HKDSE ICT Practice Paper: Computer Systems, Data Communications And Programming (12 Questions, 100 Marks, 75 Minutes)

HKDSE ICT Practice Paper: Computer Systems, Data Communications and Programming (12 Questions, 100 Marks, 75 Minutes)

2026-11-04

HKDSE ICT: Computer Systems and Data Communications

HKDSE ICT paper: 12 questions, 100 marks, 75 minutes. Binary data, image and audio sizes, transmission time, subnetting, SQL, pseudocode tracing, normalisation.

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Paper details　Level: S6 | Subject: ICT | Time: 75 minutes | Marks: 100 | Questions: 12

Section A: Computer Systems and Data Representation (32 marks)

Answer ALL questions in this section. Show all working.

#	Question	Answer and marking notes
1	(a) Convert the binary number 1011 1101 to denary. (b) Convert the same number to hexadecimal. (c) Convert the hexadecimal value 3F to an 8-bit binary number. (d) Explain why computers use binary rather than denary internally.	(a) 1011 1101 = 128 + 32 + 16 + 8 + 4 + 1 = 189. (b) Grouped in fours, 1011 is B and 1101 is D, so the value is BD. (c) 3F = 00111111. (d) A digital circuit has only two stable states, on and off, which map directly to 0 and 1; binary logic keeps the circuit simple, gives wide noise margins and suits logic gate arithmetic.
2	An uncompressed bitmap is 1920 by 1080 pixels with a colour depth of 24 bits. (a) Find its file size in KiB. (b) Express it in MiB to two decimal places. (c) If lossless compression reduces it to forty percent of the original, find the new size in bytes. (d) Explain why 8-bit indexed colour greatly reduces the file size.	(a) Size = 1920 x 1080 x 3 = 6,220,800 bytes = 6,075 KiB. (b) Dividing by 1,048,576 gives 5.93 MiB. (c) Compressed size = 6,220,800 x 0.4 = 2,488,320 bytes. (d) Indexed colour stores only an 8-bit palette index per pixel instead of 24 bits, cutting the payload to one third, and repeated indices compress better losslessly.
3	A 24 MiB file is sent over a link rated at 8 Mbps whose actual throughput is sixty percent of the rated figure. (a) Find the actual throughput in Mbps. (b) Find the transfer time in seconds. (c) Find the time if the file is first halved in size. (d) State two factors that affect actual throughput.	(a) Actual throughput = 8 x 0.6 = 4.8 Mbps. (b) Payload = 24 x 8 = 192 Mbit, so time = 192 / 4.8 = 40 s. (c) Halved payload 96 Mbit gives 96 / 4.8 = 20 s. (d) Network congestion, retransmission caused by bit errors, server response time and contention from other users on the same link.
4	A processor runs at 3.2 GHz and needs four clock cycles per instruction on average. (a) Find the number of instructions per second and express it in MIPS. (b) Find the MIPS figure at 2.6 GHz with the same average cycle count. (c) Explain how a larger cache improves real performance.	(a) Instructions per second = 3,200,000,000 / 4 = 800,000,000, that is 800 MIPS. (b) 2,600,000,000 / 4 = 650,000,000, that is 650 MIPS. (c) A cache holds frequently used instructions and data close to the processor, which shortens the average memory access time, reduces pipeline stalls and raises real throughput.

Section B: Data Communications, Networks and Security (36 marks)

Answer ALL four questions. Give units for all numerical answers.

