2025 Edexcel IGCSE Computer Science Practice Paper with Answers

In a star network topology, each workstation is connected via its own cable to a central switch/hub. If one workstation's cable is damaged, only that workstation is disconnected, while the rest of the network continues to function normally. In contrast, in a bus network, a break in the main cable can disable the entire network.

Award 1 mark for the correct answer (c). Reject any other options.

Eavesdropping (or packet sniffing) involves capturing and analyzing network packets as they transit across a network. Phishing involves deceptive emails/websites; DoS is a Denial of Service attack to crash a system; SQL Injection target databases via inputs.

Award 1 mark for identifying the threat as eavesdropping/packet sniffing (c). Reject all other options.

Shifting \(00111010\) left by 1 place results in \(01110100\). Shifting left by a second place results in \(11101000\). Vacuum positions on the right are filled with 0s.

Award 1 mark for the correct binary pattern (a). Reject other choices.

The Program Counter (PC) keeps track of the memory address of the next instruction to be fetched. The MAR holds the address of the current memory location being accessed; the MDR holds the actual data; and the Accumulator stores the results of calculations.

Award 1 mark for identifying the Program Counter (c). Reject other options.

Run-length encoding works by replacing consecutive repeated data values with a single value and a count. Sequence B (\(AAAAAABBBBCC\)) has long consecutive runs (6 A's, 4 B's, 2 C's) which can be represented compactly as 6A4B2C. The other sequences have frequent character changes, which do not compress as efficiently.

Award 1 mark for selecting the correct sequence (b). Reject others.

The combination of an Exclusive-OR (XOR) gate followed by a NOT gate is equivalent to an Exclusive-NOR (XNOR) gate, which outputs 1 when both inputs are equal and 0 when they are different.

Award 1 mark for the XNOR gate (a). Reject other options.

A logical right shift moves all bits to the right by the specified number of places. Shifting 11001000 right by 1 place gives 01100100. Shifting right by a second place gives 00110010. The empty positions on the left are filled with 0s.

1 mark for shifting the bits to the right with 0s padded on the left. 1 mark for the correct final pattern: 00110010.

Lossy compression permanently discards unnecessary data, resulting in a much larger reduction in file size compared to lossless compression. This allows more images to be stored on the device and enables faster transmission of files over mobile networks.

1 mark for stating that lossy compression produces significantly smaller file sizes (saving storage space). 1 mark for stating that it allows for faster upload, download, or sharing times.

A firewall acts as a barrier between a trusted local network and untrusted external networks. It inspects all incoming and outgoing data packets and blocks or permits them based on configured security criteria.

1 mark for explaining that it inspects/monitors incoming and/or outgoing network traffic. 1 mark for explaining that it blocks packets that do not meet pre-defined security rules.

The Program Counter (PC) is a CPU register that tracks instruction execution. It holds the address of the next instruction in memory, and once that instruction is fetched, the PC is updated to point to the following instruction.

1 mark for stating it holds the address of the next instruction to be fetched. 1 mark for stating that it increments or updates to point to the next instruction in sequence.

This is a NAND operation. For inputs (0,0), (0,1), and (1,0), the output of the AND gate is 0, so the NOT gate changes it to 1. For inputs (1,1), the output of the AND gate is 1, so the NOT gate changes it to 0. Thus, the sequence of outputs is 1, 1, 1, 0.

1 mark for identifying at least two correct outputs in the sequence. 1 mark for providing the fully correct sequence: 1, 1, 1, 0.

Electronic waste contains hazardous heavy metals like lead, mercury, and cadmium. In landfills, these can seep into ground soil and contaminate local water supplies. Additionally, throwing away devices wastes scarce raw materials that could otherwise be reclaimed.

1 mark for describing the release of toxic substances/heavy metals into the ground/water. 1 mark for describing the waste of finite raw materials or land space issues.

The sample rate defines how many times the amplitude of the analogue wave is measured per second. Raising the sample rate means more data points are recorded each second, directly increasing the total storage size of the resulting audio file.

1 mark for explaining that a higher sample rate means more samples/measurements are taken per second. 1 mark for linking this directly to producing more data and thus a larger overall file size.

