Data Representation

What are the bases of decimal, octal, binary and hexadecimal systems ?

Answer

The bases are:

1. Decimal — Base 10
2. Octal — Base 8
3. Binary — Base 2
4. Hexadecimal — Base 16

What is the common property of decimal, octal, binary and hexadecimal number systems ?

Answer

Decimal, octal, binary and hexadecimal number systems are all positional-value system.

Complete the sequence of following binary numbers : 100, 101, 110, ............... , ............... , ............... .

Answer

100, 101, 110, 111 , 1000 , 1001 .

Complete the sequence of following octal numbers : 525, 526, 527, ............... , ............... , ............... .

Answer

525, 526, 527, 530 , 531 , 532 .

Complete the sequence of following hexadecimal numbers : 17, 18, 19, ............... , ............... , ............... .

Answer

17, 18, 19, 1A , 1B , 1C .

Convert the following binary numbers to decimal and hexadecimal:

(a) 1010

(b) 111010

(c) 101011111

(d) 1100

(e) 10010101

(f) 11011100

Answer

(a) 1010

Converting to decimal:

Equivalent decimal number = 8 + 2 = 10

Therefore, (1010)₂ = (10)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1010}$

Therefore, (1010)₂ = (A)₁₆

(b) 111010

Converting to decimal:

Equivalent decimal number = 32 + 16 + 8 + 2 = 58

Therefore, (111010)₂ = (58)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0011} \quad \underlinesegment{1010}$

Therefore, (111010)₂ = (3A)₁₆

(c) 101011111

Converting to decimal:

Equivalent decimal number = 256 + 64 + 16 + 8 + 4 + 2 + 1 = 351

Therefore, (101011111)₂ = (351)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1111}$

Therefore, (101011111)₂ = (15F)₁₆

(d) 1100

Converting to decimal:

Equivalent decimal number = 8 + 4 = 12

Therefore, (1100)₂ = (12)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1100}$

Therefore, (1100)₂ = (C)₁₆

(e) 10010101

Converting to decimal:

Equivalent decimal number = 1 + 4 + 16 + 128 = 149

Therefore, (10010101)₂ = (149)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1001} \quad \underlinesegment{0101}$

Therefore, (101011111)₂ = (95)₁₆

(f) 11011100

Converting to decimal:

Equivalent decimal number = 4 + 8 + 16 + 64 + 128 = 220

Therefore, (11011100)₂ = (220)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1101} \quad \underlinesegment{1100}$

Therefore, (11011100)₂ = (DC)₁₆

Convert the following decimal numbers to binary and octal :

(a) 23

(b) 100

(c) 145

(d) 19

(e) 121

(f) 161

Answer

(a) 23

Converting to binary:

Therefore, (23)₁₀ = (10111)₂

Converting to octal:

Therefore, (23)₁₀ = (27)₈

(b) 100

Converting to binary:

Therefore, (100)₁₀ = (1100100)₂

Converting to octal:

Therefore, (100)₁₀ = (144)₈

(c) 145

Converting to binary:

Therefore, (145)₁₀ = (10010001)₂

Converting to octal:

Therefore, (145)₁₀ = (221)₈

(d) 19

Converting to binary:

Therefore, (19)₁₀ = (10011)₂

Converting to octal:

Therefore, (19)₁₀ = (23)₈

(e) 121

Converting to binary:

Therefore, (121)₁₀ = (1111001)₂

Converting to octal:

Therefore, (121)₁₀ = (171)₈

(f) 161

Converting to binary:

Therefore, (161)₁₀ = (10100001)₂

Converting to octal:

Therefore, (161)₁₀ = (241)₈

Convert the following hexadecimal numbers to binary :

(a) A6

(b) A07

(c) 7AB4

(d) BE

(e) BC9

(f) 9BC8

Answer

(a) A6

(A6)₁₆ = (10100110)₂

(b) A07

(A07)₁₆ = (101000000111)₂

(c) 7AB4

(7AB4)₁₆ = (111101010110100)₂

(d) BE

(BE)₁₆ = (10111110)₂

(e) BC9

(BC9)₁₆ = (101111001001)₂

(f) 9BC8

(9BC8)₁₆ = (1001101111001000)₂

Convert the following binary numbers to hexadecimal and octal :

(a) 10011011101

(b) 1111011101011011

(c) 11010111010111

(d) 1010110110111

(e) 10110111011011

(f) 1111101110101111

Answer

(a) 10011011101

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0100} \quad \underlinesegment{1101} \quad \underlinesegment{1101}$

Therefore, (10011011101)₂ = (4DD)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{011} \quad \underlinesegment{011} \quad \underlinesegment{101}$

Therefore, (10011011101)₂ = (2335)₈

(b) 1111011101011011

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{0111} \quad \underlinesegment{0101} \quad \underlinesegment{1011}$

