Year 11 CIE Computer Science Unit Test Mock Paper Analysis | 11年级CIE计算机单元测试模拟卷解析

📚 Year 11 CIE Computer Science: Unit Test Mock Paper Analysis | 11年级CIE计算机单元测试模拟卷解析

Welcome to our detailed walkthrough of a typical Year 11 CIE Computer Science unit test mock paper. This analysis covers essential topics including data representation, computer hardware, networking, logic gates, database queries, and algorithm design. Each question is broken down with step-by-step explanations, common pitfalls, and practical tips to help you strengthen your understanding and achieve top marks in the exam.

欢迎阅读我们精心准备的11年级CIE计算机单元测试模拟卷解析。本文涵盖了数据表示、计算机硬件、网络基础、逻辑门电路、数据库查询以及算法设计等核心考点。每道题都配有分步解析、常见错误提示和实用技巧，帮助你巩固知识点，在考试中轻松斩获高分。

1. Question 1: Binary to Denary Conversion | 问题1：二进制转十进制

The first question reads: "Convert the binary number 10110110 into denary." An 8-bit binary number uses place values from left to right: 2⁷ (128), 2⁶ (64), 2⁵ (32), 2⁴ (16), 2³ (8), 2² (4), 2¹ (2) and 2⁰ (1). Matching the bits 1, 0, 1, 1, 0, 1, 1, 0 to these place values gives:

第一题要求：“将二进制数10110110转换为十进制。”一个8位的二进制数从右到左的位值分别为：2⁷ (128)、2⁶ (64)、2⁵ (32)、2⁴ (16)、2³ (8)、2² (4)、2¹ (2) 和 2⁰ (1)。将比特位1、0、1、1、0、1、1、0与位值对应如下：

he denary equivalent of 10110110 is 182. A common mistake is miscounting the powers of two – always double-check the rightmost bit represents 2⁰ = 1.

将比特为1的位值相加：128 + 32 + 16 + 4 + 2 = 182。因此，10110110的十进制值为182。常见错误是混淆2的幂次——务必记住最右边的位代表2⁰ = 1。

2. Question 2: Denary to Hexadecimal Conversion | 问题2：十进制转十六进制

Question 2 asks: "Convert the denary number 200 into hexadecimal." 8 stays as 8. Thus, 200 in denary equals C8 in hexadecimal. We write it as C8 or 0xC8. To verify, convert back: C × 16 + 8 = 12 × 16 + 8 = 192 + 8 = 200.

第二题：“将十进制数200转换为十六进制。”将200除以16：200 ÷ 16 = 12 余 8。在十六进制中，数字10到15用字母A到F表示，因此12变为C，余数8不变。故十进制200等于十六进制C8，也可写作0xC8。反算验证：C × 16 + 8 = 12 × 16 + 8 = 192 + 8 = 200。

3. Question 3: ROM vs RAM | 问题3：只读存储器与随机存取存储器的区别

Question 3: "Explain the difference between ROM and RAM." ROM (Read-Only Memory) is non-volatile, meaning it retains data even when the computer is powered off. It stores firmware, such as the BIOS, which is essential for booting. Its contents are usually written once during manufacture and are not easily modified. RAM (Random Access Memory) is volatile – all data is lost when power is removed. It holds the operating system, applications and data currently in use, enabling fast read/write operations. The table below summarises the key differences:

第三题：“解释ROM和RAM的区别。”ROM（只读存储器）是非易失性的，即断电后数据仍能保留。它存储固件，例如启动所需的BIOS，其内容通常在制造时写入且不易修改。RAM（随机存取存储器）是易失性的——断电后所有数据立即丢失。它存放当前正在使用的操作系统、应用程序和数据，支持快速读写。下表总结了关键区别：

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4. Question 4: Logic Gates – AND Gate | 问题4：逻辑门——与门

Question 4: "Which logic gate gives output 1 only when all inputs are 1?" l multiplication. For a two-input AND gate, the output Q = A AND B. The truth table shows:

第四题：“哪种逻辑门仅在全部输入为1时输出1？”答案是AND门（与门）。AND门执行逻辑乘法。对于二输入与门，输出Q = A AND B。真值表如下：

Note that other gates have different behaviour: OR outputs 1 if any input is 1; NOT inverts its input; NAND is the opposite of AND. The question specifically describes the AND gate.

请注意其他逻辑门有不同行为：OR门任意输入为1则输出1；NOT门将输入取反；NAND门与AND门输出相反。本题描述的特征正好对应AND门。

5. Question 5: Purpose of a Router | 问题5：路由器的用途

Question 5: "Describe the purpose of a router in a network." A router is a networking device that forwards data packets between different networks. It examines the destination IP address of each packet and consults its routing table to determine the optimal path to the destination. Routers operate at the network layer (Layer 3) of the OSI model. In a home or school setting, a router connects the local area network (LAN) to the internet (WAN) and often includes Network Address Translation (NAT) and firewall features. Without a router, devices on separate networks cannot communicate directly.

第五题：“描述路由器在网络中的用途。”路由器是一种网络设备，用于在不同网络之间转发数据包。它检查每个数据包的目标IP地址，并查询路由表以确定到达目的地的最佳路径。路由器工作在OSI模型的网络层（第三层）。在家庭或学校环境中，路由器将局域网（LAN）连接到互联网（WAN），通常还具备网络地址转换（NAT）和防火墙功能。若没有路由器，不同网络中的设备无法直接通信。

6. Question 6: IP Addressing and Subnet Mask | 问题6：IP地址与子网掩码

Question 6 provides the IP address 192.168.1.10 and subnet mask 255.255.255.0, and asks: "Identify the network ID." The subnet mask uses 255 in the first three octets, indicating that those octets represent the network portion. The last octet (0) is reserved for host addresses. Therefore, the network ID is 192.168.1.0. All devices sharing the same network ID are on the same logical network segment. The host ID for the given device is .10, making its full address 192.168.1.10.

第六题给定了IP地址192.168.1

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