Symmetric encryption relies on a single shared key that both sender and receiver must keep secret. Asymmetric encryption uses a mathematically linked pair of keys: a public key that can be shared with anyone to encrypt data, and a private key kept secret by the owner to decrypt it.

1 mark for stating that symmetric encryption uses one single key for both processes. 1 mark for stating that asymmetric encryption uses two different keys (a public key and a private key).

1. Performing a logical shift right of two places on 11011000 shifts all bits two positions to the right, filling the vacant spaces on the left with 0s. This results in: 00110110. 2. Shifting right by two places is mathematically equivalent to dividing the binary integer value by 4 (integer division, discarding any remainder).

1 mark for the correct 8-bit binary pattern: 00110110. 1 mark for identifying the mathematical effect: division by 4 (accept integer division by 4).

Convert 1.5 megabytes (MB) to megabits (Mb): \(1.5 \times 8 = 12\) Mb. Calculate the transmission time: \(\text{Time} = \frac{\text{Size}}{\text{Speed}} = \frac{12\text{ Mb}}{12\text{ Mbps}} = 1.0\) second.

1 mark for converting megabytes to megabits (12 Mb) or setting up the correct division. 1 mark for the correct final answer of 1 second.

1. Vector graphics are mathematically defined, meaning they can be scaled/resized infinitely without losing quality or becoming pixelated. 2. Vector files typically have smaller file sizes than high-resolution bitmaps for simple geometric drawings because they only store instructions/formulas rather than individual pixel values.

1 mark for identifying scalability/no loss of quality when resized. 1 mark for identifying smaller file size (for simple/geometric images).

1. Run-Length Encoding counts consecutive identical characters. AAAAABBBCCCCCCDD becomes 5A3B6C2D. 2. Original character count is 16. Compressed character count is 8. Percentage reduction = \(((16 - 8) / 16) \times 100 = 50\%\).

1 mark for the correct compressed RLE string: 5A3B6C2D (accept alternative formats like A5B3C6D2). 1 mark for the correct calculation of 50% reduction.

Evaluate the expression step-by-step: 1. \(A \text{ AND } B\) becomes \(1 \text{ AND } 1 = 1\). 2. \(\text{NOT}(1) = 0\). 3. \(0 \text{ OR } C\) becomes \(0 \text{ OR } 0 = 0\). Thus, \(Q = 0\).

1 mark for correct intermediate evaluation (e.g., evaluating NOT (A AND B) as 0, or demonstrating step-by-step logic). 1 mark for the correct final output of 0 (or False).

The Program Counter (PC) stores the memory address of the next instruction to be fetched from RAM. During the fetch stage, this address is passed to the Memory Address Register (MAR) to retrieve the instruction, and the PC is then incremented by 1 to point to the next instruction in sequence.

1 mark for stating that the PC holds the memory address of the next instruction to be fetched. 1 mark for stating that the PC is incremented (by 1) to point to the next instruction.

An active attack involves the attacker interacting with, modifying, or deleting data or system resources (e.g., malware or denial-of-service). A passive attack involves intercepting, monitoring, or eavesdropping on transmitted data without altering any of the system resources (e.g., packet sniffing).

1 mark for explaining active attacks (involves altering/interfering with data or systems). 1 mark for explaining passive attacks (involves eavesdropping/intercepting traffic without alteration).

Asymmetric encryption uses two mathematically linked keys: a public key for encryption and a private key for decryption. Because the private key never needs to be shared or transmitted over the internet, there is no risk of it being intercepted. In contrast, symmetric encryption uses a single shared key that must be exchanged between parties, creating a potential point of interception.

1 mark for explaining that in asymmetric encryption, the decryption key (private key) does not need to be transmitted or shared. 1 mark for contrasting this with symmetric encryption where the shared key must be transmitted, exposing it to interception risk.

