data representation

A. Choose the Correct option.

Q1. The value of radix in binary number system is _________________.

a.   2 ✓                            b. 8                             c. 10                          d. 16

Q2. The value of radix in octal number system is _________________.

a.     2                                   b. 8 ✓          c. 10                                       d. 16

Q3. The value of radix in decimal number system is _________________.

a.     2                       b. 8                                       c. 10 ✓                                 d. 16

Q4. The value of radix in hexadecimal number system is _________________.

a.     2                                  b. 8                                         c  10                           d. 16 ✓

Q5. Which of the following are not valid symbols in octal number system?

a.     2                       b. 8 ✓                     c. 9 ✓                        d. 7

Q6. Which of the following are not valid symbols in hexadecimal number system?

a.  2                        b. 8                                       c. 9                             d. G ✓

Q7. Which of the following are not valid symbols in decimal number system?

a.   2                        b. 8                                  c. 9                               d. G ✓

Q8.  The hexadecimal digits are 1 to 0 and A to _________________.

a. E                           b. F ✓                               c. G                               d .D

Q9.  The binary equivalent of the decimal number 10 is _________________.

a. 0010                        b.10                              c.1010 ✓                 d.010

Q . 10 ASCII code is a 7 bit code for _________________.

a. letters                         b .numbers                 c. other symbol       d. all of these ✓

Q11.  How many bytes are there in 1011 1001 0110 1110 numbers?

 

a.1                                    b  2 ✓                         c.4                              d. 8

Q12. The binary equivalent of the octal Numbers 13.54 is.....

a. 1011.1011                   b.1001.1110           c. 1101.1110 ✓   d. None of these

Q13.  The octal equivalent of 111 010 is.....

a. 81                                     b. 72 ✓                     c. 71                            d. 82

Q14.  The input hexadecimal representation of 1110 is _________________.

a. 0111                                b. E ✓                         c. 15                            d.14

Q15.  Which of the following is not a binary number ?

a. 1111                                  b. 101                         c, 11E ✓                    d. 000

Q16 . Convert the hexadecimal number 2C to decimal:

a. 3A                                      b. 34                           c. 44 ✓                      d. 43

Q17.  UTF8 is a type of _________________. encoding.

a. ASCII                            b. extended ASCII          c. Unicode ✓          d. ISCII

Q18.  UTF32 is a type of _________________. encoding.

a. ASCII                                b. extended ASCII         c. Unicode ✓      d. ISCII

Q19.   Which of the following is not a valid UTF8 representation?

a. 2 octet (16 bits)    b. 3 octet (24 bits)      c. 4 octet (32 bits)   d. 8 octet (64 bits) ✓

Q20.  Which of the following is not a valid encoding scheme for characters ?

B. Fill in the blank.

Q1. The Decimal number system is composed of 10 unique symbols.

Q2. The Binary number system is composed of 2 unique symbols.

Q3. The Octal number system is composed of 8 unique symbols.

Q4. The Hexadecimal number system is composed of 16 unique symbols.

Q5. The illegal digits of octal number system are 8 and 9.

Q6. Hexadecimal number system recognizes symbols 0 to 9 and A to F.

Q7. Each octal number is replaced with 3 bits in octal to binary conversion.

Q8. Each Hexadecimal number is replaced with 4 bits in Hex to binary conversion.

Q9. ASCII is a 7 bit code while extended ASCII is a 8 bit code.

Q10. The Unicode encoding scheme can represent all symbols/characters of most languages.

Q11. The ISCII encoding scheme represents Indian Languages' characters on computers.

Q12. UTF8 can take upto 4 bytes to represent a symbol.

Q13.  UTF32 takes exactly 4 bytes to represent a symbol.

Q14. Unicode value of a symbol is called code point.

