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

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

This walkthrough covers a mock unit test designed for Year 9 students following the CIE Computer Science pathway. The paper focuses on foundational topics such as number systems, logic gates, computer architecture, algorithms, and introductory Python programming. Each question is broken down with a clear solution and explanation, helping you understand the reasoning behind correct answers and reinforcing key concepts essential for progression into IGCSE Computer Science.

本解析针对为学习 CIE 计算机科学路径的 Year 9 学生设计的单元模拟测试卷。试卷重点涵盖数制系统、逻辑门、计算机体系结构、算法以及 Python 入门编程等基础主题。每道题目均配有清晰的解答与讲解，帮助理解正确答案背后的推理，并巩固升入 IGCSE 计算机科学所必需的关键概念。

---

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

Question: Convert the binary number 1101 1010 to its denary (decimal) equivalent. Show your working.

问题：将二进制数 1101 1010 转换为等值的十进制数。写出演算过程。

Solution: er nibble. Values: 128 + 64 + 0 + 16 + 8 + 0 + 2 + 0 = 218. Therefore, 1101 1010₂ = 218₁₀.

解析：将二进制数字写在相应的位值下方，从 128(2⁷) 到 1(2⁰)。高四位为 1,1,0,1，低四位为 1,0,1,0。求和：128 + 64 + 0 + 16 + 8 + 0 + 2 + 0 = 218。因此，1101 1010₂ = 218₁₀。

---

2. Denary to Binary Conversion | 十进制转二进制

Question: Convert the denary number 195 into an 8-bit binary number.

问题：将十进制数 195 转换为 8 位二进制数。

Solution: place 0 under those. 3 - 2 = 1; place 1 under 2. 1 - 1 = 0; place 1 under 1. Result: 1100 0011₂.

解析：使用除 2 取余法或通过与位值比较。195 - 128 = 67，在 128 下记 1。67 - 64 = 3，在 64 下记 1。剩余值 3：32,16,8 无法减去，对应位记 0。3 - 2 = 1，在 2 下记 1。1 - 1 = 0，在 1 下记 1。结果为 1100 0011₂。

---

3. Hexadecimal to Denary Conversion | 十六进制转十进制

Question: Convert the hexadecimal value 2F into denary.

问题：将十六进制值 2F 转换为十进制。

Solution: he right in the 1s place. Calculation: (2 × 16) + (15 × 1) = 32 + 15 = 47. So 2F₁₆ = 47₁₀.

解析：十六进制中 F 代表 15。左边数字位于 16 的位，右边数字位于 1 的位。计算：(2 × 16) + (15 × 1) = 32 + 15 = 47。因此 2F₁₆ = 47₁₀。

---

4. Denary to Hexadecimal Conversion | 十进制转十六进制

Question: Convert the denary number 63 into hexadecimal.

问题：将十进制数 63 转换为十六进制。

Solution: The quotient 3 gives the left digit; remainder 15 corresponds to F. Therefore 63₁₀ = 3F₁₆.

解析：将 63 除以 16。63 ÷ 16 = 3 余 15。商 3 为左边数字；余数 15 对应 F。因此 63₁₀ = 3F₁₆。

---

5. Logic Gates: The AND Gate | 逻辑门：与门

Question: Complete the truth table for a two-input AND gate. Inputs are A and B, output is Q.

问题：完成双输入与门的真值表。输入为 A 和 B，输出为 Q。

A	B	Q (A AND B)
0	0	0
0	1	0
1	0	0
1	1	1

Solution: only when both inputs are 1. Hence the truth table above: Q = 1 only when A = 1 and B = 1.

解析：与门仅在两个输入均为 1 时输出 1。因此上表所示：仅当 A=1 且 B=1 时，Q=1。

---

广告 / Sponsored

6. Central Processing Unit (CPU) Components | 中央处理器部件

Question: Identify three main components inside the CPU and briefly describe the purpose of each.

