GCSE Computer Science Paper 1: Principles of Computer Science
A full length 1CP2/01 paper in five 15 mark questions: merge sort, binary search, truth tables and a trace table; two's complement, shifts, bitmaps and the Caesar cipher; hardware and software; networks and security; and the impacts of technology, ending with a 6 mark discussion of delivery drones.
Instructions
- Time allowed
- 1 hour 30 minutes
- Total marks
- 75
- Sections
- Question 1, 15 marks; Question 2, 15 marks; Question 3, 15 marks; Question 4, 15 marks; Question 5, 15 marks
- Calculator
- You must not use a calculator.
- Answer all the questions.
- The number of marks for each question is shown in brackets. Use it as a guide to how much to write.
- The clock at the top counts down from the real time allowed. When it reaches zero the paper is locked and marked, as it would be collected in the exam hall.
- In this paper 1 kibibyte (KiB) = 1024 bytes and 1 mebibyte (MiB) = 1024 KiB, as in the specification.
- There are five questions, each worth 15 marks.
- Answer all questions. Where a question says tick one box, tick only one. Show your working in calculations: the method can earn marks when the answer is wrong.
Questions
Answer every question, then finish the paper to have it marked. Marking as you go teaches you far less.
Question 1 15 marks
Computational thinking
Tick one box.
- LB Breaking a problem down into smaller problems that are easier to solve
Decomposition breaks a problem into sub-problems. Removing unnecessary detail is abstraction.
| Stage | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th | 8th |
|---|---|---|---|---|---|---|---|---|
| Pairs merged | ||||||||
| Fours merged | ||||||||
| Fully merged |
- LRow 1: 15, 38, 4, 72, 23, 56, 9, 41
- LRow 2: 4, 15, 38, 72, 9, 23, 41, 56
- LRow 3: 4, 9, 15, 23, 38, 41, 56, 72
Neighbouring single items are merged into ordered pairs: 15 38, 4 72, 23 56, 9 41. Pairs are merged into ordered fours: 4 15 38 72 and 9 23 41 56. Those two lists are merged by repeatedly taking the smaller front item: 4 9 15 23 38 41 56 72.
| Position | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| Item | 4 | 9 | 15 | 23 | 38 | 41 | 56 | 72 |
| Comparison | Item checked |
|---|---|
| 1 | |
| 2 | |
| 3 |
- L23
- L41
- L56
(0 + 7) // 2 = 3: item 23. 56 is larger, so first becomes 4. (4 + 7) // 2 = 5: item 41. 56 is larger, so first becomes 6. (6 + 7) // 2 = 6: item 56, found.
| A | B | C | P |
|---|---|---|---|
| 0 | 0 | 0 | |
| 0 | 0 | 1 | |
| 0 | 1 | 0 | |
| 0 | 1 | 1 | |
| 1 | 0 | 0 | |
| 1 | 0 | 1 | |
| 1 | 1 | 0 | |
| 1 | 1 | 1 |
- LRows 1 and 2: 0, 0
- LRows 3 and 4: 0, 1
- LRows 5 and 6: 1, 1
- LRows 7 and 8: 0, 1
A AND NOT B is 1 only when A is 1 and B is 0. B AND C is 1 only when both are 1. P is 1 when either is 1.
number = 6
result = 1
while number > 1:
result = result * number
number = number - 2
print(result)| number | result | Output |
|---|---|---|
| 6 | 1 | |
- LSecond row: 4, 6
- LThird row: 2, 24
- LFourth row: 0, 48
- LOutput: 48
Each pass multiplies result by number, then takes 2 off number: 1 x 6 = 6, 6 x 4 = 24, 24 x 2 = 48. number is then 0, so the loop ends and 48 is printed.
Question 2 15 marks
Data
- L11010011
- L-26
45 is 0010 1101. Inverting gives 1101 0010, and adding 1 gives 1101 0011. 1110 0110 is -128 + 64 + 32 + 4 + 2 = -26.
- L11101101
An arithmetic shift right copies the sign bit into the gap, keeping the number negative: 1110 1101. That is -76 divided by 4, -19.
- LThe result needs more bits than are available to store it
- Lso the extra carry bit is lost and the stored result is wrong
Overflow happens when the result of an addition is too large to fit in the number of bits available, for example a result needing 9 bits in an 8-bit register. The final carry is lost, so the stored result is wrong.