Therefore, (1111011101011011)₂ = (F75B)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{101} \quad \underlinesegment{011} \quad \underlinesegment{011}$

Therefore, (1111011101011011)₂ = (173533)₈

(c) 11010111010111

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0011} \quad \underlinesegment{0101} \quad \underlinesegment{1101} \quad \underlinesegment{0111}$

Therefore, (11010111010111)₂ = (35D7)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{011} \quad \underlinesegment{010} \quad \underlinesegment{111} \quad \underlinesegment{010} \quad \underlinesegment{111}$

Therefore, (11010111010111)₂ = (32727)₈

(d) 1010110110111

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1011} \quad \underlinesegment{0111}$

Therefore, (1010110110111)₂ = (15B7)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{110} \quad \underlinesegment{111}$

Therefore, (1010110110111)₂ = (12667)₈

(e) 10110111011011

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0010} \quad \underlinesegment{1101} \quad \underlinesegment{1101} \quad \underlinesegment{1011}$

Therefore, (10110111011011)₂ = (2DDB)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{011}$

Therefore, (10110111011011)₂ = (26733)₈

(f) 1111101110101111

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{1011} \quad \underlinesegment{1010} \quad \underlinesegment{1111}$

Therefore, (1111101110101111)₂ = (FBAF)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{101} \quad \underlinesegment{110} \quad \underlinesegment{101} \quad \underlinesegment{111}$

Therefore, (1111101110101111)₂ = (175657)₈

Encode "INDIA" and "MARCHING" into ASCII codeform.

Answer

ASCII value of I is 73.
ASCII value of N is 78.
ASCII value of D is 68.
ASCII value of I is 73.
ASCII value of A is 65.

Hence, ASCII code for the word 'INDIA' is 73 78 68 73 65.

ASCII value of M is 77.
ASCII value of A is 65.
ASCII value of R is 82.
ASCII value of C is 67.
ASCII value of H is 72.
ASCII value of I is 73.
ASCII value of N is 78.
ASCII value of G is 71.

Hence, ASCII code for the word 'MARCHING' is 77 65 82 67 72 73 78 71.

ASCII code is ............... bit code, extended ASCII is ............... bit code and Unicode is ............... bit code.

Answer

ASCII code is 7 bit code, extended ASCII is 8 bit code, and Unicode is 8, 16, 32 bit code.

What is the use of ASCII code and Unicode ?

Answer

ASCII stands for American Standard Code for Information Interchange. It uses a 7-bit code and it can represent 128 characters. ASCII code is mostly used to represent the characters of English language, standard keyboard characters as well as control characters like Carriage Return and Form Feed.

Unicode is a universal character encoding scheme that can represent different sets of characters belonging to different languages by assigning a number to each of the character.

What is the full form of ASCII and Unicode ?

Answer

The full form of ASCII is American Standard Code for Information Interchange and the full form of Unicode is Universal Character Encoding.

Name two Indian languages in Unicode along with their script names.

Answer

Two Indian languages in Unicode along with their script names are:

1. Hindi - Script: Devanagari
2. Bengali - Script: Bengali

The value of radix in binary number system is ..........

1. 2
2. 8
3. 10
4. 16

Answer

2

Reason — The base (or radix) of the binary number system is 2 because it uses only two digits: 0 and 1.

The value of radix in octal number system is ..........

1. 2
2. 8
3. 10
4. 16

Answer

8

Reason — The radix or base of the octal number system is 8 because it uses eight digits, ranging from 0 to 7, to represent numbers.

The value of radix in decimal number system is ..........

1. 2
2. 8
3. 10
4. 16

Answer

10

Reason — The radix or base of the decimal number system is 10 because it uses ten digits, from 0 to 9, to represent numbers.

The value of radix in hexadecimal number system is ..........

1. 2
2. 8
3. 10
4. 16

Answer

16

Reason — The radix or base of the hexadecimal number system is 16 because it uses sixteen symbols (0-9 and A-F) to represent numbers.

Which of the following are not valid symbols in octal number system ?

1. 2
2. 8
3. 9
4. 7

Answer

8, 9

Reason — The octal number system uses digits ranging from 0 to 7. Therefore, 8 and 9 are not valid symbols in the octal system.

Which of the following are not valid symbols in hexadecimal number system ?

1. 2
2. 8
3. 9
4. G
5. F

Answer

G

Reason — The hexadecimal number system uses digits from 0 to 9 and letters from A to F to represent values. Therefore, "G" is not a valid symbol in the hexadecimal system.

Which of the following are not valid symbols in decimal number system ?

1. 2
2. 8
3. 9
4. G
5. F

Answer

G, F

Reason — The decimal number system uses only the digits 0 to 9. Therefore, "G" and "F" are not valid symbols in the decimal system.

The hexadecimal digits are 1 to 0 and A to ..........

1. E
2. F
3. G
4. D

Answer

F

Reason — The hexadecimal number system uses the digits 0 to 9 and the letters A to F to represent values.

