Question Simulator

O Level Computer Science (2210)

Data representation Banker

16 questions · about 11% of the marks we indexed · examined in 8 of 8 sessions.

Questions by year

2026

  • 2026 May/June · Paper 12 · Q1 (b, c(i), c(ii), c(iii))4 marksBinary addition, overflow, shifts and two's complement · Data storage units and file size calculations
  • 2026 May/June · Paper 12 · Q3 (a, b, c, d, g(i), g(ii), g(iii))14 marksLossy and lossless compression · Number systems and conversions
  • 2026 May/June · Paper 12 · Q6 (a)3 marksRepresenting text, sound and images

2025

  • 2025 Oct/Nov · Paper 12 · Q1 (a, b, c, d)6 marksNumber systems and conversions
  • 2025 Oct/Nov · Paper 12 · Q2 (a, b)7 marksBinary addition, overflow, shifts and two's complement
  • 2025 May/June · Paper 12 · Q1 (c, d, e, f, g)11 marksBinary addition, overflow, shifts and two's complement · Number systems and conversions
  • 2025 May/June · Paper 12 · Q2 (b(i), b(ii), c(i), c(ii), d)8 marksData storage units and file size calculations · Lossy and lossless compression · Representing text, sound and images

2024

  • 2024 Oct/Nov · Paper 12 · Q2 (a, b, c, d(i), d(ii))7 marksData storage units and file size calculations · Lossy and lossless compression
  • 2024 Oct/Nov · Paper 12 · Q3 (a(i), a(ii), b, c, d(i), d(ii), e)13 marksBinary addition, overflow, shifts and two's complement · Number systems and conversions
  • 2024 May/June · Paper 12 · Q1 (a, b, c, d)6 marksData storage units and file size calculations
  • 2024 May/June · Paper 12 · Q3 (a, b(i), b(ii), c, d)11 marksBinary addition, overflow, shifts and two's complement · Number systems and conversions

2023

  • 2023 Oct/Nov · Paper 12 · Q2 (a, b, c, d, e)10 marksBinary addition, overflow, shifts and two's complement · Number systems and conversions
  • 2023 Oct/Nov · Paper 12 · Q5 (a, b, c, d)5 marksLossy and lossless compression · Representing text, sound and images
  • 2023 May/June · Paper 12 · Q2 (a, b(i), b(ii), c, d)10 marksBinary addition, overflow, shifts and two's complement · Number systems and conversions
  • 2023 May/June · Paper 12 · Q3 (a, b(i), b(ii), b(iii))10 marksLossy and lossless compression · Representing text, sound and images

2022

  • 2022 Oct/Nov · Paper 12 · Q1 (f(i), f(ii), f(iii))10 marksNumber systems and conversions

