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DAY 2 — Number Systems and Data Representation — Complete Study Material

Phase 1: Technical Theory | Date: 22 July 2026 | Subject: Number Systems + Data Representation | Expected Questions: 2–3

📌 Day Overview

Today you will master:

Decimal, binary, octal, and hexadecimal number systems
Decimal-to-binary, octal, and hexadecimal conversions
Binary, octal, and hexadecimal-to-decimal conversions
Binary-to-octal and binary-to-hexadecimal conversions
Octal-to-binary and hexadecimal-to-binary conversions
Octal-to-hexadecimal and hexadecimal-to-octal conversions
Conversion of fractional numbers
Binary addition and subtraction
1’s complement and 2’s complement
Subtraction using 2’s complement
ASCII, Unicode, EBCDIC, and BCD
Important character codes

Five-Hour Plan

Hour	Topic
1	Number-system theory
2	Decimal ↔ Binary, Octal, Hexadecimal
3	Binary ↔ Octal, Binary ↔ Hexadecimal, Octal ↔ Hexadecimal
4	Binary arithmetic and complements
5	Data representation and practice

1. Number Systems — Full Explanation

1.1 What Is a Number System?

A number system is a method of representing numbers using a fixed set of symbols.

For example, the decimal number system uses ten symbols:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

A number system is identified by its base, also called its radix.

The base tells us:

How many unique digits the system uses
What positional powers are used

1.2 Main Number Systems

Number System	Base	Allowed Digits	Example
Binary	2	0, 1	1011₂
Octal	8	0–7	157₈
Decimal	10	0–9	259₁₀
Hexadecimal	16	0–9, A–F	2AF₁₆

Subscript

The small number written after a value shows its base.

Examples:

1011₂ means binary 1011.
157₈ means octal 157.
259₁₀ means decimal 259.
2AF₁₆ means hexadecimal 2AF.

When subscripts cannot be written, the base may be shown like this:

(1011)₂
(157)₈
(259)₁₀
(2AF)₁₆

1.3 Positional Number System

Binary, octal, decimal, and hexadecimal are positional number systems.

In a positional system, the value of a digit depends on:

The digit itself
Its position
The base of the number system

General Expansion

For a number:

dₙdₙ₋₁...d₂d₁d₀

Its value is:

dₙ × baseⁿ + dₙ₋₁ × baseⁿ⁻¹ + ... + d₁ × base¹ + d₀ × base⁰

Remember:

Any non-zero number raised to power 0 = 1

Examples:

2⁰ = 1
8⁰ = 1
10⁰ = 1
16⁰ = 1

1.4 Place Values on Both Sides of the Radix Point

The decimal point is called a radix point when discussing different number systems.

Positions to the left use positive powers.

Positions to the right use negative powers.

... base³  base²  base¹  base⁰ . base⁻¹  base⁻²  base⁻³ ...

Decimal Example

472.35₁₀

= 4 × 10² + 7 × 10¹ + 2 × 10⁰
  + 3 × 10⁻¹ + 5 × 10⁻²

= 400 + 70 + 2 + 0.3 + 0.05

= 472.35₁₀

1.5 Decimal Number System

The decimal number system has base 10.

Digits

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Place Values

... 10⁴  10³  10²  10¹  10⁰

... 10000 1000 100 10 1

Example

Expand 5837₁₀:

5837
= 5 × 10³ + 8 × 10² + 3 × 10¹ + 7 × 10⁰
= 5000 + 800 + 30 + 7
= 5837

1.6 Binary Number System

The binary number system has base 2.

Digits

0 and 1

Each binary digit is called a bit.

Bit stands for binary digit.

Binary Place Values

Power	Value
2⁰	1
2¹	2
2²	4
2³	8
2⁴	16
2⁵	32
2⁶	64
2⁷	128
2⁸	256
2⁹	512
2¹⁰	1024
2¹¹	2048
2¹²	4096

Binary Fractional Place Values

Power	Value
2⁻¹	1/2 = 0.5
2⁻²	1/4 = 0.25
2⁻³	1/8 = 0.125
2⁻⁴	1/16 = 0.0625
2⁻⁵	1/32 = 0.03125

Example

Expand 1101₂:

1101₂
= 1 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰
= 8 + 4 + 0 + 1
= 13₁₀

Why Computers Use Binary

Digital circuits can easily represent two stable states:

On and off
High and low voltage
True and false
1 and 0

1.7 Octal Number System

The octal number system has base 8.

Digits

0, 1, 2, 3, 4, 5, 6, 7

Digits 8 and 9 are not valid in octal.

Octal Place Values

Power	Value
8⁰	1
8¹	8
8²	64
8³	512
8⁴	4096

Example

Expand 157₈:

157₈
= 1 × 8² + 5 × 8¹ + 7 × 8⁰
= 64 + 40 + 7
= 111₁₀

Binary Relationship

One octal digit corresponds to exactly three binary bits because:

8 = 2³

1.8 Hexadecimal Number System

The hexadecimal number system has base 16.

Digits

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

Letter Values

Hexadecimal Digit	Decimal Value	Binary
A	10	1010
B	11	1011
C	12	1100
D	13	1101
E	14	1110
F	15	1111

Hexadecimal Place Values

Power	Value
16⁰	1
16¹	16
16²	256
16³	4096
16⁴	65536

Example

Expand 2AF₁₆:

2AF₁₆
= 2 × 16² + A × 16¹ + F × 16⁰
= 2 × 256 + 10 × 16 + 15 × 1
= 512 + 160 + 15
= 687₁₀

Binary Relationship

One hexadecimal digit corresponds to exactly four binary bits because:

16 = 2⁴

1.9 Valid and Invalid Digits

Number System	Valid Digits	Examples of Invalid Digits
Binary	0, 1	2, 3, 8, 9
Octal	0–7	8, 9
Decimal	0–9	A, B, C
Hexadecimal	0–9, A–F	G, H, Z

Examples

101101₂ is valid.
10201₂ is invalid because binary does not contain 2.
725₈ is valid.
789₈ is invalid because octal does not contain 8 or 9.
FACE₁₆ is valid.
G15₁₆ is invalid because hexadecimal does not contain G.

2. Decimal ↔ Binary, Octal, and Hexadecimal

2.1 Conversion to Decimal Using Positional Weights

To convert binary, octal, or hexadecimal to decimal:

Write the place value of every digit.
Multiply each digit by its place value.
Add all results.

2.2 Binary to Decimal

Example 1: Convert 101101₂ to decimal

101101₂
= 1 × 2⁵ + 0 × 2⁴ + 1 × 2³
  + 1 × 2² + 0 × 2¹ + 1 × 2⁰

= 32 + 0 + 8 + 4 + 0 + 1

= 45₁₀

Example 2: Convert 111111₂ to decimal

111111₂
= 32 + 16 + 8 + 4 + 2 + 1
= 63₁₀

Shortcut for All Ones

A binary number containing n consecutive ones has decimal value:

2ⁿ − 1

Example:

111111₂ has 6 ones

= 2⁶ − 1
= 64 − 1
= 63

2.3 Binary Fraction to Decimal

Example: Convert 101.101₂ to decimal

101.101₂
= 1 × 2² + 0 × 2¹ + 1 × 2⁰
  + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³

= 4 + 0 + 1 + 0.5 + 0 + 0.125

= 5.625₁₀

2.4 Octal to Decimal

Example: Convert 347₈ to decimal

347₈
= 3 × 8² + 4 × 8¹ + 7 × 8⁰
= 3 × 64 + 4 × 8 + 7
= 192 + 32 + 7
= 231₁₀

Example with Fraction

Convert 12.4₈ to decimal:

12.4₈
= 1 × 8¹ + 2 × 8⁰ + 4 × 8⁻¹
= 8 + 2 + 4/8
= 10 + 0.5
= 10.5₁₀

2.5 Hexadecimal to Decimal

Example: Convert 3B7₁₆ to decimal

3B7₁₆
= 3 × 16² + B × 16¹ + 7 × 16⁰
= 3 × 256 + 11 × 16 + 7
= 768 + 176 + 7
= 951₁₀

Example with Fraction

Convert A.C₁₆ to decimal:

A.C₁₆
= 10 × 16⁰ + 12 × 16⁻¹
= 10 + 12/16
= 10 + 0.75
= 10.75₁₀

2.6 Decimal Integer to Another Base

Use repeated division for the integer part.

Method

To convert a decimal integer to base b:

Divide the number by b.
Record the remainder.
Divide the quotient by b again.
Continue until the quotient becomes 0.
Read the remainders from bottom to top.

2.7 Decimal to Binary

Example: Convert 45₁₀ to binary

Division	Quotient	Remainder
45 ÷ 2	22	1
22 ÷ 2	11	0
11 ÷ 2	5	1
5 ÷ 2	2	1
2 ÷ 2	1	0
1 ÷ 2	0	1

Read the remainders from bottom to top:

45₁₀ = 101101₂

Power-of-Two Shortcut

Convert 25₁₀ to binary:

25 = 16 + 8 + 1
   = 2⁴ + 2³ + 2⁰

Place 1 at powers 4, 3, and 0:

11001₂

2.8 Decimal to Octal

Example: Convert 156₁₀ to octal

Division	Quotient	Remainder
156 ÷ 8	19	4
19 ÷ 8	2	3
2 ÷ 8	0	2

Read upward:

156₁₀ = 234₈

2.9 Decimal to Hexadecimal

Example: Convert 254₁₀ to hexadecimal

Division	Quotient	Remainder
254 ÷ 16	15	14 = E
15 ÷ 16	0	15 = F

Read upward:

254₁₀ = FE₁₆

Remainder Conversion

Decimal Remainder	Hexadecimal Digit
10	A
11	B
12	C
13	D
14	E
15	F

2.10 Decimal Fraction to Another Base

Use repeated multiplication for the fractional part.

