data representation class 11 notes pdf download

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

Class 11 Computer Science Chapter-wise Notes PDF Download

Computer science class 11 notes chapter-wise with pdf download, this is according to new or latest syllabus of cbse class 11, chapter  =   getting started with python   chapter  =   python fundamental      chapter  =   data handling        chapter  =   flow of control chapter  =   conditional and iterative statements         chapter  =   string  manipulation       chapter  =   list manipulation       chapter  =   tuples      chapter  =   dictionaries     chapter  =   understanding sorting       chapter  =   computer system overview chapter  =   cyber safety chapter  =   online access and computer security chapter  =  society, law and ethics.

Notes :-  All Notes PDF are provided by http://python.mykvs.in   . You can also download notes PDF simply by clicking on above website . 

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• Introduction to Data Representation
• Computer Science

About Data Representation

Data can be anything, including a number, a name, musical notes, or the colour of an image. The way that we stored, processed, and transmitted data is referred to as data representation. We can use any device, including computers, smartphones, and iPads, to store data in digital format. The stored data is handled by electronic circuitry. A bit is a 0 or 1 used in digital data representation.

Data Representation Techniques

Classification of Computers

Computer scans are classified broadly based on their speed and computing power.

1. Microcomputers or PCs (Personal Computers): It is a single-user computer system with a medium-power microprocessor. It is referred to as a computer with a microprocessor as its central processing unit.

Microcomputer

2. Mini-Computer: It is a multi-user computer system that can support hundreds of users at the same time.

Types of Mini Computers

3. Mainframe Computer: It is a multi-user computer system that can support hundreds of users at the same time. Software technology is distinct from minicomputer technology.

Mainframe Computer

4. Super-Computer: With the ability to process hundreds of millions of instructions per second, it is a very quick computer. They  are used for specialised applications requiring enormous amounts of mathematical computations, but they are very expensive.

Supercomputer

Types of Computer Number System

Every value saved to or obtained from computer memory uses a specific number system, which is the method used to represent numbers in the computer system architecture. One needs to be familiar with number systems in order to read computer language or interact with the system. 

Types of Number System

1. Binary Number System 

There are only two digits in a binary number system: 0 and 1. In this number system, 0 and 1 stand in for every number (value). Because the binary number system only has two digits, its base is 2.

A bit is another name for each binary digit. The binary number system is also a positional value system, where each digit's value is expressed in powers of 2.

Characteristics of Binary Number System

The following are the primary characteristics of the binary system:

It only has two digits, zero and one.

Depending on its position, each digit has a different value.

Each position has the same value as a base power of two.

Because computers work with internal voltage drops, it is used in all types of computers.

Binary Number System

2. Decimal Number System

The decimal number system is a base ten number system with ten digits ranging from 0 to 9. This means that these ten digits can represent any numerical quantity. A positional value system is also a decimal number system. This means that the value of digits will be determined by their position. 

Characteristics of Decimal Number System

Ten units of a given order equal one unit of the higher order, making it a decimal system.

The number 10 serves as the foundation for the decimal number system.

The value of each digit or number will depend on where it is located within the numeric figure because it is a positional system.

The value of this number results from multiplying all the digits by each power.

Decimal Number System

Decimal Binary Conversion Table

3. octal number system.

There are only eight (8) digits in the octal number system, from 0 to 7. In this number system, each number (value) is represented by the digits 0, 1, 2, 3,4,5,6, and 7. Since the octal number system only has 8 digits, its base is 8.

Characteristics of Octal Number System:

Contains eight digits: 0,1,2,3,4,5,6,7.

Also known as the base 8 number system.

Each octal number position represents a 0 power of the base (8). 

An octal number's last position corresponds to an x power of the base (8).

Octal Number System

4. Hexadecimal Number System

There are sixteen (16) alphanumeric values in the hexadecimal number system, ranging from 0 to 9 and A to F. In this number system, each number (value) is represented by 0, 1, 2, 3, 5, 6, 7, 8, 9, A, B, C, D, E, and F. Because the hexadecimal number system has 16 alphanumeric values, its base is 16. Here, the numbers are A = 10, B = 11, C = 12, D = 13, E = 14, and F = 15.

Characteristics of Hexadecimal Number System:

A system of positional numbers.

Has 16 symbols or digits overall (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F). Its base is, therefore, 16.

