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

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

Welcome to our detailed walkthrough of a mock unit test for Year 8 Edexcel Computer Science. This article presents a series of typical questions you might meet in your assessment, together with clear explanations and key knowledge points. Use these sample solutions to reinforce your understanding and to build confidence for the real test.

欢迎阅读 Year 8 Edexcel 计算机科学单元测试模拟卷解析。本文将展示一系列在评估中可能遇到的典型题目，并配以清晰的解释和核心知识点。请利用这些样题解答来巩固理解，建立迎接真实考试的信心。

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

Question 1: Convert the binary number 1101₂ into denary. To do this, write the place values as powers of 2: 8, 4, 2, 1. Then multiply each binary digit by its place value: (1 x 8) + (1 x 4) + (0 x 2) + (1 x 1) = 8 + 4 + 0 + 1 = 13. The answer is 13.

问题 1：将二进制数 1101₂ 转换为十进制。首先写出以 2 为底的位权：8, 4, 2, 1。然后将每个二进制位乘以对应位权：(1 x 8) + (1 x 4) + (0 x 2) + (1 x 1) = 8 + 4 + 0 + 1 = 13。答案是 13。

Question 2: Convert the denary number 29 into an 8-bit binary number. One method is repeated division by 2: 29 / 2 = 14 remainder 1, 14/2 = 7 remainder 0, 7/2 = 3 remainder 1, 3/2 = 1 remainder 1, 1/2 = 0 remainder 1. Reading the remainders upwards gives 11101. As an 8-bit number with leading zeros this is 00011101₂.

问题 2：将十进制数 29 转换为 8 位二进制数。一种方法是用 2 反复取余：29 / 2 = 14 余 1，14/2 = 7 余 0，7/2 = 3 余 1，3/2 = 1 余 1，1/2 = 0 余 1。从下往上读取余数得到 11101。补足 8 位并加前导零，结果为 00011101₂。

These conversions are essential because computers store all data as binary digits (bits). Practise by picking random numbers and converting both ways.

这些转换至关重要，因为计算机将所有数据都以二进制位（比特）的形式存储。建议随机选取数字并反复练习双向转换。

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2. Binary Addition | 二进制加法

Question: Add the two 4-bit binary numbers 0101₂ and 0011₂. Line them up and add column by column from the right, remembering that 1 + 1 = 0 with a carry of 1 to the next column. So: 1 + 1 = 0 carry 1; (carry 1) + 0 + 1 = 0 carry 1; (carry 1) + 1 + 0 = 0 carry 1; (carry 1) + 0 + 0 = 1. Result: 1000₂.

问题：计算两个 4 位二进制数 0101₂ 和 0011₂ 的加法。将它们上下对齐，从右向左逐列相加，谨记 1 + 1 = 0 并向下一列进位 1。计算过程：1 + 1 = 0 进位 1；(进位 1) + 0 + 1 = 0 进位 1；(进位 1) + 1 + 0 = 0 进位 1；(进位 1) + 0 + 0 = 1。结果为 1000₂。

To check, convert to denary: 0101₂ is 5, 0011₂ is 3, 5 + 3 = 8, and 1000₂ is indeed 8. In computers an overflow can occur if the result needs more bits than available; here 4 bits can hold the answer without overflow.

验证方法：转换为十进制，0101₂ 为 5，0011₂ 为 3，5 + 3 = 8，而 1000₂ 正是 8。在计算机中，若结果所需位数超出可用位数就会发生溢出；此处 4 位可以容纳答案而无溢出。

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3. Image Representation and Pixels | 图像表示与像素

Question: Explain how a bitmap image is represented on a computer screen. A bitmap consists of a grid of tiny squares called pixels (picture elements). Each pixel is assigned a binary number that represents its colour. The number of bits used per pixel, known as colour depth, determines how many distinct colours can be displayed. For example, a 1-bit depth gives 2 colours (black and white), 8-bit depth gives 256 colours, and 24-bit depth allows around 16 million colours. Higher colour depth and larger grid size (resolution) produce a more detailed image but require more storage space.

