CIE A-Level Computer Science 9618 Paper 1 Practice Paper (Theory Fundamentals) - 75 marks, 1 hour 30 minutes, answers and mark scheme

CIE A-Level Computer Science 9618 Paper 1 Practice Paper (Theory Fundamentals) – 75 marks, 1 hour 30 minutes, answers and mark scheme

2026-10-30

TutorZone original practice paper – CIE A-Level Computer Science 9618 Paper 1 (Theory Fundamentals)

8 questions, 27 parts – 75 marks – 1 hour 30 minutes – calculator allowed

Total 75 marks · 1 h 30 min · 11 parts

Marks distribution 15 + 15 = 30

Level A2 (second year of A-Level)

Paper comparison

Paper	Time	Marks	Nature
Paper 1 (Theory Fundamentals)	1 h 30 min	75	this paper
Paper 2 (Problem-solving and Programming)	2 h	75	other AS paper
Paper 3 (Advanced Theory)	1 h 30 min	75	A Level
Paper 4 (Practical)	2 h 30 min	75	A Level

Data and formulae

Quantity	Value
1 KiB	1024 bytes
1 MiB	1024 x 1024 bytes
Image file size	width x height x colour depth (bits) / 8
Sound file size	sample rate x resolution (bits) / 8 x channels x seconds
Execution time	instructions x CPI / clock speed
Average memory access time	hit rate x hit time + (1 – hit rate) x miss time
8-bit two’s complement range	-128 to +127

Questions

Total: 75 marks · 1 h 30 min · calculator allowed

Q 1 · Information representation: two’s complement, hexadecimal, images and sound, RLE [12]

(a) Represent -45 in 8-bit two’s complement and state the range of integers that 8-bit two’s complement can represent. [2]

(b) A byte holds the hexadecimal value 9C. State its value as an unsigned integer and as a signed integer in 8-bit two’s complement. [2]

(c) An uncompressed bitmap is 800 x 600 pixels with a colour depth of 24 bits. Calculate the file size in KB (1 KB = 1000 bytes). If the same image is stored with a 256-colour palette, state the new file size and how many times smaller it is. [3]

(d) A 90-second stereo recording is sampled at 44100 Hz with a resolution of 16 bits and is not compressed. Calculate the file size in bytes and in MB (1 MB = 1024 x 1024 bytes). [2]

(e) A row of 200 pixels is 20 black, 30 white and 150 grey. Write the run-length encoded form, calculate how many values are stored and the percentage saved, and state when RLE would make a file larger. [3]

Q 2 · Data transmission and networking [10]

(a) Compare serial and parallel data transmission, giving two differences, and explain why serial is used for long distances. [3]

(b) A 12 MiB file is sent over a 20 Mbps link (1 MiB = 1024 x 1024 bytes, ignore overhead). Calculate the transfer time in seconds to 2 decimal places and explain why the real time is longer. [3]

(c) Explain the difference between a MAC address and an IP address (one use each) and state the number of bits and the number of possible addresses for IPv4 and IPv6. [4]

Q 3 · Hardware: von Neumann model, buses, storage, embedded systems [11]

(a) Name three main components of the von Neumann model and explain the purpose of the bus. [3]

(b) A processor has an address bus of 20 lines. Calculate the maximum directly addressable memory in bytes and KiB, and state the difference between the address bus and the data bus. [3]

(c) Give two advantages of a solid-state drive (SSD) over a hard disk drive (HDD). [2]

(d) Using a washing machine, explain what an embedded system is and state the role of one sensor and one actuator. [3]

Q 4 · Processor fundamentals: fetch-decode-execute, performance, cache [9]

(a) Describe the fetch-decode-execute (FDE) cycle and state the role of the PC, MAR, MDR and CIR. [3]

(b) A processor runs at 3.2 GHz with an average of 2.5 clock cycles per instruction. Calculate the instructions per second (in MIPS) and the time in milliseconds to run 5,000,000 instructions (2 decimal places). [3]

(c) A program makes 500 memory accesses and 40 of them miss the cache. A hit takes 10 ns and a miss takes 120 ns. Calculate the hit rate and the average access time, and explain how a cache improves performance. [3]

Q 5 · System software: operating system and utilities [8]

(a) State three main functions of an operating system and give an example of each. [3]

(b) Compare a compiler and an interpreter, giving two differences, and explain why an interpreter is useful during development. [3]

(c) Name two pieces of utility software and state the purpose of each. [2]

