📚 IGCSE English: Calculation Practice | IGCSE 英语：计算题专项训练
Mastering calculation questions in IGCSE Computer Science requires not only a solid grasp of mathematical operations but also the ability to understand problem statements written in English. This article provides a bilingual revision guide focused on common calculation topics, enabling you to practise reading English instructions, applying formulas, and expressing your working steps clearly. By working through data representation, storage, logic gates, and compression, you will sharpen both your computational and language skills. Each section presents key concepts in English first, followed by a Chinese explanation, ensuring you can follow the logic and terminology seamlessly.
在 IGCSE 计算机科学中攻克计算题，不仅需要熟练掌握数学运算，还要能读懂用英文描述的问题要求。本文提供一份双语专项复习指南，围绕常考的计算专题展开，帮助你练习阅读英文指令、套用公式并清晰书写解题步骤。通过数据表示、存储、逻辑门和数据压缩等内容的训练，你的计算能力和英文理解力都将得到提升。每个小节先用英文讲解核心概念，再给出对应的中文说明，让你可以顺畅跟上逻辑与专业术语。
1. Understanding Number Systems | 理解数制
Computers process data using binary, but we often work with decimal and hexadecimal representations. A solid foundation in these number systems is essential for tackling conversion exercises. In binary, each digit is a bit valued at powers of 2. The rightmost bit is the least significant, representing 2⁰. As you move left, the place values increase: 2¹, 2², 2³, and so on. Hexadecimal uses 16 symbols (0-9 and A-F), where each digit corresponds to a power of 16. Binary and hexadecimal are closely linked because a group of four binary digits exactly matches one hex digit.
计算机使用二进制处理数据，但我们更常接触十进制和十六进制。扎实掌握这些数制是应对转换类计算题的基础。在二进制中，每一位的权值是 2 的幂。最右边的位是最低有效位，表示 2⁰；向左移，各位的权值依次为 2¹、2²、2³ 等等。十六进制使用 16 个符号（0-9 和 A-F），每位对应 16 的幂。二进制和十六进制关系紧密，因为四个二进制位恰好能用一个十六进制位表示。
2. Binary to Decimal Conversion | 二进制与十进制转换
To convert a binary number to decimal, multiply each binary digit by its place value (power of 2) and sum the products. Start from the rightmost bit (2⁰). For example, convert 1101₂ to decimal: 1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8 + 4 + 0 + 1 = 13₁₀. Always label the base using subscript to avoid confusion. In exam questions, you may see a number written without a base; use context to determine if it is binary. Practise with longer strings such as 101100₂ → 44₁₀. Remember: any bit of 0 contributes zero, so you can omit those terms in your working.
将二进制数转换为十进制时，用每一位数乘以对应的权值（2 的幂），再把所有乘积相加。从最右边的位（2⁰）开始。例如，把 1101₂ 转为十进制：1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8 + 4 + 0 + 1 = 13₁₀。一定要用下标标注数制，避免混淆。考试题目中可能会写出不带下标的数，这时需要根据上下文判断是否为二进制。可练习更长的一串位，例如 101100₂ → 44₁₀。记住：0 位贡献的值为零，因此解题时可以不写这些项。
3. Decimal to Binary Conversion | 十进制转二进制
Converting a decimal number to binary involves repeated division by 2. Write down the integer part of the division and the remainder each time until the quotient becomes zero. The binary result is the sequence of remainders read from the last remainder upward. For instance, to convert 19₁₀ to binary: 19 ÷ 2 = 9 remainder 1; 9 ÷ 2 = 4 remainder 1; 4 ÷ 2 = 2 remainder 0; 2 ÷ 2 = 1 remainder 0; 1 ÷ 2 = 0 remainder 1. Reading the remainders bottom‑up gives 10011₂. Check by converting back: 16+0+0+2+1 = 19. Always verify your answer to avoid careless mistakes.
把十进制整数转换为二进制的方法是重复除以 2。每次记录商的整数部分和余数，直到商为 0。二进制结果就是从最后一个余数开始向上读取余数序列。例如，把 19₁₀ 转为二进制：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。从下往上读余数，得到 10011₂。可以反向验证：16+0+0+2+1 = 19。转换后务必验算，避免因粗心而失分。
4. Hexadecimal Conversion | 十六进制转换
Hexadecimal numbers frequently appear in computer science to shorten long binary strings. To convert hex to decimal, multiply each digit by 16ⁿ, where n starts at 0 from the right. For letters A–F, use their decimal values (A=10, B=11, C=12, D=13, E=14, F=15). Example: 2F₁₆ = 2×16¹ + 15×16⁰ = 32 + 15 = 47₁₀. To convert binary to hex, split the binary number into groups of four bits starting from the right, then replace each group with its hex equivalent. Padding with leading zeros if necessary. For instance, 11010111₂ → split as 1101 0111 → D7₁₆ (since 1101₂=13=D, 0111₂=7). Moves between hex and decimal often serve as a bridge in memory addressing questions.
