HKDSE ICT Paper 1B Data And Network Calculations Practice Paper

HKDSE ICT Paper 1B Data and Network Calculations Practice Paper

2026-11-02

HKDSE ICT

Ten original DSE ICT calculation questions on number systems, file sizes, bit rate, transfer time, subnetting, compression, parity and algorithms, each with full working.

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Paper details　Level: S4-S6 (DSE) | Subject: ICT | Time: 60 minutes | Marks: 100 | Questions: 10

Part A Number systems, files and capacity (50 marks)

Questions 1-2 Number systems and images

#	Question	Answer and marking notes
1	(Number systems) An 8-bit register stores the value 205. (a) Convert 205 to binary. (3 marks) (b) Convert the hexadecimal value 5E to decimal. (2 marks) (c) Add 10110110 and 01011010 in an 8-bit register and state whether overflow occurs. (3 marks) (d) Explain why computers represent data in binary rather than decimal. (2 marks)	(a) Repeated division by 2 gives 205 = 128 + 64 + 8 + 4 + 1, so 205 = 11001101 (that is 1100 1101). (b) 5E in hexadecimal = 5 x 16 + 14 = 80 + 14 = 94. (c) 10110110 = 182, 01011010 = 90, sum = 272. An 8-bit register holds only 0 to 255, so 272 – 256 = 16, giving 00010000. Overflow has occurred because the carry leaves the 8th bit. (d) Binary has only two states, 0 and 1, which map naturally onto two stable voltage or switch states; such circuits are simpler and more noise-resistant. Marking note: (a) must show the place values or the division method; (c) must state both the overflow and the remaining 8-bit value.
2	(Image file) An image is 1,024 x 768 pixels with 24-bit true colour per pixel. (a) Calculate the uncompressed size in bytes and in MiB (1 MiB = 1,048,576 bytes). (4 marks) (b) If 256-colour indexed colour (8 bits per pixel) is used instead, what is the file size in bytes? (3 marks) (c) Find the compression ratio of (a) to (b) in the form a to 1. (3 marks)	(a) Each pixel uses 24 bits = 3 bytes; number of pixels = 1,024 x 768 = 786,432. Size = 786,432 x 3 = 2,359,296 bytes = 2,359,296 / 1,048,576 = 2.25 MiB. (b) 786,432 x 1 = 786,432 bytes. (c) Ratio = 2,359,296 / 786,432 = 3, that is 3 to 1. Marking note: (a) must multiply the pixel count by bytes per pixel and convert to MiB correctly; (c) must be expressed as a to 1.

Questions 3-4 Audio and video

#	Question	Answer and marking notes
3	(Digital audio) A stereo track is sampled at 44,100 Hz with a sample depth of 16 bits and lasts 60 seconds. (a) Calculate the data generated per second in bytes. (3 marks) (b) Calculate the file size of the whole track in bytes and in MiB. (3 marks) (c) If mono 8-bit sampling is used, what is the total size in bytes? (2 marks) (d) Explain the effect of a higher sampling rate on file size and sound quality. (2 marks)	(a) Data per second = 44,100 x 2 bytes x 2 channels = 176,400 bytes per second. (b) 176,400 x 60 = 10,584,000 bytes = 10,584,000 / 1,048,576 = 10.09 MiB. (c) 44,100 x 1 x 1 x 60 = 2,646,000 bytes. (d) A higher sampling rate records more samples per second, so the file grows in proportion, but higher frequencies can be captured and sound quality improves. Marking note: (a) must convert the bit depth to bytes and multiply by the channel count; (d) must cover both file size and quality.
4	(Video streaming) A video is 1,920 x 1,080 with 24 bits per pixel at 30 frames per second. (a) Calculate the uncompressed data produced per second in bytes. (4 marks) (b) Convert your answer in (a) to Mbps. (3 marks) (c) If the platform actually streams at 8 Mbps, find the compression ratio in the form a to 1. (3 marks)	(a) One frame = 1,920 x 1,080 x 3 = 6,220,800 bytes; per second = 6,220,800 x 30 = 186,624,000 bytes. (b) 186,624,000 x 8 = 1,492,992,000 bits per second = 1,492.992 Mbps (about 1,493 Mbps). (c) Ratio = 1,492.992 / 8 = 186.6 to 1. Marking note: (a) must multiply by the frame rate; (b) must convert bytes to bits and divide by 1,000,000; (c) must use the same unit for both values.

Question 5 Network transfer

#	Question	Answer and marking notes
5	(Network transfer) A 700 MiB file is sent over a 100 Mbps link at an actual efficiency of 80 per cent. (a) Find the number of bits in the file (1 MiB = 1,048,576 bytes). (3 marks) (b) Find the effective throughput in Mbps and in bytes per second. (3 marks) (c) Find the time needed to finish the transfer in seconds, to 1 decimal place. (3 marks) (d) State one reason why the actual efficiency is below 100 per cent. (1 mark)	(a) 700 x 1,048,576 x 8 = 5,872,025,600 bits. (b) Effective rate = 100 x 0.8 = 80 Mbps = 80 / 8 = 10 MB per second. (c) Time = 5,872,025,600 / 80,000,000 = 73.4 seconds. (d) Any one: network congestion, protocol overhead such as headers and retransmission, limits of the server or disk, or line attenuation. Marking note: (b) must apply the efficiency before converting; (c) requires matching units.

