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

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

This walkthrough analyses a typical Year 9 CAIE Computer Science unit test mock paper, covering key topics such as binary and hexadecimal number systems, logic gates, hardware components, operating systems, network topologies, pseudocode, cybersecurity, image representation, and data processing. By working through each question with detailed bilingual explanations, you will strengthen your understanding of core concepts and develop effective exam techniques. Let's dive into the mock paper and unpack the answers step by step.

本文详细解析一份典型的 Year 9 CAIE 计算机科学单元测试模拟卷，涵盖了二进制与十六进制数制、逻辑门、硬件组件、操作系统、网络拓扑、伪代码、网络安全、图像表示和数据处理等关键主题。通过双语逐步讲解每道题目，你将巩固核心概念并掌握有效的考试技巧。下面就来逐一拆解这套模拟卷的答案吧。

1. Question 1: Binary to Denary Conversion | 第1题：二进制转十进制

Question: Convert the binary number 101101 to denary (decimal).

问题：将二进制数 101101 转换为十进制。

ar right. The binary digits 101101 correspond to: 1×2⁵ + 0×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰.

解题时，我们根据 2 的幂次给每一位分配权值，从最右边的 2⁰ 开始。二进制数 101101 对应的权值展开为：1×2⁵ + 0×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰。

ng these terms gives: 32 + 0 + 8 + 4 + 0 + 1 = 45. Therefore, the denary equivalent is 45.

计算各项：32 + 0 + 8 + 4 + 0 + 1 = 45。所以，等值的十进制数是 45。

instead of 2⁰. Always remember that the least significant bit (rightmost) is worth 2⁰ = 1.

常见的错误是权值错位，或者从 2¹ 开始计算。务必记住，最低有效位（最右边）的权值是 2⁰ = 1。

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2. Question 2: Hexadecimal to Binary and Denary | 第2题：十六进制转二进制和十进制

Question: The hexadecimal number 2F is used in a colour code. Convert 2F into an 8-bit binary number and then into denary.

问题：十六进制数 2F 用于颜色编码。请将 2F 转换为 8 位二进制数，再转换为十进制。

Each hex digit is represented by 4 bits. 2₁₆ is 0010 in binary, and F₁₆ (which is 15 in denary) is 1111. Therefore, 2F₁₆ becomes 0010 1111, or 00101111 as an 8-bit binary number.

每个十六进制数字用 4 位二进制表示。2₁₆ 对应 0010，F₁₆（十进制为 15）对应 1111。因此 2F₁₆ 转换为 0010 1111，即 8 位二进制数 00101111。

1×2³ + 1×2² + 1×2¹ + 1×2⁰. This simplifies to 32 + 8 + 4 + 2 + 1 = 47. Thus, 2F₁₆ = 47₁₀.

十进制转换同样使用权值法：0×2⁷ + 0×2⁶ + 1×2⁵ + 0×2⁴ + 1×2³ + 1×2² + 1×2¹ + 1×2⁰ = 32 + 8 + 4 + 2 + 1 = 47。所以 2F₁₆ = 47₁₀。

Always check that hex letters A–F represent the denary values 10–15. In this case, F = 15, which correctly converts to binary 1111.

务必记住十六进制字母 A–F 代表十进制数值 10–15。这里 F = 15，正确转换为二进制 1111。

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3. Question 3: Logic Gate Truth Tables | 第3题：逻辑门真值表

Question: Complete the truth tables for a two-input AND gate and a two-input OR gate. Then draw the symbol for a NOT gate and explain its function.

问题：完成一个两输入与门和一个两输入或门的真值表。然后画出非门的符号并解释其功能。

For the AND gate, the output is 1 only when both inputs A and B are 1. The OR gate outputs 1 when at least one input is 1. Here are the completed truth tables:

对于与门，只有当输入 A 和 B 都为 1 时，输出才为 1。或门在至少一个输入为 1 时输出 1。下面是完整的真值表：

A	B	AND Output	OR Output
0	0	0	0
0	1	0	1
1	0	0	1
1	1	1	1

A NOT gate has only one input and its output is the inverse of the input. Its symbol is a triangle with a small circle at the tip. If the input is 0, the NOT gate outputs 1, and vice versa.

非门只有一个输入端，其输出是输入的反相。它的符号是一个尖端带小圆圈的三角形。如果输入为 0，非门就输出 1，反之亦然。

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4. Question 4: Computer Hardware Identification | 第4题：计算机硬件识别

Question: The list contains several computer components: CPU, RAM, SSD, GPU, PSU. For each, state its main function and whether it is an input, output, processing, or storage device.

