📚 GCSE AQA Computer Science: Calculation Practice Set | GCSE AQA 计算机科学：计算题专项训练
In GCSE AQA Computer Science, calculation-based questions are a fundamental component, testing your ability to convert between number systems, determine storage requirements, evaluate file sizes, and analyse data transmission. This practice set consolidates all the essential calculation techniques into a single resource, complete with step-by-step examples and self-test exercises. By working through these topics methodically, you will sharpen your numerical skills and approach the examination with confidence.
在 GCSE AQA 计算机科学考试中，计算题是一项基本组成部分，考查你在不同数制之间转换、确定存储需求、计算文件大小以及分析数据传输方面的能力。本专项训练将所有关键计算技巧整合到一个资源中，并提供逐步示例和自测练习。通过有条理地学习这些主题，你将提升数值处理能力，自信地应对考试。
1. Binary and Decimal Conversion | 二进制与十进制转换
Binary is a base‑2 number system that uses only the digits 0 and 1. Each position in a binary number represents a power of 2, with the least significant bit on the right. To convert a binary number to decimal, multiply each bit by its place value (2ⁿ where n starts at 0 from the right) and add the results. For example, binary 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11 in decimal.
二进制是一种基数为 2 的计数系统，只使用数字 0 和 1。二进制数中的每个位置代表 2 的幂，最低有效位在最右侧。要将二进制转换为十进制，将每一位乘以其位值（从右起 n 从 0 开始）并求和。例如，二进制 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 11。
To convert decimal to binary, divide the decimal number by 2 repeatedly, recording the integer quotient and remainder each time. Stop when the quotient becomes 0. The binary equivalent is obtained by reading the remainders from the last remainder upwards. Example: convert 25 6 ÷ 2 = 3 remainder 0, 3 ÷ 2 = 1 remainder 1, 1 ÷ 2 = 0 remainder 1. Reading upwards gives 11001₂.
将十进制转换为二进制，反复除以 2，每次记录整数商和余数。当商为 0 时停止。从下往上读取余数即得到二进制数。例如，将 25 转换为二进制：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。向上读取余数得到 11001₂。
Try it yourself: Convert the binary number 110110₂ to decimal, and convert the decimal number 45 to binary.
自己试试：将二进制 110110₂ 转换为十进制，并将十进制 45 转换为二进制。
Answers: 110110₂ = 32+16+4+2 = 54; 45 ÷ 2 = 22 R1, 11 R0, 5 R1, 2 R1, 1 R0, 0 R1 → 101101₂.
答案：110110₂ = 54；45 ÷ 2 过程得到 101101₂。
2. Binary and Hexadecimal Conversion | 二进制与十六进制转换
Hexadecimal (hex) is a base‑16 system using digits 0–9 and letters A–F (A=10, B=11, … t binary to hex, split the binary sequence into groups of four bits from the right, padding with leading zeros if necessary, then convert each group. Example: 11011010₂ → 1101 1010 → D (13) and A (10) → DA₁₆.
十六进制是基数为 16 的系统，使用数字 0–9 和字母 A–F（A=10 至 F=15），它为二进制数据提供了简洁的表示方式。每个十六进制位恰好对应一组四个二进制位（一个半字节）。要将二进制转换为十六进制，从右向左每四位一组进行分组，必要时在左侧补零，然后转换每一组。例如，11011010₂ → 1101 1010 → D A → DA₁₆。
To convert hex to binary, replace each hex digit with its 4‑bit binary equivalent. For instance, 2F₁₆ → 2 = 0010, F = 1111 → 00101111₂. Hex can also be converted to decimal via powers of 16 or by first converting to binary. Example: 1A₁₆ = 1×16¹ + 10×16⁰ = 16 + 10 = 26.
要将十六进制转换为二进制，将每个十六进制位替换为其对应的四位二进制数。例如，2F₁₆ → 2 = 0010, F = 1111 → 00101111₂。十六进制也可以通过 16 的幂或先转二进制再转十进制来转换为十进制。例如 1A₁₆ = 1×16¹ + 10×16⁰ = 26。
Try it yourself: Convert 10111100₂ to hex, and convert 5D₁₆ to decimal.
自己试试：将 10111100₂ 转换为十六进制，并将 5D₁₆ 转换为十进制。
Answers: 1011 1100 → B C → BC₁₆; 5D₁₆ = 5×16 + 13 = 93.
答案：10111100₂ → BC₁₆；5D₁₆ = 93。
3. Binary Shifts | 二进制移位
Binary shifts move all bits left or right within a fixed number of bits, effectively performing multiplication or division by powers of two. A left shift by n places multiplies the binary number by 2ⁿ, filling vacated rightmost bits with zeros. A logical right shift by n places divides by 2ⁿ, discarding any remainder, and fills the leftmost positions with zeros. For example, shifting 00101100₂ (44₁₀) left by 1 yields 01011000₂ (88₁₀); shifting it right by 2 yields 00001011₂ (11₁₀), losing the fractional part from 44 ÷ 4 = 11 exactly.
二进制移位在固定位宽内将所有位向左或向右移动，实际上执行乘以或除以 2 的幂的操作。左移 n 位相当于乘以 2ⁿ，空出的右侧位用零填充。逻辑右移 n 位相当于除以 2ⁿ，丢弃任何余数，左侧位用零填充。例如，将 00101100₂（44₁₀）左移 1 位得到 01011000₂（88₁₀）；右移 2 位得到 00001011₂（11₁₀），正好 44 ÷ 4 = 11 没有余数。
When shifting, if a bit is shifted beyond the left boundary it is lost, which can lead to overflow if the result requires more bits than available. For instance, in a 4‑bit representation, 1100₂ (12) left‑shifted once becomes 1000₂ (8), which is incorrect because the most significant bit was discarded.