十六进制常用于计算机科学中，以缩短冗长的二进制串。将十六进制转为十进制时，用每位数字乘以 16ⁿ，n 从右边起从 0 开始。字母 A–F 使用对应的十进制数值（A=10, B=11, C=12, D=13, E=14, F=15）。例如：2F₁₆ = 2×16¹ + 15×16⁰ = 32 + 15 = 47₁₀。要将二进制转为十六进制，从右往左每四位二进制位分成一组，然后换成对应的十六进制符号；若最左一组不足四位，在前面补零。例如，11010111₂ → 分为 1101 0111 → D7₁₆（因为 1101₂=13=D，0111₂=7）。十六进制与十进制之间的转换常常是内存寻址题中的桥梁。
10111010₂ = BA₁₆
5. Binary Addition and Overflow | 二进制加法与溢出
Binary addition follows the same principles as decimal addition but with only two digits. The key rules are: 0+0=0, 0+1=1, 1+0=1, 1+1=0 carry 1, and 1+1+carry=1 carry 1. When adding two binary numbers, align them to the right and add column by column. An overflow occurs when the result of an addition requires more bits than the available storage size. In an 8‑bit register, adding 11111111₂ (255) and 00000001₂ (1) produces 1 00000000₂, which exceeds 8 bits. The leftmost 1 is the carry out, triggering an overflow flag. Understand how overflows affect signed and unsigned interpretations, and be able to detect them.
二进制加法遵循与十进制加法相同的原则，但只用到两个数字。核心规则是：0+0=0，0+1=1，1+0=1，1+1=0 进 1，以及 1+1+进位=1 进 1。做加法时，将两个二进制数右对齐，然后逐列相加。当加法结果所需的位数超过可用的存储位数时，就会发生溢出。在 8 位寄存器中，11111111₂（255）加 00000001₂（1）得到 1 00000000₂，这已经超出 8 位。最左边的 1 是进位输出，会触发溢出标志。你要理解溢出如何影响带符号和无符号的解释，并能识别溢出情形。
6. Storage Units and File Size Calculations | 存储单位与文件大小计算
Data size is measured in bits, bytes, kilobytes, megabytes, gigabytes and terabytes. The fundamental relationships are: 1 byte = 8 bits, 1 KB = 1024 bytes, 1 MB = 1024 KB, 1 GB = 1024 MB, 1 TB = 1024 GB. When calculating file sizes, always use the factor 1024 unless the question explicitly states otherwise (e.g. using 1000 for simplicity). Typical tasks include converting between units and finding storage requirements. For example, a 2 MB file in bytes is 2 × 1024 × 1024 = 2,097,152 bytes. Also learn to express the same size in higher or lower units and to interpret prefixes like MiB and GiB, which use power‑of‑2 definitions.
数据大小以位、字节、千字节、兆字节、千兆字节和太字节为单位。基本换算关系是：1 字节 = 8 位，1 KB = 1024 字节，1 MB = 1024 KB，1 GB = 1024 MB，1 TB = 1024 GB。计算文件大小时，除非题目明确说明（例如简化使用 1000），否则一律使用 1024 作为换算因子。常见题型包括单位换算和计算存储需求。例如，一个 2 MB 的文件换算成字节数为 2 × 1024 × 1024 = 2,097,152 字节。还需要学会把同一大小用更大或更小的单位表示，以及理解 MiB、GiB 这些基于 2 的幂的前缀。
- English: 1 GB = 1024³ bytes = 1,073,741,824 bytes.
- 中文：1 GB = 1024³ 字节 = 1,073,741,824 字节。
- English: A 500 GB hard disk can store roughly 500 × 1024³ bytes.
- 中文：一块 500 GB 的硬盘大约可存储 500 × 1024³ 字节数据。
7. Image File Size Calculation | 图像文件大小计算
The file size of an uncompressed bitmap image depends on two factors: resolution and colour depth. Resolution is the number of pixels, often given as width × height. Colour depth (bit depth) is the number of bits used to represent the colour of a single pixel. The formula is: File size (bits) = width × height × colour depth. For example, an image with dimensions 800×600 pixels and a colour depth of 24 bits requires 800 × 600 × 24 = 11,520,000 bits. Convert to bytes by dividing by 8, then to KB or MB as needed. Questions may ask you to calculate how many such images fit on a storage medium of a given capacity.
未压缩位图图像的文件大小取决于两个因素：分辨率和色深。分辨率即像素数量，通常以宽×高表示。色深（位深）是代表单个像素颜色所用的位数。计算公式为：文件大小（位）= 宽 × 高 × 色深。例如，一张分辨率为 800×600 像素、色深为 24 位的图像，需要的存储空间为 800 × 600 × 24 = 11,520,000 位。除以 8 可换算为字节，进而转换为 KB 或 MB。考题可能要求你计算在给定容量的存储介质上能存放多少张这样的图像。
8. Sound File Size Calculation | 声音文件大小计算
Sound recording size is determined by sample rate, bit depth, duration, and the number of channels. Sample rate is measured in hertz (Hz), indicating how many samples are taken per second. Bit depth is the number of bits per sample. The formula for a mono sound is: File size (bits) = sample rate × bit depth × duration (seconds). For stereo files, multiply the result by 2 (for two channels). For instance, a 30‑second stereo recording at 44,100 Hz with 16‑bit samples has a size of 44,100 × 16 × 30 × 2 = 42,336,000 bits. Always be careful with units: time must be in seconds, and you may need to convert the final answer to MB.
