IGCSE OCR Computer Science: Calculation Practice | IGCSE OCR 计算机：计算题专项训练

📚 IGCSE OCR Computer Science: Calculation Practice | IGCSE OCR 计算机：计算题专项训练

Welcome to the ultimate calculation workout for IGCSE OCR Computer Science. Mastering numerical problems is essential for success in your exams, from binary conversions to file sizes and network speeds. This article provides a structured drill covering all key calculation topics, with step-by-step examples and clear explanations. Let's sharpen your skills and boost your confidence.

欢迎来到 IGCSE OCR 计算机科学计算题专项训练。掌握数值计算题是考试成功的关键，涵盖二进制转换、文件大小到网络速度等。本文提供结构化的训练，覆盖所有关键计算主题，并配有逐步示例和清晰解释。让我们提升技能，增强信心。

1. Binary to Denary Conversion | 二进制转十进制

Convert a binary number to denary by multiplying each bit by its place value – es 8 + 4 + 0 + 1 = 13 in denary. Always write the powers above the bits to avoid mistakes.

将二进制数每位乘以对应的位权（2的幂）即可得到十进制值。例如二进制 1101₂ 计算为 (1×2³)+(1×2²)+(0×2¹)+(1×2⁰)=8+4+0+1=13。建议在每一位上方标注位权，以免出错。

1101₂ = 1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 13

, 2, 1. Bits: 1, 0, 1, 0, 1, 1. Sum = 32 + 8 + 2 + 1 = 43. Verify by recognising that each '1' contributes its weight.

再练一例：101011₂。位权从左到右为 32,16,8,4,2,1。各位为 1,0,1,0,1,1。求和 32+8+2+1=43。也可通过观察每个1的贡献来验证。

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2. Denary to Binary Conversion | 十进制转二进制

ad the remainders from bottom to top to get the binary equivalent. For 43: 43 ÷ 2 = 21 r1; 21 ÷ 2 = 10 r1; 10 ÷ 2 = 5 r0; 5 ÷ 2 = 2 r1; 2 ÷ 2 = 1 r0; 1 ÷ 2 = 0 r1. Reading remainders bottom-up gives 101011₂.

使用连续除以2取余法。将十进制数除以2，记录余数，用商重复直到商为0。余数从下往上读即为二进制。43转换：43÷2=21 余1; 21÷2=10 余1; 10÷2=5 余0; 5÷2=2 余1; 2÷2=1 余0; 1÷2=0 余1。从下往上读得 101011₂。

43₁₀ → remainder record → 101011₂

Always double-check by converting back. An 8-bit representation may require leading zeros, e.g., 43 as 00101011. In OCR questions, you may be asked to use a specific number of bits.

务必转换回去验证。在8位表示中，43 写作 00101011。OCR 考题可能要求指定位数。

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3. Hexadecimal Conversions | 十六进制转换

Hexadecimal uses base 16, with digits 0– ary to hex, group bits in fours from the right, then translate each nibble. Binary 10101111 becomes 1010 1111 → A F → AF₁₆.

十六进制使用基数为16，数字0-9和字母A (10)到F (15)。转十进制：每位乘16的幂。2F₁₆ = (2×16¹)+(15×16⁰)=47。二进制转十六进制时从右开始每4位一组：1010 1111₂ → A F → AF₁₆。

2F₁₆ = 2×16¹ + 15×16⁰ = 47₁₀

emainders. For 200: 200 ÷ 16 = 12 r8, 12 is C, so C8₁₆. Another common task: hex to binary – simply replace each hex digit with its 4-bit binary. A=1010, F=1111 → AF=10101111₂.

十进制转十六进制：反复除以16取余。200÷16=12 余8，12为C，得C8₁₆。十六进制转二进制直接将每位展开为4位二进制即可。

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4. Binary Addition and Overflow | 二进制加法与溢出

bers 11001100 and 10101010 column by column. The sum is 101110110, but in 8-bit storage we keep only the lower 8 bits (01110110) and a carry out indicates overflow. Overflow occurs when the result does not fit in the allotted number of bits, e.g., sum > 255 for unsigned 8-bit.

