📚 Year 8 AQA Computer Science: Unit Test Mock Analysis | 八年级AQA计算机：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 8 AQA Computer Science, covering topics such as binary conversion, binary addition, logic gates, flowchart symbols, pseudocode, programming concepts, hardware, networks, and cyber security. Each question is analysed with clear, step-by-step explanations to help you understand the answers and revise effectively for your upcoming assessment.
本文详细解析了一套八年级AQA计算机单元测试模拟卷，内容涵盖二进制转换、二进制加法、逻辑门、流程图符号、伪代码、编程概念、硬件、网络和网络安全等主题。每道题都配有清晰的分步解析，帮助你理解答案，为即将到来的评估高效复习。
1. Question 1: Binary to Denary Conversion | 第1题：二进制转十进制
Question: Convert the binary number 101011 into denary.
题目：将二进制数 101011 转换为十进制。
To convert a binary number to denary, we multiply each binary digit (bit) by its place value, which is a power of 2. We start from the rightmost bit, which has place value 2⁰ (1), and move left, increasing the power by one each time.
要将二进制数转换为十进制，需要把每一位数字乘以对应的位值（2的幂）。从最右侧的位开始，位值为2⁰（即1），向左每移动一位，幂次增加1。
The 6-bit binary number 101011 has bits from left to right: 1, 0, 1, 0, 1, 1. We can list the place values underneath: 2⁵ (32), 2⁴ (16), 2³ (8), 2² (4), 2¹ (2), 2⁰ (1).
6位二进制数101011从左到右各位为：1, 0, 1, 0, 1, 1。我们可以列出对应的位值：2⁵(32), 2⁴(16), 2³(8), 2²(4), 2¹(2), 2⁰(1)。
Now we multiply each bit by its place value and sum the results. Bits of 1 mean we include that power of 2; bits of 0 mean we ignore it.
接下来将每一位乘以对应的位值并求和。位为1表示包含该2的幂，位为0则忽略不计。
) + (1 × 8) + (0 × 4) + (1 × 2) + (1 × 1). This simplifies to 32 + 0 + 8 + 0 + 2 + 1 = 43.
计算过程：(1 × 32) + (0 × 16) + (1 × 8) + (0 × 4) + (1 × 2) + (1 × 1)。化简为 32 + 0 + 8 + 0 + 2 + 1 = 43。
Therefore, the denary equivalent of binary 101011 is 43.
因此，二进制101011对应的十进制数是 43。
2. Question 2: Denary to Binary Conversion | 第2题：十进制转二进制
Question: Convert the denary number 89 into an 8-bit binary number.
题目：将十进制数89转换为8位二进制数。
The most reliable method for small denary numbers is repeated division by 2, recording the remainder at each step. The binary digits are the remainders read from bottom to top.
对于较小的十进制数，最可靠的方法是反复除以2，记录每一步的余数。从下往上读取余数即得到二进制位。
ep 5: 5 ÷ 2 = 2 remainder 1. Step 6: 2 ÷ 2 = 1 remainder 0. Step 7: 1 ÷ 2 = 0 remainder 1.
步骤1：89 ÷ 2 = 44 余数1。步骤2：44 ÷ 2 = 22 余数0。步骤3：22 ÷ 2 = 11 余数0。步骤4：11 ÷ 2 = 5 余数1。步骤5：5 ÷ 2 = 2 余数1。步骤6：2 ÷ 2 = 1 余数0。步骤7：1 ÷ 2 = 0 余数1。
Now read the remainders from the last to the first: 1, 0, 1, 1, 0, 0, 1. This gives us 1011001 in binary.
现在从最后一个余数往前读：1, 0, 1, 1, 0, 0, 1。这样得到的二进制数是1011001。
Since the question asks for an 8-bit binary number, we must add leading zeros to make it 8 bits long. 1011001 has 7 bits, so we add one 0 at the front.
由于题目要求8位二进制数，我们需要在前面补零凑足8位。1011001有7位，所以在最前面加一个0。
The final 8-bit binary representation of denary 89 is 01011001.
十进制89最终的8位二进制表示为 01011001。
3. Question 3: Binary Addition and Overflow | 第3题：二进制加法与溢出
Question: Add the two 8-bit binary numbers 00110110 and 00101001. Show your working and state whether an overflow occurs.
题目：将两个8位二进制数00110110与00101001相加。写出计算过程，并说明是否发生溢出。
are: 0 + 0 = 0; 1 + 0 = 1; 1 + 1 = 0, carry 1 to the next column; 1 + 1 + 1 = 1, carry 1.
