📚 IGCSE Edexcel Computer Science: Calculation Practice | IGCSE Edexcel 计算机：计算题专项训练
This revision guide focuses on the most common calculation‑based questions in the Edexcel IGCSE Computer Science exam. Each section explains the key concept with worked examples, helping you master binary conversions, file size estimation, network transfer times, Boolean simplification and more. Follow the step‑by‑step methods to sharpen your skills and avoid careless errors under time pressure.
本文针对 Edexcel IGCSE 计算机科学考试中最常见的计算题型。每一节都通过例题讲解核心概念，帮助你掌握二进制转换、文件大小估算、网络传输时间、布尔表达式化简等内容。按照分步骤的方法进行训练，在考试时间紧张时也能有效减少粗心错误。
1. Binary and Denary Conversions | 二进制与十进制转换
Binary is a base‑2 number system using only the digits 0 and 1. Each bit position represents a power of 2, starting from 2⁰ on the right. To convert a denary (base‑10) number to binary, you can repeatedly divide by 2 and read the remainders from bottom to top, or use the place‑value table for 8‑bit numbers.
二进制是基数为 2 的数制，只使用数字 0 和 1。每一位的权值都是 2 的幂，从最右边的 2⁰ 开始。将十进制数转为二进制时，可以不断除以 2 并从下往上读取余数，也可以借助 8 位位值表进行转换。
Example: Convert 53 to an 8‑bit binary number. Using repeated division: 53 ÷ 2 = 26 r1, 26 ÷ 2 = 13 r0, 13 ÷ 2 = 6 r1, 6 ÷ 2 = 3 r0, 3 ÷ 2 = 1 r1, 1 ÷ 2 = 0 r1. Read remainders upwards: 110101, padded to 8 bits as 00110101.
例子：将 53 转为 8 位二进制数。使用除二法：53 ÷ 2 = 26 余 1，26 ÷ 2 = 13 余 0，13 ÷ 2 = 6 余 1，6 ÷ 2 = 3 余 0，3 ÷ 2 = 1 余 1，1 ÷ 2 = 0 余 1。从下往上读余数：110101，补足 8 位为 00110101。
To convert binary to denary, sum each bit multiplied by its place value. For 00110101: (0×128) + (0×64) + (1×32) + (1×16) + (0×8) + (1×4) + (0×2) + (1×1) = 32 + 16 + 4 + 1 = 53.
将二进制转为十进制时，把每一位乘以其位值后求和。对于 00110101：(0×128) + (0×64) + (1×32) + (1×16) + (0×8) + (1×4) + (0×2) + (1×1) = 32 + 16 + 4 + 1 = 53。
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 |
For negative numbers, Edexcel IGCSE uses two’s complement representation in 8 bits. The most significant bit (MSB) is the sign bit: –128 for 1. The range is −128 to +127.
对于负数，Edexcel IGCSE 使用 8 位补码表示。最高位是符号位，1 表示 −128。表示范围为 −128 到 +127。
2. Binary Addition and Overflow | 二进制加法与溢出
Binary addition follows four rules: 0+0=0, 0+1=1, 1+0=1, 1+1=0 carry 1, and 1+1+1=1 carry 1. Add column by column from the right, just like denary addition, carrying exceedances to the next higher bit.
二进制加法遵循四条规则：0+0=0，0+1=1，1+0=1，1+1=0 进 1，1+1+1=1 进 1。从右向左逐列相加，将进位传递到高位，与十进制加法类似。
Example: Add 0110 (6) and 0101 (5).
Column 0: 0+1=1 → 1
Column 1: 1+0=1 → 1
Column 2: 1+1=0 carry 1
Column 3: 0+0+carry1=1 → Result: 1011 (11).
例子：将 0110 (6) 与 0101 (5) 相加。第 0 列：0+1=1 → 1；第 1 列：1+0=1 → 1；第 2 列：1+1=0 进 1；第 3 列：0+0+进位 1=1 → 结果为 1011 (11)。
Overflow occurs when two positive numbers produce a negative result in two’s complement, or two negative numbers produce a positive result. For 8‑bit signed integers, adding 01111111 (127) and 00000001 (1) gives 10000000, which represents −128 — clearly an overflow. The exam expects you to detect when the sign bit changes incorrectly.
当两个正数相加得到负数，或两个负数相加得到正数时，就会发生溢出。对于 8 位有符号整数，将 01111111 (127) 与 00000001 (1) 相加得到 10000000，表示 −128 —— 这就是溢出。考试要求你能够检测符号位何时发生了错误改变。
3. Hexadecimal Conversions | 十六进制转换
Hexadecimal (hex) is base‑16, using digits 0–9 and letters A–F (A=10, B=11, …, F=15). Each hex digit represents exactly four binary bits, making it a compact way to write long binary sequences. Memory addresses and colour codes are often shown in hex.
