📚 Year 9 CIE Computer Science: Unit Test Mock Paper Solutions | Year 9 CIE 计算机：单元测试模拟卷解析
This article provides a full walkthrough of a mock unit test designed for Year 9 students following the Cambridge CIE Computer Science curriculum. Each section explains a typical question you might encounter in a unit test, along with clear solutions, common mistakes, and revision tips. Use this to check your understanding before the real assessment.
本文为Year 9 CIE计算机课程的学生提供一份单元检测模拟卷的详细解析。每个小节都围绕一道典型模拟题展开，给出清晰解答、常见误区以及复习提示，帮助你在线下测验前巩固知识。
1. Binary to Decimal Conversion | 二进制转十进制转换
Question: Convert the binary number 1101₂ to decimal. Show your working.
题目：将二进制数 1101₂ 转换为十进制，写出计算过程。
value and sum: 1×8 + 1×4 + 0×2 + 1×1 = 8 + 4 + 0 + 1 = 13. The decimal equivalent is 13₁₀.
解答：从右向左标出位权 8, 4, 2, 1。每一位数字乘以对应位权再求和：1×8 + 1×4 + 0×2 + 1×1 = 8 + 4 + 0 + 1 = 13。十进制结果为 13₁₀。
n the place values starting from the rightmost bit. Always begin with 2⁰ = 1 on the right.
常见错误：没有从最右边一位开始对齐位权。请牢记最右边一位是 2⁰ = 1，然后向左依次为 2¹, 2², 2³。
2. Decimal to Binary Conversion | 十进制转二进制转换
Question: Convert the decimal number 45 into an 8‑bit binary number.
题目：将十进制数 45 转换为 8 位二进制数。
1 remainder 0; 1 ÷ 2 = 0 remainder 1. Reading the remainders from bottom to top gives 101101. To make it 8‑bit, add two leading zeros: 00101101₂.
解答：采用除 2 取余法。45 ÷ 2 = 22 余 1；22 ÷ 2 = 11 余 0；11 ÷ 2 = 5 余 1；5 ÷ 2 = 2 余 1；2 ÷ 2 = 1 余 0；1 ÷ 2 = 0 余 1。从下往上读取余数得到 101101。补足 8 位，在前面加两个零：00101101₂。
: 00101101₂ = 0×128 + 0×64 + 1×32 + 0×16 + 1×8 + 1×4 + 0×2 + 1×1 = 32+8+4+1 = 45. Correct.
验证：00101101₂ = 32+8+4+1 = 45，正确。
3. Hexadecimal to Binary | 十六进制转二进制
Question: The hexadecimal number 2F is often used in computing. Convert 2F₁₆ into binary and then into decimal.
题目：十六进制数 2F 在计算机中很常见。将 2F₁₆ 转换为二进制，再转换为十进制。
Solution: Each hex digit converts to a 4‑ simply 101111₂ if leading zeros are dropped. For decimal: 2×16¹ + 15×16⁰ = 32 + 15 = 47₁₀.
解答：每一位十六进制数字对应一个 4 位二进制半字节。2₁₆ = 0010₂，F₁₆ = 15 = 1111₂。因此 2F₁₆ = 0010 1111₂，可写为 00101111₂ 或去掉前导零的 101111₂。转十进制：2×16¹ + 15×16⁰ = 32 + 15 = 47₁₀。
Remember: The letters A–F represent 10–15. F is the biggest single hex digit.
记忆：字母 A–F 分别代表 10–15，F 是最大的单数位十六进制数字。
4. Data Storage Units | 数据存储单位
Question: A file size is given as 2 KB. Express this size in bytes and in bits. Explain your calculations.
题目：一个文件大小为 2 KB。请将此大小以字节和位为单位表示，并说明计算过程。
1024 = 2048 bytes. Since 1 byte = 8 bits, the file size in bits is 2048 × 8 = 16384 bits.
解答：计算机科学中，1 KB = 1024 字节（二进制定义）。因此 2 KB = 2 × 1024 = 2048 字节。又因为 1 字节 = 8 位，所以以位为单位为 2048 × 8 = 16384 位。
rers use 1 KB = 1000 bytes, but CIE follows the binary convention unless otherwise stated.
