📚 In-depth Analysis of Past Year Exam Questions for Year 9 Edexcel Computing | Year 9 Edexcel 计算机：历年真题深度解析
This in-depth guide analyses typical past exam questions found in Year 9 Edexcel Computing assessments. It covers key topics such as binary, logic gates, algorithms, and networks, data representation, cyber security and programming, providing step-by-step explanations to help students master the concepts and improve their exam performance.
这篇深度指南分析了 Year 9 Edexcel 计算机考试中常见的典型真题。内容涵盖二进制、逻辑门、算法、网络、数据表示、网络安全和编程等关键主题，通过逐步解析帮助学生掌握概念并提升考试成绩。
1. Binary and Decimal Conversion | 二进制与十进制转换
A common exam question asks students to convert between binary (base 2) and decimal (base 10). Understanding place values is essential.
常见考题要求学生进行二进制（基数为2）与十进制（基数为10）之间的转换。理解位值至关重要。
Question: Convert the binary number 110101₂ to a decimal number. (2 marks)
问题：将二进制数 110101₂ 转换为十进制数。（2分）
To convert, write the binary digits under place values of powers of 2, starting from the rightmost as 2⁰. For 6 bits, the place values are 32, 16, 8, 4, 2 and 1.
转换时，将二进制数字写在以2的幂次为位值的下方，从最右侧的2⁰开始。对于6个比特，位值为32、16、8、4、2和1。
| Place value | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|
| Binary digit | 1 | 1 | 0 | 1 | 0 | 1 |
For 110101₂: 1×32 + 1×16 + 0×8 + 1×4 + 0×2 + 1×1.
对于110101₂：1×32 + 1×16 + 0×8 + 1×4 + 0×2 + 1×1。
1×32 + 1×16 + 0×8 + 1×4 + 0×2 + 1×1 = 32 + 16 + 0 + 4 + 0 + 1 = 53
Thus, 110101₂ = 53 in decimal.
因此，110101₂ 等于十进制中的 53。
Another typical question asks for decimal to binary. For example, convert 78 to binary.
另一种典型题目是十进制转二进制。例如，将 78 转换为二进制。
9÷2=9 R1, 9÷2=4 R1, 4÷2=2 R0, 2÷2=1 R0, 1÷2=0 R1. Read the remainders backwards: 1001110₂.
不断除以2并记录余数：78÷2=39 余0，39÷2=19 余1，19÷2=9 余1，9÷2=4 余1，4÷2=2 余0，2÷2=1 余0，1÷2=0 余1。从下往上读取余数：1001110₂。
tip: Always check your answer by reconverting to decimal. 1001110₂ = 64 + 8 + 4 + 2 = 78.
考试技巧：务必通过反向转换回十进制来检查答案。1001110₂ = 64 + 8 + 4 + 2 = 78。
2. Binary Addition and Overflow | 二进制加法与溢出
n follows simple rules: 0+0=0, 0+1=1, 1+0=1, 1+1=0 with carry 1, and 1+1+1=1 with carry 1.
学生常被要求将两个二进制数相加并检测溢出。二进制加法遵循简单规则：0+0=0，0+1=1，1+0=1，1+1=0 进位1，1+1+1=1 进位1。
Sample question: Add the two 4-bit binary numbers 1011₂ and 0110₂. Show your working and state if overflow occurs when the result is stored in 4 bits.
样题：将两个4位二进制数 1011₂ 和 0110₂ 相加。写出计算过程，并说明结果存储在4位中是否发生溢出。
most column: 1+0=1; 1+1=0 carry 1; 0+1+1=0 carry 1; 1+0+1=0 carry 1. The result is 10001₂.
将数字右对齐。从最右列开始：1+0=1；1+1=0 进位1；0+1+1=0 进位1；1+0+1=0 进位1。结果是 10001₂。
1011₂
+ 0110₂
= 10001₂
1011₂ = 11, 0110₂ = 6, and 11 + 6 = 17. 10001₂ = 16 + 1 = 17.
1011₂ = 11，0110₂ = 6，11 + 6 = 17。10001₂ = 16 + 1 = 17。
Since 4-bit unsigned storage can only hold 0000₂ to 1111₂ (0 to 15), and 17 is greater than 15, overflow occurs. The carry out of the most significant bit is 1.
由于4位无符号存储只能表示 0000₂ 到 1111₂（0 到 15），而 17 大于 15，因此发生溢出。最高有效位产生的进位为1。
Exam tip: Overflow in unsigned binary occurs when a carry goes out of the most significant bit and the result needs more bits than are available.
考试技巧：无符号二进制中，当最高有效位产生进位，并且结果需要的位数超过可用位数时，就发生溢出。
3. Logic Gates and Truth Tables | 逻辑门与真值表
Logic gates are a core topic. You must recognise symbols, complete truth tables and write Boolean expressions.
逻辑门是核心考点。你需要识别符号、完成真值表并写出布尔表达式。
AND outputs 1 only when both inputs are 1. OR outputs 1 when at least one input is 1. NOT inverts the input. NAND is NOT AND. NOR is NOT OR. XOR outputs 1 when the inputs are different.
AND 仅当两个输入均为1时输出1。OR 在至少一个输入为1时输出1。NOT 对输入取反。NAND 是 AND 取反。NOR 是 OR 取反。XOR 在输入不同时输出1。
| A | B | A AND B | A OR B | A XOR B | A NAND B | A NOR B |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 | 1 | 1 | 0 |
| 1 | 1 | 1 | 1 | 0 | 0 | 0 |
Typical question: Complete the truth table for Q = A AND (NOT B).
典型题目：完成 Q = A AND (NOT B) 的真值表。
| A | B | NOT B | Q = A AND (NOT B) |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
Exam tip: Draw intermediate
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