问题：解释位图图像如何在计算机屏幕上呈现。位图由一个称为像素（图像元素）的小方格组成的网格构成。每个像素被赋予一个二进制数表示其颜色。每像素使用的位数称为颜色深度，它决定了可显示的不同颜色数量。例如，1 位色深可显示 2 种颜色（黑与白），8 位色深提供 256 种颜色，24 位色深则能呈现约 1600 万种颜色。更高的色深和更大的网格尺寸（分辨率）能产生更精细的图像，但需要更多的存储空间。

When storing a bitmap, the computer also saves metadata such as width, height and colour depth. This is why a high-quality photograph takes up many megabytes. Understanding pixels helps you grasp why images can become blurry when scaled up – the fixed grid of pixels becomes stretched.

存储位图时，计算机还会保存宽度、高度及色深等元数据。这便是高质量照片可能占据好几兆字节的原因。理解像素有助于明白为何图像放大后变得模糊——固定的像素网格被拉伸，细节便会丢失。

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4. Flowchart Symbols and Algorithms | 流程图符号与算法

Question: Identify the standard flowchart symbol used for a decision and describe its purpose. The decision symbol is a diamond shape. It represents a point in an algorithm where a condition is tested, typically yielding a ‘Yes’ or ‘No’ branch. For example, in a program that asks for a password, a diamond would check ‘Is the correct password entered?’. If yes, the flow continues to a process; if no, it may loop back to ask again. Other common symbols include an oval for Start/End, a rectangle for a process, and a parallelogram for input/output.

问题：识别用于决策的标准流程图符号并说明其用途。决策符号是菱形。它表示算法中测试条件的分支点，通常产生 “是” 或 “否” 两条路径。例如，在一个询问密码的程序中，菱形会检查 “输入了正确的密码吗？”。若是，流程继续到处理步骤；若否，可能返回要求重新输入。其他常用符号包括：椭圆形表示开始/结束，矩形表示处理过程，平行四边形表示输入/输出。

Flowcharts allow programmers to plan the logic visually before writing code. You should be able to draw and read simple flowcharts for algorithms that include sequence, selection (decisions) and iteration (loops).

流程图使程序员能够在编写代码之前直观地规划逻辑。你应当能够绘制和阅读包含顺序、选择（决策）和迭代（循环）的简单算法流程图。

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5. Pseudocode and Variables | 伪代码与变量

Question: Write pseudocode for a program that asks the user to enter two numbers, adds them together, and displays the total. Use meaningful variable names. One possible solution:

问题：编写伪代码，要求程序让用户输入两个数字，将它们相加并显示总和。使用有意义的变量名。一种可能的解答如下：

INPUT num1
INPUT num2
total = num1 + num2
OUTPUT total

输入 num1
输入 num2
total = num1 + num2
输出 total

In this pseudocode, ‘num1’ and ‘num2’ are variables that store the user’s input. The assignment statement ‘total = num1 + num2’ calculates the sum and stores it in the variable ‘total’. Finally, OUTPUT displays the result. Pseudocode is not a real programming language but a human-readable way of describing an algorithm. It usually uses indentation to show structure and keywords like IF…THEN…ELSE, FOR, WHILE.

在这段伪代码中，’num1′ 和 ‘num2’ 是存储用户输入的变量。赋值语句 ‘total = num1 + num2’ 计算和并将其存入变量 ‘total’。最后，OUTPUT 显示结果。伪代码并非真实的编程语言，而是一种便于人们阅读的算法描述方式。它通常使用缩进表示结构，并使用 IF…THEN…ELSE、FOR、WHILE 等关键字。

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6. CPU and RAM Functions | CPU 与内存功能

Question: State the main function of the Central Processing Unit (CPU) and explain why Random Access Memory (RAM) is essential in a computer system. The CPU is the ‘brain’ of the computer that carries out instructions from programs. It performs the fetch-decode-execute cycle: it fetches an instruction from memory, decodes it to understand what operation to perform, and then executes it using the Arithmetic Logic Unit (ALU). RAM is the main volatile memory that temporarily holds the operating system, application programs and data currently in use. It is much faster than a hard drive, allowing the CPU to access data and instructions quickly. Without enough RAM, the computer would need to swap data to slower storage, causing slowdowns. RAM loses its contents when the power is turned off.