注意：部分存储厂商使用 1 KB = 1000 字节，但 CIE 考试若无特别说明，遵循二进制换算。
5. Logic Gates and Truth Tables | 逻辑门与真值表
Question: Complete the truth table for a two‑input AND gate, an OR gate, and a NOT gate applied to input A.
题目：完成一个两输入与门、一个或门以及对输入 A 施加非门的真值表。
Solution:
| A | B | A AND B | A OR B | NOT A |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 0 |
The AND gate outputs 1 only when both inputs are 1. OR outputs 1 if at least one input is 1. NOT flips the input.
与门仅在两个输入均为 1 时输出 1；或门在至少一个输入为 1 时输出 1；非门将输入取反。
Common pitfall: Confusing AND with OR. Practice by imagining a circuit: AND is like two switches in series, OR is like two switches in parallel.
常见陷阱：混淆与门和或门。可以用电路类比记忆——与门相当于两个开关串联，或门相当于两个开关并联。
6. CPU Components and the Von Neumann Architecture | CPU 组件与冯·诺依曼架构
Question: Name the three main components of the CPU in the Von Neumann architecture and briefly describe the role of each.
题目：列出冯·诺依曼架构中 CPU 的三个主要组件，并简要说明各自的作用。
Solution: The three components are the Control Unit (CU), the Arithmetic Logic Unit (ALU), and the registers. The CU directs the flow of data and instructions, fetching and decoding them. The ALU performs all arithmetic and logical operations. Registers are small, fast storage locations inside the CPU, such as the Program Counter (PC) and Accumulator.
解答：三大组件为控制单元（CU）、算术逻辑单元（ALU）和寄存器。控制单元负责指挥数据与指令的流动，完成取指与译码。算术逻辑单元执行所有算术与逻辑运算。寄存器是 CPU 内部快速的小容量存储器，例如程序计数器（PC）和累加器。
Additionally, the Immediate Access Store (IAS) or primary memory holds data and instructions currently in use, often implemented as RAM.
此外，立即存取存储器（IAS）即主存，存放当前使用的数据与指令，通常由 RAM 实现。
7. Input and Output Devices | 输入与输出设备
Question: Classify the following devices as input, output, or both: touchscreen, printer, microphone, monitor, hard disk drive.
题目：将以下设备分类为输入设备、输出设备或兼具二者功能：触摸屏、打印机、麦克风、显示器、硬盘驱动器。
Solution: Touchscreen – both (input for touch, output for display). Printer – output. Microphone – input. Monitor – output. Hard disk drive – storage, but it provides both input and output when reading and writing data; in exam classification it is normally considered a secondary storage device, not strictly I/O.
解答：触摸屏——兼具输入（触摸）和输出（显示）功能。打印机——输出。麦克风——输入。显示器——输出。硬盘驱动器——存储设备，读取时类似输入，写入时类似输出；但考试中通常将其归为辅助存储设备，不归入单纯的 I/O 设备。
Key: Always read the question carefully. Some devices like a touchscreen or a headset with microphone are both.
要点：仔细审题，有些设备如触摸屏或带麦克风的耳机兼具输入输出功能。
8. Networks: LAN vs WAN | 网络：局域网与广域网
Question: Distinguish between a Local Area Network (LAN) and a Wide Area Network (WAN). Give one example of each.
题目：区分局域网（LAN）和广域网（WAN），并各举一例。
Solution: A LAN covers a small geographical area, such as a school building or a single office, and is usually owned by one organisation. It offers high data transfer speeds and low latency. Example: the school's computer network connecting all lab PCs. A WAN spans a large geographical area, often a country or the whole world, and relies on leased telecommunication lines or satellites. The internet is the largest WAN.
解答：局域网覆盖较小的地理范围，例如一栋教学楼或一间办公室，通常由单一组织拥有，数据传输速度快、延迟低。例如学校连接所有实验室电脑的网络。广域网覆盖广阔的地理区域，可达全国或全球，通常依赖租用的电信线路或卫星。互联网是最大的广域网。
Further detail: WANs are typically slower than LANs due to longer distances and shared infrastructure. Routers and modems are essential WAN devices.
补充：由于距离远和共享基础设施，广域网通常比局域网慢。路由器和调制解调器是关键的广域网设备。
9. Algorithms and Flowcharts | 算法与流程图
Question: Draw or describe a flowchart that takes two numbers as input and outputs the larger one.
