Secondary 5 ICT Practice Paper: Binary and Data Representation, Storage and Audio, Networks and Transfer, Algorithms and Complexity, Image Files and Compression, Security

2026-11-13

ICT – Secondary 5 – Practice Paper

S5 ICT: binary and hex conversion, 8-bit two’s complement, uncompressed image and audio sizes, transfer time, search comparisons, password space. 24 parts, 100 marks.

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Paper details　S5 ICT practice paper with 24 parts on binary and hex conversion, two’s complement, image and audio file sizes, network transfer time, search algorithms and password security; full answers and marking points included.

Section A: Binary and Data Representation (16 marks)

Data 1: a system represents integers with eight bits and uses two’s complement for negatives.

#	Question	Answer and marking notes
1	(a) Convert the binary number 10110110 to decimal and to hexadecimal.	Answer: 10110110 = 128 + 32 + 16 + 4 + 2 = 182; in hexadecimal it is B6. (Marking: place values 2, decimal 1, hex 1)
2	(b) Write the integer -58 in 8-bit two’s complement form.	Answer: 58 = 00111010; invert to 11000101; add one gives 11000110 (decimal 198). (Marking: magnitude 1, inversion 2, plus one 1)
3	(c) Convert the hexadecimal number 3F2A to decimal.	Answer: 3 x 4096 + 15 x 256 + 2 x 16 + 10 = 16,170. (Marking: place values 4096 and 256 1 each, sum 1, answer 1)
4	(d) How many different bit patterns can a twelve-bit system represent?	Answer: 2 to the power 12 = 4,096 patterns. (Marking: exponent form 2, value 2)

Section B: Storage Capacity and Audio Files (16 marks)

Data 2: a camera has 12 million effective pixels, each stored with 24-bit colour and no compression.

#	Question	Answer and marking notes
5	(a) Find the file size of one photo in MB, and the total size of 1,500 photos in GB.	Answer: 12,000,000 x 3 bytes = 36,000,000 bytes = 36 MB; 1,500 x 36 = 54,000 MB = 54 GB. (Marking: bytes per pixel 1, one photo 1, total 1, unit 1)
6	(b) A three-minute stereo recording is sampled at 44.1 kHz with 16 bits. Find the uncompressed WAV size in MB, taking 1 MB as 1,000,000 bytes.	Answer: 44,100 x 2 x 2 x 180 = 31,752,000 bytes = 31.75 MB. (Marking: bytes per second 2, channels 1, time and answer 1)
7	(c) The same audio is compressed to MP3 at 128 kbps. Find the size and the compression ratio.	Answer: 128,000 / 8 = 16,000 bytes per second; x 180 = 2,880,000 bytes; ratio = 31,752,000 / 2,880,000 = about 11.0 times. (Marking: bit rate 1, size 1, ratio 2)
8	(d) A phone offers 465 GB of usable space and each photo is about 4.5 MB. How many photos fit?	Answer: 465 x 1,000,000,000 / 4,500,000 = 103,333 photos. (Marking: units 2, division 1, rounding down 1)

Section C: Networks and Data Transfer (16 marks)

Data 3: a broadband link has a bandwidth of 80 Mbps and a round-trip latency of 45 ms.

#	Question	Answer and marking notes
9	(a) In ideal conditions, what is the minimum time to send a 240 MB file?	Answer: 240 x 8 = 1,920 Mb; 1,920 / 80 = 24 s. (Marking: bit conversion 2, division 1, unit 1)
10	(b) If the actual efficiency is only 80 per cent, find the actual time needed.	Answer: 24 / 0.8 = 30 s. (Marking: efficiency 3, answer 1)
11	(c) A transfer needs twelve round trips. Find the total extra delay in seconds.	Answer: 45 ms x 12 = 540 ms = 0.54 s. (Marking: multiplication 2, ms to s 1, answer 1)
12	(d) How many MB per second does the link carry, and how long does 960 MB take?	Answer: 80 / 8 = 10 MB per second; 960 / 10 = 96 s. (Marking: rate 2, time 2)

Section D: Algorithms and Complexity (16 marks)

Data 4: one sorted array holds 4,096 elements; another holds 1,048,576 elements.

