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.

Practice paper

7 questions · 15 marks

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1Multiple choice1 markCompression Swap Remove

Identify one reason for using data compression.

[1 mark]

ATo increase the number of packets sent
BTo reduce the time taken to transmit a file
CTo improve the quality of the file
DTo make the file harder to open

Mark scheme

B To reduce the time taken to transmit a file (1)
The only correct answer is B. The correct answer is to reduce the time taken to transmit a file.
A is not correct because compression reduces file size, which can reduce the number of packets sent
C is not correct because lossy compression can reduce the quality of a file
D is not correct because encryption, not compression, can make a file harder to open

Example answers and how they were marked

Full marks1/1

B To reduce the time taken to transmit a file

Why: Correct option (1).

No marks0/1

A To increase the number of packets sent

Why: Near miss: compression reduces the number of packets sent, so no mark.

No marks0/1

C To improve the quality of the file

Why: Misconception that compression improves quality; no mark.

2Multiple choice1 markAddressing Swap Remove

A packet of data is sent across the internet to a web server. The packet has a header that stores information about the packet.

Identify the information in the packet header that a router uses to forward the packet towards its destination.

[1 mark]

AThe destination domain name
BThe destination IP address
CThe MAC address of the sender
DThe name of the file being sent

Mark scheme

B The destination IP address (1)
The only correct answer is B. The correct answer is the destination IP address.
A is not correct because a domain name is converted to an IP address by a DNS server before routing begins
C is not correct because the sender's MAC address is only used to deliver the packet on the local network
D is not correct because the file name is not used to route packets

Example answers and how they were marked

Full marks1/1

B The destination IP address

Why: Correct option (1): a router reads the destination IP address in the header to forward the packet towards its destination.

No marks0/1

A The destination domain name

Why: A domain name is converted to an IP address by a DNS server before routing begins, so it is not the routing information in the header (mark scheme line A); no mark.

No marks0/1

C The MAC address of the sender

Why: The sender's MAC address is only used to deliver the packet on the local network, so it is not used by the router to forward the packet (mark scheme line C); no mark.

3Short recall1 markThe Internet Swap Remove

Data packets are sent across the internet.

Name the device that uses IP addresses to forward a data packet from one network to another.

[1 mark]

Mark scheme

Award one mark for:

Router (1)

Allow routers (plural).

Allow minor spelling errors that are clearly router, e.g. rooter.

Do not accept switch, hub (these forward data within a network, not between networks), modem or server.

Do not accept gateway on its own.

Example answers and how they were marked

Full marks1/1

Router

Why: Correct device named (1).

No marks0/1

Modem

Why: Modem is not accepted; it is not the device that forwards packets between networks (0).

No marks0/1

Server

Why: A server is not accepted; it does not forward packets between networks (0).

4Short 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.

5Describe2 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.

6Convert2 marksDenary to Binary Conversion Swap Remove

Binary patterns can represent numbers.

Convert the denary number 199 to 8-bit binary.

[2 marks]

Mark scheme

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

1100 (1)
0111 (1)

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

Example answers and how they were marked

Full marks2/2

1100 0111

Why: 199 = 128 + 64 + 4 + 2 + 1: both nibbles are correct (2).

Some marks1/2

1100 1000

Why: Converts 200 instead (a slip in adding up the place values). The first nibble is correct (1); the second is not.

No marks0/2

1110 0011

Why: The bits are written in reverse, smallest place value first. Neither nibble is correct.

7Discuss6 marksFetch Decode Execute Cycle Swap Remove

A small robotics company is designing a new robot. The robot's computer uses von Neumann architecture. The engineers are explaining how programs are stored and run.

Discuss how the von Neumann stored program concept and main memory (RAM) are used during the fetch-decode-execute cycle.

Your answer should include:

what is meant by the stored program concept
the role of RAM in the cycle
why the stored program concept is useful.

[6 marks]

Mark scheme

Indicative content

Stored program concept

Instructions and data are stored together in the same main memory (RAM).
Instructions are fetched one at a time from memory and executed in sequence.
The program can be changed by loading different instructions into RAM.
The CPU can only execute instructions that are in main memory.

Role of RAM

RAM holds the program instructions and the data being used.
The address of the next instruction is held in the program counter (PC).
The instruction is fetched from RAM using the address bus.
The data bus carries the instruction from RAM to the CPU.
RAM is volatile, so programs are lost when power is off.
RAM is directly accessible by the CPU, unlike secondary storage.

Benefits

The same hardware can run many different programs.
Programs can be loaded and changed quickly.
No need to rewire the computer for a new task.
Makes general-purpose computers possible.

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)

Example answers and how they were marked

Full marks6/6

The von Neumann stored program concept means that both the program instructions and the data are stored together in the same main memory, RAM. During the fetch-decode-execute cycle, the CPU fetches an instruction from RAM, decodes it in the control unit, and then executes it. The program counter holds the address of the next instruction, and this address is sent along the address bus to RAM. RAM then returns the instruction along the data bus to the CPU. RAM is volatile, so it only holds the program while the robot is powered on. The stored program concept is useful because the robot's computer can run different programs without changing any hardware; new instructions can simply be loaded into RAM. This makes the computer general-purpose and flexible.

Why: Comprehensive understanding: defines stored program concept, explains RAM role with PC, address bus and data bus, applies to robot, and gives benefits. Sustained lines of reasoning. Level 3.

Some marks3/6

The stored program concept is when instructions and data are stored in RAM. The CPU fetches instructions from RAM. The program counter keeps the address of the next instruction. RAM holds the program. The benefit is that you can change the program without rewiring the computer.

Why: Adequate understanding: correct basic points about stored program concept and RAM, with some linkage, but little development or application to the robot's fetch-decode-execute cycle. Level 2.

Some marks1/6

The von Neumann stored program concept is that the CPU uses RAM to store the program. The fetch-decode-execute cycle fetches, decodes and executes. RAM is random access memory. The benefit is that it is fast.

Why: Basic independent points only; no real linkage or application to the robot. Level 1.

Edexcel GCSE Computer Science (1CP2) Exam Questions

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Questions by topic

2.1 Binary

2.2 Data representation

2.3 Data storage and compression

3.1 Hardware

3.2 Software

3.3 Programming languages

4.1 Networks

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