二进制加法规则：0+0=0, 0+1=1, 1+0=1, 1+1=0 进位1, 1+1+1=1 进位1。将两个8位数 11001100 和 10101010 逐列相加，得 101110110，但8位存储只保留低8位 (01110110)，进位输出表示溢出。当结果超出指定位数范围时发生溢出，例如无符号8位和>255。

11001100 + 10101010 = (1)01110110 → overflow

In the OCR exam, you might need to identify the carry flag and the overflow flag. For signed numbers in two' 10000000, which is an overflow for signed 8-bit because two positives produce a negative.

OCR 考试中可能需要判断进位标志和溢出标志。对于补码有符号数，溢出逻辑不同。例如 01111111 + 00000001 = 10000000，两个正数加出负数，即有符号溢出。

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5. Storage Unit Conversions | 存储单位换算

000 kB, 1 GB = 1000 MB, 1 TB = 1000 GB. To convert 2.5 GB into bytes: 2.5 × 1000³ = 2,500,000,000 bytes. Occasionally, binary-based units (kibibyte = 1024) appear, so read the question carefully.

数据存储使用字节、千字节、兆字节、吉字节、太字节。IGCSE OCR 通常采用十进制前缀：1 kB = 1000 B, 1 MB = 1000 kB, 1 GB = 1000 MB, 1 TB = 1000 GB。2.5 GB 转换为字节：2.5 × 1000³ = 2,500,000,000 字节。偶尔会出现基于1024的二进制单位，需仔细审题。

1 MB = 1000 kB = 1,000,000 bytes

stored as 1 byte (ASCII), requires 80,000 bytes ≈ 80 kB. Always keep units consistent and remember that 1 nibble = 4 bits.

文件大小计算常需用元素总数乘以单个大小。例如80,000字符的文本文件，每字符1字节，需80,000字节≈80 kB。务必保持单位一致，记住1 nibble = 4 bits。

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6. Image File Size | 图像文件大小

ith 24-bit colour gives 800 × 600 × 24 = 11,520,000 bits = 1,440,000 bytes ≈ 1.44 MB. Colour depth determines the number of colours: 1-bit = 2 colours, 8-bit = 256 colours, 24-bit ≈ 16.7 million colours.

图像大小（位）= 宽（像素）× 高（像素）× 颜色深度（位/像素）。除以8得字节数。800×600、24位色的图像大小为 800×600×24 = 11,520,000 位 = 1,440,000 字节 ≈ 1.44 MB。颜色深度决定颜色数：1位=2色，8位=256色，24位≈1670万色。

Size (bytes) = width × height × colour depth / 8

n includes metadata or file header, add that extra amount. Also, when calculating an image' ctice: 1024×768 image with 256 colours requires 1024×768×8 bits = 6,291,456 bits ≈ 786 kB.

若题目包含元数据或文件头，需加上额外大小。当给出颜色数量时，先求最小颜色深度：例如64色需6位 (2⁶=64)。练习：1024×768、256色图像需 1024×768×8 位 ≈ 786 kB。

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7. Sound File Size | 声音文件大小

s) calculates as: 44100 × 16 × 180 × 2 = 254,016,000 bits. Divide by 8 to get bytes: 31,752,000 bytes ≈ 31.75 MB. Always convert minutes to seconds and kHz to Hz.

声音文件大小（位）= 采样率 (Hz) × 位深度 × 时长（秒）× 声道数。CD 音质立体声 (44.1 kHz, 16位, 3分钟) 计算：44100 × 16 × 180 × 2 = 254,016,000 位，合 31.75 MB。记得将分钟转为秒，kHz 转为 Hz。

Size (bits) = sample rate × bit depth × time × channels

,800,000 bits ≈ 9.6 MB. Check if the question asks for an uncompressed or compressed size.