Decimal values 10, 11, 12, 13, 14, and 15 are represented by the letters A, B, C, D, E, and F, respectively.

A single digit may have a maximum value of 15. 

Each digit position corresponds to a different base power (16).

Since there are only 16 digits, any hexadecimal number can be represented in binary with 4 bits.

Hexadecimal Number System

So, we've seen how to convert decimals and use the Number System to communicate with a computer. The full character set of the English language, which includes all alphabets, punctuation marks, mathematical operators, special symbols, etc., must be supported by the computer in addition to numerical data. 

Learning By Doing

Choose the correct answer:.

1. Which computer is the largest in terms of size?

Minicomputer

Micro Computer

2. The binary number 11011001 is converted to what decimal value?

Solved Questions

1. Give some examples where Supercomputers are used.

Ans: Weather Prediction, Scientific simulations, graphics, fluid dynamic calculations, Nuclear energy research, electronic engineering and analysis of geological data.

2. Which of these is the most costly?

Mainframe computer

Ans: C) Supercomputer

FAQs on Introduction to Data Representation

1. What is the distinction between the Hexadecimal and Octal Number System?

The octal number system is a base-8 number system in which the digits 0 through 7 are used to represent numbers. The hexadecimal number system is a base-16 number system that employs the digits 0 through 9 as well as the letters A through F to represent numbers.

2. What is the smallest data representation?

The smallest data storage unit in a computer's memory is called a BYTE, which comprises 8 BITS.

3. What is the largest data unit?

The largest commonly available data storage unit is a terabyte or TB. A terabyte equals 1,000 gigabytes, while a tebibyte equals 1,024 gibibytes.

Presentation of Data Notes - Class 11 PDF Download

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

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Presentation of Data Class 11 Notes PDF (Handwritten & Short Notes)

In the Presentation of Data class 11 notes, topics and concepts are explained in a creative manner, accordingly students can also improve their creativity. With this, students can also attempt questions of Presentation of Data in an innovative way as it encourages them to engage in understanding complex topics in an ease. 

The Presentation of Data class 11 notes can be useful for both teachers and students as it is arranged in a systematic manner so they can easily evaluate their progress and quickly cover the chapter. The systematic organisation can also help students to overcome unnecessary disturbances while completing the Statistics for Economics chapter Presentation of Data. 

Presentation of Data Class 11 Notes PDF

Students can utilise the Presentation of Data class 11 notes PDF while preparing for the chapter from their comfort zone. As they can go through the PDF of class 11 notes from the Selfstudys website so that they can quickly cover Presentation of Data and can understand the chapter in a better way. 

Where Can You Find a Presentation of Data Class 11 Notes?

Students can find the Presentation of Data class 11 notes from the Selfstudys website, steps to download are: 

• Visit the Selfstudys website.
• Click NCERT Books & Solution from the navigation bar. 
• A drop down will appear, select NCERT notes from the given list.

• A new page will appear, select class 11th from the list of classes.

• Select Statistics for Economics from the list of subjects.

• Now select the chapter Presentation of Data from the given list. 

Characteristics of Presentation of Data Class 11 Notes

In the Presentation of Data class 11 notes, some of the definitions are explained in a brief way, other characteristics are: 

• Formulas are Given: The Presentation of Data notes provides important formulas for student’s understanding, accordingly one can solve all types of questions. 
• Flow Charts are Given: A flow chart is a diagram which helps students to understand tough concepts of Presentation of Data.
• Attractive Format: The Presentation of Data class 11 notes PDF are explained in an attractive format so that students can be more attentive towards the chapter. 
• Diagrams are Given: Class 11 Statistics Presentation of Data notes provides diagrams: symbolic representation, so that students can simplify complex content into easier ones. 
• For All Students: All students studying in CBSE and other state boards can look through the Presentation of Data class 11 Economics notes as it is created according to the latest syllabus. 
• Readers Friendly: The notes of Presentation of Data class 11 are user friendly so that students can easily access and learn the concepts. 

Why Should You Use a Presentation of Data Class 11 Notes PDF?