Q 6 · Security, privacy and data integrity [9]

(a) Explain the difference between symmetric and asymmetric encryption and the role of a digital certificate. [3]

(b) A site allows only 36 characters (26 lowercase letters and 10 digits) and passwords are 8 characters long. Calculate the number of possible passwords and how long a computer testing 10^9 passwords per second would need, in minutes, then suggest two ways to improve account security. [3]

(c) Describe a distributed denial of service (DDoS) attack, phishing and the purpose of a firewall. [3]

Q 7 · Databases: keys, normalisation, SQL [8]

(a) A database has Student(StudentID, Name, Class) and ClubMembership(StudentID, ClubID, JoinDate). Identify the primary keys and foreign keys and state the type of relationship between the tables. [3]

(b) An unnormalised table stores OrderID, CustomerName, CustomerAddress and several item names with unit prices in one cell, separated by commas. Explain how to normalise it to third normal form (3NF), naming one problem removed at each step. [3]

(c) Write an SQL statement that returns the number of orders in Orders(OrderID, CustomerID, Total) whose total is greater than 500, naming the result BigOrders. [2]

Q 8 · Algorithm design and data structures [8]

(a) A sorted array holds 1,000,000 elements. State the worst-case number of comparisons for a linear search and for a binary search, and state the precondition for a binary search. [3]

(b) Sort [6, 3, 9, 1, 7] into ascending order with a bubble sort that compares adjacent items and repeats full passes until no swaps occur. State the sorted array, the number of comparisons, the number of swaps and the time complexity. [3]

(c) A hash table uses h(k) = k mod 11 for the keys [15, 26, 37]. State the position of each key, explain the problem that arises and one way to resolve it; then state the difference between a stack and a queue. [2]

