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HKDSE ICT Mock Paper (10 questions, 100 marks, 90 minutes)
HKDSE ICT
HKDSE ICT mock paper: 10 questions, 100 marks, 90 minutes. Covers file size, binary conversion, spreadsheets, subnetting, algorithms, databases and IoT data, with answers.
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Paper details Level: HKDSE | Subject: ICT | Time: 1 hour 30 minutes | Marks: 100 | Questions: 10
Section A (Questions 1 to 5, 50 marks)
Answer ALL questions. A calculator may be used. Show your working, give numerical answers to the accuracy required, and state units in every final answer. All questions are original TutorZone material and this is not a past paper of any examination board.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A stereo recording is sampled at 44,100 Hz with a bit depth of 16 bits and lasts 3 minutes. (a) Find the total number of samples. (b) Find the file size in MB (1 MB = 1 000 000 bytes). (c) Find the time needed to transmit the file over a network of 8 Mbps (to 1 decimal place). (d) The sampling rate is halved and all other settings stay the same. Find the new file size. |
Answer: (a) 44,100 x 180 = 7,938,000 samples. (b) 7,938,000 x 2 bytes x 2 channels = 31,752,000 bytes = 31.75 MB. (c) 254,016,000 bits / 8 000 000 bits per second = 31.8 seconds. (d) Halving the sampling rate halves the file size: 15.88 MB. Marking: (a) 2 marks, (b) 3 marks, (c) 3 marks, (d) 2 marks, total 10 marks. |
| 2 | A 24-bit colour image has a resolution of 1,920 x 1,080 pixels. (a) Find the uncompressed file size in MB (1 MB = 1 000 000 bytes). (b) Saved as JPEG the file is 640 KB (1 KB = 1 024 bytes). Find the compression ratio as the percentage reduction in file size (to 1 decimal place). (c) How many such compressed images fit in a 4 GB store (4 000 000 000 bytes)? Give a whole number. (d) State which type of compression JPEG uses and name one kind of data lost. |
Answer: (a) 1,920 x 1,080 x 3 = 6,220,800 bytes = 6.22 MB. (b) Reduction = (1 – 655,360 / 6,220,800) x 100% = 89.5%. (c) 4 000 000 000 / 655,360 = 6103.52, so 6,103 images fit. (d) JPEG uses lossy compression; some high-frequency detail and colour data are discarded permanently, so blockiness appears when enlarged. Marking: (a) 2 marks, (b) 3 marks, (c) 3 marks, (d) 2 marks, total 10 marks. |
| 3 | (a) Convert the denary number 2026 to binary and state how many bits the answer uses. (b) Convert 2026 to hexadecimal. (c) How many different values can 10 bits represent? (d) Write -46 in 8-bit twos complement form. |
Answer: (a) 2026 = 11111101010 in binary, which uses 11 bits. (b) 2026 = 7EA in hexadecimal. (c) 2 to the power 10 = 1,024 values. (d) -46 = 256 – 46 = 210, which is 11010010. Marking: (a) 3 marks, (b) 2 marks, (c) 2 marks, (d) 3 marks, total 10 marks. |
| 4 | Cells A2:A5 hold the scores 78, 62, 45 and 91 of four students. (a) Write a formula in A6 for the total score. (b) Write a formula in A7 for the mean score and state the displayed value. (c) Write a COUNTIF formula for the number of students scoring 60 or above, and state the result. (d) Cell B2 holds =IF(A2>=80,”A”,IF(A2>=60,”B”,IF(A2>=40,”C”,”D”))). State what B2 displays. |
Answer: (a) =SUM(A2:A5), giving 276. (b) =AVERAGE(A2:A5), showing 69.0. (c) =COUNTIF(A2:A5,”>=60″), giving 3 students. (d) B2 shows B, because 78 is below 80 but at least 60. Marking: (a) 2 marks, (b) 3 marks, (c) 3 marks, (d) 2 marks, total 10 marks. |
| 5 | A 240 MB file is downloaded over a 96 Mbps broadband link. (a) Find the download time in seconds, ignoring overheads. (b) A network has the address 192.168.10.0 with a /26 subnet mask. Find the total addresses per subnet and the number of usable hosts. (c) How many /26 subnets can be made inside 192.168.10.0/24? (d) State two benefits of subnetting. |
Answer: (a) 240 MB = 1920 Mbit; time = 1920 / 96 = 20 seconds. (b) Host bits = 32 – 26 = 6; total addresses = 2 to the power 6 = 64; usable hosts = 62. (c) 26 – 24 = 2, so 2 to the power 2 = 4 subnets. (d) Less broadcast traffic, tighter security control, and easier allocation by department (any two). Marking: (a) 2 marks, (b) 3 marks, (c) 2 marks, (d) 3 marks, total 10 marks. |
Section B (Questions 6 to 10, 50 marks)
Answer ALL questions. A calculator may be used. Show your working, give numerical answers to the accuracy required, and state units in every final answer. All questions are original TutorZone material and this is not a past paper of any examination board.
