IGCSE Computer Science 0478 Paper 1 Mock | 10 Questions, 100 Marks, 1 Hour 45 Minutes

Convert the following values, showing your working.

(a) Convert denary 202 to an 8-bit binary number.
(b) Convert hexadecimal 2F to denary.
(c) Convert binary 10110110 to hexadecimal.
(d) Represent −46 in 8-bit two’s complement and explain the meaning of the most significant bit.
(e) Explain why computers use two’s complement rather than sign-and-magnitude for negative numbers.

Marking notes:
(a) 11001010 (2 marks: repeated division by 2, read upwards).
(b) 47 (2 marks: 2×16＋15).
(c) Split into 1011 and 0110 → B6 (2 marks, 1 each).
(d) 256−46＝210 → 11010010 (2 marks); the leading 1 marks a negative value and the other seven bits give the magnitude (1 mark).
(e) Addition and subtraction share one circuit and zero has a single representation, so no +0 and −0 (1 mark).

A sound is recorded at 44,100 Hz, 16-bit, 2 channels for 30 seconds; a bitmap image is 1024×768 pixels with a colour depth of 24 bits.

(a) Explain what is meant by the sample rate.
(b) Calculate the uncompressed size of the sound file in bytes and MiB (1 MiB＝1,048,576 bytes).
(c) Calculate the uncompressed size of the image in bytes and MiB.
(d) Compare the two sizes and state the ratio.
(e) Give one way to reduce the size of the sound file and state the cost of doing so.

Marking notes:
(a) The number of amplitude measurements taken each second (2 marks).
(b) 44,100×16×2×30÷8＝5,292,000 bytes (3 marks: 2 for the working, 1 for the value); ÷1,048,576＝5.05 MiB (1 mark).
(c) 1024×768×24÷8＝2,359,296 bytes＝2.25 MiB (2 marks).
(d) The sound file is 2.24 times larger (1 mark).
(e) Lower the sample rate or bit depth, or use lossy compression such as MP3; high frequencies and detail are lost and the original cannot be restored (1 mark).

A processor runs at 3.2 GHz with an average of 4 cycles per instruction (CPI) and executes 2.5×10⁹ instructions. Cache access takes about 4 ns and main memory about 60 ns.

(a) Describe what happens in each stage of the fetch–decode–execute cycle.
(b) State the role of the program counter (PC), memory address register (MAR) and memory data register (MDR).
(c) Calculate the total number of clock cycles and the time taken in seconds.
(d) Calculate how many times slower main memory is than cache.
(e) Explain why increasing cache size often improves performance more than raising the clock speed.

Marking notes:
(a) Fetch: the instruction at the address in the PC is read from memory; decode: the control unit interprets the opcode and operands; execute: the ALU performs the operation and the result is written back (3 marks).
(b) PC holds the address of the next instruction; MAR holds the address being accessed; MDR holds the data read or written (3 marks).
(c) 4×2.5×10⁹＝10000000000 cycles (1 mark); ÷3.2×10⁹＝3.125 s (2 marks).
(d) 60÷4＝15 times (1 mark).
(e) Higher clock speeds are limited by heat and power, and memory cannot keep up; a larger cache raises the hit rate and cuts the wait for main memory (1 mark).

A computer has 8 GiB of main memory with a page size of 4 KiB and a 32-bit address bus. A program needs 12 GiB of memory to run.

(a) Explain the difference between RAM and ROM.
(b) Calculate how many pages the RAM contains.
(c) With 4 KiB pages, how many bits of the 32-bit address are the offset and how many are the page number?
(d) Explain how virtual memory lets the program run and calculate how many GiB must be held on disk.
(e) Explain a page fault and thrashing.

