NCERT Computer Science Class 11 - Chapter 2 Solutions - part3

 

Chapter 2 - Encoding Schemes and Number System


NCERT Solutions


Exercise Solutions  ( continued .. )



9. Write binary equivalent of the following octal numbers.

(i) 2306

Convert to Decimal: 1222

Now convert 1222 to Binary:

1222 ÷ 2 = 611 with a remainder of 0

611 ÷ 2 = 305 with a remainder of 1

305 ÷ 2 = 152 with a remainder of 1

152 ÷ 2 = 76 with a remainder of 0

76 ÷ 2 = 38 with a remainder of 0

38 ÷ 2 = 19 with a remainder of 0

19 ÷ 2 = 9 with a remainder of 1

9 ÷ 2 = 4 with a remainder of 1

4 ÷ 2 = 2 with a remainder of 0

2 ÷ 2 = 1 with a remainder of 0

1 ÷ 2 = 0 with a remainder of 1

Binary value: 1 0 0 1 1 0 0 0 1 1 0


(ii) 5610

Convert to Decimal: 2952

Now convert 2952 to Binary:

        2952 ÷ 2 = 1476 with a remainder of 0
        1476 ÷ 2 = 738 with a remainder of 0
        738 ÷ 2 = 369 with a remainder of 0
        369 ÷ 2 = 184 with a remainder of 1
        184 ÷ 2 = 92 with a remainder of 0
        92 ÷ 2 = 46 with a remainder of 0
        46 ÷ 2 = 23 with a remainder of 0
        23 ÷ 2 = 11 with a remainder of 1
        11 ÷ 2 = 5 with a remainder of 1
        5 ÷ 2 = 2 with a remainder of 1
        2 ÷ 2 = 1 with a remainder of 0
        1 ÷ 2 = 0 with a remainder of 1


Binary value: 101110001000


(iii) 742

Convert to Decimal: 482

Now convert 482 to Binary:

        482 ÷ 2 = 241 with a remainder of 0
        241 ÷ 2 = 120 with a remainder of 1
        120 ÷ 2 = 60 with a remainder of 0
        60 ÷ 2 = 30 with a remainder of 0
        30 ÷ 2 = 15 with a remainder of 0
        15 ÷ 2 = 7 with a remainder of 1
        7 ÷ 2 = 3 with a remainder of 1
        3 ÷ 2 = 1 with a remainder of 1
        1 ÷ 2 = 0 with a remainder of 1


Binary value: 111100010


(iv) 65.203

Convert to Decimal: 53.255859375

Now convert 53.255859375 to Binary:

Integer Part:

        53 ÷ 2 = 26 with a remainder of 1
        26 ÷ 2 = 13 with a remainder of 0
        13 ÷ 2 = 6 with a remainder of 1
        6 ÷ 2 = 3 with a remainder of 0
        3 ÷ 2 = 1 with a remainder of 1
        1 ÷ 2 = 0 with a remainder of 1


Fractional Part:

        0.255859375 × 2 = 0.51171875 (integer part = 0)
        0.51171875 × 2 = 1.0234375 (integer part = 1)
        0.0234375 × 2 = 0.046875 (integer part = 0)
        0.046875 × 2 = 0.09375 (integer part = 0)
        0.09375 × 2 = 0.1875 (integer part = 0)
        0.1875 × 2 = 0.375 (integer part = 0)
        0.375 × 2 = 0.75 (integer part = 0)
        0.75 × 2 = 1.5 (integer part = 1)
        0.5 × 2 = 1.0 (integer part = 1, and we stop here as we reach zero)

Thus 65.203 in Binary is: 110101 . 010000011


10. Write binary representation of the following hexadecimal numbers:

(i) 4026

Convert to Decimal:  16422

Now convert 16422 to Binary:

        16422 ÷ 2 = 8211 with a remainder of 0
        8211 ÷ 2 = 4105 with a remainder of 1
        4105 ÷ 2 = 2052 with a remainder of 1
        2052 ÷ 2 = 1026 with a remainder of 0
        1026 ÷ 2 = 513 with a remainder of 0
        513 ÷ 2 = 256 with a remainder of 1
        256 ÷ 2 = 128 with a remainder of 0
        128 ÷ 2 = 64 with a remainder of 0
        64 ÷ 2 = 32 with a remainder of 0
        32 ÷ 2 = 16 with a remainder of 0
        16 ÷ 2 = 8 with a remainder of 0
        8 ÷ 2 = 4 with a remainder of 0
        4 ÷ 2 = 2 with a remainder of 0
        2 ÷ 2 = 1 with a remainder of 0
        1 ÷ 2 = 0 with a remainder of 1

Binary value: 100000000100110


(ii)  BCA1

Convert to Decimal: 48289

Now convert 48289 to Binary:

        48289 ÷ 2 = 24144 with a remainder of 1
        24144 ÷ 2 = 12072 with a remainder of 0
        12072 ÷ 2 = 6036 with a remainder of 0
        6036 ÷ 2 = 3018 with a remainder of 0
        3018 ÷ 2 = 1509 with a remainder of 0
        1509 ÷ 2 = 754 with a remainder of 1
        754 ÷ 2 = 377 with a remainder of 0
        377 ÷ 2 = 188 with a remainder of 1
        188 ÷ 2 = 94 with a remainder of 0
        94 ÷ 2 = 47 with a remainder of 0
        47 ÷ 2 = 23 with a remainder of 1
        23 ÷ 2 = 11 with a remainder of 1
        11 ÷ 2 = 5 with a remainder of 1
        5 ÷ 2 = 2 with a remainder of 1
        2 ÷ 2 = 1 with a remainder of 0
        1 ÷ 2 = 0 with a remainder of 1

Binary value: 1011110010100001


(iii)  98E

Convert to Decimal: 2446

Now convert 2446 to Binary:


        2446 ÷ 2 = 1223 with a remainder of 0
        1223 ÷ 2 = 611 with a remainder of 1
        611 ÷ 2 = 305 with a remainder of 1
        305 ÷ 2 = 152 with a remainder of 1
        152 ÷ 2 = 76 with a remainder of 0
        76 ÷ 2 = 38 with a remainder of 0
        38 ÷ 2 = 19 with a remainder of 0
        19 ÷ 2 = 9 with a remainder of 1
        9 ÷ 2 = 4 with a remainder of 1
        4 ÷ 2 = 2 with a remainder of 0
        2 ÷ 2 = 1 with a remainder of 0
        1 ÷ 2 = 0 with a remainder of 1


Binary value: 100110001110


(iv)  132.45

Convert to Decimal: 306.26953125


Now convert the integer part to Binary:

306 ÷ 2 = 153 with a remainder of 0
153 ÷ 2 = 76 with a remainder of 1
76 ÷ 2 = 38 with a remainder of 0
38 ÷ 2 = 19 with a remainder of 0
19 ÷ 2 = 9 with a remainder of 1
9 ÷ 2 = 4 with a remainder of 1
4 ÷ 2 = 2 with a remainder of 0
2 ÷ 2 = 1 with a remainder of 0
1 ÷ 2 = 0 with a remainder of 1


Now convert the fraction part to Binary:

0.26953125 × 2 = 0.5390625 (integer part = 0)
0.5390625 × 2 = 1.078125 (integer part = 1)
0.078125 × 2 = 0.15625 (integer part = 0)
0.15625 × 2 = 0.3125 (integer part = 0)
0.3125 × 2 = 0.625 (integer part = 0)
0.625 × 2 = 1.25 (integer part = 1)
0.25 × 2 = 0.5 (integer part = 0)
0.5 × 2 = 1.0 (integer part = 1, and we stop here as we reach zero)


Combine to get final Binary value: 100110010 . 01000101


11. How  does computer understand  the following text?   (hint: 7 bit ASCII code).

(i) HOTS

Each character in "HOTS" has a specific ASCII value:

H: ASCII = 72,     Binary = 1001000

O: ASCII = 79,     Binary = 1001111

T: ASCII = 84,     Binary = 1010100

S: ASCII = 83,      Binary = 1010011


(ii)  Main

M:  ASCII = 77,        Binary = 1001101

a:  ASCII = 97,          Binary = 1100001

i:  ASCII = 105,         Binary = 1101001

n:  ASCII = 110,        Binary = 1101110



(iii) CaSe

C:  ASCII = 67      Binary = 1000011

a:  ASCII = 97      Binary = 1100001

S: ASCII = 83       Binary = 1010011

e:  ASCII = 101    Binary = 1100101



12. The hexadecimal number system uses 16 literals (0 – 9, A– F). Write down its base value.


The base value in Hexadecimal system is: 16


13. Let X be a number system having B symbols only. Write down the base value of this number system.


In a number system where XX has BB symbols, the base value of that number system is B.

 

Example:

        In the Decimal system (base-10), there are 10 symbols (0-9), so B=10

        In the Binary system (base-2), there are 2 symbols (0 and 1), so B=2

        In the Hexadecimal system (base-16), there are 16 symbols (0-9 and A-F), so B=16.




14.  Write the equivalent hexadecimal and binary values for each character of the phrase given below.                          हम सब एक.


Representation:

: Hex = 0939,      Binary = 0000100100111001

: Hex = 092E,     Binary = 0000100100101110

: Hex = 0938,     Binary = 0000100100111000

: Hex = 092C,     Binary = 0000100100101100

: Hex = 090F,      Binary = 0000100100001111

: Hex = 0915,     Binary = 0000100100010101




15.  What is the advantage of preparing a digital content in Indian language using UNICODE font?


  1. Universal Compatibility: Ensures consistent display across various devices and operating systems without needing specific font installations.
  1. Multilingual Support: Allows seamless integration of multiple Indian languages on a single platform.
  1. Consistent Representation: Maintains uniformity in text appearance, reducing misinterpretation when sharing content.
  1. Ease of Use: Simplifies typing and displaying text without requiring special software or encoding schemes.


17.  Encode the word 'COMPUTER' using ASCII and convert the encode value into Binary value.
            

Character

ASCII Value

Binary

C

67

1000011

O

79

1001111

M

77

1001101

P

80

1010000

U

85

1010101

T

84

1010100

E

69

1000101

R

82

1010010






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