1 mark for comparison and update logic.)

题目：编写伪代码，找出整数列表中的最大值，例如 [23, 45, 12, 67, 34]。（3 分）

标准答案：

FUNCTION findMax(list)
  max ← list[0]
  FOR i ← 1 TO LENGTH(list) - 1
    IF list[i] > max THEN
      max ← list[i]
    ENDIF
  ENDFOR
  RETURN max
ENDFUNCTION

算法先将 max 初始化为第一个元素，接着遍历列表。一旦发现比当前 max 更大的元素，就更新 max。这属于线性搜索策略。（初始化 1 分，循环结构正确 1 分，比较与更新逻辑 1 分。）

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9. Cybersecurity and Data Protection | 网络安全与数据保护

Question: Identify three types of malware and describe the damage each can cause. (6 marks)

Model Answer: 1. Virus: A self-replicating program that attaches itself to legitimate files and spreads when those files are shared. It can corrupt or delete data, slow down system performance, and cause software crashes. 2. Ransomware: Malware that encrypts the victim's files and demands a ransom payment for the decryption key. It can lead to permanent data loss and severe financial damage. 3. Spyware: Software that secretly monitors user activity, capturing keystrokes, login credentials, and personal information. Damage includes identity theft and unauthorised access to online accounts. (2 marks per malware type: 1 for correct identification, 1 for damage description.)

题目：列举三种恶意软件类型，并说明它们各自造成的危害。（6 分）

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