Pre-U WJEC Computer Science: Unit Test Mock Paper Walkthrough | Pre-U WJEC 计算机：单元测试模拟卷解析

📚 Pre-U WJEC Computer Science: Unit Test Mock Paper Walkthrough | Pre-U WJEC 计算机：单元测试模拟卷解析

Mock examinations are a vital part of preparing for Pre-U WJEC Computer Science. This walkthrough dissects a typical unit test paper, covering key topics such as data representation, Boolean logic, computer architecture, networking, databases, algorithms, and compression. By analysing model answers and common pitfalls, you can refine your exam technique and deepen your conceptual understanding.

模拟考试是准备Pre-U WJEC计算机科学的关键环节。本文解析一份典型的单元测试模拟卷，涵盖数据表示、布尔逻辑、计算机体系结构、网络、数据库、算法和压缩等核心主题。通过分析模型答案和常见错误，你可以完善考试技巧并加深概念理解。

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1. Mock Paper Structure and Exam Format | 模拟卷结构与考试格式

The mock paper consists of two sections. Section A contains five short-answer questions worth a total of 50 marks, testing breadth across fundamental topics. Section B presents two extended-response questions, including a pseudocode algorithm task and an essay-style comparison, carrying 30 marks. You have 90 minutes to complete the paper, which mirrors the actual Pre-U WJEC unit test format.

模拟卷由两个部分组成。A部分包括5道简答题，共计50分，考查基础知识的广度。B部分为两道扩展回答题，包括一个伪代码算法任务和一个论文式比较题，共30分。考试时间为90分钟，真实反映Pre-U WJEC单元测试的格式。

Section	Question Types	Marks
A	Short answer: number conversion, truth table, CPU description, networking, SQL	50
B	Binary search pseudocode & efficiency; Lossy vs lossless compression	30

Understanding this structure helps you allocate time wisely — roughly 5–7 minutes per Section A question and 15–20 minutes per Section B item.

理解这一结构有助于合理分配时间——A部分每题约5–7分钟，B部分每题15–20分钟。

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2. Data Representation: Binary and Number Systems | 数据表示：二进制与数制

Sample Question 1: (a) Convert 1101.101₂ to denary. (b) Express denary 25.75 as an unsigned binary number. (c) Represent –17 in 8-bit two’s complement.

模拟题1： (a) 将二进制数 1101.101₂ 转换为十进制。(b) 将十进制数 25.75 表示为无符号二进制数。(c) 用8位二进制补码表示 –17。

For part (a), expand each bit using its place value: 1×2³ + 1×2² + 0×2¹ + 1×2⁰ + 1×2⁻¹ + 0×2⁻² + 1×2⁻³ = 8 + 4 + 0 + 1 + 0.5 + 0 + 0.125 = 13.625₁₀.

对于 (a) 部分，将各位按位权展开：1×2³ + 1×2² + 0×2¹ + 1×2⁰ + 1×2⁻¹ + 0×2⁻² + 1×2⁻³ = 8 + 4 + 0 + 1 + 0.5 + 0 + 0.125 = 13.625₁₀。

The binary point is crucial; many candidates omit negative exponents or misalign the fractional part. A quick check using a table ensures accuracy.

小数点至关重要；许多考生忽略了负指数或小数部分对齐错误。用表格快速核对可以确保准确。

Bit Position	3	2	1	0	-1	-2	-3
Value	1	1	0	1	1	0	1

For (b), convert 25 to binary as 11001₂ and 0.75 as .11₂ (since 0.75 = ½ + ¼). Thus 25.75₁₀ = 11001.11₂. Remember to multiply the fractional part by 2 repeatedly.

对于 (b)，将 25 转换为二进制 11001₂，0.75 转换为 .11₂（因为 0.75 = ½ + ¼）。因此 25.75₁₀ = 11001.11₂。记住小数部分需要连续乘以2。

Two’s complement for –17: start with +17 in 8-bit binary (0001 0001₂), invert all bits to 1110 1110₂, then add 1 to get 1110 1111₂. We are using a fixed 8-bit representation, so ignore any carry beyond the 8th bit.

补码表示 –17：先写出 +17 的8位二进制 (0001 0001₂)，全部取反得到 1110 1110₂，再加1得到 1110 1111₂。我们使用固定的8位，忽略超出第8位的进位。

Common mistake: using the wrong number of bits or forgetting to invert all bits, including leading zeros.

常见错误： 使用错误的位数或忘记将所有位取反，包括前导零。

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3. Boolean Logic and Simplification | 布尔逻辑与化简

Sample Question 2: For the Boolean expression (A ∧ B) ∨ ¬C, construct the truth table and simplify where possible using laws of Boolean algebra.

模拟题2： 对于布尔表达式 (A ∧ B) ∨ ¬C，构建真值表并在可能的情况下使用布尔代数定律进行化简。

A complete truth table must account for all combinations of A, B, C. We calculate A∧B, ¬C, and the final OR.

完整的真值表必须考虑A、B、C的所有组合。我们计算 A∧B、¬C 以及最终的或运算。

A	B	C	A ∧ B	¬C	(A∧B) ∨ ¬C
0	0	0	0	1	1
0	0	1	0	0	0
0	1	0	0	1	1
0	1	1	0	0	0
1	0	0	0	1	1
1	0	1	0	0	0
1	1	0	1	1	1
1	1	1	1	0	1

Applying Boolean algebra: the expression is already in a sum-of-products form. Using the distributive law (A∧B) ∨ ¬C ≡ (A ∨ ¬C) ∧ (B ∨ ¬C). Neither factor simplifies further without additional context, so the expression is considered minimal. A Karnaugh map confirms that no pair of adjacent 1s can be merged into a smaller term. The key is to recognise when an expression cannot be further reduced.

