data representation pdf class 11

Data Representation

Class 11 - computer science with python sumita arora, checkpoint 2.1.

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

The bases are:

• Decimal — Base 10
• Octal — Base 8
• Binary — Base 2
• Hexadecimal — Base 16

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

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

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

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

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

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

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

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

Convert the following binary numbers to decimal and hexadecimal:

(c) 101011111

(e) 10010101

(f) 11011100

Converting to decimal:

Equivalent decimal number = 8 + 2 = 10

Therefore, (1010) 2 = (10) 10

Converting to hexadecimal:

Grouping in bits of 4:

1010 undefined \underlinesegment{1010} 1010 ​

Therefore, (1010) 2 = (A) 16

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

Therefore, (111010) 2 = (58) 10

0011 undefined 1010 undefined \underlinesegment{0011} \quad \underlinesegment{1010} 0011 ​ 1010 ​

Therefore, (111010) 2 = (3A) 16

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

Therefore, (101011111) 2 = (351) 10

0001 undefined 0101 undefined 1111 undefined \underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1111} 0001 ​ 0101 ​ 1111 ​

Therefore, (101011111) 2 = (15F) 16

Equivalent decimal number = 8 + 4 = 12

Therefore, (1100) 2 = (12) 10

1100 undefined \underlinesegment{1100} 1100 ​

Therefore, (1100) 2 = (C) 16

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

Therefore, (10010101) 2 = (149) 10

1001 undefined 0101 undefined \underlinesegment{1001} \quad \underlinesegment{0101} 1001 ​ 0101 ​

Therefore, (101011111) 2 = (95) 16

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

Therefore, (11011100) 2 = (220) 10

1101 undefined 1100 undefined \underlinesegment{1101} \quad \underlinesegment{1100} 1101 ​ 1100 ​

Therefore, (11011100) 2 = (DC) 16

Convert the following decimal numbers to binary and octal :

Converting to binary:

Therefore, (23) 10 = (10111) 2

Converting to octal:

Therefore, (23) 10 = (27) 8

Therefore, (100) 10 = (1100100) 2

Therefore, (100) 10 = (144) 8

Therefore, (145) 10 = (10010001) 2

Therefore, (145) 10 = (221) 8

Therefore, (19) 10 = (10011) 2

Therefore, (19) 10 = (23) 8

Therefore, (121) 10 = (1111001) 2

Therefore, (121) 10 = (171) 8

Therefore, (161) 10 = (10100001) 2

Therefore, (161) 10 = (241) 8

Convert the following hexadecimal numbers to binary :

(A6) 16 = (10100110) 2

(A07) 16 = (101000000111) 2

(7AB4) 16 = (111101010110100) 2

(BE) 16 = (10111110) 2

(BC9) 16 = (101111001001) 2

(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

0100 undefined 1101 undefined 1101 undefined \underlinesegment{0100} \quad \underlinesegment{1101} \quad \underlinesegment{1101} 0100 ​ 1101 ​ 1101 ​

Therefore, (10011011101) 2 = (4DD) 16

Converting to Octal:

Grouping in bits of 3:

010 undefined 011 undefined 011 undefined 101 undefined \underlinesegment{010} \quad \underlinesegment{011} \quad \underlinesegment{011} \quad \underlinesegment{101} 010 ​ 011 ​ 011 ​ 101 ​

Therefore, (10011011101) 2 = (2335) 8

1111 undefined 0111 undefined 0101 undefined 1011 undefined \underlinesegment{1111} \quad \underlinesegment{0111} \quad \underlinesegment{0101} \quad \underlinesegment{1011} 1111 ​ 0111 ​ 0101 ​ 1011 ​

Therefore, (1111011101011011) 2 = (F75B) 16

001 undefined 111 undefined 011 undefined 101 undefined 011 undefined 011 undefined \underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{101} \quad \underlinesegment{011} \quad \underlinesegment{011} 001 ​ 111 ​ 011 ​ 101 ​ 011 ​ 011 ​

Therefore, (1111011101011011) 2 = (173533) 8

0011 undefined 0101 undefined 1101 undefined 0111 undefined \underlinesegment{0011} \quad \underlinesegment{0101} \quad \underlinesegment{1101} \quad \underlinesegment{0111} 0011 ​ 0101 ​ 1101 ​ 0111 ​

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

011 undefined 010 undefined 111 undefined 010 undefined 111 undefined \underlinesegment{011} \quad \underlinesegment{010} \quad \underlinesegment{111} \quad \underlinesegment{010} \quad \underlinesegment{111} 011 ​ 010 ​ 111 ​ 010 ​ 111 ​

