Year 8 CAIE Computer Science: Unit Test Mock Paper Analysis | Year 8 CAIE 计算机：单元测试模拟卷解析

📚 Year 8 CAIE Computer Science: Unit Test Mock Paper Analysis | Year 8 CAIE 计算机：单元测试模拟卷解析

Welcome to this detailed walkthrough of a Year 8 CAIE Computer Science unit test mock paper. The questions have been carefully selected to reflect the most common topics in a typical end‑of‑unit assessment, such as binary conversions, data measurement, flowcharts, search algorithms, cybersecurity basics, logic gates, and hardware identification. Each section provides a question, a clear step‑by‑step solution, and a reinforcement of the underlying concept. Use this resource to check your understanding, correct mistakes, and build confidence before the real test.

欢迎阅读 Year 8 CAIE 计算机科学单元测试模拟卷的详细解析。本卷题目精选自典型单元测评中最常见的主题，包括二进制转换、数据计量、流程图、查找算法、网络安全基础、逻辑门以及硬件识别。每个小节都给出题目、逐步解答以及对核心概念的巩固。请利用这份资料检查自己的理解情况，纠正错误，并在真正考试前建立信心。

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1. Binary to Decimal Conversion | 二进制转十进制

Question: Convert the binary number 10110₂ to its decimal equivalent. Solution: Write down the place values using powers of 2 from right to left (2⁰, 2¹, 2², 2³, 2⁴). The digits of 10110₂ are 1, 0, 1, 1, 0. Multiply each digit by its place value and sum them: (1 × 2⁴) + (0 × 2³) + (1 × 2²) + (1 × 2¹) + (0 × 2⁰) = 16 + 0 + 4 + 2 + 0 = 22. Therefore, the decimal value is 22.

题目：将二进制数 10110₂ 转换为十进制。解答：从右向左列出 2 的幂次位值（2⁰, 2¹, 2², 2³, 2⁴）。10110₂ 各位数字依次为 1, 0, 1, 1, 0。每一位乘以对应位值再求和：(1 × 2⁴) + (0 × 2³) + (1 × 2²) + (1 × 2¹) + (0 × 2⁰) = 16 + 0 + 4 + 2 + 0 = 22。因此十进制值为 22。

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2. Decimal to Binary Conversion | 十进制转二进制

Question: Convert the decimal number 45 to binary. Solution: Use the repeated division‑by‑2 method, recording the remainder each time. 45 ÷ 2 = 22 remainder 1; 22 ÷ 2 = 11 remainder 0; 11 ÷ 2 = 5 remainder 1; 5 ÷ 2 = 2 remainder 1; 2 ÷ 2 = 1 remainder 0; 1 ÷ 2 = 0 remainder 1. Read the remainders from bottom to top: 101101₂. Always stop when the quotient becomes 0. Thus, 45 in binary is 101101₂.

题目：将十进制数 45 转换为二进制。解答：采用重复除以 2 取余法。45 ÷ 2 = 22 余 1； 22 ÷ 2 = 11 余 0； 11 ÷ 2 = 5 余 1； 5 ÷ 2 = 2 余 1； 2 ÷ 2 = 1 余 0； 1 ÷ 2 = 0 余 1。从下往上读取余数：101101₂。当商为 0 时停止。因此 45 的二进制形式为 101101₂。

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3. Binary Addition | 二进制加法

Question: Add the following binary numbers: 1011₂ + 1101₂. Solution: Align the numbers to the right and add column by column, carrying over when the sum is 2 or more. Rightmost column: 1 + 1 = 2, write 0, carry 1. Next column: 1 + 0 + carry 1 = 2, write 0, carry 1. Next: 0 + 1 + carry 1 = 2, write 0, carry 1. Leftmost: 1 + 1 + carry 1 = 3 (binary 11), write 1 and carry 1. Final carry becomes the new leftmost digit. Result: 11000₂. Thus, 1011₂ + 1101₂ = 11000₂.

题目：计算二进制加法：1011₂ + 1101₂。解答：将两数右对齐，逐列相加，和达到或超过 2 时进位。最右列：1 + 1 = 2，写 0 进 1。下一列：1 + 0 + 进 1 = 2，写 0 进 1。再下一列：0 + 1 + 进 1 = 2，写 0 进 1。最左列：1 + 1 + 进 1 = 3（二进制 11），写 1 并进 1。最后的进位成为新的最左位。结果：11000₂。因此 1011₂ + 1101₂ = 11000₂。

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4. Data Units Conversion | 数据单位转换

Question: A photograph file is exactly 3.5 megabytes (MB). How many kilobytes (KB) and how many bytes does this represent? Solution: Remember the standard relationships: 1 MB = 1024 KB, and 1 KB = 1024 bytes. First convert MB to KB: 3.5 × 1024 = 3584 KB. Next convert KB to bytes: 3584 × 1024 = 3,670,016 bytes. Therefore, 3.5 MB is equal to 3584 KB and 3,670,016 bytes.

题目：一张照片文件恰好为 3.5 兆字节（MB）。它相当于多少千字节（KB）以及多少字节？解答：请牢记标准换算关系：1 MB = 1024 KB，1 KB = 1024 字节。先将 MB 转为 KB：3.5 × 1024 = 3584 KB。再将 KB 转为字节：3584 × 1024 = 3,670,016 字节。因此，3.5 MB 等于 3584 KB 以及 3,670,016 字节。

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5. Flowchart Symbol Recognition | 流程图符号识别

Question: Identify the standard flowchart symbols described below and explain what each shape represents in an algorithm. (a) An oval or rounded rectangle at the very start and end. (b) A diamond containing a question with two outgoing arrows labelled ‘Yes’ and ‘No’. (c) A rectangle containing an instruction such as ‘total = total + 1’. Answer: (a) The oval/rounded rectangle is the Start/End terminator; it shows where the algorithm begins and finishes. (b) The diamond is a Decision symbol; it tests a condition and branches the flow accordingly. (c) The rectangle is a Process symbol; it performs a calculation or data manipulation step. All three are fundamental building blocks of flowcharts used in CAIE assessments.