 

C. State True or False.

Q1.  A computer can work with Decimal number system.  False

Q2.  A computer can work with Binary number system.  True

Q3. The number of unique symbols in Hexadecimal number system is 15.False

Q4.  Number systems can also represent characters.  False

Q5. I SCII is an encoding scheme created for Indian language characters.  True

Q6.  Unicode is able to represent nearly all languages' characters.  True

Q7.  UTF8 is a fixed-length encoding scheme.   False

Q8.  UTF32 is a fixed-length encoding scheme.   True

Q 9.  UTF8 is a variable-length encoding scheme which represents characters in 1 to 4 bytes. True

Q 10.  UTF8 and UTF32 are the only encoding schemes supported by Unicode. False

D. Answer these questions in short.

Q1. What are some number systems used by computers?

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

Q2. What is the use of Hexadecimal number system on computers?

Ans. 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.

Q3. What does radix or base signifies?

Ans. 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.

Q4. What is the use of encoding schemes?

Ans. 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.

Q5. Discuss UTF-8 encoding scheme.

Ans. 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.

Q6. How is UTF-8 encoding scheme different from UTF-32 encoding scheme?

Ans. 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.

Q7. What are the most significant bit and the least significant bit in a binary code?

Ans. 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:

LSBB

MSB
              1  0  1  1  0  1  1  0​​

 

Q8. What are ASCII and extended ASCII encoding schemes?

Ans. 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.

Q9. What is the utility of ISCII encoding scheme?

Ans. 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.

Q10. What is Unicode? What is its significance?

Ans. 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.

Q11. What all encoding schemes does Unicode use to represent characters?

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

Q12. What are ASCII and ISCII? Why are these used?

Ans. 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.

Q13. What are UTF-8 and UTF-32 encoding schemes. Which one is more popular encoding scheme?

Ans.  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.

Q14. What do you understand by code point?

Ans. 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'.

Q15. What is the difference between fixed length and variable length encoding schemes ?

Ans.  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.

Power

Value

Result

1 (LSB)

20

1

1x1=1

0

21

2

0x2=0

1

22

4

1x4=4

1 (MSB)

23

8

1x8=8

 

Equivalent decimal number = 1 + 4 + 8 = 13

Therefore, (1101)2 = (13)10

(b) 111010

Answer

Binary No.

Power

Value

Result

0 (LSB)

20

1

0x1=0

1

21

2

1x2=2

0

22

4

0x4=0

1

23

8

1x8=8

1

24

16

1x16=16

1 (MSB)

25

32

1x32=32

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

Therefore, (111010)2 = (58)10

(c) 101011111

Answer

Binary No.

Power

Value

Result

1 (LSB)

20

1

1x1=1

1

21

2

1x2=2

1

22

4

1x4=4

1

23

8

1x8=8

1

24

16

1x16=16

0

25

32

0x32=0

1

26

64

1x64=64

0

27

128

0x128=0

1 (MSB)

28

256

1x256=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.

Power

Value

Result

0 (LSB)

20

1

0x1=0

0

21

2

0x2=0

1

22

4

1x4=4

1 (MSB)

23

8

1x8=8

 

Equivalent decimal number = 4 + 8 = 12

Therefore, (1100)2 = (12)10

(b) 10010101

Answer

Binary
No

Power

Value

Result

1 (LSB)

20

1

1x1=1

0

21

2

0x2=0

1

22

4

1x4=4

0

23

8

0x8=0

1

24

16

1x16=16

0

25

32

0x32=0

0

26

64

0x64=0

1 (MSB)

27

128

1x128=128

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

Therefore, (10010101)2 = (149)10

(c) 11011100

Answer

Binary
No

Power

Value

Result

0 (LSB)

20

1

0x1=0

0

21

2

0x2=0

1

22

4

1x4=4

1

23

8

1x8=8

1

24

16

1x16=16

0

25

32

0x32=0

1

26

64

1x64=64

1 (MSB)

27

128

1x128=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

2

Quotient

Remainder

2

23

1 (LSB)

2

11

1

2

5

1

2

2

0

2

1

1 (MSB)

 

0

 