Therefore, (11010111010111) 2 = (32727) 8

0001 undefined 0101 undefined 1011 undefined 0111 undefined \underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1011} \quad \underlinesegment{0111} 0001 ​ 0101 ​ 1011 ​ 0111 ​

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

001 undefined 010 undefined 110 undefined 110 undefined 111 undefined \underlinesegment{001} \quad \underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{110} \quad \underlinesegment{111} 001 ​ 010 ​ 110 ​ 110 ​ 111 ​

Therefore, (1010110110111) 2 = (12667) 8

0010 undefined 1101 undefined 1101 undefined 1011 undefined \underlinesegment{0010} \quad \underlinesegment{1101} \quad \underlinesegment{1101} \quad \underlinesegment{1011} 0010 ​ 1101 ​ 1101 ​ 1011 ​

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

010 undefined 110 undefined 111 undefined 011 undefined 011 undefined \underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{011} 010 ​ 110 ​ 111 ​ 011 ​ 011 ​

Therefore, (10110111011011) 2 = (26733) 8

1111 undefined 1011 undefined 1010 undefined 1111 undefined \underlinesegment{1111} \quad \underlinesegment{1011} \quad \underlinesegment{1010} \quad \underlinesegment{1111} 1111 ​ 1011 ​ 1010 ​ 1111 ​

Therefore, (1111101110101111) 2 = (FBAF) 16

001 undefined 111 undefined 101 undefined 110 undefined 101 undefined 111 undefined \underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{101} \quad \underlinesegment{110} \quad \underlinesegment{101} \quad \underlinesegment{111} 001 ​ 111 ​ 101 ​ 110 ​ 101 ​ 111 ​

Therefore, (1111101110101111) 2 = (175657) 8

Checkpoint 2.2

Multiple choice questions.

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

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

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

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

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

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

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

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

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

Question 10

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

• other symbol
• all of these ✓

Question 11

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

Question 12

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

• 1101.1110 ✓
• None of these

Question 13

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

Question 14

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

Question 15

Which of the following is not a binary number ?

Question 16

Convert the hexadecimal number 2C to decimal:

Question 17

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

• extended ASCII

Question 18

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

Question 19

Which of the following is not a valid UTF8 representation?

• 2 octet (16 bits)
• 3 octet (24 bits)
• 4 octet (32 bits)
• 8 octet (64 bits) ✓

Question 20

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

Fill in the Blanks

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 .

True/False Questions

A computer can work with Decimal number system. False

A computer can work with Binary number system. True

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

Number systems can also represent characters. False

ISCII is an encoding scheme created for Indian language characters. True

Unicode is able to represent nearly all languages' characters. True

UTF8 is a fixed-length encoding scheme. False

UTF32 is a fixed-length encoding scheme. True

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

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

Type A: Short Answer Questions

What are some number systems used by computers ?

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

What is the use of Hexadecimal number system on computers ?

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 ?

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 ?

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.

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 ?

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 ?

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:

1 M S B 0 1 1 0 1 1 0 L S B \begin{matrix} \underset{\bold{MSB}}{1} & 0 & 1 & 1 & 0 & 1 & 1 & \underset{\bold{LSB}}{0} \end{matrix} MSB 1 ​ ​ 0 ​ 1 ​ 1 ​ 0 ​ 1 ​ 1 ​ LSB 0 ​ ​

What are ASCII and extended ASCII encoding schemes ?

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 ?

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 ?

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:

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

What all encoding schemes does Unicode use to represent characters ?

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

What are ASCII and ISCII ? Why are these used ?

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 ?

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 ?

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 ?

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.

Type B: Application Based Questions

Convert the following binary numbers to decimal:

Equivalent decimal number = 1 + 4 + 8 = 13

Therefore, (1101) 2 = (13) 10

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

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

Convert the following binary numbers to decimal :

Equivalent decimal number = 4 + 8 = 12

(b) 10010101

(c) 11011100

Convert the following decimal numbers to binary:

Therefore, (0.25) 10 = (0.01) 2

Therefore, (122) 10 = (1111010) 2

(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:

Therefore, (122) 10 = (172) 8

Therefore, (0.675) 10 = (0.53146) 8

Convert the following hexadecimal numbers to binary:

(23D) 16 = (1000111101) 2

Convert the following binary numbers to hexadecimal:

(a) 1010110110111

(b) 10110111011011

(c) 0110101100

0001 undefined 1010 undefined 1100 undefined \underlinesegment{0001} \quad \underlinesegment{1010} \quad \underlinesegment{1100} 0001 ​ 1010 ​ 1100 ​

Therefore, (0110101100) 2 = (1AC) 16

Convert the following octal numbers to decimal:

Equivalent decimal number = 7 + 40 + 128 = 175

Therefore, (257) 8 = (175) 10

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

Therefore, (3527) 8 = (1879) 10

Equivalent decimal number = 3 + 16 + 64 = 83

Therefore, (123) 8 = (83) 10

Integral part

Fractional part.