录音文件的大小由采样率、采样深度、时长以及声道数决定。采样率以赫兹（Hz）为单位，表示每秒采集的样本数。采样深度是每个样本所用的位数。单声道声音的文件大小公式为：文件大小（位）= 采样率 × 采样深度 × 时长（秒）。对于立体声文件，需再乘以 2（两个声道）。例如，一段 30 秒、采样率 44,100 Hz、16 位采样的立体声录音，其大小为 44,100 × 16 × 30 × 2 = 42,336,000 位。务必注意单位：时间必须换算为秒，最后的结果可能需要换算成 MB。
File size (bytes) = (Sample rate × Bit depth × Duration × Channels) ÷ 8
9. Logic Gates and Truth Tables | 逻辑门与真值表
Logic gates are the building blocks of digital circuits, and calculation questions often ask you to complete truth tables or evaluate gate combinations. The common gates are: NOT (inverter, ¬A), AND (A ∧ B, output 1 only when both inputs are 1), OR (A ∨ B, output 1 when at least one input is 1), NAND (¬(A ∧ B)), and NOR (¬(A ∨ B)). You must memorise the truth table for each gate. For combined logic circuits, work through each gate step‑by‑step, labelling intermediate signals. In an exam, you might be given a Boolean expression such as Q = ¬(A ∧ B) ∨ C and need to fill in an output column for all input combinations.
逻辑门是数字电路的基本单元，计算题常要求你填写真值表或计算门电路组合的结果。常见逻辑门包括：非门（¬A）、与门（A ∧ B，只有当两个输入皆为 1 时输出才为 1）、或门（A ∨ B，至少一个输入为 1 时输出 1）、与非门（¬(A ∧ B)）和或非门（¬(A ∨ B)）。你必须熟记每种门的真值表。对于组合逻辑电路，逐步处理每个门，并标注中间信号。考试中你可能看到像 Q = ¬(A ∧ B) ∨ C 这样的布尔表达式，并需要为所有输入组合填写输出列。
| A | B | A ∧ B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
10. Data Compression Ratios | 数据压缩比
Data compression is a technique to reduce file sizes, and exam questions may ask you to calculate the compression ratio or the size saved. The compression ratio is defined as original uncompressed size divided by compressed size. A ratio of 4:1 means the compressed file is one quarter of the original. If an original file is 800 KB and the compressed version is 200 KB, the compression ratio is 800 ÷ 200 = 4:1. You might also calculate the percentage reduction: ((original – compressed) ÷ original) × 100%. When dealing with lossy and lossless compression, remember that lossy compression can achieve higher ratios at the cost of some data quality.
数据压缩是用来缩减文件大小的技术，考试题可能会要求你计算压缩比或节省的空间。压缩比定义为原始未压缩大小除以压缩后大小。若比值为 4:1，意味着压缩后文件是原始文件的四分之一。如果一个原始文件为 800 KB，压缩后为 200 KB，则压缩比为 800 ÷ 200 = 4:1。你还可能需要计算缩减百分比：((原始大小 – 压缩后大小) ÷ 原始大小) × 100%。在处理有损和无损压缩时，记住有损压缩可以取得更高的压缩比，但会牺牲部分数据质量。
Compression ratio = Original size ÷ Compressed size
Space saved (%) = [(Original – Compressed) ÷ Original] × 100
11. Practice Problem Set | 综合练习题
Translate the following English questions into your working steps and find the answers: (1) A bitmap image has a resolution of 1024×768 and a colour depth of 32 bits. Calculate its file size in megabytes. (2) Convert the decimal number 202 to binary and then to hexadecimal. (3) A 5‑minute stereo audio is recorded at 22.05 kHz with 8‑bit samples. Determine the storage requirement in MB, assuming no compression. (4) Two 8‑bit binary numbers 01101010 and 10010111 are added. State the 8‑bit result and whether an overflow occurs. (5) An uncompressed file of 3.6 MB is compressed to 1.2 MB. Find the compression ratio and the percentage of original size saved.
将以下英文题目转化为你的解题步骤并写出答案：(1) 一幅位图图像分辨率为 1024×768，色深为 32 位，计算其文件大小（以 MB 为单位）。(2) 将十进制数 202 转换为二进制，再转换为十六进制。(3) 一段 5 分钟立体声音频以 22.05 kHz 采样率、8 位采样深度录制，假设不压缩，计算所需存储空间（MB）。(4) 两个 8 位二进制数 01101010 和 10010111 相加，给出 8 位结果并说明是否发生溢出。(5) 一个未压缩的 3.6 MB 文件压缩后变为 1.2 MB，求压缩比和节省的原始大小百分比。
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