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Number systems and conversions · Core · Paper 1 style · 3 marks
  1. (a) Convert the 8-bit binary number 10110110 to denary. [1]
  2. (b) Convert the 8-bit binary number 01011011 to denary. [1]
  3. (c) Convert the denary number 77 to an 8-bit binary number. [1]
Show answer
  1. (a) 182
    128 + 32 + 16 + 4 + 2 = 182.
  2. (b) 91
    64 + 16 + 8 + 2 + 1 = 91.
  3. (c) 01001101
    77 = 64 + 8 + 4 + 1.
Q2Number systems and conversions · Core · Paper 1 style · 3 marks
  1. (a) Convert the denary number 200 to hexadecimal. [1]
  2. (b) Convert the hexadecimal number 3F to denary. [1]
  3. (c) Convert the hexadecimal number 9A to 8-bit binary. [1]
Show answer
  1. (a) C8
    200 = 12 × 16 + 8 → C8.
  2. (b) 63
    3 × 16 + 15 = 63.
  3. (c) 10011010
    9 = 1001, A = 1010.
Q3Number systems and conversions · Core · Paper 1 style · 4 marks
  1. (a) A bus ticket machine in Lahore stores the fare 156 rupees. Convert 156 to an 8-bit binary number. [1]
  2. (b) Convert 156 to hexadecimal. [1]
  3. (c) Explain why computers store data in binary. [2]
Show answer
  1. (a) 10011100
    128 + 16 + 8 + 4.
  2. (b) 9C
    1001 = 9, 1100 = C (or 9 × 16 + 12).
  3. (c) Computers are made of switches/transistors (logic gates) that have only two states, on and off, which are represented by 1 and 0.
    1 mark: circuits/transistors have two states. 1 mark: these states represent 1 and 0 / all data must be converted to binary to be processed.
Q4Number systems and conversions · Core · Paper 1 style · 4 marks
  1. (a) Convert the 4-bit binary number 1101 to denary. [1]
  2. (b) Convert the denary number 9 to a 4-bit binary number. [1]
  3. (c) Convert the binary number 11111111 to hexadecimal and to denary. [2]
Show answer
  1. (a) 13
    8 + 4 + 1.
  2. (b) 1001
    8 + 1.
  3. (c) FF; 255
    1 mark each.
Q5Number systems and conversions · Standard · Paper 1 style · 4 marks
  1. (a) A sensor in a weather station in Gilgit stores readings as 12-bit binary numbers. One reading is 1011 0110 1101.Convert this reading to hexadecimal. [1]
  2. (b) Convert the same reading to denary. Show your working. [2]
  3. (c) State the largest denary value that can be stored in 12 bits. [1]
Show answer
  1. (a) B6D
    1011 = B, 0110 = 6, 1101 = D.
  2. (b) 2925
    M1 place values 2048 + 512 + 256 + 64 + 32 + 8 + 4 + 1 (or 11 × 256 + 6 × 16 + 13); A1 2925.
  3. (c) 4095
    2¹² − 1 = 4095.
Q6Number systems and conversions · Standard · Paper 1 style · 7 marks
  1. (a) A web designer in Dubai uses the colour code #FF8C00 for an orange heading. Each pair of hexadecimal digits gives the amount of red, green and blue (0–255).Convert each pair to denary. [3]
  2. (b) Give two other uses of hexadecimal in computer science. [2]
  3. (c) Explain why programmers often use hexadecimal instead of binary. [2]
Show answer
  1. (a) Red 255, green 140, blue 0.
    1 mark each: FF = 255; 8C = 8 × 16 + 12 = 140; 00 = 0.
  2. (b) MAC addresses; error codes / memory dumps (also IPv6 addresses, assembly language, URLs).
    1 mark each for any two of: MAC addresses; IPv6 addresses; error codes; memory dumps/locations; assembly language; URL encoding.
  3. (c) Hexadecimal is shorter (one digit represents 4 bits) so it is easier to read and remember, and fewer errors are made when copying it.
    1 mark: shorter / one hex digit = 4 bits. 1 mark: easier for humans to read/remember, fewer mistakes. (Not: uses less memory.)
Q7Number systems and conversions · Standard · Paper 1 style · 4 marks
  1. (a) A small factory counts the bottles it fills each day using a 16-bit register. On Monday it fills 45 000 bottles.Convert 45 000 to a 16-bit binary number. [2]
  2. (b) Convert your answer to hexadecimal. [1]
  3. (c) State the largest number of bottles the register can count. [1]
Show answer
  1. (a) 1010 1111 1100 1000
    M1 a correct method (e.g. place values 32768 + 8192 + 2048 + 1024 + 512 + 256 + 128 + 64 + 8); A1 correct 16 bits.
  2. (b) AFC8
    1010 = A, 1111 = F, 1100 = C, 1000 = 8.
  3. (c) 65 535
    2¹⁶ − 1.
Q8Number systems and conversions · Standard · Paper 1 style · 4 marks
  1. (a) Three values are stored in a program:P = 7D (hexadecimal)Q = 01111100 (binary)R = 126 (denary)Convert P and Q to denary. [2]
  2. (b) Write the three values in order from smallest to largest. [1]