Answers and mark scheme

Q	part	Answer	Mark points	marks
1	(a)	+45 is 0010 1101; inverting the bits gives 1101 0010 and adding 1 gives 11010011. The range is -128 to +127.	1 mark: bit pattern 11010011 1 mark: range -128 to +127 (both ends)	2
1	(b)	Unsigned: 9×16 + 12 = 156. Signed: the most significant bit is 1 so the value is negative, 156 – 256 = -100.	1 mark: unsigned 156 1 mark: signed -100	2
1	(c)	Total bits = 800 x 600 x 24 = 11520000 bits, which is 1440000 bytes = 1440 KB. A palette image needs 8 bits per pixel, so 480000 bytes = 480 KB, i.e. 3 times smaller, but only 256 colours are available.	1 mark: 800x600x24 then divide by 8 1 mark: 1440 KB 1 mark: 480 KB and 3 times smaller	3
1	(d)	Bytes per second = 44100 x 2 x 2 = 176400; times 90 s gives 15876000 bytes = 15.14 MB.	1 mark: 44100x2x2x90 1 mark: 15876000 bytes / 15.14 MB	2
1	(e)	The encoded form is 20, black, 30, white, 150, grey – 6 values (3 runs x 2) instead of 200, a saving of 97.0%. If colours change constantly, most runs have length 1 and each pixel needs two values, so the file grows.	1 mark: three runs (length + colour) correct 1 mark: 6 values, 97.0% saved 1 mark: when colours alternate often / few repeated pixels, the file grows	3
2	(a)	Serial sends one bit at a time on a single line; parallel sends several bits at once on several lines. Parallel needs more wires and costs more, and bits travelling on different wires arrive at slightly different times (skew), which gets worse with distance. Serial has no skew and is cheaper over distance, so it is used for USB and network cabling.	1 mark: serial one bit at a time / parallel several bits at once 1 mark: parallel suffers skew / needs more wires 1 mark: serial has no skew over distance and is cheaper	3
2	(b)	12 MiB = 100663296 bits; dividing by 20 x 10^6 gives 5.03 s. The real time is longer because of headers, gaps between packets, acknowledgements and retransmissions, other users sharing the bandwidth and errors on the line.	1 mark: converted to 100663296 bits 1 mark: 5.03 s 1 mark: any overhead (headers / retransmission / shared bandwidth)	3
2	(c)	A MAC address is burned into the NIC by the manufacturer, is 48 bits and identifies hardware inside one network segment; an IP address is assigned by the network, can change, and is used to route packets between networks. IPv4 uses 32 bits = 2^32 = 4,294,967,296 addresses; IPv6 uses 128 bits = 2^128 addresses (about 3.4 x 10^38).	1 mark: MAC is 48 bits, fixed in hardware, local identification 1 mark: IP is changeable and used for routing 1 mark: IPv4 32 bits, 4,294,967,296 addresses 1 mark: IPv6 128 bits	4
3	(a)	Processor (containing the ALU and control unit), main memory (RAM) and input/output devices, connected by buses. A bus is a set of parallel wires that carries data, addresses and control signals between the units so the CPU can read and write memory and peripherals.	1 mark: processor / ALU / CU 1 mark: main memory or I/O 1 mark: bus carries data, address and control signals	3
3	(b)	2^20 = 1,048,576 addresses, i.e. 1,048,576 bytes = 1024 KiB. The address bus is output by the CPU to specify a memory location and its width fixes the addressable space; the data bus is bidirectional and carries the actual instruction or data, and its width sets how many bits move at once.	1 mark: 2^20 = 1,048,576 addresses 1 mark: 1024 KiB 1 mark: address unidirectional for location, data bidirectional for content	3
3	(c)	No moving parts, so it is more shock resistant, silent and uses less power; random access is faster so start-up and loading are quicker (also smaller and cooler).	1 mark: no moving parts (shock resistant / silent / low power) 1 mark: faster access	2
3	(d)	An embedded system is a computer built into a larger device to do one dedicated job, usually running firmware from ROM that the user cannot change. A sensor (water level or temperature) converts physical data into a digital signal for the processor; an actuator (motor, inlet valve) turns the processor output into physical movement.	1 mark: dedicated purpose / built in / firmware in ROM 1 mark: sensor converts physical data to digital input 1 mark: actuator converts output to physical action	3
4	(a)	Fetch: the PC value is copied to the MAR and sent on the address bus; the instruction at that address returns on the data bus into the MDR, is copied to the CIR and the PC is incremented. Decode: the control unit splits the CIR contents into opcode and operand. Execute: the ALU performs the operation or memory is read/written and the result goes to a register or memory, updating the status register.	1 mark: PC to MAR to memory to MDR to CIR, PC incremented 1 mark: control unit decodes opcode / operand 1 mark: ALU executes and result written back / status updated	3
4	(b)	Cycles per second = 3.2 x 10^9; dividing by 2.5 gives 1,280,000,000 instructions per second = 1,280 MIPS. 5,000,000 x 2.5 / (3.2 x 10^9) = 0.003906 s = 3.91 ms.	1 mark: clock / CPI = 1,280,000,000 per second (1,280 MIPS) 1 mark: instructions x CPI / clock 1 mark: 3.91 ms	3
4	(c)	460 hits, so the hit rate is 460/500 = 92%. Average = 0.92 x 10 + 0.08 x 120 = 18.80 ns. A cache is small fast memory between the CPU and main memory; frequently used instructions and data are copied into it, so most accesses are served quickly instead of waiting for main memory.	1 mark: hit rate 92% 1 mark: average 18.80 ns with working 1 mark: cache holds frequently used data and cuts main-memory latency	3
5	(a)	Memory management (allocating RAM to programs so they do not overwrite each other); process scheduling (deciding which process gets the CPU, e.g. time slicing); file management (directories, read/write, access rights), plus I/O management and the user interface.	1 mark: memory management with example 1 mark: scheduling / multitasking with example 1 mark: file or I/O management with example	3
5	(b)	A compiler translates the whole source into machine/object code once produce an executable that runs fast, but any change means recompiling. An interpreter translates and runs one line at a time, produces no executable file, and stops at the first error showing the line number, which makes debugging easier but execution slower.	1 mark: compiler translates all at once and creates an executable / interpreter runs line by line 1 mark: difference in speed or error reporting 1 mark: interpreter reports errors immediately during development	3
5	(c)	Backup software (scheduled copies to another medium for recovery) and anti-virus software (scans for and removes malware); also disk defragmenter or compression tools.	1 mark: one utility and its purpose 1 mark: a second utility and its purpose	2
6	(a)	Symmetric encryption uses one shared key for both directions: fast, but the key must be delivered securely. Asymmetric encryption uses a key pair – the public key encrypts and the private key decrypts (or signs) – which solves key distribution but is slower. A digital certificate is issued by a trusted certificate authority and binds a public key to a verified identity so the browser can confirm the server is genuine.	1 mark: symmetric one key / asymmetric key pair 1 mark: asymmetric solves key distribution but is slower 1 mark: certificate from a CA binding public key to identity	3
6	(b)	Combinations = 36^8 = 2,821,109,907,456. Time = 2,821,109,907,456 / 10^9 = 2,821.1 s = about 47.0 minutes. Improvements: allow capitals and symbols to widen the character set, use longer passwords, enforce two-factor authentication, and limit attempts or lock the account.	1 mark: 36^8 = 2,821,109,907,456 1 mark: 47.0 minutes 1 mark: two valid measures (longer / wider character set / 2FA / attempt limit)	3
6	(c)	DDoS: many compromised machines (a botnet) flood a server with requests until bandwidth or connections are exhausted and real users cannot get through. Phishing: a fake email or site impersonating a bank or organisation tricks the user into entering credentials or opening a malicious link. A firewall inspects incoming and outgoing packets and blocks unauthorised ports and addresses by rule.	1 mark: DDoS – many machines flooding to deny service 1 mark: phishing – impersonation to steal credentials 1 mark: firewall filters packets / blocks ports by rule	3
7	(a)	The primary key of Student is StudentID; the primary key of ClubMembership is the composite key (StudentID, ClubID), in which StudentID is a foreign key referencing Student. A student joins many clubs and a club has many students, so the relationship is many-to-many, implemented through the ClubMembership link table.	1 mark: primary key StudentID 1 mark: composite primary key and StudentID as foreign key 1 mark: many-to-many relationship	3
7	(b)	1NF: one value per cell – split the repeated items into many OrderLine rows (OrderID, ItemID, Qty, UnitPrice), removing the repeating group. 2NF: remove partial dependency – item name and price depend only on ItemID, so create Item(ItemID, ItemName, UnitPrice). 3NF: remove transitive dependency – customer name and address depend on CustomerID, not on OrderID, so create Customer(CustomerID, Name, Address).	1 mark: 1NF – single value per cell, repeated items split out 1 mark: 2NF – partial dependency removed (separate item table) 1 mark: 3NF – transitive dependency removed (separate customer table)	3
7	(c)	SELECT COUNT(*) AS BigOrders FROM Orders WHERE Total > 500;	1 mark: COUNT(*) with WHERE Total > 500 1 mark: AS BigOrders and valid syntax	2
8	(a)	A linear search may have to check every element: 1,000,000 comparisons. A binary search halves the range each time, so it needs 20 comparisons (2^20 = 1,048,576 > 1,000,000). The precondition is that the data must already be sorted on the search key and direct access to the middle element.	1 mark: linear 1,000,000 1 mark: binary 20 1 mark: data must be sorted (and randomly accessible)	3
8	(b)	After four full passes the array is 1 3 6 7 9; that is 16 comparisons and 5 swaps. The time complexity is O(n^2) – without an early-exit check all comparisons still run even on sorted data.	1 mark: 1 3 6 7 9 1 mark: 16 comparisons, 5 swaps 1 mark: O(n^2)	3
8	(c)	All three keys hash to position 4, which is a collision. Resolve it by chaining (a linked list of keys at that slot) or by linear probing (use the next free slot). A stack is last in first out (LIFO) and a queue is first in first out (FIFO).	1 mark: all three at 4, collision identified 1 mark: chaining / linear probing, and LIFO vs FIFO	2

