OCR GCSE Computer Science Computer systems (01), June 2025: Question 1
14 marks · Medium difficulty · Short Answer
Complete sentences on sound sampling, define a pixel, calculate the file size of 10 bitmap images in kilobytes, determine the minimum bits for 240 colours, and justify selecting optical or solid-state storage to transfer files.
Practise this questionQuestion
Question text
1 A student is creating a presentation containing images and sound.
(a) One of the files is an audio recording.
Complete the description of how a computer stores sound. Fill in the gaps using the given list of
terms. Not all terms will be used.
analogue colour depth higher repeated smaller
binary digital lower sample rate unique
bit depth height measuring sampling
byte hertz recording seconds
An … sound wave needs to be converted into a digital
sound wave.
Sound … is when the amplitude of the sound wave is
measured at set intervals.
The … is the number of times a second the sound wave
is measured. This is given in Hertz.
Each amplitude is given a … binary number. The
number of bits allocated to each sample is the …
The … the number of bits, the wider the number of
amplitudes can be measured.
[6]
(b) The student creates 10 bitmap images using a computer.
Each image has a colour depth of 8-bits and a resolution of 800 × 500 pixels.
(i) Define the term pixel.
… [1]
(ii) Calculate the total file size of the 10 images in kilobytes.
Show your working out.
File size: … kilobytes
[2]
(iii) State the minimum number of bits that will be needed to represent 240 different colours.
… [1]
(c) The student wants to use a secondary storage device to move their files to their home computer.
Identify whether the student should use an optical or solid-state type of secondary storage
device. Justify your choice.
Optical or solid-state …
Justification …
… [4]
Mark scheme
Show the mark scheme
Question Answer Mark Guidance
1 (a) 1 mark for each completed term to max 6 6
An analogue sound wave needs to be converted into a digital sound wave.
Sound sampling is when the amplitude of the sound wave is measured at set
intervals.
The sample rate is the number of times a second the sound wave is
measured. This is given in Hertz.
Each amplitude is given a unique binary number. The number of bits
allocated to each sample is the bit depth. The higher the number of bits, the
wider the number of amplitudes can be measured.
1 (b) (i) 1 mark from 1 Accept screen in place of image.
• The smallest unit/part of an image
• A (single) square/dot/diode which has one/single colour BOD block/particle etc. in place of
square.
1 (b) (ii) 1 mark for one stage of working: 2 For MP1 accept 8 * 5 BOD.
• 800 * 500 (= 400 000)
• 400 10 = 4 000 000 Accept any method of doing
• 4 000 000 * 8 (= 32 000 000 bits) calculations e.g. statements, grids,
• 32 000 000 / 8 = 4 000 000 calculations.
• 4 000 000 / 1000 = 4 000
Accept division by 1024 instead of
1 mark for answer 1000. Final answer will be 3906.25
4000 kilobytes kilobytes.
If no answer in the final answer space,
look for answer clearly identified in
working.
1 (b) (iii) 8 1 Allow calculation that equates to 8
1 (c) No mark for choice 4 Accept type by example.
1 mark per bullet to max 4 for matching justification
No choice – check justification for
Solid-state choice e.g. clearly stated choice and then award
• Durable/robust // Less likely to be damaged/break 7 justification. No clear choice then 0
• … no moving parts // because it does not get scratched like a disk marks.
• Larger capacity (than optical) // store more data
• … needed because the files could be very large // there could be many files Allow justification marks for using one,
to transfer // store large number of files or not using the other.
• Portable
• … small in (physical) size // lightweight SS – BOD faster to transfer the data
• Fast to read/write/access data for read/write/access
SS – BOD more ‘efficient’ to
• More compatible
read/write/access.
• … no additional device/drive is needed
SS – accept optical may only be able
to be written to once.
Optical choice e.g.
• Large (enough) capacity // store sufficient data // BOD larger capacity
• … needed because the files could be very large // there could be many files Do not accept longevity/reliability.
to transfer // store large number of files
• Portable
• … small in (physical) size // lightweight
• Cost per unit is less // Cost for the same amount of storage is less
• The fast access/read/write speed is not required
• … files are being copied not run direct from the storage
How to answer it
Data Representation & Secondary Storage Guide
This multi-part question assesses core knowledge across Topic 1.2 (Memory & Storage) and Topic 2.4 (Data Representation):
- Sound representation: Analogue-to-digital conversion, sampling, sample rate, bit depth, and unique values.
- Bitmap images: Definition of a pixel, calculating file sizes in kilobytes, and determining colour depth from a palette size.
- Secondary storage characteristics: Evaluating optical vs solid-state media using capacity, durability, portability, speed, and compatibility.
Part (a) Storing Sound on a Computer
Fill-in-the-blanks: Analogue to Digital Conversion [6 marks]
✅ Correct Answer (Completed Text)
An analogue sound wave needs to be converted into a digital sound wave.