Therefore, (23)10 = (10111)2

(b) 100

Answer

2

Quotient

Remainder

2

100

0 (LSB)

2

50

0

2

25

1

2

12

0

2

6

0

2

3

1

2

1

1 (MSB)

 

0

 

Therefore, (100)10 = (1100100)2

(c) 145

Answer

2

Quotient

Remainder

2

145

1 (LSB)

2

72

0

2

36

0

2

18

0

2

9

1

2

4

0

2

2

0

2

1

1 (MSB)

 

0

 

Therefore, (145)10 = (10010001)2

(d) 0.25

Answer

Multiply

=

Resultant

Carry

0.25 x 2

=

0.5

0

0.5 x 2

=

0

1

Therefore, (0.25)10 = (0.01)2

Question 4

Convert the following decimal numbers to binary:

(a) 19

Answer

2

Quotient

Remainder

2

19

1 (LSB)

2

9

1

2

4

0

2

2

0

2

1

1 (MSB)

 

0

 

Therefore, (19)10 = (10011)2

(b) 122

Answer

2

Quotient

Remainder

2

122

0 (LSB)

2

61

1

2

30

0

2

15

1

2

7

1

2

3

1

2

1

1 (MSB)

 

0

 

Therefore, (122)10 = (1111010)2

(c) 161

Answer

2

Quotient

Remainder

2

161

1 (LSB)

2

80

0

2

40

0

2

20

0

2

10

0

2

5

1

2

2

0

2

1

1 (MSB)

 

0

 

Therefore, (161)10 = (10100001)2

(d) 0.675

Answer

Multiply

=

Resultant

Carry

0.675 x 2

=

0.35

1

0.35 x 2

=

0.7

0

0.7 x 2

=

0.4

1

0.4 x 2

=

0.8

0

0.8 x 2

=

0.6

1

(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

8

Quotient

Remainder

8

19

3 (LSB)

8

2

2 (MSB)

 

0

 

Therefore, (19)10 = (23)8

(b) 122

Answer

8

Quotient

Remainder

8

122

2 (LSB)

8

15

7

8

1

1 (MSB)

 

0

 

Therefore, (122)10 = (172)8

(c) 161

Answer

Answer

8

Quotient

Remainder

8

161

1 (LSB)

8

20

4

8

2

2 (MSB)

 

0

 

Therefore, (161)10 = (241)8

(d) 0.675

Answer

Multiply

=

Resultant

Carry

0.675 x 8

=

0.4

5

0.4 x 8

=

0.2

3

0.2 x 8

=

0.6

1

0.6 x 8

=

0.8

4

0.8 x 8

=

0.4

6

Therefore, (0.675)10 = (0.53146)8

Question 6

Convert the following hexadecimal numbers to binary:

(a) A6

Answer

Hexadecimal
Number

Binary
Equivalent

6

0110

A (10)

1010

(A6)16 = (10100110)2

(b) A07

Answer

Hexadecimal
Number

Binary
Equivalent

7

0111

0

0000

A (10)

1010

(A07)16 = (101000000111)2

(c) 7AB4

Answer

Hexadecimal
Number

Binary
Equivalent

4

0100

B (11)

1011

A (10)

1010

7

0111

(7AB4)16 = (111101010110100)2

Question 7

Convert the following hexadecimal numbers to binary:

(a) 23D

Answer

Hexadecimal
Number

Binary
Equivalent

D (13)

1101

3

0011

2

0010

(23D)16 = (1000111101)2

(b) BC9

Answer

Hexadecimal
Number

Binary
Equivalent

9

1001

C (12)

1100

B (11)

1011

(BC9)16 = (101111001001)2

(c) 9BC8

Answer

Hexadecimal
Number

Binary
Equivalent

8

1000

C (12)

1100

B (11)

1011

9

1001

(9BC8)16 = (1001101111001000)2

Question 8

Convert the following binary numbers to hexadecimal:

(a) 10011011101

Answer

Grouping in bits of 4:

0100undefined0100​1101​1101​

Binary
Number

Equivalent
Hexadecimal

1101

D (13)

1101

D (13)

0100

4

Therefore, (10011011101)2 = (4DD)16

(b) 1111011101011011

Answer

Grouping in bits of 4:

1111undefined1111​0111​0101​1011​

Binary
Number

Equivalent
Hexadecimal

1011

B (11)

0101

5

0111

7

1111

F (15)

Therefore, (1111011101011011)2 = (F75B)16

(c) 11010111010111

Answer

Grouping in bits of 4:

0011undefined0011​0101​1101​0111​

Binary
Number

Equivalent
Hexadecimal

0111

7

1101

D (13)

0101

5

0011

3

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

Question 9

Convert the following binary numbers to hexadecimal:

(a) 1010110110111

Answer

Grouping in bits of 4:

0001undefined0001​0101​1011​0111​

Binary
Number

Equivalent
Hexadecimal

0111

7

1011

B (11)

0101

5

0001

1

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

(b) 10110111011011

Answer

Grouping in bits of 4:

0010undefined0010​1101​1101​1011​

Binary
Number

Equivalent
Hexadecimal

1011

B (11)

1101

D (13)

1101

D (13)

0010

2

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

(c) 0110101100

Answer

Grouping in bits of 4:

0001undefined0001​1010​1100​

Binary
Number

Equivalent
Hexadecimal

1100

C (12)

1010

A (10)

0001

1

Therefore, (0110101100)2 = (1AC)16

Question 10

Convert the following octal numbers to decimal:

(a) 257

Answer

Octal
No

Power

Value

Result

7 (LSB)

80

1

7x1=7

5

81

8

5x8=40

2 (MSB)

82

64

2x64=128

Equivalent decimal number = 7 + 40 + 128 = 175

Therefore, (257)8 = (175)10

(b) 3527

Answer

Octal
No

Power

Value

Result

7 (LSB)

80

1

7x1=7

2

81

8

2x8=16

5

82

64

5x64=320

3 (MSB)

83

512

3x512=1536

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

Therefore, (3527)8 = (1879)10

(c) 123

Answer

Octal
No

Power

Value

Result

3 (LSB)

80

1

3x1=3

2

81

8

2x8=16

1 (MSB)

82

64

1x64=64

Equivalent decimal number = 3 + 16 + 64 = 83

Therefore, (123)8 = (83)10

(d) 605.12

Answer

Integral part

Octal
No

Power

Value

Result

5

80

1

5x1=5

0

81

8

0x8=0

6

82

64

6x64=384

Fractional part

Octal
No

Power

Value

Result

1

8-1

0.125

1x0.125=0.125

2

8-2

0.0156

2x0.0156=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

Power

Value

Result

6

160

1

6x1=6

A (10)

161

16

10x16=160

Equivalent decimal number = 6 + 160 = 166

Therefore, (A6)16 = (166)10

(b) A13B

Answer

Hexadecimal
Number

Power

Value

Result

B (11)

160

1

11x1=11

3

161

16

3x16=48

1

162

256

1x256=256

A (10)

163

4096

10x4096=40960

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

Therefore, (A13B)16 = (41275)10

(c) 3A5

Answer

Hexadecimal
Number

Power

Value

Result

5

160

1

5x1=5

A (10)

161

16

10x16=160

3

162

256

3x256=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

Power

Value

Result

9

160

1

9x1=9

E (14)

161

16

14x16=224

Equivalent decimal number = 9 + 224 = 233

Therefore, (E9)16 = (233)10

(b) 7CA3

Answer

Hexadecimal
Number

Power

Value

Result

3 (11)

160

1

3x1=3

A (10)

161

16

10x16=160

C (12)

162

256

12x256=3072

7

163

4096

7x4096=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

16

Quotient

Remainder

16

132

4

16

8

8

 

0

 

