📚 Year 10 CIE Computer Science: Unit Test Mock Paper Walkthrough | 十年级 CIE 计算机：单元测试模拟卷解析
This article presents a comprehensive walkthrough of a mock unit test for Year 10 CIE Computer Science. Designed to cover the core syllabus topics, it provides step-by-step explanations for each question type you are likely to encounter. Use this guide to reinforce your understanding and boost your exam confidence.
本文全面解析了一份针对十年级CIE计算机科学的单元测试模拟卷。模拟卷覆盖核心大纲主题，针对你可能遇到的各类题目提供逐步解析。请利用本指南巩固理解并提升应考信心。
1. Binary to Denary Conversion | 二进制转十进制转换
Question: Convert the binary number 10110110 into denary. Show all your working.
题目：将二进制数 10110110 转换为十进制。列出所有计算步骤。
128) + (0 × 64) + (1 × 32) + (1 × 16) + (0 × 8) + (1 × 4) + (1 × 2) + (0 × 1) = 128 + 0 + 32 + 16 + 0 + 4 + 2 + 0 = 182. The denary value is 182.
将每个二进制位按位权从最低位 (1) 到 128 排列。每一位乘以其位权后求和： (1×32) + (1×16) + (0×8) + (1×4) + (1×2) + (0×1) = 128 + 0 + 32 + 16 + 0 + 4 + 2 + 0 = 182。十进制结果为 182。
A common mistake is forgetting that a 0 bit still contributes a place value of zero; always include it in the expansion to avoid misalignment. Another pitfall is reversing the bit order – remember the least significant bit is on the right.
常见错误是忽略了0位仍会贡献零值；务必在展开式中包含它以防止错位。另一个陷阱是颠倒位序——记住最低有效位在右侧。
2. Denary to Hexadecimal Conversion | 十进制转十六进制转换
Question: Convert the denary number 182 into hexadecimal. Use the successive division method.
题目：将十进制数 182 转换为十六进制。请使用连续除法。
r 6. The quotient 11 corresponds to the hex digit B, and the remainder 6 remains 6. Reading from the last quotient to the first remainder gives B6. Therefore, 182 in hexadecimal is B6.
182 除以 16：182 ÷ 16 = 11 余 6。商 11 对应十六进制数字 B，余数 6 保留为 6。从最后的商读到第一个余数，得到 B6。因此 182 的十六进制为 B6。
Many learners confuse the order: always write the final quotient first, then the remainders in reverse order of calculation. Also be vigilant with hex digits for 10–15 (A–F).
许多同学会混淆顺序：务必先写最终商，然后按计算逆序写余数。同时要注意 10–15 对应的十六进制字母 (A–F)。
3. Data Storage Units | 数据存储单位
Question: II. Calculate the file size in kilobytes. Use the convention that 1 kilobyte = 1000 bytes.
题目：一个文本文件包含 2000 个字符，每个字符用 8 位 ASCII 编码。计算该文件大小，以千字节 (KB) 为单位。使用 1 KB = 1000 字节的约定。
its, total bytes = 16 000 ÷ 8 = 2000 bytes. Then convert to kilobytes: 2000 bytes ÷ 1000 = 2 KB.
总位数 = 2000 字符 × 8 位 = 16 000 位。因为 1 字节 = 8 位，总字节数 = 16 000 ÷ 8 = 2000 字节。再转为千字节：2000 字节 ÷ 1000 = 2 KB。
or memory addressing, but for file sizes, 1000 is more common unless instructed otherwise.
务必留意题目指定的换算约定；CIE 有时在内存寻址中规定 1 KB = 1024 字节，但对于文件大小通常用 1000，除非另有说明。
4. CPU Components and the Fetch-Decode-Execute Cycle | CPU 部件与取指-译码-执行周期
Question: Describe the role of the Control Unit (CU) during the fetch stage of the fetch-decode-execute cycle.
题目：描述控制单元 (CU) 在取指-译码-执行周期中取指阶段的作用。
During the fetch stage, the Control Unit sends a signal to read the next instruction from the memory address held in the Program Counter (PC). It places the fetched instruction into the Current Instruction Register (CIR) and increments the PC to point to the next instruction. The CU coordinates all these data movements by activating the necessary control lines.