题目：绘制或描述一个流程图，要求输入两个数，输出其中较大的那个。
Solution: The flowchart would start with a 'Start' oval, then a parallelogram for input 'Read A, B'. Next a decision diamond asks 'Is A > B?'. If yes, output A; if no, output B. Finally an 'End' oval. The process can be described in pseudocode as well: INPUT A, B; IF A > B THEN PRINT A ELSE PRINT B.
解答：流程图以开始椭圆框为起点，然后输入平行四边形框“输入 A, B”。接着判断菱形框“A > B?”，若为是则输出 A，否则输出 B。最后以结束椭圆框结束。伪代码可表示为：INPUT A, B; IF A > B THEN PRINT A ELSE PRINT B.
Key symbols: oval for start/end, parallelogram for input/output, rectangle for process, diamond for decision. Flow lines show direction.
关键符号：椭圆表示开始/结束，平行四边形表示输入/输出，矩形表示处理步骤，菱形表示判断，带箭头的流程线指示方向。
10. Programming Concepts: Trace Tables | 编程概念：跟踪表
Question: Study the following pseudocode and determine the final value of 'count':
题目：阅读以下伪代码，确定变量 count 的最终值：
count ← 0
FOR i ← 1 TO 5
count ← count + i
ENDFOR
; i=3: count=3+3=6; i=4: count=6+4=10; i=5: count=10+5=15. After the loop, count holds 15.
解答：跟踪循环过程。i=1: count=1; i=2: count=3; i=3: count=6; i=4: count=10; i=5: count=15。循环结束后 count 的值为 15。
Trace tables help you visualise each iteration. Always update the variable step by step.
跟踪表有助于直观展现每一次迭代，务必逐步更新变量的值。
11. Cybersecurity Threats | 网络安全威胁
Question: Explain what phishing is and describe one way users can protect themselves. Also name one type of malware and its effect.
题目：解释什么是网络钓鱼，并说明用户自我保护的一种方法。另举出一种恶意软件及其危害。
Solution: Phishing is a social engineering attack where criminals send fraudulent emails or messages that appear to be from trusted sources, attempting to trick recipients into revealing personal information like passwords or bank details. To protect themselves, users should verify the sender's address, avoid clicking suspicious links, and enable multi‑factor authentication. An example of malware is ransomware, which encrypts a victim's files and demands payment for the decryption key.
解答：网络钓鱼是一种社会工程攻击，犯罪分子发送看似来自可信来源的欺骗性邮件或消息，诱骗收件人泄露密码或银行信息等个人资料。用户可通过核实发件人地址、不点击可疑链接以及启用多因素身份验证来保护自己。恶意软件的例子如勒索软件，它会加密受害者的文件并要求支付赎金才能解密。
Other common malware: viruses (self‑replicate and corrupt files), worms (spread across networks), trojans (disguised as legitimate software).
其他常见恶意软件：病毒（自我复制并破坏文件）、蠕虫（通过网络传播）、特洛伊木马（伪装成合法软件）。
12. File Compression: Lossy vs Lossless | 文件压缩：有损与无损
Question: Compare lossy and lossless compression, giving one advantage of each and one suitable file type for each.
题目：比较有损压缩与无损压缩，各说出一条优点并举出一种适合的文件类型。
Solution: Lossless compression reduces file size without losing any data; the original file can be perfectly reconstructed. It is essential for text and program files. Advantage: no quality loss. Examples: PNG images, ZIP archives. Lossy compression permanently removes some data to achieve much smaller file sizes, and is suitable for multimedia where perfect accuracy is not essential. Advantage: significantly smaller files. Examples: JPEG images, MP3 audio.
解答：无损压缩减小文件大小但不丢失任何数据，原文件可被完美重建，适合文本和程序文件。优点：无质量损失，例如 PNG 图像、ZIP 压缩包。有损压缩永久性移除部分数据以换取更小的文件体积，适用于无需完美精度的多媒体文件。优点：文件体积显著缩小，例如 JPEG 图像、MP3 音频。
Revision tip: Remember that repeated patterns allow lossless compression (run‑length encoding), while human perception limits enable lossy compression (removing sounds we can't hear).
复习提示：无损压缩利用重复模式（如行程编码），有损压缩则利用人类感知局限（如移除听不见的声音）。
Published by TutorHao | Computer Science Revision Series | aleveler.com
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