二进制加法遵循与十进制加法相同的规则，但只使用数字0和1。关键规则是：0+0=0；1+0=1；1+1=0，向下一位进1；1+1+1=1，进1。
Let us align the numbers and add column by column from the rightmost bit. 00110110 represents denary 54, and 00101001 represents denary 41.
我们将两个数对齐，从最右侧的位开始逐列相加。00110110表示十进制54，00101001表示十进制41。
ry 1. Next: 0 + 0 + carry 1 = 1. Leftmost: 0 + 0 = 0. The resulting 8-bit sum is 01011111.
最右列：0+1=1，无进位。下一列：1+0=1。下一列：1+0=1。下一列：0+1=1。下一列：1+0=1。下一列：1+1=0，进1。下一列：0+0+进位1=1。最左列：0+0=0。得到的8位和为01011111。
is well within range. No carry out from the most significant bit occurred, so no overflow.
检查是否溢出。在无符号8位二进制中，范围是0到255。我们的和为0+64+0+16+8+4+2+1 = 95，完全在范围内。没有从最高位产生进位，因此无溢出。
The final answer is 01011111, and overflow does not occur.
最终答案是 01011111，没有发生溢出。
4. Question 4: Logic Gate Truth Table | 第4题：逻辑门真值表
Question: Draw the truth table for the logic expression A AND (NOT B).
题目：画出逻辑表达式 A AND (NOT B) 的真值表。
4 rows. We first work out the NOT B column, then combine it with A using the AND operator.
真值表列出所有可能的输入组合及其对应的输出。对于两个输入A和B，共有2²=4行。我们先算出NOT B列，再将其与A用AND运算符结合。
A AND (NOT B) gives an output of 1 only when A is 1 and B is 0.
A AND (NOT B) 仅在A为1且B为0时输出为1。
| A | B | NOT B | A AND (NOT B) |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
is 1 only in the third row, where A=1 and B=0. This combination is often used to create a 'gate' that opens only when a certain condition is false.
因此，输出只在第3行为1，即A=1，B=0的组合。这种组合常被用来创建一个“门”，仅在某个条件为假时打开。
5. Question 5: Flowchart Symbols | 第5题：流程图符号
Question: Match the following flowchart symbol shapes to their meanings: oval, rectangle, diamond, parallelogram.
题目：将下列流程图符号形状与其含义进行匹配：椭圆形、矩形、菱形、平行四边形。
Flowcharts use standard symbols so that anyone can read the steps of an algorithm. Each shape has a specific purpose.
流程图使用标准符号以便任何人都能读懂算法的步骤。每种形状都有特定的用途。
Oval: Represents the Start or End of a process. It indicates where the algorithm begins and finishes.
椭圆形：表示流程的开始或结束。指示算法从哪里开始，在哪里结束。
Rectangle: Represents a process or an action. This is the most common symbol and can contain instructions like 'add 1 to total' or 'calculate average'.
矩形：表示一个处理或动作。这是最常见的符号，可以包含如“total加1”或“计算平均值”等指令。
Diamond: Represents a decision. It always contains a question that can be answered with 'Yes' or 'No', leading to different branches in the flow.
菱形：表示一个判断。它总是包含一个可用“是”或“否”回答的问题，引导流程走向不同的分支。
Parallelogram: Represents input or output of data, such as reading a value from the user or displaying a result on screen.
平行四边形：表示数据的输入或输出，例如从用户读取一个值或在屏幕上显示结果。
: Oval → Start/End; Rectangle → Process; Diamond → Decision; Parallelogram → Input/Output.
在考试中，你只需将每个形状与其正确的功能连线：椭圆形→开始/结束；矩形→处理；菱形→判断；平行四边形→输入/输出。
6. Question 6: Writing Pseudocode | 第6题：编写伪代码
Question: Write pseudocode that inputs three numbers and outputs the largest one.
题目：编写一段伪代码，输入三个数字并输出其中最大的一个。
Pseudocode uses simple English-like statements to describe an algorithm without worrying about strict programming syntax. It should be clear, logical, and use standard constructs like INPUT, OUTPUT, IF...THEN...ELSE.
伪代码使用类似英语的简单语句来描述算法，无需担心严格的编程语法。它应清晰、逻辑性强，并使用如INPUT、OUTPUT、IF...THEN...ELSE等标准结构。
One possible solution is as follows:
一种可能的解法如下：
INPUT a, b, c
IF a > b AND a > c THEN
OUTPUT a
ELSE IF b > c THEN
OUTPUT b
ELSE
OUTPUT c
INPUT a, b, c
IF a > b AND a > c THEN
OUTPUT a
ELSE IF b > c THEN
OUTPUT b
ELSE
OUTPUT c
This works by comparing the first number with the other two. If it is greater than both, it is the largest. Otherwise, we compare the second with the third to decide between them. The algorithm uses nested selection to cover all possibilities.
