A-Level CIE Computer Science Unit Test: Information Representation | A-Level CIE 计算机：信息表示单元测试卷

📚 A-Level CIE Computer Science Unit Test: Information Representation | A-Level CIE 计算机：信息表示单元测试卷

This unit test covers key concepts from the CIE A-Level Computer Science syllabus (9618) topic ‘Information Representation’. It includes multiple-choice and structured questions designed to assess understanding of binary, hexadecimal, two’s complement, floating-point representation, and data compression. Use this test to check your knowledge and identify areas for revision.

本单元测试覆盖CIE A-Level计算机科学大纲(9618)中“信息表示”主题的核心概念。试卷包含选择题和简答题，旨在评估对二进制、十六进制、补码、浮点表示和数据压缩的理解。请用此测试来检查你的知识掌握情况并找出需要复习的薄弱环节。

1. Section A: Multiple Choice Questions | 选择题部分

Question 1: What is the denary number 178 in binary? A) 10110010 B) 10110011 C) 10110110 D) 10100110

问题1：十进制数178的二进制形式是哪一个？

Question 2: Which hexadecimal value represents the binary number 1101 1011? A) DB B) BD C) C9 D) DA

问题2：十六进制值DB所对应的二进制数是1101 1011吗？

Question 3: The two’s complement representation of –23 in an 8-bit register is: A) 11101001 B) 11101000 C) 10010111 D) 11101010

问题3：在8位寄存器中，–23的补码表示是哪一个？

Question 4: How many different values can be represented with 5 bits in unsigned binary? A) 10 B) 16 C) 25 D) 32

问题4：使用5位无符号二进制可以表示多少个不同的值？

in binary as 0.1011 × 2⁻² in normalised floating-point. Its denary value is approximately: A) 0.171875 B) 0.34375 C) 0.6875 D) 1.375

问题5：一个实数以二进制存储为规格化浮点数0.1011 × 2⁻²，其十进制值约是多少？

Question 6: Which sampling rate, according to Nyquist theorem, is the minimum required to accurately reproduce a 20 kHz sound frequency? A) 10 kHz B) 20 kHz C) 30 kHz D) 40 kHz

问题6：根据奈奎斯特定理，要精确再现20 kHz声音频率所需的最低采样率是？

Question 7: A colour image uses 24-bit colour depth. What is the maximum number of distinct colours that can be displayed? A) 8 B) 256 C) 16 777 216 D) 4 294 967 296

问题7：一张彩色图片使用24位色深，最多可以显示多少种不同的颜色？

Question 8: Which of the following is a lossless compression method? A) JPEG B) MP3 C) Run-length encoding D) MPEG

问题8：下列哪一种是无损压缩方法？

Question 9: The binary addition of 0101 and 0111 yields: A) 1100 with no overflow B) 1100 with overflow C) 1010 with overflow D) 1100 with carry out of 0

问题9：二进制数0101与0111相加的结果是？

Question 10: In BCD (binary-coded decimal), the number 59 is represented as: A) 0101 1001 B) 0011 1011 C) 111011 D) 00111001

问题10：在BCD码中，数字59表示为？

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2. Section B: Structured Questions | 简答题部分

Question 11: A computer uses a normalised floating-point representation with a 12-bit mantissa and a 4-bit exponent, both in two’s complement. The mantissa is stored using the form m.xxxx and the exponent is an integer power of 2. (a) Explain why normalisation is important in floating-point representation. (b) Show the representation of the denary number –7.25 in this format. Give your answer in binary.

问题11：某计算机使用规格化浮点表示，12位尾数和4位阶码，均采用补码。尾数以m.xxxx形式存储，阶码是2的整数次幂。 (a) 解释浮点表示中规格化的重要性。 (b) 写出十进制数–7.25在这种格式下的二进制表示。

Question 12: A bitmap image is 800 pixels wide and 600 pixels high, with a colour depth of 8 bits. Calculate the file size in kilobytes. Show your working. If run-length encoding is applied and consecutive identical pixels are encoded as (count, colour value), state one situation where this method would increase the file size instead of reducing it.

问题12：一幅位图图像宽800像素，高600像素，色深8位。计算其文件大小（以千字节为单位），写出计算步骤。如果采用游程编码，将连续相同像素编码为(计数, 颜色值)的形式，请指出一种这种编码方法反而会增加文件大小的情况。

Question 13: A sound file is recorded at a sampling rate of 44.1 kHz with a sample resolution of 16 bits for a duration of 3 minutes. The audio is in stereo. (a) Calculate the raw file size in megabytes. (b) If the file is compressed with a ratio of 4:1, what is the compressed file size? (c) Explain why increasing the sampling rate and sample resolution improves sound quality but increases file size.

问题13：一段声音文件以44.1 kHz的采样率、16位采样精度录制，时长3分钟，立体声。 (a) 计算原始文件大小（以兆字节计）。 (b) 如果该文件以4:1的比例压缩，压缩后的文件大小是多少？ (c) 解释为什么提高采样率和采样精度会提升音质但使文件变大。

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3. Answers and Explanations | 答案与解析

44→22 r0, 22→11 r0, 11→5 r1, 5→2 r1, 2→1 r0, 1→0 r1. Reading remainders upwards: 10110010.

