📚 Year 9 OCR Computer Science: Unit Test Mock Walkthrough | Year 9 OCR 计算机科学：单元测试模拟卷解析
This walkthrough breaks down a sample Year 9 OCR Computer Science unit test, covering key topics from the KS3 curriculum such as data representation, computer architecture, networking, programming, and cybersecurity. Each section presents a typical exam-style question, followed by a model answer and detailed explanation to help you understand the marking scheme and common pitfalls.
本文解析一份典型的 Year 9 OCR 计算机科学单元测试模拟卷，涵盖 KS3 课程中的数据表示、计算机体系结构、网络、编程和网络安全等重点主题。每个部分包含一道典型试题、参考答案与详细解析，帮助你理解评分标准与常见易错点。
1. Paper Structure and Mark Allocation | 试卷结构与分值分布
A typical Year 9 OCR unit test lasts 45 minutes and carries 40 marks. It is divided into three sections: multiple-choice questions (5 marks), short-answer theory (20 marks), and a programming/pseudocode task (15 marks). Knowing the structure helps you manage revision time and exam pace.
典型的 Year 9 OCR 单元测试时长45分钟，满分40分。试卷分为三部分：选择题（5分）、简答题（20分）和编程/伪代码题（15分）。了解结构有助于你安排复习和把控考试节奏。
Common topics assessed include binary and hexadecimal conversion, logic gates, computer components, network topologies, Python loops, sorting algorithms, and online threats. In this walkthrough, we work through questions from each of these areas and highlight exactly what examiners look for.
常考知识点包括二进制与十六进制转换、逻辑门、计算机组件、网络拓扑、Python循环、排序算法和在线威胁。本文将逐一带你解析这些领域的题目，并明确指出评分标准。
2. Binary to Denary Conversions | 二进制与十进制转换
Question: Convert the 8‑bit binary number 11010110₂ into denary. Show all your working. (3 marks)
题目：将8位二进制数 11010110₂ 转换为十进制。写出完整计算过程。（3分）
+ 1×4 + 1×2 + 0×1. Summing gives 128 + 64 + 16 + 4 + 2 = 214. Always show the expanded sum – one mark is often awarded just for writing the correct place values.
位权从左到右依次为128、64、32、16、8、4、2、1。每位与位权相乘：1×128 + 1×64 + 0×32 + 1×16 + 0×8 + 1×4 + 1×2 + 0×1。求和得 128 + 64 + 16 + 4 + 2 = 214。务必将展开的加法写出，通常写出正确位权就能得到1分。
A second common question reverses the process: convert denary 89 to an 8‑ r0; 1 ÷ 2 = 0 r1. Reading remainders upwards gives 1011001, padded to 8 bits as 01011001₂.
另一种常见考法是反向转换：将十进制89转换为8位二进制。反复除以2并记录余数。89 ÷ 2 = 44 余1；44 ÷ 2 = 22 余0；22 ÷ 2 = 11 余0；11 ÷ 2 = 5 余1；5 ÷ 2 = 2 余1；2 ÷ 2 = 1 余0；1 ÷ 2 = 0 余1。余数从下往上读得1011001，补足8位为 01011001₂。
3. Hexadecimal and Data Units | 十六进制与数据单位
Question: Convert denary 200 into a 2‑digit hexadecimal number. Show your working. (2 marks)
题目：将十进制200转换为两位十六进制数，写出计算过程。（2分）
r 8. Decimal 12 corresponds to the hex digit C, and 8 stays as 8. Therefore, the answer is C8₁₆. Always use capital letters for hex digits A‑F in your answer.
200除以16：200 ÷ 16 = 12 余 8。十进制12对应十六进制数字C，8保持不变。因此答案是 C8₁₆。答案中十六进制字母 A‑F 必须大写。
e = 8 bits, 1 kilobyte (KB) = 1024 bytes, 1 megabyte (MB) = 1024 KB. A file size of 2048 KB is exactly 2 MB. Thinking in powers of 2 avoids unit mistakes.
数据单位的理解同样重要。考官可能要求你在比特、字节、千字节和兆字节之间转换。记牢：1 字节 = 8 比特，1 千字节（KB）= 1024 字节，1 兆字节（MB）= 1024 KB。例如文件大小2048 KB 正好是 2 MB。用2的幂次去思考能避免单位换算错误。
4. Logic Gate Combinations | 逻辑门组合
Question: te : Complete the truth table for Q = (A AND B) OR (NOT C). Inputs A, B, C are each 0 or 1. (4 marks)
题目：补全 Q = (A AND B) OR (NOT C) 的真值表。输入 A、B、C 取值为 0 或 1。（4分）
Calculate step by step: first evaluate A AND B, then NOT C, finally OR the two. Complete all eight rows. The truth table is shown below.