Method

To convert a decimal fraction to base b:

Multiply the fraction by b.
Record the integer part of the result.
Multiply the new fractional part by b.
Continue until the fraction becomes 0 or enough digits are obtained.
Read integer parts from top to bottom.

2.11 Decimal Fraction to Binary

Example: Convert 0.625₁₀ to binary

Multiplication	Integer Part	Fractional Part
0.625 × 2 = 1.250	1	0.250
0.250 × 2 = 0.500	0	0.500
0.500 × 2 = 1.000	1	0

Read the integer parts from top to bottom:

0.625₁₀ = 0.101₂

Mixed Number Example

Convert 13.625₁₀ to binary:

Integer part:

13₁₀ = 1101₂

Fractional part:

0.625₁₀ = 0.101₂

Combined result:

13.625₁₀ = 1101.101₂

2.12 Decimal Fraction to Octal

Convert 0.6875₁₀ to octal:

Multiplication	Integer Part	Fractional Part
0.6875 × 8 = 5.5	5	0.5
0.5 × 8 = 4.0	4	0

Therefore:

0.6875₁₀ = 0.54₈

2.13 Decimal Fraction to Hexadecimal

Convert 0.8125₁₀ to hexadecimal:

0.8125 × 16 = 13.0

13 = D

Therefore:

0.8125₁₀ = 0.D₁₆

3. Binary ↔ Octal, Binary ↔ Hexadecimal, Octal ↔ Hexadecimal

3.1 Binary-to-Octal Table

Each octal digit equals three binary bits.

Binary	Octal
000	0
001	1
010	2
011	3
100	4
101	5
110	6
111	7

3.2 Binary to Octal

Method

Start from the radix point.
For the integer part, group bits in sets of three from right to left.
For the fractional part, group bits in sets of three from left to right.
Add zeros at the outer ends if required.
Replace each group with one octal digit.

Example 1

Convert 1101011₂ to octal:

Group from right:

1 101 011

Add leading zeros:

001 101 011

Convert:

001 = 1
101 = 5
011 = 3

Therefore:

1101011₂ = 153₈

Example 2: Fraction

Convert 101101.011₂ to octal:

101 101 . 011
  5   5 .   3

Therefore:

101101.011₂ = 55.3₈

3.3 Octal to Binary

Replace each octal digit with exactly three binary bits.

Example

Convert 572₈ to binary:

5 = 101
7 = 111
2 = 010

Therefore:

572₈ = 101111010₂

Fraction Example

Convert 6.35₈ to binary:

6 = 110
3 = 011
5 = 101

Therefore:

6.35₈ = 110.011101₂

3.4 Binary-to-Hexadecimal Table

Each hexadecimal digit equals four binary bits.

Binary	Hex	Binary	Hex
0000	0	1000	8
0001	1	1001	9
0010	2	1010	A
0011	3	1011	B
0100	4	1100	C
0101	5	1101	D
0110	6	1110	E
0111	7	1111	F

3.5 Binary to Hexadecimal

Method

Group the integer part into sets of four from right to left.
Group the fractional part into sets of four from left to right.
Add outer zeros if needed.
Replace every group with one hexadecimal digit.

Example 1

Convert 10111101₂ to hexadecimal:

1011 1101
   B    D

Therefore:

10111101₂ = BD₁₆

Example 2

Convert 1110111₂ to hexadecimal:

0111 0111
   7    7

Therefore:

1110111₂ = 77₁₆

Fraction Example

Convert 1010.1101₂ to hexadecimal:

1010 . 1101
   A .    D

Therefore:

1010.1101₂ = A.D₁₆

3.6 Hexadecimal to Binary

Replace each hexadecimal digit with exactly four binary bits.

Example

Convert 3AF₁₆ to binary:

3 = 0011
A = 1010
F = 1111

Therefore:

3AF₁₆ = 001110101111₂

Leading zeros may be removed:

3AF₁₆ = 1110101111₂

Fraction Example

Convert C.7₁₆ to binary:

C = 1100
7 = 0111

Therefore:

C.7₁₆ = 1100.0111₂

3.7 Octal to Hexadecimal

There is no simple direct grouping rule between octal and hexadecimal.

Use binary as the bridge:

Octal → Binary → Hexadecimal

Example

Convert 725₈ to hexadecimal.

Step 1: Octal to binary

7 = 111
2 = 010
5 = 101

725₈ = 111010101₂

Step 2: Group into four bits

0001 1101 0101

Step 3: Convert to hexadecimal

0001 = 1
1101 = D
0101 = 5

Therefore:

725₈ = 1D5₁₆

3.8 Hexadecimal to Octal

Use binary as the bridge:

Hexadecimal → Binary → Octal

Example

Convert 2F3₁₆ to octal.

Step 1: Hexadecimal to binary

2 = 0010
F = 1111
3 = 0011

2F3₁₆ = 001011110011₂

Step 2: Group into three bits

001 011 110 011

Step 3: Convert to octal

Therefore:

2F3₁₆ = 1363₈

3.9 Universal Conversion Routes

Starting Base	Target Base	Best Method
Binary	Decimal	Positional weights
Octal	Decimal	Positional weights
Hexadecimal	Decimal	Positional weights
Decimal	Binary	Repeated division by 2
Decimal	Octal	Repeated division by 8
Decimal	Hexadecimal	Repeated division by 16
Binary	Octal	Groups of 3 bits
Octal	Binary	Replace each digit with 3 bits
Binary	Hexadecimal	Groups of 4 bits
Hexadecimal	Binary	Replace each digit with 4 bits
Octal	Hexadecimal	Use binary as a bridge
Hexadecimal	Octal	Use binary as a bridge

4. Binary Arithmetic

4.1 Binary Addition Rules

Operation	Sum	Carry
0 + 0	0	0
0 + 1	1	0
1 + 0	1	0
1 + 1	0	1
1 + 1 + 1	1	1

Remember:

1 + 1 = 10₂

1 + 1 + 1 = 11₂

4.2 Binary Addition Example 1

Add 1011₂ and 0110₂:

Column calculation from right:

1 + 0 = 1
1 + 1 = 10: write 0, carry 1
0 + 1 + carry 1 = 10: write 0, carry 1
1 + 0 + carry 1 = 10: write 0, carry 1
Write final carry 1

Therefore:

1011₂ + 0110₂ = 10001₂

Decimal check:

11 + 6 = 17

4.3 Binary Addition Example 2

Add 1101₂ and 1011₂:

Therefore:

1101₂ + 1011₂ = 11000₂

Decimal check:

13 + 11 = 24

4.4 Binary Subtraction Rules

Operation	Result	Borrow
0 − 0	0	0
1 − 0	1	0
1 − 1	0	0
0 − 1	1	Borrow 1

When 1 is borrowed from the next binary column:

10₂ − 1₂ = 1₂

The borrowed 1 has the value 2 in the current column.

4.5 Direct Binary Subtraction Example

Subtract 0101₂ from 1101₂:

Therefore:

1101₂ − 0101₂ = 1000₂

Decimal check:

13 − 5 = 8

4.6 1’s Complement

The 1’s complement of a binary number is obtained by changing:

Every 0 to 1
Every 1 to 0

Example 1

Find the 1’s complement of 101100₂:

Original:       101100
1’s complement: 010011

Example 2

Find the 1’s complement of 00010111₂:

Original:       00010111
1’s complement: 11101000

Important Rule

The number of bits must remain the same.

4.7 2’s Complement

The 2’s complement of a binary number is obtained by:

Finding its 1’s complement.
Adding 1.

Formula

2’s complement = 1’s complement + 1

Example 1

Find the 2’s complement of 101100₂:

Original:       101100
1’s complement: 010011
Add 1:          000001
                ------
2’s complement: 010100

Example 2

Find the 2’s complement of 00010111₂:

Original:       00010111
1’s complement: 11101000
Add 1:          00000001
                --------
2’s complement: 11101001

Right-to-Left Shortcut

To find the 2’s complement:

Start from the right.
Copy all bits up to and including the first 1.
Complement all remaining bits on the left.

Example:

Original:       101100
From right, copy 100
Complement remaining 101 to 010

Result:         010100

4.8 Subtraction Using 2’s Complement

To calculate:

A − B

Use this process:

Make both numbers the same bit length.
Find the 2’s complement of B.
Add it to A.
If an extra carry is produced, discard the carry.
The remaining bits are the positive answer.
If no carry is produced, find the 2’s complement of the result and attach a negative sign.

4.9 Positive Result Using 2’s Complement

Calculate 13 − 5 using 4 bits.

Convert to binary:

13 = 1101
 5 = 0101

Find 2’s complement of 0101:

0101
1’s complement = 1010
Add 1           = 1011

Add to 1101:

   1101
 + 1011
 ------
  11000

Discard the extra carry:

1000₂ = 8₁₀

Therefore:

13 − 5 = 8

4.10 Negative Result Using 2’s Complement

Calculate 5 − 9 using 4 bits.

Convert to binary:

5 = 0101
9 = 1001

Find 2’s complement of 1001:

1001
1’s complement = 0110
Add 1           = 0111

Add:

  0101
+ 0111
------
  1100

No extra carry was produced.

Find the 2’s complement of 1100:

1’s complement = 0011
Add 1           = 0100

Therefore:

Result = −0100₂ = −4₁₀

4.11 1’s Complement Subtraction

For A − B using 1’s complement:

Find the 1’s complement of B.
Add it to A.
If an end carry occurs, add it to the least significant bit.
If no end carry occurs, complement the result and attach a negative sign.

