Year 11 CAIE Computer Science: Unit Test Mock Paper Walkthrough | CAIE IGCSE 计算机科学单元测试模拟卷解析

📚 Year 11 CAIE Computer Science: Unit Test Mock Paper Walkthrough | CAIE IGCSE 计算机科学单元测试模拟卷解析

Mock unit tests are one of the most effective ways to prepare for IGCSE Computer Science. This walkthrough analyses a representative set of questions you are likely to meet in a Year 11 CAIE test, covering data representation, logic, networking, programming concepts and more. Each section breaks down a typical problem, explains the correct answer and highlights common pitfalls.

单元模拟测试是备考 IGCSE 计算机科学最有效的方式之一。本文解析一套典型试题，涵盖数据表示、逻辑电路、网络、编程概念等 Year 11 CAIE 常见考点。每一节都将拆解一道经典题目，给出正确答案并指出常见失分点。

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1. Binary to Hexadecimal & Denary | 二进制、十六进制与十进制转换

Question: Convert the binary number 10111011 to (i) hexadecimal and (ii) denary. Show all your working.

题目：将二进制数 10111011 转换为 (i) 十六进制和 (ii) 十进制。写出全部步骤。

Answer (i): Split the byte into two nibbles: 1011 and 1011. 1011 in binary equals 11 in denary, which is B in hexadecimal. So the whole number is BB₁₆.

答案 (i)：将字节拆分为两个半字节：1011 和 1011。1011 对应十进制 11，也就是十六进制的 B。因此整个数为 BB₁₆。

lace the bits: 1×128 + 0×64 + 1×32 + 1×16 + 1×8 + 0×4 + 1×2 + 1×1 = 128+32+16+8+2+1 = 187.

答案 (ii)：列出位权重 128,64,32,16,8,4,2,1。对应位相乘：1×128+0×64+1×32+1×16+1×8+0×4+1×2+1×1=128+32+16+8+2+1=187。

Common pitfall: Switching nibbles the wrong way round or forgetting that hexadecimal letters A–F represent 10–15.

常见错误：半字节顺序颠倒，或忘记十六进制字母 A–F 对应 10～15。

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2. Logic Expression & Truth Table | 逻辑表达式与真值表

OT (A AND B) OR (B AND C), complete the truth table for all input combinations of A, B, C.

题目：对于逻辑表达式 Q = NOT (A AND B) OR (B AND C)，完成 A、B、C 所有组合的真值表。

Answer: Work column by column. First compute A AND B, then NOT of that, then B AND C, finally OR the two intermediate results. The truth table is shown below.

答案：逐列计算。先求 A AND B，再取反；计算 B AND C；最后将两个中间结果进行 OR 运算。真值表如下。

A	B	C	A AND B	NOT(A AND B)	B AND C	Q
0	0	0	0	1	0	1
0	0	1	0	1	0	1
0	1	0	0	1	0	1
0	1	1	0	1	1	1
1	0	0	0	1	0	1
1	0	1	0	1	0	1
1	1	0	1	0	0	0
1	1	1	1	0	1	1

er cases the output is 1. Students often misplace the NOT operation or omit a combination.

注意到只有当 A=1、B=1、C=0 时 Q 为 0，其余情况输出均为 1。学生常把非运算的位置弄错，或漏掉某个输入组合。

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3. File Size Calculation | 文件大小计算

r depth of 16 bits. Calculate the minimum file size in kibibytes (KiB). Show your working.

题目：一张位图的分辨率为 800×600 像素，颜色深度为 16 位。计算最小文件大小，以 kibibytes (KiB) 为单位。写出步骤。

80,000 ÷ 8 = 960,000 bytes. 1 KiB = 1024 bytes, so file size = 960,000 ÷ 1024 ≈ 937.5 KiB.

答案：总像素 = 800 × 600 = 480,000 像素。总位数 = 480,000 × 16 = 7,680,000 位。8 位 = 1 字节，所以总字节 = 7,680,000 ÷ 8 = 960,000 字节。1 KiB = 1024 字节，文件大小 = 960,000 ÷ 1024 ≈ 937.5 KiB。

A frequent mistake is using 1000 instead of 1024 for the conversion to KiB, or forgetting to multiply by the colour depth.

常见失误：在转换为 KiB 时使用了 1000 而非 1024，或忘记乘以颜色深度。

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4. IP Addresses vs MAC Addresses | IP 地址与 MAC 地址对比

Question: State two differences between an IP address and a MAC address.

