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HKDSE ICT Practice Paper: Computer Systems, Data Communication and Programming
HKDSE ICT – 12 questions – 100 marks – 75 minutes
Original HKDSE ICT practice paper: number systems, file sizes, CPU performance, bandwidth, subnetting, compression, program tracing and algorithms, with full worked solutions.
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Paper details HKDSE ICT practice paper on number systems, file sizes, bandwidth, subnetting, compression and program tracing, with a full marking scheme.
Section A: Data Representation and Computer Systems (Q1-Q4)
Show all working; answers alone earn no method marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A computer stores integers in 8-bit two’s complement form. (a) Convert the binary number 10110110 to hexadecimal and to denary. (4 marks) (b) Convert the denary number 202 to an 8-bit binary number and state the value of the most significant bit (MSB). (4 marks) |
(a) Split 10110110 into nibbles: 1011 = B and 0110 = 6, so the hexadecimal value is B6. Denary = 128 + 32 + 16 + 4 + 2 = 182. (2 marks each) (b) 202 = 128 + 64 + 8 + 2 = 11001010, and the most significant bit (leftmost) is 1, representing the value 128. (3 marks conversion, 1 mark MSB) |
| 2 | A camera sensor has 4000 x 3000 pixels and uses 24 bits per pixel. The image is stored without compression. (a) Calculate the file size of one photograph in MB (1 MB = 1 000 000 bytes). (3 marks) (b) Calculate the total capacity needed for 1 000 photographs in GB. (2 marks) (c) State two ways to reduce the file size. (3 marks) |
(a) Pixels = 4,000 x 3,000 = 12,000,000. At 24 bits per pixel that is 3 bytes per pixel, so the file is 12,000,000 x 3 = 36,000,000 bytes = 36.0 MB. (b) 1 000 photographs = 36.0 MB x 1 000 = 36.0 GB. (c) Any two: lower the resolution, reduce the colour depth, use lossy compression such as JPEG, or save at a higher compression setting. (1.5 marks each) |
| 3 | A 750 MB file is transferred over a 100 Mbps link at 80 per cent of the theoretical rate. (a) Calculate the effective transfer rate in Mbps and in MB per second. (3 marks) (b) Calculate the transfer time in seconds. (3 marks) (c) If a 1 Gbps link is used at the same efficiency, what is the new transfer time? (2 marks) |
(a) Effective rate = 100 x 0.8 = 80 Mbps, which is 10 MB per second. (3 marks) (b) Time = 750 / 10 = 75.0 seconds. (3 marks) (c) At 1 Gbps and the same efficiency the rate is 800 Mbps = 100 MB per second, so time = 750 / 100 = 7.5 seconds. (2 marks) |
| 4 | An image of 1024 x 768 pixels is stored at 8-bit and at 16-bit colour depth. (a) Calculate the number of colours available at each depth. (2 marks) (b) Calculate both file sizes in KB (1 KB = 1 024 bytes). (4 marks) (c) Explain why professional photo editing usually works at 16 bits. (2 marks) |
(a) 8-bit: 2 to the power 8 = 256 colours; 16-bit: 2 to the power 16 = 65,536 colours. (1 mark each) (b) Pixels = 786,432. At 8 bits the file is 786,432 x 1 byte = 786,432 bytes = 768 KB; at 16 bits it is 786,432 x 2 bytes = 1,572,864 bytes = 1,536 KB. (2 marks each) (c) A 16-bit image records far more levels per channel, so tonal transitions are smoother and heavy adjustments such as brightening or curves produce less banding. (2 marks) |
Section B: Data Communication, Networking and Compression (Q5-Q8)
Use 1 MB = 1 000 000 bytes throughout.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A processor runs at 3.2 GHz with an average CPI of 4.0. (a) Calculate its performance in MIPS. (3 marks) (b) Calculate the time needed to execute 2,000,000,000 instructions. (3 marks) (c) If CPI falls to 2.5 at the same clock rate, what is the new execution time? (3 marks) |
(a) Instructions per second = 3.2 x 10 to the power 9 / 4.0 = 800,000,000, that is 800 MIPS. (3 marks) (b) Time = 2,000,000,000 / 800,000,000 = 2.5 seconds. (3 marks) (c) At CPI 2.5 the rate is 3.2 x 10 to the power 9 / 2.5 = 1,280,000,000 instructions per second, so time = 2,000,000,000 / 1,280,000,000 = 1.5625 seconds, about 1.56 seconds, a saving of 0.9 seconds. (3 marks) |
| 6 | A stereo audio clip is sampled at 44,100 Hz with 16 bits per sample and lasts 3 minutes. It is stored uncompressed. (a) Calculate the bit rate in kbps. (3 marks) (b) Calculate the file size in MB (1 MB = 1 000 000 bytes). (3 marks) (c) Explain why streaming services still compress audio. (3 marks) |
