Mark Scheme

Section A — Structured Questions

1. (a) 0110 1010 = 64 + 32 + 8 + 2 = 106 (M1 for place values, A1). [2] (b) 75 = 64 + 8 + 2 + 1 = 0100 1011 (M1 A1). [2] (c) 0101 1001 + 0010 0110 = 0111 1111 (M1 for method/carrying, A1 correct answer; accept 127 in denary check). [3] (d) (up to 1) Because electronic components/transistors have two states (on/off, high/low voltage) which represent 1 and 0; binary is reliable to store and process. [1] (e) 1101 1110 → 1101 = D, 1110 = E → DE (M1 for splitting into nibbles, A1). [2]

2. (a) (up to 2) The CPU processes instructions and data / carries out the fetch–decode–execute cycle; it controls and coordinates the operations of the computer. [2] (b) (up to 4) Two differences: RAM is volatile (loses contents when power is off), ROM is non-volatile; RAM can be read from and written to, ROM is (usually) read-only; RAM holds programs/data currently in use, ROM holds start-up instructions (e.g. the bootloader/BIOS). 2 clear differences. [4] (c) (up to 3) Cache is a small, very fast memory close to the CPU; it stores frequently used instructions/data so the CPU can access them quickly instead of fetching from slower RAM, which speeds up processing. [3]

3. (a) (up to 2) A LAN covers a small geographical area (e.g. one building/site), usually owned by one organisation; a WAN covers a large geographical area (e.g. between cities/countries), often using third-party infrastructure. [2] (b) (up to 3) In a star topology, all devices are connected to a central switch/hub (2); advantage — if one cable/device fails the rest keep working / good performance as each device has its own connection (1). [3] (c) (up to 4) A router connects different networks together and directs/forwards data (packets) between them, e.g. connecting a LAN to the internet (2); protocols are needed because they are agreed rules that let devices (which may be different) communicate in a standard way, so data is sent and understood correctly (2). [4]

4. (a) Outputs top to bottom: 0, 0, 0, 1 (all four correct = 3; two or three correct = 2; one correct = 1). [3] (b) (up to 5) Linear search checks each item in turn from the start until the target is found (works on any list); binary search repeatedly checks the middle item and halves the search area, so it is much faster for large lists (up to 4). Condition: the list must be sorted/in order (1). [5]

5. (a) (up to 2) Malware = malicious software designed to damage or gain unauthorised access to a system (1); example — virus, worm, Trojan, spyware, ransomware (1). [2] (b) (up to 4) Two methods explained: strong passwords / user access levels (limit who can log in and what they can access); firewall (controls network traffic); anti-malware software; encryption of data. 1 + 1 development each. [4] (c) (up to 2) One issue: privacy — personal data could be misused, sold or stolen; data-protection law requires it to be used fairly and kept secure; consent. [2]

Section B — Extended Response

6. (a) (up to 6) Fetch–decode–execute cycle: Fetch — the next instruction is copied from memory (RAM) to the CPU (using the program counter/address); Decode — the control unit interprets/decodes the instruction to work out what to do; Execute — the instruction is carried out (e.g. a calculation in the ALU, or data moved); the cycle then repeats for the next instruction. 2 marks per stage. [6] (b) (up to 4) Two factors: clock speed (more cycles per second = faster); number of cores (more cores can process more instructions at once); cache size (more cache reduces slow trips to RAM). 1 + 1 development each. [4]

7. (up to 6) Discussion of environmental and ethical impacts, both sides:

• Negative environmental: energy use of devices/data centres; e-waste from discarded devices; resources used in manufacture.
• Positive environmental: technology can reduce travel (video calls), cut paper use, and help monitor/manage the environment.
• Ethical: privacy and data misuse; the digital divide (unequal access); effects on jobs/automation; but also improved access to information and communication. Levels: 5–6 = balanced discussion of both environmental and ethical impacts with benefits and drawbacks; 3–4 = several points, one area stronger; 1–2 = basic. [6]

Sample Answers with Examiner Commentary

Question 1(e) — Sample Answers

Grade A response. "To convert 1101 1110 to hexadecimal, I split the binary number into two nibbles (groups of four bits): 1101 and 1110. The first nibble, 1101, is 8 + 4 + 1 = 13 in denary, which is D in hex. The second nibble, 1110, is 8 + 4 + 2 = 14 in denary, which is E in hex. So the answer is DE."