The returned carry is called an end-around carry.

Example: 7 − 5 Using 4 Bits

7 = 0111
5 = 0101

1’s complement of 0101:

Add:

  0111
+ 1010
------
 10001

Add end-around carry:

0001 + 1 = 0010

Therefore:

7 − 5 = 2

For most exam questions, 2’s complement subtraction is more common.

4.12 Signed Binary Numbers

A signed number can represent positive and negative values.

Common methods include:

Sign-magnitude
1’s complement
2’s complement

Sign-Magnitude

The leftmost bit represents the sign:

0 = Positive
1 = Negative

For an 8-bit sign-magnitude number:

00000101 = +5
10000101 = −5

Sign-magnitude has two representations of zero:

00000000 = +0
10000000 = −0

1’s Complement Representation

A negative number is represented by complementing all bits of its positive form.

It also has two representations of zero:

00000000 = +0
11111111 = −0

2’s Complement Representation

A negative number is represented using the 2’s complement of its positive form.

It has only one representation of zero.

This is the most common method used by modern computers for signed integers.

Range of an n-Bit 2’s Complement Number

−2ⁿ⁻¹ to 2ⁿ⁻¹ − 1

Examples:

Bits	Range
4	−8 to +7
8	−128 to +127
16	−32768 to +32767

4.13 Binary Overflow

Overflow occurs when a result is outside the range that can be represented using the available number of bits.

For 4-bit signed 2’s complement:

Range = −8 to +7

Example:

5 + 4 = 9

The value 9 cannot be represented as a positive 4-bit signed 2’s complement number, so signed overflow occurs.

5. Data Representation

5.1 Why Character Encoding Is Needed

Computers store data as binary.

Letters and symbols must therefore be assigned numeric codes.

A character encoding is a system that assigns numbers to:

Letters
Digits
Punctuation marks
Control characters
Special symbols

Important encoding systems include:

ASCII
Unicode
EBCDIC

BCD is mainly a decimal-digit representation system.

5.2 ASCII

ASCII stands for:

American Standard Code for Information Interchange

ASCII assigns numeric codes to characters.

Standard ASCII

Standard ASCII uses 7 bits.

Number of possible codes:

2⁷ = 128

Code range:

0 to 127

Extended ASCII

The term “extended ASCII” is commonly used for various 8-bit character sets.

Number of possible 8-bit patterns:

2⁸ = 256

Code range:

0 to 255

There is no single universal extended-ASCII table.

5.3 ASCII Character Groups

Character Group	Decimal Range
Digits 0–9	48–57
Uppercase A–Z	65–90
Lowercase a–z	97–122

Essential Codes

Character	Decimal ASCII	Hexadecimal
0	48	30
1	49	31
9	57	39
A	65	41
B	66	42
Z	90	5A
a	97	61
b	98	62
z	122	7A
Space	32	20
New line/LF	10	0A
Carriage return	13	0D

Must Memorize

A = 65
a = 97
0 = 48

Useful Differences

a − A = 97 − 65 = 32

Therefore, the ASCII codes of corresponding uppercase and lowercase English letters differ by 32.

Examples:

B = 66
b = 98

98 − 66 = 32

Sequential Codes

The letters and digits have consecutive codes.

Examples:

A = 65
B = 66
C = 67
D = 68

0 = 48
1 = 49
2 = 50
3 = 51

Find a Character Code

Find ASCII of M:

A = 65
M is the 13th letter

M = 65 + 12
  = 77

Find ASCII of m:

a = 97
m = 97 + 12
  = 109

5.4 Unicode

Unicode is a universal character-encoding standard designed to represent characters from many languages and writing systems.

It includes codes for:

English
Indian scripts
Arabic
Chinese
Mathematical symbols
Currency symbols
Emoji
Historical scripts

Code Point

A code point is the unique numerical value assigned to a Unicode character.

Unicode code points are commonly written with:

U+

Examples:

Character	Unicode Code Point
A	U+0041
a	U+0061
0	U+0030
₹	U+20B9

Unicode Encoding Formats

Unicode code points can be stored using encoding formats such as:

UTF-8
UTF-16
UTF-32

UTF stands for Unicode Transformation Format.

Format	Basic Code-Unit Size
UTF-8	8 bits
UTF-16	16 bits
UTF-32	32 bits

Important Points

Unicode is not limited to 256 characters.
Unicode supports a very large collection of scripts and symbols.
The first 128 Unicode code points match ASCII.
UTF-8 uses one to four bytes for a valid Unicode code point.
Unicode improves text exchange between different systems and languages.

5.5 ASCII vs Unicode

Feature	ASCII	Unicode
Full form	American Standard Code for Information Interchange	Universal character standard called Unicode
Original size	7 bits	Large code space
Standard code count	128	Supports a very large number of characters
Language support	Mainly English characters and controls	Many languages, symbols, and emoji
Examples	A, B, 0, @	English, Indian scripts, Arabic, emoji
Relationship	Basic older character set	First 128 code points match ASCII

5.6 EBCDIC

EBCDIC stands for:

Extended Binary Coded Decimal Interchange Code

Features

It is an 8-bit character encoding.
It was developed by IBM.
It is mainly associated with IBM mainframe and midrange systems.
It can represent 256 bit patterns in an 8-bit code page.

ASCII vs EBCDIC

Feature	ASCII	EBCDIC
Developed by	Standards community; based on ASA/ANSI standardization	IBM
Standard basic size	7 bits	8 bits
Main association	General computer and communication systems	IBM mainframe and midrange systems
Character arrangement	Letters and digits are arranged differently from EBCDIC	Uses IBM-specific code-page arrangements
Full form	American Standard Code for Information Interchange	Extended Binary Coded Decimal Interchange Code

Exam Point

If the question says:

“Which character code is mainly associated with IBM mainframes?”

The answer is:

EBCDIC

5.7 BCD

BCD stands for:

Binary-Coded Decimal

In BCD, each decimal digit is represented separately using four binary bits.

This common form is called 8421 BCD because the bit weights are:

8, 4, 2, 1

5.8 BCD Table

Decimal Digit	8421 BCD
0	0000
1	0001
2	0010
3	0011
4	0100
5	0101
6	0110
7	0111
8	1000
9	1001

The patterns 1010 to 1111 are invalid for one decimal digit in standard 8421 BCD.

5.9 Decimal to BCD

Convert every decimal digit separately.

Example: Convert decimal 59 to BCD

5 = 0101
9 = 1001

Therefore:

59₁₀ in BCD = 0101 1001

Example: Convert decimal 407 to BCD

4 = 0100
0 = 0000
7 = 0111

Therefore:

407₁₀ in BCD = 0100 0000 0111

5.10 BCD to Decimal

Separate the bits into groups of four and convert each group into one decimal digit.

Example

Convert 0010 1001 BCD to decimal:

0010 = 2
1001 = 9

Therefore:

0010 1001 BCD = 29₁₀

5.11 BCD vs Pure Binary

BCD and ordinary binary are not the same.

Example: Decimal 25

Pure binary:

25₁₀ = 11001₂

BCD:

2 = 0010
5 = 0101

25₁₀ = 0010 0101 in BCD

Comparison

Feature	Pure Binary	BCD
Method	Converts complete number to base 2	Converts each decimal digit separately
Decimal 25	11001	0010 0101
Storage efficiency	More efficient	Less efficient
Decimal display conversion	Requires conversion	Convenient digit-by-digit representation
Valid 4-bit values	0000–1111	Only 0000–1001 for one decimal digit

5.12 BCD Addition Rule

When adding two BCD digits:

Add them as binary values.
If the 4-bit result is greater than 1001, or a carry occurs, add 0110.
Carry to the next decimal digit if required.

Example: 7 + 5 in BCD

1100 is not a valid single BCD digit.

Add 0110:

  1100
+ 0110
------
1 0010

Therefore:

0001 0010 BCD = 12₁₀

🔑 Quick Revision Sheet

Bases and Digits

System	Base	Digits
Binary	2	0, 1
Octal	8	0–7
Decimal	10	0–9
Hexadecimal	16	0–9, A–F

Powers to Memorize

Powers of 2

Power	Value
2⁰	1
2¹	2
2²	4
2³	8
2⁴	16
2⁵	32
2⁶	64
2⁷	128
2⁸	256
2⁹	512
2¹⁰	1024

Powers of 8

8⁰ = 1
8¹ = 8
8² = 64
8³ = 512

Powers of 16

16⁰ = 1
16¹ = 16
16² = 256
16³ = 4096

Conversion Rules

Conversion	Rule
Any base to decimal	Multiply digits by positional powers and add
Decimal integer to another base	Repeated division
Decimal fraction to another base	Repeated multiplication
Binary to octal	Group 3 bits
Octal to binary	Replace each digit with 3 bits
Binary to hexadecimal	Group 4 bits
Hexadecimal to binary	Replace each digit with 4 bits
Octal to hexadecimal	Octal → Binary → Hexadecimal
Hexadecimal to octal	Hexadecimal → Binary → Octal

Binary Arithmetic

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10
1 + 1 + 1 = 11

0 − 0 = 0
1 − 0 = 1
1 − 1 = 0
0 − 1 = 1 with borrow

Complements

1’s complement: Invert every bit

2’s complement: 1’s complement + 1

2’s Complement Subtraction

A − B = A + 2’s complement of B

Carry produced: Discard carry; result is positive.
No carry: Take 2’s complement of result and add a negative sign.

