Year 11 WJEC Computer Science: Unit Test Mock Paper Walkthrough | WJEC 计算机单元测试模拟卷精讲

📚 Year 11 WJEC Computer Science: Unit Test Mock Paper Walkthrough | WJEC 计算机单元测试模拟卷精讲

Mock exams are your secret weapon for mastering WJEC Computer Science. This walkthrough unpacks a representative Year 11 unit test, blending theory, programming, and problem-solving exactly as the board expects. Work through each sample question and expert commentary to sharpen your answers and avoid the traps that lose marks.

模拟考是攻克 WJEC 计算机科学的秘密武器。这份精讲逐题拆解一份典型的 Year 11 单元测试卷，涵盖理论、编程与问题解决，完全贴合考纲要求。通过每道样题和专家解析打磨你的答题技巧，避开失分陷阱。

1. CPU Architecture and the Fetch-Decode-Execute Cycle | 处理器架构与取指–译码–执行周期

Sample question: Explain the role of the Program Counter (PC) during the Fetch-Decode-Execute cycle. [2 marks]

样题：解释程序计数器 (PC) 在取指–译码–执行周期中的作用。[2 分]

The Program Counter holds the memory address of the next instruction to be processed. During the fetch stage, the content of the PC is copied to the Memory Address Register (MAR); immediately afterwards the PC is incremented by 1 so it points to the following instruction.

程序计数器保存下一条要处理的指令的内存地址。在取指阶段，PC 的内容被复制到存储器地址寄存器 (MAR)；紧接着 PC 加 1，指向下一条指令的地址。

To secure both marks you must mention the increment step. Candidates who only say 'it stores the address of the next instruction' earn just 1 mark. Always link the PC to the MAR transfer.

要拿满两分，必须提到自增操作。只说“它存储下一条指令的地址”的考生只能得 1 分。务必把 PC 与 MAR 的数据传输联系起来。

Quick fact: The Current Instruction Register (CIR) holds the actual instruction after it has been fetched, and the Control Unit (CU) decodes it.

速记：当前指令寄存器 (CIR) 存放取来的实际指令，由控制单元 (CU) 完成译码。

---

2. Primary Memory, Virtual Memory and Storage | 主存、虚拟内存与存储器

Sample question: A computer has 4 GB of RAM. A user opens ten large video files and the system becomes slow. Explain why this happens and how virtual memory helps. [4 marks]

样题：某计算机配有 4 GB 内存。用户同时打开十个大型视频文件，系统运行变慢。解释为什么会这样，并说明虚拟内存如何缓解问题。[4 分]

RAM is volatile, fast-access memory that holds currently running programs and data. When ten large video files fill the 4 GB of RAM, the operating system must transfer idle data to the hard disk (or SSD) in a process called paging. The area of storage used for this overflow is virtual memory.

RAM 是易失性快速存储器，存放当前运行的程序和数据。当十个大型文件塞满 4 GB 内存后，操作系统必须通过分页把暂时不用的数据转移到硬盘或固态硬盘，这块溢写区域就是虚拟内存。

Because storage is much slower than RAM, the system spends a lot of time swapping data back and forth, causing noticeable slowdown. The answer gains full marks if it links slowness to the speed difference between RAM and secondary storage.

由于辅存远比内存慢，系统会花费大量时间来回搬移数据，造成明显卡顿。若能把卡顿与内存和辅存的速度差异联系，即可获满分。

Vocabulary check: s data when power is off. Non-volatile = keeps data. RAM is volatile; SSD is non-volatile.

术语检查：Volatile 指断电丢失数据，non-volatile 指数据保留。RAM 易失，固态硬盘非易失。

---

3. Number Systems and Image Representation | 数制转换与图像表示

Sample question: Convert the denary number 156 into an 8-bit binary number and then into hexadecimal. Show your working. [3 marks]

样题：将十进制数 156 转换成 8 位二进制，再转换为十六进制。写出步骤。[3 分]

2 = 4 rem 1, 4 ÷ 2 = 2 rem 0, 2 ÷ 2 = 1 rem 0, 1 ÷ 2 = 0 rem 1. Reading remainders bottom‑up gives 10011100. Splitting into nibbles: 1001 (9) and 1100 (C) → 9C in hex.

