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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 markCharacter Sets Swap Remove

A new character set must be able to represent 70 different characters.

Identify the minimum number of bits needed to represent each character.

[1 mark]
  • A6
  • B7
  • C8
  • D70
Mark scheme
  • B 7 (1)
  • The only correct answer is B. The correct answer is 7.
  • A is not correct because 6 bits give only 64 combinations, which is not enough for 70 characters
  • C is not correct because 7 bits already give 128 combinations, so 8 bits is not the minimum needed
  • D is not correct because 70 is the number of characters needed, not a number of bits
Example answers and how they were marked
Full marks1/1

B 7

Why: The only correct answer: 7 bits give 128 combinations, enough for 70 characters, and it is the smallest number of bits that does (1).

No marks0/1

C 8

Why: Distractor, no credit: a single-tick item has no partial credit, and 8 bits is not the minimum because 7 bits already give 128 combinations, so the scheme's C line gives no mark.

No marks0/1

A 6

Why: Misconception, no credit: 6 bits give only 64 combinations, which cannot represent 70 different characters, so the scheme's A line gives no mark.

2Multiple choice1 markUnits & Conversion Swap Remove

Units of data storage can be binary or decimal.

Identify the unit of storage that is equivalent to 1024 gibibytes.

[1 mark]
  • AGibibyte
  • BMebibyte
  • CTebibyte
  • DTerabyte
Mark scheme
  • C Tebibyte (1)
  • The only correct answer is C. The correct answer is tebibyte.
  • A is not correct because a gibibyte is 1024 mebibytes
  • B is not correct because a mebibyte is 1024 kibibytes
  • D is not correct because a terabyte is 1000 gigabytes
Example answers and how they were marked
Full marks1/1

C Tebibyte

Why: Correct: 1024 gibibytes make one tebibyte (1).

No marks0/1

A Gibibyte

Why: Near miss: a gibibyte is 1024 mebibytes, not 1024 gibibytes (0).

No marks0/1

D Terabyte

Why: Typical misconception: uses the decimal terabyte (1000 gigabytes) instead of the binary tebibyte (0).

3Short recall2 marksLANs & WANs Swap Remove

A school has computers in every classroom.

Define the term 'local area network (LAN)'.

[2 marks]
Mark scheme

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

  • A network that covers a small geographical area / is on a single site / is within one building (1)
  • (Two or more) computers/devices connected together (1)
  • So that the connected devices can share data/files/resources such as printers or an internet connection (1)
  • Uses cabling/infrastructure owned by the organisation (1)

Do not accept a network that covers a large geographical area / connects networks in different countries, as this describes a WAN.

Do not accept 'a network in a school' on its own, as this repeats the stem without a LAN feature such as a single site or connected devices.

Do not accept 'a network that connects to the internet' without a LAN feature such as a single site or connected devices.

Do not accept 'owned by one organisation' on its own, as a WAN can also be owned by one organisation.

Do not accept 'local' on its own without saying what it means.

The share mark needs the sharing to be linked to the connected devices/network, e.g. 'share files' or 'share printers'.

Example answers and how they were marked
Full marks2/2

A LAN is a group of computers connected together on one site, such as a school, so that they can share files and printers.

Why: Connected together (1) and single site (1); the sharing point is also made but the maximum is two marks.

Some marks1/2

A network that is in one building.

Why: Covers a single site/one building (1) but does not say that devices are connected together or what they share.

No marks0/2

A network that covers a large area and connects different countries.

Why: Describes a WAN, not a LAN, so no creditworthy LAN point is made.

4Explain2 marksFile Management Swap Remove

An operating system stops a new file from replacing an existing file with the same name in the same folder.

Explain one reason why the operating system does this.

[2 marks]
Mark scheme

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

  • Existing data would be lost (1) because the new file would overwrite/replace its contents (1)
  • The original file could not be recovered (1) because its contents would be destroyed/replaced (1)
  • A program that uses the existing file would fail (1) because the file's data would have been changed/removed (1)
  • The user could lose important work (1) because the old file is replaced by the new one (1)

For both marks, the expansion must follow/associate with the statement.

Do not accept 'so files do not have the same name' without a reason linked to loss of data.

Do not accept 'to keep files tidy' or 'so the computer does not crash' without a reason linked to the existing file's contents being lost.

Example answers and how they were marked
Full marks2/2

The user would lose the existing file (1) because the new file would overwrite its contents (1).

Why: Linked point (1) and reason (1).

Some marks1/2

The old file would be lost (1).

Why: Point only; no linked reason stating that the new file would overwrite its contents.

No marks0/2

So two files do not have the same name (0) because each file needs a different name.

Why: Repeats the rule about names and does not explain the loss of existing data.

5Explain2 marksTranslators Swap Remove

An interpreter translates and executes high-level code.

Explain why an interpreted program must be translated again each time it is run.

[2 marks]
Mark scheme

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

  • No object code/machine code file is stored/saved (1) so there is no translated program to run on the next occasion (1)
  • The translation is only temporary/is lost when the program ends (1) because each line is translated as the program runs (1)
  • No executable file is created/saved (1) so the whole program has to be translated again every time it is run (1)
  • The interpreter translates each line while the program is running (1) so nothing is kept that could be run without the interpreter (1)

For both marks, the expansion must follow/associate with the statement.

Do not accept 'it is interpreted' or 'it is not compiled' without saying that no object code/executable is stored/saved.

