S5 ICT Paper: Data Representation, Networks, Databases, Algorithms And Information Security (100 Marks)

S5 ICT Paper: Data Representation, Networks, Databases, Algorithms and Information Security (100 marks)

2026-11-13

S5 ICT – computing and systems

S5 ICT paper: 6 questions, 24 parts, 100 marks, 60 minutes; covering hex conversion, image and audio size, bandwidth, database keys, sorting and search counts.

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Paper details　S5 ICT practice paper with six questions on hex conversion, image and audio file size, network bandwidth and throughput, database keys and indexing, sorting and binary search comparisons, password space and web caching; every figure is calculated.

Part A: Hexadecimal, image and audio size (16 marks)

An uncompressed 24-bit image at 1920 by 1080 pixels, and a 3-minute uncompressed stereo audio clip at 44,100 Hz and 16 bits.

#	Question	Answer and marking notes
1	(a) Convert the hexadecimal number 3A5F to decimal.	Answer: 3 times 4,096 plus 10 times 256 plus 5 times 16 plus 15 gives 14,943.
2	(b) Find the uncompressed image size in MB, where 1 MB is 1,048,576 bytes.	Answer: There are 2,073,600 pixels and 6,220,800 bytes, so the size is 5.93 MB.
3	(c) Find the uncompressed audio size in MB.	Answer: Each second takes 176,400 bytes, so 180 seconds take 31,752,000 bytes, that is 30.28 MB.
4	(d) If the audio is compressed at a ratio of 8 to 1, find the compressed size.	Answer: The compressed size is 3.79 MB, a saving of about 26.50 MB.

Part B: Bandwidth, delay and throughput (16 marks)

A 750 MB file is sent over a 100 Mbps link at 80 percent efficiency. The link is then upgraded to 1 Gbps with the same efficiency.

#	Question	Answer and marking notes
5	(a) Find the transfer time on the original link.	Answer: The file is 6,000 Mb and the effective rate is 80 Mbps, so it takes 75 seconds.
6	(b) Four round trips each add 25 ms of delay; find the delay as a percentage of total time.	Answer: The delay is 100 ms, so it is 0.13 percent of the total time.
7	(c) Find the new transfer time and the speed-up.	Answer: The effective rate is 800 Mbps, so the time is 7.50 seconds, which is 10 times faster.
8	(d) A router handles 12,000 packets per second at 1,250 bytes each; find throughput in MB/s and Mbps.	Answer: Throughput is 15,000,000 bytes per second, that is 15 MB/s or 120 Mbps.

Part C: Database keys, indexing and normalisation (16 marks)

A student table holds 1,200 records with student number, name and class; the school has 40 classes of equal size and the class value repeats in every record.

#	Question	Answer and marking notes
9	(a) State which field suits the primary key and why.	Answer: The student number, because it is unique for every student; names and classes may repeat and cannot be unique.
10	(b) Find the average number of students per class.	Answer: 1,200 divided by 40 gives 30 students.
11	(c) With an index on class, records scanned fall to the class average; find the percentage cut.	Answer: The cut is (1,200 – 30) divided by 1,200, which is 97.50 percent.
12	(d) Explain how moving class into its own table reduces repeated data.	Answer: Class is stored 1,200 times before; after splitting it appears only 40 times, a fall of 96.67 percent.

Part D: Comparison counts in sorting and searching (16 marks)

A bubble sort program needs the maximum number of comparisons when data is in reverse order. Another 1,000 sorted items are available for searching.

#	Question	Answer and marking notes
13	(a) Find the worst-case comparison count for 8 items.	Answer: n times n minus one over two gives 8 times 7 divided by 2, that is 28 comparisons.
14	(b) Find the worst-case comparison count for 64 items.	Answer: 64 times 63 divided by 2 gives 2,016 comparisons.
15	(c) Find the maximum comparisons for a binary search on 1 000 sorted items.	Answer: 2 to the power 10 is 1,024, which covers all 1,001 positions, so at most 10 comparisons are needed.
16	(d) Compare with the average for a linear search and find how many times faster binary search is.	Answer: A linear search averages 500 comparisons against 10 for binary search, so it is 50 times faster.

Part E: Password space and brute-force time (18 marks)

A system requires 8-character passwords using 26 lowercase letters, 26 uppercase letters and 10 digits. An attacker can try one billion combinations per second.

#	Question	Answer and marking notes
17	(a) Find the total number of 8-character combinations.	Answer: Each position has 62 possibilities, so the total is 62 to the power 8, that is 218,340,105,584,896.
18	(b) If only lowercase letters are allowed, find the count and the ratio to (a).	Answer: 26 to the power 8 is 208,827,064,576, only one in 1,046 of (a), a fall of 99.90 percent.
19	(c) At one billion tries per second, how many hours to break (a)?	Answer: 218,340,105,584,896 divided by one billion gives 218,340 seconds, which is 60.65 hours.
20	(d) If the password grows to 10 characters, how many years are needed, taking a year as 31,536,000 seconds?	Answer: The count is 62 to the power 10, so the time is 26.61 years.

Part F: Web traffic and caching benefit (18 marks)

A website serves 50,000 requests a day at 320 KB each, over a 30-day month. After caching is enabled, 60 percent of requests are served from the cache.

#	Question	Answer and marking notes
21	(a) Find the daily traffic in GB, where 1 GB is 1,000,000 KB.	Answer: 50,000 times 320 gives 16,000,000 KB, that is 16 GB a day.
22	(b) Find the monthly traffic.	Answer: 16 times 30 gives 480 GB.
23	(c) Find the daily origin traffic after caching.	Answer: Origin traffic is 40 percent of the total, that is 6.40 GB a day.
24	(d) Find the monthly saving.	Answer: The monthly saving is 192 GB, which is 40 percent of the original 480 GB.

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A TutorZone practice paper; original questions and passages, not publisher material.