📚 IGCSE Edexcel Computer Science: Mastering Calculation Questions | IGCSE Edexcel 计算机科学：计算题专项训练
Calculation questions form a vital part of the IGCSE Edexcel Computer Science examination. They assess your ability to apply quantitative reasoning to technical concepts, from binary conversions and data storage to file sizes, transmission times, and processor performance. This guide provides focused practice on all major calculation topic areas, with step-by-step explanations and bilingual commentary to ensure you approach every numerical problem with confidence.
计算题是IGCSE Edexcel计算机科学考试的重要组成部分。这些题目考查你是否能将定量推理应用于技术概念，涵盖二进制转换、数据存储、文件大小、传输时间以及处理器性能等方面。本文针对所有主要计算专题进行集中训练，配有逐步解析和中英双语说明，帮助你充满信心地应对每一道数值问题。
1. Binary and Hexadecimal Conversions | 二进制与十六进制转换
Understanding how to convert between decimal (denary), binary, and hexadecimal is fundamental. A binary digit (bit) is either 0 or 1. A group of 4 bits can be represented by one hexadecimal digit (0–9, A–F). To convert binary to hexadecimal, split the binary number into groups of 4 from the right, adding leading zeros if needed, then convert each group. To convert decimal to binary, repeatedly divide by 2 and record the remainders.
掌握十进制、二进制与十六进制之间的转换是基础。一个二进制位只能取0或1。4个二进制位为一组，可以用一个十六进制数字（0–9、A–F）表示。将二进制转换为十六进制时，从右侧开始每四位分成一组，必要时左侧补零，再分别转换。十进制转二进制则不断除以2并记录余数。
ng remainders from bottom to top gives 11001101₂. Splitting into 1100 1101: 1100₂ = C₁₆, 1101₂ = D₁₆, so 205₁₀ = CD₁₆.
示例：将205₁₀转换为二进制和十六进制。205 ÷ 2 = 102 余1; 102 ÷ 2 = 51 余0; 51 ÷ 2 = 25 余1; 25 ÷ 2 = 12 余1; 12 ÷ 2 = 6 余0; 6 ÷ 2 = 3 余0; 3 ÷ 2 = 1 余1; 1 ÷ 2 = 0 余1。从下往上读余数得到11001101₂。拆分为1100 1101：1100₂ = C₁₆，1101₂ = D₁₆，因此205₁₀ = CD₁₆。
2. Data Storage Units and Capacity | 数据存储单位与容量
B = 1024 bytes; 1 MB = 1024 KB; 1 GB = 1024 MB; 1 TB = 1024 GB. When calculating how many files can fit on a storage device, always ensure both capacities are in the same unit.
数据大小以位、字节、千字节、兆字节、吉字节、太字节和拍字节衡量。请使用换算关系：1字节 = 8位；1 KB = 1024字节；1 MB = 1024 KB；1 GB = 1024 MB；1 TB = 1024 GB。计算存储设备能容纳多少文件时，务必保证两者的容量单位一致。
Worked example: A USB drive has a capacity of 16 GB. How many high‑resolution photos, each 8 MB, can be stored? st, ? First, convert 16 GB to MB: 16 × 1024 = 16384 MB. Number of photos = 16384 ÷ 8 = 2048 photos.
解题示例：一个U盘容量为16 GB。每张高分辨率照片大小为8 MB，可以存储多少张照片？首先将16 GB转换为MB：16 × 1024 = 16384 MB。照片数 = 16384 ÷ 8 = 2048张。
3. Image File Size Calculation | 图像文件大小计算
n resolution (width × height in pixels) and colour depth (bits per pixel). The formula is: File size (bits) = width × height × colour depth. Often you need to convert the result to bytes, KB, or MB. For a 1024×768 image with 24‑ . In bytes: 18,874,368 ÷ 8 = 2,359,296 bytes, which is approximately 2.25 MB (÷1024÷1024).
图像文件大小取决于分辨率（宽度×高度，以像素为单位）和色彩深度（每像素的位数）。公式为：文件大小（位）= 宽度 × 高度 × 色彩深度。通常需要将结果转换为字节、KB或MB。对于一幅1024×768、24位色彩深度的图像，大小 = 1024 × 768 × 24 = 18,874,368位。转换为字节：18,874,368 ÷ 8 = 2,359,296字节，大约是2.25 MB (÷1024÷1024)。
If metadata is included, add a fixed overhead such as 54 bytes for BMP format. Always check if the question asks for the raw pixel data or the overall file size.