问题：下表列出了几个计算机组件：CPU、RAM、SSD、GPU、PSU。请说明每个组件的主要功能，并指出它是输入、输出、处理还是存储设备。

The CPU (Central Processing Unit) executes instructions and processes data; it is a processing device. RAM (Random Access Memory) holds data and instructions that are currently in use; it is primary storage (volatile). An SSD (Solid State Drive) provides long-term, non-volatile storage for files and software.

CPU（中央处理器）负责执行指令和处理数据，属于处理设备。RAM（随机存取存储器）保存当前正在使用的数据和指令，属于主存储器（易失性）。SSD（固态硬盘）为文件和软件提供长期、非易失性存储。

The GPU (Graphics Processing Unit) specialises in rendering images and video; it is an output device when sending signals to a display, but it also processes graphics data. The PSU (Power Supply Unit) converts mains electricity to the correct voltages for the computer’s components and thus is not classified as input, output, or storage – it is a support component providing power.

GPU（图形处理器）专门负责渲染图像和视频，向显示器发送信号时属于输出设备，但它也处理图形数据。PSU（电源供应单元）将市电转换为计算机组件所需的正确电压，因此不属于输入、输出或存储设备——它是提供电力的支持组件。

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5. Question 5: Operating Systems | 第5题：操作系统

Question: Define an operating system and give two examples. Explain one task it performs in managing hardware resources.

问题：定义操作系统并给出两个例子。解释它在管理硬件资源方面执行的一项任务。

An operating system (OS) is system software that manages computer hardware, runs applications, and provides a user interface. It acts as an intermediary between the user and the hardware. Common examples include Microsoft Windows and Linux.

操作系统 (OS) 是管理计算机硬件、运行应用程序并提供用户界面的系统软件。它作为用户与硬件之间的中介。常见的例子有 Microsoft Windows 和 Linux。

One crucial hardware management task is memory management. The OS allocates RAM space to different programs, keeps track of what memory is free, and ensures that one process does not overwrite another’s data. It also handles virtual memory when physical RAM is full.

一项关键的硬件管理任务是内存管理。操作系统为不同程序分配 RAM 空间，记录哪些内存是空闲的，并确保一个进程不会覆盖另一个进程的数据。当物理 RAM 写满时，它还会处理虚拟内存。

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6. Question 6: Network Topologies | 第6题：网络拓扑

Question: Compare the star and bus network topologies. State one advantage and one disadvantage of each.

问题：比较星型拓扑和总线型拓扑。分别说明每种拓扑的一个优点和一个缺点。

In a star topology, all devices connect to a central switch or hub. An advantage is that if one cable fails, only that device is affected; the rest of the network remains operational. However, a disadvantage is the dependency on the central node – if the central switch fails, the entire network goes down.

在星型拓扑中，所有设备都连接到一个中央交换机或集线器。优点是一根电缆发生故障时只影响该设备，网络其余部分仍可运行。缺点是依赖于中心节点，如果中央交换机故障，整个网络就会瘫痪。

A bus topology has all devices connected along a single backbone cable. It is cheaper to install because less cable is needed. On the downside, if the backbone cable breaks, the whole network fails, and performance degrades as more devices are added due to data collisions.

总线型拓扑中，所有设备沿单一主干电缆连接。它的安装成本较低，因为需要的电缆较少。缺点是如果主干电缆断裂，整个网络就会崩溃，并且随着设备增多，数据碰撞会导致性能下降。

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7. Question 7: Pseudocode Loop | 第7题：伪代码循环

Question: Write pseudocode that uses a loop to output the numbers from 1 to 10. Then modify it to output only the even numbers.

问题：编写一段伪代码，使用循环输出数字 1 到 10。然后修改它以只输出偶数。

Using a FOR loop, the simplest solution is:

FOR i ← 1 TO 10

 OUTPUT i

ENDFOR

使用 FOR 循环，最简单的方案是：

FOR i ← 1 TO 10

 OUTPUT i

ENDFOR

To output only even numbers, we can use an IF statement inside the loop to check when i MOD 2 equals 0. The modified pseudocode becomes:

FOR i ← 1 TO 10

 IF i MOD 2 = 0 THEN

  OUTPUT i

 ENDIF

ENDFOR

要只输出偶数，我们可以在循环内使用 IF 语句检查 i MOD 2 是否等于 0。修改后的伪代码如下：

FOR i ← 1 TO 10

 IF i MOD 2 = 0 THEN

  OUTPUT i

 ENDIF

ENDFOR

Note the use of the assignment arrow ← and the MOD operator for remainder. Pseudocode must be clear and language-agnostic; indentation shows the block structure.