The binary equivalent of the decimal number 10 is ..........

1. 0010
2. 10
3. 1010
4. 010

Answer

1010

Reason —

Therefore, (10)₁₀ = (1010)₂.

ASCII code is a 7 bit code for ..........

1. letters
2. numbers
3. other symbols
4. all of these

Answer

all of these

Reason — The ASCII (American Standard Code for Information Interchange) code is a 7-bit code that represents a set of characters including letters, numbers, and various symbols.

How many bytes are there in 1011 1001 0110 1110 numbers?

1. 1
2. 2
3. 4
4. 8

Answer

2

Reason — To calculate the number of bytes:

1. The given number is 1011 1001 0110 1110, which has 16 bits.
2. 1 byte = 8 bits.

To find the number of bytes:

Number of bytes = Total number of bits / Bits per byte

Number of bytes = 16 bits / 8 bits per byte = 2 bytes

So, the number of bytes is 2.

The binary equivalent of the octal Numbers 13.54 is.....

1. 1011.1011
2. 1001.1110
3. 1101.1110
4. None of these

Answer

1101.1110

Reason —

Therefore, (13.54)₈ = ($\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{011}}.\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{100}}$)₂.

The octal equivalent of 111 010 is.....

1. 81
2. 72
3. 71
4. 82

Answer

72

Reason —

Grouping in bits of 3:

$\underlinesegment{111} \quad \underlinesegment{010}$

Therefore, (111010)₂ = (72)₈

The input hexadecimal representation of 1110 is ..........

1. 0111
2. E
3. 15
4. 14

Answer

E

Reason —

Grouping in bits of 4:

$\underlinesegment{1110}$

Therefore, (1110)₂ = (E)₁₆.

Which of the following is not a binary number ?

1. 1111
2. 101
3. 11E
4. 000

Answer

11E

Reason — A binary number system uses only the digits 0 and 1. Therefore, the letter E is not a binary number. The other options (1111, 101, and 000) are valid binary numbers.

Convert the hexadecimal number 2C to decimal:

1. 3A
2. 34
3. 44
4. 43

Answer

44

Reason —

Equivalent decimal number = 32 + 12 = 44

Therefore, (2C)₁₆ = (44)₁₀.

UTF8 is a type of .......... encoding.

1. ASCII
2. extended ASCII
3. Unicode
4. ISCII

Answer

Unicode

Reason — UTF-8 (Unicode Transformation Format - 8-bit) is a type of Unicode encoding. It is a variable-width encoding that can represent every character in Unicode character set.

UTF32 is a type of .......... encoding.

1. ASCII
2. extended ASCII
3. Unicode
4. ISCII

Answer

Unicode

Reason — UTF-32 (Unicode Transformation Format - 32-bit) is a type of Unicode encoding. It is a fixed length encoding scheme that uses exactly 4 bytes to represent all Unicode code points.

Which of the following is not a valid UTF8 representation?

1. 2 octet (16 bits)
2. 3 octet (24 bits)
3. 4 octet (32 bits)
4. 8 octet (64 bits)

Answer

8 octet (64 bits)

Reason — UTF-8 is a variable-width encoding that uses 1 to 4 bytes (octets) to represent each character. The valid UTF-8 representations are:

• 1 octet (8 bits)
• 2 octets (16 bits)
• 3 octets (24 bits)
• 4 octets (32 bits)

Which of the following is not a valid encoding scheme for characters ?

1. ASCII
2. ISCII
3. Unicode
4. ESCII

Answer

ESCII

Reason — ASCII (American Standard Code for Information Interchange), ISCII (Indian Script Code for Information Interchange), and Unicode are well-known character encoding schemes used to represent text in computers and other devices.

The Decimal number system is composed of 10 unique symbols.

The Binary number system is composed of 2 unique symbols.

The Octal number system is composed of 8 unique symbols.

The Hexadecimal number system is composed of 16 unique symbols.

The illegal digits of octal number system are 8 and 9.

Hexadecimal number system recognizes symbols 0 to 9 and A to F.

Each octal number is replaced with 3 bits in octal to binary conversion.

Each Hexadecimal number is replaced with 4 bits in Hex to binary conversion.

ASCII is a 7 bit code while extended ASCII is a 8 bit code.

The Unicode encoding scheme can represent all symbols/characters of most languages.

The ISCII encoding scheme represents Indian Languages' characters on computers.

UTF8 can take upto 4 bytes to represent a symbol.

UTF32 takes exactly 4 bytes to represent a symbol.

Unicode value of a symbol is called code point.

A computer can work with Decimal number system.

Answer

False

Reason — Computers work with the binary number system (base-2), which uses only 0s and 1s for processing data.

A computer can work with Binary number system.

Answer

True

Reason — Computers work with the binary number system (base-2), which uses only 0s and 1s for processing data.

The number of unique symbols in Hexadecimal number system is 15.