Part B Networks, algorithms and applications (50 marks)

Questions 6-7 Subnetting and compression

#	Question	Answer and marking notes
6	(Subnetting) The network 192.168.10.0 uses a /26 subnet mask. (a) Write the subnet mask in dotted decimal form. (2 marks) (b) How many subnets can be created and how many usable host addresses does each have? (3 marks) (c) Write the broadcast address of the first subnet. (2 marks) (d) Write the usable host address range of the third subnet. (3 marks)	(a) /26 means the first 26 bits are 1: 255.255.255.192, that is 255.255.255.192. (b) Six host bits remain, so each block holds 64 addresses; 256 / 64 = 4 subnets, each with 64 – 2 = 62 usable hosts. (c) The first subnet runs from 192.168.10.0 to 192.168.10.63, so its broadcast address is 192.168.10.63. (d) The third subnet runs from 192.168.10.128 to 192.168.10.191, so usable addresses are 192.168.10.129 – 192.168.10.190. Marking note: (b) must subtract the network and broadcast addresses; (d) must exclude .128 and .191.
7	(Compression) The string AAABBBCCCCCDD is compressed with run-length encoding (RLE), each run written as count then character. (a) Write the compressed string. (3 marks) (b) How many characters are there before and after compression? (2 marks) (c) Find the percentage saved, to 1 decimal place. (2 marks) (d) Is RLE lossless or lossy? Give one type of data on which RLE performs poorly. (3 marks)	(a) The runs are A x 3, B x 3, C x 5 and D x 2, so the compressed string is 3A3B5C2D. (b) 13 characters before and 8 after. (c) Saving = (13 – 8) / 13 x 100 = 38.5 per cent. (d) RLE is lossless: decompression restores the original exactly. It performs poorly on data without long runs, such as scanned photographs or random text, where the file can even grow. Marking note: (a) must use count then character; (d) must state lossless and give a real example.

Questions 8-9 Error detection and algorithms

#	Question	Answer and marking notes
8	(Error detection) The data bits 1011001 use even parity. (a) Find the value of the parity bit. (2 marks) (b) Write the 8 bits actually transmitted. (2 marks) (c) A receiver gets 10110010 and 10110011. Which one is rejected as erroneous? Explain. (3 marks) (d) State one type of error that parity checking cannot detect. (3 marks)	(a) 1011001 contains 4 ones, an even number, and even parity requires an even total, so the parity bit is 0. (b) 10110010 is transmitted. (c) 10110010 has 4 ones, an even number, so it passes; 10110011 has 5 ones, an odd number, so it is rejected. (d) Parity detects only an odd number of flipped bits. If two bits flip, the number of ones stays even and the error goes undetected. Marking note: (c) must count the ones; (d) must name the two-bit error case.
9	(Algorithm efficiency) A sorted array holds 1,000 records. (a) How many comparisons does linear search take on average and in the worst case? (3 marks) (b) How many comparisons does binary search take in the worst case? (Hint: 2 to the power n must be at least 1,001.) (3 marks) (c) Can binary search be used directly on an unsorted array? Explain. (2 marks) (d) In the worst case, how many times faster is binary search than linear search? (2 marks)	(a) Linear search takes (1,000 + 1) / 2 = 500.5 comparisons on average and 1,000 in the worst case. (b) Since 2 to the power 9 = 512 is below 1,001 and 2 to the power 10 = 1,024 is not, the worst case needs 10 comparisons. (c) No. Binary search decides whether to go left or right from the middle element, which requires sorted data; otherwise the target may be discarded. (d) 1,000 / 10 = 100 times. Marking note: (a) must use (n+1)/2; (d) must compare worst cases.

Question 10 Cloud backup

#	Question	Answer and marking notes
10	(Cloud backup) 250 photos average 4.2 MB each. The upload link is 20 Mbps at 90 per cent efficiency. (a) Find the total data volume in MB and in GiB (1 GiB = 1,024 MB). (3 marks) (b) Find the effective upload rate in MB per second. (3 marks) (c) Find the time needed for the backup in seconds and in minutes, to 1 decimal place. (3 marks) (d) Suggest one way to shorten the backup time. (1 mark)	(a) 250 x 4.2 = 1,050 MB; 1,050 / 1,024 = 1.025 GiB. (b) Effective rate = 20 x 0.9 = 18 Mbps = 18 / 8 = 2.25 MB per second. (c) Time = 1,050 / 2.25 = 466.7 seconds = 7.8 minutes. (d) Any one: a faster upload link, compressing or shrinking the photos, a wired or more stable connection, or uploading in off-peak hours. Marking note: (a) requires dividing by 1,024 for GiB; (c) requires both seconds and minutes.

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