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:

Equivalent decimal number = 6 + 160 = 166

Therefore, (A6) 16 = (166) 10

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

Therefore, (A13B) 16 = (41275) 10

Equivalent decimal number = 5 + 160 + 768 = 933

Therefore, (3A5) 16 = (933) 10

Equivalent decimal number = 9 + 224 = 233

Therefore, (E9) 16 = (233) 10

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

Therefore, (7CA3) 16 = (31907) 10

Convert the following decimal numbers to hexadecimal:

Therefore, (132) 10 = (84) 16

Therefore, (2352) 10 = (930) 16

Therefore, (122) 10 = (7A) 16

Therefore, (0.675) 10 = (0.ACCCC) 16

Therefore, (206) 10 = (CE) 16

Therefore, (3619) 10 = (E23) 16

Convert the following hexadecimal numbers to octal:

(38AC) 16 = (11100010101100) 2

011 undefined   100 undefined   010 undefined   101 undefined   100 undefined \underlinesegment{011}\medspace\underlinesegment{100}\medspace\underlinesegment{010}\medspace\underlinesegment{101}\medspace\underlinesegment{100} 011 ​ 100 ​ 010 ​ 101 ​ 100 ​

(38AC) 16 = (34254) 8

(7FD6) 16 = (111111111010110) 2

111 undefined   111 undefined   111 undefined   010 undefined   110 undefined \underlinesegment{111}\medspace\underlinesegment{111}\medspace\underlinesegment{111}\medspace\underlinesegment{010}\medspace\underlinesegment{110} 111 ​ 111 ​ 111 ​ 010 ​ 110 ​

(7FD6) 16 = (77726) 8

(ABCD) 16 = (1010101111001101) 2

001 undefined   010 undefined   101 undefined   111 undefined   001 undefined   101 undefined \underlinesegment{001}\medspace\underlinesegment{010}\medspace\underlinesegment{101}\medspace\underlinesegment{111}\medspace\underlinesegment{001}\medspace\underlinesegment{101} 001 ​ 010 ​ 101 ​ 111 ​ 001 ​ 101 ​

(ABCD) 16 = (125715) 8

Convert the following octal numbers to binary:

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

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

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

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

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

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

Convert the following binary numbers to octal

111 undefined 010 undefined \underlinesegment{111} \quad \underlinesegment{010} 111 ​ 010 ​

Therefore, (111010) 2 = (72) 8

(b) 110110101

110 undefined 110 undefined 101 undefined \underlinesegment{110} \quad \underlinesegment{110} \quad \underlinesegment{101} 110 ​ 110 ​ 101 ​

Therefore, (110110101) 2 = (665) 8

(c) 1101100001

001 undefined 101 undefined 100 undefined 001 undefined \underlinesegment{001} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \underlinesegment{001} 001 ​ 101 ​ 100 ​ 001 ​

Therefore, (1101100001) 2 = (1541) 8

011 undefined 001 undefined \underlinesegment{011} \quad \underlinesegment{001} 011 ​ 001 ​

Therefore, (11001) 2 = (31) 8

(b) 10101100

010 undefined 101 undefined 100 undefined \underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100} 010 ​ 101 ​ 100 ​

Therefore, (10101100) 2 = (254) 8

(c) 111010111

111 undefined 010 undefined 111 undefined \underlinesegment{111} \quad \underlinesegment{010} \quad \underlinesegment{111} 111 ​ 010 ​ 111 ​

Therefore, (111010111) 2 = (727) 8

Add the following binary numbers:

(i) 10110111 and 1100101

1 1 0 1 1 1 0 1 1 1 1 1 1 + 1 1 0 0 1 0 1 1 0 0 0 1 1 1 0 0 \begin{matrix} & & \overset{1}{1} & \overset{1}{0} & 1 & 1 & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & 1 \\ + & & & 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} + ​ 1 ​ 1 1 0 ​ 0 1 1 0 ​ 1 1 0 ​ 1 0 1 ​ 0 1 0 1 ​ 1 1 1 1 ​ 1 1 0 0 ​ 1 1 0 ​ ​

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

(ii) 110101 and 101111

1 1 1 1 0 1 1 1 0 1 1 + 1 0 1 1 1 1 1 1 0 0 1 0 0 \begin{matrix} & & \overset{1}{1} & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & \overset{1}{0} & 1 \\ + & & 1 & 0 & 1 & 1 & 1 & 1 \\ \hline & \bold{1} & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{0} \end{matrix} + ​ 1 ​ 1 1 1 1 ​ 1 1 0 0 ​ 0 1 1 0 ​ 1 1 1 1 ​ 0 1 1 0 ​ 1 1 0 ​ ​