Answer

False

Reason — The hexadecimal number system has 16 unique symbols: 0–9 and A–F.

Number systems can also represent characters.

Answer

False

Reason — Number systems are used to represent numbers, not characters. Characters are represented using encoding schemes such as ASCII or Unicode, which map numbers to characters.

ISCII is an encoding scheme created for Indian language characters.

Answer

True

Reason — ISCII (Indian Script Code for Information Interchange) is an encoding scheme developed to represent various Indian language scripts, such as Devanagari, Tamil, Telugu, and others, in digital systems.

Unicode is able to represent nearly all languages' characters.

Answer

True

Reason — Unicode is a universal character encoding standard that can represent characters from almost all languages around the world.

UTF8 is a fixed-length encoding scheme.

Answer

False

Reason — UTF-8 is a variable-width encoding that uses 1 to 4 bytes (octets) to represent each character.

UTF32 is a fixed-length encoding scheme.

Answer

True

Reason — UTF-32 (Unicode Transformation Format - 32-bit) is a type of Unicode encoding. It is a fixed length encoding scheme that uses exactly 4 bytes to represent all Unicode code points.

UTF8 is a variable-length encoding scheme and can represent characters in 1 through 4 bytes.

Answer

True

Reason — UTF-8 is a variable-width encoding that uses 1 to 4 bytes (octets) to represent each character.

UTF8 and UTF32 are the only encoding schemes supported by Unicode.

Answer

False

Reason — Unicode supports multiple encoding schemes, including UTF-8, UTF-16, and UTF-32.

Assertion. A 3-digit binary number can have maximum 8 unique binary values.

Reason. Maximum number of values possible for an n-digit binary number are 2ⁿ.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

Both A and R are true and R is the correct explanation of A.

Explanation
A 3-digit binary number can represent up to 2³=8 unique values, as each digit in a binary number can be either 0 or 1. The maximum number of values possible for an n-digit binary number is 2ⁿ, which confirms that a 3-digit binary number can represent up to 8 unique values.

Assertion. The decimal number system has base 10.

Reason. The decimal number system is composed of 10 unique symbols.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

Both A and R are true and R is the correct explanation of A.

Explanation
The decimal number system has a base of 10 because it uses 10 unique symbols (0 to 9).

Assertion. Some similar looking numbers of different number systems have the same value, e.g., 7₈ = 7₁₀.

Reason. Not all the similar looking numbers of different number systems have the same value, e.g., 23₈ ≠ 23₁₀.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

Both A and R are true but R is not the correct explanation of A.

Explanation
In some cases, numbers that look the same in different number systems do represent the same value. For instance, 7₈ (base-8) and 7₁₀ (base-10) both represent the decimal value 7. However, not all numbers that appear similar in different number systems are equivalent. For example, 23₈ ​(base-8) is not equal to 23₁₀ (base-10). In base-8, 23₈ represents 19 in decimal. In contrast, 23₁₀ represents the decimal value 23. Thus, while the number 23 looks the same, its value differs depending on the number system's base.

Assertion. 029₈ is a valid octal number.

Reason. An octal number cannot have digits 8 and 9 as these are invalid digits in octal number system.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

A is false but R is true.

Explanation
The number 029₈ is invalid octal number because the digit 9 is not allowed in the octal system. An octal number system uses only the digits 0 to 7. Digits 8 and 9 are not valid in this system.

Assertion. An octal digit can be directly converted to binary by placing 3 digit binary equivalent of each octal digit.

Reason. Binary number system has base 2 and octal number system has base 8 and 2³ = 8, hence 3 binary digits for an octal digit.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

Both A and R are true and R is the correct explanation of A.

Explanation
Each octal digit can be directly converted to a 3-digit binary number. For example, the octal digit 7 converts to the binary number 111. This is because the relationship between the octal and binary systems is that 2³ = 8, meaning each octal digit (base-8) maps exactly to 3 binary digits (base-2).

Assertion. 'BED' is a legal hexadecimal value while 'BEDS' is not.

Reason. Hexadecimal number system does not recognize plural items.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

A is true but R is false.

Explanation
BED' is a legal hexadecimal value because the hexadecimal number system uses the digits 0–9 and the letters A–F. On the other hand, 'BEDS' is not a valid hexadecimal number because it includes the letter S, which is not recognized in the hexadecimal system.

Assertion. The 16 unique symbols of hexadecimal can be decided by the OS.

Reason. Unicode has decided the base symbols for hexadecimal as 0-9, A-F.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

A is false but R is true.

Explanation
The 16 unique symbols used in the hexadecimal number system (0-9 and A-F) are standard and defined by the hexadecimal numbering system itself. Unicode specifies these symbols as 0-9 and A-F for hexadecimal notation.

Assertion. Unicode can represent nearly all known characters of all known and scripted languages.