Character Codes

Code	Full Form	Key Point
ASCII	American Standard Code for Information Interchange	Standard ASCII uses 7 bits
Unicode	Universal character standard	Supports many languages and symbols
EBCDIC	Extended Binary Coded Decimal Interchange Code	Mainly associated with IBM systems
BCD	Binary-Coded Decimal	Each decimal digit uses 4 bits

ASCII Codes to Memorize

A = 65
Z = 90
a = 97
z = 122
0 = 48
9 = 57
Space = 32

BCD Rule

Convert each decimal digit separately into four bits.

Example:

29₁₀ = 0010 1001 BCD

📝 Practice Questions

Conversion Practice: Questions 1–20

Question 1

What is the decimal value of 101101₂?

A. 41
B. 43
C. 45
D. 47

Question 2

What is the decimal value of 1110101₂?

A. 115
B. 117
C. 119
D. 121

Question 3

What is the decimal value of 101.101₂?

A. 5.125
B. 5.5
C. 5.625
D. 6.25

Question 4

What is the binary equivalent of 25₁₀?

A. 10101₂
B. 11001₂
C. 11010₂
D. 11100₂

Question 5

What is the binary equivalent of 78₁₀?

A. 1001110₂
B. 1010110₂
C. 1011110₂
D. 1100110₂

Question 6

What is the binary equivalent of 13.625₁₀?

A. 1101.011₂
B. 1101.101₂
C. 1110.101₂
D. 1011.101₂

Question 7

What is the decimal value of 347₈?

A. 223
B. 229
C. 231
D. 239

Question 8

What is the octal equivalent of 156₁₀?

A. 224₈
B. 234₈
C. 244₈
D. 264₈

Question 9

What is the decimal value of 12.4₈?

A. 10.25
B. 10.5
C. 12.4
D. 12.5

Question 10

What is the decimal value of 2AF₁₆?

A. 671
B. 675
C. 687
D. 697

Question 11

What is the hexadecimal equivalent of 254₁₀?

A. EF₁₆
B. FE₁₆
C. F4₁₆
D. EE₁₆

Question 12

What is the decimal value of A.C₁₆?

A. 10.12
B. 10.5
C. 10.75
D. 12.10

Question 13

What is the octal equivalent of 1101011₂?

A. 143₈
B. 153₈
C. 163₈
D. 173₈

Question 14

What is the binary equivalent of 572₈?

A. 101101010₂
B. 101110010₂
C. 101111010₂
D. 111101010₂

Question 15

What is the hexadecimal equivalent of 10111101₂?

A. AD₁₆
B. BC₁₆
C. BD₁₆
D. CD₁₆

Question 16

What is the binary equivalent of 3AF₁₆?

A. 001110101111₂
B. 001101101111₂
C. 001111101010₂
D. 001110111101₂

Question 17

What is the hexadecimal equivalent of 725₈?

A. 1C5₁₆
B. 1D5₁₆
C. 2D5₁₆
D. 1E5₁₆

Question 18

What is the octal equivalent of 2F3₁₆?

A. 1353₈
B. 1363₈
C. 1373₈
D. 1463₈

Question 19

What is the hexadecimal equivalent of 1110111₂?

A. 67₁₆
B. 77₁₆
C. 7E₁₆
D. E7₁₆

Question 20

What is the binary equivalent of 6.35₈?

A. 110.011101₂
B. 110.101011₂
C. 101.011101₂
D. 110.011110₂

Binary Arithmetic and Data Representation: Questions 21–30

Question 21

What is 1011₂ + 0110₂?

A. 10001₂
B. 10010₂
C. 11001₂
D. 11111₂

Question 22

What is 1101₂ + 1011₂?

A. 10110₂
B. 11000₂
C. 11010₂
D. 11100₂

Question 23

What is 1101₂ − 0101₂?

A. 0110₂
B. 0111₂
C. 1000₂
D. 1010₂

Question 24

What is the 1’s complement of 101100₂?

A. 010011₂
B. 010100₂
C. 101011₂
D. 110011₂

Question 25

What is the 2’s complement of 101100₂?

A. 010011₂
B. 010100₂
C. 101101₂
D. 110100₂

Question 26

Using 4-bit 2’s complement subtraction, what is 1101₂ − 0101₂?

A. 0110₂
B. 0111₂
C. 1000₂
D. 1010₂

Question 27

What is the decimal ASCII code of uppercase A?

A. 48
B. 65
C. 90
D. 97

Question 28

Which encoding is mainly associated with IBM mainframe systems?

A. ASCII
B. Unicode
C. EBCDIC
D. BCD

Question 29

What is the BCD representation of decimal 59?

A. 0011 1011
B. 0101 1001
C. 0011 1010
D. 0101 1010

Question 30

Which 4-bit pattern is invalid as one decimal digit in 8421 BCD?

A. 0011
B. 0111
C. 1001
D. 1100

✅ Answer Key + Direct Solutions

Q.	Answer
1	C
2	B
3	C
4	B
5	A
6	B
7	C
8	B
9	B
10	C
11	B
12	C
13	B
14	C
15	C
16	A
17	B
18	B
19	B
20	A
21	A
22	B
23	C
24	A
25	B
26	C
27	B
28	C
29	B
30	D

Direct Solutions

1. Answer: C

101101₂
= 32 + 8 + 4 + 1
= 45₁₀

2. Answer: B

1110101₂
= 64 + 32 + 16 + 4 + 1
= 117₁₀

3. Answer: C

101.101₂
= 4 + 1 + 1/2 + 1/8
= 5.625₁₀

4. Answer: B

25 = 16 + 8 + 1
25₁₀ = 11001₂

5. Answer: A

78 = 64 + 8 + 4 + 2
78₁₀ = 1001110₂

6. Answer: B

13₁₀ = 1101₂

0.625 × 2 = 1.250 → 1
0.250 × 2 = 0.500 → 0
0.500 × 2 = 1.000 → 1

13.625₁₀ = 1101.101₂

7. Answer: C

347₈
= 3 × 64 + 4 × 8 + 7
= 192 + 32 + 7
= 231₁₀

8. Answer: B

156 ÷ 8 = 19 remainder 4
 19 ÷ 8 = 2  remainder 3
  2 ÷ 8 = 0  remainder 2

156₁₀ = 234₈

9. Answer: B

12.4₈
= 1 × 8 + 2 × 1 + 4 × 1/8
= 8 + 2 + 0.5
= 10.5₁₀

10. Answer: C

2AF₁₆
= 2 × 256 + 10 × 16 + 15
= 512 + 160 + 15
= 687₁₀

11. Answer: B

254 ÷ 16 = 15 remainder 14 = E
 15 ÷ 16 = 0  remainder 15 = F

254₁₀ = FE₁₆

12. Answer: C

A.C₁₆
= 10 + 12/16
= 10 + 0.75
= 10.75₁₀

13. Answer: B

1101011₂
= 001 101 011
= 1   5   3

= 153₈

14. Answer: C

5 = 101
7 = 111
2 = 010

572₈ = 101111010₂

15. Answer: C

1011 1101
   B    D

10111101₂ = BD₁₆

16. Answer: A

3 = 0011
A = 1010
F = 1111

3AF₁₆ = 001110101111₂

17. Answer: B

725₈
= 111 010 101₂
= 0001 1101 0101₂
= 1D5₁₆

18. Answer: B

2F3₁₆
= 0010 1111 0011₂
= 001 011 110 011₂
= 1363₈

19. Answer: B

1110111₂
= 0111 0111₂
= 77₁₆

20. Answer: A

6 = 110
3 = 011
5 = 101

6.35₈ = 110.011101₂

21. Answer: A

   1011
 + 0110
 ------
  10001

22. Answer: B

   1101
 + 1011
 ------
  11000

23. Answer: C

   1101
 - 0101
 ------
   1000

24. Answer: A

Original:       101100
1’s complement: 010011

25. Answer: B

Original:       101100
1’s complement: 010011
Add 1:          000001
                ------
2’s complement: 010100

26. Answer: C

2’s complement of 0101:

0101 → 1010 + 1 = 1011

   1101
 + 1011
 ------
  11000

Discard carry:

1000₂

27. Answer: B

ASCII of A = 65

28. Answer: C

IBM mainframe encoding = EBCDIC

29. Answer: B

5 = 0101
9 = 1001

59₁₀ = 0101 1001 BCD

30. Answer: D

Valid BCD digits:
0000 to 1001

1100 is invalid for one BCD digit.

🎯 DAY 2 Completion Checklist

Can convert between ANY two number systems fluently
Can do Binary addition and 2's complement subtraction
Know ASCII code ranges (A=65, a=97, 0=48)
Solved 20 conversion practice problems

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DAY 2 — Number Systems and Data Representation — Complete Study Material

Phase 1: Technical Theory | Date: 22 July 2026 | Subject: Number Systems + Data Representation | Expected Questions: 2–3

📌 Day Overview

Today you will master:

Decimal, binary, octal, and hexadecimal number systems
Decimal-to-binary, octal, and hexadecimal conversions
Binary, octal, and hexadecimal-to-decimal conversions
Binary-to-octal and binary-to-hexadecimal conversions
Octal-to-binary and hexadecimal-to-binary conversions
Octal-to-hexadecimal and hexadecimal-to-octal conversions
Conversion of fractional numbers
Binary addition and subtraction
1’s complement and 2’s complement
Subtraction using 2’s complement
ASCII, Unicode, EBCDIC, and BCD
Important character codes

Five-Hour Plan

Hour	Topic
1	Number-system theory
2	Decimal ↔ Binary, Octal, Hexadecimal
3	Binary ↔ Octal, Binary ↔ Hexadecimal, Octal ↔ Hexadecimal
4	Binary arithmetic and complements
5	Data representation and practice

1. Number Systems — Full Explanation

1.1 What Is a Number System?

A number system is a method of representing numbers using a fixed set of symbols.

