📚 Year 9 CAIE Computer Science: Unit Test Mock Paper Walkthrough | CAIE Year 9 计算机 单元测试模拟卷解析
This walkthrough breaks down a full Year 9 CAIE Computer Science unit test mock paper, covering number systems, logic, data representation, pseudocode, networking, and programming. Each question is explained step by step so you can see exactly where marks are earned and how to avoid common mistakes.
本文全面解析一套 CAIE Year 9 计算机单元测试模拟卷，涵盖数制、逻辑、数据表示、伪代码、网络以及编程。每题均配有逐步讲解，帮助你明确得分点并避免典型错误。
1. Mock Paper Overview | 模拟卷概览
The paper contains eight questions worth a total of 50 marks. Topics tested include binary and hexadecimal conversions, data storage units, Boolean logic and logic gates, pseudocode tracing, networking fundamentals, and a Python programming task. Questions follow the style of CAIE Lower Secondary and IGCSE Computer Science assessments, focusing on application rather than mere recall.
本卷共八道大题，总分50分。考查内容包括二进制与十六进制转换、数据存储单位、布尔逻辑与逻辑门、伪代码跟踪、网络基础以及一道 Python 编程题。题型贴近 CAIE 初中及 IGCSE 计算机科学测评风格，侧重实际应用而非简单记忆。
2. Question 1: Number Systems – Binary & Denary | 第1题：数制——二进制与十进制
Question (a) – 2 marks: Convert 1011₂ to denary.
题目(a)（2分）： 将二进制数 1011₂ 转换为十进制。
is 1×8, the next 0×4, then 1×2, and the last 1×1. Adding these gives 8 + 0 + 2 + 1 = 11₁₀.
将二进制数按位值 8, 4, 2, 1 对位。1011₂：最左位为 1×8，次位 0×4，再是 1×2，最后 1×1。相加得 8 + 0 + 2 + 1 = 11₁₀。
Answer: 11
答案： 11
Question (b) – 2 marks: Convert the denary number 29 to binary.
题目(b)（2分）： 将十进制数 29 转换为二进制。
3 ÷ 2 = 1 remainder 1; 1 ÷ 2 = 0 remainder 1. Reading the remainders upwards gives 11101₂.
反复除以 2 记录余数：29 ÷ 2 = 14 余 1；14 ÷ 2 = 7 余 0；7 ÷ 2 = 3 余 1；3 ÷ 2 = 1 余 1；1 ÷ 2 = 0 余 1。从下往上读余数得 11101₂。
Answer: 11101
答案： 11101
Question (c) – 1 mark: vert 0011₂ to denary, treating it as a 4-bit number with the most significant bit as 1×2³.
题目(c)（1分）： 将 0011₂ 视为 4 位二进制数转换为十进制（最高位为 1×2³）。
change the value, the place values are still 8,4,2,1. 0011₂ = 0×8 + 0×4 + 1×2 + 1×1 = 3₁₀.
尽管前导零不影响数值，位值仍为 8,4,2,1。0011₂ = 0×8 + 0×4 + 1×2 + 1×1 = 3₁₀。
Answer: 3
答案： 3
3. Question 2: Hexadecimal Conversions | 第2题：十六进制转换
Question (a) – 1 mark: Convert A3₁₆ to denary.
题目(a)（1分）： 将 A3₁₆ 转换为十进制。
presents 10, and the digit 3 is in the units place. So A3₁₆ = 10×16 + 3 = 160 + 3 = 163₁₀.
十六进制数 A 表示 10，个位 3。因此 A3₁₆ = 10×16 + 3 = 160 + 3 = 163₁₀。
Answer: 163
答案： 163
Question (b) – 2 marks: Convert the denary number 255 to hexadecimal.
题目(b)（2分）： 将十进制数 255 转换为十六进制。
r 15. The quotient 15 is F in hex and the remainder 15 is also F. Hence the hexadecimal is FF₁₆.
将 255 除以 16：255 ÷ 16 = 15 余 15。商 15 在十六进制中为 F，余数 15 也为 F。因此十六进制为 FF₁₆。
Answer: FF
答案： FF
Question (c) – 1 mark: Give one reason why hexadecimal is used in computing.
题目(c)（1分）： 给出计算机中使用十六进制的一个原因。
Hexadecimal is more compact than binary: one hex digit represents four bits, so it is easier for humans to read and write memory addresses or colour codes.
