📚 Year 12 WJEC Computer Science: Unit Test Mock Paper Analysis | Year 12 WJEC 计算机：单元测试模拟卷解析
Mock papers are one of the most effective revision tools for WJEC AS Computer Science. They let you practise the exact style of questions you will face in the real Unit 1 exam, build time-management skills, and highlight knowledge gaps. This article walks you through a typical full-length mock paper, breaking down key questions, model answers, and common mistakes. Use it alongside your notes to boost your confidence and exam technique.
模拟卷是 WJEC AS 计算机科学最高效的复习工具之一。它们能让你练习真实单元 1 考试中会遇到的问题风格，培养时间管理技能，并突显知识薄弱点。本文带你走一遍一份典型完整模拟卷，拆解关键题目、标准答案和常见错误。配合你的笔记使用，可增强你的信心和考试技巧。
1. Introduction to the Mock Paper | 模拟卷介绍
Our mock paper is designed to mirror the WJEC Unit 1 (Fundamentals of Computer Science) exam. It includes a mix of short-answer questions worth 1–4 marks and longer structured questions that test both knowledge and application. The topics covered are number systems, Boolean logic, data structures, algorithms, pseudocode, computer architecture, networks, and system software – all core Year 12 content.
我们的模拟卷旨在模拟 WJEC 单元 1（计算机科学基础）考试。它包含 1–4 分的简短回答题以及测试知识与应用的结构化长问题。涵盖的主题包括数制系统、布尔逻辑、数据结构、算法、伪代码、计算机体系结构、网络和系统软件——均为 Year 12 的核心内容。
Before diving into each question, try to attempt the paper yourself under timed conditions (1 hour 30 minutes). Then compare your answers with the walkthrough below.
在深入每道题之前，尝试在计时条件下（1 小时 30 分钟）自己完成试卷。然后将你的答案与下面的解析进行对比。
2. Question 1: Number Systems and Binary Arithmetic | 问题 1：数制系统与二进制运算
Question 1(a): Convert the binary number 10110111₂ to hexadecimal. Show your working. [2 marks]
问题 1(a)： 将二进制数 10110111₂ 转换为十六进制。写出步骤。 [2 分]
To convert, split the binary digits into groups of four starting from the right: 1011 0111. The left group 1011₂ equals 11 in decimal, which is B in hexadecimal. The right group 0111₂ is 7. Therefore, the answer is B7₁₆. A common mistake is to group from the left; always group from the least significant bit (right).
转换时，从右开始每四位分组：1011 0111。左组 1011₂ 等于十进制 11，即十六进制 B。右组 0111₂ 为 7。因此答案是 B7₁₆。常见错误是从左边分组；务必从最低有效位（右侧）分组。
Question 1(b): Perform the binary addition 0110₂ + 0101₂. State whether an overflow occurs if the numbers are stored in 4-bit registers. [3 marks]
问题 1(b)： 执行二进制加法 0110₂ + 0101₂。若数字存储在 4 位寄存器中，说明是否发生溢出。 [3 分]
result is 1011₂. In 4-bit registers, 0110 (6) and 0101 (5) are positive if unsigned, giving 11, which fits in 4 bits – no overflow for unsigned. However, in two' esents -5, indicating an overflow as the sum exceeds 7. The question ideally specifies unsigned, so no overflow. Always check the context.
逐位相加：0+1=1，1+0=1，1+1=0 进位 1，0+0 加进位 1 得 1。结果为 1011₂。在 4 位寄存器中，若无符号，0110 (6) 和 0101 (5) 相加得 11，刚好在 4 位范围内，无溢出。但在二进制补码有符号表示中，6+5=1011 表示 -5，因和超过 7 而溢出。题目通常指定无符号，故无溢出。注意结合语境判断。
Question 1(c): Represent the decimal number -5 in 4-bit two's complement. [2 marks]
问题 1(c)： 用 4 位二进制补码表示十进制数 -5。 [2 分]
Start with +5 in 4 bits: 0101. Invert all bits: 1010. Add 1: 1011. So -5 is 1011₂. It is easy to forget the 'add 1' step, which would give the wrong representation -6 in one's complement.