(a) Bit rate = 44,100 x 16 x 2 = 1,411,200 bps = 1411.2 kbps. (3 marks) (b) Bytes per second = 1,411,200 / 8 = 176,400; file size = 176,400 x 180 = 31,752,000 bytes = 31.75 MB. (3 marks) (c) The uncompressed rate of about 1411.2 kbps is far above the stable bandwidth of most mobile connections, causing buffering and heavy data use; compression allows smooth playback within limited bandwidth and saves data. (3 marks) |
| 7 | A network uses the address block 192.168.10.0/26. (a) Write down the subnet mask. (3 marks) (b) Calculate the number of addresses in each subnet and the number of usable hosts. (3 marks) (c) How many /26 subnets can be created, and what is the network address of the second subnet? (3 marks) |
(a) /26 means the first 26 bits are the network part, giving the mask 255.255.255.192. (3 marks) (b) Host bits = 6, so addresses = 2 to the power 6 = 64 and usable hosts = 64 – 2 = 62 after removing the network and broadcast addresses. (3 marks) (c) 256 / 64 = 4 subnets, and the second subnet is 192.168.10.64. (3 marks) |
| 8 | A 48 MB data file is reduced to 12 MB by lossless compression. (a) Calculate the compression ratio of original to compressed size. (3 marks) (b) Calculate the percentage saved. (3 marks) (c) Distinguish lossless from lossy compression and give one example of each. (3 marks) |
(a) Compression ratio = 48 to 12 = 4 to 1. (3 marks) (b) Saved = (48 – 12) / 48 x 100 = 75.0 per cent. (3 marks) (c) Lossless compression restores the original data exactly and suits text, source code and executables, for example ZIP and PNG; lossy compression discards detail the eye or ear barely notices for a much higher ratio, for example JPEG and MP3. (3 marks) |
Section C: Programming and Information Processing (Q9-Q12)
Code is given in Python syntax; range(a, b) excludes b.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Study the following code: total = 0 for i in range(1, 11): if i %% 3 == 0: total = total + i print(total) (a) State the output of the program. (3 marks) (b) How many times does the loop iterate? (2 marks) (c) What is the output if range(1, 101) is used instead? (3 marks) |
(a) i runs from 1 to 10 and only 3, 6 and 9 are divisible by 3, so total = 3 + 6 + 9 = 18 and the program prints 18. (3 marks) (b) range(1, 11) yields 1 to 10, so the loop iterates 10 times. (2 marks) (c) Between 1 and 100 the multiples of 3 run from 3 to 99, 33 of them, so total = 3 x (1 + 2 + … + 33) = 3 x 561 = 1683. (3 marks) |
| 10 | Column A of a spreadsheet holds the marks of five students: 60, 72, 85, 90, 68. (a) Calculate the mean mark. (3 marks) (b) Calculate the number and percentage of students scoring 80 or above. (3 marks) (c) Calculate the range of the marks. (2 marks) |
(a) Total = 60 + 72 + 85 + 90 + 68, so the mean = 375 / 5 = 75.0 marks. (3 marks) (b) 2 students scored 80 or above (85 and 90), which is 2 / 5 x 100 = 40.0 per cent. (3 marks) (c) Range = 90 – 60 = 30 marks. (2 marks) |
| 11 | A product barcode has the first 11 digits 40063812000. The check rule is: multiply digits in odd positions by 1 and those in even positions by 3, take the sum modulo 10 and subtract it from 10 (a remainder of 0 gives a check digit of 0). (a) Calculate the weighted sum. (3 marks) (b) Calculate the check digit. (3 marks) (c) Write the complete 12-digit barcode. (2 marks) |
(a) Weighted sum = 4×1 + 0x3 + 0x1 + 6×3 + 3×1 + 8×3 + 1×1 + 2×3 + 0x1 + 0x3 + 0x1 = 4 + 18 + 3 + 24 + 1 + 6 = 56. (3 marks) (b) 56 mod 10 = 6 and 10 – 6 = 4, so the check digit is 4. (3 marks) (c) The complete barcode is 400638120004. (2 marks) |
| 12 | A target key is searched in 1,000 records already sorted by key, using either linear search or binary search. (a) Calculate the worst-case number of comparisons for each method. (3 marks) (b) State two preconditions for binary search. (2 marks) (c) If the data were unsorted, what must be done before binary search, and what is the extra cost? (3 marks) |
(a) Linear search may examine all 1,000 records, while binary search halves the range each time, so the worst case is log2(1,000) rounded up = 10 comparisons. (3 marks) (b) Any two: the data must be sorted; the structure must allow random access such as an array, so it is unsuitable for sequential-only storage. (2 marks) (c) The data must first be sorted, for example by merge sort or quicksort, which costs O(n log n), around 1,000 x 10 = 10,000 comparisons in the worst case; that one-off cost only pays off when the same sorted set is searched repeatedly. (3 marks) |
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