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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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7 questions · 16 marks

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1Multiple choice1 markAddressing Swap Remove

A technician is checking the IP addresses of the devices on a network.

Identify the valid IPv4 address.

[1 mark]
  • A192.168.1
  • B192.168.1.1
  • C192.168.1.1.1
  • D192.168.1.256
Mark scheme
  • B 192.168.1.1 (1)
  • The only correct answer is B. The correct answer is 192.168.1.1.
  • A is not correct because it is an incomplete address with only three numbers, not four
  • C is not correct because it is an address with five numbers, not four
  • D is not correct because it contains 256, which is outside the range 0 to 255
Example answers and how they were marked
Full marks1/1

B 192.168.1.1

Why: Correct option (1): an IPv4 address is four numbers, each from 0 to 255, separated by dots.

No marks0/1

D 192.168.1.256

Why: Near miss: it has four numbers but 256 is outside the range 0 to 255 (mark scheme line D), so no mark.

No marks0/1

A 192.168.1

Why: It has only three numbers, so it is not a complete IPv4 address (mark scheme line A), so no mark.

2Convert1 markBinary to Denary Conversion Swap Remove

Computers manipulate binary patterns.

Convert the binary pattern 1010 0100 to denary.

[1 mark]
Mark scheme

Award one mark for:

  • 164 (1)
Example answers and how they were marked
Full marks1/1

164

Why: Correct: 128 + 32 + 4 = 164 (1).

No marks0/1

37

Why: Reads the place values from the wrong end (the left-hand bit as 1), so the value is wrong.

No marks0/1

3

Why: Counts the 1s instead of adding their place values.

3Short recall2 marksCharacter Sets Swap Remove

ASCII is a character set that uses 7 bits to represent each character.

Give two limitations of 7-bit ASCII when it is used to represent text.

[2 marks]
Mark scheme

Award one mark for any of the following up to a maximum of two marks:

  • (Only) 128 different characters can be represented / 7 bits give 2^7 = 128 different codes (1)
  • It cannot represent characters from languages other than English / non-English alphabets, e.g. Greek, Chinese (1)
  • It cannot represent accented characters, e.g. é, ñ (1)
  • It cannot represent symbols or emoji that are not in the ASCII character set (1)

Award a maximum of one mark for the last three bullets, as they are all examples of characters that fall outside the 128 ASCII codes.

Allow 'it can only represent 128 characters, so some characters are missing' for the first bullet.

Do not accept 'it uses 7 bits' / 'it only uses 7 bits' on its own, without saying what this limits.

Do not accept 'it can only store a limited amount of text' / 'it can only store short messages', as 7 bits limits the number of characters available, not the length of a message.

Do not accept 'it can only represent the characters in the ASCII character set', as this repeats the question.

Do not accept 'it is old/out of date' or 'it cannot store numbers'.

Example answers and how they were marked
Full marks2/2

Only 128 different characters can be represented, so it cannot store characters from other languages such as Greek.

Why: 128 character limit (1) and characters from other languages (1); two separate creditworthy limitations.

Some marks1/2

It cannot show accented characters like é and it only uses 7 bits.

Why: Accented characters (1); 'only uses 7 bits' is not qualified, so scores 0.

No marks0/2

It only uses 7 bits, so it can only store a small amount of text.

Why: Scores 0: text length is not a limitation of 7-bit ASCII, and 'only uses 7 bits' is unqualified; both Do not accept lines apply.

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.

6Convert2 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.

7Discuss6 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.

Questions by topic

2.1 Binary

Build a paper on this topic 58
Adding Binary Numbers 6 questions Binary and Hexadecimal Basics 10 questions Binary Shifts 6 questions Binary to Denary Conversion 6 questions Binary to Hexadecimal Conversion 6 questions Denary to Binary Conversion 6 questions Denary to Hexadecimal Conversion 6 questions Denary to Negative Binary Conversion 6 questions Hexadecimal to Binary Conversion 6 questions

2.2 Data representation

Build a paper on this topic 56
Character Sets 18 questions Images 19 questions Sound 19 questions

2.3 Data storage and compression

Build a paper on this topic 24
Compression 18 questions Units & Conversion 6 questions

3.1 Hardware

Build a paper on this topic 77
Embedded Systems 17 questions Fetch Decode Execute Cycle 15 questions Primary Memory 15 questions Secondary Memory 15 questions Stored Program Computers 15 questions

3.2 Software

Build a paper on this topic 124
File Management 18 questions Memory Management 16 questions Peripheral Management 11 questions Process Management 12 questions Purpose & Function of Operating Systems 17 questions Software Basics 13 questions User Management 18 questions Utility Software 19 questions

3.3 Programming languages

Build a paper on this topic 37
Low and High Level Languages 19 questions Translators 18 questions

4.1 Networks

Build a paper on this topic 138
Addressing 16 questions LANs & WANs 16 questions Network Speed 5 questions Packet Switching 16 questions Protocols & Layers 17 questions The Internet 16 questions Topologies 17 questions Wired Connections 17 questions Wireless Connections 18 questions

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