For example, the decimal number system uses ten symbols:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

A number system is identified by its base, also called its radix.

The base tells us:

How many unique digits the system uses
What positional powers are used

1.2 Main Number Systems

Number System	Base	Allowed Digits	Example
Binary	2	0, 1	1011₂
Octal	8	0–7	157₈
Decimal	10	0–9	259₁₀
Hexadecimal	16	0–9, A–F	2AF₁₆

Subscript

The small number written after a value shows its base.

Examples:

1011₂ means binary 1011.
157₈ means octal 157.
259₁₀ means decimal 259.
2AF₁₆ means hexadecimal 2AF.

When subscripts cannot be written, the base may be shown like this:

(1011)₂
(157)₈
(259)₁₀
(2AF)₁₆

1.3 Positional Number System

Binary, octal, decimal, and hexadecimal are positional number systems.

In a positional system, the value of a digit depends on:

The digit itself
Its position
The base of the number system

General Expansion

For a number:

dₙdₙ₋₁...d₂d₁d₀

Its value is:

dₙ × baseⁿ + dₙ₋₁ × baseⁿ⁻¹ + ... + d₁ × base¹ + d₀ × base⁰

Remember:

Any non-zero number raised to power 0 = 1

Examples:

2⁰ = 1
8⁰ = 1
10⁰ = 1
16⁰ = 1

1.4 Place Values on Both Sides of the Radix Point

The decimal point is called a radix point when discussing different number systems.

Positions to the left use positive powers.

Positions to the right use negative powers.

... base³  base²  base¹  base⁰ . base⁻¹  base⁻²  base⁻³ ...

Decimal Example

472.35₁₀

= 4 × 10² + 7 × 10¹ + 2 × 10⁰
  + 3 × 10⁻¹ + 5 × 10⁻²

= 400 + 70 + 2 + 0.3 + 0.05

= 472.35₁₀

1.5 Decimal Number System

The decimal number system has base 10.

Digits

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Place Values

... 10⁴  10³  10²  10¹  10⁰

... 10000 1000 100 10 1

Example

Expand 5837₁₀:

5837
= 5 × 10³ + 8 × 10² + 3 × 10¹ + 7 × 10⁰
= 5000 + 800 + 30 + 7
= 5837

1.6 Binary Number System

The binary number system has base 2.

Digits

0 and 1

Each binary digit is called a bit.

Bit stands for binary digit.

Binary Place Values

Power	Value
2⁰	1
2¹	2
2²	4
2³	8
2⁴	16
2⁵	32
2⁶	64
2⁷	128
2⁸	256
2⁹	512
2¹⁰	1024
2¹¹	2048
2¹²	4096

Binary Fractional Place Values

Power	Value
2⁻¹	1/2 = 0.5
2⁻²	1/4 = 0.25
2⁻³	1/8 = 0.125
2⁻⁴	1/16 = 0.0625
2⁻⁵	1/32 = 0.03125

Example

Expand 1101₂:

1101₂
= 1 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰
= 8 + 4 + 0 + 1
= 13₁₀

Why Computers Use Binary

Digital circuits can easily represent two stable states:

On and off
High and low voltage
True and false
1 and 0

1.7 Octal Number System

The octal number system has base 8.

Digits

0, 1, 2, 3, 4, 5, 6, 7

Digits 8 and 9 are not valid in octal.

Octal Place Values

Power	Value
8⁰	1
8¹	8
8²	64
8³	512
8⁴	4096

Example

Expand 157₈:

157₈
= 1 × 8² + 5 × 8¹ + 7 × 8⁰
= 64 + 40 + 7
= 111₁₀

Binary Relationship

One octal digit corresponds to exactly three binary bits because:

8 = 2³

1.8 Hexadecimal Number System

The hexadecimal number system has base 16.

Digits

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

Letter Values

Hexadecimal Digit	Decimal Value	Binary
A	10	1010
B	11	1011
C	12	1100
D	13	1101
E	14	1110
F	15	1111

Hexadecimal Place Values

Power	Value
16⁰	1
16¹	16
16²	256
16³	4096
16⁴	65536

Example

Expand 2AF₁₆:

2AF₁₆
= 2 × 16² + A × 16¹ + F × 16⁰
= 2 × 256 + 10 × 16 + 15 × 1
= 512 + 160 + 15
= 687₁₀

Binary Relationship

One hexadecimal digit corresponds to exactly four binary bits because:

16 = 2⁴

1.9 Valid and Invalid Digits

Number System	Valid Digits	Examples of Invalid Digits
Binary	0, 1	2, 3, 8, 9
Octal	0–7	8, 9
Decimal	0–9	A, B, C
Hexadecimal	0–9, A–F	G, H, Z

Examples

101101₂ is valid.
10201₂ is invalid because binary does not contain 2.
725₈ is valid.
789₈ is invalid because octal does not contain 8 or 9.
FACE₁₆ is valid.
G15₁₆ is invalid because hexadecimal does not contain G.

2. Decimal ↔ Binary, Octal, and Hexadecimal

2.1 Conversion to Decimal Using Positional Weights

To convert binary, octal, or hexadecimal to decimal:

Write the place value of every digit.
Multiply each digit by its place value.
Add all results.

2.2 Binary to Decimal

Example 1: Convert 101101₂ to decimal

101101₂
= 1 × 2⁵ + 0 × 2⁴ + 1 × 2³
  + 1 × 2² + 0 × 2¹ + 1 × 2⁰

= 32 + 0 + 8 + 4 + 0 + 1

= 45₁₀

Example 2: Convert 111111₂ to decimal

111111₂
= 32 + 16 + 8 + 4 + 2 + 1
= 63₁₀

Shortcut for All Ones

A binary number containing n consecutive ones has decimal value:

2ⁿ − 1

Example:

111111₂ has 6 ones

= 2⁶ − 1
= 64 − 1
= 63

2.3 Binary Fraction to Decimal

Example: Convert 101.101₂ to decimal

101.101₂
= 1 × 2² + 0 × 2¹ + 1 × 2⁰
  + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³

= 4 + 0 + 1 + 0.5 + 0 + 0.125

= 5.625₁₀

2.4 Octal to Decimal

Example: Convert 347₈ to decimal

347₈
= 3 × 8² + 4 × 8¹ + 7 × 8⁰
= 3 × 64 + 4 × 8 + 7
= 192 + 32 + 7
= 231₁₀

Example with Fraction

Convert 12.4₈ to decimal:

12.4₈
= 1 × 8¹ + 2 × 8⁰ + 4 × 8⁻¹
= 8 + 2 + 4/8
= 10 + 0.5
= 10.5₁₀

2.5 Hexadecimal to Decimal

Example: Convert 3B7₁₆ to decimal

3B7₁₆
= 3 × 16² + B × 16¹ + 7 × 16⁰
= 3 × 256 + 11 × 16 + 7
= 768 + 176 + 7
= 951₁₀

Example with Fraction

Convert A.C₁₆ to decimal:

A.C₁₆
= 10 × 16⁰ + 12 × 16⁻¹
= 10 + 12/16
= 10 + 0.75
= 10.75₁₀

2.6 Decimal Integer to Another Base

Use repeated division for the integer part.

Method

To convert a decimal integer to base b:

Divide the number by b.
Record the remainder.
Divide the quotient by b again.
Continue until the quotient becomes 0.
Read the remainders from bottom to top.

2.7 Decimal to Binary

Example: Convert 45₁₀ to binary

Division	Quotient	Remainder
45 ÷ 2	22	1
22 ÷ 2	11	0
11 ÷ 2	5	1
5 ÷ 2	2	1
2 ÷ 2	1	0
1 ÷ 2	0	1

Read the remainders from bottom to top:

45₁₀ = 101101₂

Power-of-Two Shortcut

Convert 25₁₀ to binary:

25 = 16 + 8 + 1
   = 2⁴ + 2³ + 2⁰

Place 1 at powers 4, 3, and 0:

11001₂

2.8 Decimal to Octal

Example: Convert 156₁₀ to octal

Division	Quotient	Remainder
156 ÷ 8	19	4
19 ÷ 8	2	3
2 ÷ 8	0	2

Read upward:

156₁₀ = 234₈

2.9 Decimal to Hexadecimal

Example: Convert 254₁₀ to hexadecimal

Division	Quotient	Remainder
254 ÷ 16	15	14 = E
15 ÷ 16	0	15 = F

Read upward:

254₁₀ = FE₁₆

Remainder Conversion

Decimal Remainder	Hexadecimal Digit
10	A
11	B
12	C
13	D
14	E
15	F

2.10 Decimal Fraction to Another Base

Use repeated multiplication for the fractional part.

Method

To convert a decimal fraction to base b:

Multiply the fraction by b.
Record the integer part of the result.
Multiply the new fractional part by b.
Continue until the fraction becomes 0 or enough digits are obtained.
Read integer parts from top to bottom.