  3. (c) Convert R to hexadecimal. [1]
Show answer
  1. (a) P = 125, Q = 124
    1 mark each: 7 × 16 + 13 = 125; 64 + 32 + 16 + 8 + 4 = 124.
  2. (b) Q (124), P (125), R (126)
    B1 Q, P, R.
  3. (c) 7E
    126 = 7 × 16 + 14.
Q9Number systems and conversions · Challenge · Paper 1 style · 5 marks
  1. (a) A printer displays the error code 1E7 (hexadecimal). Convert this code to denary. Show your working. [2]
  2. (b) Convert the error code to 12-bit binary. [1]
  3. (c) The printer manual lists error codes from 100 to 2FF in hexadecimal. Calculate how many different error codes this range contains. [2]
Show answer
  1. (a) 487
    M1 1 × 256 + 14 × 16 + 7; A1 487.
  2. (b) 0001 1110 0111
    1 = 0001, E = 1110, 7 = 0111.
  3. (c) 512
    M1 2FF − 100 + 1 = 767 − 256 + 1; A1 512.
Q10Number systems and conversions · Challenge · Paper 1 style · 5 marks
  1. (a) A value is stored in a 16-bit register as 0000 0011 1110 1000.(i) Convert it to denary.(ii) Convert it to hexadecimal. [3]
  2. (b) State how many hexadecimal digits are needed to represent any 16-bit binary number, and explain why. [2]
Show answer
  1. (a) (i) 1000 (ii) 03E8
    (i) M1 512 + 256 + 128 + 64 + 32 + 8; A1 1000. (ii) B1 03E8 (accept 3E8).
  2. (b) 4 digits, because each hexadecimal digit represents exactly 4 bits and 16 ÷ 4 = 4.
    1 mark: 4. 1 mark: one hex digit = 4 bits (a nibble).
Q11Number systems and conversions · Challenge · Paper 1 style · 4 marks
  1. (a) Bilal converts the hexadecimal number A5 to denary and writes 105. Explain the mistake he made and give the correct answer. [2]
  2. (b) Convert the hexadecimal number 2A3 to 12-bit binary. [1]
  3. (c) Convert 2A3 to denary. [1]
Show answer
  1. (a) He treated A as 10 and wrote it next to 5 (or used place value 10 instead of 16); correct answer is 10 × 16 + 5 = 165.
    1 mark: identifies error – A not multiplied by 16 / treated as denary digits '10' and '5'. 1 mark: 165.
  2. (b) 0010 1010 0011
    2 = 0010, A = 1010, 3 = 0011.
  3. (c) 675
    2 × 256 + 10 × 16 + 3 = 675.
Q12Binary addition, overflow, shifts and two's complement · Core · Paper 1 style · 3 marks
  1. (a) Add the 8-bit binary numbers 01011010 and 00110111. Give your answer as an 8-bit binary number. [2]
  2. (b) Convert your answer to denary to check it. [1]
Show answer
  1. (a) 10010001
    M1 correct carries shown; A1 10010001 (check: 90 + 55 = 145).
  2. (b) 145
    128 + 16 + 1.
Q13Binary addition, overflow, shifts and two's complement · Core · Paper 1 style · 4 marks
  1. (a) The register holds 00010110. Perform a logical left shift of two places. Give the 8-bit result. [1]
  2. (b) Convert the original value and the result to denary. [2]
  3. (c) State the effect of a logical left shift of two places on a positive binary number. [1]
Show answer
  1. (a) 01011000
    Bits move two places left; zeros enter on the right.
  2. (b) 22 and 88
    1 mark each.
  3. (c) It multiplies the number by 4 (2²), provided no 1s are lost.
    B1 multiply by 4.
Q14Binary addition, overflow, shifts and two's complement · Standard · Paper 1 style · 4 marks
  1. (a) Add the 8-bit binary numbers 10110101 and 01101100. Show your working. [2]
  2. (b) Explain why the result cannot be stored correctly in an 8-bit register. [2]
Show answer
  1. (a) 1 0010 0001 – the result needs 9 bits; the 8-bit register holds 00100001.
    M1 working with carries; A1 (1)00100001 (181 + 108 = 289).
  2. (b) The answer 289 is larger than 255, the largest value 8 bits can hold, so a ninth bit is needed; this is an overflow error and the carry bit is lost.
    1 mark: result greater than 255 / needs 9 bits. 1 mark: overflow error – the extra bit is lost so the stored value (33) is wrong.
Q15Binary addition, overflow, shifts and two's complement · Standard · Paper 1 style · 6 marks
  1. (a) Perform a logical right shift of three places on 11010000. Give the 8-bit result and its denary value. [2]
  2. (b) Perform a logical right shift of two places on 10110111. Give the 8-bit result and its denary value. [2]
  3. (c) Explain why the result in (b) is not exactly the original value divided by 4. [2]
Show answer
  1. (a) 00011010 = 26
    1 mark for 00011010; 1 mark for 26 (208 ÷ 8).
  2. (b) 00101101 = 45
    1 mark for 00101101; 1 mark for 45.
