📚 In-depth Analysis of Past Exam Questions for Year 10 OCR Computer Science | Year 10 OCR 计算机：历年真题深度解析
Past papers provide invaluable insight into exam structure, question types, and marking expectations for OCR GCSE Computer Science (J277). This article dissects real exam-style questions, offering step-by-step explanations and pinpointing common mistakes, equipping Year 10 students with the skills to excel.
历年真题为考生提供了宝贵的考纲结构、题型与评分要求洞察。本文深入剖析真实考试风格题目，逐步解析并指出常见错误，帮助 Year 10 学生掌握高分技能。
1. CPU Architecture & Fetch Cycle | CPU架构与取指周期
A typical 6-mark question asks: "Describe the fetch-decode-execute cycle, referring to the roles of the Program Counter (PC), Memory Address Register (MAR), Memory Data Register (MDR), and Current Instruction Register (CIR)."
典型的6分题会问："描述取指-解码-执行周期，并说明程序计数器（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）的作用。"
For high marks, your answer must trace the flow of data and control signals in strict order, using the correct register names and bus terminology.
要获得高分，答案必须严格按顺序追踪数据与控制信号的流动，并使用正确的寄存器名称及总线术语。
FETCH: The PC holds the address of the next instruction. This address is copied to the MAR and sent along the address bus. The control unit issues a read signal. The instruction stored at that memory location is transferred to the MDR via the data bus, then copied to the CIR. The PC increments to point to the next instruction.
取指：PC保存下一条指令的地址。该地址被复制到MAR，并通过地址总线发送。控制单元发出读信号。该内存地址存储的指令通过数据总线传输到MDR，然后复制到CIR。PC递增以指向下一条指令。
DECODE: The instruction in the CIR is decoded by the control unit, determining what operation to perform and which parts of the CPU are needed.
解码：CIR中的指令由控制单元解码，确定执行什么操作以及需要CPU的哪些部分。
EXECUTE: The operation is carried out. This could involve the ALU performing a calculation or data being moved between registers/memory. The cycle then repeats.
执行：执行操作。这可能涉及ALU执行计算，或者在寄存器/内存之间移动数据。然后周期重复。
Common mistakes include confusing MAR with MDR or forgetting to mention that the PC is incremented during the fetch step. Examiners look for precise language: "address bus" for addresses and "data bus" for data.
常见错误包括混淆MAR和MDR，或忘记说明PC在取指阶段递增。考官寻找精确的表述："地址总线"用于地址，"数据总线"用于数据。
2. Binary Arithmetic & Overflow | 二进制算术与溢出
Binary addition questions often carry 2-4 marks. Consider: "Perform the binary addition 10101101₂ + 01110011₂ on an 8-bit register and state whether overflow occurs."
二进制加法题目通常占2-4分。例如："在8位寄存器中计算 10101101₂ + 01110011₂ 的二进制加法，并说明是否发生溢出。"
A good approach is to align the bits and add column by column, showing carry bits. The result is 1 00011100₂ (9 bits). Since only 8 bits are available, the leftmost bit is discarded, leaving 00011100₂. Overflow is detected because the carry into the most significant bit (MSB) differs from the carry out.
较好的方法是按位对齐并逐列相加，显示进位。结果为 1 00011100₂（9位）。由于只有8位可用，最左边的位被丢弃，留下 00011100₂。溢出被检测到，因为进入最高有效位（MSB）的进位与从MSB出去的进位不同。
When two positive numbers sum to a result that appears negative in two's complement, overflow has occurred. Always mention that an 8-bit register cannot hold a 9-bit result.
当两个正数相加得到的结果在补码表示中看似负数时，就发生了溢出。一定要指出8位寄存器无法容纳9位结果。
3. Memory & Storage Devices | 存储器与存储设备
A frequently examined topic is the need for virtual memory. A typical question: "Explain why virtual memory is used and describe one disadvantage of relying on it." (4 marks)
常考主题之一是虚拟内存的必要性。典型问题："解释为什么要使用虚拟内存，并描述依赖它的一个缺点。"（4分）
Virtual memory is a section of the hard disk or SSD used as an extension of RAM. It allows the computer to run more programs than physical RAM could otherwise support by swapping inactive data to disk when RAM is full.