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | An array of 1,024 elements is sorted in ascending order. Linear search and binary search are each used to find a target that exists in the array. (a) Find the maximum number of comparisons made by binary search in the worst case. (b) Find the maximum number of comparisons made by linear search in the worst case. (c) Find the average number of comparisons for a successful linear search. (d) Find how many times more comparisons linear search needs in the worst case (to 1 decimal place) and state one precondition of binary search. |
Answer: (a) 2 to the power 10 = 1 024, so at most 10 comparisons. (b) At most 1,024 comparisons. (c) Average = (1 024 + 1) / 2 = 512.5. (d) 1,024 / 10 = 102.4 times; the data must already be sorted. Marking: (a) 3 marks, (b) 2 marks, (c) 2 marks, (d) 3 marks, total 10 marks. |
| 7 | Study the code below. total = 0 for i in range(1, 11): if i % 2 == 0: total = total + i * i print(total) (a) Write the value printed. (b) How many times is the condition inside the loop true? (c) What is printed if i % 2 == 0 is changed to i % 2 == 1? (d) State one main difference between a for loop and a while loop. |
Answer: (a) 4 + 16 + 36 + 64 + 100 = 220. (b) i is 2, 4, 6, 8 and 10, so 5 times. (c) 1 + 9 + 25 + 49 + 81 = 165. (d) A for loop repeats a known number of times over a sequence; a while loop repeats while a condition holds, so the count may be unknown in advance. Marking: (a) 3 marks, (b) 2 marks, (c) 3 marks, (d) 2 marks, total 10 marks. |
| 8 | Table Student holds four records: (S01, 78), (S02, 62), (S03, 45) and (S04, 91), with columns StudentID and Score. (a) Write SQL that lists the StudentID of students scoring 75 or above, and state the result. (b) Write SQL for the mean score of all students. (c) Calculate that mean score (2 decimal places). (d) Class details are to be stored separately. Explain why a second table is needed and which kind of key links the two tables. |
Answer: (a) SELECT StudentID FROM Student WHERE Score >= 75; this returns S01 and S04, 2 rows. (b) SELECT AVG(Score) FROM Student; (c) 276 / 4 = 69.00. (d) Repeating class data in every student record causes redundancy and update anomalies; a separate Class table with ClassID as primary key is used, and Student carries ClassID as a foreign key. Marking: (a) 3 marks, (b) 2 marks, (c) 2 marks, (d) 3 marks, total 10 marks. |
| 9 | A password has 8 characters, each chosen from 26 upper-case letters, 26 lower-case letters or 10 digits (62 in total). (a) Find the total number of 8-character combinations. (b) Express that number of combinations in bits (1 decimal place). (c) The password is lengthened to 12 characters. Find its information content in bits (1 decimal place) and how many times more combinations it gives than the 8-character password. (d) Explain how salting improves the security of stored passwords. |
Answer: (a) 62 to the power 8 = 218,340,105,584,896 combinations. (b) log2(218,340,105,584,896) = 47.6 bits. (c) 62 to the power 12 = 3,226,266,762,397,899,821,056, giving 71.5 bits, which is 62 to the power 4 = 14,776,336 times more combinations. (d) A random salt is added to each account before hashing, so identical passwords give different hashes; rainbow tables and bulk comparison fail and each account must be attacked separately. Marking: (a) 2 marks, (b) 2 marks, (c) 3 marks, (d) 3 marks, total 10 marks. |
| 10 | An Internet of Things temperature sensor takes a reading every 30 seconds and records 16 bytes per reading, running continuously for 24 hours a day. (a) Find the number of readings per day. (b) Find the data produced per day in bytes. (c) Find the total data for 30 days in MB (1 MB = 1 000 000 bytes, 2 decimal places). (d) Compressing before transmission cuts the data to 40 per cent. Find the 30-day compressed volume in MB (2 decimal places) and state two benefits of compressing at the sensor. |
Answer: (a) 86 400 / 30 = 2,880 readings. (b) 2,880 x 16 = 46,080 bytes. (c) 86,400 x 16 = 1,382,400 bytes = 1.38 MB. (d) 1.38 MB x 0.4 = 0.55 MB; benefits include less bandwidth and lower cost, less storage, longer battery life and lower network delay (any two). Marking: (a) 2 marks, (b) 2 marks, (c) 3 marks, (d) 3 marks, total 10 marks. |
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A TutorZone practice paper; original questions and passages, not publisher material.
