Class 11 - Computer Science with Python Sumita Arora - Chapter 2 - Data Representation

 

Chapter 2

Data Representation

Class 11 - Computer Science with Python Sumita Arora

Multiple Choice Questions

Question 1

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

  1. 2 ✓
  2. 8
  3. 10
  4. 16

Question 2

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

  1. 2
  2. 8 ✓
  3. 10
  4. 16

Question 3

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

  1. 2
  2. 8
  3. 10 ✓
  4. 16

Question 4

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

  1. 2
  2. 8
  3. 10
  4. 16 ✓

Question 5

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

  1. 2
  2. 8 ✓
  3. 9 ✓
  4. 7

Question 6

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

  1. 2
  2. 8
  3. 9
  4. G ✓
  5. F

Question 7

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

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

Question 8

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

  1. E
  2. F ✓
  3. G
  4. D

Question 9

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

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

Question 10

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

  1. letters
  2. numbers
  3. other symbol
  4. all of these ✓

Question 11

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

  1. 1
  2. 2 ✓
  3. 4
  4. 8

Question 12

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

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

Question 13

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

  1. 81
  2. 72 ✓
  3. 71
  4. 82

Question 14

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

  1. 0111
  2. E ✓
  3. 15
  4. 14

Question 15

Which of the following is not a binary number ?

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

Question 16

Convert the hexadecimal number 2C to decimal:

  1. 3A
  2. 34
  3. 44 ✓
  4. 43

Question 17

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

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

Question 18

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

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

Question 19

Which of the following is not a valid UTF8 representation?

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

Question 20

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

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

Fill in the Blanks

Question 1

The Decimal number system is composed of 10 unique symbols.

Question 2

The Binary number system is composed of 2 unique symbols.

Question 3

The Octal number system is composed of 8 unique symbols.

Question 4

The Hexadecimal number system is composed of 16 unique symbols.

Question 5

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

Question 6

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

Question 7

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

Question 8

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

Question 9

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

Question 10

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

Question 11

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

Question 12

UTF8 can take upto 4 bytes to represent a symbol.

Question 13

UTF32 takes exactly 4 bytes to represent a symbol.

Question 14

Unicode value of a symbol is called code point.

True/False Questions

Question 1

A computer can work with Decimal number system.
False

Question 2

A computer can work with Binary number system.
True

Question 3

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

Question 4

Number systems can also represent characters.
False

Question 5

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

Question 6

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

Question 7

UTF8 is a fixed-length encoding scheme.
False

Question 8

UTF32 is a fixed-length encoding scheme.
True

Question 9

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

Question 10

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

Type A: Short Answer Questions

Question 1

What are some number systems used by computers ?

Answer

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

Question 2

What is the use of Hexadecimal number system on computers ?

Answer

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

Question 3

What does radix or base signify ?

Answer

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

Question 4

What is the use of encoding schemes ?

Answer

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

Question 5

Discuss UTF-8 encoding scheme.

Answer

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

Question 6

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

Answer

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

Question 7

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

Answer

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

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

Question 8

What are ASCII and extended ASCII encoding schemes ?

Answer

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

Question 9

What is the utility of ISCII encoding scheme ?

Answer

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

Question 10

What is Unicode ? What is its significance ?

Answer

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

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

Question 11

What all encoding schemes does Unicode use to represent characters ?

Answer

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

Question 12

What are ASCII and ISCII ? Why are these used ?

Answer

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

Question 13

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

Answer

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

Question 14

What do you understand by code point ?

