📚 Year 10 Eduqas Computer Science: Unit Test Practice Paper Walkthrough | 十年级Eduqas计算机单元测试模拟卷解析
Welcome to our detailed walkthrough of a typical Year 10 Eduqas GCSE Computer Science unit test. This practice paper is designed to help you consolidate the key concepts from both Unit 1 (Understanding Computer Science) and Unit 2 (Computational Thinking and Programming). Each section below presents a mock exam question, followed by a step-by-step explanation, common pitfalls, and revision tips. By working through these examples, you will sharpen your problem-solving skills and feel much more confident when sitting your real assessment.
欢迎阅读我们为十年级Eduqas GCSE计算机科学单元测试准备的详细模拟卷解析。这份练习卷旨在帮助你巩固第一单元（理解计算机科学）和第二单元（计算思维与编程）的核心概念。以下每个小节都展示了一道模拟考题，并附有逐步解析、常见错误和复习建议。通过逐一练习这些例题，你将提升解题技巧，在真正面对考试时也会更加自信。
1. Number Systems – Binary, Denary and Hexadecimal Conversions | 数制系统：二进制、十进制与十六进制转换
Mock question: 'Convert the denary number 182 into an 8-bit binary number, and then convert this binary value into hexadecimal. Show all your working.'
模拟题：“将十进制数 182 转换为 8 位二进制数，再将此二进制值转换为十六进制。请展示全部计算步骤。”
32 leaves 22, subtract 16 leaves 6, and then 6 can be made of 4 + 2. So we set a 1 under 128, 32, 16, 4, and 2; all other positions get 0. This gives us the binary pattern 10110110₂.
我们先找出哪些 2 的幂相加等于 182。一个 8 位字节的位置值分别是 128、64、32、16、8、4、2、1。182 − 128 = 54，再从 54 中取出 32 剩下 22，减去 16 剩下 6，而 6 正好由 4 + 2 组成。因此在 128、32、16、4 和 2 的位置填 1，其余位置填 0，得到的二进制数为 10110110₂。
To convert this binary number to hexadecimal, we split it into two nibbles (groups of four bits): 1011 and 0110. The nibble 1011 corresponds to denary 11, which is the hexadecimal digit B. The nibble 0110 corresponds to denary 6, which remains 6 in hex. Therefore, 182 in hexadecimal is B6₁₆.
将二进制数转换为十六进制时，我们先将其分为两个四位组（半字节）：1011 和 0110。半字节 1011 对应的十进制值为 11，即十六进制数字 B；0110 对应的十进制值为 6，十六进制仍为 6。因此，182 的十六进制表示为 B6₁₆。
182₁₀ = 10110110₂ = B6₁₆
A typical mistake is to misalign the nibbles or to start with the least significant bit group incorrectly. Always group from the right for hexadecimal conversion, and make sure you pad the leftmost nibble with leading zeros if needed.
一个常见错误是四位分组时没有从右向左对齐，或者忘记在最左侧补充前导零。进行十六进制转换时一定要从最低有效位开始分组，并在必要时用零补足最左边的半字节。
2. Logic Gates and Truth Tables | 逻辑门与真值表
Mock question: 'Complete the truth table for the logic gate combination that represents NOT (A OR B). Then identify the equivalent single gate.'
模拟题：“补全逻辑门组合 NOT (A OR B) 的真值表，并写出与之等效的单一逻辑门。”
An OR gate outputs 1 when at least one input is 1. By placing a NOT gate after the OR gate, we invert the result. This combination is exactly the NOR gate. The truth table for two inputs A and B needs four rows. We first compute A OR B, then apply NOT.
或门在至少一个输入为 1 时输出 1。在或门后加上一个非门，我们就把输出反转了。这个组合恰好就是或非门（NOR）。两个输入 A 和 B 的真值表共有四行，我们先计算 A OR B，再对其取反。
| A | B | A OR B | NOT(A OR B) |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 |
The final column shows that the output is 1 only when both inputs are 0. This is the defining behaviour of the NOR gate. Therefore, the circuit is equivalent to a single NOR gate.
从最后一列可以看到，只有当两个输入都为 0 时输出才为 1，这正是或非门的特征。因此，该电路等效于一个单独的或非门。
When drawing truth tables in the exam, always list all input combinations in ascending binary order: 00, 01, 10, 11. This prevents missing a row and makes it easier for the examiner to follow your reasoning.
考试中绘制真值表时，请务必按照二进制递增顺序列出所有输入组合：00、01、10、11。这样可以避免遗漏行数，也便于阅卷人理解你的推理过程。
3. Sorting Algorithms – Bubble Sort Walkthrough | 排序算法：冒泡排序步骤解析
Mock question: ' is to be sorted into ascending order using bubble sort. Show how the array changes after each complete pass.'