Students should use Presentation of Data class 11 notes PDF so that they can provided with major benefits, those benefits are: 

• Helps in Organisation: The notes of Presentation of Data class 11 keeps students organised about their preparation of the chapter, accordingly they can increase their productivity. 
• Provides an Outline: The notes on Presentation of Data provides an outline for students so that they can get a clarity about concepts of the chapter. 
• Helps in Revising: It is important for students to revise concepts to score good marks, they can recall from the class 11 statistics notes Presentation of Data. 
• Improves Attention Span: The class 11 Statistics notes Presentation of Data helps students to improve their attention span towards the chapter so that they can learn the concepts in a better way. 
• Get an Idea About High Weightage Topics: With the help of Presentation of Data class 11 notes, students can get an idea about the high weightage topics so that they can prepare accordingly. However, these details are not given directly in the notes, but one can refer to the previous year question papers to understand them better.
• Improves Memory: The Presentation of Data class 11 notes PDF helps students to improve their memorisation skills so that they can solve problems within the given time limit. 

When is the Right Time to Go Through The Presentation of Data Class 11 Notes?

Generally, students can go through the Presentation of Data class 11 notes after completing the chapter from NCERT book as it is considered to be the right time. Because in the class 11 notes, topics and concepts of Presentation of Data are explained in a better approach so that students can solve all their doubts as well as confusions.

When to Use the Presentation of Data Class 11 Notes While Preparation? 

Students can totally depend on the Presentation of Data class 11 notes while preparing for the exam, crucial ways that one can depend on the notes are: 

• Class Room Study: Students can go through the Presentation of Data notes during classroom study as it provides a general plan of topics and concepts.
• Weekly Revision: Students can look through the Presentation of Data class 11 notes PDF during weekly revision so that they can clarify doubts on a weekly basis. 
• During Board Exam: Before the board exam, students have to revise all topics so that they can also focus on weaker topics. Accordingly, students can improvise their marks in questions related to Presentation of Data. 
• During Weekly Test: Some students regularly take weekly tests to evaluate their preparation, and can utilise the notes on Presentation of Data so that they can remember the important formulas on their fingertips. 
• Last Minute Preparation: Students generally have the habit of completing the topics during the last minute, so it is better for them to use the Presentation of Data class 11 notes PDF as it can ease their preparation. 

Tips to Prepare for Presentation of Data With The Help of Class 11 Statistics for Economics Notes

Students can follow strategy tips to prepare for Presentation of Data with the help of Class 11 Statistics for Economics Notes , those important tips are: 

• Go Through the NCERT Book: To start the preparation, students need to first go through the Presentation of Data chapter in NCERT book and can also take the help of class 11 Statistics for Economics notes to cover the lesson. 
• Use Mannerism: Students need to create their own tricks: flow charts, diagrams and mannerism to complete Presentation of Data so that they can improve their creativity to complete the notes. 
• Practise Questions: Students are advised to practise questions of the chapter, they can take the help of Presentation of Data notes so that they can understand the concepts in a better way. 
• Make Important Points: While completing the chapter, students can take the help of Presentation of Data notes to make a note of important points in their notebook. 
• Select Study Schedule: Every student has their own comfortable time to complete the notes on Presentation of Data in which their productivity is high.
• Don’t Cram: Students don’t need to cram the topics, it is important for them to understand each and every topic, they can take the help of notes of Presentation of Data class 11.

How to Make the Preparation Effective While Completing Presentation of Data Class 11 Notes?

Effective preparation with the help of Presentation of Data is the key to good score, tips to make the preparation effective are: 

• Form a Proper Routine: Students need to form a proper routine to complete Presentation of Data notes so that they can complete quickly and can also increase their efficiency. 
• Take Intervals: Students are advised to take frequent intervals: listen to music, short walk, snack time to complete Presentation of Data class 11 notes PDF so that they can promote efficiency. 
• Concentration on One Topic: While completing class 11 Economics notes, students need to concentrate on just one topic, this can improve their efficiency to complete Presentation of Data. 
• Update Learning Style: Students need to update their learning style: way of learning on a regular basis to complete class 11 statistics Presentation of Data notes so that they can improve their efficiency. 
• Utilise the Time: Students need to complete the notes of Presentation of Data class 11 without wasting much time in other things can help increase the efficiency. 
• Solve Doubts: It is normal for students to come up with some doubts regarding Presentation of Data as it is important for them to clear their doubts so that they can improve their productivity. 

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Presentation of Data Class 11 Notes: CBSE 11th Economics Chapter 4, Download PDF

Cbse class 11 presentation of data   revision   notes: find here handwritten revision notes for cbse class 11 economics chapter 4, presentation of data. a pdf download link is also available for the same, at the bottom of the article..