Answer

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

Question 15

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

Answer

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

Type B: Application Based Questions

Question 1

Convert the following binary numbers to decimal:

(a) 1101

Answer

Binary
No		PowerValueResult
1 (LSB)2011x1=1
02120x2=0
12241x4=4
1 (MSB)2381x8=8

Equivalent decimal number = 1 + 4 + 8 = 13

Therefore, (1101)2 = (13)10

(b) 111010

Answer

Binary
No		PowerValueResult
0 (LSB)2010x1=0
12121x2=2
02240x4=0
12381x8=8
124161x16=16
1 (MSB)25321x32=32

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

Therefore, (111010)2 = (58)10

(c) 101011111

Answer

Binary
No		PowerValueResult
1 (LSB)2011x1=1
12121x2=2
12241x4=4
12381x8=8
124161x16=16
025320x32=0
126641x64=64
0271280x128=0
1 (MSB)282561x256=256

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

Therefore, (101011111)2 = (351)10

Question 2

Convert the following binary numbers to decimal :

(a) 1100

Answer

Binary
No		PowerValueResult
0 (LSB)2010x1=0
02120x2=0
12241x4=4
1 (MSB)2381x8=8

Equivalent decimal number = 4 + 8 = 12

Therefore, (1100)2 = (12)10

(b) 10010101

Answer

Binary
No		PowerValueResult
1 (LSB)2011x1=1
02120x2=0
12241x4=4
02380x8=0
124161x16=16
025320x32=0
026640x64=0
1 (MSB)271281x128=128

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

Therefore, (10010101)2 = (149)10

(c) 11011100

Answer

Binary
No		PowerValueResult
0 (LSB)2010x1=0
02120x2=0
12241x4=4
12381x8=8
124161x16=16
025320x32=0
126641x64=64
1 (MSB)271281x128=128

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

Therefore, (11011100)2 = (220)10

Question 3

Convert the following decimal numbers to binary:

(a) 23

Answer

2QuotientRemainder
2231 (LSB)
2111
251
220
211 (MSB)
 0 

Therefore, (23)10 = (10111)2

(b) 100

Answer

2QuotientRemainder
21000 (LSB)
2500
2251
2120
260
231
211 (MSB)
 0 

Therefore, (100)10 = (1100100)2

(c) 145

Answer

2QuotientRemainder
21451 (LSB)
2720
2360
2180
291
240
220
211 (MSB)
 0 

Therefore, (145)10 = (10010001)2

(d) 0.25

Answer

Multiply=ResultantCarry
0.25 x 2=0.50
0.5 x 2=01

Therefore, (0.25)10 = (0.01)2

Question 4

Convert the following decimal numbers to binary:

(a) 19

Answer

2QuotientRemainder
2191 (LSB)
291
240
220
211 (MSB)
 0 

Therefore, (19)10 = (10011)2

(b) 122

Answer

2QuotientRemainder
21220 (LSB)
2611
2300
2151
271
231
211 (MSB)
 0 

Therefore, (122)10 = (1111010)2

(c) 161

Answer

2QuotientRemainder
21611 (LSB)
2800
2400
2200
2100
251
220
211 (MSB)
 0 

Therefore, (161)10 = (10100001)2

(d) 0.675

Answer

Multiply=ResultantCarry
0.675 x 2=0.351
0.35 x 2=0.70
0.7 x 2=0.41
0.4 x 2=0.80
0.8 x 2=0.61

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

Therefore, (0.675)10 = (0.10101)2

Question 5

Convert the following decimal numbers to octal:

(a) 19

Answer

8QuotientRemainder
8193 (LSB)
822 (MSB)
 0 

Therefore, (19)10 = (23)8

(b) 122

Answer

8QuotientRemainder
81222 (LSB)
8157
811 (MSB)
 0 

Therefore, (122)10 = (172)8

(c) 161

Answer

Answer

8QuotientRemainder
81611 (LSB)
8204
822 (MSB)
 0 

Therefore, (161)10 = (241)8

(d) 0.675

Answer

Multiply=ResultantCarry
0.675 x 8=0.45
0.4 x 8=0.23
0.2 x 8=0.61
0.6 x 8=0.84
0.8 x 8=0.46

Therefore, (0.675)10 = (0.53146)8

Question 6

Convert the following hexadecimal numbers to binary:

(a) A6

Answer

Hexadecimal
Number		Binary
Equivalent
60110
A (10)1010

(A6)16 = (10100110)2

(b) A07

Answer

Hexadecimal
Number		Binary
Equivalent
70111
00000
A (10)1010

(A07)16 = (101000000111)2

(c) 7AB4

Answer

Hexadecimal
Number		Binary
Equivalent
40100
B (11)1011
A (10)1010
70111

(7AB4)16 = (111101010110100)2

Question 7

Convert the following hexadecimal numbers to binary:

(a) 23D

Answer

Hexadecimal
Number		Binary
Equivalent
D (13)1101
30011
20010

(23D)16 = (1000111101)2

(b) BC9

Answer

Hexadecimal
Number		Binary
Equivalent
91001
C (12)1100
B (11)1011

(BC9)16 = (101111001001)2

(c) 9BC8

Answer

Hexadecimal
Number		Binary
Equivalent
81000
C (12)1100
B (11)1011
91001

(9BC8)16 = (1001101111001000)2

Question 8

Convert the following binary numbers to hexadecimal:

(a) 10011011101

Answer

Grouping in bits of 4:

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

Binary
Number		Equivalent
Hexadecimal
1101D (13)
1101D (13)
01004

Therefore, (10011011101)2 = (4DD)16

(b) 1111011101011011

Answer

Grouping in bits of 4:

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

Binary
Number		Equivalent
Hexadecimal
1011B (11)
01015
01117
1111F (15)

Therefore, (1111011101011011)2 = (F75B)16

(c) 11010111010111

Answer

Grouping in bits of 4:

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

Binary
Number		Equivalent
Hexadecimal
01117
1101D (13)
01015
00113

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

Question 9

Convert the following binary numbers to hexadecimal:

(a) 1010110110111

Answer

Grouping in bits of 4:

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

Binary
Number		Equivalent
Hexadecimal
01117
1011B (11)
01015
00011

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

(b) 10110111011011

Answer

Grouping in bits of 4:

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

Binary
Number		Equivalent
Hexadecimal
1011B (11)
1101D (13)
1101D (13)
00102

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

(c) 0110101100

Answer

Grouping in bits of 4:

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

Binary
Number		Equivalent
Hexadecimal
1100C (12)
1010A (10)
00011

Therefore, (0110101100)2 = (1AC)16

Question 10

Convert the following octal numbers to decimal:

(a) 257

Answer

Octal
No		PowerValueResult
7 (LSB)8017x1=7
58185x8=40
2 (MSB)82642x64=128

Equivalent decimal number = 7 + 40 + 128 = 175

Therefore, (257)8 = (175)10

(b) 3527

Answer

Octal
No		PowerValueResult
7 (LSB)8017x1=7
28182x8=16
582645x64=320
3 (MSB)835123x512=1536

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

Therefore, (3527)8 = (1879)10

(c) 123

Answer

Octal
No		PowerValueResult
3 (LSB)8013x1=3
28182x8=16
1 (MSB)82641x64=64

Equivalent decimal number = 3 + 16 + 64 = 83

Therefore, (123)8 = (83)10

(d) 605.12

Answer

Integral part
Octal
No		PowerValueResult
58015x1=5
08180x8=0
682646x64=384
Fractional part
Octal
No		PowerValueResult
18-10.1251x0.125=0.125
28-20.01562x0.0156=0.0312

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

Therefore, (605.12)8 = (389.1562)10

Question 11

Convert the following hexadecimal numbers to decimal:

(a) A6

Answer

Hexadecimal
Number		PowerValueResult
616016x1=6
A (10)1611610x16=160

Equivalent decimal number = 6 + 160 = 166

Therefore, (A6)16 = (166)10

(b) A13B

Answer

Hexadecimal
Number		PowerValueResult
B (11)160111x1=11
3161163x16=48
11622561x256=256
A (10)163409610x4096=40960

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

Therefore, (A13B)16 = (41275)10

(c) 3A5

Answer

Hexadecimal
Number		PowerValueResult
516015x1=5
A (10)1611610x16=160
31622563x256=768

