Full Chapter 1 Exam Practice

📖 How to use this exam practice

Work through every question as you would in an exam. Write your answers, then reveal the mark scheme and tick only the marks you genuinely earned. Your running score updates live at the top.

💡 For the effects table (Q14) — tick which boxes apply, then reveal to see which are correct. For the 8-mark discussion (Q19) — write a full answer before revealing the mark points.

Questions

Total marks

~55

Minutes

Sections

🔢 Section A — Number Systems | 20 marks

Question 1

2 marks

All data needs to be converted to binary so it can be processed by a computer. Explain why a computer can only process binary data.

Mark Scheme — 2 marks

Computers are made from transistors / electronic circuits / logic gates that can only be in two states: on or off / high voltage or low voltage [1]

These two states correspond directly to the two binary digits 0 (off/low) and 1 (on/high) — binary is the only number system that maps directly to this physical reality [1]

Do not accept: "because computers use switches" without explaining WHY two states = binary. Do not accept: "binary is easier for computers" without the physical explanation.

Question 2

3 marks

Convert these three denary numbers to 8-bit binary.

Mark Scheme — 3 marks (1 each)

50 = 00110010 (32+16+2=50 ✓) [1]

101 = 01100101 (64+32+4+1=101 ✓) [1]

213 = 11010101 (128+64+16+4+1=213 ✓) [1]

Question 3

3 marks

Add these two 8-bit binary values. Give your answer in binary. Show all working including the carry row.

00111001 + 01001010 ──────────

Mark Scheme — 3 marks

Carry row correctly shown: 01111000 [1]

At least 6 of 8 result bits correct / correct method shown [1]

Correct final answer: 10000011 (= 57+74 = 131 ✓) [1]

carry: 0 1 1 1 1 0 0 0 A: 0 0 1 1 1 0 0 1 (= 57) B: 0 1 0 0 1 0 1 0 (= 74) ───────────────── R: 1 0 0 0 0 0 1 1 (= 131) No overflow ✓

Question 4

4 marks

Part of a MAC address is 97–5C. Each pair of hex digits is stored as binary in an 8-bit register. Complete both registers. Show your nibble working.

Mark Scheme — 4 marks

97: high nibble 9=1001 correctly placed in bits 7–4 [1]

97: low nibble 7=0111 in bits 3–0 → full: 10010111 [1]

5C: high nibble 5=0101 correctly placed in bits 7–4 [1]

5C: low nibble C=12=1100 in bits 3–0 → full: 01011100 [1]

Question 5

2 marks

Using two's complement, represent the denary value −54 as an 8-bit binary number. Show all working.

Mark Scheme — 2 marks

Method shown: +54 converted to binary (00110110) then inverted (11001001) [1]

Correct final answer: 11001010 Verify: −128+64+8+2=−54 ✓ [1]

+54 = 00110110 Invert: 11001001 Add 1: 11001010 ← answer

Question 6

3 marks

Two 8-bit binary values are added. The denary result is 301. This generates an error when stored in an 8-bit register.
State the name of this error and explain why it occurs.

Mark Scheme — 3 marks

Error name: Overflow [1]

301 exceeds the maximum value (255) that can be stored in an 8-bit register / result requires more than 8 bits [1]

A carry bit is generated beyond the MSB and is lost / the stored result is therefore incorrect / the carry cannot be stored [1]

Question 7

2 marks

HTML colour codes and MAC addresses are two uses of hexadecimal. Give two other uses of hexadecimal in computer science.

Mark Scheme — 2 marks (1 each)

First valid use — e.g. memory addresses / error codes / assembly/machine code debugging / IPv6 / Unicode code points [1]

Second valid use, different from first [1]

Do not accept: HTML colours or MAC addresses (already given in question)

Question 8

2 marks

Give the correct denary value for the 12-bit binary value 000101010111. Show all working.

