S6 ICT Practice Paper: Data Representation, Multimedia, Networks, Databases, Algorithms And Security (100 Marks)

S6 ICT Practice Paper: Data Representation, Multimedia, Networks, Databases, Algorithms and Security (100 marks)

2026-11-12

S6 ICT – Data, Networks and Security

S6 ICT practice paper: 6 questions, 24 parts, 100 marks in 60 minutes. Covers binary conversion, file sizes, networks, SQL, algorithms and password security.

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Paper details　S6 ICT practice paper on data representation, multimedia sizes, networks, databases, algorithms and security, with answers.

Part 1: Data representation (18 marks)

Four questions. Show all conversion steps.

#	Question	Answer and marking notes
1	(a) Convert the denary value 202 into an 8-bit binary number and into hexadecimal.	Answer: 202 = 128 + 64 + 8 + 2, so the 8-bit binary form is 11001010. Grouping into fours gives 1100 = C and 1010 = A, so the hexadecimal form is CA. (3 marks binary, 2 marks hexadecimal)
2	(b) Write -46 in 8-bit two complement form and state the meaning of the most significant bit.	Answer: 8-bit two complement covers -128 to 127, so -46 is stored as 256 – 46 = 210, that is 11010010. The most significant bit is 1, which marks a negative value. (3 marks form, 1 mark meaning)
3	(c) A bitmap of resolution 1024 x 768 uses a colour depth of 8 bits. Find its uncompressed file size in bytes and in KiB.	Answer: number of pixels = 1024 x 768 = 786432. At 8 bits per pixel each pixel takes 1 byte, so the size is 786,432 bytes, which is 768 KiB after dividing by 1024. (2 marks pixels, 2 marks conversion)
4	(d) How many bytes does the string ICT2026 take in ASCII and in UTF-16? Also state the size of 4 Chinese characters in UTF-8.	Answer: ASCII uses one byte per character, so ICT2026 takes 7 bytes; UTF-16 uses two bytes per character, so it takes 14 bytes. A common Chinese character takes 3 bytes in UTF-8, so 4 characters take 12 bytes. (1 to 2 marks each)

Part 2: Multimedia and compression (18 marks)

Four questions. Use 1024 as the base and give answers to two decimal places.

#	Question	Answer and marking notes
5	(a) A photo of 4032 x 3024 pixels uses 24-bit colour. Find its uncompressed size in MiB.	Answer: pixels = 4032 x 3024 = 12,192,768. At 3 bytes per pixel the file holds 36,578,304 bytes, which is 34.88 MiB.
6	(b) A stereo recording sampled at 44100 Hz with 16 bits lasts 3 minutes. Find its uncompressed size in MiB.	Answer: bytes per second = 44100 x 16 / 8 x 2 = 176,400. With 180 seconds the file holds 31,752,000 bytes, that is 30.28 MiB. (2 marks rate, 1 mark channels, 2 marks conversion)
7	(c) If the photo in (a) is compressed to 4.36 MiB by JPEG, find the compression ratio and state the cost of lossy compression.	Answer: ratio = 34.88 / 4.36 = 8.00, about 8 to 1. Lossy compression discards detail the eye barely detects, so the original pixels cannot be restored and repeated compression accumulates error.
8	(d) A video has a bit rate of 2.5 Mbps and lasts 12 minutes. Find its file size in MiB.	Answer: total bits = 2,500,000 x 12 x 60 = 1,800,000,000. Dividing by 8 gives bytes, and dividing twice by 1024 gives 214.58 MiB.

Part 3: Networks and data transmission (16 marks)

Four questions. State units in every answer.

#	Question	Answer and marking notes
9	(a) A 750 MiB file is sent over a 100 Mbps link. Ignoring overhead, find the time needed.	Answer: 1 MiB is 8 Mib, so the data is 750 x 8 = 6,000 Mib. Time = 6,000 / 100 = 60 seconds, that is one minute.
10	(b) A packet passes 4 routers, each adding 2 ms of processing delay, with 15 ms one-way propagation delay. Find the one-way delay and the round trip time.	Answer: one-way delay = 4 x 2 + 15 = 23 ms; round trip time = 23 x 2 = 46 ms. (2 marks each)
11	(c) Host 192.168.10.130 uses the subnet mask /26. Find the subnet address, the broadcast address and the number of usable hosts.	Answer: /26 leaves 6 host bits, so each subnet holds 2 to the power 6 = 64 addresses. 192.168.10.130 falls in the block starting 192.168.10.128, so the subnet address is 192.168.10.128 and the broadcast address is 192.168.10.191; usable hosts = 64 – 2 = 62. (1 to 2 marks each)
12	(d) A 5,000-byte message crosses a network whose MTU is 1500 bytes with a 40-byte header. Find the number of packets and the payload of the last packet.	Answer: payload per packet = 1500 – 40 = 1460 bytes. 5,000 / 1460 needs 4 packets: the first 3 carry 1460 bytes each and the last carries 620 bytes. (2 marks each)

