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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 514 exam-style questions, each with its mark scheme and example answers. Print it, or answer it on paper and then open the mark schemes. Or browse the questions by topic.

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8 questions · 17 marks

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1Multiple choice1 markCompression Swap Remove

Identify one reason for using data compression.

[1 mark]
  • ATo increase the number of packets sent
  • BTo reduce the time taken to transmit a file
  • CTo improve the quality of the file
  • DTo make the file harder to open
Mark scheme
  • B To reduce the time taken to transmit a file (1)
  • The only correct answer is B. The correct answer is to reduce the time taken to transmit a file.
  • A is not correct because compression reduces file size, which can reduce the number of packets sent
  • C is not correct because lossy compression can reduce the quality of a file
  • D is not correct because encryption, not compression, can make a file harder to open
Example answers and how they were marked
Full marks1/1

B To reduce the time taken to transmit a file

Why: Correct option (1).

No marks0/1

A To increase the number of packets sent

Why: Near miss: compression reduces the number of packets sent, so no mark.

No marks0/1

C To improve the quality of the file

Why: Misconception that compression improves quality; no mark.

2Multiple choice1 markAddressing Swap Remove

A packet of data is sent across the internet to a web server. The packet has a header that stores information about the packet.

Identify the information in the packet header that a router uses to forward the packet towards its destination.

[1 mark]
  • AThe destination domain name
  • BThe destination IP address
  • CThe MAC address of the sender
  • DThe name of the file being sent
Mark scheme
  • B The destination IP address (1)
  • The only correct answer is B. The correct answer is the destination IP address.
  • A is not correct because a domain name is converted to an IP address by a DNS server before routing begins
  • C is not correct because the sender's MAC address is only used to deliver the packet on the local network
  • D is not correct because the file name is not used to route packets
Example answers and how they were marked
Full marks1/1

B The destination IP address

Why: Correct option (1): a router reads the destination IP address in the header to forward the packet towards its destination.

No marks0/1

A The destination domain name

Why: A domain name is converted to an IP address by a DNS server before routing begins, so it is not the routing information in the header (mark scheme line A); no mark.

No marks0/1

C The MAC address of the sender

Why: The sender's MAC address is only used to deliver the packet on the local network, so it is not used by the router to forward the packet (mark scheme line C); no mark.

3Short recall1 markThe Internet Swap Remove

Data packets are sent across the internet.

Name the device that uses IP addresses to forward a data packet from one network to another.

[1 mark]
Mark scheme

Award one mark for:

  • Router (1)

Allow routers (plural).

Allow minor spelling errors that are clearly router, e.g. rooter.

Do not accept switch, hub (these forward data within a network, not between networks), modem or server.

Do not accept gateway on its own.

Example answers and how they were marked
Full marks1/1

Router

Why: Correct device named (1).

No marks0/1

Modem

Why: Modem is not accepted; it is not the device that forwards packets between networks (0).

No marks0/1

Server

Why: A server is not accepted; it does not forward packets between networks (0).

4Short recall2 marksUnits & Conversion Swap Remove

A solid state drive has a capacity of 5 gibibytes.

Give an expression to calculate the number of bits in 5 gibibytes. You do not need to do the calculation.

[2 marks]
Mark scheme

Award one mark for each of:

  • 5 × 1024^3 / 5 × 1024 × 1024 × 1024 / 5 × 2^30, for the number of bytes (1)
  • A fully correct expression with × 8 to convert bytes to bits, in any order, e.g. 5 × 1024^3 × 8, 8 × 5 × 2^30 (1)

Accept 5 × 1 073 741 824 × 8.

Accept 5 × 2^33 or 42 949 672 960 as a fully correct expression (2 marks).

Do not accept 5 × 1000^3 × 8, which uses decimal multiples.

Do not accept ÷ 8, which converts bits to bytes.

Example answers and how they were marked
Full marks2/2

8 × 5 × 2^30

Why: 5 × 2^30 is the number of bytes (1) and × 8 converts bytes to bits (1).

Some marks1/2

5 × 2^30

Why: Correct number of bytes using the binary multiple 2^30 (1), but no conversion to bits.

No marks0/2

5 × 1000^3 × 8

Why: Uses decimal 1000^3 instead of binary 1024^3, so the expression is not fully correct; no creditworthy point.

5Describe2 marksUser Management Swap Remove

An operating system manages the users of a computer.

Describe how an operating system checks that a user is allowed to log in.

[2 marks]
Mark scheme

Award up to two marks for a linked description such as:

  • The user enters a username/user ID and password/PIN (1) which is compared with the account details the operating system has stored (1)
  • The operating system keeps a file/list/database of users and their passwords (1) to check the login details the user enters against (1)
  • The operating system compares the entered details with the details stored for that user (1) and if they do not match, access is refused/denied (1)
  • The entered details are checked against the access rights stored for that user (1) so access is only allowed when the details are correct (1)

Do not award for 'it asks for a password' on its own, without saying what the password is checked against.

A stored list/file of users and passwords with no check of the login details earns one mark only.

Do not accept responses about file management, such as finding or opening the user's files.

Example answers and how they were marked
Full marks2/2

The user types in a username and password. The operating system compares these with the details it has stored for that account, and if they match the user is allowed to log in.

Why: Username and password entered (1) which is compared with the account details the operating system has stored (1).

Some marks1/2

The operating system has a list of all the users and their passwords.

Why: The stored list of users and passwords (1), but no check of the login details against it, so the development mark is not earned (0); the mark scheme states a list alone earns one mark only.

No marks0/2

The operating system finds the user's files on the hard drive and opens the last one they saved.

Why: Describes file management, not checking a user's login details; the mark scheme says not to accept responses about file management, so there is no creditworthy point.