在取指阶段，控制单元发出信号，从程序计数器 (PC) 保存的内存地址处读取下一条指令。它将取出的指令放入当前指令寄存器 (CIR)，并递增 PC 指向下一条指令。CU 通过激活必要的控制线来协调所有这些数据移动。
A full answer should mention the PC, MAR (Memory Address Register), MDR (Memory Data Register) and CIR, showing how the CU orchestrates the flow. Many candidates lose marks by omitting the increment step.
完整答案应提及 PC、MAR（存储器地址寄存器）、MDR（存储器数据寄存器）和 CIR，体现 CU 如何编排数据流。许多考生因漏掉递增步骤而失分。
5. RAM vs ROM | 随机存取存储器与只读存储器对比
Question: State two differences between RAM and ROM.
题目：列出 RAM 和 ROM 的两点区别。
Difference 1: RAM is volatile memory – its contents are lost when the power is turned off. ROM is non-volatile and retains data without power.
Difference 2: RAM is typically used to store programs and data currently in use by the CPU, allowing read-write operations. ROM mainly stores firmware and boot instructions, and under normal operation it can only be read, not written to.
区别一：RAM 是易失性存储器——断电后内容丢失。ROM 是非易失性存储器，无需电源也能保存数据。
区别二：RAM 通常用于存储 CPU 当前正在使用的程序和数据，支持读写操作。ROM 主要存储固件和引导指令，正常操作时只能读取，不能写入。
Be precise: 'volatile' and 'non-volatile' are the technical terms examiners expect. Avoid vague statements like 'RAM is faster' which, although sometimes true, is not a defining functional difference for this syllabus.
用词要精确：“易失性”和“非易失性”是考官期望的专业术语。避免使用诸如“RAM 更快”等模糊表述，虽然有时属实，但并非本大纲定义的功能性区别。
6. Input Devices and Their Applications | 输入设备及其应用
Question: A supermarket checkout system needs to read product barcodes quickly and accurately. Identify the most suitable input device and justify your choice.
题目：超市收银系统需快速准确读取商品条码。指出最合适的输入设备并说明理由。
The most suitable device is a barcode scanner (or barcode reader). It uses a laser or LED light to detect the pattern of black and white stripes. The reflected light is converted into a digital code that the computer can match to a product database. This method is fast, reduces human error, and works reliably even when the barcode is slightly damaged.
最合适的设备是条码扫描器。它利用激光或 LED 光检测黑白条纹图案，反射光被转换为数字代码，计算机可将代码与商品数据库匹配。该方法速度快、减少人为错误，且条码轻微损坏时仍能可靠工作。
Alternative devices like RFID readers are sometimes used, but barcode scanners remain the primary choice at conventional checkouts because of their low cost and compatibility with existing packaging.
RFID 阅读器等替代设备虽有时使用，但条码扫描器因成本低且与现有包装兼容，仍是传统收银台的首选。
7. Logic Gate Circuits and Truth Tables | 逻辑门电路与真值表
Question: expression X = A AND (NOT B). Complete the truth table for this circuit (click to expand).
题目：某逻辑电路有两个输入 A 和 B，输出 X 由表达式 X = A AND (NOT B) 定义。完成该电路的真值表（点击展开）。
| A | B | NOT B | X = A AND NOT B |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
nd NOT B. The output is 1 only when A = 1 AND (NOT B) = 1, which occurs when A = 1, B = 0.
完成真值表时，首先将 B 取反得到中间列 NOT B。然后对 A 和 NOT B 施以 AND 运算。只有当 A = 1 且 NOT B = 1 (即 A=1, B=0) 时输出才为 1。
Many students mistakenly apply the AND operation before negation. Always follow the order of operations indicated by the parentheses or the circuit diagram. Practice drawing the logic gate symbol for AND and NOT to reinforce understanding.