Topic breakdown

Q	Topic	Marks
1	Information representation: two’s complement, hexadecimal, images and sound, RLE	12
2	Data transmission and networking	10
3	Hardware: von Neumann model, buses, storage, embedded systems	11
4	Processor fundamentals: fetch-decode-execute, performance, cache	9
5	System software: operating system and utilities	8
6	Security, privacy and data integrity	9
7	Databases: keys, normalisation, SQL	8
8	Algorithm design and data structures	8

Common mistakes

Writing the range of 8-bit two’s complement as -127 to 127.
Forgetting the number of channels when calculating sound file size, or mixing bits with bytes.
Omitting the CPI when calculating execution time; not weighting cache access time by the hit rate.
Normalisation answers that say only “split the tables” without naming the dependency removed.

How to use this paper Sit the paper closed-book in 90 minutes, then mark point by point with the answer booklet; redo every lost mark and note whether it was the formula, the substitution or the unit.

Syllabus reference Format follows the CIE 9618 Computer Science syllabus (2025-2027) Paper 1: 1 h 30 min, 75 marks, theory fundamentals.

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Paper 1 lasts 1 hour 30 minutes and carries 75 marks; it is an AS paper made up of structured questions only.

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A model answer and point-by-point mark scheme for every part, plus a data and formulae sheet and a list of common mistakes.

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