逐步计算：先算 A AND B，再算 NOT C，最后将两者进行 OR 运算。完整列出八种组合。真值表如下：
| 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 |
Examiners look for correct intermediate columns and will award marks even if the final Q column has a minor slip, provided your working is logical. Always fill in the AND and NOT columns first.
考官关注中间计算列是否正确；只要推导过程合理，即便最终的 Q 列偶有小错，仍可能拿到大部分分数。务必先填写 AND 和 NOT 列。
5. Inside the CPU | CPU 内部结构
Question: Describe the role of the Control Unit (CU) and the Arithmetic Logic Unit (ALU) in the fetch‑decode‑execute cycle. (4 marks)
题目：描述控制单元（CU）和算术逻辑单元（ALU）在取指-解码-执行周期中的作用。（4分）
The Control Unit directs the flow of data inside the CPU. It fetches the next instruction from RAM, decodes it to determine what operation must be performed, and then sends control signals to other components to carry out the instruction.
控制单元负责指挥 CPU 内部的数据流动。它从内存中取出下一条指令，解码以确定需要执行什么操作，然后向其他部件发送控制信号以执行该指令。
The ALU performs all arithmetic operations (addition, subtraction, multiplication, division) and logical operations (AND, OR, NOT). During the execute stage, the ALU processes the data according to the decoded instruction and stores the result in a register.
算术逻辑单元执行所有算术运算（加、减、乘、除）和逻辑运算（AND、OR、NOT）。在执行阶段，ALU 根据已解码的指令处理数据，并将结果存入寄存器。
Other key registers include the Program Counter (PC) which holds the address of the next instruction, and the Accumulator (ACC) that temporarily holds results from the ALU. Using correct technical vocabulary scores higher marks.
其他关键寄存器包括程序计数器（PC，存放下一条指令地址）和累加器（ACC，临时存放 ALU 运算结果）。使用规范的技术词汇能获得更高分数。
6. Wired and Wireless Networks | 有线与无线网络
Question: Compare a star network topology with a mesh topology. Give one advantage and one disadvantage of each. (4 marks)
题目：比较星型网络拓扑与网状网络拓扑，各给出一个优点和一个缺点。（4分）
In a star topology, all devices connect to a central switch. Advantage: if one cable fails, only that device is affected. Disadvantage: the whole network goes down if the central switch fails. It is simple to troubleshoot and widely used in schools.
星型拓扑中所有设备连接到一个中央交换机。优点：若一根电缆故障，仅影响该设备。缺点：中央交换机故障会导致全网瘫痪。其故障排查简单，广泛用于学校网络。
A mesh topology connects every device to several others, often used in wireless mesh networks. Advantage: even if one link breaks, data can take an alternative path – it offers high reliability. Disadvantage: cabling and configuration are much more expensive due to the number of connections.
网状拓扑中每个设备与多个其他设备相连，常用于无线网状网络。优点：即使一条链路中断，数据可经备选路径传输，可靠性高。缺点：由于连接数量多，布线和配置成本高得多。
You could also be asked about protocols. For a quick revision: Ethernet governs wired LANs, Wi‑Fi (IEEE 802.11) handles wireless connections, TCP/IP manages reliable data delivery on the internet, and HTTP/HTTPS delivers web pages.
也可能考到协议。快速复习：以太网管理有线局域网，Wi‑Fi（IEEE 802.11）处理无线连接，TCP/IP 负责互联网上可靠的数据传输，HTTP/HTTPS 负责网页传输。
7. Python Programming: Trace the Loop | Python 编程：循环追踪
Question: State the exact output of the following Python code. (3 marks)
题目：写出下列 Python 代码的准确输出。（3分）
total = 0
for i in range(1, 6):
if i % 2 == 0:
total = total + i
print(total)
The loop runs for i = 1, 2, 3, 4, 5. The condition i % al becomes 2, then 6. The print statement executes only once after the loop, displaying 6.
循环变量 i 依次为 1, 2, 3, 4, 5。条件 i % 2 == 0 在 i 为偶数时成立。因此只有 2 和 4 被加到 total 中。从 0 开始，total 先变成 2，再变成 6。print 语句只在循环结束后执行一次，输出 6。
A common mistake is placing print(total) inside the loop, which would print 0, 2, 2, 6, 6 across five lines. Always check indentation to determine whether a statement belongs to the loop. Examiners deduct marks for missing the final value or adding extra outputs.
