📚 Year 10 Cambridge Computer Science Unit Test Mock Paper Walkthrough | 剑桥10年级计算机科学单元测试模拟卷解析
This comprehensive walkthrough examines a mock unit test designed for Year 10 Cambridge Computer Science students. It covers core topics including data representation, logic gates, computer architecture, networking, and algorithm design. Each question is followed by a detailed explanation to help you understand the underlying concepts and avoid common pitfalls.
本篇文章深入解析一份为剑桥Year 10计算机科学设计的单元测试模拟卷，涵盖数据表示、逻辑门、计算机架构、网络和算法设计等核心主题。每道题目后附有详细解释，帮助你理解背后的概念并避免常见错误。
1. Binary to Decimal Conversion and Binary Addition | 二进制转十进制与二进制加法
Question: Convert the 8-bit binary number 10110110₂ into its decimal equivalent. Then add the binary number 00101101₂ to it using 8-bit addition, showing all carries. Determine whether an overflow error occurs and explain why.
问题：将 8 位二进制数 10110110₂ 转换为十进制。然后使用 8 位加法将其与 00101101₂ 相加，写出所有进位。判断是否发生溢出错误，并解释原因。
+ 2 = 182₁₀. The binary addition proceeds as follows: 10110110 + 00101101 = 11100011₂. Let's verify in decimal: 182 + 45 = 227. Since 227 is within the range 0– flow occurs. An overflow flag would only be raised if the result exceeded 11111111₂ (255).
答案： + (0×64) + (1×32) + (1×16) + (0×8) + (1×4) + (1×2) + (0×1) = 128 + 32 + 16 + 4 + 2 = 182₁₀。二进制加法过程：10110110 + 00101101 = 11100011₂。用十进制验证：182 + 45 = 227。由于 227 在 8 位无符号整数的范围 0–255 之内，没有发生溢出。只有结果超过 11111111₂ (255) 时才会触发溢出标志。
2. Hexadecimal to Binary and Decimal | 十六进制转二进制与十进制
Question: The hexadecimal number 5F₁₆ is given. Convert it into an 8-bit binary number and then into decimal, showing each step of the conversion process.
问题：已知十六进制数 5F₁₆。先将其转换为 8 位二进制数，再转换为十进制数，并写出每一步转换过程。
: (0×128) + (1×64) + (0×32) + (1×16) + (1×8) + (1×4) + (1×2) + (1×1) = 64 + 16 + 8 + 4 + 2 + 1 = 95₁₀. Therefore, 5F₁₆ = 01011111₂ = 95₁₀.
答案：分别处理每个十六进制位。5₁₆ → 0101₂（因为 5 = 4+1）。F₁₆ → 1111₂（F = 15）。组合后得到 01011111₂。十进制值计算： (1×64) + (0×32) + (1×16) + (1×8) + (1×4) + (1×2) + (1×1) = 64 + 16 + 8 + 4 + 2 + 1 = 95₁₀。因此，5F₁₆ = 01011111₂ = 95₁₀。
3. Image File Size and Colour Depth | 图像文件大小与颜色深度
Question: A digital image has a width of 800 pixels and a height of 600 pixels, with a colour depth of 24 bits per pixel. Calculate the uncompressed file size in kilobytes (KB). Explain what would happen to the file size and the visual quality if the colour depth were reduced to 8 bits per pixel.
问题：一张数字图像的宽为 800 像素、高为 600 像素，颜色深度为每像素 24 位。计算未压缩文件的大小，单位为千字节 (KB)。解释当颜色深度降低到每像素 8 位时，文件大小和视觉质量会如何变化。
tes. Convert to kilobytes: 1,440,000 ÷ 1024 = 1406.25 KB. If the colour depth is reduced to 8 bits per pixel, the number of colours decreases from 2²⁴ (about 16.7 million) to 2⁸ (256 colours). The file size becomes one-third of the original (800 × 600 × 8 = 3,840,000 bits = 468.75 KB), but the image quality drops significantly, leading to visible banding and loss of smooth colour transitions.
答案：总像素数 = 800 × 600 = 480,000 像素。总位数 = 480,000 × 24 = 11,520,000 位。转换为字节：11,520,000 ÷ 8 = 1,440,000 字节。转换为千字节：1,440,000 ÷ 1024 = 1406.25 KB。若颜色深度降为每像素 8 位，可显示的颜色数从 2²⁴（约 1670 万种）降至 2⁸（256 种）。文件大小变为原来的三分之一（800 × 600 × 8 = 3,840,000 位 = 468.75 KB），但图像质量明显下降，会出现可见的色带和平滑过渡的丢失。
4. Sound Sampling and Digital Audio | 声音采样与数字音频
Question: Describe the process by which an analogue sound wave is converted into a digital form using sampling. Identify two factors that determine the quality of the digital recording and explain how each factor affects the file size.
