📚 Year 8 WJEC Computing: Unit Test Mock Paper Walkthrough | Year 8 WJEC 计算机：单元测试模拟卷解析
This walkthrough breaks down a WJEC-style Year 8 computing unit test mock paper. We cover binary, data types, logic gates, flowcharts, programming basics, hardware, networks, cybersecurity, and debugging – all key topics from the Year 8 curriculum. Each question is explained step by step so you can understand common pitfalls and strengthen your revision.
本文逐步解析一套符合WJEC风格的八年级计算机单元测试模拟卷。我们涵盖二进制、数据类型、逻辑门、流程图、编程基础、硬件、网络、网络安全和调试等所有八年级课程的关键主题。每个题目都会逐步讲解，帮助你理解常见失分点并巩固复习。
1. Binary and Denary Conversions | 二进制与十进制转换
Question 1 (6 marks): (a) Convert the denary number 93 into an 8-bit binary number. Show your working. (b) Convert the binary number 01011010 into denary.
题目1（6分）：(a) 将十进制数93转换为8位二进制数，写出过程。(b) 将二进制数01011010转换为十进制。
For part (a), use the division-by-two method. 93 ÷ 2 = 46 remainder 1; 46 ÷ 2 = 23 remainder 0; 23 ÷ 2 = 11 remainder 1; 11 ÷ 2 = 5 remainder 1; 5 ÷ 2 = 2 remainder 1; 2 ÷ 2 = 1 remainder 0; 1 ÷ 2 = 0 remainder 1. Read remainders from bottom to top: 1011101. To make it 8-bit, add a leading zero: 01011101.
第(a)小题使用除2取余法。93 ÷ 2 = 46 余 1；46 ÷ 2 = 23 余 0；23 ÷ 2 = 11 余 1；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：1011101。补足8位，在最前面加0，得到01011101。
Part (b) uses place values: 128, 64, 32, 16, 8, 4, 2, 1. 01011010 has 0×128 + 1×64 + 0×32 + 1×16 + 1×8 + 0×4 + 1×2 + 0×1 = 64 + 16 + 8 + 2 = 90. So the denary value is 90.
第(b)小题使用位权值表：128、64、32、16、8、4、2、1。01011010 对应 0×128 + 1×64 + 0×32 + 1×16 + 1×8 + 0×4 + 1×2 + 0×1 = 64 + 16 + 8 + 2 = 90。因此十进制值为90。
2. Data Types and Variables | 数据类型与变量
Question 2 (4 marks): State the most appropriate data type (integer, real, string, Boolean) for each scenario: (a) storing a student’s age, (b) storing a student’s name, (c) flagging whether homework is completed, (d) storing the price of a pencil in pounds (£0.75).
题目2（4分）：为以下情况选择最合适的数据类型（整数、实数、字符串、布尔型）：(a) 存储学生年龄，(b) 存储学生姓名，(c) 标记作业是否完成，(d) 存储铅笔的价格（单位英镑，如0.75）。
(a) Age should be an integer because it is a whole number. (b) A student’s name is text, so it needs a string data type. (c) A yes/no or true/false flag is a Boolean. (d) A price includes a decimal fraction, so real (or float) is the correct choice.
(a) 年龄应为整数，因为它是整数。(b) 学生姓名是文本，因此需要使用字符串类型。(c) 是/否或真/假的标记属于布尔型。(d) 价格含有小数部分，因此实数（或浮点数）是正确的选择。
3. Logic Gates and Truth Tables | 逻辑门与真值表
Question 3 (6 marks): Complete the truth table for the logic circuit Q = NOT (A AND B). Draw the circuit symbol for a NOT gate.
题目3（6分）：补全逻辑电路 Q = NOT (A AND B) 的真值表，并画出非门（NOT gate）的电路符号。
First recall the AND truth table: outputs 1 only when both inputs are 1. The NOT gate inverts the signal. So when A=0, B=0, A AND B =0, NOT gives 1. A=0, B=1 → 0 → 1. A=1, B=0 → 0 → 1. A=1, B=1 → 1 → 0. This results in a NAND truth table.
首先回顾与门真值表：仅当两个输入均为1时输出1。非门将信号取反。因此当A=0，B=0时，A AND B = 0，NOT后得1。A=0，B=1 → 0 → 1。A=1，B=0 → 0 → 1。A=1，B=1 → 1 → 0。最终得到的是与非门（NAND）的真值表。
The NOT gate symbol is a triangle pointing right with a small circle at the tip. Label the input A and the output Q.
非门符号是一个指向右侧的三角形，顶端加一个小圆圈。标记输入端为A，输出端为Q。
4. Algorithms and Flowchart Symbols | 算法与流程图符号
Question 4 (8 marks): Draw a flowchart for an algorithm that asks a user to enter a password. If the password is ‘ALevel2025’, display ‘Access granted’ and stop. Otherwise, display ‘Try again’ and let the user retry, up to a maximum of 3 attempts. If all attempts fail, display ‘Locked out’.
题目4（8分）：画一个流程图，实现以下算法：要求用户输入密码。如果密码为’ALevel2025’，显示’Access granted’并停止。否则显示’Try again’并允许重试，最多尝试3次。如果全部失败，显示’Locked out’。
Start with an oval ‘Start’ symbol. Then use a process rectangle to initialise a counter variable (attempts = 0). Enter a loop: a parallelogram for input ‘Enter password’. A decision diamond checks if input equals ‘ALevel2025’. If yes, output ‘Access granted’ in a parallelogram and go to End. If no, increase attempts by 1 in a rectangle. Another decision diamond asks if attempts = 3; if false, output ‘Try again’ and loop back to input; if true, output ‘Locked out’ and go to End.
以椭圆形’开始’符号开始。接着使用矩形处理框初始化一个计数器变量（attempts = 0）。进入循环：用平行四边形输入’Enter password’。用菱形判断输入是否等于’ALevel2025’。如果是，用平行四边形输出’Access granted’后转至结束。如果否，在矩形框内将attempts增加1。再用一个菱形判断 attempts 是否等于3；若不等于，用平行四边形输出’Try again’并返回输入步骤；若等于，输出’Locked out’并转至结束。
Remember the standard symbols: oval for Start/End, rectangle for process, parallelogram for input/output, diamond for decision, arrows for flow direction. You would be expected to draw these clearly in your answer.
记住标准符号：椭圆形表示开始/结束，矩形表示处理，平行四边形表示输入/输出，菱形表示判断，箭头表示流程方向。你在答题时应该清晰画出这些符号。
5. Programming Concepts – Loops and Conditions | 编程概念 – 循环
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