📚 Year 7 AQA Computer Science: Mock Test Walkthrough | Year 7 AQA 计算机科学：单元测试模拟卷解析
This walkthrough provides a detailed breakdown of a typical Year 7 AQA Computer Science unit test. Each question is presented with a clear explanation, helping you understand the core concepts and avoid common mistakes. Use this guide to check your knowledge and build confidence for your real assessment.
这份模拟卷解析针对典型的 Year 7 AQA 计算机科学单元测试，逐题提供详细讲解，帮助你理解核心概念并避免常见错误。用这份指南来检验你的知识，树立信心，迎接真实评估。
1. Hardware vs. Software | 硬件与软件
Question: Tick (√) the correct box to show whether each item is hardware or software.
问题：在下表中为每个项目勾选正确的栏目，表明它是硬件还是软件。
| Item | Hardware | Software |
|---|---|---|
| Monitor | √ | |
| Scratch | √ | |
| Printer | √ | |
| Windows OS | √ |
Hardware refers to the physical parts of a computer that you can touch, such as a monitor, keyboard, or printer. Software is a set of instructions that tells the hardware what to do; examples include operating systems, games, and applications like Scratch.
硬件是指计算机的物理部件，你可以触摸到，例如显示器、键盘或打印机。软件是一组指令，告诉硬件该做什么；例子包括操作系统、游戏以及像 Scratch 这样的应用程序。
Common mistake: some students think that anything shown on screen is software. Remember, the screen itself is hardware, but the programs displayed are software.
常见错误：有些学生认为屏幕上显示的任何东西都是软件。请记住，屏幕本身是硬件，但显示的程序是软件。
2. Input and Output Devices | 输入与输出设备
Question: Identify each device as an input, output, or storage device. Circle the correct classification.
问题：识别每个设备属于输入设备、输出设备还是存储设备，在正确的类别上画圈。
- Mouse → Input / Output / Storage
- Speakers → Input / Output / Storage
- USB flash drive → Input / Output / Storage
- Microphone → Input / Output / Storage
Answers: Mouse – Input, Speakers – Output, USB flash drive – Storage, Microphone – Input. Input devices send data into the computer; output devices receive data from the computer; storage devices keep data for later use.
答案：鼠标——输入，扬声器——输出，U 盘——存储，麦克风——输入。输入设备将数据送入计算机；输出设备从计算机接收数据；存储设备保存数据供以后使用。
A touchscreen is a special case: it functions as both input (you touch it) and output (it displays information). In Year 7 tests, it is often accepted as an input device or both.
触摸屏是一个特殊例子：它既作为输入设备（你触摸它）也作为输出设备（它显示信息）。在 Year 7 测试中，通常可被接受为输入设备或两者皆是。
3. Binary Numbers | 二进制数
Question: Convert the decimal number 13 into binary. Show your working.
问题：将十进制数 13 转换成二进制。写出你的计算过程。
place a 1 under 8, 1 under 4, 0 under 2, and 1 under 1. The binary representation is 1101.
我们使用 8-4-2-1 位值法：8 + 4 + 1 = 13。因此在 8 下面放 1，4 下面放 1，2 下面放 0，1 下面放 1。二进制表示为 1101。
the sum matches the original decimal number: (1×8) + (1×4) + (0×2) + (1×1) = 8+4+0+1 = 13.
务必记住二进制数最左侧是最高有效位，并检查总和是否等于原十进制数：(1×8) + (1×4) + (0×2) + (1×1) = 8+4+0+1 = 13。
4. Data Units | 数据单位
Question: Put these data sizes in order from smallest to largest: 1 MB, 1 KB, 1 GB, 1 Byte, 1 Bit.
问题：将下列数据大小从小到大排列：1 MB、1 KB、1 GB、1 Byte、1 Bit。
te). A bit is the smallest unit; 8 bits make 1 byte. Then 1024 bytes make 1 KB, 1024 KB make 1 MB, and 1024 MB make 1 GB.
正确顺序为：比特 → 字节 → KB（千字节） → MB（兆字节） → GB（吉字节）。比特是最小单位；8 个比特组成 1 字节。然后 1024 字节构成 1 KB，1024 KB 构成 1 MB，1024 MB 构成 1 GB。
Examiners often check whether you confuse bits with bytes. Always note the capital 'B' for Byte and lowercase 'b' for bit when writing abbreviations.
考官经常检查你是否混淆比特和字节。书写缩写时务必注意大写 'B' 表示字节，小写 'b' 表示比特。
5. CPU, RAM, and ROM | 中央处理器、随机存取存储器与只读存储器
Question: Match each component to its description. Write the correct letter next to each term.
问题：将每个部件与其描述配对。在术语旁写上正确的字母。
| CPU | A. Stores the boot-up instructions permanently |
| RAM | B. Executes instructions and performs calculations |
| ROM | C. Temporarily holds data and programs currently in use |
Answer: CPU – B, RAM – C, ROM – A. The Central Processing Unit (CPU) is the brain of the computer that processes instructions. RAM is volatile memory that loses its content when the power is off, used for running programs. ROM is non-volatile and stores the firmware needed to start the computer.
答案：CPU – B，RAM – C，ROM – A。中央处理器（CPU）是计算机的大脑，负责处理指令。RAM 是易失性存储器，断电后内容丢失，用于运行程序。ROM 是非易失性存储器，存储启动计算机所需的固件。
A typical mistake is to say RAM is used for long-term storage like saving files; that is actually the job of the hard drive or SSD.
