Year 9 AQA Computer Science: Unit Test Mock Exam Walkthrough | Year 9 AQA 计算机：单元测试模拟卷解析

📚 Year 9 AQA Computer Science: Unit Test Mock Exam Walkthrough | Year 9 AQA 计算机：单元测试模拟卷解析

Welcome to this detailed walkthrough of a Year 9 AQA Computer Science unit test mock paper. We will break down each question type, provide model answers, and explain the underlying concepts to help you prepare effectively.

欢迎阅读这篇关于 Year 9 AQA 计算机科学单元测试模拟卷的详细解析。我们将逐一分解每种题型，提供参考答案，并解释背后的概念，帮助你有效备考。

1. Mock Paper Overview | 模拟卷概览

This mock exam simulates a typical end-of-unit assessment for Year 9 students following the AQA Computer Science curriculum. It covers fundamental topics such as data representation, computer architecture, algorithms, programming basics, logic gates, and networking. The paper includes a mix of multiple-choice, short-answer, and code-based questions. Understanding the structure will help you manage your time and focus during the actual test.

本模拟卷模拟了遵循 AQA 计算机科学课程大纲的 Year 9 学生的典型单元末评估。它涵盖了数据表示、计算机体系结构、算法、编程基础、逻辑门和网络等基本主题。试卷包含选择题、简答题和代码题等多种题型。了解试卷结构将有助于你在实际考试中管理时间并集中注意力。

2. Question 1: Binary Conversions | 问题1：二进制转换

The question: “Convert the denary number 189 to an 8-bit binary number. Show your working.” The standard method is to find the largest power of 2 that fits into the number, subtract, and repeat, using the 8 bit positions (128, 64, 32, 16, 8, 4, 2, 1).

题目：“将十进制数 189 转换为 8 位二进制数。写出演算过程。” 标准方法是找出不超过该数的最大 2 的幂，减去后重复，使用 8 个位权（128、64、32、16、8、4、2、1）。

189 – 128 = 61 remainder -> bit 7 = 1; 61 – 32 = 29 -> bit 5 = 1; 29 – 16 = 13 -> bit 4 = 1; 13 – 8 = 5 -> bit 3 = 1; 5 – 4 = 1 -> bit 2 = 1; 1 – 2 = not possible -> bit 1 = 0; 1 – 1 = 0 -> bit 0 = 1. The binary is 10111101.

189 – 128 = 61，余数 -> 位 7 = 1；61 – 32 = 29 -> 位 5 = 1；29 – 16 = 13 -> 位 4 = 1；13 – 8 = 5 -> 位 3 = 1；5 – 4 = 1 -> 位 2 = 1；1 – 2 不够减 -> 位 1 = 0；1 – 1 = 0 -> 位 0 = 1。二进制为 10111101。

Another sub-question: “Convert the binary 01011010 to denary.” Place values: 128,64,32,16,8,4,2,1. Bits: 0,1,0,1,1,0,1,0. Sum: 64+16+8+2 = 90.

另一个子问题：“将二进制 01011010 转换为十进制。” 位权：128、64、32、16、8、4、2、1。各位：0、1、0、1、1、0、1、0。计算总和：64+16+8+2 = 90。

3. Question 2: Units of Storage | 问题2：存储单位

Question: “Arrange the following storage sizes from smallest to largest: 2 GB, 512 MB, 5000 KB, 0.5 TB.” To compare, convert all to a common unit such as megabytes. 1 KB = 1024 bytes, 1 MB = 1024 KB, 1 GB = 1024 MB, 1 TB = 1024 GB.

题目：“将以下存储容量从小到大排列：2 GB, 512 MB, 5000 KB, 0.5 TB。” 为了比较，需要将所有值转换为同一单位，比如兆字节。1 KB = 1024 字节，1 MB = 1024 KB，1 GB = 1024 MB，1 TB = 1024 GB。

Convert 5000 KB to MB: 5000 ÷ 1024 ≈ 4.88 MB. 512 MB remains 512 MB. 2 GB = 2 × 1024 = 2048 MB. 0.5 TB = 0.5 × 1024 = 512 GB = 512 × 1024 = 524288 MB. Therefore, the correct order from smallest to largest is: 5000 KB, 512 MB, 2 GB, 0.5 TB.

转换 5000 KB 为 MB：5000 ÷ 1024 ≈ 4.88 MB。512 MB 保持不变。2 GB = 2048 MB。0.5 TB = 512 GB = 524288 MB。因此，从小到大排列为：5000 KB, 512 MB, 2 GB, 0.5 TB。

4. Question 3: CPU Components | 问题3：CPU 组件

Question: “Match each CPU component to its function: (a) Memory Data Register (MDR), (b) Memory Address Register (MAR), (c) Arithmetic Logic Unit (ALU), (d) Control Unit.” Functions: i) Stores the address of the next instruction or data to be fetched; ii) Performs calculations and logical operations; iii) Temporarily holds data that has been fetched from or is to be written to memory; iv) Decodes instructions and sends control signals.

题目：“将每个 CPU 组件与其功能匹配：(a) 内存数据寄存器 (MDR)，(b) 内存地址寄存器 (MAR)，(c) 算术逻辑单元 (ALU)，(d) 控制单元。” 功能：i) 存储待取指令或数据的地址；ii) 执行计算和逻辑操作；iii) 临时存放从内存读取或待写入内存的数据；iv) 译码指令并发送控制信号。

Correct matches: MDR -> iii (holds data); MAR -> i (holds address); ALU -> ii (calculations); Control Unit -> iv (decode and control

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