连续除以 2 取余：156 除 2 得 78 余 0，78 除 2 得 39 余 0，39 除 2 得 19 余 1，19 除 2 得 9 余 1，9 除 2 得 4 余 1，4 除 2 得 2 余 0，2 除 2 得 1 余 0，1 除 2 得 0 余 1。余数从下往上读得 10011100。分割为半字节：1001 对应 9，1100 对应 C，得到十六进制 9C。

Sample question: tmap image uses a colour depth of 16 bits. Calculate the file size in kilobytes. [2 marks]

样题：一幅 800×600 像素的位图使用 16 位色深，计算其文件大小（千字节）。[2 分]

divide by 1024 for kibibytes unless the question explicitly asks for the decimal (1000) definition.

总比特数 = 800 × 600 × 16 = 7,680,000 位。转为字节：7,680,000 ÷ 8 = 960,000 字节。转为千字节：960,000 ÷ 1024 = 937.5 KB。除非题目明确要求用十进制 (1000) 计算，否则一律除以 1024。

---

4. Network Topologies and Protocols | 网络拓扑与协议

Sample question: Compare a star network topology with a bus topology. Give one advantage and one disadvantage of each. [4 marks]

样题：比较星形拓扑与总线拓扑，各给出一个优点和一个缺点。[4 分]

A star network connects all nodes to a central switch. Advantage: if one cable fails, only that node is affected. Disadvantage: if the central switch fails, the entire network goes down.

星形拓扑将所有节点连至中央交换机。优点：一根线缆故障只影响该节点。缺点：中央交换机若故障则全网瘫痪。

A bus topology uses a single backbone cable. Advantage: it is cheaper to install because less cable is needed. Disadvantage: a break in the backbone cable brings down the whole network, and data collisions can slow performance.

总线拓扑使用单根主干线缆。优点：安装成本低，所需线缆少。缺点：主干断裂会导致全网瘫痪，并且数据碰撞会降低性能。

Sample question: Name the four layers of the TCP/IP model and state the purpose of the Transport Layer. [2 marks]

样题：列出 TCP/IP 模型的四层结构，并说明传输层的作用。[2 分]

The layers are Application, Transport, Internet (or Network), and Link (or Network Access). The Transport Layer segments data, ensures reliable delivery via TCP (or fast delivery via UDP), and adds port numbers to direct data to the correct application.

四层分别为应用层、传输层、互联网层（或称网络层）和链路层（或称网络接入层）。传输层负责数据分段，通过 TCP 保证可靠交付（或通过 UDP 实现快速传输），并添加端口号将数据交给正确的应用程序。

---

5. SQL and Database Queries | SQL 与数据库查询

Sample question: The table Students has the fields StudentID, FirstName, Surname, YearGroup and TutorGroup. Write an SQL query to display the first name and surname of all students in Year 11 who are in tutor group 11A. [3 marks]

样题：表 Students 包含字段 StudentID、FirstName、Surname、YearGroup 和 TutorGroup。写出 SQL 查询，显示所有 Year 11 且导师组为 11A 的学生的名字和姓氏。[3 分]

The expected SQL is:
SELECT FirstName, Surname FROM Students WHERE YearGroup = 11 AND TutorGroup = '11A';
Because TutorGroup contains letters, it must be treated as a string and enclosed in single quotes. Marks are awarded for the correct SELECT clause, the correct WHERE clause with both conditions, and the use of AND.

