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Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 693 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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3 questions · 6 marks

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1Short recall2 marksPrimary Memory Swap Remove

A computer is designed using 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:

  • The program instructions and data are stored together in main memory/RAM (1)
  • The same memory is used for both instructions and data (1)
  • The instructions are fetched from main memory and executed by the CPU (1)
  • The instructions are fetched and executed one at a time / in sequence (1)
  • The program can be changed/replaced by loading new instructions into memory, without changing the hardware (1)

Do not accept the program is stored in secondary storage / on the hard disk.

Do not accept the program is stored in the CPU / in registers.

Example answers and how they were marked
Full marks2/2

The program instructions and the data are stored together in main memory, and the CPU fetches the instructions one at a time and executes them.

Why: Creditworthy point that instructions and data are stored in the same main memory (1) and the linked point that the CPU fetches and executes them from memory (1).

Some marks1/2

The program is held in the main memory (RAM) so that the computer can run it.

Why: Earns the point that the program instructions are stored in main memory (1), but does not say that the CPU fetches and executes them, so the second mark is not earned.

No marks0/2

The program is stored on the hard disk and loaded into the CPU when it is run.

Why: Both parts are misconceptions named in the Do not accept lines: the stored program concept keeps the instructions in main memory, not on secondary storage or in the CPU.

2Describe2 marksEmbedded Systems Swap Remove

A greenhouse uses an embedded system to control the growing conditions for plants.

Describe how the embedded system controls the growing conditions in the greenhouse.

[2 marks]
Mark scheme

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

  • It reads/monitors the temperature/humidity/light level using sensors (1) to switch the heaters/vents/lamps/watering system on or off when the reading is outside the set range (1)
  • It compares the readings from the sensors with the set values/thresholds stored in its memory (1) to decide whether the conditions need to be changed (1)
  • It monitors the light level using a light sensor (1) and turns the lamps on when the light falls below the level the plants need (1)
  • It monitors the soil moisture using a soil moisture sensor (1) and turns the water pump/irrigation on when the soil is too dry (1)

For the second mark, the action/response must be linked to the sensor reading given in the first point.

Allow named sensors (temperature/humidity/light/soil moisture) and named outputs (heater, fan, vent/window, lamp, irrigation/water pump).

Do not accept answers that only name a sensor or only name an output, e.g. 'it has a temperature sensor and a heater', as there is no link between the reading and the control action.

Do not accept answers where a person switches the equipment on or off, as the embedded system controls the conditions automatically.

Example answers and how they were marked
Full marks2/2

The embedded system reads the temperature using a sensor and switches the heater on when the temperature is below the set value.

Why: Sensor reading point (1) and linked control action (1).

Some marks1/2

The embedded system compares the readings from the sensors with the set values stored in its memory.

Why: First point only (1) - the comparison with stored values - with no linked action.

No marks0/2

The greenhouse has a temperature sensor and a heater.

Why: Only names a sensor and an output, which the mark scheme says is not accepted.

3Explain2 marksSecondary Memory Swap Remove

A Blu-ray disc and a CD are the same physical size.

Explain why a Blu-ray disc can store more data than a CD.

[2 marks]
Mark scheme

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

  • The pits and lands on a Blu-ray disc are smaller/closer together (1) because the blue laser has a shorter wavelength than the infrared laser used by a CD (1)
  • More pits and lands can be stored in the same length of track (1) because the blue laser has a shorter wavelength, so it can focus on a smaller area/read smaller details (1)
  • A blue laser can focus on a smaller area (1) so smaller pits and lands can be used, allowing more data to be stored on the disc (1)

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

Do not accept 'a Blu-ray is newer/better' or 'a Blu-ray has more space/bigger storage' without a linked reason about the laser/pits and lands.

Do not accept answers about read/write speed, for example 'a Blu-ray spins faster', because the question asks about the amount of data stored, not how quickly it is read.

Do not award 'a Blu-ray disc can have more layers' on its own as the reason; it must be linked to the smaller pits and lands/shorter wavelength of the laser.

Do not accept 'red laser' for a CD; a CD uses an infrared laser (red is used by DVD).

Example answers and how they were marked
Full marks2/2

A Blu-ray is read with a blue laser which has a shorter wavelength so the pits and lands on the disc can be smaller and more of them fit on the disc.

Why: Linked point (1) shorter wavelength of the blue laser with reason (1) smaller pits and lands so more data fits on the same disc.

Some marks1/2

The pits and lands on a Blu-ray are smaller than the ones on a CD so it can store more data.

Why: Point (1) that the pits and lands are smaller, but the second part repeats the question and does not give the linked reason (shorter wavelength of the blue laser/infrared CD laser), so only one mark.

No marks0/2

A Blu-ray spins faster than a CD so more data can be read from it in the same time.

Why: Do not accept answers about read/write speed; the question asks about the amount of data stored, not how quickly it is read.

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