Equivalent decimal number = 5 + 160 + 768 = 933

Therefore, (3A5)16 = (933)10

Question 12

Convert the following hexadecimal numbers to decimal:

(a) E9

Answer

Hexadecimal
Number		PowerValueResult
916019x1=9
E (14)1611614x16=224

Equivalent decimal number = 9 + 224 = 233

Therefore, (E9)16 = (233)10

(b) 7CA3

Answer

Hexadecimal
Number		PowerValueResult
3 (11)16013x1=3
A (10)1611610x16=160
C (12)16225612x256=3072
716340967x4096=28672

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

Therefore, (7CA3)16 = (31907)10

Question 13

Convert the following decimal numbers to hexadecimal:

(a) 132

Answer

16QuotientRemainder
161324
1688
 0 

Therefore, (132)10 = (84)16

(b) 2352

Answer

16QuotientRemainder
1623520
161473
1699
 0 

Therefore, (2352)10 = (930)16

(c) 122

Answer

16QuotientRemainder
16122A (10)
1677
 0 

Therefore, (122)10 = (7A)16

(d) 0.675

Answer

Multiply=ResultantCarry
0.675 x 16=0.8A (10)
0.8 x 16=0.8C (12)
0.8 x 16=0.8C (12)
0.8 x 16=0.8C (12)
0.8 x 16=0.8C (12)

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

Therefore, (0.675)10 = (0.ACCCC)16

Question 14

Convert the following decimal numbers to hexadecimal:

(a) 206

Answer

16QuotientRemainder
16206E (14)
1612C (12)
 0 

Therefore, (206)10 = (CE)16

(b) 3619

Answer

16QuotientRemainder
1636193
162262
1614E (14)
 0 

Therefore, (3619)10 = (E23)16

Question 15

Convert the following hexadecimal numbers to octal:

(a) 38AC

Answer

Hexadecimal
Number		Binary
Equivalent
C (12)1100
A (10)1010
81000
30011

(38AC)16 = (11100010101100)2

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
1004
1015
0102
1004
0113

(38AC)16 = (34254)8

(b) 7FD6

Answer

Hexadecimal
Number		Binary
Equivalent
60110
D (13)1101
F (15)1111
70111

(7FD6)16 = (111111111010110)2

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
1106
0102
1117
1117
1117

(7FD6)16 = (77726)8

(c) ABCD

Answer

Hexadecimal
Number		Binary
Equivalent
D (13)1101
C (12)1100
B (11)1011
A (10)1010

(ABCD)16 = (1010101111001101)2

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
1015
0011
1117
1015
0102
0011

(ABCD)16 = (125715)8

Question 16

Convert the following octal numbers to binary:

(a) 123

Answer

Octal
Number		Binary
Equivalent
3011
2010
1001

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

(b) 3527

Answer

Octal
Number		Binary
Equivalent
7111
2010
5101
3011

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

(c) 705

Answer

Octal
Number		Binary
Equivalent
5101
0000
7111

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

Question 17

Convert the following octal numbers to binary:

(a) 7642

Answer

Octal
Number		Binary
Equivalent
2010
4100
6110
7111

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

(b) 7015

Answer

Octal
Number		Binary
Equivalent
5101
1001
0000
7111

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

(c) 3576

Answer

Octal
Number		Binary
Equivalent
6110
7111
5101
3011

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

(d) 705

Answer

Octal
Number		Binary
Equivalent
5101
0000
7111

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

Question 18

Convert the following binary numbers to octal

(a) 111010

Answer

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
0102
1117

Therefore, (111010)2 = (72)8

(b) 110110101

Answer

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
1015
1106
1106

Therefore, (110110101)2 = (665)8

(c) 1101100001

Answer

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
0011
1004
1015
0011

Therefore, (1101100001)2 = (1541)8

Question 19

Convert the following binary numbers to octal

(a) 11001

Answer

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
0011
0113

Therefore, (11001)2 = (31)8

(b) 10101100

Answer

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
1004
1015
0102

Therefore, (10101100)2 = (254)8

(c) 111010111

Answer

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
1117
0102
1117

Therefore, (111010111)2 = (727)8

Question 20

Add the following binary numbers:

(i) 10110111 and 1100101

Answer

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

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

(ii) 110101 and 101111

Answer

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

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

(iii) 110111.110 and 11011101.010

Answer

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

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

(iv) 1110.110 and 11010.011

Answer

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

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

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 ?