Mark Scheme — 2 marks

Correct method: identifying active bits (256, 64, 16, 4, 2, 1) from the 12-bit place value row [1]

Denary value = 343 (256+64+16+4+2+1=343 ✓) [1]

12-bit: 2048|1024|512|256|128|64|32|16|8|4|2|1 Bits: 0 | 0 | 0 | 1 | 0 |1 | 0| 1|0|1|1|1 Active: 256+64+16+4+2+1 = 343

📘 Section B — Text, Images, Sound & Data | 29 marks

Question 9

1 mark

The Unicode character set is used to represent text typed into a computer. One disadvantage of Unicode over ASCII is that it takes up more storage space.
Give one reason why it takes up more storage space.

Mark Scheme — 1 mark

Unicode uses more bits per character than ASCII (e.g. 16 or 32 bits vs 7 or 8 bits) to represent a larger range of characters [1]

Allow: "each character requires more bits/bytes to store" as equivalent. Must reference more bits per character — not just "more characters".

Question 10

1 mark

Jack has an MP3 file stored on his computer. Tick (✓) to show which type of data is stored in an MP3 file.

AVideo

BSound

CImage

Mark Scheme — 1 mark

Sound ✓ — MP3 (MPEG-1 Audio Layer III) is a lossy compressed audio format. It stores sound, not video or images. [1]

Question 11

2 marks

A student compresses a sound file. The compression reduces the sample rate and the sample resolution. State what is meant by the sample rate and sample resolution.

Mark Scheme — 2 marks

Sample rate: the number of samples taken per second when recording the sound / measured in Hz or kHz / how frequently the analogue wave is measured [1]

Sample resolution (bit depth): the number of bits used to store each sample / determines how many amplitude levels are possible / the precision of each measurement [1]

Do not accept: confusing sample rate (frequency of measurement) with sample resolution (precision per measurement)

Question 12

1 mark

A logo is stored as a bitmap image. Each square in the image represents one pixel. State what is meant by the term image resolution.

Mark Scheme — 1 mark

The number of pixels in a digital image / stated as width × height (e.g. 1920 × 1080) / the total number of picture elements that make up the image [1]

Allow: "number of pixels per unit area" / "number of pixels wide and tall". Do not accept: "quality of the image" alone — must reference pixels.

Question 13

2 marks

The Unicode character set is used to represent text typed into a computer. Describe what is meant by a character set.

Mark Scheme — 2 marks

A character set is a defined collection / list of characters (letters, digits, symbols, punctuation, control codes) that a computer system can represent and use [1]

Each character in the set is assigned a unique numerical code (binary value) — the computer stores the number and looks up the character when it needs to display it [1]

Award [1] for the collection of characters idea. Award [1] for the unique code/number mapping. Both points needed for 2 marks.

Question 14

3 marks

The ASCII code for 'T' is 84.
(a) Give the ASCII code for 'W'. [1]
(b) Give the ASCII code for lowercase 't'. [1]
(c) Write the ASCII code for 'T' as an 8-bit binary number. [1]

Mark Scheme — 3 marks

(a) W is 3 letters after T → 84 + 3 = 87 [1]

(b) Lowercase = uppercase + 32 → 84 + 32 = 116 [1]

Question 15

3 marks

An image has a resolution of 1920 × 1080 pixels and a colour depth of 24 bits. Calculate the file size in megabytes (MB). Show all working.

Mark Scheme — 3 marks

Total pixels = 1920 × 1080 = 2,073,600 pixels [1]

Total bits = 2,073,600 × 24 = 49,766,400 bits → ÷ 8 = 6,220,800 bytes [1]

6,220,800 ÷ 1024 = 6,075 KB → ÷ 1024 = ≈ 5.93 MB [1]

Question 16

3 marks

A student records a podcast about computer science. Describe how an analogue sound wave is converted into digital form.

Mark Scheme — 3 marks

The analogue sound wave is sampled at regular intervals — the amplitude (height) of the wave is measured at each sample point / the ADC (Analogue-to-Digital Converter) takes the measurements [1]

Each sample value is quantised — rounded to the nearest available amplitude level / the number of levels depends on the bit depth (e.g. 8-bit = 256 levels) [1]

Each quantised value is stored as a binary number — the sequence of binary values represents the digital audio / the number of bits per sample = bit depth [1]

Award marks for: sampling described ✓ | quantisation / binary conversion ✓ | stored as binary / bit depth referenced ✓

Question 17

3 marks

Tick (✓) one or more boxes on each row to identify the effect(s) that each change will have on the sound file.