有时直接给出比特率：大小 = 比特率 × 时间。例如 320 kbps 的 MP3，时长4分钟：320,000 × 240 = 76,800,000 位 ≈ 9.6 MB。注意题中是压缩还是未压缩大小。

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8. Network Data Transfer Time | 网络传输时间

00 Megabits. On a 100 Mbps connection, time = 800 / 100 = 8 seconds. Be careful with units – 'Mbps' is Megabits per second, 'MB' is Megabytes. Always write out conversions: 1 byte = 8 bits.

时间(秒) = 数据量(位) ÷ 传输速率(bps)。将所有数据量转为位：100 MB 文件 = 100 × 8 = 800 兆比特。在 100 Mbps 连接下，时间 = 800 / 100 = 8 秒。注意单位区分：Mbps 为兆比特/秒，MB 为兆字节。牢记 1 字节 = 8 比特。

t = (file size in bits) / (transfer speed in bps)

hen overheads or protocol headers are included, add those percentages. For instance, if 10% ider parallel downloads. Practice: 500 MB file on 200 Mbps takes 500×8 / 200 = 20 seconds.

若含协议开销，需考虑百分比。如10%开销，有效速率降低。另外可能计算多文件下载或并行下载。练习：500 MB 文件在 200 Mbps 下需 500×8/200 = 20 秒。

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9. Compression Ratio | 压缩比

sses to 100 KB, the ratio is 500/100 = 5, i.e., 5:1. The percentage reduction is (original – compressed)/original × 100% = (500–100)/500 × 100% = 80%. Both metrics appear in OCR questions.

压缩比 = 原始大小 ÷ 压缩后大小，常表示为 5:1。500 KB 图像压缩到 100 KB，压缩比为 500/100 = 5，即 5:1。减小百分比 = (原始-压缩)/原始 × 100% = 80%。OCR 试题中两种指标都可能出现。

Compression ratio = original : compressed

provide both uncompressed and compressed sizes. You might also be asked to find the compressed size from a given ratio: compressed = original / ratio.

有损压缩通常得到更高压缩比但损失质量。计算时使用题中所给实际大小；有时给出压缩比求压缩后大小：压缩后 = 原始 / 比。

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10. Logic Gate Calculations | 逻辑门计算

D B = 0, NOT C = 0, then 0 OR 0 = 0. Build a truth table listing all 2³ = 8 combinations of A, B, C. For each row, compute intermediate values and final Q. This method proves essential for exam questions.

逻辑电路计算将二进制输入代入布尔表达式。电路 Q = (A AND B) OR (NOT C)，若 A=1, B=0, C=1：A AND B = 0, NOT C = 0, 然后 0 OR 0 = 0。为所有8种组合建立真值表，逐行计算中间值和最终 Q。此方法对考试至关重要。

A	B	C	A AND B	NOT C	Q
0	0	0	0	1	1
0	0	1	0	0	0
0	1	0	0	1	1
0	1	1	0	0	0
1	0	0	0	1	1
1	0	1	0	0	0
1	1	0	1	1	1
1	1	1	1	0	1

A = A, or draw circuits from a truth table. Always label gates clearly and be systematic.

可能还需运用布尔代数规则简化表达式，例如 A AND A = A，或根据真值表画出电路。始终系统化地标注逻辑门。

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11. Processor Performance Calculations | 处理器性能计算

akes 4 cycles on average, the instructions per second (IPS) = clock speed / CPI = 3×10⁹ / 4 = 750 million instructions per second. Multiply by time to get total instructions.

时钟速度表示 CPU 每秒执行的周期数。若 CPU 主频 3 GHz (3×10⁹ 周期/秒)，平均每条指令需4周期，则每秒指令数 (IPS) = 时钟速度 / CPI = 3×10⁹/4 = 7

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