Form 4 ICT Practice Paper: Data Representation, Computer Systems And Networks

Form 4 ICT Practice Paper: Data Representation, Computer Systems and Networks

2026-11-05

Form 4 ICT – 12 questions – 100 marks – 75 minutes

Form 4 ICT paper: 12 questions, 100 marks, 75 minutes on number bases, two’s complement, ASCII, file sizes, subnetting, TCP and UDP, encryption and logic gates.

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Paper details　Form 4 ICT practice paper with 12 structured questions on binary and hexadecimal conversion, two’s complement, ASCII codes, audio and image file sizes, subnetting and encryption.

Section A: Number Systems and Coding (Q1-Q4)

Answer all questions; show the expression, substitution and unit for every calculation.

#	Question	Answer and marking notes
1	Convert the decimal number 205 into 8-bit binary and write its hexadecimal form.	Answer: 205 = 128 + 64 + 8 + 4 + 1, so the binary form is 11001101; grouping the bits into nibbles gives 1100 1101, that is CD in hexadecimal. Marks: 5 for the binary value, 4 for the hexadecimal value.
2	8-bit two’s complement: if 10110110 is read as an unsigned integer, what is its decimal value? If it is read as a signed two’s complement integer, what is its value?	Answer: as an unsigned integer it is 182 (128 + 32 + 16 + 4 + 2); the leading 1 marks a negative number, so the two’s complement value is 182 – 256 = -74. Marks: 4 unsigned, 4 signed, with the subtraction shown.
3	In ASCII the letter A has the decimal value 65. Find the decimal value of the letter h and its hexadecimal form.	Answer: upper-case H is 65 + 7 = 72 and a lower-case letter is 32 higher, so h = 104; in hexadecimal that is 68. Marks: 4 for the decimal value, 4 for the hexadecimal form.
4	An uncompressed stereo audio clip has a sample rate of 44 100 Hz, a bit depth of 16 bits, two channels and a length of 30 seconds. Find the bit rate in bps and the file size in MB (1 MB = 1 000 000 bytes).	Answer: bit rate = 44 100 x 16 x 2 = 1 411 200 bps; number of bytes = 1 411 200 / 8 x 30 = 5 292 000, which is 5.292 MB. Marks: 4 for the bit rate, 4 for the file size, with the expression and unit shown.

Section B: Multimedia, Memory and Networks (Q5-Q8)

Data: 1 MB = 1 000 000 bytes and 1 GB = 1 024 MB; a /26 subnet mask leaves 64 addresses per block.

#	Question	Answer and marking notes
5	An uncompressed bitmap is 1024 x 768 pixels at 24-bit colour. What is its file size in MB (1 MB = 1 000 000 bytes)? How does the size change if it is stored as an 8-bit 256-colour indexed image?	Answer: 1024 x 768 = 786 432 pixels, at 3 bytes each that is 2 359 296 bytes, about 2.36 MB; a 256-colour indexed image needs 1 byte per pixel, about 786 432 bytes or 0.79 MB, roughly one third of the original. Marks: 6 for the calculation, 3 for the comparison.
6	Compare RAM, ROM and cache memory in terms of speed, whether data survives power-off and their use. How many bytes are there in 4 GB of memory (1 GB = 1 024 MB, 1 MB = 1 024 KB)?	Answer: RAM is readable and writable but loses its contents at power-off and acts as main memory; ROM is read-only and keeps its contents, storing firmware; cache is the fastest and smallest, holding frequently used instructions and data. 4 GB = 4 x 1024 cubed = 4 294 967 296 bytes. Marks: 6 for the comparison, 3 for the byte count.
7	A host has the IP address 192.168.1.130 with the subnet mask 255.255.255.192 (/26). Find the network address, the broadcast address and the number of usable host addresses.	Answer: a /26 block holds 64 addresses and 130 falls in the 128 to 191 block, so the network address is 192.168.1.128 and the broadcast address is 192.168.1.191; usable host addresses = 64 – 2 = 62. Marks: 3 network address, 3 broadcast address, 2 host count.
8	Compare the transport layer protocols TCP and UDP and give one suitable use for each.	Answer: TCP sets up a connection, acknowledges and retransmits, and guarantees order and completeness, which suits web browsing and file download; UDP is connectionless and does not retransmit, so it has low overhead and low delay, which suits video streaming, online games and DNS queries. Marks: 4 TCP, 4 UDP, each needing a property and an example.

Section C: Programming, Security and Logic (Q9-Q12)

Give numerical answers to three significant figures and answer concept questions in full sentences.

#	Question	Answer and marking notes
9	What is printed after the following pseudocode runs? total = 0 for i from 1 to 5: total = total + i * i print total	Answer: the values 1, 4, 9, 16 and 25 are added in turn, so the output is 55. Marks: 6 for the trace table, which must show total after each pass, 3 for the answer.
10	How do symmetric and asymmetric encryption differ in their use of keys? Why do websites store passwords as hashes rather than as encrypted text?	Answer: symmetric encryption uses one shared key, which is fast but hard to distribute securely; asymmetric encryption uses a public key to encrypt and a private key to decrypt, which solves distribution but is slower. A hash is one-way and irreversible, so a stolen database still does not reveal the passwords, and verification only compares hash values. Marks: 5 for the comparison, 4 for the hashing reason.
11	A logic circuit has the output Y = (A AND NOT B) OR (NOT A AND B). Find Y when A = 1 and B = 0 and name the gate this expression describes.	Answer: with A = 1 and B = 0, A AND NOT B = 1 and NOT A AND B = 0, so Y = 1. The expression is exclusive OR (XOR), which outputs 1 only when the two inputs differ. Marks: 5 for the substitution, 3 for the name.
12	An uncompressed image is 2 359 296 bytes and becomes 480 000 bytes after JPEG compression. Find the compression ratio and the percentage saved, and state whether JPEG is lossy or lossless.	Answer: ratio = 2 359 296 / 480 000 = 4.92, about 4.9 to 1; saving = (2 359 296 – 480 000) / 2 359 296 = 79.7 per cent. JPEG is lossy compression and discards detail the eye is unlikely to notice. Marks: 3 ratio, 3 percentage, 2 for the lossy statement.

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