模拟题：“数组 numbers = [23, 15, 42, 8, 16] 需要使用冒泡排序按升序排列。请展示每一趟完整遍历后数组的变化。”
Bubble sort works by repeatedly stepping through the list, comparing adjacent items and swapping them if they are in the wrong order. The pass finishes when we reach the end of the list. After each pass, the largest unsorted element 'bubbles' to its correct position at the end.
冒泡排序通过反复遍历列表、比较相邻元素并在顺序错误时交换它们来工作。当走到列表末端时一趟遍历结束。每完成一趟，当前未排序部分的最大元素就会“冒泡”到正确的末尾位置。
swap → [15, 23, 8, 42, 16]; compare 42 and 16 → swap → [15, 23, 8, 16, 42]. End of pass 1.
初始数组：[23, 15, 42, 8, 16]。第一趟：比较 23 和 15 → 交换 → [15, 23, 42, 8, 16]；比较 23 和 42 → 不交换；比较 42 和 8 → 交换 → [15, 23, 8, 42, 16]；比较 42 和 16 → 交换 → [15, 23, 8, 16, 42]。第一趟结束。
16, 42], 23 and 16 (swap) → [15, 8, 16, 23, 42]. The last element 42 is already in place.
第二趟：[15, 23, 8, 16, 42] → 比较 15 和 23（不交换），23 和 8（交换）→ [15, 8, 23, 16, 42]，23 和 16（交换）→ [15, 8, 16, 23, 42]。最后一个元素 42 已经就绪。
nd 8 (swap) → [8, 15, 16, 23, 42]; 15 and 16 (no swap); 16 and 23 (no swap). Array sorted.
第三趟：[15, 8, 16, 23, 42] → 比较 15 和 8（交换）→ [8, 15, 16, 23, 42]；15 和 16（不交换）；16 和 23（不交换）。数组完成排序。
Some students forget to reduce the range of comparison after each pass, or they keep swapping unnecessarily. Mentioning that the sorted portion grows from the right earns marks for showing algorithmic understanding.
有些学生忘记每趟之后应缩小比较范围，或者继续进行不必要的交换。如果能指出已排序部分从右端逐步扩大，可以展示你对算法原理的理解，从而获得相应分数。
4. Data Structures – Arrays and Indexing | 数据结构：数组与索引
Mock question: 'Consider the Python list names = ['Alice', 'Bob', 'Charlie', 'Dana']. What is the value of names[2]? Explain how indexing works and why your answer is correct.'
模拟题：“考虑 Python 列表 names = ['Alice', 'Bob', 'Charlie', 'Dana']。请问 names[2] 的值是什么？解释索引的运作方式，并说明为什么你的答案是正确的。”
In Python, lists are zero-indexed. This means the first item is at position 0, the second at position 1, and so on. Therefore, names[0] is 'Alice', names[1] is 'Bob', names[2] is 'Charlie', and names[3] is 'Dana'. The index 2 refers to the third element in the list, which is 'Charlie'.
在 Python 中，列表采用从零开始的索引方式，即第一个元素位于位置 0，第二个在位置 1，以此类推。因此 names[0] 是 'Alice'，names[1] 是 'Bob'，names[2] 是 'Charlie'，names[3] 是 'Dana'。索引 2 指向列表中的第三个元素，即 'Charlie'。
A common mistake is to assume indexing begins at 1, which would incorrectly give 'Bob' as the answer. Always check whether the question uses a zero-based or one-based convention. In most programming languages and all Eduqas pseudocode references, indexing starts at 0 unless stated otherwise.
一个常见错误是以为索引从 1 开始，这样就会错误地认为答案是 'Bob'。务必先确认题目采用的是从零计数还是从一开始的约定。在绝大多数编程语言以及所有 Eduqas 伪代码中，除非另有说明，索引均从 0 开始。
5. The CPU – Fetch-Decode-Execute Cycle | 中央处理器：取指-译码-执行周期
Mock question: 'Describe the fetch-decode-execute cycle and explain the role of the Program Counter (PC) and the Memory Address Register (MAR) within it.'
模拟题：“描述取指-译码-执行周期，并解释程序计数器 (PC) 和内存地址寄存器 (MAR) 在其中扮演的角色。”
The fetch-decode-execute cycle is the fundamental process by which the CPU executes instructions stored in main memory. In the fetch stage, the Program Counter holds the address of the next instruction to be fetched. This address is copied into the MAR, which then sends it along the address bus to memory. The instruction is retrieved and placed in the Memory Data Register (MDR). The PC then increments to point to the next instruction.
取指-译码-执行周期是 CPU 执行存放在主存中的指令的基本过程。在取指阶段，程序计数器保存着下一个待取指令的地址。该地址被复制到内存地址寄存器 (MAR)，然后 MAR 通过地址总线将地址发往内存。取出的指令放入内存数据寄存器 (MDR)，随后程序计数器自增，指向下一条指令。
During the decode stage, the Control Unit (CU) interprets the instruction fetched. In the execute stage, the CU signals other components – such as the Arithmetic Logic Unit (ALU) or memory – to carry out the operation. Throughout the cycle, the PC always holds the address of the instruction that will be fetched next, ensuring sequential execution unless a jump occurs.