虚拟内存是硬盘或固态硬盘的一部分，用作RAM的扩展。当RAM已满时，它通过将不活动的数据交换到磁盘，使计算机能够运行比物理RAM原本能支持更多的程序。
However, disk access is much slower than RAM, so excessive swapping causes disk thrashing and noticeably reduces performance. Many students forget to mention that the processor prioritises active processes in RAM, moving less urgent data to virtual memory.
然而，磁盘访问比RAM慢得多，因此过度的交换会导致磁盘颠簸并显著降低性能。许多学生忘记说明处理器优先将活跃进程保留在RAM中，将不太紧急的数据移动到虚拟内存。
4. Network Protocols & Layers | 网络协议与分层
"Describe the role of each layer in the TCP/IP model" is a classic 4-mark question. The TCP/IP model has four layers: Application, Transport, Internet, and Link (or Network Access).
"描述TCP/IP模型中每一层的作用"是经典的4分题。TCP/IP模型有四层：应用层、传输层、互联网层和链路层（或网络接入层）。
| Layer | Role |
|---|---|
| Application | Provides network services to user applications, e.g. HTTP, FTP, SMTP. |
| Transport | Ensures reliable data transfer, using TCP for error checking or UDP for speed. |
| Internet | Addresses and routes packets via IP; determines the best path between networks. |
| Link | Handles physical transmission of data over hardware, e.g. Ethernet, Wi-Fi. |
A common mistake is to assign protocol examples to the wrong layer — for instance, placing IP in the Transport layer. Use the table above to remember the correct mapping.
常见错误是将协议示例分配到错误的层——例如，将IP放在传输层。使用上表记住正确的对应关系。
5. Cybersecurity Threats & Prevention | 网络安全威胁与预防
Phishing frequently appears in 2-4 mark questions. "Describe how a phishing attack works and explain two methods by which a user can protect themselves."
网络钓鱼频繁出现在2-4分题中。"描述网络钓鱼攻击如何运作，并解释用户保护自己的两种方法。"
Phishing involves a fraudulent email or message that impersonates a trusted organisation to trick the recipient into revealing sensitive information, such as passwords or bank details. The attacker often creates a sense of urgency to prompt hasty action.
网络钓鱼涉及欺诈性电子邮件或消息，冒充可信任的组织，诱骗收件人透露密码或银行详细信息等敏感信息。攻击者通常制造紧迫感，促使用户匆忙行动。
Protection methods: 1) Check the sender's email address carefully for misspellings or suspicious domains. 2) Never click on links in unsolicited emails; instead, type the known URL directly into the browser.
保护方法：1）仔细检查发件人的电子邮件地址是否有拼写错误或可疑域名。2）绝不要点击垃圾邮件中的链接；而应直接在浏览器中输入已知的URL。
Many candidates miss the mark by giving generic advice like "use antivirus software" without linking it specifically to phishing. Tie each defence directly to the attack vector.
许多考生因给出"使用防病毒软件"之类的泛泛建议而失分，却没有将其与网络钓鱼具体联系起来。每种防御措施都必须直接针对攻击途径。
6. Linear vs Binary Search | 线性搜索与二分搜索
Comparison questions ask: "Explain the differences between linear search and binary search, including the precondition for binary search." (4-6 marks)
比较题会问："解释线性搜索和二分搜索之间的区别，包括二分搜索的前提条件。"（4-6分）
| Aspect | Linear Search | Binary Search |
|---|---|---|
| Data requirement | Works on unsorted data | Requires sorted data |
| Method | Checks each element sequentially | Repeatedly divides the search interval in half |
| Worst-case complexity | O(n) — proportional to list size | O(log n) — much faster for large lists |
If the data is unsorted, you cannot apply binary search. Many students fail to mention the need for a sorted list as a mandatory precondition, losing a crucial mark.
如果数据未排序，就无法应用二分搜索。许多学生未能说明排序列表是必要前提条件，从而丢失关键分数。
7. Bubble Sort Algorithm | 冒泡排序算法
OCR expects you to trace bubble sort on a small list. A 4-mark question: "Show the passes of bubble sort on the list [6, 2, 8, 1, 3] and state the total number of comparisons."