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

(iii) 110111.110 and 11011101.010

0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 . 1 1 1 0 + 1 1 0 1 1 1 0 1 . 0 1 0 1 0 0 0 1 0 1 0 1 . 0 0 0 \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 \\ + & & 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} + ​ 1 ​ 0 1 1 0 ​ 0 1 1 0 ​ 1 1 0 0 ​ 1 1 1 1 ​ 0 1 1 0 ​ 1 1 1 1 ​ 1 1 0 0 ​ 1 1 1 1 ​ . . . ​ 1 1 0 0 ​ 1 1 0 ​ 0 0 0 ​ ​

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

(iv) 1110.110 and 11010.011

0 1 1 1 1 1 1 0 1 . 1 1 1 0 + 1 1 0 1 0 . 0 1 1 1 0 1 0 0 1 . 0 0 1 \begin{matrix} & & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & 1 & \overset{1}{0} & . & \overset{1}{1} & 1 & 0 \\ + & & 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} + ​ 1 ​ 0 1 1 0 ​ 1 1 1 1 ​ 1 1 0 0 ​ 1 1 0 ​ 0 1 0 1 ​ . . . ​ 1 1 0 0 ​ 1 1 0 ​ 0 1 1 ​ ​

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

Question 21

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 ?

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.

Question 22

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

(i) 100101.101

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

100 undefined 101 undefined . 101 undefined \underlinesegment{100} \quad \underlinesegment{101} \quad \bold{.} \quad \underlinesegment{101} 100 ​ 101 ​ . 101 ​

Therefore, (100101.101) 2 = (45.5) 8

Hexadecimal Conversion

0010 undefined 0101 undefined   .   1010 undefined \underlinesegment{0010} \quad \underlinesegment{0101} \medspace . \medspace \underlinesegment{1010} 0010 ​ 0101 ​ . 1010 ​

Therefore, (100101.101) 2 = (25.A) 16

(ii) 10101100.01011

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

Therefore, (10101100.01011) 2 = (172.34375) 10

010 undefined 101 undefined 100 undefined . 010 undefined 110 undefined \underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \bold{.} \quad \underlinesegment{010} \quad \underlinesegment{110} 010 ​ 101 ​ 100 ​ . 010 ​ 110 ​

Therefore, (10101100.01011) 2 = (254.26) 8

1010 undefined 1100 undefined   .   0101 undefined   1000 undefined \underlinesegment{1010} \quad \underlinesegment{1100} \medspace . \medspace \underlinesegment{0101} \medspace \underlinesegment{1000} 1010 ​ 1100 ​ . 0101 ​ 1000 ​

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

Decimal Conversion:

Equivalent decimal number = 2 + 8 = 10

001 undefined 010 undefined \underlinesegment{001} \quad \underlinesegment{010} 001 ​ 010 ​

Therefore, (1010) 2 = (12) 8

(iv) 10101100.010111

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

Therefore, (10101100.010111) 2 = (172.359375) 10

010 undefined 101 undefined 100 undefined . 010 undefined 111 undefined \underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \bold{.} \quad \underlinesegment{010} \quad \underlinesegment{111} 010 ​ 101 ​ 100 ​ . 010 ​ 111 ​

Therefore, (10101100.010111) 2 = (254.27) 8

1010 undefined 1100 undefined   .   0101 undefined   1100 undefined \underlinesegment{1010} \quad \underlinesegment{1100} \medspace . \medspace \underlinesegment{0101} \medspace \underlinesegment{1100} 1010 ​ 1100 ​ . 0101 ​ 1100 ​

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

Textbook - Data Representation, Computer Science (Python), Class 11 PDF Download

Faqs on textbook - data representation, computer science (python), class 11, computer science (python), mock tests for examination, video lectures, previous year questions with solutions, textbook - data representation, sample paper, objective type questions, shortcuts and tricks, extra questions, viva questions, study material, semester notes, important questions, practice quizzes, past year papers.

Textbook - Data Representation, Computer Science (Python), Class 11 Free PDF Download

Importance of textbook - data representation, computer science (python), class 11, textbook - data representation, computer science (python), class 11 notes, textbook - data representation, computer science (python), class 11 class 11 questions, study textbook - data representation, computer science (python), class 11 on the app, welcome back, create your account for free.

Forgot Password

Unattempted tests, change country, practice & revise.

