Cambridge IGCSE Computer Science 0478/12 is the May/June 2026 Paper 1 (Computer Systems), variant 2. It lasts 1 hour 45 minutes, is worth 75 marks and is sat by every 0478 candidate. It had seven compulsory questions and no calculator, and Question 3, on a video streaming service, carried 22 marks on its own. This guide goes through the paper question by question and shows what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of both 0478 papers from this series, see IGCSE Computer Science 0478 May/June 2026 past papers: what came up and how it was marked.

0478/12 at a glance

Question	Context and topic	Marks	Main command word
1	Smartwatch: input devices, the bit, two’s complement, sensors	6	Tick, give, identify
2	The CPU: its role, registers, the MAR and MDR in the fetch, clock speed	8	State, explain, describe
3	Video streaming service: hexadecimal, packet switching, storage, compression	22	Convert, give, explain
4	Odd parity, other error detection methods, encryption	10	Give, explain, state
5	Cyber security threats, access levels	10	Complete, explain
6	Flower-identifying app: images in binary, AI, cloud storage	10	Explain, give
7	Vending machine: low-level languages, assemblers, IDE functions	9	State, identify, give
	Total	75

Question-by-question walkthrough

Question 1: a smartwatch (6 marks)

• Microphone (1 mark): an input device (box A).
• Smallest unit of data (1 mark): the bit.
• Two’s complement base (1 mark): base 2 (box A). Ticking more than one box scores nothing, here and on every tick-box part of the paper.
• 32 degrees in 8-bit two’s complement (1 mark): 00100000. The scheme says leading zeros must be present, so “100000” scores nothing.
• Minus 14 degrees (1 mark): 11110010. Write 14 as 00001110, flip every bit to get 11110001, then add 1. If the answer line is blank, a clearly correct answer in the working space can still earn the mark.
• A sensor in a smartwatch (1 mark): accepted answers included accelerometer, light, temperature, proximity, infra-red, acoustic, magnetic field, moisture, humidity and pressure. Motion, sound, touch and heart rate were all marked too vague, and describing what a sensor does (“it measures light”) instead of naming it was not accepted.

Question 2: the CPU (8 marks)

• Role of the CPU (1 mark): to process data or execute instructions, or to carry out the fetch-decode-execute cycle. Writing “tasks”, “programs” or “information” in place of instructions was too vague.
• Two other registers (2 marks): program counter, current instruction register and accumulator. The MAR and MDR were given in the question, so they scored nothing. Correct registers outside the syllabus, such as an index or status register, were accepted.
• The MAR and MDR in a fetch (3 marks): the address of the value is sent to the MAR; it is the address of the data to be fetched from RAM; the value itself is then copied into the MDR. Saying “instruction” instead of “value” was given benefit of the doubt.
• A 3 GHz clock speed (2 marks): 3 billion (3 × 10⁹ was fine) and that this is the number of fetch-decode-execute cycles, or clock cycles, carried out per second. You need both the number and what it counts.

Question 3: a video streaming service (22 marks)

• The incorrect statement about hexadecimal (1 mark): D, that hexadecimal uses less storage space than binary. The computer still stores the code as binary; hexadecimal only makes it shorter and easier for people to read and debug.
• Hex to binary (2 marks): 152 is 0001 0101 0010 and 30F is 0011 0000 1111. Convert each hex digit to its own 4-bit group. Missing leading zeros were penalised only once across the two answers.
• Binary to hex (2 marks): 010001110101 is 475 and 101011101000 is AE8. Split into groups of four from the right.
• Hex to denary (2 marks): 0A2 is 162 (10 × 16 + 2) and 104 is 260 (256 + 4).
• Parts of a packet (2 marks): the header and the trailer, alongside the payload.
• The router in packet switching (3 marks): the scheme listed eight points. The router reads the packet’s destination IP address, forwards the packet towards its destination, decides the route each packet takes, often the fastest or least busy, and uses and updates a routing table. It can also check a packet’s hop count and discard packets that reach their limit. Two notes matter. A general statement such as “each packet takes a different path” scores nothing unless the router is doing the choosing. And one sentence such as “it chooses the fastest route” earns two points, one for choosing the route and one for “fastest”, but “best path” alone was too vague.
• Two features of magnetic storage (2 marks): it uses electromagnets, it has platters, the platters are divided into tracks and sectors, and it has a read/write head. “Uses magnets” was too vague.
• Faster storage (1 mark): solid-state storage (SSD).
• Why lossless was chosen (2 marks): the original video can be recovered exactly; quality is not reduced; storage space is not a concern. Just restating what lossless does (“no data is permanently removed”) was not enough, and “closer to the original” was marked wrong, because lossless gives back the original exactly.
• One advantage of lossy (1 mark): smaller files, less storage space, faster transmission, less bandwidth, or less buffering when streaming.
• How lossless compression works (4 marks): a compression algorithm is used, such as run-length encoding; repeated frames or pixels (or repeated samples in the sound) are identified and indexed; redundant data is removed, for example by recording only the differences between one frame and the next. “Repeating data” or “repeating patterns” was too vague: say what repeats. Other correct lossless methods such as Huffman encoding were also accepted.