- LCorrect answer, full marks: 2.25 MiB
- LMethod: Multiplies width by height by colour depth
- LMethod: Divides by 8, then by 1024 twice
1024 x 768 = 786,432 pixels. 786,432 x 24 = 18,874,368 bits. / 8 = 2,359,296 bytes. / 1024 = 2304 KiB. / 1024 = 2.25 MiB.
- LFDE
- LHELLO
C, A, B each move three letters on: F, D, E. Moving K, H, O, O, R three letters back gives HELLO.
- LLossless compression lets the original data be restored exactly, with nothing lost
- LLossy compression permanently removes some data, usually giving a smaller file
Lossless compression reduces the file size in a way that can be reversed, so the original data is restored exactly. Lossy compression permanently removes data that is judged less important, so it makes files smaller but the original can never be fully restored.
- L7
ASCII is a 7-bit code, which gives 128 characters.
- L63
- LAC
3F = 3 x 16 + 15 = 63. 1010 is A and 1100 is C: AC.
Question 3 15 marks
Computers
| Defragmentation | Backup | Compression | Anti-malware | |
|---|---|---|---|---|
| Rearranges the parts of files on a hard disk so they are stored together | ||||
| Makes copies of files so they can be restored if the originals are lost | ||||
| Reduces the size of files | ||||
| Detects and removes malicious software |
Tick one box in each row.
- LRearranges the parts of files on a hard disk so they are stored together: Defragmentation
- LMakes copies of files so they can be restored if the originals are lost: Backup
- LReduces the size of files: Compression
- LDetects and removes malicious software: Anti-malware
Defragmentation reorganises, backup copies, compression shrinks and anti-malware protects.
- LThe address bus carries the address of the memory location to be read or written, from the CPU to memory
- LThe data bus carries data and instructions between the CPU and memory, in both directions
The address bus carries the address of the memory location the CPU wants to use, one way, from the CPU to memory. The data bus carries the data or instruction itself between the CPU and memory, in either direction.
Tick one box.
- LA The controller in a microwave oven
An embedded system is a computer built into another device to do one dedicated job.
- LRAM holds the programs and data in use, and can be changed, but it is volatile
- LROM is non-volatile, so it keeps the start-up instructions (the boot program) when the power is off
RAM is where the programs and data currently in use are held while the computer runs, and its contents change constantly, but it is volatile. ROM is non-volatile, so it keeps the instructions needed to start the computer even when the power is off.
| Compiler | Interpreter | Assembler | |
|---|---|---|---|
| Translates the whole program into an executable file before it is run | |||
| Translates and runs a high-level program one line at a time | |||
| Translates low-level mnemonics into machine code |
Tick one box in each row.
- LTranslates the whole program into an executable file before it is run: Compiler
- LTranslates and runs a high-level program one line at a time: Interpreter
- LTranslates low-level mnemonics into machine code: Assembler
A compiler produces an executable, an interpreter works line by line, and an assembler translates assembly language mnemonics.
- LLow-level code gives direct control of the hardware, such as registers and memory locations
- Lso the code can run faster and use less memory, which matters in an embedded system
A low-level language gives the programmer direct control of the processor's registers and memory, so the code can be made to run very fast and use very little memory, which matters for device drivers and embedded systems.
Tick one box.
- LC Managing the files stored on secondary storage
File management is one of the operating system's jobs, along with memory, process, peripheral and user management.
Question 4 15 marks
Networks
Tick one box.
- LD It covers a large geographical area
A WAN connects networks over a large area, often using infrastructure owned by other companies.
- LAdvantage: if one cable or device fails, the rest of the network keeps working
- LDisadvantage: if the central switch fails the whole network fails, and it needs a lot of cable
Each device has its own cable to the central switch, so if one cable fails only that device is affected. But everything depends on the switch: if it fails, the whole network goes down, and all that cable makes it expensive to install.
| POP3 | SMTP | HTTPS | FTP | |
|---|---|---|---|---|
| Downloads emails from a server, usually deleting them from it | ||||
| Sends emails | ||||
| Transfers web pages securely | ||||
| Transfers files between computers |
Tick one box in each row.
- LDownloads emails from a server, usually deleting them from it: POP3
- LSends emails: SMTP
- LTransfers web pages securely: HTTPS
- LTransfers files between computers: FTP
POP3 downloads mail, SMTP sends it, HTTPS is secure web traffic and FTP transfers files.
| Application | Transport | Internet | Link | |
|---|---|---|---|---|
| HTTP | ||||
| TCP | ||||
| IP | ||||
| Ethernet |
Tick one box in each row.