移位时，如果某一位移出了左边界则会丢失，当结果所需的位数超过可用位宽时就会发生溢出。例如，在 4 位表示中，1100₂（12）左移 1 位变成 1000₂（8），结果不正确，因为最高位被丢弃了。
Try it yourself: Apply a left shift of 3 to 00001010₂ (10₁₀) and state the decimal result. Then apply a right shift of 4 to 10100000₂ (160₁₀) and predict both the binary and decimal outcomes.
自己试试：将 00001010₂（10）左移 3 位，写出十进制结果；再将 10100000₂（160）右移 4 位，预测二进制和十进制结果。
Answers: Left shift 3: 10 × 2³ = 80, binary 01010000₂ (8‑bit). Right shift 4: 160 ÷ 16 = 10, binary 00001010₂.
答案：左移3位：80；右移4位：10，二进制 00001010₂。
4. Binary Addition and Overflow | 二进制加法与溢出
arting from the rightmost column, add bits along with any incoming carry. For example, add 0110₂ (6) and 0011₂ ghtmost 0+1=1; next 1+1=0 carry 1; next 1+0+carry=0 carry 1; leftmost 0+0+carry=1 → result 1001₂ (9).
二进制加法遵循简单规则：0+0=0，1+0=1，0+1=1，1+1=0 进位 1，以及 1+1+1（输入进位）= 1 进位 1。从最右列开始，将各位与进来的进位相加。例如，将 0110₂（6）和 0011₂（3）相加：最右 0+1=1；次列 1+1=0 进位 1；次列 1+0+进位=0 进位 1；最左 0+0+进位=1 → 结果 1001₂（9）。
Overflow occurs when the result of an addition exceeds the maximum value that can be stored in the given number of bits. In an 8‑bit register, the largest unsigned number is 255 (11111111₂). Adding 10000000₂ (128) and 10000000₂ (128) yields 1 00000000₂, which requires 9 bits. The extra 1 is lost, leaving 00000000₂, an overflow error. CPUs typically set a flag to signal overflow.
当加法的结果超出给定位数所能存储的最大值时，就会发生溢出。在一个 8 位寄存器中，最大无符号数是 255（11111111₂）。将 10000000₂（128）和 10000000₂（128）相加得到 1 00000000₂，需要 9 位，多余的 1 丢失，留下 00000000₂，即溢出错误。CPU 通常会置位一个标志来表示溢出。
Try it yourself: Perform the binary addition of 0101₂ and 0111₂ in a 4‑bit system. Identify whether overflow occurs and state the decimal result if no overflow.
自己试试：在 4 位系统中计算 0101₂ 和 0111₂ 的加法。判断是否发生溢出，如果没有溢出则写出十进制结果。
Answer: 0101 (5) + 0111 (7) = 1100 (12); 12 fits in 4 bits (max 15), so no overflow.
答案：5+7=12，4 位表示 1100，没有溢出。
5. Character Encoding and Text Size | 字符编码与文本大小
Computers store text using character encoding standards. ASCII typically uses 7 or 8 bits per character, while Unicode (often UTF‑16 in GCSE examples) uses 16 bits (2 bytes) per character. The size of a plain text file is calculated as: number of characters × bits/character, then converted to bytes by dividing by 8. For instance, a 40‑character message encoded in ASCII (8‑bit) occupies 40 × 8 = 320 bits = 40 bytes. The same message in Unicode (16‑bit) would need 40 × 16 = 640 bits = 80 bytes.
计算机使用字符编码标准存储文本。ASCII 通常每个字符使用 7 或 8 位，而 Unicode（GCSE 示例中常为 UTF‑16）每个字符使用 16 位（2 字节）。纯文本文件的大小计算为：字符数 × 每字符位数，然后除以 8 转换为字节。例如，一条 40 个字符的 ASCII（8 位）消息占用 320 位 = 40 字节。同一条消息若采用 Unicode（16 位）则需要 640 位 = 80 字节。
iB = 1024 bytes), but exam boards usually expect the decimal convention unless instructed otherwise.
计算字符数时要包括空格和标点符号。文件大小通常以字节、千字节（1 KB = 1000 字节）或 kibibyte（1 KiB = 1024 字节）表示，但考试局通常要求使用十进制换算，除非另有说明。
Try it yourself: A text file contains the string 'Hello, World!' (13 characters). Calculate its size in bytes for ASCII (8‑bit) and for Unicode (16‑bit). Also express the Unicode size in kilobytes (decimal).
自己试试：一个文本文件包含字符串 'Hello, World!'（13 个字符）。分别计算其 ASCII（8 位）和 Unicode（16 位）制式下的字节大小，并将 Unicode 大小用十进制千字节表示。
Answers: ASCII: 13 × 8 = 104 bits = 13 bytes. Unicode: 13 × 16 = 208 bits = 26 bytes = 0.026 KB.
答案：ASCII 13 字节；Unicode 26 字节 = 0.026 KB。
6. Image File Size | 图像文件大小
A bitmap image is stored as a grid of pixels, each assigned a colour represented by a binary number. The colour depth is the number of bits used for each pixel. For example, a colour depth of
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