Computer Science 9618/32 October/November 2025: question paper and mark scheme

Mark scheme

1(a)(i) DECLARE Car1 : Car 1 1(a)(ii) One mark for each correct answer 2 Example answer Car1.Colour ← "Blue" Car1.IntoStock ← 21/10/2025 1(b)(i) One mark per mark point (Max 2) 2 MP1 TYPE Body = MP2 (Convertible, Hatchback, Saloon, SUV) Example answer TYPE Body = (Convertible, Hatchback, Saloon, SUV) 1(b)(ii) DECLARE BodyStyle : Body 1

Mark scheme

2(a) Two marks for working 3 • correct calculation/application of exponent seen • correct method to find the final answer Working: Exponent = 8 + 2 + 1 = 11 // =0.111100101 x 2 11 // =11110010100.0 (moving bp 11 places to right) Method to find the answer // 1024 + 512 + 256 + 128 + 16 + 4 One mark for correct answer Denary value: 1940 Example of solution using fractions: 2(b) One mark per mark point (Max 3) 3 MP1 correct method to find the binary number MP2 correct use of exponent MP3 correct answer in the space provided Working: 26.6875 converted to binary (0)11010.1011 // 16+8+2+0.5+0.125+0.0625 movement of binary point by 5 places Mantissa Exponent 0 1 1 0 1 0 1 0 1 1 0 0 0 1 0 1

Mark scheme

3(a) One mark per mark point (Max 2) 2 • HTTP to send and receive / transfer web pages / hypertext / images / videos • IMAP allows users to access/read their emails from any device without removing the message from the mail server // synchronises emails on any device 3(b) One mark per mark point (Max 4) 4 MP1 Files are shared over a peer-to-peer network MP2 A small file, a torrent (descriptor file), is initially created by a peer MP3 The torrent (descriptor file) contains metadata about the file to be shared MP4 The whole file must initially be located on at least one peer MP5 The file is broken into equal sized pieces MP6 Other peers who wish to download the file first obtain the torrent (descriptor file) and connect to a tracker MP7 A tracker acts like a server that holds all the data about all the computers connected to it / the swarm MP8 Each peer downloads parts of the file from other peers until the entire file has been downloaded MP9 As each peer receives a piece of the file, they become a source for that piece of the file MP10 A peer who has a complete file and is sharing it is a seed // Once a peer has completely downloaded the file and made the file available to others in the swarm, they become a seed.

Mark scheme

4(a) One mark for a benefit of circuit switching (Max 1) 2 MP1 Once the connection is made, it is available until the end of the transmission (suitable for long continuous transmission) MP2 The dedicated path ensures a steady rate of data transmission, once connection has been made (because the whole bandwidth is available) MP3 The dedicated path ensures data is less likely to be lost. MP4 No intermediate delays once the circuit is established, enables real time transmission MP5 No additional time/delay to reorder needed because data arrives in the order it was sent / data sent as a continuous stream One mark for a benefit of packet switching (Max 1) MP6 No need to tie up a communication line // path available to multiple users // bandwidth can be shared by multiple users MP7 Possible to overcome failed / faulty / busy lines by re-routing packets MP8 Individual packets can be resent if lost/damaged MP9 Users only charged for the duration of connectivity MP10 High rate of data transmission possible MP11 Always uses digital networks so data is transmitted directly to the destination MP12 Good security because all packets can take different routes // A more secure method because all packets can take different routes 4(b) One mark per difference (Max 2) 2 MP1 Circuit switching requires a dedicated line to be connected before the data transfer takes place, in packet switching, data transfer commences directly/ straight away MP2 In circuit switching, each data unit knows the entire path address, but in packet switching, each data unit only knows the final address (routers decide intermediate paths) MP3 Circuit switching uses the whole bandwidth of the transmission path, but packet switching shares it with other users // circuit switching uses a constant high bandwidth. Packet switching has variable bandwidth MP4 In circuit switching each data unit follows the same route, but in packet switching, packets can follow any route MP5 In circuit switching data arrives in the correct order but in packet switching it has to be reordered MP6 Data remains intact / a continuous stream in circuit switching but is segmented in packet switching MP7 Lost data in circuit switching requires the whole transmission to be resent, but individual packets can be resent in packet switching MP8 Circuit switching doesn’t suffer data loss but packets can be lost in packet switching.

Mark scheme

5(a) One mark per mark point (Max 2) 2 MP1 The low-level scheduler manages the handling of interrupts based on priority MP2 … ensuring that critical events are handled without delay MP3 It uses an Interrupt Vector Table (IVT) / Interrupt Descriptor/Despatch Table (IDT) / Interrupt Service Routine (ISR) lookup MP4 … to map the interrupt to the specific handling routine / ISR. 5(b) One mark per correct answer (Max 3) 3 Process state Reason running CPU time has been allocated and the process is being executed. The process is waiting (in the ready queue) for a slice of CPU time. It could run. // It has been displaced ready by a higher priority process and could otherwise still run. blocked The process is waiting for an I/O operation / some event to take place / be completed.