2.11 Decimal Fraction to Binary

Example: Convert 0.625₁₀ to binary

Multiplication	Integer Part	Fractional Part
0.625 × 2 = 1.250	1	0.250
0.250 × 2 = 0.500	0	0.500
0.500 × 2 = 1.000	1	0

Read the integer parts from top to bottom:

0.625₁₀ = 0.101₂

Mixed Number Example

Convert 13.625₁₀ to binary:

Integer part:

13₁₀ = 1101₂

Fractional part:

0.625₁₀ = 0.101₂

Combined result:

13.625₁₀ = 1101.101₂

2.12 Decimal Fraction to Octal

Convert 0.6875₁₀ to octal:

Multiplication	Integer Part	Fractional Part
0.6875 × 8 = 5.5	5	0.5
0.5 × 8 = 4.0	4	0

Therefore:

0.6875₁₀ = 0.54₈

2.13 Decimal Fraction to Hexadecimal

Convert 0.8125₁₀ to hexadecimal:

0.8125 × 16 = 13.0

13 = D

Therefore:

0.8125₁₀ = 0.D₁₆

3. Binary ↔ Octal, Binary ↔ Hexadecimal, Octal ↔ Hexadecimal

3.1 Binary-to-Octal Table

Each octal digit equals three binary bits.

Binary	Octal
000	0
001	1
010	2
011	3
100	4
101	5
110	6
111	7

3.2 Binary to Octal

Method

Start from the radix point.
For the integer part, group bits in sets of three from right to left.
For the fractional part, group bits in sets of three from left to right.
Add zeros at the outer ends if required.
Replace each group with one octal digit.

Example 1

Convert 1101011₂ to octal:

Group from right:

1 101 011

Add leading zeros:

001 101 011

Convert:

001 = 1
101 = 5
011 = 3

Therefore:

1101011₂ = 153₈

Example 2: Fraction

Convert 101101.011₂ to octal:

101 101 . 011
  5   5 .   3

Therefore:

101101.011₂ = 55.3₈

3.3 Octal to Binary

Replace each octal digit with exactly three binary bits.

Example

Convert 572₈ to binary:

5 = 101
7 = 111
2 = 010

Therefore:

572₈ = 101111010₂

Fraction Example

Convert 6.35₈ to binary:

6 = 110
3 = 011
5 = 101

Therefore:

6.35₈ = 110.011101₂

3.4 Binary-to-Hexadecimal Table

Each hexadecimal digit equals four binary bits.

Binary	Hex	Binary	Hex
0000	0	1000	8
0001	1	1001	9
0010	2	1010	A
0011	3	1011	B
0100	4	1100	C
0101	5	1101	D
0110	6	1110	E
0111	7	1111	F

3.5 Binary to Hexadecimal

Method

Group the integer part into sets of four from right to left.
Group the fractional part into sets of four from left to right.
Add outer zeros if needed.
Replace every group with one hexadecimal digit.

Example 1

Convert 10111101₂ to hexadecimal:

1011 1101
   B    D

Therefore:

10111101₂ = BD₁₆

Example 2

Convert 1110111₂ to hexadecimal:

0111 0111
   7    7

Therefore:

1110111₂ = 77₁₆

Fraction Example

Convert 1010.1101₂ to hexadecimal:

1010 . 1101
   A .    D

Therefore:

1010.1101₂ = A.D₁₆

3.6 Hexadecimal to Binary

Replace each hexadecimal digit with exactly four binary bits.

Example

Convert 3AF₁₆ to binary:

3 = 0011
A = 1010
F = 1111

Therefore:

3AF₁₆ = 001110101111₂

Leading zeros may be removed:

3AF₁₆ = 1110101111₂

Fraction Example

Convert C.7₁₆ to binary:

C = 1100
7 = 0111

Therefore:

C.7₁₆ = 1100.0111₂

3.7 Octal to Hexadecimal

There is no simple direct grouping rule between octal and hexadecimal.

Use binary as the bridge:

Octal → Binary → Hexadecimal

Example

Convert 725₈ to hexadecimal.

Step 1: Octal to binary

7 = 111
2 = 010
5 = 101

725₈ = 111010101₂

Step 2: Group into four bits

0001 1101 0101

Step 3: Convert to hexadecimal

0001 = 1
1101 = D
0101 = 5

Therefore:

725₈ = 1D5₁₆

3.8 Hexadecimal to Octal

Use binary as the bridge:

Hexadecimal → Binary → Octal

Example

Convert 2F3₁₆ to octal.

Step 1: Hexadecimal to binary

2 = 0010
F = 1111
3 = 0011

2F3₁₆ = 001011110011₂

Step 2: Group into three bits

001 011 110 011

Step 3: Convert to octal

Therefore:

2F3₁₆ = 1363₈

3.9 Universal Conversion Routes

Starting Base	Target Base	Best Method
Binary	Decimal	Positional weights
Octal	Decimal	Positional weights
Hexadecimal	Decimal	Positional weights
Decimal	Binary	Repeated division by 2
Decimal	Octal	Repeated division by 8
Decimal	Hexadecimal	Repeated division by 16
Binary	Octal	Groups of 3 bits
Octal	Binary	Replace each digit with 3 bits
Binary	Hexadecimal	Groups of 4 bits
Hexadecimal	Binary	Replace each digit with 4 bits
Octal	Hexadecimal	Use binary as a bridge
Hexadecimal	Octal	Use binary as a bridge

4. Binary Arithmetic

4.1 Binary Addition Rules

Operation	Sum	Carry
0 + 0	0	0
0 + 1	1	0
1 + 0	1	0
1 + 1	0	1
1 + 1 + 1	1	1

Remember:

1 + 1 = 10₂

1 + 1 + 1 = 11₂

4.2 Binary Addition Example 1

Add 1011₂ and 0110₂:

Column calculation from right:

1 + 0 = 1
1 + 1 = 10: write 0, carry 1
0 + 1 + carry 1 = 10: write 0, carry 1
1 + 0 + carry 1 = 10: write 0, carry 1
Write final carry 1

Therefore:

1011₂ + 0110₂ = 10001₂

Decimal check:

11 + 6 = 17

4.3 Binary Addition Example 2

Add 1101₂ and 1011₂:

Therefore:

1101₂ + 1011₂ = 11000₂

Decimal check:

13 + 11 = 24

4.4 Binary Subtraction Rules

Operation	Result	Borrow
0 − 0	0	0
1 − 0	1	0
1 − 1	0	0
0 − 1	1	Borrow 1

When 1 is borrowed from the next binary column:

10₂ − 1₂ = 1₂

The borrowed 1 has the value 2 in the current column.

4.5 Direct Binary Subtraction Example

Subtract 0101₂ from 1101₂:

Therefore:

1101₂ − 0101₂ = 1000₂

Decimal check:

13 − 5 = 8

4.6 1’s Complement

The 1’s complement of a binary number is obtained by changing:

Every 0 to 1
Every 1 to 0

Example 1

Find the 1’s complement of 101100₂:

Original:       101100
1’s complement: 010011

Example 2

Find the 1’s complement of 00010111₂:

Original:       00010111
1’s complement: 11101000

Important Rule

The number of bits must remain the same.

4.7 2’s Complement

The 2’s complement of a binary number is obtained by:

Finding its 1’s complement.
Adding 1.

Formula

2’s complement = 1’s complement + 1

Example 1

Find the 2’s complement of 101100₂:

Original:       101100
1’s complement: 010011
Add 1:          000001
                ------
2’s complement: 010100

Example 2

Find the 2’s complement of 00010111₂:

Original:       00010111
1’s complement: 11101000
Add 1:          00000001
                --------
2’s complement: 11101001

Right-to-Left Shortcut

To find the 2’s complement:

Start from the right.
Copy all bits up to and including the first 1.
Complement all remaining bits on the left.

Example:

Original:       101100
From right, copy 100
Complement remaining 101 to 010

Result:         010100

4.8 Subtraction Using 2’s Complement

To calculate:

A − B

Use this process:

Make both numbers the same bit length.
Find the 2’s complement of B.
Add it to A.
If an extra carry is produced, discard the carry.
The remaining bits are the positive answer.
If no carry is produced, find the 2’s complement of the result and attach a negative sign.

4.9 Positive Result Using 2’s Complement

Calculate 13 − 5 using 4 bits.

Convert to binary:

13 = 1101
 5 = 0101

Find 2’s complement of 0101:

0101
1’s complement = 1010
Add 1           = 1011

Add to 1101:

   1101
 + 1011
 ------
  11000

Discard the extra carry:

1000₂ = 8₁₀

Therefore:

13 − 5 = 8

4.10 Negative Result Using 2’s Complement

Calculate 5 − 9 using 4 bits.

Convert to binary:

5 = 0101
9 = 1001

Find 2’s complement of 1001:

1001
1’s complement = 0110
Add 1           = 0111

Add:

  0101
+ 0111
------
  1100

No extra carry was produced.

Find the 2’s complement of 1100:

1’s complement = 0011
Add 1           = 0100

Therefore:

Result = −0100₂ = −4₁₀

4.11 1’s Complement Subtraction

For A − B using 1’s complement:

Find the 1’s complement of B.
Add it to A.
If an end carry occurs, add it to the least significant bit.
If no end carry occurs, complement the result and attach a negative sign.

The returned carry is called an end-around carry.

Example: 7 − 5 Using 4 Bits

7 = 0111
5 = 0101

1’s complement of 0101:

Add:

  0111
+ 1010
------
 10001

Add end-around carry:

0001 + 1 = 0010

Therefore:

7 − 5 = 2

For most exam questions, 2’s complement subtraction is more common.

4.12 Signed Binary Numbers

A signed number can represent positive and negative values.

Common methods include:

Sign-magnitude
1’s complement
2’s complement

Sign-Magnitude

The leftmost bit represents the sign:

0 = Positive
1 = Negative

For an 8-bit sign-magnitude number:

00000101 = +5
10000101 = −5

Sign-magnitude has two representations of zero:

00000000 = +0
10000000 = −0

1’s Complement Representation

A negative number is represented by complementing all bits of its positive form.