• Andhra Pradesh
• Chhattisgarh
• West Bengal
• Madhya Pradesh
• Maharashtra
• Jammu & Kashmir
• NCERT Books 2022-23
• NCERT Solutions
• NCERT Notes
• NCERT Exemplar Books
• NCERT Exemplar Solution
• States UT Book
• School Kits & Lab Manual
• NCERT Books 2021-22
• NCERT Books 2020-21
• NCERT Book 2019-2020
• NCERT Book 2015-2016
• RD Sharma Solution
• TS Grewal Solution
• TR Jain Solution
• Selina Solution
• Frank Solution
• ML Aggarwal Solution
• Lakhmir Singh and Manjit Kaur Solution
• I.E.Irodov solutions
• ICSE - Goyal Brothers Park
• ICSE - Dorothy M. Noronhe
• Sandeep Garg Textbook Solution
• Micheal Vaz Solution
• S.S. Krotov Solution
• Evergreen Science
• KC Sinha Solution
• ICSE - ISC Jayanti Sengupta, Oxford
• ICSE Focus on History
• ICSE GeoGraphy Voyage
• ICSE Hindi Solution
• ICSE Treasure Trove Solution
• Thomas & Finney Solution
• SL Loney Solution
• SB Mathur Solution
• P Bahadur Solution
• Narendra Awasthi Solution
• MS Chauhan Solution
• LA Sena Solution
• Integral Calculus Amit Agarwal Solution
• IA Maron Solution
• Hall & Knight Solution
• Errorless Solution
• Pradeep's KL Gogia Solution
• OP Tandon Solutions
• Sample Papers
• Previous Year Question Paper
• Value Based Questions
• CBSE Syllabus
• CBSE MCQs PDF
• Assertion & Reason
• New Revision Notes
• Revision Notes
• HOTS Question
• Marks Wise Question
• Toppers Answer Sheets
• Exam Paper Aalysis
• Concept Map
• CBSE Text Book
• Additional Practice Questions
• Vocational Book
• CBSE - Concept
• KVS NCERT CBSE Worksheets
• Formula Class Wise
• Formula Chapter Wise
• JEE Crash Course
• JEE Previous Year Paper
• Important Info
• JEE Mock Test
• JEE Sample Papers
• SRM-JEEE Mock Test
• VITEEE Mock Test
• BITSAT Mock Test
• Manipal Engineering Mock Test
• AP EAMCET Previous Year Paper
• COMEDK Previous Year Paper
• GUJCET Previous Year Paper
• KCET Previous Year Paper
• KEAM Previous Year Paper
• Manipal Previous Year Paper
• MHT CET Previous Year Paper
• WBJEE Previous Year Paper
• AMU Previous Year Paper
• TS EAMCET Previous Year Paper
• SRM-JEEE Previous Year Paper
• VITEEE Previous Year Paper
• BITSAT Previous Year Paper
• UPSEE Previous Year Paper
• CGPET Previous Year Paper
• CUSAT Previous Year Paper
• AEEE Previous Year Paper
• Crash Course
• Previous Year Paper
• NCERT Based Short Notes
• NCERT Based Tests
• NEET Sample Paper
• Previous Year Papers
• Quantitative Aptitude
• Numerical Aptitude Data Interpretation
• General Knowledge
• Mathematics
• Agriculture
• Accountancy
• Business Studies
• Political science
• Enviromental Studies
• Mass Media Communication
• Teaching Aptitude
• NAVODAYA VIDYALAYA
• SAINIK SCHOOL (AISSEE)
• Mechanical Engineering
• Electrical Engineering
• Electronics & Communication Engineering
• Civil Engineering
• Computer Science Engineering
• CBSE Board News
• Scholarship Olympiad
• School Admissions
• Entrance Exams
• All Board Updates
• Miscellaneous
• State Wise Books
• Engineering Exam

NCERT Solutions for Class 11 Computer Science Data Representation PDF Download

Most students feel confused while preparing for Data Representation; for them, the NCERT Solutions for Class 11 Computer Science Data Representation can be considered as the best study material. Going through the Data Representation questions as well as answers can help students to remove confusion and accordingly, they can be prepared for tests or any kind of exams. 

NCERT Solutions for Class 11 Computer Science Data Representation PDF

The answers in the NCERT Solutions for Class 11 Computer Science Data Representation PDF provide accurate explanations so that students can attempt the questions in an accurate way. Students can refer to the Computer Science Data Representation questions through the Selfstudys website from their own comfort zone. By solving Data Representation questions from the website, students can improve their confidence level. 

Where Can Students Find the NCERT Solutions for Class 11 Computer Science Data Representation?

Students can find the NCERT Solutions for Class 11 Computer Science Data Representation from the Selfstudys website; the steps to refer to are discussed below: 

• Open the Selfstudys website. 

• Click the navigation button, now click NCERT Books & Solutions from the list. 

• A drop-down menu will appear, select NCERT Solutions from the list. 