Click cells to tick them, then reveal the mark scheme to check your answers.

Change	File size increases	File size decreases	Accuracy increases	Accuracy decreases
Duration changes from 10 minutes → 20 minutes
Sample rate changes from 44 kHz → 8 kHz
Bit depth changes from 8 bits → 16 bits

Mark Scheme — 3 marks (1 per correct row)

Duration 10→20 min: File size increases ✓ only. More data recorded = larger file. No effect on accuracy per sample. [1]

Sample rate 44kHz→8kHz: File size decreases ✓ AND Accuracy decreases ✓. Fewer samples per second = smaller file but less faithful representation of the wave. [1]

Bit depth 8→16 bits: File size increases ✓ AND Accuracy increases ✓. More bits per sample = larger file but more amplitude levels = more precise. [1]

Award 1 mark per row where ALL correct boxes are ticked and NO incorrect boxes are ticked.

Question 18

2 marks

A student is creating an image file for a website. State one advantage of using lossy compression and one advantage of using lossless compression for this image.

Mark Scheme — 2 marks

Lossy advantage: produces a much smaller file size / faster to download/transmit / requires less storage / pages load faster for website visitors [1]

Lossless advantage: original image perfectly preserved / no data permanently lost / important for logos or images where exact colours/edges must be maintained (e.g. company branding) [1]

Question 19

8 marks

Discuss the trade-offs between image quality and file size when choosing a bitmap image's resolution and colour depth. Provide examples to illustrate your answer.

This is an 8-mark extended response. Write a full, structured answer covering both resolution and colour depth, with examples and calculations.

Mark Scheme — 8 marks (award up to 8 from the points below)

Resolution points:

Resolution is the number of pixels in a digital image, stated as width × height (e.g. 1920 × 1080) [1]

Increasing resolution increases file size — more pixels means more data to store (file size = pixels × colour depth) [1]

Increasing resolution improves image quality — more pixels capture finer detail, edges are sharper, the image is more realistic [1]

Resolution example: a 1920×1080 image has 9× as many pixels as a 640×360 image — the file is 9× larger but captures much more detail [1]

Colour depth points:

Colour depth is the number of bits per pixel — it determines how many colours are possible (2ⁿ possible colours) [1]

Increasing colour depth increases file size — more bits stored per pixel makes each pixel larger in memory [1]

Increasing colour depth improves image quality — more colours available means smoother gradients, more realistic skin tones, and less banding / posterisation [1]

Colour depth example with calculation: 1920×1080 at 24-bit (True Colour) = 5.93 MB; at 8-bit (256 colours) = 1.98 MB — same resolution but very different quality and file size. Or: 24-bit has 16,777,216 colours vs 8-bit with just 256 colours. [1]

For full 8 marks: must cover both resolution AND colour depth (at least 3 points each), must include at least one numerical example or calculation, must reference the trade-off (larger file = better quality; smaller file = lower quality).

Do not award: vague statements like "better quality" without explaining how/why. Do not award: repeating the same point reworded.

Model answer structure for 8/8: 1. Define resolution (pixels, width × height) 2. Higher resolution → more detail, sharper image 3. Higher resolution → more pixels → larger file 4. Resolution example + file size comparison 5. Define colour depth (bits per pixel, 2^n colours) 6. Higher colour depth → more colours, smoother gradients 7. Higher colour depth → more bits → larger file 8. Colour depth example with calculation (24-bit vs 8-bit) Key phrase needed: "trade-off between quality and file size"

📊 Your Performance

Tick marks above as you go — score updates automatically.

out of 49 marks

Complete the questions above to see your grade

80%+ (40–49)

Excellent — Grade A

60–79% (30–39)

Good — Grade B/C

40–59% (20–29)

Developing

Below 40% (0–19)

Foundation — revisit

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