Presentation of Data Class 11 Revision Notes: In this article, students can find complete Revision Notes for CSBE Class 11 Economics Chapter 4, Presentation of data. Entire notes have been prepared according to the updated CBSE Syllabus 2023-2024.

Presentation of data teaches students about the power of visualization. How data can be represented in a manner such that it becomes understandable, informative, and interesting at the same time. Economists use multiple forms of representation to showcase different types of data. Each of the forms is explained in detail here so that it becomes easier for students to present data in different forms, as per the requirement.

This article also presents links to other important study materials for Class 11 students. Links to the updated CBSE Class 11 Syllabus are also attached below for your reference.

CBSE Class 11 Economics Syllabus 2023-2024(PDF)

CBSE Class 11 Economics Deleted Syllabus 2023-2024(PDF)

Revision Notes for CBSE Class 11 Economics Chapter 1 2023-2024(PDF)

Revision Notes for CBSE Class 11 Economics Chapter 2 2023-2024(PDF)

Revision Notes for CBSE Class 11 Economics Chapter 3 2023-2024(PDF)

Revision Notes for Class 11 Economics Chapter 4 are presented below:

Presentation of Data - Since data is complex and voluminous, it has to be presented in a manner such that the data becomes understandable and presentable.

Forms of Presentation :

1.Textual or Descriptive Presentation - In this form of presentation, data are described within the text. It is suitable when the quantity of data is not too large. The benefit of using this form is that it gives room for explanation of all relevant data and a disadvantage is that person has to go through every single line for conclusion.

• Qualitative Classification - When classification is done according to attributes, such as social status, physical status, nationality, etc., it is called qualitative classification.
• Quantitative Classification - When classification is done on the basis of characteristics that are quantitative in nature is called quantitative classification.
• Temporal Classification : In this classification, time becomes the classifying variable and data are categorized according to time. Time may be in hours, days, weeks, months, years, etc.
• Spatial Classification : When classification is done on the basis of place, it is called spatial classification. The place may be a village/tow
• Table Number - A table number is assigned to a table for identification purposes. If more than one table is presented, it is the table number that distinguishes one table from another. It is given at the top or at the beginning of the title of the table. Generally, table numbers are whole numbers in ascending order if there are many tables in a book.
• Title - It narrates the contents of the table. It has to be clear, brief, and carefully worded so that the interpretations made from the table are clear and free from ambiguity. It finds a place at the head of the table succeeding the table number or just below it.
• Captions or Column Headings - At the top of each column in a table a column designation is given to explain the figures of the column. This is called a caption or column heading.
• Stubs or Row Headings - Like a caption or column heading, each row of the table has to be given a heading. The designations of the rows are also called stubs or stub items, and the complete left column is known as the stub column.
• Body - The body of a table is the main part and it contains the actual data. The location of any one figure/data in the table is fixed and determined by the row and column of the table.\
• Unit of Measurement - Units of measurement must be stated along with the title. If different units are there for rows or columns of the table, these units must be stated along with ‘stubs’ or ‘captions.
• Source - It is a brief statement or phrase indicating the source of data presented in the table. If more than one source is there, all the sources are to be written in the source. The source is generally written at the bottom of the table.
• Note - Note is the last part of the table. It explains the specific feature of the data content of the table which is not self-explanatory and has not been explained earlier.
• Geometric diagram - A bar diagram and a pie diagram are the two types of geometric diagrams. The bar diagram comprises a group of equispaced and equiwidth rectangular bars for each class or category of data. Height or length of the bar reads the magnitude of data. Bar diagrams are further categorized into simple bar diagrams, multiple bar diagrams, and component bar diagram. A pie diagram is also a component diagram, but unlike a bar diagram, here it is a circle whose area is proportionally divided among the components.
• Frequency diagram - Data in the form of grouped frequency distributions are generally represented by frequency diagrams like histograms, frequency polygons, frequency curves, and ogive.
• Arithmetic line graph - An arithmetic line graph is also called a time series graph. In this graph, time(hour, day/date, week, month, year, etc.) is plotted along the x-axis and the value of the variable (time series data) is along the y-axis.

This chapter is comparatively short and easier to grasp. These notes shall be enough for your preparation for the annual examinations.

CBSE Class 11 Syllabus 2023-24 (All Subjects)

CBSE Class 11 Deleted Syllabus 2023-24 (All Subjects)

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