十六进制是基数为 16 的数制，使用数字 0–9 和字母 A–F（A=10, B=11, …, F=15）。每个十六进制位恰好对应四位二进制，因此是书写长二进制串的紧凑方式。内存地址和颜色代码常用十六进制表示。
To convert denary to hex, divide by 16 repeatedly. For 253: 253 ÷ 16 = 15 remainder 13 (D), then 15 ÷ 16 = 0 remainder 15 (F). Reading from bottom: FD.
将十进制转为十六进制时，反复除以 16。对于 253：253 ÷ 16 = 15 余 13 (D)；15 ÷ 16 = 0 余 15 (F)。从下往上读得 FD。
To convert binary to hex, split the binary into groups of 4 bits from the right. Add leading zeros if needed. For 11011010, split into 1101 (D) and 1010 (A) → DA. Converting hex back to binary just involves writing the 4‑bit equivalent of each hex digit.
将二进制转为十六进制时，从右向左每四位一组进行划分，不够的左侧补零。对于 11011010，分为 1101 (D) 和 1010 (A) → DA。将十六进制转回二进制，只需把每个十六进制位替换为对应的四位二进制即可。
4. Logic Gates and Truth Tables | 逻辑门与真值表
Boolean algebra uses TRUE (1) and FALSE (0) values. The basic gates are AND (output 1 only if all inputs are 1), OR (output 1 if at least one input is 1), NOT (inverts the input), and XOR (output 1 if inputs are different). Truth tables list all possible input combinations and the resulting output.
布尔代数使用 TRUE (1) 和 FALSE (0) 两种值。基本门电路有 AND（仅当所有输入为 1 时输出 1）、OR（至少一个输入为 1 时输出 1）、NOT（将输入取反）和 XOR（输入不同时输出 1）。真值表列出所有输入组合及其对应的输出。
Example: Build a truth table for Q = (A AND B) OR (NOT C). Start by listing the 2³ = 8 combinations of A, B, C. Then compute NOT C, then A AND B, and finally the OR of those two columns. For row A=0, B=1, C=1: NOT C=0, A AND B=0 → Q=0. For row A=1, B=1, C=0: NOT C=1, A AND B=1 → Q=1.
例子：为 Q = (A AND B) OR (NOT C) 建立真值表。首先列出 A、B、C 的 2³ = 8 种组合。然后计算 NOT C，再计算 A AND B，最后将这两列进行 OR 运算。对于 A=0, B=1, C=1：NOT C=0，A AND B=0 → Q=0。对于 A=1, B=1, C=0：NOT C=1，A AND B=1 → Q=1。
| A | B | C | NOT C | A AND B | Q |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 1 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 | 1 |
5. Simplifying Boolean Expressions | 布尔表达式化简
Exam questions may ask you to simplify a Boolean expression using the laws of Boolean algebra. Common identities include the identity law (A OR 0 = A), null law (A AND 0 = 0), idempotent law (A OR A = A), absorption law (A OR (A AND B) = A), and De Morgan’s laws: NOT(A AND B) = (NOT A) OR (NOT B) and NOT(A OR B) = (NOT A) AND (NOT B).
考试中可能会要求你用布尔代数定律化简表达式。常见的恒等式包括：同一律 (A OR 0 = A)、零律 (A AND 0 = 0)、幂等律 (A OR A = A)、吸收律 (A OR (A AND B) = A) 以及德摩根定律：NOT(A AND B) = (NOT A) OR (NOT B)，NOT(A OR B) = (NOT A) AND (NOT B)。
Example: Simplify Q = (A AND B) OR (A AND NOT B). Factor A out: A AND (B OR NOT B). Since (B OR NOT B) = 1, the expression reduces to A AND 1 = A. Therefore Q = A. You can verify this by constructing truth tables for both sides — they must be identical.
例子：化简 Q = (A AND B) OR (A AND NOT B)。提取公因子 A：A AND (B OR NOT B)。由于 (B OR NOT B) = 1，表达式化简为 A AND 1 = A。因此 Q = A。你可以通过为两边建立真值表来验证——它们必须完全相同。
6. File Size Calculations: Text and Images | 文件大小计算：文本与图像
The size of a text file depends on the number of characters and the encoding. In ASCII, each character uses 1 byte (8 bits). In extended Unicode, a character may use 2 bytes. For N characters, file size = N × bytes per character.