注意使用了赋值箭头 ← 和求余运算符 MOD。伪代码必须清晰且与语言无关；缩进用于展示代码块结构。

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8. Question 8: Phishing Attacks | 第8题：网络钓鱼攻击

Question: Explain what a phishing attack is and suggest two ways users can protect themselves against such threats.

问题：解释什么是钓鱼攻击，并提出两种用户保护自己免受此类威胁的方法。

Phishing is a social engineering attack where criminals send deceptive emails or messages pretending to be from reputable organisations to trick recipients into revealing sensitive information, such as passwords or credit card numbers. These messages often contain urgent language and links to fake websites that look genuine.

钓鱼攻击是一种社会工程攻击，犯罪分子发送伪装成来自信誉良好机构的欺骗性邮件或消息，诱骗收件人泄露密码或信用卡号等敏感信息。这些消息通常带有紧迫性语言和指向看似真实的虚假网站的链接。

To stay safe, users should first verify the sender’s email address and avoid clicking on suspicious links. Instead, they can manually type the known website address into the browser. Second, enable two-factor authentication (2FA) on important accounts, which adds an extra layer of security even if a password is compromised.

安全防护方面，用户首先应核实发件人的邮箱地址，避免点击可疑链接，可以手动在浏览器中输入已知的网站地址。其次，在重要账户上启用双重认证 (2FA)，这样即使密码泄露也能增加一层额外的安全保障。

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9. Question 9: Image Resolution and Colour Depth | 第9题：图像分辨率与色深

pth of 24 bits per pixel. Calculate the file size in kilobytes (KB), stating any assumptions. Explain how reducing the resolution or colour depth affects the file size and image quality.

问题：一幅位图图像的分辨率为 800 × 600 像素，色深为每像素 24 位。计算其文件大小（以千字节 KB 为单位），并说明假设条件。解释降低分辨率或色深对文件大小和图像质量的影响。

KB = 1024 bytes, file size ≈ 1,440,000 ÷ 1024 ≈ 1406.25 KB. (Assumption: no compression.)

首先将比特转换为字节：1 字节 = 8 比特。总比特数 = 800 × 600 × 24 = 11,520,000 比特。除以 8 得到 1,440,000 字节。因 1 KB = 1024 字节，文件大小 ≈ 1,440,000 ÷ 1024 ≈ 1406.25 KB。（假设无压缩。）

Reducing the resolution decreases the number of pixels, so fewer data are stored and the file size shrinks. However, the image becomes less detailed and may appear pixelated when enlarged. Reducing colour depth lowers the number of bits per pixel, making the file smaller but also reducing the range of colours that can be displayed, which can cause banding or posterisation.

降低分辨率会减少像素数量，因此存储的数据变少，文件缩小，但图像细节减少，放大时可能出现像素化。降低色深则减少每个像素的比特数，文件同样变小，但可显示的颜色范围变窄，可能导致色带或色调分离。

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10. Question 10: Data Processing Cycle | 第10题：数据处理循环

Question: A supermarket barcode scanner is used at the checkout. Describe the stages of the data processing cycle involved, from scanning to displaying the price, and name the hardware components used.

问题：超市收银台使用条码扫描器。描述从扫描到显示价格所涉及的数据处理循环各阶段，并说出所用到的硬件组件。

The data processing cycle has four main stages: input, processing, output, and storage. When a barcode is scanned, the input stage occurs – the barcode scanner (an input device) reads the barcode and sends the numeric code to the computer.

数据处理循环有四个主要阶段：输入、处理、输出和存储。扫描条码时，输入阶段开始——条码扫描器（输入设备）读取条码并将数字编码发送给计算机。

During processing, the CPU uses the code to look up the product details and price in a database stored on a hard drive or SSD (storage). The output stage then displays the product name and price on a monitor (output device) and may print a receipt using a receipt printer. Storage is also used to update inventory records after the sale.

在处理阶段，CPU 利用该编码在硬盘或 SSD（存储器）上存储的数据库中查找产品详情和价格。输出阶段则将产品名称和价格显示在显示器（输出设备）上，并可能通过收据打印机打印小票。销售完成后，存储还会用于更新库存记录。

This cycle illustrates how input, processing, output, and storage components work together seamlessly in a point-of-sale system, a typical example of a real-world computer application.

这个循环展示了输入、处理、输出和存储组件如何在一个销售点系统中无缝协作，是真实世界计算机应用的典型例子。

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

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