十六进制比二进制更紧凑：一个十六进制数字表示四位二进制，便于人们读写内存地址或颜色编码。
4. Question 3: Data Storage Units | 第3题：数据存储单位
Question (a) – 1 mark: How many bits are in one byte?
题目(a)（1分）： 一个字节包含多少位？
One byte is 8 bits.
一个字节等于 8 位。
Answer: 8
答案： 8
Question (b) – 2 marks: Convert 2048 KB to MB.
题目(b)（2分）： 将 2048 KB 转换为 MB。
1 MB = 1024 KB. So divide 2048 by 1024: 2048 ÷ 1024 = 2. Therefore, 2048 KB = 2 MB.
1 MB = 1024 KB。将 2048 除以 1024：2048 ÷ 1024 = 2。因此 2048 KB = 2 MB。
Answer: 2 MB
答案： 2 MB
Question (c) – 2 marks: A file size is 2.5 MB. Express this in bytes, showing your working.
题目(c)（2分）： 某文件大小为 2.5 MB，请将其表示为字节数并给出计算过程。
1,048,576 = 2,621,440 bytes. Alternatively, 2.5 × 2²⁰ ≈ 2.5 × 1,048,576 = 2,621,440 bytes.
1 MB = 1024 × 1024 = 1,048,576 字节。于是 2.5 × 1,048,576 = 2,621,440 字节。也可写为 2.5 × 2²⁰ ≈ 2.5 × 1,048,576 = 2,621,440 字节。
Answer: 2,621,440 bytes
答案： 2,621,440 字节
Question (d) – 1 mark: Explain why 1 KB is often taken as 1024 bytes, not 1000 bytes.
题目(d)（1分）： 解释为什么 1 KB 常视为 1024 字节而非 1000 字节。
Computers use binary (base‑ ns with binary addressing; using 1000 would not match the natural power-of-2 organisation.
计算机采用二进制寻址，因此存储容量采用 2 的幂次。2¹⁰ = 1024，与二进制地址结构一致；若取 1000 则无法匹配二的幂次组织方式。
5. Question 4: Boolean Logic and Truth Tables | 第4题：布尔逻辑与真值表
Question (a) – 2 marks: Complete the truth table for A AND B.
题目(a)（2分）： 补全 A AND B 的真值表。
The AND operation outputs 1 only when both inputs are 1. The completed table is shown below.
当且仅当两个输入均为 1 时 AND 输出 1。完整真值表如下。
| A | B | A AND B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
真值表显示，只有 (1,1) 组合得到输出 1，其余情况输出 0。
Question (b) – 2 marks: Write the Boolean expression represented by 'X = NOT A OR (B AND C)' and draw the equivalent logic diagram (description only).
题目(b)（2分）： 写出表达式 'X = NOT A OR (B AND C)' 并描述其逻辑图。
The expression uses a NOT gate on A, an AND gate for B and C, then an OR gate combining the two. The diagram would show A going into a NOT, B and C into an AND, and the outputs feeding an OR gate producing X.
该表达式对 A 取反，B 与 C 相与，再将两者结果进行或运算。逻辑图中 A 通过非门，B 和 C 通过与门，两个输出送入或门得到 X。
Question (c) – 2 marks: Evaluate X when A=1, B=0, C=1.
题目(c)（2分）： 设 A=1, B=0, C=1，求 X 的值。
Step 1: NOT A = NOT 1 = 0. Step 2: B AND C = 0 AND 1 = 0. Step 3: X = 0 OR 0 = 0.
第一步：NOT A = NOT 1 = 0。第二步：B AND C = 0 AND 1 = 0。第三步：X = 0 OR 0 = 0。
Answer: X = 0
答案： X = 0
6. Question 5: Logic Gates Circuit | 第5题：逻辑门电路
Question – 5 marks: T C). Describe the circuit using AND, OR, NOT gates and give a truth table for all input combinations.
题目（5分）： 设计 X = (A AND B) OR (NOT C) 的逻辑电路。用与、或、非门描述电路，并给出所有输入组合的真值表。
The circuit has two branches: (1) A and B feed into an AND gate; (2) C passes through a NOT gate. The outputs of the AND gate and the NOT gate go into an OR gate, producing X.
电路分为两条支路：第一条 A 和 B 送入与门；第二条 C 通过非门。与门和非门的输出送入或门，产生 X。
The truth table for all combinations of A, B, C (8 rows) is shown below.
输入 A, B, C 共有 8 种组合，真值表如下。
| A | B | C | NOT C | A AND B | X |
|---|---|---|---|---|---|
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