6Describe2 marksLow and High Level Languages Swap Remove

A programmer has written a program in assembly language.

Describe why it is necessary to translate the program into machine code before it can be run.

[2 marks]
Mark scheme

Award up to two marks for a linked description such as:

  • The processor/CPU can only execute machine code/binary instructions (1) so the assembly instructions must be converted into machine code before the program can run (1)
  • Each assembly instruction/mnemonic matches/corresponds to one machine code instruction (1) so each line must be replaced with the machine code the processor executes (1)
  • The program is written using mnemonics (1) that must be translated into machine code for the processor to carry out the instructions (1)
  • The hardware/processor does not understand assembly language directly (1) so it must be translated into machine code that the processor can execute (1)

Do not accept 'assembly language is a low-level language'.

Do not accept 'it is hard to understand' or 'it is not English'.

Do not accept benefits, such as 'machine code runs faster' or 'the program will run more quickly'.

Do not accept 'an assembler/a translator converts it' on its own, with no reference to the processor only being able to execute machine code/binary.

Do not accept responses about high-level languages or about compilers/interpreters.

Example answers and how they were marked
Full marks2/2

The processor can only execute machine code, so the assembly instructions must be converted into machine code before the program can run.

Why: States what the processor can execute (1) and links it to converting the program's instructions into machine code (1).

Some marks1/2

Each assembly instruction matches one machine code instruction.

Why: Correct point about the one-to-one match (1) but no development saying that this is because the processor only executes machine code.

No marks0/2

Because an assembler translates assembly language into machine code.

Why: Names the translator but gives no reason tied to the processor only being able to execute machine code, so no mark; this is in the 'Do not accept' guidance.

7Convert2 marksDenary to Binary Conversion Swap Remove

Binary patterns can represent numbers.

Convert the denary number 199 to 8-bit binary.

[2 marks]
Mark scheme

Award one mark for each correct nibble in the correct order:

  • 1100 (1)
  • 0111 (1)

0 marks if the response is not 8-bit binary.

Example answers and how they were marked
Full marks2/2

1100 0111

Why: 199 = 128 + 64 + 4 + 2 + 1: both nibbles are correct (2).

Some marks1/2

1100 1000

Why: Converts 200 instead (a slip in adding up the place values). The first nibble is correct (1); the second is not.

No marks0/2

1110 0011

Why: The bits are written in reverse, smallest place value first. Neither nibble is correct.

8Discuss6 marksFetch Decode Execute Cycle Swap Remove

A small robotics company is designing a new robot. The robot's computer uses von Neumann architecture. The engineers are explaining how programs are stored and run.

Discuss how the von Neumann stored program concept and main memory (RAM) are used during the fetch-decode-execute cycle.

Your answer should include:

  • what is meant by the stored program concept
  • the role of RAM in the cycle
  • why the stored program concept is useful.
[6 marks]
Mark scheme

Indicative content

Stored program concept

  • Instructions and data are stored together in the same main memory (RAM).
  • Instructions are fetched one at a time from memory and executed in sequence.
  • The program can be changed by loading different instructions into RAM.
  • The CPU can only execute instructions that are in main memory.

Role of RAM

  • RAM holds the program instructions and the data being used.
  • The address of the next instruction is held in the program counter (PC).
  • The instruction is fetched from RAM using the address bus.
  • The data bus carries the instruction from RAM to the CPU.
  • RAM is volatile, so programs are lost when power is off.
  • RAM is directly accessible by the CPU, unlike secondary storage.

Benefits

  • The same hardware can run many different programs.
  • Programs can be loaded and changed quickly.
  • No need to rewire the computer for a new task.
  • Makes general-purpose computers possible.

Levels

  • Level 0 (0 marks): No rewardable content.
  • Level 1 (1–2 marks): Basic, independent points are made, showing elements of understanding of key concepts/principles of computer science. (AO1) The discussion will contain basic information with little linkage between points made or application to the context. (AO2)
  • Level 2 (3–4 marks): Demonstrates adequate understanding of key concepts/principles of computer science. (AO1) The discussion shows some linkages and lines of reasoning with some structure and application to the context. (AO2)
  • Level 3 (5–6 marks): Demonstrates comprehensive understanding of key concepts/principles of computer science to support the discussion being presented. (AO1) The discussion is well developed, with sustained lines of reasoning that are coherent and logically structured, and which clearly apply to the context. (AO2)
Example answers and how they were marked
Full marks6/6

The von Neumann stored program concept means that both the program instructions and the data are stored together in the same main memory, RAM. During the fetch-decode-execute cycle, the CPU fetches an instruction from RAM, decodes it in the control unit, and then executes it. The program counter holds the address of the next instruction, and this address is sent along the address bus to RAM. RAM then returns the instruction along the data bus to the CPU. RAM is volatile, so it only holds the program while the robot is powered on. The stored program concept is useful because the robot's computer can run different programs without changing any hardware; new instructions can simply be loaded into RAM. This makes the computer general-purpose and flexible.

Why: Comprehensive understanding: defines stored program concept, explains RAM role with PC, address bus and data bus, applies to robot, and gives benefits. Sustained lines of reasoning. Level 3.

Some marks3/6

The stored program concept is when instructions and data are stored in RAM. The CPU fetches instructions from RAM. The program counter keeps the address of the next instruction. RAM holds the program. The benefit is that you can change the program without rewiring the computer.

Why: Adequate understanding: correct basic points about stored program concept and RAM, with some linkage, but little development or application to the robot's fetch-decode-execute cycle. Level 2.

Some marks1/6

The von Neumann stored program concept is that the CPU uses RAM to store the program. The fetch-decode-execute cycle fetches, decodes and executes. RAM is random access memory. The benefit is that it is fast.

Why: Basic independent points only; no real linkage or application to the robot. Level 1.

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