OCR要求在小列表上追踪冒泡排序。一道4分题："展示对列表[6, 2, 8, 1, 3]进行冒泡排序的各趟过程，并说明总的比较次数。"
1,8,3] → (swap 8,3) [2,6,1,3,8]. End of pass 1, largest element 8 in place. 4 comparisons.
第1趟：比较并交换相邻元素：[6,2,8,1,3] → (交换6,2) [2,6,8,1,3] → [2,6,8,1,3] → (交换8,1) [2,6,1,8,3] → (交换8,3) [2,6,1,3,8]。第1趟结束，最大元素8就位。4次比较。
[1,2,3,6,8]. 2 comparisons. Pass 4: 1 comparison, no swaps. Total comparisons = 4+3+2+1 = 10. Highlight that the algorithm can stop early if no swaps occur in a pass.
第2趟：3次比较。第3趟：2次比较。第4趟：1次比较，无交换。总比较次数=10。要强调如果一趟中没有发生交换，算法可以提前终止。
8. Programming: Sequence, Selection & Iteration | 编程中的顺序、选择与迭代
Pseudocode questions test structured programming. A typical task: "Write pseudocode to print all numbers from 1 to 50 that are multiples of 3."
伪代码题测试结构化编程。典型任务："编写伪代码，输出从1到50中所有3的倍数。"
A solution must use a loop (iteration) and an IF statement (selection):
解决方案必须使用循环（迭代）和IF语句（选择）：
FOR i ← 1 TO 50
IF i MOD 3 = 0 THEN
OUTPUT i
ENDIF
ENDFOR
The arrow (← es assignment, and MOD is used for the remainder. Common mistakes include using assignment '=' instead of comparison, or placing the OUTPUT outside the loop.
箭头（←）表示赋值，MOD用于求余数。常见错误包括用赋值'='代替比较，或将OUTPUT放在循环外部。
9. SQL Queries & Data Retrieval | SQL查询与数据检索
SQL forms part of the databases topic. A 3-mark question: "Write an SQL query to retrieve the name and age of all students in the 'Year10' table who are aged 15 or older."
SQL是数据库主题的一部分。一道3分题："编写SQL查询，从'Year10'表中检索所有年龄在15岁或以上的学生的姓名和年龄。"
The answer must use SELECT...FROM...WHERE with correct syntax:
答案必须使用正确的SELECT...FROM...WHERE语法：
SELECT Name, Age
FROM Year10
WHERE Age >= 15;
d the logical operator >=. Many candidates forget the semicolon or misspell the table name — precise spelling matters.
注意用逗号分隔列名，使用正确的表名以及逻辑运算符 >=。许多考生忘记分号或拼错表名——精确的拼写至关重要。
10. Ethical & Legal Impacts of Technology | 技术的伦理与法律影响
The 8-mark extended response often tackles ethical dilemmas. "Discuss the ethical and legal issues associated with the collection of personal data by companies."
8分拓展题常涉及伦理困境。"讨论与公司收集个人数据相关的伦理和法律问题。"
Legally, the UK's Data Protection Act 2018 and GDPR enforce principles that data must be collected fairly, used only for specified purposes, kept secure, and not kept longer than necessary. Companies must obtain clear consent.
法律上，英国的《2018年数据保护法》和GDPR强制执行以下原则：数据必须公平收集，仅用于特定目的，保持安全，且保存时间不超过必要期限。公司必须获得明确同意。
Ethically, issues arise when companies sell data to third parties without explicit user knowledge, or use algorithms to manipulate behaviour. The imbalance of power between corporations and individuals raises questions of autonomy and transparency.
伦理上，当公司在用户不知情的情况下将数据出售给第三方，或使用算法操纵行为时，问题就出现了。企业与个人之间的权力失衡引发了关于自主性和透明度的质疑。
Achieve top marks by giving specific examples, such as fitness trackers sharing health data with insurers, and by discussing the role of anonymisation and the right to erasure.
要获得最高分，需要给出具体例子，例如健身追踪器与保险公司共享健康数据，并讨论匿名化以及被遗忘权的作用。
Published by TutorHao | Computer Science Revision Series | aleveler.com
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