Sound sampling is when the amplitude of the sound wave is measured at set intervals.
The sample rate is the number of times a second the sound wave is measured. This is given in Hertz.
Each amplitude is given a unique binary number. The number of bits allocated to each sample is the bit depth .
The higher the number of bits, the wider the number of amplitudes can be measured.
💡 Key Knowledge
- Analogue sound: Continuous, real-world sound waves.
- Sampling: Taking snapshots of wave height (amplitude) at regular time intervals.
- Sample rate: Measured in Hertz (Hz) — 1 Hz = 1 sample per second.
- Bit depth (sample resolution): More bits per sample allow a wider range of discrete amplitude values to be represented, improving dynamic range and quality.
🧠 Exam Technique
Cross off terms from the given word bank as you use them. Notice that distracting terms like colour depth, hertz, and byte were included to test precise terminology.
❌ Common Errors
- Confusing bit depth with sample rate.
- Writing colour depth instead of bit depth for audio.
- Writing lower instead of higher for dynamic range.
Part (b)(i) Defining a Pixel
Definition of basic image units [1 mark]
✅ Correct Answers (Any one)
- The smallest unit / part of an image (or screen).
- A single square / dot which has one single colour.
🧠 Exam Technique
Keep definitions concise. The key word examiners look for is smallest or single colour. Just saying "a dot on a screen" is often too vague unless you specify it has a single colour or is the smallest addressable picture element.
Part (b)(ii) Calculating Image File Size
Resolution: 800 × 500, Colour Depth: 8 bits, 10 Images [2 marks]
📐 Step-by-Step Calculation
- Pixels per image: 800 × 500 = 400,000 pixels
- Total pixels across all 10 images: 400,000 × 10 = 4,000,000 pixels
- Total bits: 4,000,000 × 8 bits = 32,000,000 bits
- Convert bits to bytes: 32,000,000 ÷ 8 = 4,000,000 bytes
(Shortcut: Since colour depth is 8 bits, 1 pixel = 1 byte!) - Convert bytes to kilobytes:
Using metric (OCR standard): 4,000,000 ÷ 1,000 = 4,000 kB
(Note: Dividing by 1024 to give 3906.25 kB is also accepted).
✅ Final Answer
4,000 kilobytes (or 3,906.25 kilobytes)
❌ Common Errors
- Forgetting to multiply by 10: Calculating for 1 image (giving 400 kB) instead of all 10 images.
- Unit conversion errors: Dividing by 8 to get bytes, but then forgetting to divide by 1000/1024 to reach kilobytes.
- Not showing clear working: Even if the arithmetic fails, 1 working mark is awarded for 800 × 500 .
Part (b)(iii) Minimum Bits for Colour Depth
Representing 240 different colours [1 mark]
✅ Correct Answer
8 bits
💡 Key Knowledge: Powers of 2
- Formula: Colours = 2ⁿ (where n = number of bits)
- 7 bits: 2⁷ = 128 colours (Too few! 128 < 240)
- 8 bits: 2⁸ = 256 colours (Sufficient! 256 ≥ 240)
❌ Common Error
Choosing 7 bits because 128 is close, or attempting to give a fractional bit count. Bits must be a whole integer.
Part (c) Selecting Secondary Storage
Optical vs Solid-State Justification [4 marks]
🧠 Exam Technique: "Identify & Justify"
The choice itself (optical or solid-state) gets 0 marks. All 4 marks come from the justifications. You must pair a storage characteristic with why it matters in this scenario (moving presentation files between home and school).
Option A: Choosing Solid-State (Recommended)
- Durability / Robustness: Has no moving parts, so it won't break or get scratched when carried in a bag.
- Portability: Small physical size and lightweight to carry easily between school and home.
- Capacity: Typically larger capacity than optical media, easily holding all audio and image files.
- Speed: Faster read/write access speeds to quickly transfer files.
- Compatibility: Modern PCs all have USB ports; many modern laptops lack optical disc drives.
Option B: Choosing Optical (Alternative)
- Sufficient capacity: A CD/DVD has enough storage for 10 images and sound.
- Portability: Discs are thin, flat, and lightweight.
- Cost: Very cheap per disc / unit cost is low.
- Speed is sufficient: Fast access speeds are not strictly required because files are just copied, not streamed live from disc.
❌ Common Misconceptions to Avoid
- Do NOT cite longevity or reliability: The mark scheme explicitly states: "Do not accept longevity/reliability."
- Generic statements: Simply writing "it is fast" without linking to copying files or read/write speeds loses marks.
- Failing to choose: If no device type is stated or clearly implied, 0 marks are awarded even if good points are made.
Topics
1.2 Memory and storage · 1.2.2 Secondary storage · 1.2.3 Units · 1.2.4 Data storage
Question and mark scheme from the OCR GCSE Computer Science examination, Computer systems (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.