Mark scheme

6(a) • One mark for working, (all four columns P, Q, R and S) 3 • One mark for first four rows of column Z • One mark for second four rows of column Z Working space A B C P Q R S Z 0 0 0 1 1 0 0 1 0 0 1 1 1 1 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 1 0 1 0 1 1 1 0 0 1 1 0 1 1 1 1 0 1 0 0 0 6(b)(i) Two marks if no errors present 2 One mark if only one error present BC 00 01 11 10 A 0 0 1 1 0 1 0 1 0 1 6(b)(ii) One mark for each correct loop (Max 2) 2 BC 00 01 11 10 A 0 0 1 1 0 1 0 1 0 1 6(b)(iii) One mark for each mark point (Max 2) 2 • One correct Boolean term with a + / OR sign • All Boolean terms and operators correct and no other terms present _ _ _ A.C + B.C + A.B.C

Mark scheme

7(a) One from 1 Symmetric (cryptography / encryption) Quantum (cryptography) 7(b) One mark per mark point (Max 4) 4 MP1 The two keys held by the organisation are a private key and a public key MP2 The organisation makes the public key available to anyone who wishes to send them secure transmissions // The sender obtains the organisation’s public key MP3 The sender uses the organisation’s public key to encrypt the message / plain text // The sender uses the organisation’s public key to turn the message into cipher text MP4 The organisation uses its private key to decrypt the message.

Mark scheme

8(a) To improve the code by making it use minimum resources (CPU, memory/storage, time) // Answer by example: To improve 1 the code by • minimising program storage • minimising CPU time • minimising memory use • minimising program execution time (includes peripheral use as well as CPU) 8(b) One mark (a – b + c) 3 One mark * (c - a) One mark / d Complete answer (a – b + c) * (c - a) / d 8(c) One mark per ring (Max 4) 4 6 3 16 16 10 16 9 9 3 3 3 3 30 30 46

Mark scheme

9(a)(i) One from: 1 e.g. • Virtual assistants // AI Agents • Chatbots • Composing music • Self-driving cars • Natural language processing • Fraud detection • Image colouring // photograph enhancement • Visual/Image recognition/facial recognition • Text mining • Computer assisted translation • Large language models (LLMs) • Playing chess • Grading student exams • Weather forecasting 9(a)(ii) Use / introduce more hidden layers 1 9(b) One mark per mark point (Max 4) 4 MP1 Initial outputs are compared to expected outputs MP2 … weightings are adjusted to minimise the difference between actual and expected outputs MP3 Calculus is used to find the error gradient in the obtained outputs MP4 … the results are fed back into the neural network MP5 … weightings of each neuron / node are adjusted as a result of the feedback MP6 … the process repeats until results are more accurate

Mark scheme

10(a) One mark per mark point (Max 2) 2 MP1 Use an exception handling routine (to respond to unwanted / unexpected events when the program is running) MP2 Use of try … except / catch // Generate some form of error message 10(b) One mark per mark point (Max 2) 2 MP1 Coding errors MP2 User errors MP3 Hardware failure // Losing connection to a device e.g. a printer

Mark scheme

11 One mark per mark point (Max 7) 7 MP1 LDD 300 seen MP2 Correct use of STO seen (at least once) MP3 Correct use of LDD 420 seen MP4 Correct use of LDI B MP5 Correct use of ADD A Opcode Operand LDD 300 STO A LDD 420 STO B LDI B ADD A STO Answer MP6 Correct labelling of three addresses A:, B: and Answer: MP7 Correct contents in A: and B: MP8 Correct value in Answer: Label Contents A: 86 B: 150 Answer: 112

Mark scheme

12(a)(i) One mark for each correctly completed line (Max 4) 4 PROCEDURE Push(NewData : STRING) IF Top < Max – 1 THEN Top ← Top + 1 StackArray[Top] ← NewData ELSE OUTPUT "Stack full; new data cannot be added" ENDIF ENDPROCEDURE 12(a)(ii) One mark per mark point (Max 2) 2 MP1 Input with variable, with or without prompt MP2 Procedure call for Push with parameter used matching input variable Example answer INPUT MyData CALL Push(MyData) 12(b) One mark per mark point (Max 3) 3 MP1 Stacks store data in Last In First Out (LIFO) / First In Last Out (FILO) order MP2 Each time a recursive algorithm calls itself data is pushed onto the stack MP3 When the recursive algorithm reaches its base case / starts to unwind MP4 … data is popped from the stack in the reverse order to which it was pushed onto it.