It also has two representations of zero:

00000000 = +0
11111111 = −0

2’s Complement Representation

A negative number is represented using the 2’s complement of its positive form.

It has only one representation of zero.

This is the most common method used by modern computers for signed integers.

Range of an n-Bit 2’s Complement Number

−2ⁿ⁻¹ to 2ⁿ⁻¹ − 1

Examples:

Bits	Range
4	−8 to +7
8	−128 to +127
16	−32768 to +32767

4.13 Binary Overflow

Overflow occurs when a result is outside the range that can be represented using the available number of bits.

For 4-bit signed 2’s complement:

Range = −8 to +7

Example:

5 + 4 = 9

The value 9 cannot be represented as a positive 4-bit signed 2’s complement number, so signed overflow occurs.

5. Data Representation

5.1 Why Character Encoding Is Needed

Computers store data as binary.

Letters and symbols must therefore be assigned numeric codes.

A character encoding is a system that assigns numbers to:

Letters
Digits
Punctuation marks
Control characters
Special symbols

Important encoding systems include:

ASCII
Unicode
EBCDIC

BCD is mainly a decimal-digit representation system.

5.2 ASCII

ASCII stands for:

American Standard Code for Information Interchange

ASCII assigns numeric codes to characters.

Standard ASCII

Standard ASCII uses 7 bits.

Number of possible codes:

2⁷ = 128

Code range:

0 to 127

Extended ASCII

The term “extended ASCII” is commonly used for various 8-bit character sets.

Number of possible 8-bit patterns:

2⁸ = 256

Code range:

0 to 255

There is no single universal extended-ASCII table.

5.3 ASCII Character Groups

Character Group	Decimal Range
Digits 0–9	48–57
Uppercase A–Z	65–90
Lowercase a–z	97–122

Essential Codes

Character	Decimal ASCII	Hexadecimal
0	48	30
1	49	31
9	57	39
A	65	41
B	66	42
Z	90	5A
a	97	61
b	98	62
z	122	7A
Space	32	20
New line/LF	10	0A
Carriage return	13	0D

Must Memorize

A = 65
a = 97
0 = 48

Useful Differences

a − A = 97 − 65 = 32

Therefore, the ASCII codes of corresponding uppercase and lowercase English letters differ by 32.

Examples:

B = 66
b = 98

98 − 66 = 32

Sequential Codes

The letters and digits have consecutive codes.

Examples:

A = 65
B = 66
C = 67
D = 68

0 = 48
1 = 49
2 = 50
3 = 51

Find a Character Code

Find ASCII of M:

A = 65
M is the 13th letter

M = 65 + 12
  = 77

Find ASCII of m:

a = 97
m = 97 + 12
  = 109

5.4 Unicode

Unicode is a universal character-encoding standard designed to represent characters from many languages and writing systems.

It includes codes for:

English
Indian scripts
Arabic
Chinese
Mathematical symbols
Currency symbols
Emoji
Historical scripts

Code Point

A code point is the unique numerical value assigned to a Unicode character.

Unicode code points are commonly written with:

U+

Examples:

Character	Unicode Code Point
A	U+0041
a	U+0061
0	U+0030
₹	U+20B9

Unicode Encoding Formats

Unicode code points can be stored using encoding formats such as:

UTF-8
UTF-16
UTF-32

UTF stands for Unicode Transformation Format.

Format	Basic Code-Unit Size
UTF-8	8 bits
UTF-16	16 bits
UTF-32	32 bits

Important Points

Unicode is not limited to 256 characters.
Unicode supports a very large collection of scripts and symbols.
The first 128 Unicode code points match ASCII.
UTF-8 uses one to four bytes for a valid Unicode code point.
Unicode improves text exchange between different systems and languages.

5.5 ASCII vs Unicode

Feature	ASCII	Unicode
Full form	American Standard Code for Information Interchange	Universal character standard called Unicode
Original size	7 bits	Large code space
Standard code count	128	Supports a very large number of characters
Language support	Mainly English characters and controls	Many languages, symbols, and emoji
Examples	A, B, 0, @	English, Indian scripts, Arabic, emoji
Relationship	Basic older character set	First 128 code points match ASCII

5.6 EBCDIC

EBCDIC stands for:

Extended Binary Coded Decimal Interchange Code

Features

It is an 8-bit character encoding.
It was developed by IBM.
It is mainly associated with IBM mainframe and midrange systems.
It can represent 256 bit patterns in an 8-bit code page.

ASCII vs EBCDIC

Feature	ASCII	EBCDIC
Developed by	Standards community; based on ASA/ANSI standardization	IBM
Standard basic size	7 bits	8 bits
Main association	General computer and communication systems	IBM mainframe and midrange systems
Character arrangement	Letters and digits are arranged differently from EBCDIC	Uses IBM-specific code-page arrangements
Full form	American Standard Code for Information Interchange	Extended Binary Coded Decimal Interchange Code

Exam Point

If the question says:

“Which character code is mainly associated with IBM mainframes?”

The answer is:

EBCDIC

5.7 BCD

BCD stands for:

Binary-Coded Decimal

In BCD, each decimal digit is represented separately using four binary bits.

This common form is called 8421 BCD because the bit weights are:

8, 4, 2, 1

5.8 BCD Table

Decimal Digit	8421 BCD
0	0000
1	0001
2	0010
3	0011
4	0100
5	0101
6	0110
7	0111
8	1000
9	1001

The patterns 1010 to 1111 are invalid for one decimal digit in standard 8421 BCD.

5.9 Decimal to BCD

Convert every decimal digit separately.

Example: Convert decimal 59 to BCD

5 = 0101
9 = 1001

Therefore:

59₁₀ in BCD = 0101 1001

Example: Convert decimal 407 to BCD

4 = 0100
0 = 0000
7 = 0111

Therefore:

407₁₀ in BCD = 0100 0000 0111

5.10 BCD to Decimal

Separate the bits into groups of four and convert each group into one decimal digit.

Example

Convert 0010 1001 BCD to decimal:

0010 = 2
1001 = 9

Therefore:

0010 1001 BCD = 29₁₀

5.11 BCD vs Pure Binary

BCD and ordinary binary are not the same.

Example: Decimal 25

Pure binary:

25₁₀ = 11001₂

BCD:

2 = 0010
5 = 0101

25₁₀ = 0010 0101 in BCD

Comparison

Feature	Pure Binary	BCD
Method	Converts complete number to base 2	Converts each decimal digit separately
Decimal 25	11001	0010 0101
Storage efficiency	More efficient	Less efficient
Decimal display conversion	Requires conversion	Convenient digit-by-digit representation
Valid 4-bit values	0000–1111	Only 0000–1001 for one decimal digit

5.12 BCD Addition Rule

When adding two BCD digits:

Add them as binary values.
If the 4-bit result is greater than 1001, or a carry occurs, add 0110.
Carry to the next decimal digit if required.

Example: 7 + 5 in BCD

1100 is not a valid single BCD digit.

Add 0110:

  1100
+ 0110
------
1 0010

Therefore:

0001 0010 BCD = 12₁₀

🔑 Quick Revision Sheet

Bases and Digits

System	Base	Digits
Binary	2	0, 1
Octal	8	0–7
Decimal	10	0–9
Hexadecimal	16	0–9, A–F

Powers to Memorize

Powers of 2

Power	Value
2⁰	1
2¹	2
2²	4
2³	8
2⁴	16
2⁵	32
2⁶	64
2⁷	128
2⁸	256
2⁹	512
2¹⁰	1024

Powers of 8

8⁰ = 1
8¹ = 8
8² = 64
8³ = 512

Powers of 16

16⁰ = 1
16¹ = 16
16² = 256
16³ = 4096

Conversion Rules

Conversion	Rule
Any base to decimal	Multiply digits by positional powers and add
Decimal integer to another base	Repeated division
Decimal fraction to another base	Repeated multiplication
Binary to octal	Group 3 bits
Octal to binary	Replace each digit with 3 bits
Binary to hexadecimal	Group 4 bits
Hexadecimal to binary	Replace each digit with 4 bits
Octal to hexadecimal	Octal → Binary → Hexadecimal
Hexadecimal to octal	Hexadecimal → Binary → Octal

Binary Arithmetic

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10
1 + 1 + 1 = 11

0 − 0 = 0
1 − 0 = 1
1 − 1 = 0
0 − 1 = 1 with borrow

Complements

1’s complement: Invert every bit

2’s complement: 1’s complement + 1

2’s Complement Subtraction

A − B = A + 2’s complement of B

Carry produced: Discard carry; result is positive.
No carry: Take 2’s complement of result and add a negative sign.

Character Codes

Code	Full Form	Key Point
ASCII	American Standard Code for Information Interchange	Standard ASCII uses 7 bits
Unicode	Universal character standard	Supports many languages and symbols
EBCDIC	Extended Binary Coded Decimal Interchange Code	Mainly associated with IBM systems
BCD	Binary-Coded Decimal	Each decimal digit uses 4 bits

ASCII Codes to Memorize

A = 65
Z = 90
a = 97
z = 122
0 = 48
9 = 57
Space = 32

BCD Rule

Convert each decimal digit separately into four bits.

Example:

29₁₀ = 0010 1001 BCD

📝 Practice Questions

Conversion Practice: Questions 1–20

Question 1

What is the decimal value of 101101₂?

A. 41
B. 43
C. 45
D. 47

Question 2

What is the decimal value of 1110101₂?

A. 115
B. 117
C. 119
D. 121

Question 3

What is the decimal value of 101.101₂?