Part 4: Relational databases and SQL (18 marks)

Four questions. Write complete SQL clauses.

#	Question	Answer and marking notes
13	(a) The student table has 6 rows and the club table has 4 rows, and 5 students have joined a club. Find the rows of the unrestricted product and of an inner join on club number.	Answer: the unrestricted product gives 6 x 4 = 24 rows. The inner join keeps only matching keys, so it returns 5 rows and the student without a club is dropped.
14	(b) Explain the difference between a primary key and a candidate key, and state how many candidate keys a student table needs so that student numbers can be changed.	Answer: the primary key is the candidate key chosen to identify each row uniquely, and there is only one per table; candidate keys are all keys that are unique and not null. To change student numbers without losing identification the table needs at least 2 candidate keys, so another key can take over. (2 marks each)
15	(c) Write SQL that lists the names of Form 6 students scoring above 70, sorted by score from high to low, and state the number of result rows.	Answer: SELECT name FROM student WHERE form = 6 AND score > 70 ORDER BY score DESC; from the given data the matching rows are three students, so 3 rows are returned. (3 marks clause, 1 mark rows)
16	(d) A denormalised table holds 1,200 records and repeats 45 bytes of club data in each. How many bytes and KiB are saved by moving club data to a separate table?	Answer: saving = 1,200 x 45 = 54,000 bytes, which is 52.73 KiB. The cost is that queries need an extra join. (3 marks bytes, 2 marks KiB and cost)

Part 5: Algorithms and programming (16 marks)

Four questions. State the number of steps performed.

#	Question	Answer and marking notes
17	(a) A loop adds the integers from 1 to 50. Find the final total and the number of additions.	Answer: total = 50 x 51 / 2 = 1,275; the addition runs 50 times. (2 marks each)
18	(b) Bubble sort is applied to 12 items. Find the number of comparisons in the worst case.	Answer: the first pass makes 11 comparisons and each later pass makes one fewer, so the total is 12 x 11 / 2 = 66. (2 marks each)
19	(c) A binary search runs on 1,000 sorted items. Find the number of comparisons in the worst case.	Answer: each comparison halves the range; 2 to the power 10 is 1,024, which covers 1,000 items, so at most 10 comparisons are needed.
20	(d) The variable x starts at 1 and the statement x = 2x + k runs five times with k from 1 to 5. Find x after each run.	Answer: the values are 3, 8, 19, 42 and 89 in order, so the final value is 89. (1 mark per step)

Part 6: Information systems and security (14 marks)

Four questions. Justify every answer.

#	Question	Answer and marking notes
21	(a) A password uses 62 possible characters and is 8 characters long. Find the number of combinations, and the time to try all of them at one billion attempts per second, in hours and days.	Answer: combinations = 62 to the power 8 = 2.183e+14. Time = 2.183e+14 / 1,000,000,000 = 218,340 seconds, which is 60.65 hours or about 2.53 days. (2 marks each)
22	(b) Compare symmetric and asymmetric encryption in key management and give one use of each.	Answer: symmetric encryption uses one shared key, which is fast but hard to distribute; asymmetric encryption uses a public key to encrypt and a private key to decrypt, which solves distribution but is slow, so it is used for digital signatures and for exchanging a symmetric key. (2 marks each)
23	(c) A system is backed up nightly and needs 4 hours to restore. State its recovery point objective and recovery time objective in numbers.	Answer: the recovery point objective is 24 hours, the worst-case loss of a day of data, and the recovery time objective is 4 hours. Shortening the first needs more frequent backups.
24	(d) Explain why passwords should be stored with salted hashing rather than encryption, and state what the salt does.	Answer: a hash is one-way, so the site never needs the original and a leak does not reveal passwords, while encryption is reversible and a stolen key exposes everything. A salt adds a random value per account so identical passwords hash differently, blocking rainbow tables and bulk matching.

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