Mark: 2/2. Examiner commentary: A clear, correct conversion using the standard and most reliable method: splitting the byte into two 4-bit nibbles and converting each to a single hex digit. Showing the denary value of each nibble (13 → D, 14 → E) makes the working transparent and would protect the method mark even if a slip occurred. Splitting into nibbles is exactly the technique examiners want to see, as it avoids the errors that come from converting the whole number at once.

Grade C response. "1101 1110 is 222 in denary. So in hex it is... I divided 222 by 16 and got about 13, so the answer is 13 or D something."

Mark: 0/2. Examiner commentary: The candidate's denary conversion of the whole byte (222) is actually correct, but the hex conversion then breaks down: dividing 222 by 16 gives 13 remainder 14, which the candidate does not complete, and the final answer "13 or D something" is not a valid hex number. The straightforward method is to split the binary into nibbles (1101 = D, 1110 = E) and read off DE directly, avoiding the division method altogether. No marks can be awarded without a correct final value, but the fix — nibble splitting — is simple.

Question 6(a) — Sample Answers

Grade A response. "The fetch–decode–execute cycle is how the CPU processes instructions. First, in the fetch stage, the CPU uses the program counter to find the address of the next instruction and copies that instruction from memory (RAM) into the CPU. Next, in the decode stage, the control unit interprets the instruction to work out what operation needs to be done and what data is needed. Finally, in the execute stage, the instruction is carried out — for example, the ALU performs a calculation, or data is moved or stored. The cycle then repeats for the next instruction, and this happens millions of times per second."

Mark: 6/6. Examiner commentary: A complete, correctly ordered description with two marks' worth of detail for each stage. The candidate names the components involved (program counter, control unit, ALU) and explains what actually happens at each step rather than just listing "fetch, decode, execute". Noting that the cycle repeats completes the picture. This depth — the role of each stage and the hardware involved — is what separates a full-mark answer from a bare list of the three words.

Grade C response. "The CPU fetches the instruction, then decodes it, then executes it. Then it does the next one."

Mark: 3/6. Examiner commentary: The candidate correctly names the three stages in the right order and knows the cycle repeats, which earns three marks. However, the question asks the candidate to describe the cycle, and there is no detail about what happens at each stage — fetching the instruction from memory using the program counter, the control unit decoding it, and the instruction being carried out (e.g. in the ALU) during execution. Adding a sentence explaining each stage would double the mark.

Question 4(b) — Sample Answers

Grade A response. "A linear search checks each item in a list one at a time, starting at the beginning, until it either finds the item it is looking for or reaches the end of the list. It works on any list, sorted or not, but it can be slow for a long list because it may have to check every item. A binary search is much faster: it looks at the middle item of the list, and if that is not the target, it discards the half of the list the target cannot be in and repeats the process on the remaining half, halving the search area each time. However, a binary search can only be used if the list is already sorted into order, because it relies on being able to decide which half to discard."

Mark: 5/5. Examiner commentary: A full-mark answer. The candidate explains both algorithms with enough detail to show a real understanding of how each works — checking items one by one for linear, and repeatedly halving the search area for binary — rather than just naming them. Crucially, it states the essential condition for binary search (the list must be sorted) and even explains why that condition is needed, which many candidates omit. Contrasting the speed of the two searches adds further quality.

Grade C response. "A linear search goes through the list one by one. A binary search is faster. The list has to be sorted for a binary search."

Mark: 3/5. Examiner commentary: The candidate correctly describes a linear search, states that binary search is faster, and knows the key condition that the list must be sorted, earning three marks. What is missing is any description of how a binary search actually works — checking the middle item and repeatedly halving the search area. "Faster" states the outcome without explaining the method. Adding a sentence describing the halving process would gain the remaining marks.