HAIDER NOTES | Computer Science Paper 1 | O Levels 2210 / IGCSE 0478
HAIDER NOTES
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COMPUTER SCIENCE
Paper 1: Computer Systems
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APPLICABLE COURSES & QUALIFICATIONS
Qualification Course Code Level / Year Group
Cambridge O Levels 2210 Grade 9–11 (Ages 14–17)
Cambridge IGCSE 0478 Grade 9–11 (Ages 14–17)
Cambridge IGCSE (9–1) 0984 Grade 9–11 (Ages 14–17)
AS / A Level (prep) 9618 (foundation) Grade 11–12 (Ages 16–18)
FBISE / Pakistan Board Equivalent Matric / SSC Level (Class 9–10)
AQA / Edexcel GCSE (UK) Computer Science Year 10–11 (Ages 14–16)
IB MYP Computer Science MYP 4–5 Grade 9–10 (Ages 14–16)
Syllabus Edition: 2023 – 2025
Prepared by: Haider | All Rights Reserved
© Haider Notes — All Rights Reserved | Page 1 of 43
,HAIDER NOTES | Computer Science Paper 1 | O Levels 2210 / IGCSE 0478
UNIT 1: DATA REPRESENTATION
1.1 Uses of the Binary System
➤ To process data in logic gates / transistors
➤ To store data in registers
➤ To process data on a computer
1.2 Number Systems — Denary, Binary & Hexadecimal
Computers use three main number systems. Understanding each one and the differences
between them is essential.
Denary
Feature Binary Hexadecimal
(Decimal)
Base Base 10 Base 2 Base 16
Digits used 0–9 0 and 1 only 0–9 and A–F
Used in Daily counting & Storing data in MAC addresses, IP
calculations registers; logic gates; addresses, HTML colour
CPU processing codes, error codes
Digit count (same Fewer digits More digits Fewer digits
value)
Why is Hexadecimal Beneficial?
➤ Used in colour codes, error codes, IP addresses and MAC addresses
➤ More easily readable/understandable by humans than long binary strings
➤ Easier to debug memory dumps and machine code
➤ Takes less screen space compared to binary
➤ Each hex digit represents exactly 4 bits (one nibble) — easy conversion
1.3 Data Conversions
Binary → Denary
Write the positional values (128, 64, 32, 16, 8, 4, 2, 1) above each bit and add the values where
the bit is 1.
Example: Convert 11101110 to denary
128 64 32 16 8 4 2 1
1 1 1 0 1 1 1 0
11101110 = 128 + 64 + 32 + 8 + 4 + 2 = 238
© Haider Notes — All Rights Reserved | Page 2 of 43
,HAIDER NOTES | Computer Science Paper 1 | O Levels 2210 / IGCSE 0478
Denary → Binary (Two Methods)
Method 1 — Subtract the largest possible power of 2 repeatedly:
Convert 142: 142−128=14 → 14−8=6 → 6−4=2 → 2−2=0 → Result: 10001110
Method 2 — Successive Division by 2 (read remainders bottom-up):
DIV MOD
142 0
71 1 ↑
35 1 ↑
17 1 ↑
8 0 ↑
4 0 ↑
2 0 ↑
1 0 ↑
0 1 ↑ (MSB)
Result: 142 = 10001110
Binary → Hexadecimal
Split binary into groups of 4 bits from right, then convert each group.
Example: 101111100001 → 1011 | 1110 | 0001 → B | E | 1 → BE1
Hexadecimal → Binary
Convert each hex digit to its 4-bit binary equivalent.
4 5 A
0100 0101 1010
45A = 010001011010
Hexadecimal → Denary
Multiply each digit by its positional value (256, 16, 1) and sum.
Example: 45A → (4×256) + (5×16) + (10×1) = 1024 + 80 + 10 = 1114
Denary → Hexadecimal
Successive division by 16; read remainders bottom-up. (13 = D, 10 = A, etc.)