#	Question	Answer and marking notes
13	(a) Find the worst-case number of comparisons for linear search and for binary search, and the ratio between them.	Answer: linear 4,096; binary log2 4,096 = 12; ratio = 341. (Marking: linear 1, logarithm 2, ratio 1)
14	(b) For a sorted array of 1,048,576 elements, how many comparisons does binary search need in the worst case?	Answer: 2 to the power 20 = 1,048,576, so 20 comparisons. (Marking: exponent form 2, answer 2)
15	(c) Bubble sort on 500 elements makes n(n-1)/2 comparisons in the worst case. Find that number.	Answer: 500 x 499 / 2 = 124,750 comparisons. (Marking: substitution 2, working 1, answer 1)
16	(d) An O(n squared) program takes 0.02 s for 100 items. Find the time for 400 items.	Answer: input grows 4 times, so time grows 4 squared = 16 times; 0.02 x 16 = 0.32 s. (Marking: square growth 2, factor 1, answer 1)

Section E: Image Files and Compression (18 marks)

Data 5: a 1,920 x 1,080 pixel image is stored once in 24-bit full colour and once in 8-bit indexed colour.

#	Question	Answer and marking notes
17	(a) Find the size of the 24-bit version in MB, taking 1 MB as 1,000,000 bytes.	Answer: 1,920 x 1,080 x 3 = 6,220,800 bytes = 6.22 MB. (Marking: pixel count 1, bytes per pixel 2, answer 1)
18	(b) Find the size of the 8-bit indexed version and the percentage of space saved.	Answer: 1,920 x 1,080 x 1 = 2,073,600 bytes = 2.07 MB; saving = 66.7 per cent. (Marking: size 2, percentage formula 1, answer 1)
19	(c) If width and height are halved to 960 x 540, what is the 24-bit file size and its ratio to the original?	Answer: 960 x 540 x 3 = 1,555,200 bytes, which is 25 per cent of the original. (Marking: pixels 2, bytes 1, ratio 1)
20	(d) A row of 100 identical pixels is stored with run-length encoding at two bytes per run: find the compression ratio. The same picture takes 4.2 MB as PNG and 0.9 MB as JPEG: find the percentage saved.	Answer: run-length encoding gives 100 / 2 = 50.0 times; (4.2 – 0.9) / 4.2 x 100 per cent = 78.6 per cent. (Marking: run-length ratio 3; PNG to JPEG difference 1, divide by base 1, answer 1)

Section F: Information Security (18 marks)

Data 6: a password may use 26 upper-case letters, 26 lower-case letters and 10 digits, 62 symbols in all; an attacker tries 1,000,000,000 combinations per second.

#	Question	Answer and marking notes
21	(a) Find the total number of eight-character passwords.	Answer: 62 to the power 8 = 218,340,105,584,896 combinations. (Marking: base 2, exponent form 2, value 1)
22	(b) Using the previous result, how many hours on average does the attacker need to try every combination?	Answer: 218,340,105,584,896 / 1,000,000,000 = 218,340 s; / 3,600 = about 60.65 hours. (Marking: division 2, conversion 1, answer 1)
23	(c) With sixteen characters the space is about 4.767e+28 combinations. How many times larger is it than the eight-character version?	Answer: 62 to the 16 divided by 62 to the 8 equals 62 to the 8, about 2.18e+14 times, so each extra character multiplies the space by 62. (Marking: exponent division 2, simplification 1, factor 1, explanation 1)
24	(d) Compare symmetric and asymmetric encryption in the use of keys, giving one use of each.	Answer: symmetric encryption uses one shared key for both directions and is fast, so it suits bulk file encryption; asymmetric encryption uses a public key to encrypt and a private key to decrypt, and suits digital signatures and key exchange. (Marking: shared key 1, key pair 1, one use each 1)

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