#	Question	Answer and marking notes
5	The network 192.168.10.0/24 must be split into four equal subnets. (a) Find the prefix length of each subnet. (b) Find the number of usable host addresses per subnet. (c) Write the subnet mask. (d) Write the network address and the broadcast address of the first subnet.	(a) Two host bits are borrowed, so the prefix is /26. (b) Each subnet has 64 addresses; after removing the network and broadcast addresses, 62 are usable. (c) The subnet mask is 255.255.255.192. (d) The first subnet is 192.168.10.0 with broadcast address 192.168.10.63.
6	A 30 second stereo recording uses a 44.1 kHz sampling rate with 16 bit quantisation and no compression. (a) Find the file size in bytes. (b) Express it in MiB to two decimal places. (c) Find the size as a percentage of the original if it becomes mono with half the sampling rate. (d) Explain the trade-off of lossy audio compression.	(a) Size = 44,100 x 2 x 2 x 30 = 5,292,000 bytes. (b) Dividing by 1,048,576 gives 5.05 MiB. (c) Mono and half the sampling rate together leave 25 percent of the original. (d) Lossy compression discards masked and high frequency detail that listeners rarely notice, which shrinks files sharply, at the cost of an unrecoverable original and cumulative artefacts.
7	Trace the pseudocode and state the final value of total, listing the value after each iteration. total ← 0; for i ← 1 to 5; total ← total + i * i; end for; output total.	The loop adds 1, 4, 9, 16 and 25 in turn. After each iteration total is 1、5、14、30、55, so the final value is 55.
8	The table Orders has columns order_id, customer and amount, with rows (1, A, 120), (2, B, 680), (3, C, 450), (4, D, 900) and (5, E, 300). (a) Write an SQL statement that counts rows whose amount exceeds 500. (b) State the count. (c) Find the total amount of those rows. (d) Explain the role of a primary key.	(a) SELECT COUNT(*) FROM Orders WHERE amount > 500; (b) Rows 2 and 4 qualify, so the count is 2. (c) 680 + 900 = 1,580. (d) A primary key identifies each row uniquely, forbids duplicates and nulls, and can be referenced by foreign keys, which enforces entity integrity.

Section C: Programming, Databases and System Development (32 marks)

Answer ALL questions. Show each step of any trace.

#	Question	Answer and marking notes
9	Compare symmetric and asymmetric encryption, and explain how a digital signature provides both authentication and data integrity. Your answer should state the key arrangement and the performance difference of the two families.	(a) Symmetric encryption uses one shared key for both encryption and decryption; it is fast and suits bulk data but key distribution is hard. (b) Asymmetric encryption uses a public key to encrypt and a private key to decrypt; it solves key distribution but is much slower. (c) A digital signature hashes the message (SHA-256, for example) and encrypts the digest with the sender private key; the receiver verifies with the public key, which authenticates the sender and fails if the content changed. (d) In practice asymmetric encryption exchanges a short lived symmetric key, which then protects the payload.
10	A sorted array holds 1000 elements. (a) How many comparisons does a linear search need in the worst case? (b) How many comparisons does a binary search need at most? (c) Explain why binary search requires sorted data.	(a) Every element may be examined, so 1,000 comparisons. (b) Each comparison halves the range, so at most 10 comparisons, since two to the power 10 is 1,024 and covers 1,000 elements. (c) Binary search decides which half can hold the target by comparing with the middle element; without sorted order that decision is invalid and the target range may be discarded.
11	Describe the five main stages of the system development life cycle, and state one data collection activity commonly carried out during systems analysis together with its purpose.	(a) The five stages are systems analysis, systems design, implementation, testing and maintenance; some models add a feasibility study before analysis. (b) Interviews and questionnaires are commonly used in systems analysis to record problems of the existing system and user requirements, which feed the requirements specification. (c) The stages are not purely sequential, because defects found in testing or maintenance send the project back to design or analysis, so the process iterates.
12	State the problem removed by first, second and third normal form, and explain how IaaS, PaaS and SaaS divide responsibility between provider and customer.	(a) First normal form requires one value per cell, removing repeating groups and multi-valued fields. (b) Second normal form requires every non key column to depend on the whole composite key, removing partial dependency. (c) Third normal form removes transitive dependency between non key columns. (d) IaaS supplies virtual machines and storage while the customer manages the system and application; PaaS also supplies the runtime and middleware so the customer deploys only the application; SaaS delivers a ready application and the customer manages only accounts and data.

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