Marking notes:
(a) RAM is read/write and loses its contents when power is removed; ROM is read-only and keeps its contents (2 marks).
(b) 8×1,024×1,024÷4＝2,097,152 pages (2 marks).
(c) Offset 12 bits (2^12＝4,096 bytes, 1 mark); page number 20 bits (2 marks).
(d) The operating system keeps some pages on disk and loads them when needed, so only the working pages must be in RAM (1 mark); 12−8＝4 GiB stays on disk (2 marks).
(e) A page fault occurs when the required page is not in RAM and must be loaded from disk; thrashing is constant swapping when memory is too small, which lowers CPU utilisation (1 mark).

A 10-bit ADC digitises a sensor signal from 0 to 5 V at a sample rate of 8,000 Hz.

(a) Name two input devices and two output devices (actuators).
(b) Calculate the number of levels a 10-bit ADC can represent.
(c) Calculate the voltage represented by each level in mV, to two decimal places.
(d) If a 12-bit ADC were used, what would the number of levels and the voltage per level be, and what is the benefit?
(e) Calculate the number of samples taken in 5 seconds and explain why too low a sample rate causes distortion.

Marking notes:
(a) Input: keyboard, mouse, temperature sensor, LDR (any two, 2 marks); output: monitor, printer, motor, buzzer (any two, 2 marks).
(b) 2^10＝1,024 levels (1 mark).
(c) 5÷(1024−1)＝0.00489 V ≈ 4.89 mV (2 marks).
(d) 2^4096＝4,096 levels and 5÷4095＝0.00122 V ≈ 1.22 mV (2 marks); resolution is higher and quantisation error smaller (1 mark).
(e) 8,000×5＝40,000 samples (1 mark); when the sample rate is below twice the highest frequency, high frequencies are aliased to low ones and the signal is distorted (1 mark).

A school network uses 192.168.10.0/24 and is split into /26 subnets. One computer has the address 192.168.10.130.

(a) Give one advantage and one disadvantage of star topology compared with mesh.
(b) Write the subnet mask for /26.
(c) Calculate the number of addresses and the number of usable hosts in each subnet.
(d) State the network address and the broadcast address of the subnet the computer belongs to.
(e) Explain the difference between a MAC address and an IP address, and what ARP does.

Marking notes:
(a) Star: one faulty node does not affect the others, but a failed central switch stops the whole network (2 marks); mesh: many paths give high fault tolerance but cabling and port costs are high (awarded as the comparison, 1 mark).
(b) 255.255.255.192 (2 marks).
(c) 64 addresses per subnet (1 mark); 62 usable hosts (2 marks: network and broadcast addresses excluded).
(d) Network address 192.168.10.128 (1 mark); broadcast address 192.168.10.191 (1 mark).
(e) A MAC address is burned in by the manufacturer and used within one segment, while an IP address is assigned by the network and routed between networks (1 mark); ARP resolves an IP address to a MAC address (1 mark).
(A /24 can hold 4 /26 subnets; credit any correct statement of this.)

A 5 MiB file is sent over a TCP/IP network. Each packet carries at most 1,460 bytes of payload and a 40-byte header.

(a) Describe the process of packet switching.
(b) Calculate the minimum number of packets needed.
(c) Calculate the extra data added by all the headers in bytes and as a percentage, to two decimal places.
(d) Compare TCP and UDP with one use for each.
(e) Explain why packets may arrive out of order and how the receiver reassembles them.

Marking notes:
(a) The file is split into packets, each with source and destination addresses and a sequence number, and routers forward them hop by hop, so packets may take different routes (2 marks).
(b) 5×1,024×1,024＝5,242,880 bytes; ÷1,460＝3,592 packets (2 marks, rounded up).
(c) 3,592×40＝143,680 bytes (2 marks); ÷5,242,880×100＝2.74% (2 marks).
(d) TCP: web pages, email, downloads, needing reliable ordered delivery; UDP: streaming, online games, DNS, where low latency matters more than a few lost packets (1 mark).
(e) Packets take different routes and meet congestion, so they arrive in any order; the receiver uses sequence numbers to reorder them, discards duplicates and requests retransmission of missing packets (1 mark).

