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Edexcel GCSE Computer Science (1CP2) Exam Questions

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4 questions · 10 marks

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1Short recall3 marksFetch Decode Execute Cycle Swap Remove

The fetch-decode-execute cycle is used by the CPU.

State three things that happen during the fetch stage.

[3 marks]
Mark scheme

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

  • The address (of the next instruction) is sent from the CPU to main memory / placed on the address bus (1)
  • The (next) instruction is fetched/read from main memory (RAM) (1)
  • The instruction is brought/copied into the CPU / received into the MDR (1)
  • The program counter (PC) is incremented/updated to hold the address of the next instruction (1)

Allow 'the PC/MAR is placed on the address bus' for the address being sent to main memory.

Allow 'the MDR receives the instruction' for the instruction being brought into the CPU.

Do not award a second mark for the instruction being fetched from main memory and the instruction being brought into the CPU; these are the same point.

Do not award 'the instruction is transferred along the data bus' on its own; this is the same point as the instruction being fetched from main memory.

Do not accept answers about decoding the instruction, e.g. 'the instruction is fetched from RAM and decoded', as decoding happens in the decode stage.

Do not accept answers about the execute stage, e.g. 'the ALU carries out the calculation' or 'the result of a calculation is stored in a register', as these happen in the execute stage.

Example answers and how they were marked
Full marks3/3

The address of the next instruction is sent to main memory, the instruction is fetched from RAM, and the program counter is incremented.

Why: Address sent to main memory (1), instruction fetched from RAM (1) and program counter incremented (1).

Some marks1/3

The instruction is fetched from main memory and then decoded.

Why: Fetching the instruction from main memory (1); decoding happens in the decode stage, so it is not creditworthy in the fetch stage.

No marks0/3

The ALU carries out the calculation and the result is stored in a register.

Why: The ALU carrying out a calculation and storing a result happen in the execute stage, not the fetch stage, so no point is creditworthy.

2Short 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.

3Describe2 marksLow and High Level Languages Swap Remove

A device driver is software that allows the operating system to control a piece of hardware. Device drivers are often written in a low-level programming language.

Describe one reason why a low-level programming language is suitable for writing a device driver.

[2 marks]
Mark scheme

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

  • Instructions can manipulate/control the hardware/memory/registers directly (1) so the driver can control the specific device it is written for (1)
  • The code is written for one specific processor/architecture (1) so it works with the exact hardware it will control (1)
  • The programmer controls exactly how memory/registers are used (1) so the code is efficient and needs little memory, which suits a device that has limited memory (1)
  • Each instruction maps to one machine code instruction (1) so the driver code is small enough to fit in the memory of the device (1)

Extra detail, such as naming the device the driver controls, is not needed to earn the marks.

Do not accept 'it is closer to machine code' on its own: it must be developed to show how this helps the driver, e.g. it can control the hardware directly.

Do not accept 'it is faster/quicker' with no link to the hardware/device, e.g. the device responds quickly.

Do not accept 'it is easier to learn', 'it is easier to debug' or 'it takes less time to write'.

Example answers and how they were marked
Full marks2/2

Low-level instructions can manipulate the hardware and memory directly (1) so the driver can control the specific device it was written for, such as a sound card (1).

Why: Point about direct control of hardware/memory (1) linked to controlling the specific device (1). The named device is extra detail and is not needed.

Some marks1/2

Low-level instructions can manipulate the hardware directly.

Why: Correct point (1) but no linked development saying how this helps the device driver.

No marks0/2

It is closer to machine code so it runs faster than a high-level language.

Why: 'Closer to machine code' is not developed into how this helps the driver, and 'runs faster' is not linked to the hardware/device, so no mark.

4Calculate3 marksAdding Binary Numbers Swap Remove

Binary patterns can represent numbers.

(a)

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

[2 marks]
1286432168421
11011011
11011100
Mark scheme

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

  • 1011 (1)
  • 0111 (1)

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

Ignore a carry shown beyond the eighth bit.

Example answers and how they were marked
Full marks2/2

1011 0111

Why: 219 + 220 = 439, which needs 9 bits; the 8 bits that are kept are both correct (2).

Some marks1/2

0000 0111

Why: Forgets to carry the 1s. The second nibble is correct (1); the first is not.

No marks0/2

1101 1111

Why: Never carries. Neither nibble is correct.

(b)

State the name of the error that occurs when this addition is carried out.

[1 mark]
Mark scheme

Award one mark for:

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

Overflow

Why: The result needs more bits than are available (1).

No marks0/1

Underflow

Why: A near miss: the result is too big for 8 bits, not too small.

No marks0/1

A syntax error

Why: A syntax error is a mistake in a program's code, not in arithmetic.

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