先用 4 位表示 +5：0101。按位取反：1010。加 1：1011。因此 -5 为 1011₂。很容易忘记“加 1”这一步，导致给出反码（-6）的错误结果。
3. Question 2: Boolean Logic and Simplification | 问题 2：布尔逻辑与化简
Question 2(a): Simplify the Boolean expression (A ∧ B) ∨ (A ∧ ¬B). Show each step. [3 marks]
问题 2(a)： 化简布尔表达式 (A ∧ B) ∨ (A ∧ ¬B)。写出每一步。 [3 分]
Factor out A: A ∧ (B ∨ ¬B). The expression (B ∨ ¬B) is always TRUE (1). So the result is A ∧ 1, which simplifies to A. This uses the distributive law and the complement law.
提取公因式 A：A ∧ (B ∨ ¬B)。表达式 (B ∨ ¬B) 恒为 TRUE (1)。故得 A ∧ 1，化简为 A。这里运用了分配律和互补律。
Question 2(b): Construct a truth table for XOR (A ⊕ B). [2 marks]
问题 2(b)： 构造异或 (A ⊕ B) 的真值表。 [2 分]
| A | B | A ⊕ B |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
XOR gives 1 only when inputs differ. In logic circuits, it can be expressed as (A ∧ ¬B) ∨ (¬A ∧ B). A common pitfall is confusing XOR with OR, where 1+1=1 in OR but 0 in XOR.
异或仅在输入不同时输出 1。在逻辑电路中可表示为 (A ∧ ¬B) ∨ (¬A ∧ B)。常见误区是将异或与或混淆，或运算 1+1=1，而异或为 0。
Question 2(c): Draw a logic circuit for A ∨ (B ∧ C). (You may describe the gates in words.) [2 marks]
问题 2(c)： 画出 A ∨ (B ∧ C) 的逻辑电路图。（可用文字描述门电路。） [2 分]
First, B and C feed into an AND gate. The output of that AND gate is then combined with A using an OR gate. So A goes directly to the OR gate; B and C go to the AND gate, whose output is the second OR input. The final output is A OR (B AND C).
首先，B 和 C 输入一个与门。与门输出与 A 一起进入或门。因此 A 直接连到或门；B 和 C 进入与门，其输出作为或门的第二个输入。最终输出为 A OR (B AND C)。
4. Question 3: Stacks and Queues | 问题 3：栈与队列
Question 3: An empty stack undergoes these operations: Push(15), Push(22), Pop(), Push(8), Push(31), Pop(). Show the contents of the stack after each operation. [5 marks]
问题 3： 一个空栈依次执行以下操作：Push(15), Push(22), Pop(), Push(8), Push(31), Pop()。写出每次操作后栈的内容。 [5 分]
Stack uses LIFO. After Push(15): [15]. Push(22): [15, 22]. Pop(): removes 22, stack [15]. Push(8): [15, 8]. Push(31): [15, 8, 31]. Pop(): removes 31, stack [15, 8]. Remember to show the top of the stack on the right in your answer; marks are often awarded for correct order.
栈遵循后进先出。Push(15) 后：[15]。Push(22) 后：[15, 22]。Pop() 移除 22，栈 [15]。Push(8)：[15, 8]。Push(31)：[15, 8, 31]。Pop() 移除 31，栈 [15, 8]。答案中记得把栈顶写在右侧；给分点通常依据正确顺序。
Extension: If the same operations were applied to a queue (Enqueue instead of Push, Dequeue instead of Pop), the outcome would be very different. Queue is FIFO: initial Enqueue(15), Enqueue(22): [15, 22]. Dequeue removes 15 leaving [22]. Enqueue(8): [22, 8]. Enqueue(31): [22, 8, 31]. Dequeue removes 22, leaving [8, 31].
拓展： 若相同操作应用于队列（入队代替压入，出队代替弹出），结果截然不同。队列先进先出：初始 Enqueue(15), Enqueue(22) 为 [15, 22]。Dequeue 移走 15，留下 [22]。Enqueue(8)：[22, 8]。Enqueue(31)：[22, 8, 31]。Dequeue 移走 22，剩下 [8, 31]。
5. Question 4: Sorting Algorithms | 问题 4：排序算法
Question 4: Show the first two passes of bubble sort on the list [7, 2, 9, 1, 5]. Indicate any swaps. [4 marks]
问题 4： 写出对列表 [7, 2, 9, 1, 5] 进行冒泡排序的前两趟过程。标出所有交换。 [4 分]
s 1, largest element 9 bubbled to the end. Pass 2: Compare 2 and 7 → no swap. Compare 7 and 1 → swap → [2, 1, 7, 5, 9]. Compare 7 and 5 → swap → [2, 1, 5, 7, 9]. No need to compare last element as it is sorted. After two passes the list is not fully sorted, but the algorithm continues.