Reason. There are multiple character encoding schemes in Unicode, e.g., UTF-8.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

Both A and R are true but R is not the correct explanation of A.

Explanation
Unicode is developed as a universal character set with the aim of defining all the characters needed for writing the majority of known languages in use on computers in one place. It includes multiple encoding systems to represent these characters, such as UTF-8, UTF-16, and UTF-32.

Assertion. The memory consumed to represent a character/value varies in UTF-8 while it is the same for all characters/values for UTF-32.

Reason. UTF-8 is a variable length encoding scheme and UTF-32 is a fixed length encoding scheme.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

Both A and R are true and R is the correct explanation of A.

Explanation
UTF-8 is a variable-width encoding that uses 1 to 4 bytes (octets) to represent each character, depending on the character's code point. UTF-32, on the other hand, is a fixed-width encoding scheme that uses exactly 4 bytes for every Unicode code point, providing a consistent length for all characters.

What are some number systems used by computers ?

Answer

The most commonly used number systems are decimal, binary, octal and hexadecimal number systems.

What is the use of Hexadecimal number system on computers ?

Answer

The Hexadecimal number system is used in computers to specify memory addresses (which are 16-bit or 32-bit long). For example, a memory address 1101011010101111 is a big binary address but with hex it is D6AF which is easier to remember. The Hexadecimal number system is also used to represent colour codes. For example, FFFFFF represents White, FF0000 represents Red, etc.

What does radix or base signify ?

Answer

The radix or base of a number system signifies how many unique symbols or digits are used in the number system to represent numbers. For example, the decimal number system has a radix or base of 10 meaning it uses 10 digits from 0 to 9 to represent numbers.

What is the use of encoding schemes ?

Answer

Encoding schemes help Computers represent and recognize letters, numbers and symbols. It provides a predetermined set of codes for each recognized letter, number and symbol. Most popular encoding schemes are ASCI, Unicode, ISCII, etc.

Discuss UTF-8 encoding scheme.

Answer

UTF-8 is a variable width encoding that can represent every character in Unicode character set. The code unit of UTF-8 is 8 bits called an octet. It uses 1 to maximum 6 octets to represent code points depending on their size i.e. sometimes it uses 8 bits to store the character, other times 16 or 24 or more bits. It is a type of multi-byte encoding.

How is UTF-8 encoding scheme different from UTF-32 encoding scheme ?

Answer

UTF-8 is a variable length encoding scheme that uses different number of bytes to represent different characters whereas UTF-32 is a fixed length encoding scheme that uses exactly 4 bytes to represent all Unicode code points.

What is the most significant bit and the least significant bit in a binary code ?

Answer

In a binary code, the leftmost bit is called the most significant bit or MSB. It carries the largest weight. The rightmost bit is called the least significant bit or LSB. It carries the smallest weight. For example:

$\begin{matrix} \underset{\bold{MSB}}{1} & 0 & 1 & 1 & 0 & 1 & 1 & \underset{\bold{LSB}}{0} \end{matrix}$

What are ASCII and extended ASCII encoding schemes ?

Answer

ASCII encoding scheme uses a 7-bit code and it represents 128 characters. Its advantages are simplicity and efficiency. Extended ASCII encoding scheme uses a 8-bit code and it represents 256 characters.

What is the utility of ISCII encoding scheme ?

Answer

ISCII or Indian Standard Code for Information Interchange can be used to represent Indian languages on the computer. It supports Indian languages that follow both Devanagari script and other scripts like Tamil, Bengali, Oriya, Assamese, etc.

What is Unicode ? What is its significance ?

Answer

Unicode is a universal character encoding scheme that can represent different sets of characters belonging to different languages by assigning a number to each of the character. It has the following significance:

1. It defines all the characters needed for writing the majority of known languages in use today across the world.
2. It is a superset of all other character sets.
3. It is used to represent characters across different platforms and programs.

What all encoding schemes does Unicode use to represent characters ?

Answer

Unicode uses UTF-8, UTF-16 and UTF-32 encoding schemes.

What are ASCII and ISCII ? Why are these used ?

Answer

ASCII stands for American Standard Code for Information Interchange. It uses a 7-bit code and it can represent 128 characters. ASCII code is mostly used to represent the characters of English language, standard keyboard characters as well as control characters like Carriage Return and Form Feed. ISCII stands for Indian Standard Code for Information Interchange. It uses a 8-bit code and it can represent 256 characters. It retains all ASCII characters and offers coding for Indian scripts also. Majority of the Indian languages can be represented using ISCII.

What are UTF-8 and UTF-32 encoding schemes? Which one is more popular encoding scheme ?

Answer

UTF-8 is a variable length encoding scheme that uses different number of bytes to represent different characters whereas UTF-32 is a fixed length encoding scheme that uses exactly 4 bytes to represent all Unicode code points. UTF-8 is the more popular encoding scheme.

What do you understand by code point ?