  3. (c) The original value is 183 and 183 ÷ 4 = 45.75; the two right-most bits (1 and 1) are shifted out and lost, so the fractional part is lost and the answer is rounded down.
    1 mark: 1s shifted off the right end are lost. 1 mark: so precision is lost / result 45 instead of 45.75.
Q16Binary addition, overflow, shifts and two's complement · Standard · Paper 1 style · 4 marks
  1. (a) Convert the denary number −45 to an 8-bit two's complement binary number. Show your working. [2]
  2. (b) Convert the 8-bit two's complement number 10101110 to denary. [2]
Show answer
  1. (a) 11010011
    M1 +45 = 00101101, invert = 11010010, add 1; A1 11010011.
  2. (b) −82
    M1 −128 + 32 + 8 + 4 + 2; A1 −82.
Q17Binary addition, overflow, shifts and two's complement · Challenge · Paper 1 style · 5 marks
  1. (a) State the range of denary values that can be stored in an 8-bit two's complement register. [1]
  2. (b) The two's complement numbers 01001011 and 00111100 are added in an 8-bit register. Give the 8-bit result. [1]
  3. (c) Convert the 8-bit result to denary as a two's complement number, and explain why it is not the correct answer to the addition. [3]
Show answer
  1. (a) −128 to +127
    B1.
  2. (b) 10000111
    B1.
  3. (c) −121; the correct total is 75 + 60 = 135, which is larger than +127, so overflow has occurred and the result appears negative.
    1 mark: −121. 1 mark: correct sum is 135. 1 mark: exceeds +127, so overflow / the sign bit has been changed.
Q18Binary addition, overflow, shifts and two's complement · Challenge · Paper 1 style · 5 marks
  1. (a) An 8-bit register holds 01101001. Perform a logical left shift of two places and give the 8-bit result. [1]
  2. (b) Convert the original and the result to denary. [2]
  3. (c) Explain why the result is not four times the original value. [2]
Show answer
  1. (a) 10100100
    B1.
  2. (b) 105 and 164
    1 mark each.
  3. (c) 105 × 4 = 420, which is larger than 255, the maximum for 8 bits; the two left-most bits (0 and 1) were shifted out and lost, so the 1 that was shifted out changes the value.
    1 mark: 420 too large for 8 bits / exceeds 255. 1 mark: most significant bit(s) (a 1) lost during the shift.
Q19Lossy and lossless compression · Core · Paper 1 style · 6 marks
  1. (a) Explain the difference between lossy and lossless compression. [2]
  2. (b) State whether lossy or lossless compression should be used for each file, and give a reason:(i) a spreadsheet of exam marks(ii) a holiday photograph being posted on social media [4]
Show answer
  1. (a) Lossy compression permanently removes some data, so the original file cannot be recovered exactly; lossless compression reduces file size without losing any data, so the original can be restored exactly.
    1 mark: lossy – data permanently removed, original cannot be restored. 1 mark: lossless – no data lost, original can be reconstructed exactly.
  2. (b) (i) Lossless – every number must be exactly correct; any lost data would corrupt the file. (ii) Lossy – the photo can be made much smaller and the small loss of quality is not noticeable to viewers.
    Each part: 1 mark for the type, 1 mark for the reason.
Q20Lossy and lossless compression · Core · Paper 1 style · 3 marks
  1. (a) State why compressing a computer program (an .exe file) must use lossless compression. [1]
  2. (b) Give two benefits of compressing files before emailing them. [2]
Show answer
  1. (a) If any data were lost, the program instructions would be corrupted and the program would not run.
    B1 every bit must be kept / lost data would stop the program working.
  2. (b) Smaller files transfer faster; they may fit within the email attachment size limit; less bandwidth/storage used.
    1 mark each for any two of: faster to send/upload; less bandwidth; fits attachment limits; less storage space used.
Q21Lossy and lossless compression · Standard · Paper 1 style · 5 marks
  1. (a) A text file contains this string of 26 characters:GGGGGGGGRRRRRYYYYYYYYYYYGGIt is compressed using run-length encoding (RLE), storing each run as a count followed by the character.Write the RLE version of the string. [2]
  2. (b) Each count and each character is stored in 1 byte. Calculate the size of the original string and the size of the compressed data, in bytes. [2]
  3. (c) State whether RLE is lossy or lossless. [1]