To convert 109 to 8-bit binary, find the place values that sum to 109:
\(109 = 64 + 32 + 8 + 4 + 1\)
Placing 1s under these values in an 8-bit table (128, 64, 32, 16, 8, 4, 2, 1) gives:
- 128: 0
- 64: 1
- 32: 1
- 16: 0
- 8: 1
- 4: 1
- 2: 0
- 1: 1
Result: \(01101101\)

Award 1 mark for showing correct working (e.g., decomposing the number as \(64 + 32 + 8 + 4 + 1\) or demonstrating successive division by 2 with remainders).
Award 1 mark for the correct 8-bit binary value: 01101101.

1. Calculate total number of pixels: \(400 \times 300 = 120,000\) pixels.
2. Since the colour depth is 8 bits (which equals 1 byte), the file size in bytes is: \(120,000 \times 1\text{ byte} = 120,000\text{ bytes}\).
3. Convert bytes to kilobytes (KB) by dividing by 1000: \(120,000 / 1000 = 120\text{ KB}\).

Award 1 mark for showing correct working (e.g., calculating 120,000 bytes, or writing the formula \(\frac{400 \times 300 \times 8}{8 \times 1000}\)).
Award 1 mark for the correct final numerical answer: 120 (KB is already specified in the question).

1. Convert the file size from megabytes to megabits: \(15\text{ MB} \times 8 = 120\text{ megabits (Mb)}\).
2. Calculate the download time: \(\text{Time} = \frac{\text{File size in Mb}}{\text{Transmission speed in Mbps}} = \frac{120}{24} = 5\text{ seconds}\).

Award 1 mark for showing the conversion of MB to megabits (e.g., \(15 \times 8 = 120\)), or demonstrating a mathematically equivalent expression like \(\frac{15 \times 8}{24}\).
Award 1 mark for the correct final answer of 5 (seconds).

1. Evaluate the first sub-expression: \(\text{NOT } A = \text{NOT } 0 = 1\).
2. Evaluate the second sub-expression: \(B \text{ OR } C = 0 \text{ OR } 1 = 1\).
3. Evaluate the final expression: \(Q = 1 \text{ AND } 1 = 1\) (or True).

Award 1 mark for correctly evaluating both intermediate sub-expressions (\(\text{NOT } A = 1\) AND \(B \text{ OR } C = 1\)).
Award 1 mark for the correct final output value of 1 (accept 'True').

During the fetch stage, the CPU needs to retrieve the next instruction from memory. The Program Counter (PC) keeps track of the sequence of execution by holding the memory address of the next instruction. This address is then copied to the Memory Address Register (MAR), which physically interfaces with the address bus to locate and read the instruction from RAM.

Award 1 mark for describing the role of the Program Counter (e.g., it stores/holds the address of the next instruction to be fetched).
Award 1 mark for describing the role of the Memory Address Register (e.g., it holds the address of the specific location in memory currently being accessed/read).

Virtualisation allows a single physical server to host multiple virtual machines (VMs) running separate operating systems and services. This drastically increases the utilisation rate of hardware, meaning fewer physical servers need to be manufactured, powered, and cooled. Consequently, the data centre consumes much less electrical power and produces less electronic waste (e-waste) over time.

Award 1 mark for identifying that virtualisation reduces the need for physical hardware/servers by consolidatng them onto fewer machines.
Award 1 mark for connecting this reduction to a clear environmental benefit (e.g., lower power consumption for running/cooling hardware, or reduced carbon emissions/e-waste).

A packet-filtering firewall works by analyzing packet headers (IP, ports, protocols). It compares these headers to access control lists (rules). Finally, it allows or drops the packet.

1 mark for identifying that it inspects/analyzes packet headers (containing IP addresses or port numbers). 1 mark for explaining that it compares this information against a pre-defined set of rules / access control list (ACL). 1 mark for stating that it either permits (forwards) or blocks (drops) the packet based on the outcome of the rule comparison.

A binary search works by comparing the middle element and deciding whether to search the left or right sub-list. This decision is based on order (e.g. alphabetical or numerical). If the list is unsorted, this logical elimination is impossible. The programmer must first sort the list using a sorting algorithm.

1 mark for explaining that binary search works by comparing the target with the midpoint and discarding half the items. 1 mark for stating that if the list is unsorted, the comparison cannot determine which half of the list contains the target (rendering the elimination step useless). 1 mark for stating that the list must first be sorted (e.g., using a merge or bubble sort) before the binary search can be applied.