• A new page will appear, select Class 11 from the list of classes. 
• Now select Computer Science from the list of subjects.
• Again a new page will appear, select Data Representation from the list of chapters. 

Attributes of NCERT Solutions for Class 11 Computer Science Data Representation

The attributes of NCERT Solutions for Class 11 Computer Science Data Representation are considered to the special quality; which each student needs to know before solving or referring to, some of the attributes are discussed below: 

• Explained in Para Wise Manner: Questions in the NCERT Solutions for Class 11 Computer Science Data Representation revision are explained in a para-wise manner so that by referring to it students can improve their comprehension skills. 
• All Topics are Covered: In the NCERT Solutions for Class 11 Computer Science Data Representation theory, all topics are covered so that by solving it, students can understand all the topics. 
• Explained in PointWise Manner: Some questions in the NCERT Solutions for Class 11 Computer Science Data Representation PDF are explained in a point-wise manner so that students can reinforce their memorisation skills. 
• Free Accessibility: The NCERT Solutions of Class 11 Computer Science questions are available free of cost so that students can access at any time without paying any amount. 
• Based on the Latest Syllabus: The Computer Science Data Representation questions from the NCERT Class 11 Solutions are based on the latest syllabus so students can have updated knowledge of all the questions. 
• Available in the PDF: The NCERT Class 11 Computer Science Solutions are available in the portable document format so that students can practise Data Representation questions from their own comfort zone. 

How Can NCERT Solutions for Class 11 Computer Science Data Representation Help Students?

The NCERT Solutions for Class 11 Computer Science Data Representation can be helpful in various ways; those ways are: 

• Helps to Understand the Concepts: Regular solving questions from the NCERT Solutions for Class 11 Computer Science Data Representation revision can help students to understand the concepts in a better way. The answers of Data Representation questions also provide detailed analysis of the concepts. 
• Helps to Clarify Doubts: By referring to the NCERT Solutions for Class 11 Computer Science Data Representation theory, students can easily clarify the doubts; accordingly they can build a strong foundation. Students can easily solve doubts as the answers of Data Representation are explained in a para wise or point wise way according to the need. 
• Helps in Exam Preparation: The NCERT Solutions for Class 11 Computer Science Data Representation PDF can help students in exam preparation as it provides practise questions and its answers. 
• Helps in Self Study: The Data Representation questions from the NCERT Class 11 Computer Science Solutions can help students in the process of self study. Students studying Data Representation by themselves can help them to boost their self esteem which can be useful in further chapters of Computer Science. 
• Helps to Improve Performance: By solving Computer Science Data Representation questions from the Class 11 NCERT Solutions, students can develop their problem solving skills and accordingly they can improve their performance level. 
• Saves Time: Students don’t need to search for Data Representation Computer Science questions here and there as the NCERT Class 11 Solutions are already available in the PDF. 

When Should NCERT Solutions for Class 11 Computer Science Data Representation Be Used?

Students can use the NCERT Solutions for Class 11 Computer Science Data Representation according to the below scenarios, those are: 

• To Complete the Chapter: Students can prefer utilising the NCERT Solutions for Class 11 Computer Science Data Representation revision so that they can complete the chapter in a proper way. 
• To Complete Homework Assignments: The NCERT Solutions for Class 11 Computer Science Data Representation theory can be used to complete the homework assignments as it provides answers to each question. 
• To Prepare for Tests: To prepare well for Computer Science Data Representation test, students need to solve more questions; for that they can refer to the NCERT Class 11 Solutions. 
• To Solve Confusions: If students are struggling to complete the concepts or to solve the question, then they can refer to the NCERT Solutions for Class 11 Computer Science Data Representation PDF. 
• During the Class: Students can utilise the Data Representation questions from the Class 11 Computer Science NCERT Solutions during the class as an additional supplement. By using it as an additional supplement, students can understand the concepts of Data Representation in a better way. 
• To Improve Confidence Level: To improve the confidence level while attempting questions of Data Representation, students can prefer utilising the NCERT Class 11 Computer Science Solutions. 

A Step-by-Step Guide to Solve Questions from NCERT Solutions for Class 11 Computer Science Data Representation

A step by step guide is the outline process to solve questions from NCERT Solutions for Class 11 Computer Science Data Representation; those steps are discussed below: 

• Read the Chapter Properly: Read the Data Representation properly to understand the concepts and theories covered in the Class 11 Computer Science. Before solving questions from the NCERT Solutions for Class 11 Computer Science Data Representation revision, students should make sure that they have proper understanding of the concepts. 
• Identify the Easy Questions: Students are advised to identify the easy questions from NCERT Solutions for Class 11 Computer Science Data Representation theory, before actually attempting it. Students should identify the easy questions of Data Representation according to what is being asked in it. 
• Analyse the Question: It is advisable for students to analyse the question from the NCERT Solutions for Class 11 Computer Science Data Representation PDF so that they can easily determine the approach. 
• Use Relevant Theories and Concepts: To attempt Data Representation questions from the NCERT Solutions for Class 11 Computer Science effectively, students need to use the relevant theories and concepts. 
• Provide Examples: Students need to use the required examples to support the answer of Data Representation Computer Science questions from the NCERT Class 11 Solutions so that they can demonstrate the understanding with relevance to real life theory. 
• Check the Answer: Checking the answers is the final step of solving Data Representation questions from the NCERT Class 11 Computer Science Solutions; accordingly, students can match their answers. 