Therefore, (132)10 = (84)16

(b) 2352

Answer

16

Quotient

Remainder

16

2352

0

16

147

3

16

9

9

 

0

 

Therefore, (2352)10 = (930)16

(c) 122

Answer

16

Quotient

Remainder

16

122

A (10)

16

7

7

 

0

 

Therefore, (122)10 = (7A)16

(d) 0.675

Answer

Multiply

=

Resultant

Carry

0.675 x 16

=

0.8

A (10)

0.8 x 16

=

0.8

C (12)

0.8 x 16

=

0.8

C (12)

0.8 x 16

=

0.8

C (12)

0.8 x 16

=

0.8

C (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

16

Quotient

Remainder

16

206

E (14)

16

12

C (12)

 

0

 

Therefore, (206)10 = (CE)16

(b) 3619

Answer

16

Quotient

Remainder

16

3619

3

16

226

2

16

14

E (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

8

1000

3

0011

(38AC)16 = (11100010101100)2

Grouping in bits of 3:

011undefined 100undefined 010undefined 101undefined 100undefined011​100​010​101​100​

Binary
Number

Equivalent
Octal

100

4

101

5

010

2

100

4

011

3

(38AC)16 = (34254)8

(b) 7FD6

Answer

Hexadecimal
Number

Binary
Equivalent

6

0110

D (13)

1101

F (15)

1111

7

0111

(7FD6)16 = (111111111010110)2

Grouping in bits of 3:

111undefined 111undefined 111undefined 010undefined 110undefined111​111​111​010​110​

Binary
Number

Equivalent
Octal

110

6

010

2

111

7

111

7

111

7

(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:

001undefined 010undefined 101undefined 111undefined 001undefined 101undefined001​010​101​111​001​101​

Binary
Number

Equivalent
Octal

101

5

001

1

111

7

101

5

010

2

001

1

(ABCD)16 = (125715)8

Question 16

Convert the following octal numbers to binary:

(a) 123

Answer

Octal
Number

Binary
Equivalent

3

011

2

010

1

001

Therefore, (123)8 = (001undefined 010undefined 011undefined001​010​011​)2

(b) 3527

Answer

Octal
Number

Binary
Equivalent

7

111

2

010

5

101

3

011

Therefore, (3527)8 = (011undefined 101undefined 010undefined 111undefined011​101​010​111​)2

(c) 705

Answer

Octal
Number

Binary
Equivalent

5

101

0

000

7

111

Therefore, (705)8 = (111undefined 000undefined 101undefined111​000​101​)2

Question 17

Convert the following octal numbers to binary:

(a) 7642

Answer

Octal
Number

Binary
Equivalent

2

010

4

100

6

110

7

111

Therefore, (7642)8 = (111undefined 110undefined 100undefined 010undefined111​110​100​010​)2

(b) 7015

Answer

Octal
Number

Binary
Equivalent

5

101

1

001

0

000

7

111

Therefore, (7015)8 = (111undefined 000undefined 001undefined 101undefined111​000​001​101​)2

(c) 3576

Answer

Octal
Number

Binary
Equivalent

6

110

7

111

5

101

3

011

Therefore, (3576)8 = (011undefined 101undefined 111undefined 110undefined011​101​111​110​)2

(d) 705

Answer

Octal
Number

Binary
Equivalent

5

101

0

000

7

111

Therefore, (705)8 = (111undefined 000undefined 101undefined111​000​101​)2

Question 18

Convert the following binary numbers to octal

(a) 111010

Answer

Grouping in bits of 3:

111undefined111​010​

Binary
Number

Equivalent
Octal

010

2

111

7

Therefore, (111010)2 = (72)8

(b) 110110101

Answer

Grouping in bits of 3:

110undefined110​110​101​

Binary
Number

Equivalent
Octal

101

5

110

6

110

6

Therefore, (110110101)2 = (665)8

(c) 1101100001

Answer

Grouping in bits of 3:

001undefined001​101​100​001​

Binary
Number

Equivalent
Octal

001

1

100

4

101

5

001

1

Therefore, (1101100001)2 = (1541)8

Question 19

Convert the following binary numbers to octal

(a) 11001

Answer

Grouping in bits of 3:

011undefined011​001​

Binary
Number

Equivalent
Octal

001

1

011

3

Therefore, (11001)2 = (31)8

(b) 10101100

Answer

Grouping in bits of 3:

010undefined010​101​100​

Binary
Number

Equivalent
Octal

100

4

101

5

010

2

Therefore, (10101100)2 = (254)8

(c) 111010111

Answer

Grouping in bits of 3:

111undefined111​010​111​

Binary
Number

Equivalent
Octal

111

7

010

2

111

7

Therefore, (111010111)2 = (727)8

Question 20

Add the following binary numbers:

(i) 10110111 and 1100101

Answer

1101110111111+1100101100011100+​1​110​0110​110​101​0101​1111​1100​110​​

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

(ii) 110101 and 101111

Answer

11110111011+1011111100100+​1​1111​1100​0110​1111​0110​110​​

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

(iii) 110111.110 and 11011101.010

Answer

0101111101111111.1110+11011101.010100010101.000+​1​0110​0110​1100​1111​0110​1111​1100​1111​...​1100​110​000​​

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

(iv) 1110.110 and 11010.011

Answer

011111101.1110+11010.011101001.001+​1​0110​1111​1100​110​0101​...​1100​110​011​​

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:

2

Quotient

Remainder

2

65

1 (LSB)

2

32

0

2

16

0

2

8

0

2

4

0

2

2

0

2

1

1 (MSB)

 

0

 

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

Converting 65 to Hexadecimal:

16

Quotient

Remainder

16

65

1

16

4

4

 

0

 

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

Similarly, converting 97 to binary:

2

Quotient

Remainder

2

97

1 (LSB)

2

48

0

2

24

0

2

12

0

2

6

0

2

3

1

2

1

1 (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

Power

Value

Result

1

20

1

1x1=1

0

21

2

0x2=0

1

22

4

1x4=4

0

23

8

0x8=0

0

24

16

0x16=0

1

25

32

1x32=32

Decimal Conversion of fractional part:

Binary
No

Power

Value

Result

1

2-1

0.5

1x0.5=0.5

0

2-2

0.25

0x0.25=0

1

2-3

0.125

1x0.125=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:

100undefined100​101​.101​

Binary
Number

Equivalent
Octal

101

5

100

4

.

.

101

5

Therefore, (100101.101)2 = (45.5)8

Hexadecimal Conversion

Grouping in bits of 4:

0010undefined0010​0101​.1010​

Binary
Number

Equivalent
Hexadecimal

0101

5

0010

2

.

 

1010

A (10)

Therefore, (100101.101)2 = (25.A)16

(ii) 10101100.01011

Answer

Decimal Conversion of integral part:

Binary
No

Power

Value

Result

0

20

1

0x1=0

0

21

2

0x2=0

1

22

4

1x4=4

1

23

8

1x8=8

0

24

16

0x16=0

1

25

32

1x32=32

0

26

64

0x64=0

1

27

128

1x128=128

Decimal Conversion of fractional part:

Binary
No

Power

Value

Result

0

2-1

0.5

0x0.5=0

1

2-2

0.25

1x0.25=0.25

0

2-3

0.125

0x0.125=0

1

2-4

0.0625

1x0.0625=0.0625

1

2-5

0.03125

1x0.03125=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:

010undefined010​101​100​.010​110​

Binary
Number

Equivalent
Octal

100

4

101

5

010

2

.

.

010

2

110

6

Therefore, (10101100.01011)2 = (254.26)8

Hexadecimal Conversion

Grouping in bits of 4:

1010undefined1010​1100​.0101​1000​

Binary
Number

Equivalent
Hexadecimal

1100

C (12)

1010

A (10)

.