Example: 2004 ÷ 16 = 125 r 4 → 125 ÷ 16 = 7 r 13(D) → 7 ÷ 16 = 0 r 7 → Result: 7D4
1.4 Binary Addition & Overflow
Binary addition rules: 0+0=0, 0+1=1, 1+0=1, 1+1=10 (carry 1), 1+1+1=11 (carry 1)
© Haider Notes — All Rights Reserved | Page 3 of 43
, HAIDER NOTES | Computer Science Paper 1 | O Levels 2210 / IGCSE 0478
Example: 126 + 62
CARR
128 64 32 16 8 4 2 1
Y
1 0 1 0 0 0 1 0
0 0 0 1 1 0 1 0
SUM 1 0 1 1 1 1 0 0
10111100₂ = 188₁₀ ✓
Overflow Error
An overflow error occurs when the result of a calculation is too large to be stored in the allocated
word size (register).
➤ An 8-bit register can store a maximum value of 255
➤ If the result exceeds 255, a carry bit extends beyond the register — this is overflow
➤ Example: 110 + 222 = 332 — this exceeds 255, causing an overflow error
📌 An overflow error is detected by an extra carry bit appearing beyond the most
significant bit (MSB) of the register.
1.5 Logical Binary Shifts
Shift LEFT by n places = Multiply by 2ⁿ
Shift RIGHT by n places = Divide by 2ⁿ
Shift Direction Effect Error Condition
Left by n places Multiplies the value by 2ⁿ Error if 1-bits are shifted out (lost)
from the left — result is incorrect
Right by n places Divides the value by 2ⁿ Error if 1-bits are shifted out (lost)
from the right — precision lost
Example 1: Shift 00010101 (21) two places LEFT → 01010100 = 84 = 21 × 2² ✓
Example 2: Shift 01100100 (100) two places RIGHT → 00011001 = 25 = 100 ÷ 2² ✓
1.6 Two's Complement (Negative Binary Numbers)
➤ Used to represent negative binary numbers in computers
➤ The leftmost (most significant) bit represents a negative value
➤ If the MSB is 1, the number is negative; if 0, it is positive
Negative Binary → Denary
Use the same positional values but treat the MSB as negative.
-128 64 32 16 8 4 2 1
1 0 0 1 0 0 1 1
© Haider Notes — All Rights Reserved | Page 4 of 43
HAIDER NOTES
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
COMPUTER SCIENCE
Paper 1: Computer Systems
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
APPLICABLE COURSES & QUALIFICATIONS
Qualification Course Code Level / Year Group
Cambridge O Levels 2210 Grade 9–11 (Ages 14–17)
Cambridge IGCSE 0478 Grade 9–11 (Ages 14–17)
Cambridge IGCSE (9–1) 0984 Grade 9–11 (Ages 14–17)
AS / A Level (prep) 9618 (foundation) Grade 11–12 (Ages 16–18)
FBISE / Pakistan Board Equivalent Matric / SSC Level (Class 9–10)
AQA / Edexcel GCSE (UK) Computer Science Year 10–11 (Ages 14–16)
IB MYP Computer Science MYP 4–5 Grade 9–10 (Ages 14–16)
Syllabus Edition: 2023 – 2025
Prepared by: Haider | All Rights Reserved
© Haider Notes — All Rights Reserved | Page 1 of 43
,HAIDER NOTES | Computer Science Paper 1 | O Levels 2210 / IGCSE 0478
UNIT 1: DATA REPRESENTATION
1.1 Uses of the Binary System
➤ To process data in logic gates / transistors
➤ To store data in registers
➤ To process data on a computer
1.2 Number Systems — Denary, Binary & Hexadecimal
Computers use three main number systems. Understanding each one and the differences
between them is essential.
Denary
Feature Binary Hexadecimal
(Decimal)
Base Base 10 Base 2 Base 16
Digits used 0–9 0 and 1 only 0–9 and A–F
Used in Daily counting & Storing data in MAC addresses, IP
calculations registers; logic gates; addresses, HTML colour
CPU processing codes, error codes
Digit count (same Fewer digits More digits Fewer digits
value)
Why is Hexadecimal Beneficial?