问题：指出 CPU 内部的三个主要部件，并简要描述各自的用途。

Solution:
1. Control Unit (CU) – directs the flow of data and coordinates all processor activities by sending control signals.
2. Arithmetic Logic Unit (ALU) – performs arithmetic operations (addition, subtraction) and logical comparisons (AND, OR, NOT).
3. Registers – small, high-speed storage locations that temporarily hold data, instructions, or addresses currently being used by the CPU.

解析：
1. 控制单元 (CU) – 通过发送控制信号引导数据流动并协调处理器的所有活动。
2. 算术逻辑单元 (ALU) – 执行算术运算（加、减）和逻辑比较（AND、OR、NOT）。
3. 寄存器 – 小型高速存储位置，暂时存放 CPU 当前使用的数据、指令或地址。

---

7. Classifying Input and Output Devices | 输入输出设备分类

Question: From the following list, classify each device as input, output, or both: keyboard, touchscreen, speaker, scanner, projector, microphone.

问题：从下列清单中，将每个设备分类为输入设备、输出设备或两者兼具：键盘、触摸屏、扬声器、扫描仪、投影仪、麦克风。

Solution:
- Keyboard – input
- Touchscreen – both (input via touch, output via display)
- Speaker – output
- Scanner – input
- Projector – output
- Microphone – input

解析：
- 键盘 – 输入设备
- 触摸屏 – 二者兼具（通过触摸输入，通过显示输出）
- 扬声器 – 输出设备
- 扫描仪 – 输入设备
- 投影仪 – 输出设备
- 麦克风 – 输入设备

---

8. Flowchart Algorithm Analysis | 流程图算法分析

Question: The flowchart below reads two numbers X and Y, then outputs the larger one. Describe the decision process in plain English.

问题：以下流程图读取两个数 X 和 Y，然后输出其中较大的一个。用通俗英语描述其判断过程。

Solution: The algorithm first checks if X is greater than Y. If the condition is true, X is output as the largest. If the condition is false, Y must be greater or they are equal; in either situation Y is used as the larger number (for equality, displaying Y is acceptable). This is an example of a simple selection structure.

解析：算法首先检查 X 是否大于 Y。如果条件为真，则输出 X 作为最大值。如果条件为假，则 Y 必然更大或二者相等；在这两种情况下均使用 Y 作为较大者（相等时显示 Y 也可）。这是一个简单选择结构的例子。

---

9. Python Code Debugging | Python 代码调试

Question: The following Python code is supposed to print a message when the variable age is exactly 15. Identify and correct the error.

if age = 15
print('teenager')

问题：以下 Python 代码本应在变量 age 恰好为 15 时打印一条消息。找出并更正错误。

if age = 15
print('teenager')

Solution: There are two errors. First, a single equals sign '=' is the assignment operator; the equality comparison requires two equals signs '==' e containing print must be indented to become part of the if-block. The corrected code is:

if age == 15:
    print('teenager')

解析：存在两处错误。首先，单个等号 '=' 是赋值运算符；相等比较需使用双等号 '=='。其次，包含 print 的行必须缩进才能成为 if 语句块的一部分。修正后的代码为：

if age == 15:
    print('teenager')

---

10. Data Measurement Units | 数据计量单位

Question: How many bytes are there in 2 kilobytes (KB)? Show your reasoning and express the result as a power of two if possible.

问题：2 千字节（KB）中有多少字节？写出推理过程，并尽可能将结果表示为 2 的幂。

Solution: 1024 bytes. Therefore, 2 KB = 2 × 1024 = 2048 bytes. As a power of two, 2048 = 2¹¹ bytes.

解析：在计算机科学中，1 KB 通常定义为 2¹⁰ 字节 = 1024 字节。因此，2 KB = 2 × 1024 = 2048 字节。表示为 2 的幂，2048 = 2¹¹ 字节。

---

Published by TutorHao | Computer Science Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）