Answer

Code point refers to a code from a code space that represents a single character from the character set represented by an encoding scheme. For example, 0x41 is one code point of ASCII that represents character 'A'.

What is the difference between fixed length and variable length encoding schemes ?

Answer

Variable length encoding scheme uses different number of bytes or octets (set of 8 bits) to represent different characters whereas fixed length encoding scheme uses a fixed number of bytes to represent different characters.

Convert the following binary numbers to decimal:

(a) 1101

Answer

Equivalent decimal number = 1 + 4 + 8 = 13

Therefore, (1101)₂ = (13)₁₀

(b) 111010

Answer

Equivalent decimal number = 2 + 8 + 16 + 32 = 58

Therefore, (111010)₂ = (58)₁₀

(c) 101011111

Answer

Equivalent decimal number = 1 + 2 + 4 + 8 + 16 + 64 + 256 = 351

Therefore, (101011111)₂ = (351)₁₀

Convert the following binary numbers to decimal :

(a) 1100

Answer

Equivalent decimal number = 4 + 8 = 12

Therefore, (1100)₂ = (12)₁₀

(b) 10010101

Answer

Equivalent decimal number = 1 + 4 + 16 + 128 = 149

Therefore, (10010101)₂ = (149)₁₀

(c) 11011100

Answer

Equivalent decimal number = 4 + 8 + 16 + 64 + 128 = 220

Therefore, (11011100)₂ = (220)₁₀

Convert the following decimal numbers to binary:

(a) 23

Answer

Therefore, (23)₁₀ = (10111)₂

(b) 100

Answer

Therefore, (100)₁₀ = (1100100)₂

(c) 145

Answer

Therefore, (145)₁₀ = (10010001)₂

(d) 0.25

Answer

Therefore, (0.25)₁₀ = (0.01)₂

Convert the following decimal numbers to binary:

(a) 19

Answer

Therefore, (19)₁₀ = (10011)₂

(b) 122

Answer

Therefore, (122)₁₀ = (1111010)₂

(c) 161

Answer

Therefore, (161)₁₀ = (10100001)₂

(d) 0.675

Answer

(We stop after 5 iterations if fractional part doesn't become 0)

Therefore, (0.675)₁₀ = (0.10101)₂

Convert the following decimal numbers to octal:

(a) 19

Answer

Therefore, (19)₁₀ = (23)₈

(b) 122

Answer

Therefore, (122)₁₀ = (172)₈

(c) 161

Answer

Therefore, (161)₁₀ = (241)₈

(d) 0.675

Answer

Therefore, (0.675)₁₀ = (0.53146)₈

Convert the following hexadecimal numbers to binary:

(a) A6

Answer

(A6)₁₆ = (10100110)₂

(b) A07

Answer

(A07)₁₆ = (101000000111)₂

(c) 7AB4

Answer

(7AB4)₁₆ = (111101010110100)₂

Convert the following hexadecimal numbers to binary:

(a) 23D

Answer

(23D)₁₆ = (1000111101)₂

(b) BC9

Answer

(BC9)₁₆ = (101111001001)₂

(c) 9BC8

Answer

(9BC8)₁₆ = (1001101111001000)₂

Convert the following binary numbers to hexadecimal:

(a) 10011011101

Answer

Grouping in bits of 4:

$\underlinesegment{0100} \quad \underlinesegment{1101} \quad \underlinesegment{1101}$

Therefore, (10011011101)₂ = (4DD)₁₆

(b) 1111011101011011

Answer

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{0111} \quad \underlinesegment{0101} \quad \underlinesegment{1011}$

Therefore, (1111011101011011)₂ = (F75B)₁₆

(c) 11010111010111

Answer

Grouping in bits of 4:

$\underlinesegment{0011} \quad \underlinesegment{0101} \quad \underlinesegment{1101} \quad \underlinesegment{0111}$

Therefore, (11010111010111)₂ = (35D7)₁₆

Convert the following binary numbers to hexadecimal:

(a) 1010110110111

Answer

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1011} \quad \underlinesegment{0111}$

Therefore, (1010110110111)₂ = (15B7)₁₆

(b) 10110111011011

Answer

Grouping in bits of 4:

$\underlinesegment{0010} \quad \underlinesegment{1101} \quad \underlinesegment{1101} \quad \underlinesegment{1011}$

Therefore, (10110111011011)₂ = (2DDB)₁₆

(c) 0110101100

Answer

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{1010} \quad \underlinesegment{1100}$

Therefore, (0110101100)₂ = (1AC)₁₆

Convert the following octal numbers to decimal:

(a) 257

Answer

Equivalent decimal number = 7 + 40 + 128 = 175

Therefore, (257)₈ = (175)₁₀

(b) 3527

Answer

Equivalent decimal number = 7 + 16 + 320 + 1536 = 1879

Therefore, (3527)₈ = (1879)₁₀

(c) 123

Answer

Equivalent decimal number = 3 + 16 + 64 = 83

Therefore, (123)₈ = (83)₁₀

(d) 605.12

Answer

Equivalent decimal number = 5 + 384 + 0.125 + 0.0312 = 389.1562

Therefore, (605.12)₈ = (389.1562)₁₀

Convert the following hexadecimal numbers to decimal:

(a) A6

Answer

Equivalent decimal number = 6 + 160 = 166

Therefore, (A6)₁₆ = (166)₁₀

(b) A13B

Answer

Equivalent decimal number = 11 + 48 + 256 + 40960 = 41275

Therefore, (A13B)₁₆ = (41275)₁₀

(c) 3A5

Answer

Equivalent decimal number = 5 + 160 + 768 = 933

Therefore, (3A5)₁₆ = (933)₁₀

Convert the following hexadecimal numbers to decimal:

(a) E9

Answer

Equivalent decimal number = 9 + 224 = 233

Therefore, (E9)₁₆ = (233)₁₀

(b) 7CA3

Answer

Equivalent decimal number = 3 + 160 + 3072 + 28672 = 31907

Therefore, (7CA3)₁₆ = (31907)₁₀

Convert the following decimal numbers to hexadecimal:

(a) 132

Answer

Therefore, (132)₁₀ = (84)₁₆

(b) 2352

Answer

Therefore, (2352)₁₀ = (930)₁₆

(c) 122

Answer

Therefore, (122)₁₀ = (7A)₁₆

(d) 0.675

Answer

(We stop after 5 iterations if fractional part doesn't become 0)

Therefore, (0.675)₁₀ = (0.ACCCC)₁₆

Convert the following decimal numbers to hexadecimal:

(a) 206

Answer

Therefore, (206)₁₀ = (CE)₁₆

(b) 3619

Answer

Therefore, (3619)₁₀ = (E23)₁₆

Convert the following hexadecimal numbers to octal:

(a) 38AC

Answer

(38AC)₁₆ = (11100010101100)₂

Grouping in bits of 3:

$\underlinesegment{011}\medspace\underlinesegment{100}\medspace\underlinesegment{010}\medspace\underlinesegment{101}\medspace\underlinesegment{100}$

(38AC)₁₆ = (34254)₈

(b) 7FD6

Answer

(7FD6)₁₆ = (111111111010110)₂

Grouping in bits of 3:

$\underlinesegment{111}\medspace\underlinesegment{111}\medspace\underlinesegment{111}\medspace\underlinesegment{010}\medspace\underlinesegment{110}$

(7FD6)₁₆ = (77726)₈

(c) ABCD

Answer

(ABCD)₁₆ = (1010101111001101)₂

Grouping in bits of 3:

$\underlinesegment{001}\medspace\underlinesegment{010}\medspace\underlinesegment{101}\medspace\underlinesegment{111}\medspace\underlinesegment{001}\medspace\underlinesegment{101}$

(ABCD)₁₆ = (125715)₈

Convert the following octal numbers to binary:

(a) 123

Answer

Therefore, (123)₈ = ($\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{011}}$)₂

(b) 3527

Answer

Therefore, (3527)₈ = ($\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{111}}$)₂

(c) 705

Answer

Therefore, (705)₈ = ($\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{101}}$)₂

Convert the following octal numbers to binary:

(a) 7642

Answer

Therefore, (7642)₈ = ($\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{110}}\medspace\bold{\underlinesegment{100}}\medspace\bold{\underlinesegment{010}}$)₂

(b) 7015

Answer

Therefore, (7015)₈ = ($\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{101}}$)₂

(c) 3576

Answer

Therefore, (3576)₈ = ($\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{110}}$)₂

(d) 705

Answer

Therefore, (705)₈ = ($\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{101}}$)₂

Convert the following binary numbers to octal

(a) 111010

Answer

Grouping in bits of 3:

$\underlinesegment{111} \quad \underlinesegment{010}$

Therefore, (111010)₂ = (72)₈

(b) 110110101

Answer

Grouping in bits of 3:

$\underlinesegment{110} \quad \underlinesegment{110} \quad \underlinesegment{101}$

Therefore, (110110101)₂ = (665)₈

(c) 1101100001

Answer

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \underlinesegment{001}$

Therefore, (1101100001)₂ = (1541)₈

Convert the following binary numbers to octal

(a) 11001

Answer

Grouping in bits of 3:

$\underlinesegment{011} \quad \underlinesegment{001}$

Therefore, (11001)₂ = (31)₈

(b) 10101100

Answer

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100}$

Therefore, (10101100)₂ = (254)₈

(c) 111010111

Answer

Grouping in bits of 3:

$\underlinesegment{111} \quad \underlinesegment{010} \quad \underlinesegment{111}$

Therefore, (111010111)₂ = (727)₈

Add the following binary numbers:

(i) 10110111 and 1100101

Answer

$\begin{matrix} & & \overset{1}{1} & \overset{1}{0} & 1 & 1 & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & 1 \</p> <ul> <li>& & & 1 & 1 & 0 & 0 & 1 & 0 & 1 \\ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{0} & \bold{1} & \bold{1} & \bold{1} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (10110111)₂ + (1100101)₂ = (100011100)₂

(ii) 110101 and 101111

Answer

$\begin{matrix} & & \overset{1}{1} & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & \overset{1}{0} & 1 \</p> <ul> <li>& & 1 & 0 & 1 & 1 & 1 & 1 \\ \hline & \bold{1} & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (110101)₂ + (101111)₂ = (1100100)₂

(iii) 110111.110 and 11011101.010

Answer

$\begin{matrix} & & \overset{1}{0} & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & \overset{1}{1} & . & \overset{1}{1} & 1 & 0 \</p> <ul> <li>& & 1 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & . & 0 & 1 & 0 \\ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{.} & \bold{0} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (110111.110)₂ + (11011101.010)₂ = (100010101)₂

(iv) 1110.110 and 11010.011

Answer

$\begin{matrix} & & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & 1 & \overset{1}{0} & . & \overset{1}{1} & 1 & 0 \</p> <ul> <li>& & 1 & 1 & 0 & 1 & 0 & . & 0 & 1 & 1 \\ \hline & \bold{1} & \bold{0} & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{.} & \bold{0} & \bold{0} & \bold{1} \end{matrix}$

Therefore, (1110.110)₂ + (11010.011)₂ = (101001.001)₂

Given that A's code point in ASCII is 65, and a's code point is 97. What is the binary representation of 'A' in ASCII ? (and what's its hexadecimal representation). What is the binary representation of 'a' in ASCII ?

Answer

Binary representation of 'A' in ASCII will be binary representation of its code point 65.

Converting 65 to binary:

Therefore, binary representation of 'A' in ASCII is 1000001.

Converting 65 to Hexadecimal:

Therefore, hexadecimal representation of 'A' in ASCII is (41)₁₆.

Similarly, converting 97 to binary:

Therefore, binary representation of 'a' in ASCII is 1100001.

Convert the following binary numbers to decimal, octal and hexadecimal numbers.

(i) 100101.101

Answer

Decimal Conversion of integral part:

Decimal Conversion of fractional part:

Equivalent decimal number = 1 + 4 + 32 + 0.5 + 0.125 = 37.625

Therefore, (100101.101)₂ = (37.625)₁₀

Octal Conversion

Grouping in bits of 3:

$\underlinesegment{100} \quad \underlinesegment{101} \quad \bold{.} \quad \underlinesegment{101}$

Therefore, (100101.101)₂ = (45.5)₈

Hexadecimal Conversion

Grouping in bits of 4:

$\underlinesegment{0010} \quad \underlinesegment{0101} \medspace . \medspace \underlinesegment{1010}$

Therefore, (100101.101)₂ = (25.A)₁₆

(ii) 10101100.01011

Answer

Decimal Conversion of integral part:

Decimal Conversion of fractional part:

Equivalent decimal number = 4 + 8 + 32 + 128 + 0.25 + 0.0625 + 0.03125 = 172.34375

Therefore, (10101100.01011)₂ = (172.34375)₁₀

Octal Conversion

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \bold{.} \quad \underlinesegment{010} \quad \underlinesegment{110}$

Therefore, (10101100.01011)₂ = (254.26)₈

Hexadecimal Conversion

Grouping in bits of 4:

$\underlinesegment{1010} \quad \underlinesegment{1100} \medspace . \medspace \underlinesegment{0101} \medspace \underlinesegment{1000}$

Therefore, (10101100.01011)₂ = (AC.58)₁₆

(iii) 1010

Answer

Decimal Conversion:

Equivalent decimal number = 2 + 8 = 10

Therefore, (1010)₂ = (10)₁₀

Octal Conversion

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{010}$

Therefore, (1010)₂ = (12)₈

Hexadecimal Conversion

Grouping in bits of 4:

$\underlinesegment{1010}$

Therefore, (1010)₂ = (A)₁₆

(iv) 10101100.010111

Answer

Decimal Conversion of integral part:

Decimal Conversion of fractional part:

Equivalent decimal number = 4 + 8 + 32 + 128 + 0.25 + 0.0625 + 0.03125 + 0.015625 = 172.359375

Therefore, (10101100.010111)₂ = (172.359375)₁₀

Octal Conversion

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \bold{.} \quad \underlinesegment{010} \quad \underlinesegment{111}$

Therefore, (10101100.010111)₂ = (254.27)₈

Hexadecimal Conversion

Grouping in bits of 4:

$\underlinesegment{1010} \quad \underlinesegment{1100} \medspace . \medspace \underlinesegment{0101} \medspace \underlinesegment{1100}$

Therefore, (10101100.010111)₂ = (AC.5C)₁₆