问题：描述通过采样将模拟声波转换为数字形式的过程。指出决定数字录音质量的两个因素，并解释每个因素如何影响文件大小。
Answer: An analogue sound wave is continuous. To digitise it, the amplitude of the wave is measured at regular intervals – this is sampling. Each sample is quantised to the nearest value according to the bit depth. The two critical factors are the sampling rate (e.g., 44.1 kHz) and the bit depth (e.g., 16 bits). A higher sampling rate captures more detail per second, improving fidelity but increasing the number of samples, hence a larger file. A greater bit depth allows more precise amplitude measurements, giving a wider dynamic range and less quantisation noise, but each sample uses more bits, further enlarging the file.
答案：模拟声波是连续的。为了数字化，需要在固定间隔测量声波的振幅——这就是采样。每个样本根据位深度量化为最接近的数值。两个关键因素是采样率（例如 44.1 kHz）和位深度（例如 16 位）。更高的采样率每秒捕捉更多细节，能提高保真度，但会增加样本数量，从而导致文件变大。更大的位深度能进行更精细的振幅测量，提供更宽的动态范围和更少的量化噪声，但每个样本占用更多位，进一步增大了文件。
5. Logic Gate Truth Table and Evaluation | 逻辑门真值表与求值
le with inputs A, B, C. Using your table, determine the output X when A = 1, B = 0, C = 1.
问题：对于逻辑表达式 X = (A AND B) OR (NOT C)，列出输入 A、B、C 的完整真值表。利用真值表，求出当 A = 1, B = 0, C = 1 时的输出 X。
. The output column X is calculated by first evaluating (A AND B), then (NOT C), and finally OR-ing the two.
答案：真值表有 2³ = 8 行。输出列 X 的计算步骤为：先求 (A AND B)，再求 (NOT C)，最后将两者进行 OR 运算。
| A | B | C | A AND B | NOT C | X |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 |
From the table, when A = 1, B = 0, C = 1: (A AND B) = 0, NOT C = 0, so X = 0 OR 0 = 0.
查表可知，当 A = 1, B = 0, C = 1 时：(A AND B) = 0，NOT C = 0，因此 X = 0 OR 0 = 0。
6. CPU Components and the Fetch-Decode-Execute Cycle | CPU 组件与取指-解码-执行周期
Question: Explain the roles of the Control Unit (CU) and the Arithmetic Logic Unit (ALU) inside the Central Processing Unit (CPU). Then describe in detail what happens during the fetch, decode, and execute phases of the von Neumann machine cycle.
问题：解释中央处理器 (CPU) 内部的控制单元 (CU) 和算术逻辑单元 (ALU) 的作用。然后详细描述冯·诺依曼架构机器周期中取指、解码、执行三个阶段所发生的事情。
Answer: The Control Unit coordinates all activities of the CPU. It decodes instructions, generates control signals to direct data flow between registers, memory, and the ALU, and manages the program counter. The ALU performs arithmetic operations (addition, subtraction) and logical operations (AND, OR, NOT), as well as comparisons. In the fetch phase, the address in the Program Counter (PC) is placed on the address bus, the instruction is read from memory into the Memory Data Register (MDR), and then transferred to the Current Instruction Register (CIR), with the PC incremented. In the decode phase, the CU interprets the bit pattern in the CIR to determine the operation and the addressing mode. During execution, the CU activates the relevant components; if an arithmetic operation is required, operands are fetched from registers or memory, the ALU performs the calculation, and the result is stored back, typically in the Accumulator or a general-purpose register. The cycle then repeats.
答案：控制单元负责协调 CPU 的所有活动。它解码指令，产生控制信号以引导寄存器、内存和 ALU 之间的数据流动，并管理程序计数器。ALU 执行算术运算（加、减）和逻辑运算（AND、OR、NOT）以及比较操作。取指阶段：程序计数器 (PC) 中的地址发到地址总线上，指令从内存读入内存数据寄存器 (MDR)，然后传送到当前指令寄存器 (CIR)，同时 PC 加 1。解码阶段：CU 解析 CIR 中的位模式，确定操作码和寻址方式。执行阶段：CU 激活相关组件；如果需要算术运算，就从寄存器或内存中取得操作数，ALU 完成计算，结果通常存回累加器或通用寄存器中。之后循环再次开始。
7. Comparing RAM and ROM | RAM 与 ROM 的比较
Question: Compare Random Access Memory (RAM) and Read-Only Memory (ROM) in terms of volatility, typical usage in a personal computer, and whether their contents can be altered during normal operation.
问题：从易失性、个人计算机中的典型用途以及正常运行时内容是否可修改这三个方面，比较随机存取存储器 (RAM) 和只读存储器 (ROM)。
Answer: RAM is volatile – its data is lost when the power is switched off. It holds the operating system, applications, and data currently in use, allowing fast read and write operations. ROM is non-volatile; it retains data even without power. It typically stores the BIOS or firmware needed to boot the computer. RAM can
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