问题：说明中央处理器（CPU）的主要功能，并解释为何随机存取存储器（RAM）在计算机系统中必不可少。CPU 是计算机的 “大脑”，负责执行程序指令。它执行取指-解码-执行周期：从内存取出指令，解码以了解要执行的操作，然后利用算术逻辑单元（ALU）执行。RAM 是易失性主存储器，临时存放操作系统、应用程序及当前使用的数据。它的速度远快于硬盘，使 CPU 能够快速访问数据和指令。若 RAM 不足，计算机需频繁与较慢的存储设备交换数据，导致运行缓慢。断电后 RAM 的内容会丢失。

Exam questions often ask you to compare RAM with ROM (Read-Only Memory). Remember, ROM is non-volatile and stores firmware/boot instructions, while RAM is volatile and used for active tasks.

考题常要求比较 RAM 与 ROM（只读存储器）。记住，ROM 是非易失性的，存储固件/启动指令，而 RAM 是易失性的，用于活动任务。

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7. Input and Output Devices | 输入与输出设备

Question: Give two examples of input devices and two examples of output devices. Also explain why a touchscreen is often described as both an input and output device. Input devices send data into the computer: keyboard, mouse, microphone, scanner. Output devices present data from the computer to the user: monitor, printer, speakers, headphones. A touchscreen acts as an input device because it senses the user’s touch (position, gestures). At the same time it is an output device because it displays visual information on the screen. Combining input and output in one unit saves space and makes interfaces more intuitive, which is why smartphones and tablets use them.

问题：举出两种输入设备和两种输出设备的例子。并解释为什么触摸屏常被形容为既输入又输出的设备。输入设备将数据送入计算机：键盘、鼠标、麦克风、扫描仪。输出设备将计算机中的数据呈现给用户：显示器、打印机、扬声器、耳机。触摸屏充当输入设备是因为它能感应用户的触摸（位置、手势）；同时它也是输出设备，因为它在屏幕上显示视觉信息。将输入与输出集于一身节省了空间并使界面更加直观，因此智能手机和平板电脑广泛采用触摸屏。

You may also encounter questions about sensors (input) and actuators (output) in control systems. Be ready to classify devices correctly.

你还可能遇到有关控制系统中传感器（输入）和执行器（输出）的问题。务必能正确分类设备。

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8. LAN vs Internet | 局域网与互联网

Question: Compare a Local Area Network (LAN) with the Internet. A LAN is a network that connects computers within a small geographical area, such as a school or an office building. It is typically owned, set up and managed by a single organisation. The Internet, on the other hand, is a global network of interconnected networks that links millions of private, public, academic and government networks. No single entity owns the Internet; it relies on shared protocols (TCP/IP) and infrastructure. A LAN offers high data transfer speeds, low latency, and is often more secure because it is private. The Internet covers the entire planet but speeds can vary, and extra security measures like firewalls are needed.

问题：比较局域网（LAN）与互联网。局域网是连接小地理范围内（如学校或办公楼）计算机的网络。它通常由单一组织拥有、搭建和管理。互联网则是连接全球数百万私有、公共、学术及政府网络的网络之网络。没有任何实体拥有互联网；它依赖共享协议（TCP/IP）和基础设施。局域网提供高数据传输速率、低延迟，且因其私密性往往更安全。互联网覆盖全球，但速度可能不一，且需要防火墙等额外的安全措施。

To help memorise differences, a table can be useful:

为了方便记忆，可以使用对比表格：

Feature	LAN	Internet
Geographic coverage	Small (building/campus)	Global
Ownership	Single organisation	No central owner
Typical hardware	Switches, cables, Wi-Fi access points	Routers, undersea cables, satellites
Security	Easier to secure; private	Public; requires encryption and firewalls

In your exam, you might be asked to state benefits of networking, such as sharing files, printers and internet connections. These concepts build on the LAN/Internet distinction.

考试中可能要求你陈述网络的好处，如共享文件、打印机和互联网连接。这些概念都基于局域网与互联网的区别。

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