Edexcel GCSE Computer Science (1CP2) exam questions

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Practice paper

8 questions · 16 marks

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1Multiple choice1 markCompression Swap Remove

Identify the definition of lossy compression.

[1 mark]

AData is removed only while the file is being transmitted
BData is permanently removed to reduce the file size
CData is restored exactly when it is decompressed
DData is stored using a dictionary of repeated patterns

Mark scheme

B Data is permanently removed to reduce the file size (1)
The only correct answer is B. The correct answer is data is permanently removed to reduce the file size.
A is not correct because the data removed by a lossy algorithm is not restored after transmission
C is not correct because that describes lossless compression, where the original file is rebuilt exactly
D is not correct because dictionary encoding is a lossless method that records repeated patterns once

Example answers and how they were marked

Full marks1/1

B Data is permanently removed to reduce the file size

Why: Correct option (1).

No marks0/1

C Data is restored exactly when it is decompressed

Why: Near miss: this describes lossless compression, so no mark.

No marks0/1

A Data is removed only while the file is being transmitted

Why: Misconception that compression only applies during transmission; the data is removed permanently, so 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).

4Convert1 markBinary to Denary Conversion Swap Remove

Computers manipulate binary patterns.

Convert the binary pattern 1110 0000 to denary.

[1 mark]

Mark scheme

Award one mark for:

224 (1)

Example answers and how they were marked

Full marks1/1

224

Why: Correct: 128 + 64 + 32 = 224 (1).

No marks0/1

Why: Reads the place values from the wrong end (the left-hand bit as 1), so the value is wrong.

No marks0/1

Why: Counts the 1s instead of adding their place values.

5Short recall2 marksCharacter Sets Swap Remove

ASCII is a character set that uses 7 bits to represent each character.

Give two limitations of 7-bit ASCII when it is used to represent text.

[2 marks]

Mark scheme

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

(Only) 128 different characters can be represented / 7 bits give 2^7 = 128 different codes (1)
It cannot represent characters from languages other than English / non-English alphabets, e.g. Greek, Chinese (1)
It cannot represent accented characters, e.g. é, ñ (1)
It cannot represent symbols or emoji that are not in the ASCII character set (1)

Award a maximum of one mark for the last three bullets, as they are all examples of characters that fall outside the 128 ASCII codes.

Allow 'it can only represent 128 characters, so some characters are missing' for the first bullet.

Do not accept 'it uses 7 bits' / 'it only uses 7 bits' on its own, without saying what this limits.

Do not accept 'it can only store a limited amount of text' / 'it can only store short messages', as 7 bits limits the number of characters available, not the length of a message.

Do not accept 'it can only represent the characters in the ASCII character set', as this repeats the question.

Do not accept 'it is old/out of date' or 'it cannot store numbers'.

Example answers and how they were marked

Full marks2/2

Only 128 different characters can be represented, so it cannot store characters from other languages such as Greek.

Why: 128 character limit (1) and characters from other languages (1); two separate creditworthy limitations.

Some marks1/2

It cannot show accented characters like é and it only uses 7 bits.

Why: Accented characters (1); 'only uses 7 bits' is not qualified, so scores 0.

No marks0/2

It only uses 7 bits, so it can only store a small amount of text.

Why: Scores 0: text length is not a limitation of 7-bit ASCII, and 'only uses 7 bits' is unqualified; both Do not accept lines apply.

6Short 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.

7Convert2 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.

8Discuss6 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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Practice paper

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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