Show answer
  1. (a) 8G 5R 11Y 2G
    M1 correct runs identified; A1 all four counts correct and in order.
  2. (b) Original 26 bytes; compressed 8 bytes.
    1 mark each: 26 × 1 = 26; 4 runs × 2 bytes = 8.
  3. (c) Lossless.
    B1 – the original string can be rebuilt exactly.
Q22Lossy and lossless compression · Standard · Paper 1 style · 6 marks
  1. (a) A music-streaming app in Karachi compresses songs using lossy compression. Describe how lossy compression reduces the size of a sound file. [3]
  2. (b) Describe two ways lossy compression can reduce the size of an image file. [2]
  3. (c) Give one reason why the app compresses the songs. [1]
Show answer
  1. (a) Perceptual music shaping removes sounds that most people cannot hear, such as very high or low frequencies and quiet sounds played at the same time as louder ones; the sample rate and/or sample resolution can also be reduced.
    1 mark each for any three of: perceptual music shaping; removes frequencies outside the range of human hearing; removes quieter sounds masked by louder sounds; reduces sample rate; reduces sample resolution/bit depth; data removed permanently.
  2. (b) Reduce the resolution (fewer pixels); reduce the colour depth (fewer bits per pixel / fewer colours).
    1 mark each: reduce resolution / number of pixels; reduce colour depth. Accept: group similar colours together.
  3. (c) Smaller files stream faster and use less of the user's mobile data and storage.
    B1 faster download/streaming; less bandwidth/data used; less storage needed on the server or device.
Q23Lossy and lossless compression · Challenge · Paper 1 style · 5 marks
  1. (a) One row of a simple image is 32 pixels wide. Each pixel is stored in 1 byte. The row contains: 12 white, 3 black, 9 white, 1 red, 7 white pixels, in that order.It is compressed with RLE, where each run is stored as 1 byte for the count and 1 byte for the colour.Calculate the size of the row before and after compression, and the percentage reduction. [3]
  2. (b) Another row contains 8 pixels that alternate white, black, white, black, … Explain why RLE would make this row larger, not smaller. [2]
Show answer
  1. (a) Before: 32 bytes. After: 5 runs × 2 = 10 bytes. Reduction: 22 bytes = 68.75%.
    B1 32 bytes; B1 10 bytes; B1 68.75% (accept 68.8%).
  2. (b) Every run has length 1, so each pixel needs 2 bytes (count + colour) instead of 1: 16 bytes instead of 8.
    1 mark: no repeated adjacent values / each run is only 1 pixel. 1 mark: each 1-byte pixel becomes 2 bytes, so file doubles (8 → 16 bytes).
Q24Data storage units and file size calculations · Core · Paper 1 style · 3 marks
  1. (a) An image is 800 pixels wide and 600 pixels high, with a colour depth of 24 bits. Calculate the file size in bytes. Show your working. [2]
  2. (b) Convert your answer to kibibytes (KiB). [1]
Show answer
  1. (a) 1 440 000 bytes
    M1 800 × 600 × 24 ÷ 8; A1 1 440 000.
  2. (b) 1406.25 KiB
    1 440 000 ÷ 1024.
Q25Data storage units and file size calculations · Standard · Paper 1 style · 3 marks
  1. (a) A 30-second mono voice recording for a language app is sampled at 44 100 Hz with a sample resolution of 16 bits. Calculate the file size in bytes. [2]
  2. (b) Convert the file size to mebibytes (MiB). Give your answer to 2 decimal places. [1]
Show answer
  1. (a) 2 646 000 bytes
    M1 44 100 × 16 × 30 ÷ 8; A1 2 646 000.
  2. (b) 2.52 MiB
    2 646 000 ÷ 1024 ÷ 1024 = 2.523…
Q26Data storage units and file size calculations · Standard · Paper 1 style · 5 marks
  1. (a) Put these units in order from smallest to largest: GiB, KiB, TiB, byte, MiB, nibble, bit, PiB. [2]
  2. (b) Convert 3 GiB to MiB. [1]
  3. (c) A camera's photos are each 6 MiB. Calculate how many complete photos fit on a 16 GiB memory card. Show your working. [2]
Show answer
  1. (a) bit, nibble, byte, KiB, MiB, GiB, TiB, PiB
    2 marks fully correct; 1 mark with one pair swapped.
  2. (b) 3072 MiB
    3 × 1024.
  3. (c) 2730
    M1 16 × 1024 ÷ 6 = 2730.67; A1 2730 (rounded down).
Q27Data storage units and file size calculations · Challenge · Paper 1 style · 6 marks
  1. (a) A drone in a crop survey in Sindh takes images of 4000 × 3000 pixels with a colour depth of 24 bits. The images are not compressed. Calculate the size of one image in bytes. [2]