题目：列举 IP 地址与 MAC 地址的两个区别。

Answer: (1) An IP address is a logical address assigned by a network administrator or DHCP server, whereas a MAC address is a physical address hard-coded into the NIC by the manufacturer. (2) IP addresses can be changed or reassigned; MAC addresses are permanent and unique to each device.

答案：(1) IP 地址是由网络管理员或 DHCP 服务器分配的逻辑地址，而 MAC 地址是制造商固化在网卡中的物理地址。(2) IP 地址可以更改或重新分配；MAC 地址是永久性的，每个设备唯一。

For full marks, also note that IP addresses are used for routing across networks (Layer 3), while MAC addresses are used within a local network (Layer 2).

为获得满分，还应指出 IP 地址用于跨网络路由（第 3 层），而 MAC 地址用于局域网内部（第 2 层）。

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5. HTML Structure & Hyperlinks | HTML 结构与超链接

Question: Write HTML code to display a level-1 heading saying 'Welcome' and a hyperlink that opens 'gallery.html' with the clickable text 'View Gallery'. The link should open in a new tab.

题目：编写 HTML 代码，显示一个一级标题“Welcome”和一个超链接，点击文字“View Gallery”后在新标签页中打开 gallery.html。

Answer:

答案：

<h1>Welcome</h1>
<a href="gallery.html" target="_blank">View Gallery</a>

The 'target' attribute with value '_blank' forces the link to open in a new tab. Missing the attribute or misspelling 'href' is a common source of lost marks.

使用值为“_blank”的 target 属性可使链接在新标签页打开。漏写该属性或拼错 href 是常见的丢分点。

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6. Pseudocode: Count-Controlled Loop & Condition | 伪代码：计数循环与条件判断

Question: Write a pseudocode algorithm that asks the user to input 10 numbers and outputs the largest number entered.

题目：编写伪代码算法，要求用户输入 10 个数字，输出其中最大的一个。

Answer:

SET largest TO -9999
FOR count ← 1 TO 10
   INPUT num
   IF num > largest THEN
      largest ← num
   ENDIF
NEXT count
OUTPUT largest

Initialising 'largest' to a very small value (or the first input) is essential. Without initialisation the comparison fails. Many students forget to give an initial value or introduce an off-by-one error in the loop.

将“largest”初始化为一个极小的值（或首次输入值）至关重要。不初始化会导致比较出错。许多同学忘记赋初值，或循环次数出现差一错误。

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7. Error Detection: Parity Checks | 错误检测：奇偶校验

Question: A system uses even parity. The received byte is 10110101, where the leftmost bit is the parity bit. Determine whether an error has occurred and explain why.

题目：某系统采用偶校验。接收到的字节为 10110101，其中最左位为校验位。判断是否发生错误并解释原因。

is 1, which would make the total number of 1s in the whole byte equal to 5 (odd). For even parity the total should be even, so an error has occurred. Alternatively, with even parity the parity bit should have been 0, so the mismatch indicates an error.

答案：除去校验位后的数据位为 0110101，其中 1 的个数为 4（偶数）。校验位是 1，使得整个字节中 1 的总数为 5（奇数）。偶校验要求总数为偶数，因此发生了错误。或者，使用偶校验时校验位本应为 0，不匹配说明出错。

Be careful to identify the correct parity bit position as stated in the question; some questions put the parity bit on the right.

注意根据题目描述确认校验位的位置；有些题目把校验位放在最右边。

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8. Encryption: Symmetric Key Weakness | 加密：对称密钥的弱点

Question: Describe one major weakness of symmetric encryption and suggest how it can be overcome.

题目：描述对称加密的一个主要弱点，并提出克服方法。

Answer: The main weakness is key distribution: the same secret key must be shared between sender and receiver securely before communication. If the key is intercepted during transmission, an attacker can decrypt all messages. This can be overcome by using asymmetric encryption to exchange the symmetric key, combining the speed of symmetric encryption with the security of asymmetric key exchange.

答案：主要弱点是密钥分发问题：发送方和接收方在通信前必须以安全的方式共享同一个密钥。如果密钥在传输中被截获，攻击者就能解密所有消息。可以通过使用非对称加密来交换对称密钥的方式克服，从而结合对称加密的速度与非对称密钥交换的安全性。

You should also mention that symmetric encryption is faster than asymmetric, which is why hybrid systems are common.