第一趟：比较 7 和 2 → 交换 → [2, 7, 9, 1, 5]。比较 7 和 9 → 无交换。比较 9 和 1 → 交换 → [2, 7, 1, 9, 5]。比较 9 和 5 → 交换 → [2, 7, 1, 5, 9]。第一趟结束，最大元素 9 浮到最后。第二趟：比较 2 和 7 → 无交换。比较 7 和 1 → 交换 → [2, 1, 7, 5, 9]。比较 7 和 5 → 交换 → [2, 1, 5, 7, 9]。已不用比较最后一个已排序元素。两趟后列表未完全有序，但算法继续。
Many students lose marks by forgetting to show when no swap occurs or by continuing comparisons beyond the unsorted portion. Always mark the end of each pass clearly.
许多学生因忘记在无交换时注明，或对已排序部分仍继续比较而失分。务必清楚标明每趟的结束。
6. Question 5: Pseudocode and Trace Tables | 问题 5：伪代码与跟踪表
Question 5: Consider the following pseudocode:
sum ← 0
FOR i ← 1 TO 5
sum ← sum + i * 3
ENDFOR
OUTPUT sum
Complete a trace table showing the values of i and sum at each iteration. What is the final output? [4 marks]
问题 5： 考虑以下伪代码：完成一个显示每次迭代 i 和 sum 值的跟踪表。最终输出是什么？ [4 分]
Trace table:
| Iteration | i | sum (before update) | sum (after sum + i*3) |
| 1 | 1 | 0 | 3 |
| 2 | 2 | 3 | 9 |
| 3 | 3 | 9 | 18 |
| 4 | 4 | 18 | 30 |
| 5 | 5 | 30 | 45 |
missing the initial state of sum as 0 or forgetting that i increments after the loop body.
最终输出：45。每次计算 i * 3：3, 6, 9, 12, 15，总和为 45。常见错误包括遗漏 sum 初值 0，或忘记循环体后 i 自增。
7. Question 6: Computer Architecture – CPU and Memory | 问题 6：计算机体系结构 – CPU 与存储器
Question 6(a): Describe the fetch-decode-execute cycle. In your answer refer to the program counter (PC), memory address register (MAR), memory data register (MDR), and the current instruction register (CIR). [5 marks]
问题 6(a)： 描述取指-译码-执行周期。在你的回答中提及程序计数器 (PC)、存储器地址寄存器 (MAR)、存储器数据寄存器 (MDR) 和当前指令寄存器 (CIR)。 [5 分]
Fetch: The PC holds the address of the next instruction; this address is copied to the MAR and sent via the address bus to memory. The instruction at that address is placed into the MDR via the data bus and then copied to the CIR. The PC is incremented. Decode: The control unit decodes the instruction in the CIR. Execute: The instruction is carried out; this may involve the ALU for arithmetic or address buses for data movement. The cycle repeats.
取指：PC 保存下一条指令的地址；该地址复制到 MAR，通过地址总线发送到存储器。该地址处的指令经数据总线放入 MDR，再复制到 CIR。PC 递增。译码：控制单元对 CIR 中的指令进行译码。执行：执行指令，可能涉及 ALU 进行算术运算或地址总线传输数据。周期重复。
Question 6(b): Explain the difference between RAM and ROM. Give one use of each in a computer system. [2 marks]
问题 6(b)： 解释 RAM 和 ROM 的区别。各举一个在计算机系统中的用途。 [2 分]
RAM is volatile memory used to store data and programs currently in use; its contents are lost when power is off. ROM is non-volatile, containing permanent instructions such as the BIOS/bootstrap loader. Example: RAM holds the operating system after booting; ROM stores the firmware to start the computer.
RAM 是易失性存储器，用于存放当前使用的数据与程序，断电后内容丢失。ROM 为非易失性存储器，包含永久指令，如 BIOS/引导加载程序。例如：RAM 在启动后保存操作系统；ROM 存储启动计算机的固件。
8. Question 7: Networks and Communication | 问题 7：网络与通信
Question 7(a): Compare a star network topology with a
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