What are the Challenges Faced While Solving Questions from the NCERT Solutions for Class 11 Computer Science Data Representation?

The challenges are those forces which make the students do a lot of effort to achieve; the same is faced while solving questions from the NCERT Solutions for Class 11 Computer Science Data Representation, some of the challenges are discussed below: 

• Lack of Understanding: If students are not able to understand the concepts of Data Representation, then they may face difficulty in solving questions from the NCERT Class 11 Computer Science Solutions; this may be mostly faced if students are new to the subject Computer Science. 
• To Apply the Concepts: Even after understanding the concepts of Data Representation, students may face difficulty in applying the relevant concepts while attempting questions from the NCERT Class 11 Computer Science Solutions. 
• Lack of Practice: If students don’t have enough time to practise questions then they may struggle to solve questions from NCERT Solutions for Class 11 Computer Science Data Representation revision. 
• Time Management: Students may find it challenging to utilise the time efficiently to practise questions from the NCERT Solutions for Class 11 Computer Science Data Representation theory. Some students spend much time on one single question of Data Representation. 
• Difficulty in Interpreting Questions: The questions in the NCERT Solutions for Class 11 Computer Science Data Representation PDF require correct interpretation and understanding; students may find it challenging to interpret questions. 
• Fear of Making Mistakes: The fear of making mistakes can hold students from attempting Data Representation questions of Class 11 NCERT Computer Science Solutions; this can be a very challenging one. 

• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 10 Maths
• CBSE Syllabus 2023-24
• Social Media Channels
• Login Customize Your Notification Preferences

One Last Step...

• Second click on the toggle icon

Provide prime members with unlimited access to all study materials in PDF format.

Allow prime members to attempt MCQ tests multiple times to enhance their learning and understanding.

Provide prime users with access to exclusive PDF study materials that are not available to regular users.

NCERT Books and Solutions for all classes

Data Representation Class 11 Computer Science Exam Questions

Please refer to Data Representation Class 11 Computer Science Exam Questions provided below. These questions and answers for Class 11 Computer Science have been designed based on the past trend of questions and important topics in your class 11 Computer Science books. You should go through all Class 11 Computer Science Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

Class 11 Computer Science Exam Questions Data Representation

Class 11 Computer Science students should read and understand the important questions and answers provided below for Data Representation which will help them to understand all important and difficult topics.

Very Short answer Type Questions

Question: Covert the following Decimal numbers to binary – (a) 23 (b) 100 (c) 161 (d) 145 Answer:  (a) 10111 (b) 1100100 (c) 10100001 (d) 10010001

Question: Convert 111110111101012 to octal. Answer:  37365

Question: Covert the following Octal numbers to Binary – (a) 456 (b) 26 (c) 751 (d) 777 Answer:  (a) 100101110 (b) 010110 (c) 111101001 (d)111111111

Question: Covert the following binary numbers to Hexadecimal – (a)101000001 (b) 11100011 (c) 10101111 (d) 101101111 Answer:  (a) 141 (b) E3 (c) AF (d)16F

Question: Covert the following binary numbers to decimal – (a)1010 (b) 111000 (c) 10101111 (d) 10110 Answer:  (a) 10 (b) 56 (c) 175 (d) 22

Question: Add the binary numbers (a) 110101 and 101111 (b) 10110 and 1101 Answer:  (a) 1100100 (b) 100011

Question:  Covert the following Hexadecimal numbers to Binary – (a) BE (b) BC9 (c) A07 (d) 7AB4 Answer:  (a)10111110 (b) 101111001001 (c) 101000000111 (d) 0111101010110100

Question: Convert the following: (a) 4468 to ( )16 (b) 47.58 to ( )10 (c) 45.910 to ( )2 Answer:  (a) 126 (b) 39.625 (c) 101101.1110

Short Answer Type Questions  

Question: How UTF-8 encoding scheme different from UTF-32 encoding scheme? Answer:  UTF-8: Variable-width encoding, backwards compatible with ASCII. ASCII characters (U+0000 to U+007F) take 1 byte, code points U+0080 to U+07FF take 2 bytes, code points U+0800 to U+FFFF take 3 bytes, code points U+10000 to U+10FFFF take 4 bytes. Good for English text, not so good for Asian text. UTF-32 uses 32-bit values for each character. That allows them to use a fixed-width code for every character. UTF-32 is opposite, it uses the most memory (each character is a fixed 4 bytes wide), but on the other hand, you know that every character has this precise length, so string manipulation becomes far simpler. You can compute the number of characters in a string simply from the length in bytes of the string. You can’t do that with UTF-8.