应用布尔代数：该表达式已经是积之和形式。使用分配律 (A∧B) ∨ ¬C ≡ (A ∨ ¬C) ∧ (B ∨ ¬C)。两个因子在没有额外上下文的情况下都无法进一步化简，因此该表达式被认为是最简形式。卡诺图证实没有相邻的1可以合并成更小的项。关键是识别哪些表达式无法进一步化简。

Exam tip: Always show all steps in simplification, even if you reach the original form — this demonstrates understanding of Boolean laws.

考试技巧：始终展示化简的所有步骤，即使最终形式与原来相同——这展示了你对布尔定律的理解。

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4. Computer Architecture: The Fetch-Decode-Execute Cycle | 计算机体系结构：取指-译码-执行周期

Sample Question 3: Describe the fetch-decode-execute (FDE) cycle and explain the roles of the Program Counter (PC) and Memory Address Register (MAR).

模拟题3：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）的作用。

The FDE cycle is the fundamental process by which the CPU performs instructions. It repeats billions of times per second.

FDE周期是CPU执行指令的基本过程，每秒重复数十亿次。

• Fetch: The address from the PC is copied to the MAR; the PC increments. The control unit sends a read signal, and the instruction at that address is fetched from memory into the Memory Data Register (MDR), then transferred to the Current Instruction Register (CIR).
• 中文：取指：将PC中的地址复制到MAR；PC自增。控制单元发出读信号，该地址处的指令从内存取出放入MDR，然后传送到当前指令寄存器（CIR）。
• Decode: The control unit interprets the bit pattern in the CIR, determining the opcode and operands.
• 中文：译码：控制单元解读CIR中的位模式，确定操作码和操作数。
• Execute: The appropriate functional unit (ALU, etc.) carries out the operation, which may involve reading from or writing to memory or I/O.
• 中文：执行：相应的功能单元（ALU等）执行操作，可能涉及对内存或I/O的读写。

The PC holds the address of the next instruction to be fetched; its increment ensures sequential execution, unless a jump occurs. The MAR temporarily stores the address being accessed on the address bus, enabling the memory unit to locate the correct cell.

PC保存下一条待取指令的地址；其递增确保了顺序执行，除非发生跳转。MAR暂存地址总线上正在访问的地址，使内存单元能够定位正确的存储单元。

Common errors: confusing MAR with MDR or describing the cycle without mentioning the necessary registers. Using a diagram (drawn in the exam) adds clarity.

常见错误：混淆MAR和MDR，或描述周期时未提及必要的寄存器。考试中画出示意图会增加条理性。

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5. Networking Concepts and Switching Techniques | 网络概念与交换技术

Sample Question 4: Define LAN, WAN, and protocol. Compare circuit switching and packet switching, giving an advantage of each.

模拟题4：定义LAN、WAN和协议。比较电路交换和分组交换，并分别给出一个优点。

A LAN (Local Area Network) connects computers in a small geographical area, typically using Ethernet. A WAN (Wide Area Network) spans larger distances, often relying on leased lines or satellite links. A protocol is a set of rules governing data communication, such as TCP/IP.

LAN（局域网）连接小地理范围内的计算机，通常使用以太网。WAN（广域网）跨越更大距离，通常依赖专线或卫星链路。协议是管理数据通信的一组规则，例如TCP/IP。

Circuit switching establishes a dedicated physical path between two nodes for the duration of the communication (e.g., traditional telephone networks). Packet switching breaks data into packets, each independently routed through the network (e.g., the internet).

电路交换为通信期间在两节点之间建立专用的物理路径（如传统电话网络）。分组交换将数据分成数据包，每个包独立路由通过网络（如互联网）。

Aspect	Circuit Switching	Packet Switching
Path	Dedicated end-to-end	Shared, dynamic routing
Resource usage	Reserved, even if idle	Efficient, statistical multiplexing
Delay	Constant, low jitter	Variable, may queue
Advantage	Guaranteed bandwidth for real-time applications	Robustness and efficient use of links

Always link your comparison to practical examples; for instance, video conferencing may prefer circuit switching for consistent quality, whereas email benefits from packet switching’s flexibility.

始终将比较与实际例子相联系；例如，视频会议可能更倾向于电路交换以获得稳定的质量，而电子邮件则受益于分组交换的灵活性。

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6. Databases and SQL Queries | 数据库与SQL查询

Sample Question 5: Given a table STUDENTS with columns (id, name, grade, score), write an SQL statement to retrieve the names of students with a score above 80, ordered by grade in descending order.

模拟题5：给定一个包含列（id, name, grade, score）的表 STUDENTS，编写SQL语句检索分数高于80分的学生姓名，并按grade降序排列。

The expected solution is:

预期解决方案是：

SELECT name FROM STUDENTS WHERE score > 80 ORDER BY grade DESC;

Each clause must be placed correctly: SELECT identifies the column, FROM specifies the table, WHERE filters rows, and ORDER BY sorts the result. DESC ensures Z comes before A, or numerically high before low.

每个子句必须正确放置：SELECT 指定列，FROM 指定表，WHERE 过滤行，ORDER BY 对结果排序。DESC 确保Z在A之前，或数值从高到低。

Common mistake: writing ‘ORDER BY grade DESC’ without realising that grade may be a character field — in a mock paper, assume it is a letter grade; descending sort works as expected. Another pitfall is selecting the wrong columns or misspelling table names. Always use a semicolon to terminate the statement.

常见错误： 编写 ‘ORDER BY grade DESC’ 时未

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