1. 答案：A) 10110010。将178连续除以2：178→89余0，89→44余1，44→22余0，22→11余0，11→5余1，5→2余1，2→1余0，1→0余1。从下往上读余数得到10110010。

swer: A) DB. Hex D = 1101, hex B = 1011, so correct. BD would be 1011 1101, not the given.

2. 答案：A) DB。十六进制D = 1101，B = 1011，所以DB正确。BD对应1011 1101，与题目不符。

3. Answer: A) 11101001. +23 in 8-bit = 00010111. Invert: 11101000, add 1: 11101001.

3. 答案：A) 11101001。+23的8位二进制为00010111。取反得11101000，加1得11101001。

4. Answer: D) 32. 2⁵ = 32.

4. 答案：D) 32。2的5次方等于32。

1 × 2⁻⁴ = 0.5 + 0.125 + 0.0625 = 0.6875. Multiply by 2⁻² (0.25): 0.6875 × 0.25 = 0.171875.

5. 答案：A) 0.171875。0.1011₂ = 1×2⁻¹ + 0×2⁻² + 1×2⁻³ + 1×2⁻⁴ = 0.5 + 0.125 + 0.0625 = 0.6875。乘以2⁻² (0.25)：0.6875 × 0.25 = 0.171875。

swer: D) 40 kHz. Nyquist theorem: sampling rate >= 2 × max frequency. 2 × 20 kHz = 40 kHz.

6. 答案：D) 40 kHz。奈奎斯特定理：采样率 ≥ 最大频率的两倍。2 × 20 kHz = 40 kHz。

7. Answer: C) 16 777 216. 24 bits allow 2²⁴ = 16 777 216 combinations.

7. 答案：C) 16 777 216。24位可表示2²⁴ = 16 777 216种组合。

8. Answer: C) Run-length encoding. JPEG and MPEG use lossy compression, MP3 is lossy. Run-length encoding is lossless.

8. 答案：C) 游程编码。JPEG和MPEG使用有损压缩，MP3也是有损。游程编码是无损的。

: B) 1100 with overflow. 0101 (5) + 0111 (7) = 1100, but in 4-bit two’s complement 1100 is –4, so overflow occurred because result is 12, outside range –8 to 7.

9. 答案：B) 1100 有溢出。0101 (5) + 0111 (7) = 1100，但在4位补码中1100代表–4，因此发生了溢出，因为实际结果为12，超出了–8到7的范围。

CD codes each decimal digit separately: 5→0101, 9→1001. No conversion of the whole number.

10. 答案：A) 0101 1001。BCD码对每位十进制数字单独编码：5→0101，9→1001。不将整个数字转换。

ng leading zeros and giving a unique representation for each number. It allows the largest number of significant bits to be stored. (b) –7.25 = –111.01₂ = –0.11101 × 2³. With two’s complement mantissa 12 bits and exponent 4 bits: 0 for positive fraction). Negative: invert all bits 1.00010111111, add 1 → 1.00011000000. Exponent: +3 in 4-bit two’s complement = 0011. Representation: mantissa 100011000000, exponent 0011.

11. (a) 规格化通过确保尾数的最高有效位为1（正数）或0（负数）来最大化精度，消除前导零，并为每个数字提供唯一表示。这样可以存储尽可能多的有效位。 (b) –7.25 = –111.01₂ = –0.11101 × 2³。12位补码尾数和4位阶码： 尾数：正数值0.11101000000（12位，第一位为0表示正小数）。取负：所有位取反1.00010111111，加1得1.00011000000。 阶码：+3的4位补码为0011。 表示：尾数100011000000，阶码0011。

Convert to bytes: 3 840 000 / 8 = 480 000 bytes. To kilobytes: 480 000 / 1024 = 468.75 KB. r every pixel adds the count overhead, doubling the size if originally one byte per pixel.

12. 文件大小 = 宽 × 高 × 色深（位）。800 × 600 × 8 = 3 840 000位。换算成字节：3 840 000 / 8 = 480 000 字节。转换为千字节：480 000 / 1024 = 468.75 KB。 当连续重复像素很少时，游程编码会增加文件大小；每个像素存储为(1, 颜色)会添加计数开销，如果原本每个像素一个字节，文件大小可能翻倍。

000 bits. Bytes = 254 016 000 / 8 = 31 752 000 bytes. MB = 31 752 000 / (1024 × 1024) ≈ 30.28 MB. (b) Compressed size = 30.28 MB / 4 = 7.57 MB. pth) allows finer amplitude levels, reducing quantisation error. Both increase the amount of data per second, therefore file size grows.

13. (a) 总采样数 = 采样率 × 时间 × 声道数 = 44 100 × (3 × 60) × 2 = 44 100 × 180 × 2 = 15 876 000次。总位数 = 15 876 000 × 16 = 254 016 000位。字节数 = 254 016 000 / 8 = 31 752 000字节。MB = 31 752 000 / (1024 × 1024) ≈ 30.28 MB。 (b) 压缩后大小 = 30.28 MB / 4 = 7.57 MB。 (c) 更高的采样率能捕获更多频率细节，更接近原始模拟波形，减少混叠；更高的采样精度（位深度）允许更精细的振幅等级，减少量化误差。两者都增大了每秒的数据量，因此文件大小增加。

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