 

0101

5

1000

8

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

(iii) 1010

Answer

Decimal Conversion:

Binary
No

Power

Value

Result

0

20

1

0x1=0

1

21

2

1x2=2

0

22

4

0x4=0

1

23

8

1x8=8

Equivalent decimal number = 2 + 8 = 10

Therefore, (1010)2 = (10)10

Octal Conversion

Grouping in bits of 3:

001undefined001​010​

Binary
Number

Equivalent
Octal

010

2

001

1

Therefore, (1010)2 = (12)8

Hexadecimal Conversion

Grouping in bits of 4:

1010undefined1010​

Binary
Number

Equivalent
Hexadecimal

1010

A (10)

Therefore, (1010)2 = (A)16

(iv) 10101100.010111

Answer

Decimal Conversion of integral part:

Binary
No

Power

Value

Result

0

20

1

0x1=0

0

21

2

0x2=0

1

22

4

1x4=4

1

23

8

1x8=8

0

24

16

0x16=0

1

25

32

1x32=32

0

26

64

0x64=0

1

27

128

1x128=128

Decimal Conversion of fractional part:

Binary
No

Power

Value

Result

0

2-1

0.5

0x0.5=0

1

2-2

0.25

1x0.25=0.25

0

2-3

0.125

0x0.125=0

1

2-4

0.0625

1x0.0625=0.0625

1

2-5

0.03125

1x0.03125=0.03125

1

2-6

0.015625

1x0.015625=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:

010undefined010​101​100​.010​111​

Binary
Number

Equivalent
Octal

100

4

101

5

010

2

.

.

010

2

111

7

Therefore, (10101100.010111)2 = (254.27)8

Hexadecimal Conversion

Grouping in bits of 4:

1010undefined1010​1100​.0101​1100​

Binary
Number

Equivalent
Hexadecimal

1100

C (12)

1010

A (10)

.

 

0101

5

1100

C (12)

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

 


Comments

Post a Comment

Popular posts from this blog

The Road Not Taken

 The Road Not Taken  The Road Not Taken -       Robert Frost Introduction The poem ‘The Road Not Taken’ is based on common day to day life experience, when we have to make one choice out of many choices. At this point in time, we need to make the fullest use of our wisdom to choose the right option. In this poem, the two roads symbolize the choices that are presented by the life to us. Our future and the way of living our life totally depend on the decisions taken by us in the present time. Later on, sometimes, we look back and become thoughtful about those decisions we took in the past. Peeping in the past, we also wonder how our life would have been completely different had we chosen the other options. Title The title of the poem suggests us that it is indicating about absence. It is about the road that was “Not Taken” by the poet. And it gives him time and space to rethink about his decision. But on the other hand, the title may suggest that the poe...
Read more

Chapter 4 The Basic Writing Skills

  Chapter 4 Basic Writing Skills 1. Tick () the correct option. a. Which of the following field(s) use writing as a medium of communication? i. Medicine                                       ii    Education      iii. Technology                               iv All of these   b. Which of the following is necessary for effective writing? i Grammar and spellings                 ii. Concept and idea iii. Environment                      ...
Read more

The Fun They Had

Chapter 1 The Fun They Had Additional Questions: Q1. What are the reactions of Margie and Tommy to the real book? Q2. How were Tommy and Margie assessed in their subjects? Q3. Which is, according to you, an ideal school – the one where students study together in a special building or the room at home with a mechanical teacher (computer) in it? Q5. Read the following extract and answer the questions that follow. He was round little man with a red face and a whole box of tools with dials and wires. He smiled at Margie and gave her an apple, then took the  teacher  apart. Margie had hoped he wouldn’t know how to put it together again, but he knew how all right, and, after an hour or so, there it was again, large and black and ugly, with a big screen on which all the lessons were shown and the questions were asked. That wasn’t so bad. (a)    Who was the round little man in the above lines? (b)    Why did he come with the box of tools with dials and wires? (c) ...
Read more