在译码阶段，控制单元 (CU) 对取来的指令进行解释。在执行阶段，控制单元发信号给其他部件——如算术逻辑单元 (ALU) 或内存——以执行具体操作。在整个周期中，PC 始终保存着即将被取出的下一条指令的地址，从而保证指令的顺序执行，除非发生跳转。
When answering this type of question, using precise register names and the address/data bus terminology is essential. A sketch or a clear sequence of steps will help demonstrate your understanding of how data moves between components.
回答这类题目时，准确使用寄存器名称以及地址/数据总线的术语非常关键。画一张简图或列出清晰的步骤顺序，都有助于展示你对数据在部件之间流动过程的理解。
6. Networks – IP Addressing and Protocols | 网络：IP地址与协议
Mock question: 'State two differences between IPv4 and IPv6. Give an example of each address.'
模拟题：“说明 IPv4 和 IPv6 之间的两个区别，并分别给出一个地址示例。”
IPv4 uses 32-bit addresses, typically written as four decimal octets separated by dots, for example 192.168.1.10. IPv6 uses 128-bit addresses, expressed as eight groups of four hexadecimal digits separated by colons, such as 2001:0db8:0000:0042:0000:8a2e:0370:7334. The larger address space of IPv6 solves the IPv4 address exhaustion problem and allows for many more unique addresses.
IPv4 使用 32 位地址，通常写作由点号分隔的四个十进制八位组，例如 192.168.1.10。IPv6 使用 128 位地址，表示为冒号分隔的八组四位十六进制数字，例如 2001:0db8:0000:0042:0000:8a2e:0370:7334。IPv6 拥有更大的地址空间，解决了 IPv4 地址枯竭的问题，能够提供数量极多的唯一地址。
Another key difference is that IPv6 has built-in support for IPsec (Internet Protocol Security), which provides better authentication and encryption. In contrast, IPv4 requires additional configuration for similar security features. IPv6 also simplifies the header format, allowing routers to process packets more efficiently.
另一个重要区别是 IPv6 内置了对 IPsec（互联网协议安全）的支持，提供了更好的身份验证与加密能力，而 IPv4 则需要额外配置才能实现类似的安全特性。此外，IPv6 简化了报头格式，使路由器能够更高效地处理数据包。
For exam answers, memorising one realistic example for each address type and being able to discuss address length and security are sufficient to gain full marks.
在考试回答中，只要能记住每种地址类型的一个代表性示例，并能讨论地址长度与安全性差异，就足以获得满分。
7. Software Classification – System vs Application Software | 软件分类：系统软件与应用软件
Mock question: 'Distinguish between system software and application software. Provide two examples of each.'
模拟题：“区分系统软件和应用软件，并各举出两个例子。”
System software is designed to manage and control the computer hardware, providing a platform for other software to run. Examples include the operating system (such as Windows, Linux) and utility software (such as antivirus or disk defragmenter). Application software, on the other hand, helps users perform specific tasks that are not directly related to managing the computer itself. Examples are word processors (like Microsoft Word) and web browsers (like Google Chrome).
系统软件用于管理和控制计算机硬件，为其他软件的运行提供平台。例如，操作系统（如 Windows、Linux）和实用工具软件（如防病毒软件或磁盘碎片整理程序）。应用软件则帮助用户完成与计算机管理本身无直接关系的具体任务。例如，文字处理软件（如 Microsoft Word）和网页浏览器（如 Google Chrome）。
A good way to remember the distinction is that system software runs in the background and is essential for the computer to function, whereas application software is optional and user-focused. When listing examples, always be as specific as possible – for instance, name a particular word processor rather than just 'word processing software'.
一个简单的记忆方法是：系统软件在后台运行，对于计算机的运行至关重要；而应用软件是可选的，面向用户具体需求。在列举例子时，务必尽可能具体——例如，写出具体的文字处理软件名称，而不只是笼统地说“文字处理软件”。
8. Cybersecurity Threats and Prevention | 网络安全威胁与防范
Mock question: 'Explain what a phishing attack is. Describe two measures an individual can take to reduce the risk of falling victim to phishing.'
模拟题：“解释什么是网络钓鱼攻击。描述个人可以采取的两项措施，以降低遭受网络钓鱼攻击的风险。”
Phishing is a social engineering attack in which criminals send fraudulent communications, often emails or text messages, that appear to come from a reputable source. The aim is to trick the recipient into revealing sensitive information such as passwords, credit card details, or personal identification numbers. The message typically contains a link to a fake website that closely imitates a legitimate one.
网络钓鱼是一种社会工程攻击，犯罪分子发送通常伪造成可信来源的欺诈性通信（多为电子邮件或短信），目的在于诱骗收件人透露密码
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