期望的 SQL 语句为：
SELECT FirstName, Surname FROM Students WHERE YearGroup = 11 AND TutorGroup = '11A';
因为 TutorGroup 含有字母，必须作为字符串处理并用单引号括起。评分点包括正确的 SELECT 子句、包含两个条件的 WHERE 子句以及正确使用 AND。

Always write SQL keywords in uppercase to help the examiner read your answer quickly. If the query needed all fields, you would write SELECT *.

SQL 关键字一律大写有助于考官快速审阅。如果需要查询所有字段，则应写 SELECT *。

---

6. Algorithms: Flowcharts and Pseudocode | 算法：流程图与伪代码

Sample question: A program stores three numbers. Draw a flowchart that outputs the largest number. [5 marks]

样题：某程序存储了三个数字。绘制流程图，输出其中的最大值。[5 分]

Your flowchart should start with input of three values A, B, C. Use a decision diamond to compare A and B, then compare the larger with C. Output the final largest value. Use correct symbols: oval for start/stop, parallelogram for input/output, rhombus for decisions, and arrows to show flow. Missing a decision symbol or final output often loses marks.

流程图应从输入 A、B、C 开始，用菱形判断框比较 A 和 B，再将较大者与 C 比较，最后输出最大值。使用正确的图形：椭圆表示开始/结束，平行四边形表示输入/输出，菱形表示判断，箭头表示流向。缺少判断符号或最终输出是常见失分点。

Pseudocode version:
INPUT A, B, C
IF A > B THEN max = A ELSE max = B
IF C > max THEN max = C
OUTPUT max

伪代码版本：
INPUT A, B, C
IF A > B THEN max = A ELSE max = B
IF C > max THEN max = C
OUTPUT max

WJEC accepts pseudocode in any clear style, but you must be consistent with indentation and keywords. Exam boards never penalise a particular syntax, provided the logic is correct.

WJEC 接受任何书写清晰的伪代码风格，但缩进和关键字必须一致。逻辑正确就不会因具体语法而扣分。

---

广告 / Sponsored

7. Programming with Python (On-Screen Test) | Python 编程实战（屏幕化考试）

Sample question: an integer N and prints a right-angled triangle of asterisks with N rows. Example for N=3:
*
**
*** [4 marks]

样题：编写 Python 程序，要求用户输入整数 N，然后输出 N 行星号构成的直角三角形。如 N=3 输出：
*
**
*** [4 分]

A correct solution:
N = int(input('Enter a number: '))
for i in range(1, N+1):
print('*' * i)
This loop uses the repetition operator * to repeat the asterisk i times. Marks are earned for correct input conversion to int, a loop that runs N times, and printing the correct pattern.

正确解答：
N = int(input('Enter a number: '))
for i in range(1, N+1):
print('*' * i)
该循环用重复运算符 * 将星号重复 i 次。得分点包括正确将输入转换为整数、循环运行 N 次以及输出正确图案。

Common mistake: Writing range(N) instead of range(1, N+1) makes i start at 0, printing an extra blank line and missing the top row. Always test your code with a small value like 2.

常见错误：写成 range(N) 而非 range(1, N+1) 会导致 i 从 0 开始，多打一行空白并缺少第一行。编完代码后一定要用 N=2 这样的小数测试。

---

8. Boolean Logic and Logic Gate Circuits | 布尔逻辑与逻辑门电路

Sample question: (A AND B) OR (NOT C). Then complete the truth table for all input combinations. [5 marks]

样题：画出布尔表达式 Q = (A AND B) OR (NOT C) 的逻辑电路，并完成所有输入组合的真值表。[5 分]

The circuit requires one AND gate with inputs A and B, one NOT gate with input C, and one OR gate combining the AND output and the NOT output. Begin by drawing A and B feeding into the AND, C feeding into the NOT, then both outputs going into the OR. Label all gates clearly.

电路需要一个与门（输入 A、B），一个非门（输入 C），再通过或门将两者的输出合并。先画 A、B 接入与门，C 接入非门，二者输出共同输入或门。所有门电路必须清晰标注。

Truth table:

A	B	C	A AND B	NOT C	Q
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

Complete the intermediate columns first to avoid mistakes. In an exam, writing the table neatly in logical order (e.g., binary count 000, 001, 010...) saves time.