Question: What are ASCII and extended ASCII schemes? Answer:The standard ASCII  character set uses just 7 bits for each character. There are severallarger character sets that use 8 bits, which gives them 128 additional characters. The extra characters are used to represent non-English characters, graphics symbols, and mathematical symbols.

The extended ASCII  character set uses 8 bits, which gives it an additional 128 characters. The extra characters represent characters from foreign languages and special symbols for drawing pictures.

Question: What is the use of encoding schemes? Answer:  A character encoding provides a key to unlock (ie. crack) the code. It is a set of mappingsbetween the bytes in the computer and the characters in the character set. Without the key, thedata looks like garbage. So, when you input text using a keyboard or in some other way, the character encoding mapscharacters you choose to specific bytes in computer memory, and then to display the text itreads the bytes back into characters. Unfortunately, there are many different character sets and character encodings, ie. many different ways of mapping between bytes, code points and characters. The section Additional information provides a little more detail for those who are interested.

Question: What is Unicode? What is its significance? Answer:  Unicode is a character encoding standard that has widespread acceptance. Microsoft software uses Unicode at its core. Whether you realize it or not, you are using Unicode already! Basically, ―computers just deal with numbers. They store letters and other characters by assigning a number for each one. Before Unicode was invented, there were hundreds of different encoding systems for assigning these numbers. No single encoding could contain enough characters.1‖ This has been the problem we, in SIL, have often run into. If you are using a legacy encodingyour font conflicts with the font someone in another area of the world uses. You might have an in your font while someplace else someone used a at the same codepoint. Your files are  incompatible. Unicode provides a unique number for every character and so you do not have this problem if you use Unicode. If your document calls for U+0289 it will be clear to any computer program what the character should be

Question: Discuss UTF-8 encoding Scheme. Answer:  UTF-8 is a compromise character encoding that can be as compact as ASCII (if the file is just plain English text) but can also contain any unicode characters (with some increase in file size). UTF stands for Unicode Transformation Format. The ‘8’ means it uses 8-bit blocks to represent a character.

Question: What is the utility of ISCII encoding schemes? Prepared By: Sanjeev Bhadauria & Neha Tyagi Answer:  ISCII is a bilingual character encoding (not glyphs) scheme. Roman characters and punctuation marks as defined in the standard lower-ASCII take up the first half the character set (first 128 slots). Characters for indie languages are allocated to the upper slots (128-255). T

Question: What are ASCII and ISCII? Why are these used? Answer:  ASCII uses a 7-bit encoding and ISCII uses an 8-bit which is an extension of ASCII. These are encoding schemes to represent character set in s computer system.

Question: What do you understand by code point and code unit? Answer:  A code point is the atomic unit of information. … Each code point is a number which is given meaning by the Unicode standard. A code unit is the unit of storage of a part of anencoded code point. In UTF-8 this means 8-bits, in UTF-16 this means 16-bits.

Question: What is code space? How is it related to code point? Answer:  In computing, Code space may refer to: In memory address space: code space, where machine code is stored. For a character encoding: code space (or code space), the range of code points.

Question: Compare UTF-8 and UTF-32 encoding schemes. Which one is most popular scheme? Answer: UTF-8:  Variable-width encoding, backwards compatible with ASCII. ASCII characters (U+0000 to U+007F) take 1 byte, code points U+0080 to U+07FF take 2 bytes, code points U+0800 to U+FFFF take 3 bytes, code points U+10000 to U+10FFFF take 4 bytes. Good for English text, not so good for Asian text. UTF-32  uses 32-bit values for each character. That allows them to use a fixed-width code for every character. UTF-32 is opposite, it uses the most memory (each character is a fixed 4 bytes wide), but on the other hand, you know that every character has this precise length, so string manipulation becomes far simpler. You can compute the number of characters in a string simply from the length in bytes of the string. You can’t do that with UTF-8.

We hope you liked the above provided Data Representation Class 11 Computer Science Exam Questions. If you have any further questions please put them in the comments section below so that our teachers can provide you an answer.

Related Posts:

Related Posts

Plant growth and development class 11 biology exam questions, transport in plants class 11 biology exam questions.

Table Creation And Data Manipulation Commands Class 11 Computer Science Exam Questions