A company encrypts data with a 128-bit (2¹²⁸) symmetric key. Assume an attacker can test 10⁹ keys per second.

(a) Name three common network threats and give an example of each.
(b) Explain the role of a firewall and of a proxy server in protecting a network.
(c) Calculate the size of the key space and estimate the time to try every key, in years (scientific notation is acceptable).
(d) Explain why passwords should be stored as salted hashes rather than encrypted.
(e) Explain how two-factor authentication (2FA) reduces the risk when a password is stolen.

Marking notes:
(a) Phishing emails that steal login details; DDoS flooding a server with requests; man-in-the-middle interception; malware or ransomware (any three, 3 marks).
(b) A firewall filters traffic by rules and blocks unauthorised ports (1 mark); a proxy server makes requests for internal users, hides internal IP addresses and can cache and filter content (1 mark).
(c) 2¹²⁸ ≈ 3.4×10³⁸ keys (1 mark); ÷10⁹ ≈ 3.4×10²⁹ s ≈ 1.1×10²² years (2 marks).
(d) A hash cannot be reversed, so a stolen database does not reveal passwords; a salt makes identical passwords hash differently and defeats rainbow tables and bulk comparison (2 marks).
(e) A one-time code or biometric is also required, so the stolen password alone cannot be used to log in (1 mark).

A company uploads a 20 GiB backup to the cloud over a 50 Mbps link; cloud storage costs $0.023 per GB per month. A blockchain requires each block hash to begin with three hexadecimal zeros.

(a) Give two advantages and one disadvantage of cloud storage.
(b) Calculate the upload time in minutes and seconds.
(c) Calculate the cost of storing 500 GB for one year in US dollars.
(d) Calculate the average number of attempts needed to find a valid hash and explain proof of work.
(e) Explain why a blockchain is hard to alter and give two benefits of a distributed ledger.

Marking notes:
(a) Advantages: scales on demand, access from anywhere, no own hardware, provider handles backups (any two, 2 marks); disadvantage: depends on the internet, recurring fees, data subject to a third party’s policies and privacy law (any one, 1 mark).
(b) 20×1,024³×8＝171,798,691,840 bits; ÷50×10⁶＝3,436.0 s (2 marks)＝57 min 16 s (1 mark).
(c) 500×$0.023＝$11.50 per month (1 mark); ×12＝$138.00 per year (2 marks).
(d) 16³＝4,096 attempts (2 marks); miners keep changing the nonce until the hash meets the difficulty target, proving computational work (1 mark).
(e) Each block stores the previous block’s hash, so changing one block invalidates every later hash and a majority of nodes would have to be controlled (1 mark); a distributed ledger has no single point of failure and is verified by many parties (1 mark).

A piece of text has 250 characters, of which 60 are Chinese and the rest ASCII. The system stores it in UTF-8 (1 byte per ASCII character, 3 bytes per Chinese character).

(a) Explain the difference between a character set and a character encoding.
(b) Calculate the number of bytes the text occupies in UTF-8.
(c) Write the ASCII codes of ‘A’, ‘a’ and ‘0’, and calculate the difference between ‘a’ and ‘A’.
(d) How many characters can 7-bit ASCII represent, and how large is the Unicode code space?
(e) Explain why ASCII alone cannot handle multilingual documents and how UTF-8 stays compatible with older systems.

Marking notes:
(a) A character set is the collection of available characters with a code for each; an encoding is how those codes are stored as binary (2 marks).
(b) 190×1＋60×3＝370 bytes (3 marks: 2 for the working, 1 for the value).
(c) ‘A’＝65, ‘a’＝97, ‘0’＝48 (2 marks); ‘a’−’A’＝32 (1 mark).
(d) 2⁷＝128 characters (1 mark); Unicode has 0x110000＝17,907,986 code points (1 mark).
(e) ASCII has only 128 characters and cannot represent Chinese, Japanese or Arabic (1 mark); UTF-8 uses the same single-byte codes for ASCII, so old files and programs still read them (1 mark).