- L1 mark for both: HTTP: Application; TCP: Transport
- L1 mark for both: IP: Internet; Ethernet: Link
HTTP is an application protocol, TCP works at the transport layer, IP at the internet layer and Ethernet at the link layer.
- LIt checks the data (packets) coming into and going out of the network
- Lagainst a set of rules, blocking anything that is not allowed
A firewall inspects every packet entering or leaving the network and compares it against a set of rules, blocking any traffic that is not allowed, such as connections from unknown addresses or to unauthorised ports.
- LEmails or messages that pretend to come from a trusted organisation, to trick people into giving personal details or clicking a malicious link
- LProtection: training users to spot suspicious messages, email filtering, or checking the sender and links
Phishing is sending emails or texts that pretend to come from a trusted organisation, such as a bank, to trick people into clicking a link or giving away personal details. Staff can be trained to recognise suspicious messages, and email filters can block many of them.
Tick two boxes.
- LB It is less affected by interference
- LC It usually transfers data faster
- LNo marks if more than 2 boxes are ticked
Cables give faster, more reliable connections. Freedom to move and no cabling are advantages of wireless.
Question 5 15 marks
Issues and impact
| Computer Misuse Act 1990 | Data Protection Act 2018 | Copyright, Designs and Patents Act 1988 | |
|---|---|---|---|
| A company sells its customers' email addresses without their consent | |||
| A student uploads a film to a website without permission | |||
| A hacker guesses a teacher's password and changes students' grades |
Tick one box in each row.
- LA company sells its customers' email addresses without their consent: Data Protection Act 2018
- LA student uploads a film to a website without permission: Copyright, Designs and Patents Act 1988
- LA hacker guesses a teacher's password and changes students' grades: Computer Misuse Act 1990
Misusing personal data breaks data protection law, sharing a film breaks copyright, and unauthorised access to change data breaks the Computer Misuse Act.
- LBenefit: it is usually free, and the code can be changed to suit the business
- LDrawback: there may be no official support, or fewer features and less compatibility
Open-source software is usually free, which saves the business money, and its source code can be modified to suit its needs. However, there is often no official support line, so the business may struggle to get help when something goes wrong.
- LA computer program improves its performance at a task by learning from data
- LIt finds patterns itself, rather than being explicitly programmed with rules
Machine learning is a type of artificial intelligence in which a program improves at a task by being trained on data: it finds the patterns itself instead of being explicitly programmed with rules.
- LThe system is trained on data that is unrepresentative or reflects past prejudice
- Lso its decisions unfairly favour or disadvantage some groups of people
An AI system learns from the data it is trained on. If that data is unrepresentative, or reflects unfair decisions made in the past, the system learns the same patterns, so its decisions unfairly favour some groups of people over others.
- LEnvironmental benefit: battery drones use less fuel and produce fewer emissions than vans
- LEnvironmental cost: making and charging drones and their batteries uses resources and energy, creating e-waste
- LNoise and disturbance to people and wildlife
- LDelivery drivers may lose their jobs
- LDrones with cameras raise privacy concerns as they fly over homes and gardens
- LLegal: personal data such as addresses and camera footage must be handled under the Data Protection Act; flights must follow aviation rules
- LSafety and liability: who is responsible if a drone crashes or injures someone
- LReaches a balanced conclusion
Drone delivery could cut emissions, because small battery drones use far less energy per parcel than delivery vans, and they keep vans off congested roads. But the drones and their batteries must be manufactured from mined materials, charged and eventually disposed of, and hundreds of drones overhead would add noise that disturbs people and wildlife. Ethically, delivery drivers could lose their jobs, and drones carrying cameras raise privacy concerns as they fly over homes and gardens. Legally, the retailer must handle customers' addresses and any camera footage under the Data Protection Act 2018, and flights must follow aviation regulations about where and how high drones may fly. It must also be clear who is liable if a drone crashes and injures someone. On balance drones may suit light parcels in rural areas, but only with strict rules on safety, noise and privacy.
This is an original paper modelled on the real Pearson Edexcel 1CP2/01 past papers: the same structure, the same balance of topics, the same command words and mark tariffs, 75 marks in 1 hour 30 minutes, exactly as the real paper allows. It is not a past paper. Real past papers are the copyright of the exam board and cannot be republished here, so links to the official papers on the board's own site are on the exam papers page.