先填好中间列可减少失误。考试时按二进制顺序（000, 001, 010 …）整齐列出真值表既省时又不易遗漏。

---

9. Legal, Ethical and Environmental Issues | 法律、伦理与环境议题

Sample question: A company collects customers' names, addresses and shopping habits. Explain two responsibilities the company has under the Data Protection Act (GDPR). [4 marks]

样题：一家公司收集了顾客的姓名、地址与购物习惯。依据《数据保护法》（GDPR），解释该公司需要承担的两项责任。[4 分]

First, the company must state clearly why the data is being collected and only use it for that purpose. This is the 'purpose limitation' principle. Second, it must keep the data secure from unauthorised access, e.g., by encrypting databases and training staff on data handling.

第一，公司必须明确说明收集数据的目的，并且仅用于该目的，即“目的限制”原则。第二，必须保障数据安全，防止未经授权的访问，例如加密数据库并对员工进行数据处理培训。

You could also mention the right of access: customers can request a copy of their data. Links to real-world regulations like GDPR always impress examiners.

也可提及访问权：顾客有权索取个人数据副本。将答案与现实中的 GDPR 等法规挂钩总能赢得考官好感。

Sample question: Describe one ethical concern about facial recognition technology used in a school. [2 marks]

样题：阐述在学校使用人脸识别技术所涉及的一项伦理关切。[2 分]

Facial recognition can lead to constant surveillance, making students feel they are being watched all the time, which could harm their mental wellbeing and erode trust. Another concern is bias in the algorithm: if the system is less accurate for certain ethnic groups, it may result in unfair treatment.

人脸识别可能造成持续监控，使学生感到时刻被监视，可能损害心理健康并侵蚀信任。另一关切是算法偏见：如果系统对某些族群的准确率较低，可能导致不公平对待。

---

10. Testing, Trace Tables and Error Types | 测试、跟踪表与错误类型

Sample question: A programmer has written a function that divides two numbers. Explain, using examples, what test data should be used to thoroughly test the function. [4 marks]

样题：某程序员编写了一个两数相除的函数。举例说明应使用哪些测试数据充分测试该函数。[4 分]

ery large numbers, or dividing by 1), and erroneous data (e.g., dividing by zero, or entering text instead of a number). A robust test plan also includes decimal inputs to verify the handling of floating-point results.

测试数据必须包含正常数据（如 10 ÷ 2 = 5）、边界数据（如极大数字或除以 1）以及异常数据（如除数为零或输入文本）。完善的测试计划还应包含小数输入，以验证浮点结果的处理。

Using a trace table during algorithm design helps identify logic errors. Show the examiner that you can walk through an algorithm with sample values and spot where a variable becomes incorrect.

在算法设计阶段使用跟踪表有助于发现逻辑错误。向考官展示你能用样例值模拟算法运行，并指出变量在何处出错。

---

11. Exam Technique and Final Tips | 应试技巧与最后叮嘱

Before the test, make sure you can effortlessly convert between denary, binary and hex, write basic SQL SELECT statements, and hand-trace loops. Read the command word: 'explain' requires a reason, 'describe' needs facts, 'compare' demands both similarities and differences. Manage your time so that 3-mark questions don't eat up half the paper.

考前请确保能轻松完成十进制、二进制和十六进制转换，能写出基本的 SQL SELECT 语句，能手跑循环。看清指令词：'explain' 要给出原因，'describe' 需要事实陈述，'compare' 必须同时提及相同点与不同点。合理分配时间，别让 3 分题吃掉半张试卷。

In on-screen programming tasks, always add comments to explain your logic – they can earn marks even if the code has a small bug. Finally, if you are stuck on a logic gate diagram, build a truth table on scrap paper first

Published by TutorHao | Year 11 Computer Science Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）