Answer

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

Converting 65 to binary:

2QuotientRemainder
2651 (LSB)
2320
2160
280
240
220
211 (MSB)
 0 

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

Converting 65 to Hexadecimal:

16QuotientRemainder
16651
1644
 0 

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

Similarly, converting 97 to binary:

2QuotientRemainder
2971 (LSB)
2480
2240
2120
260
231
211 (MSB)
 0 

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

Answer

Decimal Conversion of integral part:

Binary
No		PowerValueResult
12011x1=1
02120x2=0
12241x4=4
02380x8=0
024160x16=0
125321x32=32

Decimal Conversion of fractional part:

Binary
No		PowerValueResult
12-10.51x0.5=0.5
02-20.250x0.25=0
12-30.1251x0.125=0.125

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

Therefore, (100101.101)2 = (37.625)10

Octal Conversion

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
1015
1004
..
1015

Therefore, (100101.101)2 = (45.5)8

Hexadecimal Conversion

Grouping in bits of 4:

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

Binary
Number		Equivalent
Hexadecimal
01015
00102
. 
1010A (10)

Therefore, (100101.101)2 = (25.A)16

(ii) 10101100.01011

Answer

Decimal Conversion of integral part:

Binary
No		PowerValueResult
02010x1=0
02120x2=0
12241x4=4
12381x8=8
024160x16=0
125321x32=32
026640x64=0
1271281x128=128

Decimal Conversion of fractional part:

Binary
No		PowerValueResult
02-10.50x0.5=0
12-20.251x0.25=0.25
02-30.1250x0.125=0
12-40.06251x0.0625=0.0625
12-50.031251x0.03125=0.03125

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

Therefore, (10101100.01011)2 = (172.34375)10

Octal Conversion

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
1004
1015
0102
..
0102
1106

Therefore, (10101100.01011)2 = (254.26)8

Hexadecimal Conversion

Grouping in bits of 4:

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

Binary
Number		Equivalent
Hexadecimal
1100C (12)
1010A (10)
. 
01015
10008

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

(iii) 1010

Answer

Decimal Conversion:

Binary
No		PowerValueResult
02010x1=0
12121x2=2
02240x4=0
12381x8=8

Equivalent decimal number = 2 + 8 = 10

Therefore, (1010)2 = (10)10

Octal Conversion

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
0102
0011

Therefore, (1010)2 = (12)8

Hexadecimal Conversion

Grouping in bits of 4:

\underlinesegment{1010}

Binary
Number		Equivalent
Hexadecimal
1010A (10)

Therefore, (1010)2 = (A)16

(iv) 10101100.010111

Answer

Decimal Conversion of integral part:

Binary
No		PowerValueResult
02010x1=0
02120x2=0
12241x4=4
12381x8=8
024160x16=0
125321x32=32
026640x64=0
1271281x128=128

Decimal Conversion of fractional part:

Binary
No		PowerValueResult
02-10.50x0.5=0
12-20.251x0.25=0.25
02-30.1250x0.125=0
12-40.06251x0.0625=0.0625
12-50.031251x0.03125=0.03125
12-60.0156251x0.015625=0.015625

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

Therefore, (10101100.010111)2 = (172.359375)10

Octal Conversion

Grouping in bits of 3:

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

Binary
Number		Equivalent
Octal
1004
1015
0102
..
0102
1117

Therefore, (10101100.010111)2 = (254.27)8

Hexadecimal Conversion

Grouping in bits of 4:

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

Binary
Number		Equivalent
Hexadecimal
1100C (12)
1010A (10)
. 
01015
1100C (12)

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

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