Edexcel GCSE Computer Science (1CP2) Exam Questions

• 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

• 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

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'.

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.

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

• 5 (1)
• A (1)

Answer: 5A

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

• 0010 (1)
• 0100 (1)

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

Ignore a carry shown beyond the eighth bit.

Award one mark for:

• overflow (error) (1)

Indicative content

the role of RAM in the stored program concept

• In the von Neumann stored program concept, a program's instructions and the data it uses are held together in main memory.
• RAM holds the instructions and data of the program that the CPU is currently processing.
• The CPU fetches each instruction from RAM during the fetch-decode-execute cycle, so a program must be in RAM before it can be run.
• RAM is directly accessible by the CPU, so instructions can be fetched quickly enough for the CPU to keep working.

what happens to a program when it is opened

• When a program is opened, a copy of its instructions and data is transferred from secondary storage, such as the hard drive, into RAM.
• The operating system loads the program into RAM so that the CPU can start fetching and executing its instructions.
• When the program is closed, it is removed from RAM, and the space it used is free for another program.
• If a computer does not have enough RAM, the program may not open, or another open program has to be closed first.

why the CPU cannot execute a program directly from secondary storage

• Secondary storage, such as a hard disk drive, is much slower to access than RAM.
• The CPU can process instructions much faster than secondary storage can supply them, so the CPU would have to wait and would not be working at full speed.
• RAM is fast electronic memory with no moving parts, so data can be read from it in a fraction of the time taken by a magnetic hard disk drive.
• Running the school's programs directly from secondary storage would make the computers in the IT room very slow to use.

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)

Additional guidance

• Level 3 responses must cover both the role of RAM in the stored program concept and the speed difference between RAM and secondary storage, with a reason such as the CPU having to wait for instructions.
• Answers that make only basic, independent points, such as what RAM is, that a program is copied into RAM when it is opened, or that RAM is faster than a hard drive, without linking them to explain why the CPU would have to wait for instructions, are limited to Level 1.
• Do not accept statements that RAM is permanent storage, that RAM replaces secondary storage, or that secondary storage is read-only.

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.