许多学生误将 AND 运算放在取反之前。务必遵循括号或电路图指示的运算顺序。练习绘制 AND 和 NOT 的逻辑门符号以加深理解。
8. IP Addressing vs MAC Addressing | IP 地址与 MAC 地址对比
Question: Distinguish between an IP address and a MAC address.
题目：区分 IP 地址和 MAC 地址。
An IP address (Internet Protocol address) is a logical address assigned to a device on a network, used for routing data across different networks. It can be dynamic (changed by DHCP) or static. A MAC address (Media Access Control address) is a unique physical address hard-coded into a network interface card (NIC) by the manufacturer. MAC addresses operate at the data link layer and are used for communication within the same local network segment.
IP 地址是分配给网络设备的逻辑地址，用于跨网络路由数据，可以是动态的 (DHCP 分配) 或静态的。MAC 地址是由制造商硬编码在网络接口卡中的唯一物理地址，工作在数据链路层，用于同一局域网段内的通信。
The key difference is that an IP address tells you where a device is located on a network (or the internet), while a MAC address identifies which specific device it is. IP addresses can change, MAC addresses generally do not.
关键区别在于 IP 地址告诉你设备在网络中的位置，而 MAC 地址标识具体是哪一台设备。IP 地址可变，MAC 地址通常不变。
9. Phishing and Data Security | 网络钓鱼与数据安全
Question: Explain what is meant by 'phishing' and describe one method users can adopt to protect themselves.
题目：解释什么是“网络钓鱼”，并描述一种用户可采取的防护方法。
Phishing is a social engineering attack where a malicious party sends fraudulent emails or messages that appear to come from a legitimate source (e.g., a bank or online service). The goal is to trick the recipient into revealing sensitive information such as passwords, credit card numbers, or personal details. A common protection method is to never click on suspicious links and instead manually type the official website address into the browser. Additionally, checking for HTTPS and the padlock icon can help verify authenticity, but users must remain cautious as some phishing sites also use HTTPS.
网络钓鱼是一种社会工程攻击，恶意方发送看似来自合法来源（如银行或在线服务）的欺诈电子邮件或消息，诱骗收件人泄露密码、信用卡号等敏感信息。一种常见的防护方法是切勿点击可疑链接，而应在浏览器中手动输入官方网站地址。此外，检查 HTTPS 和挂锁图标有助于验证真实性，但用户仍需保持警惕，因为部分钓鱼网站也使用 HTTPS。
Other effective measures include enabling two-factor authentication (2FA), using spam filters, and regularly updating software. For the exam, be sure to use the term 'social engineering' and give a concrete, specific precaution – not just 'be careful'.
其他有效措施包括启用双重认证 (2FA)、使用垃圾邮件过滤器以及定期更新软件。在考试中，务必使用“社会工程”这一术语，并给出具体明确的预防措施，而非只是“小心一点”。
10. Pseudocode Trace and Algorithmic Thinking | 伪代码跟踪与算法思维
Question: Study the following pseudocode.
Total ← 0
FOR Count ← 1 TO 5
Total ← Total + Count
NEXT Count
OUTPUT Total
What value is displayed?
题目：分析以下伪代码：Total ← 0; FOR Count ← 1 TO 5; Total ← Total + Count; NEXT Count; OUTPUT Total。显示的值是多少？
becomes 0+1=1, 1+2=3, 3+3=6, 6+4=10, 10+5=15. After the loop, the output statement prints 15.
跟踪循环：初始 Total = 0。Count 依次取值 1,2,3,4,5。Total 变为 0+1=1, 1+2=3, 3+3=6, 6+4=10, 10+5=15。循环结束后输出语句打印 15。
Be careful with loop boundaries: the loop runs exactly 5 times because the upper bound is inclusive in this syntax. A common error is to stop at 4 or include 0. Trace tables are a powerful tool to avoid such off-by-one errors.
注意循环边界：该语法中上限为闭区间，循环恰好执行 5 次。常见错误是在 4 处停止或包括 0。使用跟踪表是避免此类差一错误的有力工具。
Published by TutorHao | Computer Science Revision Series | aleveler.com
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