常见错误是将 print(total) 写在循环内部，那样会逐行输出 0、2、2、6、6。务必通过缩进判断语句是否属于循环体。答案漏掉最终值或多输出了额外内容都会被扣分。
8. Bubble Sort Algorithm | 冒泡排序算法
Question: Show the contents of the list after the first pass of a bubble sort on [5, 1, 4, 2, 8]. The sort should arrange numbers in ascending order. (3 marks)
题目：写出对列表 [5, 1, 4, 2, 8] 进行第一趟冒泡排序后的列表内容，要求升序排列。（3分）
p → [1, 4, 5, 2, 8]. Compare 5 and 2 → swap → [1, 4, 2, 5, 8]. Compare 5 and 8 → no swap. After one pass the list is [1, 4, 2, 5, 8]. Note that the largest number (8) has bubbled to the end.
冒泡排序依次比较相邻两个元素，若顺序错误则交换。第一趟：比较 5 和 1 → 交换 → [1, 5, 4, 2, 8]；比较 5 和 4 → 交换 → [1, 4, 5, 2, 8]；比较 5 和 2 → 交换 → [1, 4, 2, 5, 8]；比较 5 和 8 → 不交换。一趟结束后列表变为 [1, 4, 2, 5, 8]。注意最大数 8 已“冒泡”到最右端。
In the next pass the unsorted part shrinks because the last element is already in its final position. OCR Year 9 questions do not require full code but expect you to trace the algorithm step by step, clearly showing each swap.
下一趟排序时未排序部分缩短，因为最后一个元素已经就位。OCR Year 9 不要求写出完整代码，但希望你能逐步追踪算法，清晰展示每一次交换。
9. Cybersecurity: Phishing and Firewalls | 网络安全：钓鱼与防火墙
Question: Identify three signs of a phishing email and describe how a firewall helps protect a school network. (5 marks)
题目：指出钓鱼邮件的三个特征，并描述防火墙如何保护学校网络。（5分）
Phishing emails often have a generic greeting like 'Dear Customer,' contain suspicious links or attachments, and create a sense of urgency, e.g., 'Your account will be closed in 24 hours!' They may also have grammatical errors and a sender address that looks slightly altered from the genuine one.
钓鱼邮件通常使用“尊敬的客户”等泛泛的称呼，包含可疑链接或附件，并制造紧迫感，例如“您的账户将在24小时内关闭！”它们还可能存在语法错误，发件人地址与真实地址仅有细微差异。
A firewall monitors incoming and outgoing network traffic based on predetermined security rules. It blocks unauthorised access attempts and prevents malicious data packets from reaching internal devices. In a school network, it filters web traffic, blocks known malware sites, and stops pupils from accessing inappropriate content, acting as a barrier between the trusted internal network and the untrusted internet.
防火墙根据预设的安全规则监控进出网络流量。它能阻止未授权的访问尝试，并防止恶意数据包抵达内部设备。在学校网络中，防火墙过滤网页流量、阻断已知恶意软件网站、阻止学生访问不当内容，在可信的内部网络与不可信的外部互联网之间充当屏障。
10. Common Errors and Revision Tips | 常见错误与复习技巧
Mistake 1: Forgetting that binary place values double from right to left, starting at 1. Always label the positions clearly. Mistake 2: Mixing up logic gate symbols – an AND gate has a D‑shaped curve, OR has a curved back, NOT is a triangle with a small circle. Draw them several times in your revision notes.
错误一：忘记二进制位权从右向左倍增，起始值为1。务必清晰标注。错误二：混淆逻辑门符号——AND 门是 D 形曲线，OR 门背面弯曲，NOT 门是三角形加小圆圈。复习时多画几遍。
tracing, confusing = (assignment) with == (comparison). Mistake 4: In networking, writing 'hub' instead of 'switch' – o the intended recipient. Modern networks use switches, so using the correct term matters.
错误三：在 Python 追踪中，混淆 =（赋值）与 ==（比较）。错误四：在网络题中误写“集线器”而非“交换机”——集线器向所有端口广播，交换机只向目标设备发送数据。现代网络使用交换机，因此用词准确很重要。
Tip: practise past questions under timed conditions, write out full workings even when not explicitly asked, and always check unit conversions. Spending five minutes reviewing the mark scheme for each question you attempt will dramatically improve your exam technique.
建议：限时练习历年真题，即使题目未明确要求也写出完整推导过程，并始终检查单位换算。每做完一题花五分钟细读评分标准，能显著提升答题技巧。
Published by TutorHao | Computer Science Revision Series | aleveler.com
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