Menu

Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 604 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.

Build a paper

Topics (none ticked: all of them)
Question types (none ticked: all of them)

Practice paper

4 questions · 6 marks

Pick again
Print
1Multiple choice1 markPrimary Memory Swap Remove

A computer is running an application. The application's data and instructions are transferred from secondary storage into RAM before the CPU processes them.

Identify the reason for this transfer.

[1 mark]
  • ARAM is non-volatile
  • BRAM is faster to access than secondary storage
  • CRAM has a larger capacity than secondary storage
  • DRAM is cheaper than secondary storage
Mark scheme
  • B RAM is faster to access than secondary storage (1)
  • The only correct answer is B. The correct answer is RAM is faster to access than secondary storage.
  • A is not correct because RAM is volatile
  • C is not correct because RAM has a smaller capacity than secondary storage
  • D is not correct because RAM is more expensive than secondary storage
Example answers and how they were marked
Full marks1/1

B RAM is faster to access than secondary storage

Why: Selects the correct option: RAM is faster to access than secondary storage, so data and instructions can reach the CPU quickly (1).

No marks0/1

A RAM is non-volatile

Why: Near miss: RAM is volatile; this does not explain the transfer into RAM.

No marks0/1

C RAM has a larger capacity than secondary storage

Why: Misconception: assumes primary memory has a larger capacity, but RAM has a smaller capacity than secondary storage.

2Short recall2 marksStored Program Computers Swap Remove

Computers use the von Neumann stored program concept.

Define the term stored program concept.

[2 marks]
Mark scheme

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

  • Program instructions are stored in main memory/RAM (1)
  • The data used by the program is stored in main memory/RAM (1)
  • The CPU fetches/reads instructions from main memory to execute them (1)
  • The program can be run without the instructions being input again (1)

Allow 'the program is stored in memory' for one mark.

Do not accept answers about programs being stored on secondary storage, such as a hard disk or USB memory stick.

Do not accept answers that only describe the von Neumann architecture (e.g. having a CPU and memory) without stating that instructions are stored in main memory.

Example answers and how they were marked
Full marks2/2

Both the program instructions and the data the program uses are held in the same main memory, and the CPU fetches these instructions from memory to execute them.

Why: Instructions stored in main memory (1) and the CPU fetches instructions from memory to execute them (1).

Some marks1/2

Program instructions are stored in main memory.

Why: Instructions stored in main memory (1). No second point about data being stored in the same memory or about the CPU fetching instructions.

No marks0/2

The program is saved on the hard disk so that it can be loaded into RAM when the user runs it.

Why: A near-miss about secondary storage, which the mark scheme says not to accept; the stored program concept is about instructions being held in main memory.

3Short recall1 markUnits & Conversion Swap Remove

A file has a size of 2 kibibytes.

State the number of bytes in the file.

[1 mark]
Mark scheme

Award one mark for:

  • 2048 (1)
  • 2 × 1024 (1)
  • 2^11 (1)

Allow 2048 bytes.

Do not accept 2000, which is the number of bytes in two decimal kilobytes.

Do not accept 1024, which is the number of bytes in one kibibyte.

Do not accept 16, which is the number of bits in two bytes.

Example answers and how they were marked
Full marks1/1

2048

Why: 2048 is 2 × 1024, the number of bytes in 2 kibibytes (1).

No marks0/1

2000

Why: 2000 uses the decimal kilobyte, not the binary kibibyte; the mark scheme does not accept it.

No marks0/1

16

Why: 16 is the number of bits in two bytes, confusing bits with bytes; the mark scheme does not accept it.

4Calculate2 marksAdding Binary Numbers Swap Remove

Computer systems manipulate bit patterns.

Complete the table to show the result of the 8-bit binary addition.

[2 marks]
1286432168421
10111000
00011111
Mark scheme

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

  • 1101 (1)
  • 0111 (1)

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

Example answers and how they were marked
Full marks2/2

1101 0111

Why: 184 + 31 = 215, carrying where a column adds to 2: both nibbles are correct (2).

Some marks1/2

1010 0111

Why: Writes 0 for 1 + 1 but forgets to carry the 1. The second nibble is correct (1); the first is not.

No marks0/2

1011 1111

Why: Writes 1 in every column with a 1 and never carries. Neither nibble is correct.

Questions by topic

More ways to revise Edexcel GCSE Computer Science (1CP2)

Teachers: build a paper, print it with the mark schemes at the end, and sign your school up to set your classes activities on every topic.

Fetch Decode Explain isn't affiliated with or endorsed by Pearson Edexcel.