Question 4: parity, error detection and encryption (10 marks)

• Odd parity bits (3 marks): all three bytes needed a parity bit of 1. Count the 1s in each 7-bit byte (4, 6 and 2, all even) and add a 1 to make the total odd.
• Why an error can be missed (2 marks): bits have swapped places, or an even number of bits have flipped, so the count of 1s is still odd and the byte still passes the check. “An even number of errors” was too vague, and “the number of 1s is correct” was not enough on its own.
• Two other error detection methods (2 marks): parity block check, even parity, checksum, echo check and automatic repeat request (ARQ). A check digit was rejected, because it checks data entry, not transmission. A cyclic redundancy check was accepted.
• A method of encryption (1 mark): symmetric or asymmetric. Protocols that use encryption, such as SSL, were not accepted.
• An encryption key (1 mark): a string of characters or numbers used by the encryption algorithm to encrypt or decrypt data.
• The name for encrypted data (1 mark): ciphertext.

Question 5: cyber security (10 marks)

• Threats table (5 marks, one per gap): the missing terms were brute-force attack, distributed denial of service (DDoS) and data interception (packet sniffing); the missing descriptions were for hacking (a person gains unauthorised access to a device, account or system) and spyware (malware that records the user’s key presses). Two traps: describing hacking as “stealing or changing data without permission” was too vague because it doesn’t mention gaining access, and saying spyware just “tracks the user” was too vague unless you named what it records, such as key presses, webcam footage or GPS location.
• How access levels help (5 marks): they restrict each user’s access to data by giving different permissions (read-only, read/write, no access, administrator); an attacker who gets into an account can then only reach what that user could reach; this stops them deleting, editing or corrupting data, installing malware without administrator rights, or viewing confidential data. “Different access levels” was too vague for the permissions point: name a permission. A named example of confidential data, such as company financial accounts, helps; “personal data” alone was too vague.

Question 6: a flower-identifying app (10 marks)

• Image to binary (3 marks): the image is broken down into pixels, each pixel is one colour, and each colour has a binary value. Simply saying the image “has pixels” was not enough. Resolution and colour depth earned nothing here, because they don’t explain the conversion.
• How AI improves identification (4 marks): machine learning, where the app adapts its own rules or data; image recognition that finds patterns in different flowers; feedback from users to store which identifications were right or wrong; and training on pictures of flowers to learn more of their features. “It can learn and adapt” on its own was not enough, and “learns from past mistakes” was too vague.
• Cloud storage advantage (1 mark): the data doesn’t take up space on the phone, it doesn’t need downloading again on a new phone, or it is still there if the phone breaks or is lost. “Can be accessed anywhere” was too vague, since data on the phone can be too. Answers about unlimited storage, maintenance or security were not relevant to this scenario.
• Two disadvantages (2 marks): you need an internet connection (a slow one makes the app hard to use), the app fails if the cloud service is down, you rely on a third party to keep your data secure, and access may be slower than local storage. Cost was not accepted, and “Wi-Fi connection” was too vague.

Question 7: low-level languages and IDEs (9 marks)

• What a low-level language is (1 mark): one that is closer to the language a computer processes, or to machine code. “Understood by the computer” was not accepted, nor “uses mnemonics” (not all low-level languages do), a benefit such as controlling hardware, or a description of what it is not.
• Two examples (2 marks): assembly language, machine code (binary) and object code.
• Translator (1 mark): an assembler.
• Why choose low-level (2 marks): the program doesn’t need to be portable; it can directly control the hardware; it runs faster; it can be written for specific hardware; it uses memory more efficiently; it takes less storage space. “Can communicate with hardware” was too vague, because a high-level language can too; the point is that the control is direct.
• Three IDE functions from a list (3 marks): auto-correction, run-time environment and prettyprint. The other five options, such as IP and MAC addresses, were distractors.