这个算法首先将第一个数与其他两个比较。如果它同时大于两者，则为最大数。否则，将第二个数与第三个数比较，从中选出较大者。该算法使用嵌套选择来涵盖所有可能。
Marks would be awarded for using INPUT/OUTPUT, correct logic, appropriate indentation, and clear naming. Remember that pseudocode does not need to compile; it just needs to be understood by a human.
评分时会关注是否正确使用了INPUT/OUTPUT、逻辑是否正确、缩进是否恰当以及命名是否清晰。记住伪代码不需要能够编译运行，只需让人理解即可。
7. Question 7: Variables and Constants | 第7题：变量与常量
Question: A program calculates the total cost of purchasing apples. The price of one apple is fixed at £0.30, and the user enters how many apples they want. Identify one variable and one constant in this program and explain the difference.
题目：一个程序用于计算购买苹果的总价。每个苹果的单价固定为0.30英镑，用户输入想要的苹果数量。请指出该程序中的一个变量和一个常量，并解释两者的区别。
In programming, a constant is a named memory location whose value does not change while the program is running. A variable is a named memory location that can hold different values at different times.
在编程中，常量是一个命名的内存位置，程序运行时其值不会改变。变量也是一个命名的内存位置，可以在不同时间保存不同的值。
so it should be stored as a constant. We might call it PRICE_PER_APPLE and set it to 0.30.
在这个场景中，每个苹果的单价是固定的，因此应存储为常量。我们可以将其命名为PRICE_PER_APPLE并设置为0.30。
The number of apples entered by the user is unknown until the program runs, so it must be stored in a variable. It could be called quantity and would change each time a different user runs the program.
用户输入的苹果数量在程序运行前是未知的，因此必须存储在变量中。我们可以将其命名为quantity，每次不同用户运行程序时其值都可能改变。
The key difference is immutability: constants are set once and never change, making the code more readable and less error-prone. Variables can be updated as the program executes.
关键区别在于不可变性：常量一经设定便永不改变，使得代码更易读且不易出错。变量则可以在程序执行过程中被更新。
Therefore, the constant is PRICE_PER_APPLE and the variable is quantity.
因此，常量是PRICE_PER_APPLE，变量是quantity。
8. Question 8: CPU Components – The ALU | 第8题：CPU组件——算术逻辑单元
Question: Name the component inside the Central Processing Unit (CPU) that performs arithmetic calculations and logical operations.
题目：请说出中央处理器 (CPU) 中负责执行算术运算和逻辑操作的组件名称。
The CPU contains several key parts, including the Control Unit (CU), the Arithmetic Logic Unit (ALU), and registers like the Accumulator and Program Counter.
CPU内部包含几个关键部件，包括控制单元 (CU)、算术逻辑单元 (ALU)，以及累加器和程序计数器等寄存器。
The component that directly handles arithmetic (addition, subtraction, multiplication, division) and logical operations (AND, OR, NOT, comparisons) is the Arithmetic Logic Unit, or ALU.
直接处理算术运算（加、减、乘、除）和逻辑运算（与、或、非、比较）的组件是 算术逻辑单元，简称ALU。
It receives instructions from the Control Unit, takes data from registers, performs the required operation, and stores the result back into a register. All mathematical and decision-making processes inside a computer rely on the ALU.
它从控制单元接收指令，从寄存器获取数据，执行所需操作，并将结果存回寄存器。计算机内部所有的数学计算和决策过程都依赖于ALU。
Thus, the answer is the ALU (Arithmetic Logic Unit).
因此，答案是 ALU（算术逻辑单元）。
9. Question 9: Switch vs Router | 第9题：交换机与路由器
Question: Explain the difference between a network switch and a router.
题目：解释网络交换机和路由器的区别。
Both switches and routers are devices that connect network nodes, but they operate at different layers and serve distinct purposes.
交换机和路由器都是连接网络节点的设备，但它们工作在不同的网络层次，服务于不同的目的。
A switch operates within a Local Area Network (LAN). It connects devices like computers, printers, and servers together. A switch uses MAC addresses to forward data frames to the specific destination device on the same network.
交换机工作在局域网(LAN)内部，将计算机、打印机和服务器等设备连接在一起。交换机使用MAC地址将数据帧转发到同一网络内的特定目标设备。
Published by TutorHao | Year 8 Computer Science Revision Series | aleveler.com
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