一个典型错误是说 RAM 用于长期存储如保存文件；那实际上是硬盘或固态硬盘的工作。
6. Algorithms and Flowcharts | 算法与流程图
Question: Draw a flowchart that asks the user to enter their age. If age is 13 or over, display "You can sign up". Otherwise, display "Sorry, too young".
问题：绘制一个流程图，要求用户输入年龄。如果年龄大于或等于 13，显示"You can sign up"；否则显示"Sorry, too young"。
nput/output, a diamond for decision, and rectangles for processes. The decision box checks "age >= 13?". If true, output "You can sign up"; if false, output "Sorry, too young".
你的流程图应包含椭圆形表示开始/结束，平行四边形表示输入/输出，菱形表示判断，矩形表示处理。判断框检查"age >= 13?"。如果为真，输出"You can sign up"；如果为假，输出"Sorry, too young"。
In an exam, always use the correct symbols and ensure the flow lines have arrows. Missing arrows can lose marks. The logical flow must be clear and the condition must have exactly two exit paths (Yes/No).
在考试中，务必使用正确的符号并确保流程线带有箭头。缺少箭头会丢分。逻辑流程必须清晰，条件必须有且仅有两条出口路径（是/否）。
7. Programming Basics – Variables and Loops | 编程基础——变量与循环
Question: What will be the output of this pseudocode?
total ← 0
FOR count FROM 1 TO 5
total ← total + count
ENDFOR
OUTPUT total
问题：以下伪代码的输出是什么？（代码如上）
0 and accumulates: 0+1=1, 1+2=3, 3+3=6, 6+4=10, 10+5=15. Therefore the final output is 15.
循环运行，count 取值 1, 2, 3, 4, 5。变量 total 从 0 开始累加：0+1=1, 1+2=3, 3+3=6, 6+4=10, 10+5=15。因此最终输出为 15。
It is essential to trace the value of the variable on paper during the test. Write down each step to avoid off-by-one errors, which are very common when students miscount the number of iterations.
测试时在纸上跟踪变量的值至关重要。写下每一步，避免 off-by-one 错误，这种错误在学生数错迭代次数时非常常见。
8. E-Safety and Network Etiquette | 网络安全与网络礼仪
Question: List three personal details you should never share online in a public chatroom. Explain why for each.
问题：列出三项你不应该在公共聊天室分享的个人信息，并逐一解释原因。
Answers may include: full name, home address, phone number, school name, or password. Sharing your full name and address can allow strangers to locate you in the real world, posing a safety risk. A phone number can be used for unwanted contact or scams. A password gives others access to your accounts, leading to identity theft or cyberbullying.
答案可以包括：全名、家庭地址、电话号码、学校名称或密码。分享全名和地址可能让陌生人在现实中找到你，构成安全风险。电话号码可被用于骚扰或诈骗。密码会让别人访问你的账户，导致身份盗用或网络欺凌。
Remember that even a photo with a school logo in the background can reveal your location. Think before you post and always apply the 'grandparent rule': if you wouldn't want your grandparents to see it, don't share it.
请记住，即使是背景中有学校标志的照片也可能暴露你的位置。发帖前要三思，并始终遵守"祖父母规则"：如果你不想让你的祖父母看到，那就不要分享。
9. Logic Gates – AND, OR, NOT | 逻辑门——与、或、非
truth table for the logic gate shown, where inputs are A and B, and output is Q = A AND B.
问题：完成所示逻辑门的真值表，输入为 A 和 B，输出为 Q = A AND B。
| A | B | Q (A AND B) |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
An AND gate outputs 1 only when both inputs are 1. In all other cases the output is 0. The OR gate outputs 1 if at least one input is 1, while the NOT gate simply inverts the single input.
与门仅在两个输入都为 1 时输出 1。其他情况下输出均为 0。或门只要至少一个输入为 1 就输出 1，而非门只是将单一输入取反。
Learn to recognise the distinctive shapes: the AND gate has a flat back and curved front; the OR gate has a curved back and pointed front; the NOT gate is a triangle with a small circle at the tip.
学会识别独特的形状：与门背面平直前面弯曲；或门背面弯曲前面尖；非门是一个三角形，尖端有一个小圆圈。
10. Sorting Algorithms – Bubble Sort | 排序算法——冒泡排序
Question: Show the first pass of a bubble sort on this list: [7, 3, 9, 1]. Describe what happens after each comparison.
问题：展示对列表 [7, 3, 9, 1] 进行冒泡排序的第一趟过程。描述每次比较后发生了什么。
p → [3, 7, 9, 1]. Compare 9 and 1 → swap → [3, 7, 1, 9]. After the first pass, the largest number (9) has bubbled to the end.
冒泡排序比较相邻元素，如果顺序错误就交换它们。第一趟：比较 7 和 3 → 交换 → [3, 7, 9, 1]。比较 7 和 9 → 不交换 → [3, 7, 9, 1]。比较 9 和 1 → 交换 → [3, 7, 1, 9]。第一趟结束后，最大的数 (9) 已经冒泡到末尾。
In the exam, ensure you show the list after every swap to gain full marks. A common error is to continue swapping beyond the first pass without resetting; the next pass only needs to check the first three elements, as the last is already sorted.
在考试中，确保每次交换后都展示列表状态以获得满分。一个常见错误是在第一趟后不重置而继续交换；下一趟只需检查前三个元素，因为最后一个已经排好序。
Published by TutorHao | Computer Science Revision Series | aleveler.com
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