  2. (b) Calculate how many complete images fit on a 32 GiB memory card. Show your working. [3]
  3. (c) The colour depth is reduced to 8 bits. State the effect on the file size of each image. [1]
Show answer
  1. (a) 36 000 000 bytes
    M1 4000 × 3000 × 24 ÷ 8; A1 36 000 000.
  2. (b) 954
    M1 32 × 1024 × 1024 × 1024 = 34 359 738 368 bytes; M1 ÷ 36 000 000 = 954.4…; A1 954.
  3. (c) It becomes one third of the size (12 000 000 bytes).
    B1 reduced to a third / 12 000 000 bytes.
Q28Representing text, sound and images · Core · Paper 1 style · 6 marks
  1. (a) Explain what is meant by a character set. [2]
  2. (b) Explain why a messaging app used in Pakistan stores text in Unicode rather than ASCII. [3]
  3. (c) State one drawback of using Unicode. [1]
Show answer
  1. (a) A list of all the characters a computer can represent, with a unique binary code for each character.
    1 mark: a defined list of characters recognised by the computer. 1 mark: each has a unique binary value/code.
  2. (b) ASCII uses 7 (or 8) bits so it can only represent 128 (or 256) characters, mostly English letters; Unicode uses more bits per character so it can represent far more characters, including Urdu script and emojis.
    1 mark each: ASCII limited to 128/256 characters; Unicode uses more bits per character / represents many more characters; so includes Urdu/other language characters and emojis.
  3. (c) Each character uses more bits, so text files are larger.
    B1 larger file size / more storage.
Q29Representing text, sound and images · Standard · Paper 1 style · 7 marks
  1. (a) Describe how sound is converted into digital data. [3]
  2. (b) Define sample rate and sample resolution. [2]
  3. (c) Explain the effect of increasing the sample rate on a recording. [2]
Show answer
  1. (a) The sound wave (analogue) is sampled at regular time intervals; the amplitude (height) of the wave is measured at each sample; each measurement is stored as a binary number.
    1 mark each: wave sampled at regular intervals; amplitude measured at each sample; value converted to/stored as binary (ADC).
  2. (b) Sample rate: the number of samples taken per second. Sample resolution: the number of bits used to store each sample.
    1 mark each.
  3. (c) The digital sound is closer to the original analogue sound (better accuracy/quality), but the file size increases.
    1 mark: more accurate / higher quality. 1 mark: larger file size.
Q30Representing text, sound and images · Challenge · Paper 1 style · 7 marks
  1. (a) Define the terms image resolution and colour depth. [2]
  2. (b) An image uses a colour depth of 6 bits. Calculate how many different colours each pixel can be. [1]
  3. (c) A designer changes an image from 8-bit to 16-bit colour depth without changing its resolution. Calculate how many different colours are now possible, and state the effect on the file size. [2]
  4. (d) Explain why increasing the resolution of an image from 1280 × 720 to 1920 × 1080 increases the file size. [2]
Show answer
  1. (a) Resolution: the number of pixels in the image (width × height). Colour depth: the number of bits used to represent the colour of each pixel.
    1 mark each.
  2. (b) 64
    2⁶ = 64.
  3. (c) 65 536 colours; the file size doubles.
    1 mark: 2¹⁶ = 65 536. 1 mark: file size doubles (twice as many bits per pixel).
  4. (d) There are more pixels (2 073 600 instead of 921 600), and each pixel needs data to be stored, so more bits are needed.
    1 mark: more pixels. 1 mark: each pixel's colour is stored so more data/bits overall (2.25 times as many).

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Number systems and conversions Banker8 questions · 8/8 sessions
Binary addition, overflow, shifts and two's complement Banker7 questions · 7/8 sessions
Lossy and lossless compression 5 questions · 5/8 sessions
Data storage units and file size calculations 4 questions · 4/8 sessions
Representing text, sound and images 4 questions · 4/8 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Number systems and conversions Banker538/8
Binary addition, overflow, shifts and two's complement Banker357/8
Lossy and lossless compression 205/8
Data storage units and file size calculations 144/8
Representing text, sound and images 134/8

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