文本文件的大小取决于字符数量和编码方式。在 ASCII 编码中，每个字符占用 1 字节（8 位）；在扩展 Unicode 中，每个字符可能占用 2 字节。对于 N 个字符，文件大小 = N × 每字符字节数。
Example: A plain text message contains 2400 characters in ASCII. Its size is 2400 bytes, which divided by 1024 gives approximately 2.34 KiB.
例子：一条纯文本消息包含 2400 个 ASCII 字符。它的大小为 2400 字节，除以 1024 后约为 2.34 KiB。
For bitmap images, resolution (width × height in pixels) and colour depth (bits per pixel) dictate the raw file size. Image file size (bits) = width × height × colour depth.
对于位图图像，分辨率（以像素为单位的宽 × 高）和颜色深度（每像素位数）决定了原始文件大小。图像文件大小（位）= 宽 × 高 × 颜色深度。
Example: An image of 800×600 pixels with 16‑bit colour depth: size = 800 × 600 × 16 = 7,680,000 bits. Convert to bytes: 7,680,000 ÷ 8 = 960,000 bytes. Convert to KiB: 960,000 ÷ 1024 ≈ 937.5 KiB.
例子：一幅 800×600 像素、16 位颜色深度的图像：大小 = 800 × 600 × 16 = 7,680,000 位。转换为字节：7,680,000 ÷ 8 = 960,000 字节。转换为 KiB：960,000 ÷ 1024 ≈ 937.5 KiB。
Always remember that file size calculations in the IGCSE exam use binary prefixes: 1 KiB = 1024 bytes, 1 MiB = 1024 KiB. Don’t use 1000 unless the question explicitly specifies decimal units.
请牢记，IGCSE 考试中的文件大小计算使用二进制前缀：1 KiB = 1024 bytes，1 MiB = 1024 KiB。除非题目明确指定十进制单位，否则不要使用 1000。
7. Calculating Sound File Sizes | 声音文件大小计算
Uncompressed sound file size depends on the sample rate (samples per second), bit depth (bits per sample), number of channels (1 for mono, 2 for stereo) and duration in seconds. The total number of bits = sample rate × bit depth × channels × time.
未经压缩的声音文件大小取决于采样率（每秒样本数）、采样精度（每样本位数）、声道数（1 为单声道，2 为立体声）和以秒为单位的时长。总位数 = 采样率 × 采样精度 × 声道数 × 时长。
Example: 3 minutes of stereo audio recorded at 44.1 kHz with 16‑bit resolution. Time = 3 × 60 = 180 seconds. Bits = 44,100 × 16 × 2 × 180 = 254,016,000 bits. In bytes: 254,016,000 ÷ 8 = 31,752,000 bytes. In MiB: 31,752,000 ÷ (1024 × 1024) ≈ 30.28 MiB.
例子：一段 3 分钟的立体声音频，采样率 44.1 kHz，16 位采样精度。时长 = 3 × 60 = 180 秒。位数 = 44,100 × 16 × 2 × 180 = 254,016,000 位。字节数：254,016,000 ÷ 8 = 31,752,000 字节。MiB：31,752,000 ÷ (1024 × 1024) ≈ 30.28 MiB。
A common mistake is to forget the channel multiplier or to mix seconds and minutes. Always express time in seconds before multiplying.
一个常见错误是忘记乘以声道数，或混淆秒与分钟。务必先将时间转换为秒再相乘。
8. Data Transfer Time | 数据传输时间
Data transfer time = amount of data ÷ transfer rate. Attention must be paid to units: network speeds are usually given in bits per second (bps) using decimal prefixes (1 kbps = 1000 bps, 1 Mbps = 1,000,000 bps), while file sizes often use binary prefixes. Convert everything to bits for consistent calculation.
数据传输时间 = 数据量 ÷ 传输速率。注意单位：网络速率通常以每秒位数 (bps) 表示，使用十进制前缀（1 kbps = 1000 bps，1 Mbps = 1,000,000 bps），而文件大小常使用二进制前缀。为计算一致，将所有数据都转换为位。
Example: A 10 MiB file is to be uploaded over a 2 Mbps connection. Data bits = 10 × 1024 × 1024 × 8 = 83,886,080 bits. Transfer rate = 2 × 1,000,000 = 2,000,000 bps. Time = 83,886,080 ÷ 2,000,000 ≈ 41.94 seconds.
例子：通过 2 Mbps 连接上传一个 10 MiB 文件。数据位数 = 10 × 1024 × 1024 × 8 = 83,886,080 位。传输速率 = 2 × 1,000,000 = 2,000,000 bps。时间 = 83,886,080 ÷ 2,000,000 ≈ 41.94 秒。
Some IGCSE questions may ask you to calculate the transfer time for a
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