题目：识别下面描述的标准流程图符号，并说明每种形状在算法中代表什么。(a) 算法开头和结尾处出现的椭圆形或圆角矩形。 (b) 包含一个问题的菱形，并带有标有“是”和“否”的两条出口箭头。 (c) 包含一条指令的矩形，例如 ‘total = total + 1’。解答：(a) 椭圆/圆角矩形是开始/结束符，表明算法开始和结束的位置。 (b) 菱形是判断符号，用于测试条件并据此分流。 (c) 矩形是处理符号，执行计算或数据处理步骤。这三种符号是 CAIE 考题中流程图的基石。

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6. Linear Search Algorithm | 线性搜索算法

Question: A list of numbers is [8, 15, 3, 22, 3, 9]. Using a linear search, how would you locate the first occurrence of the value 3? Describe each step and state the index where it is found (assume indexing starts at 0). Answer: A linear search checks each element in order. Step 1: Check index 0 (value 8) – not equal to 3. Step 2: Check index 1 (value 15) – not equal. Step 3: Check index 2 (value 3) – match found. The algorithm stops and returns index 2. Therefore, the first occurrence of 3 is at position 2.

题目：已知列表 [8, 15, 3, 22, 3, 9]。如何使用线性搜索找到数值 3 的第一个出现位置？请描述每一步，并指出它所在的索引（假设索引从 0 开始）。解答：线性搜索将按顺序逐个检查元素。第 1 步：检查索引 0（值 8）– 不等于 3。第 2 步：检查索引 1（值 15）– 不等于。第 3 步：检查索引 2（值 3）– 找到匹配。算法停止并返回索引 2。因此，3 第一次出现的位置是索引 2。

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7. Understanding Firewalls | 理解防火墙

Question: Explain what a firewall is and give one example of how it protects a school computer network. Answer: A firewall is a network security system, either hardware or software, that monitors incoming and outgoing traffic based on predefined security rules. It acts as a barrier between a trusted internal network and an untrusted external network (like the Internet). For example, a school’s firewall may block access to social media websites during lessons, preventing students from being distracted and stopping potentially harmful content from entering the network.

题目：解释什么是防火墙，并给出一个它在学校计算机网络中起保护作用的示例。解答：防火墙是一种网络安全系统，可以是硬件或软件，它能根据预定义的安全规则监控进出的网络流量。它相当于在可信内部网络与不可信外部网络（如互联网）之间设置了一道屏障。例如，学校的防火墙可以在上课期间屏蔽社交媒体网站，既能避免学生分心，又能阻止潜在有害内容进入网络。

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8. AND Gate Truth Table | 与门真值表

Question: Complete the truth table for a 2‑input AND gate. The inputs are labelled A and B, and the output is Q. Use the binary digits 0 and 1. Answer: An AND gate outputs 1 only when both inputs are 1. The completed truth table is shown below.

题目：填写一个两输入与门的真值表。输入标记为 A 和 B，输出为 Q。请使用二进制数字 0 和 1。解答：与门只有在两个输入均为 1 时才输出 1。完整的真值表如下所示。

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

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9. Input and Output Devices | 输入与输出设备

Question: Give two examples of input devices and two examples of output devices commonly found in a classroom, and briefly explain the job of each. Answer: Input devices send data into the computer. Examples: a keyboard (allows the user to type text and commands) and a mouse (allows pointing, clicking and selecting items on screen). Output devices display or present data from the computer. Examples: a monitor (displays text, images and video) and a printer (produces a physical paper copy of digital documents). Together they enable human‑computer interaction.

题目：请列举教室中常见的两种输入设备和两种输出设备，并简要说明每种设备的功能。解答：输入设备负责将数据送入计算机。例如：键盘（用于输入文字和命令）和鼠标（用于指向、点击和选择屏幕上的项目）。输出设备负责显示或呈现来自计算机的数据。例如：显示器（显示文字、图像和视频）和打印机（将数字文档打印成纸质副本）。它们共同实现了人机交互。

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10. Pseudocode: Counting Loop | 伪代码：计数循环

Question: Write a short pseudocode algorithm that uses a loop to output the numbers 1, 2, 3, 4, 5 each on a new line. Then explain how the loop works. Answer: The pseudocode could be:

FOR count ← 1 TO 5
    OUTPUT count
NEXT count

Explanation: The FOR loop initializes a variable ‘count’ to 1. Before each iteration, it checks whether ‘count’ is less than or equal to 5. If true, it executes the body (outputs the current value of ‘count’) and then increments ‘count’ by 1. This repeats until ‘count’ becomes 6, at which point the loop stops.

题目：写一段简短的伪代码算法，使用循环在新的一行输出数字 1 到 5。然后解释循环如何工作。解答：伪代码如下：

FOR count ← 1 TO 5
    OUTPUT count
NEXT count

解释：FOR 循环将变量 ‘count’ 初始化为 1。每次迭代前，它会检查 ‘count’ 是否小于等于 5。如果为真，则执行循环体（输出当前 ‘count’ 的值），然后将 ‘count’ 增加 1。这一过程持续进行，直到 ‘count’ 变为 6，此时循环停止。

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