还应该提到对称加密比非对称加密快，这正是混合系统普遍使用的原因。

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9. Database Query (SQL) | 数据库查询 (SQL)

Question: A table 'Students' has fields: StudentID, Name, Grade, Class. Write an SQL statement to select the Name and Grade of all students in Class '11A' where the Grade is greater than 80.

题目：表“Students”包含字段：StudentID, Name, Grade, Class。编写一条 SQL 语句，选择班号为“11A”且 Grade 大于 80 的所有学生的姓名和 Grade。

Answer: The SQL would be:

答案：SQL 语句为：

SELECT Name, Grade
FROM Students
WHERE Class = '11A' AND Grade > 80;

Remember the semi-colon at the end, the correct use of quotes around the text value '11A', and the logical operator AND to combine conditions.

注意末尾的分号、文字值“11A”需用引号括起来，以及用逻辑与运算符 AND 连接条件。

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10. Data Protection & Ethics | 数据保护与伦理

Question: A company collects personal data from customers. Name one principle from the data protection act and explain how the company must apply it.

题目：一家公司收集客户个人数据。说出数据保护法中的一个原则，并解释该公司应如何遵守。

Answer: One principle is 'Data must be processed fairly and lawfully'. The company must inform customers how their data will be used, obtain clear consent, and not use the data for purposes beyond what was stated. It should also have a clear privacy notice.

答案：一个原则是“数据必须公平、合法地处理”。公司必须告知客户其数据将如何被使用，获得明确同意，并且不得将数据用于超出声明范围之外的用途。公司还应提供清晰的隐私声明。

Other acceptable principles: kept secure, accurate and up to date, not kept longer than necessary. Always link the principle to concrete actions.

其他可接受的原则：保持数据安全、准确和最新，保存时间不超过必要期限。一定要把原则与具体措施联系起来。

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11. Fetch-Decode-Execute Cycle | 取指-译码-执行周期

Question: Describe the steps of the fetch stage and decode stage in the CPU.

题目：描述 CPU 在取指阶段和译码阶段的步骤。

Answer: During the fetch stage, the address in the Program Counter (PC) is placed on the address bus; the Control Unit sends a read signal, and the instruction from that memory address is fetched via the data bus into the Current Instruction Register (CIR). Then the PC is incremented to point to the next instruction. During the decode stage, the Control Unit interprets the instruction in the CIR and prepares the necessary control signals for the execute stage.

答案：在取指阶段，程序计数器 (PC) 中的地址被放在地址总线上；控制单元发出读信号，该内存地址的指令通过数据总线被取出并存入当前指令寄存器 (CIR)。随后 PC 递增，指向下一条指令。在译码阶段，控制单元译码 CIR 中的指令，并为执行阶段准备必要的控制信号。

Many marks are lost for not mentioning the roles of the buses or the increment of PC. A clear step-by-step description earns full credit.

许多学生因未提及总线的作用或 PC 的递增而丢分。清晰的分步描述才可获得满分。

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12. Sensors & Control System | 传感器与控制系统

Question: A greenhouse uses a microprocessor-controlled window that opens automatically when the temperature exceeds 30°C. Describe how the system operates, naming the input and output components.

题目：温室中使用微处理器控制的窗户，当温度超过 30°C 时自动打开。描述该系统的工作过程，并说出输入和输出组件名称。

Answer: The input component is a temperature sensor, which sends an analogue signal to an analogue-to-digital converter (ADC). The microprocessor reads the digital temperature value, compares it with the stored threshold (30°C), and if the temperature > 30°C, sends a digital signal to open the window. The output component is an actuator (or motor) connected to the window.

答案：输入组件是温度传感器，它将模拟信号发送给模数转换器 (ADC)。微处理器读取数字温度值，与存储的阈值 (30°C) 进行比较，若温度 > 30°C，则发送数字信号打开窗户。输出组件是连接窗户的执行器（或电机）。

The system operates in a continuous feedback loop; the sensor keeps monitoring so the window can close again when the temperature drops below 30°C.

系统工作在一个持续的反馈循环中；传感器持续监测，当温度回落到 30°C 以下时窗户可再次关闭。

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Published by TutorHao | CAIE IGCSE Computer Science Revision Series | aleveler.com

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