During fetch, the PC holds the memory location of the current instruction. It sends this address to the MAR. Once copied, the PC increments by 1.

1 mark for stating that the PC holds the address of the next instruction to be executed. 1 mark for stating that this address is copied/transferred to the MAR (so the bus can retrieve the instruction). 1 mark for explaining that the PC is then incremented (by one/to point to the next instruction) so it is ready for the next cycle.

RLE compresses consecutive identical elements. Large uniform areas compress well; highly diverse non-repeating data compresses poorly.

1 mark for explaining that RLE is effective with data containing long sequences of repeating/identical values (e.g., simple icons, black-and-white images). 1 mark for explaining that RLE is ineffective when adjacent data values rarely repeat (e.g., natural photographs, complex texts). 1 mark for explaining that in the ineffective scenario, storing the frequency count alongside every single unique byte/character would actually increase the total file size (or cause data expansion).

Data centres require massive power for computation and cooling, and generate significant electronic waste when upgraded.

1 mark for identifying one valid environmental concern (e.g., high electricity consumption, water usage for cooling, or carbon emissions from fossil-fuel-powered grids). 1 mark for explaining the impact of this concern (e.g., high power use leads to increased carbon footprint/global warming, or heavy water usage depletes local reservoirs). 1 mark for identifying/explaining a second distinct concern (e.g., electronic waste / e-waste generated when servers are frequently upgraded/replaced, which may contain toxic chemicals that pollute landfills).

For inputs (0,0), both give 0. For inputs (1,0) or (0,1), both give 1. For inputs (1,1), OR gives 1 and XOR gives 0.

1 mark for stating that both gates output 0 when both inputs are 0 (or that both output 1 when only one input is 1). 1 mark for stating that the OR gate outputs 1 when both inputs are 1. 1 mark for stating that the XOR gate outputs 0 when both inputs are 1.

(a)
- First part: Input A and Input B combined with AND gives: \(A \text{ AND } B\).
- Second part: Input B is inverted with NOT (\(\text{NOT } B\)) and combined with C using AND gives: \((\text{NOT } B) \text{ AND } C\).
- Third part: Both intermediate outputs are combined with OR, resulting in: \(Q = (A \text{ AND } B) \text{ OR } (\text{NOT } B \text{ AND } C)\).

(b) Substituting the values A = 1, B = 0, C = 1 into the expression:
- \(A \text{ AND } B = 1 \text{ AND } 0 = 0\)
- \(\text{NOT } B = \text{NOT } 0 = 1\)
- \(\text{NOT } B \text{ AND } C = 1 \text{ AND } 1 = 1\)
- \(Q = 0 \text{ OR } 1 = 1\)

Award up to 4 marks:
- 1 mark for the correct first term: \(A \text{ AND } B\) (accept equivalent notation like \(A \cdot B\)).
- 1 mark for the correct second term: \(\text{NOT } B \text{ AND } C\) (accept equivalent notation like \(\bar{B} \cdot C\)).
- 1 mark for combining both terms correctly with an OR operator to form the complete expression for Q.
- 1 mark for the correct final output of 1 with supporting substitution working.

(a) Following the rules of the tree construction:
- 'A' is on the left branch from the root: Code = 0 (1 bit)
- To reach 'B': Root -> right (1) -> right (1): Code = 11 (2 bits)
- To reach 'C': Root -> right (1) -> left (0) -> left (0): Code = 100 (3 bits)
- To reach 'D': Root -> right (1) -> left (0) -> right (1): Code = 101 (3 bits)

(b) To encode the message 'AABBCD':
- 'A' occurs 2 times: \(2 \times 1 \text{ bit} = 2 \text{ bits}\)
- 'B' occurs 2 times: \(2 \times 2 \text{ bits} = 4 \text{ bits}\)
- 'C' occurs 1 time: \(1 \times 3 \text{ bits} = 3 \text{ bits}\)
- 'D' occurs 1 time: \(1 \times 3 \text{ bits} = 3 \text{ bits}\)
Total bits = \(2 + 4 + 3 + 3 = 12 \text{ bits}\).