A. 5.125
B. 5.5
C. 5.625
D. 6.25

Question 4

What is the binary equivalent of 25₁₀?

A. 10101₂
B. 11001₂
C. 11010₂
D. 11100₂

Question 5

What is the binary equivalent of 78₁₀?

A. 1001110₂
B. 1010110₂
C. 1011110₂
D. 1100110₂

Question 6

What is the binary equivalent of 13.625₁₀?

A. 1101.011₂
B. 1101.101₂
C. 1110.101₂
D. 1011.101₂

Question 7

What is the decimal value of 347₈?

A. 223
B. 229
C. 231
D. 239

Question 8

What is the octal equivalent of 156₁₀?

A. 224₈
B. 234₈
C. 244₈
D. 264₈

Question 9

What is the decimal value of 12.4₈?

A. 10.25
B. 10.5
C. 12.4
D. 12.5

Question 10

What is the decimal value of 2AF₁₆?

A. 671
B. 675
C. 687
D. 697

Question 11

What is the hexadecimal equivalent of 254₁₀?

A. EF₁₆
B. FE₁₆
C. F4₁₆
D. EE₁₆

Question 12

What is the decimal value of A.C₁₆?

A. 10.12
B. 10.5
C. 10.75
D. 12.10

Question 13

What is the octal equivalent of 1101011₂?

A. 143₈
B. 153₈
C. 163₈
D. 173₈

Question 14

What is the binary equivalent of 572₈?

A. 101101010₂
B. 101110010₂
C. 101111010₂
D. 111101010₂

Question 15

What is the hexadecimal equivalent of 10111101₂?

A. AD₁₆
B. BC₁₆
C. BD₁₆
D. CD₁₆

Question 16

What is the binary equivalent of 3AF₁₆?

A. 001110101111₂
B. 001101101111₂
C. 001111101010₂
D. 001110111101₂

Question 17

What is the hexadecimal equivalent of 725₈?

A. 1C5₁₆
B. 1D5₁₆
C. 2D5₁₆
D. 1E5₁₆

Question 18

What is the octal equivalent of 2F3₁₆?

A. 1353₈
B. 1363₈
C. 1373₈
D. 1463₈

Question 19

What is the hexadecimal equivalent of 1110111₂?

A. 67₁₆
B. 77₁₆
C. 7E₁₆
D. E7₁₆

Question 20

What is the binary equivalent of 6.35₈?

A. 110.011101₂
B. 110.101011₂
C. 101.011101₂
D. 110.011110₂

Binary Arithmetic and Data Representation: Questions 21–30

Question 21

What is 1011₂ + 0110₂?

A. 10001₂
B. 10010₂
C. 11001₂
D. 11111₂

Question 22

What is 1101₂ + 1011₂?

A. 10110₂
B. 11000₂
C. 11010₂
D. 11100₂

Question 23

What is 1101₂ − 0101₂?

A. 0110₂
B. 0111₂
C. 1000₂
D. 1010₂

Question 24

What is the 1’s complement of 101100₂?

A. 010011₂
B. 010100₂
C. 101011₂
D. 110011₂

Question 25

What is the 2’s complement of 101100₂?

A. 010011₂
B. 010100₂
C. 101101₂
D. 110100₂

Question 26

Using 4-bit 2’s complement subtraction, what is 1101₂ − 0101₂?

A. 0110₂
B. 0111₂
C. 1000₂
D. 1010₂

Question 27

What is the decimal ASCII code of uppercase A?

A. 48
B. 65
C. 90
D. 97

Question 28

Which encoding is mainly associated with IBM mainframe systems?

A. ASCII
B. Unicode
C. EBCDIC
D. BCD

Question 29

What is the BCD representation of decimal 59?

A. 0011 1011
B. 0101 1001
C. 0011 1010
D. 0101 1010

Question 30

Which 4-bit pattern is invalid as one decimal digit in 8421 BCD?

A. 0011
B. 0111
C. 1001
D. 1100

✅ Answer Key + Direct Solutions

Q.	Answer
1	C
2	B
3	C
4	B
5	A
6	B
7	C
8	B
9	B
10	C
11	B
12	C
13	B
14	C
15	C
16	A
17	B
18	B
19	B
20	A
21	A
22	B
23	C
24	A
25	B
26	C
27	B
28	C
29	B
30	D

Direct Solutions

1. Answer: C

101101₂
= 32 + 8 + 4 + 1
= 45₁₀

2. Answer: B

1110101₂
= 64 + 32 + 16 + 4 + 1
= 117₁₀

3. Answer: C

101.101₂
= 4 + 1 + 1/2 + 1/8
= 5.625₁₀

4. Answer: B

25 = 16 + 8 + 1
25₁₀ = 11001₂

5. Answer: A

78 = 64 + 8 + 4 + 2
78₁₀ = 1001110₂

6. Answer: B

13₁₀ = 1101₂

0.625 × 2 = 1.250 → 1
0.250 × 2 = 0.500 → 0
0.500 × 2 = 1.000 → 1

13.625₁₀ = 1101.101₂

7. Answer: C

347₈
= 3 × 64 + 4 × 8 + 7
= 192 + 32 + 7
= 231₁₀

8. Answer: B

156 ÷ 8 = 19 remainder 4
 19 ÷ 8 = 2  remainder 3
  2 ÷ 8 = 0  remainder 2

156₁₀ = 234₈

9. Answer: B

12.4₈
= 1 × 8 + 2 × 1 + 4 × 1/8
= 8 + 2 + 0.5
= 10.5₁₀

10. Answer: C

2AF₁₆
= 2 × 256 + 10 × 16 + 15
= 512 + 160 + 15
= 687₁₀

11. Answer: B

254 ÷ 16 = 15 remainder 14 = E
 15 ÷ 16 = 0  remainder 15 = F

254₁₀ = FE₁₆

12. Answer: C

A.C₁₆
= 10 + 12/16
= 10 + 0.75
= 10.75₁₀

13. Answer: B

1101011₂
= 001 101 011
= 1   5   3

= 153₈

14. Answer: C

5 = 101
7 = 111
2 = 010

572₈ = 101111010₂

15. Answer: C

1011 1101
   B    D

10111101₂ = BD₁₆

16. Answer: A

3 = 0011
A = 1010
F = 1111

3AF₁₆ = 001110101111₂

17. Answer: B

725₈
= 111 010 101₂
= 0001 1101 0101₂
= 1D5₁₆

18. Answer: B

2F3₁₆
= 0010 1111 0011₂
= 001 011 110 011₂
= 1363₈

19. Answer: B

1110111₂
= 0111 0111₂
= 77₁₆

20. Answer: A

6 = 110
3 = 011
5 = 101

6.35₈ = 110.011101₂

21. Answer: A

   1011
 + 0110
 ------
  10001

22. Answer: B

   1101
 + 1011
 ------
  11000

23. Answer: C

   1101
 - 0101
 ------
   1000

24. Answer: A

Original:       101100
1’s complement: 010011

25. Answer: B

Original:       101100
1’s complement: 010011
Add 1:          000001
                ------
2’s complement: 010100

26. Answer: C

2’s complement of 0101:

0101 → 1010 + 1 = 1011

   1101
 + 1011
 ------
  11000

Discard carry:

1000₂

27. Answer: B

ASCII of A = 65

28. Answer: C

IBM mainframe encoding = EBCDIC

29. Answer: B

5 = 0101
9 = 1001

59₁₀ = 0101 1001 BCD

30. Answer: D

Valid BCD digits:
0000 to 1001

1100 is invalid for one BCD digit.

🎯 DAY 2 Completion Checklist

Can convert between ANY two number systems fluently
Can do Binary addition and 2's complement subtraction
Know ASCII code ranges (A=65, a=97, 0=48)
Solved 20 conversion practice problems

Q.	Answer
1	C
2	B
3	C
4	B
5	A
6	B
7	C
8	B
9	B
10	C
11	B
12	C
13	B
14	C
15	C
16	A
17	B
18	B
19	B
20	A
21	A
22	B
23	C
24	A
25	B
26	C
27	B
28	C
29	B
30	D

Q.	Answer
1	C
2	B
3	C
4	B
5	A
6	B
7	C
8	B
9	B
10	C
11	B
12	C
13	B
14	C
15	C
16	A
17	B
18	B
19	B
20	A
21	A
22	B
23	C
24	A
25	B
26	C
27	B
28	C
29	B
30	D

Q.	Answer
1	C
2	B
3	C
4	B
5	A
6	B
7	C
8	B
9	B
10	C
11	B
12	C
13	B
14	C
15	C
16	A
17	B
18	B
19	B
20	A
21	A
22	B
23	C
24	A
25	B
26	C
27	B
28	C
29	B
30	D

Q.	Answer
1	C
2	B
3	C
4	B
5	A
6	B
7	C
8	B
9	B
10	C
11	B
12	C
13	B
14	C
15	C
16	A
17	B
18	B
19	B
20	A
21	A
22	B
23	C
24	A
25	B
26	C
27	B
28	C
29	B
30	D

Q.	Answer
1	C
2	B
3	C
4	B
5	A
6	B
7	C
8	B
9	B
10	C
11	B
12	C
13	B
14	C
15	C
16	A
17	B
18	B
19	B
20	A
21	A
22	B
23	C
24	A
25	B
26	C
27	B
28	C
29	B
30	D

Q.	Answer
1	C
2	B
3	C
4	B
5	A
6	B
7	C
8	B
9	B
10	C
11	B
12	C
13	B
14	C
15	C
16	A
17	B
18	B
19	B
20	A
21	A
22	B
23	C
24	A
25	B
26	C
27	B
28	C
29	B
30	D