Do not accept 'the source code is high-level so it must be translated' or 'the source code has to be turned into machine code' on its own, as this is true of any translation.

Do not accept answers about the source code being deleted.

Example answers and how they were marked
Full marks2/2

The interpreter does not store an object code file, so there is no machine code file to run on the next occasion.

Why: No object code stored (1) linked to no translated program for the next run (1).

Some marks1/2

No object code is saved.

Why: Point that nothing is stored (1) but no linked reason saying why this means the program must be translated again.

No marks0/2

It has to be translated again because the interpreter deletes the source code after each run.

Why: Incorrect: the interpreter does not delete the source code; no mark scheme point.

6Convert2 marksDenary to Hexadecimal Conversion Swap Remove

Computers manipulate binary patterns.

Convert the denary number 90 to hexadecimal.

[2 marks]
Mark scheme

Award one mark for each correct digit in the correct position:

  • 5 (1)
  • A (1)

Answer: 5A

Example answers and how they were marked
Full marks2/2

5A

Why: 90 ÷ 16 = 5 remainder 10, so the digits are 5 and A: both digits are correct (2).

Some marks1/2

510

Why: One digit is correct (1), but a digit over 9 is written in denary instead of as a letter.

No marks0/2

A5

Why: The digits are the wrong way round, so neither is in the correct position.

7Discuss6 marksLow and High Level Languages Swap Remove

A company is developing the firmware for a robotic vacuum cleaner. The firmware must read the distance sensors and control the motors. The vacuum cleaner has a small amount of memory.

Discuss the characteristics of a low-level programming language that make it suitable for writing the firmware.

Your answer should include:

  • control of the hardware
  • the efficiency of the code
  • the link between the language and the processor.
[6 marks]
Mark scheme

Indicative content

Control of the hardware

  • Low-level languages can read and write the contents of registers and memory addresses directly, so the firmware can take the values from the distance sensors without calling a library routine.
  • Assembly language instructions can read from and write to specific input/output ports on the vacuum cleaner's processor, so the motors can be switched on and off directly.
  • The programmer decides exactly which processor operations are carried out and in which order, which matters when the vacuum must reverse away from a step or an obstacle.
  • A low-level language puts nothing between the program and the hardware, so the programmer can control the hardware of the vacuum cleaner precisely.

Efficiency of the code

  • Assembly language code can be hand-optimised instruction by instruction, so the code that checks a sensor is as short as possible.
  • The programmer can choose which registers hold values that are used again and again, which cuts down the number of memory accesses the firmware needs.
  • A hand-written low-level program can be made compact, so it is more likely to fit in the small amount of memory the vacuum cleaner has.
  • Because each instruction does one defined job, the time taken to react to a sensor reading is predictable, which suits firmware that must not overrun.

Link between the language and the processor

  • Low-level languages are processor-specific, so the instructions used are those of the exact processor in the vacuum cleaner rather than general statements that need translating.
  • Each assembly language instruction maps to one machine code instruction, so the programmer knows precisely which processor operation will be carried out by each line of code.
  • An assembler turns each mnemonic into its machine code equivalent, so the machine code running on the vacuum cleaner's processor is exactly what the programmer wrote.
  • Code written for a different processor would have to be rewritten, which is acceptable here because the vacuum cleaner always uses the same processor.

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)

Within Level 1, award 1 mark for a single basic point and 2 marks for two or more independent basic points; Level 2 requires the points to be linked to the vacuum cleaner; Level 3 requires both control of the hardware and efficiency of the code, and a few concise, linked sentences can reach Level 3, so extra length is not required.

Accept registers, memory addresses, input/output ports, machine code and mnemonics as ways of describing access to the hardware, and accept 'a low-level language is faster' as a basic point on its own. Do not award 'binary is a low-level language' (machine code is written in binary, but binary is not a language), or characteristics of high-level languages such as being easy to read or portable, as characteristics of a low-level language.

Example answers and how they were marked
Full marks6/6

A low-level language can read and write to registers and to input/output ports directly, so the firmware can take the distance sensor readings and switch the motors without needing library routines. It can be hand-optimised instruction by instruction, so the code that checks a sensor is short and the time it takes is predictable, and a compact program is more likely to fit in the vacuum cleaner's small memory. The language is processor-specific and each assembly instruction maps to one machine code instruction, so the programmer knows exactly which operation the vacuum's processor will carry out, and the assembled machine code is exactly what they wrote.

Why: Covers all three bullets: direct control of registers and ports applied to the sensors and motors, hand-optimised compact code with predictable timing applied to the small memory, and the one-to-one link to the processor's own instruction set. Linked and applied throughout, so Level 3.

Some marks2/6

A low-level language can control the hardware directly. It can read the sensors and control the motors. It is also faster than a high-level language, so the vacuum responds quickly. This is useful because it has little memory.

Why: Three basic, independent points: direct control of the hardware applied to the sensors and motors, faster execution, and suitability for limited memory. They are asserted without development (no registers, ports or predictable timing) and the link between the language and the processor is missing, so Level 1, two or more basic points (2).

Some marks1/6

A low-level language is written in binary, so the computer can understand it directly. It is faster than a high-level language.

Why: 'Faster than a high-level language' is a creditworthy basic point (1). 'Written in binary' is not awarded, as binary itself is not a language and the do-not-award rule applies. Only one undeveloped point with no application to the vacuum cleaner, so Level 1, single basic point (1).

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