如果包含元数据，则需要加上固定开销，例如BMP格式为54字节。一定要看清楚题目问的是原始像素数据还是整体文件大小。
4. Sound File Size Calculation | 声音文件大小计算
Sound file size = sample rate (Hz) × sample resolution (bits) × duration (seconds) × number of channels. For example, a 3‑minute stereo recording at 44.1 kHz with 16‑ = 44100 × 16 × 180 × 2 = 254,016,000 bits. In MB: 254,016,000 ÷ 8 ÷ 1024 ÷ 1024 ≈ 30.3 MB.
声音文件大小 = 采样率（Hz）× 采样分辨率（位）× 时长（秒）× 声道数。例如，一段3分钟、44.1 kHz、16位立体声录音：时长 = 3 × 60 = 180秒。大小（位）= 44100 × 16 × 180 × 2 = 254,016,000位。转换为MB：254,016,000 ÷ 8 ÷ 1024 ÷ 1024 ≈ 30.3 MB。
Hz: 1 kHz = 1000 Hz. Always convert minutes to seconds before inserting into the formula.
注意kHz与Hz的区别：1 kHz = 1000 Hz。代入公式前一定要将分钟转换为秒。
5. Compression Ratio and File Size Reduction | 压缩比与文件瘦身
The compression ratio expresses how much a file has been reduced. Compression ratio = uncompressed size ÷ compressed size ze: compressed size = original size ÷ ratio. If a 500 KB file is compressed with a ratio of 4:1, the new size = 500 ÷ 4 = 125 KB.
压缩比表示文件缩减的程度。压缩比 = 未压缩大小 ÷ 压缩后大小。10:1的压缩比意味着原文件是压缩后的十倍。已知压缩比和原大小求压缩后大小：压缩后大小 = 原大小 ÷ 压缩比。如果一个500 KB的文件以4:1压缩，新大小为500 ÷ 4 = 125 KB。
Lossy compression typically achieves higher ratios than lossless compression, but quality degrades. Calculations are identical regardless of type.
有损压缩通常比无损压缩能达到更高的压缩比，但质量会下降。无论哪种类型，计算方法相同。
6. Logic Gate Truth Tables and Boolean Expressions | 逻辑门真值表与布尔表达式
Although not always strictly numerical, logic gate calculations involve analysing inputs and determining outputs. For a Boolean expression such as Q = (A AND B) OR (NOT C) ws). Calculate intermediate columns and then the final Q. This helps in circuit simplification and checking equivalence.
虽然逻辑门并非总是纯粹数值计算，但分析输入确定输出的过程也属于计算范畴。对于布尔表达式 Q = (A AND B) OR (NOT C)，构建包含A、B、C所有可能组合（2³ = 8行）的真值表。先计算中间列，再得出最终Q。这有助于电路化简和等价性检查。
ing the output high. Always use systematic binary counting for inputs: 000, 001, 010, etc.
在考试计算题中，可能会要求你数出Q = 1的行数，代表输出为高电平。输入必须按系统二进制计数排列：000, 001, 010等。
7. Data Transmission Time | 数据传输时间
Time (seconds) = amount of data (bits) ÷ transfer rate (bits per second) bits; 100 Mbps = 100,000,000 bps. Time = 629,145,600 ÷ 100,000,000 ≈ 6.29 seconds. (Note: some networks use 1 Mbps = 1,000,000 bps, check exam context.)
时间（秒）= 数据量（位）÷ 传输速率（位/秒）。确保单位对齐：将文件大小转换为位，速率转换为bps。如果一个75 MB的文件通过100 Mbps网络传输：75 MB = 75 × 1024 × 1024 × 8 位 = 629,145,600 位；100 Mbps = 100,000,000 bps。时间 = 629,145,600 ÷ 100,000,000 ≈ 6.29秒。（注意：有些网络将1 Mbps定义为1,000,000 bps，请根据考题上下文判断。）
Often you must also account for protocol overheads (e.g., 10% extra for headers) or consider that transfer rates are in bytes per second (Bps) instead of bps. Watch out for uppercase ‘B’ (byte) vs lowercase ‘b’ (bit).
很多时候还需要考虑协议开销（例如额外10%用于头部信息），或者传输速率以字节/秒（Bps）而非bps给出。务必区分大写B（字节）和小写b（位）。
8. Processor Performance Calculations | 处理器性能计算
Clock speed dictates how many cycles per second the CPU can execute. Execution time for one instruction = 1 ÷ clock speed oseconds. With multiple cores, the theoretical throughput can be multiplied, but not all tasks parallelise perfectly.
时钟频率决定CPU每秒可执行多少个周期。一条指令的执行时间 = 1 ÷ 时钟频率（假设每指令1个周期）。对于3.5 GHz的处理器：每秒时钟周期 = 3,500,000,000。每个周期用时 = 1/3,500,000,000 ≈ 0.2857纳秒。多核处理器理论上可将吞吐量翻倍，但并非所有任务都能完美并行。
essor A: 2.8 GHz × 2 IPC = 5.6 billion IPS; Processor B: 3.2 GHz × 1.8 IPC = 5.76 billion IPS — B is slightly faster.