Where the marks are on 0478/12

By syllabus area:

Area	Questions	Marks
Data representation, including images and compression	1(b), 1(c), 3(a)–(d), 3(g), 6(a)	21
Data transmission, error detection and encryption	3(e), 4	15
Hardware and storage, including cloud storage	1(a), 1(d), 2, 3(f), 6(c)	16
Software: languages, translators, IDEs	7	9
Cyber security	5	10
Artificial intelligence	6(b)	4
Total		75

• “Explain” parts carried 24 marks: 2(c), 3(e)(ii), 3(g)(iii), 4(b)(i), 5(b), 6(a) and 6(b). These are point-marked, so each separate, specific point is a mark.
• Conversions and binary answers carried 11 marks with no calculator: 1(c)(ii) and (iii), 3(b), 3(c), 3(d) and 4(a). These should be full marks with practice.
• Question 3 alone is 22 marks, almost a third of the paper, and spans four topics.

How to self-mark 0478/12 with the mark scheme

1. Learn the symbols. A single slash separates alternative words within a point, a double slash separates alternative answers, underlined words must appear, and “max” caps a list. Examiner notes use TV for too vague, NE for not enough and BOD for benefit of the doubt.
2. Check whether a part says “mark first answer given”. For those parts, such as the registers in 2(b) and the sensor in 1(d), only your first answer on each line counts, so extra answers can’t rescue a wrong one.
3. Read the guidance column for TV and NE notes. Many answers that sound right, such as “motion sensor” or “repeating data”, are listed as too vague.
4. Check your binary answers digit by digit. Missing leading zeros cost the 1(c) marks; in 3(b) they cost one mark only.
5. For the explain parts, tick each numbered point you made and stop at the maximum. Two points that say the same thing earn one mark.
6. Log each lost mark as missing knowledge, a vague answer or a misread question, and redo those parts a week later.

Sit 0478/12 as a timed mock

The paper allows 105 minutes for 75 marks, about 1.4 minutes per mark. A workable plan:

Block	Marks	Time
Question 1	6	7 minutes
Question 2	8	11 minutes
Question 3	22	30 minutes
Question 4	10	14 minutes
Question 5	10	14 minutes
Question 6	10	14 minutes
Question 7	9	11 minutes
Check conversions and tick boxes	–	4 minutes
Total	75	105 minutes

Do the conversions in the working space and copy them onto the answer line, so a slip in copying still leaves a correct answer for the examiner to find.

Other May/June 2026 0478 papers

• 0478 Paper 22 (Algorithms, Programming and Logic)

Practise and download

• Download 0478/12 and its mark scheme on the Learning Portal →
• All 0478 May/June 2026 papers: overview and marking patterns
• Try a free IGCSE Computer Science quiz

Need help with a 0478 topic? Book a free trial lesson with an IGCSE Computer Science tutor →

Frequently asked questions

What topics were on 0478 Paper 12 May/June 2026? +

Seven questions: a smartwatch (input devices, the bit, two's complement and sensors); the CPU and its registers; a video streaming service (hexadecimal, packet switching, storage and compression); odd parity, error detection and encryption; cyber security threats and access levels; a flower-identifying app (images in binary, AI and cloud storage); and low-level languages and IDEs.

How many marks is 0478/12 and how long is it? +

It is out of 75 marks and lasts 1 hour 45 minutes, with no calculator allowed. Question 3, on a video streaming service, is the biggest at 22 marks.

What were the two's complement answers on 0478/12 2026? +

32 degrees is 00100000 and minus 14 degrees is 11110010. Both answers needed all eight bits, so leading zeros had to be written.

What were the hexadecimal conversion answers in 0478/12? +

152 is 0001 0101 0010 and 30F is 0011 0000 1111 in binary; 010001110101 is 475 and 101011101000 is AE8 in hexadecimal; 0A2 is 162 and 104 is 260 in denary.

Which answers were too vague on 0478/12 2026? +

Examples include motion, sound, touch or heart rate as a smartwatch sensor, 'uses magnets' for magnetic storage, 'repeating data' for lossless compression, 'learn from past mistakes' for AI and 'can be accessed anywhere' for cloud storage.

Where can I download 0478/12 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge IGCSE Computer Science 0478 past papers page.

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