Award up to 4 marks:
- 1 mark for correctly identifying the codes for 'A' (0) and 'B' (11).
- 1 mark for correctly identifying the codes for 'C' (100) and 'D' (101).
- 1 mark for showing correct working for the bit calculation of the message 'AABBCD' (multiplying frequencies by code lengths).
- 1 mark for the correct total of 12 bits.

To achieve 5-6 marks, the candidate must produce a well-structured response that covers environmental issues, ethical issues, and realistic ways to minimize these impacts, linking them to the context of the school.

**Example Answer:**
Replacing 30 computers creates significant environmental and ethical challenges. Environmentally, if the school sends these computers to a landfill, toxic chemicals such as lead and mercury can leach into local ecosystems, poisoning soil and water. This also contributes to the rising global volume of e-waste. Ethically, exporting this waste to developing nations often exposes vulnerable workers to health hazards during unsafe salvage processes. Furthermore, discarding functioning machines worsens the digital divide, as poorer communities are deprived of technology.

To minimize this impact, the school should first consider if upgrading components, such as RAM or storage, would make the current machines usable. If they must be replaced, the school can donate them to local charities or lower-income schools to bridge the digital divide. Finally, any completely unusable machines should be sent to a certified e-waste recycler that guarantees safe extraction of materials, rather than dumping them.

**Level 1 (1–2 marks):**
* The response identifies basic environmental or ethical issues (e.g., "landfills get full" or "it is bad for the environment").
* Suggests simple ways to minimize impact with little or no explanation.
* The response lacks structure and computer science terminology is used loosely.

**Level 2 (3–4 marks):**
* The response explains some environmental and/or ethical issues, showing an understanding of e-waste or recycling risks.
* Suggests practical ways to minimize impact (e.g., donating, recycling safely).
* The response is mostly structured, and some computer science terminology is used appropriately.

**Level 3 (5–6 marks):**
* The response provides a balanced and detailed discussion of both environmental and ethical issues.
* Offers a range of realistic solutions tailored to a school scenario (e.g., donation, upgrade, certified recycling).
* The response is well-structured, coherent, and uses precise terminology throughout.

To achieve 5-6 marks, the candidate must compare both algorithms across all three aspects (time, space, and dataset suitability) and provide a justified final recommendation.

**Example Answer:**
The developer should definitely choose merge sort over bubble sort. In terms of time efficiency, bubble sort has an average and worst-case time complexity of \( O(n^2) \). For 1,000,000 names, this would take hundreds of billions of operations, making it extremely slow and completely impractical. In contrast, merge sort is a divide-and-conquer algorithm with a time complexity of \( O(n \log n) \), meaning it can sort the names in seconds.

When considering space efficiency, bubble sort is superior because it sorts in-place, requiring \( O(1) \) auxiliary memory. Merge sort requires \( O(n) \) auxiliary memory to hold temporary lists as they are split and merged. However, 1,000,000 names would only occupy a few megabytes of RAM, which modern computers can easily handle. Therefore, the time saving of merge sort far outweighs the memory drawback, making merge sort the only viable option for this dataset.

**Level 1 (1–2 marks):**
* Identifies simple facts about bubble sort (e.g., "it swaps adjacent items") and/or merge sort (e.g., "it splits lists in half").
* Makes a simple recommendation with basic or no justification.

**Level 2 (3–4 marks):**
* Explains the difference in time efficiency (e.g., mentioning that merge sort is faster or referencing algorithmic complexity like \( O(n^2) \) vs \( O(n \log n) \)).
* Mentions space efficiency or memory usage for at least one of the algorithms.
* Provides a logical recommendation based on one of these factors.

**Level 3 (5–6 marks):**
* Provides a detailed, balanced comparison of both algorithms covering time efficiency, space complexity, and dataset size suitability.
* Uses correct terminology (e.g., in-place, divide-and-conquer, auxiliary memory, Big O notation).
* Reaches a fully justified conclusion pointing out why merge sort is chosen despite its higher memory usage.