➤ Used in colour codes, error codes, IP addresses and MAC addresses
➤ More easily readable/understandable by humans than long binary strings
➤ Easier to debug memory dumps and machine code
➤ Takes less screen space compared to binary
➤ Each hex digit represents exactly 4 bits (one nibble) — easy conversion
1.3 Data Conversions
Binary → Denary
Write the positional values (128, 64, 32, 16, 8, 4, 2, 1) above each bit and add the values where
the bit is 1.
Example: Convert 11101110 to denary
128 64 32 16 8 4 2 1
1 1 1 0 1 1 1 0
11101110 = 128 + 64 + 32 + 8 + 4 + 2 = 238
© Haider Notes — All Rights Reserved | Page 2 of 43
,HAIDER NOTES | Computer Science Paper 1 | O Levels 2210 / IGCSE 0478
Denary → Binary (Two Methods)
Method 1 — Subtract the largest possible power of 2 repeatedly:
Convert 142: 142−128=14 → 14−8=6 → 6−4=2 → 2−2=0 → Result: 10001110
Method 2 — Successive Division by 2 (read remainders bottom-up):
DIV MOD
142 0
71 1 ↑
35 1 ↑
17 1 ↑
8 0 ↑
4 0 ↑
2 0 ↑
1 0 ↑
0 1 ↑ (MSB)
Result: 142 = 10001110
Binary → Hexadecimal
Split binary into groups of 4 bits from right, then convert each group.
Example: 101111100001 → 1011 | 1110 | 0001 → B | E | 1 → BE1
Hexadecimal → Binary
Convert each hex digit to its 4-bit binary equivalent.
4 5 A
0100 0101 1010
45A = 010001011010
Hexadecimal → Denary
Multiply each digit by its positional value (256, 16, 1) and sum.
Example: 45A → (4×256) + (5×16) + (10×1) = 1024 + 80 + 10 = 1114
Denary → Hexadecimal
Successive division by 16; read remainders bottom-up. (13 = D, 10 = A, etc.)
Example: 2004 ÷ 16 = 125 r 4 → 125 ÷ 16 = 7 r 13(D) → 7 ÷ 16 = 0 r 7 → Result: 7D4
1.4 Binary Addition & Overflow
Binary addition rules: 0+0=0, 0+1=1, 1+0=1, 1+1=10 (carry 1), 1+1+1=11 (carry 1)
© Haider Notes — All Rights Reserved | Page 3 of 43
, HAIDER NOTES | Computer Science Paper 1 | O Levels 2210 / IGCSE 0478
Example: 126 + 62
CARR
128 64 32 16 8 4 2 1
Y
1 0 1 0 0 0 1 0
0 0 0 1 1 0 1 0
SUM 1 0 1 1 1 1 0 0
10111100₂ = 188₁₀ ✓
Overflow Error
An overflow error occurs when the result of a calculation is too large to be stored in the allocated
word size (register).
➤ An 8-bit register can store a maximum value of 255
➤ If the result exceeds 255, a carry bit extends beyond the register — this is overflow
➤ Example: 110 + 222 = 332 — this exceeds 255, causing an overflow error
📌 An overflow error is detected by an extra carry bit appearing beyond the most
significant bit (MSB) of the register.
1.5 Logical Binary Shifts
Shift LEFT by n places = Multiply by 2ⁿ
Shift RIGHT by n places = Divide by 2ⁿ
Shift Direction Effect Error Condition
Left by n places Multiplies the value by 2ⁿ Error if 1-bits are shifted out (lost)
from the left — result is incorrect
Right by n places Divides the value by 2ⁿ Error if 1-bits are shifted out (lost)
from the right — precision lost
Example 1: Shift 00010101 (21) two places LEFT → 01010100 = 84 = 21 × 2² ✓
Example 2: Shift 01100100 (100) two places RIGHT → 00011001 = 25 = 100 ÷ 2² ✓
1.6 Two's Complement (Negative Binary Numbers)
➤ Used to represent negative binary numbers in computers
➤ The leftmost (most significant) bit represents a negative value
➤ If the MSB is 1, the number is negative; if 0, it is positive
Negative Binary → Denary
Use the same positional values but treat the MSB as negative.
-128 64 32 16 8 4 2 1
1 0 0 1 0 0 1 1
© Haider Notes — All Rights Reserved | Page 4 of 43