比较两款处理器时，用时钟频率乘以每周期指令数（IPC）计算出每秒指令数（IPS）。处理器A：2.8 GHz × 2 IPC = 56亿IPS；处理器B：3.2 GHz × 1.8 IPC = 57.6亿IPS——B稍快一些。
9. Storage Space on a Magnetic Disk | 磁盘存储空间计算
A magnetic hard disk’s capacity = surfaces per platter × number of tracks per surface × sectors per track × bytes per sector es per sector: capacity = 8 × 16000 × 512 × 512 = 33,554,432,000 bytes, which is approximately 31.25 GB (dividing by 1024³).
磁盘容量 = 盘片数 × 每盘片表面数 × 每表面磁道数 × 每磁道扇区数 × 每扇区字节数。例如，一个磁盘有4个盘片（每个有2个读写头，因此8个表面），每表面16,000磁道，每磁道512扇区，每扇区512字节：容量 = 8 × 16000 × 512 × 512 = 33,554,432,000字节，约等于31.25 GB（除以1024³）。
Some questions may ask for unformatted capacity vs formatted capacity; often a percentage reduction is applied for gaps and sector overheads. Always check the specification.
有些题目会问未格式化容量与格式化容量的区别；通常会因间隙和扇区开销减去一个百分比。务必阅读题目说明。
10. Network Transmission of Packets and Bandwidth Sharing | 数据包网络传输与带宽共享
ackets, each packet carries payload data plus header information. Total data transmitted = number of packets × (payload size + header size). If a 1 MB file is split into 512‑byte payloads with 64‑ 648 bytes, approximately 1.125 MB. The transmission time increases due to header overhead.
文件被拆分成数据包时，每个数据包包含有效载荷数据和头部信息。传输的总数据量 = 数据包数量 ×（有效载荷大小 + 头部大小）。如果一个1 MB文件以512字节有效载荷和64字节头部进行分包，所需数据包数量 = 1 MB ÷ 512字节 = 1024 × 1024 ÷ 512 = 2048个包。总传输量 = 2048 × (512 + 64) = 1,179,648字节，约1.125 MB。由于头部开销，传输时间相应增加。
Bandwidth sharing can be calculated by dividing total available bandwidth by the number of devices equally sharing it, assuming no quality of service (QoS).
带宽共享计算可用总带宽除以均享的设备数量，前提是没有服务质量控制。
11. Error Detection and Checksum/Parity Calculations | 差错检测与校验和/奇偶校验计算
Parity bytes: count the number of 1s in a data unit. Even parity requires an even total count; the parity bit is set accordingly. A block check character (BCC) is often calculated using XOR across all bytes in a transmission block. Practice XOR calculations bit‑by‑bit. Checksums involve summing data values and taking the least significant byte of the sum as the checksum, while a CRC uses polynomial division to produce a remainder that is appended.
奇偶校验字节：数一数数据单元中1的个数。偶校验要求总数为偶数，据此设置校验位。块校验字符（BCC）通常通过对传输块所有字节进行逐位异或运算来计算。请练习逐位异或。校验和则是将数据值相加，取和的最低有效字节作为校验和；而循环冗余校验（CRC）使用多项式除法生成余数并附加在后面。
In IGCSE Edexcel, you may be asked to calculate a simple checksum or parity bit for a given binary sequence. Always show your working clearly.
在IGCSE Edexcel考试中，你可能会被要求计算给定二进制序列的简单校验和或奇偶位。务必清晰展示计算步骤。
12. Memory Addressing and Address Bus Width | 内存寻址与地址总线宽度
The size of memory that can be addressed is determined by the width of the address bus: Addressable memory = 2ⁿ bytes, where n is the number of address lines. A 32‑ ress bus width to address a given amount of memory, take the logarithm base 2. For 16 GB of memory: 16 GB = 2³⁴ bytes (since 16 × 1024 × 1024 × 1024 = 2³⁴), so a 34‑bit address bus is needed.
可寻址的内存大小由地址总线宽度决定：可寻址内存 = 2ⁿ字节，其中n为地址线数量。32位地址总线可寻址2³²个独立位置。若每个存储单元存放1字节，则2³²字节 = 4 GB内存。要计算寻址一定内存所需的地址总线宽度，取以2为底的对数。对于16 GB内存：16 GB = 2³⁴字节（因为16 × 1024 × 1024 × 1024 = 2³⁴），因此需要34位地址总线。
Keep in mind that some architectures use word‑based addressing where each address points to a word (e.g., 4 bytes); in that case multiply accordingly.
注意有些架构采用字寻址，每个地址指向一个字（如4字节）；此时需相应乘以字长倍数。
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