NCERT Computer Science Class #11 – Chapter #2 Solutions

Chapter 2 – Encoding Schemes and Number System

Exercise Solutions

1. Write base values of binary, octal and hexadecimal number system.

The base values of different number systems are as follows:

These systems use different sets of symbols to represent values, with binary using 0 and 1, octal using digits 0-7, and hexadecimal using digits 0-9 and letters A-F.

2. Give full form of ASCII and ISCII.

3. Try the following conversions:

(i) (514)₈ = (?)₁₀

To convert the octal number (514)₈ to decimal, we can use the positional value of each digit in the octal system, which is base 8. Each digit is multiplied by 8 raised to the power of its position (counting from right to left, starting at 0).

Conversion Steps:

– Identify the digits:  5, 1, 4

-Assign positional values:

• 5 is in the 8² place
• 1 is in the 8¹ place
• 4 is in the 8⁰ place

–  Calculate the decimal value:

(514)₈  =  5×8²  +  1×8¹  +  4×8⁰

            =  5×64  +  1×8  +  4×1

            =  320  +  8  +  4

            =  332

(ii) (220)₈   =   (?)₂

Lets convert from Octal to Decimal first:

(220)₈​  =   2×8²   +   2×8¹   +   0×8⁰

            =   2×64   +  2×8     +   0×1

            =   128     +  16       +      0

            =   144

Now, convert to Binary:

To convert the decimal number 144 to binary, you can use the method of successive division by 2.

Here’s how it works:

– Divide the number by 2 and record the quotient and the remainder.

– Continue dividing the quotient by 2 until the quotient becomes 0.

– The binary equivalent is the remainders read in reverse order (from bottom to top).

Now, read the remainders from bottom to top:  10010000

(iii) (76F)₁₆   =   (?)₁₀

Follow below steps:

Identify the values of each digit:

• 7 corresponds to 7
• 6 corresponds to 6
• F corresponds to 15

Apply the positional values:

Each digit in a hexadecimal number is multiplied by 16ⁿ, where n is the position of the digit from right to left, starting at zero:

• For 7

                7 × 16²  =  7 × 256  =  1792

• For 6

                6 × 16¹  = 6 × 16 = 96

• For F

                15 × 16⁰  =  15 × 1  =  15

Sum the results

                  1792  +  96  +  15  =  1903

Thus, the decimal equivalent of (76F)₁₆​ is 1903

(iv)  (4D9)₁₆    =   (?)₁₀

Identify the values of each digit:

• 4 corresponds to 44
• D corresponds to 13
• 9 corresponds to 9

Apply the positional values:

Each digit in a hexadecimal number is multiplied by 16ⁿ, where n is the position of the digit from right to left, starting at zero:

• For 4

            4  ×  16²  =  4  ×  256  =  1024

• For D

            13  ×  16¹  =  13 × 16  =  208

• For 9

            9  ×  16⁰  =  9×1  =  9

Sum the results

            1024  +  208  +  9  =  1241

Thus, the decimal equivalent of (4D9)₁₆​ is 1241.

(v) (11001010)₂   =   (?)₁₀

Step 1: Identify the values of each digit

Each digit in a binary number represents a power of 2, starting from the rightmost digit (which is 2⁰).

Step 2: Calculate the decimal value

        1  ×  2⁷  =  1  ×  128  =  128

        1  ×  2⁶  =  1  ×  64  =  64

        0  ×  2⁵  =  0  ×  32  =  0

        0  ×  2⁴  =  0  ×  16  =  0

        1  ×  2³  =  1  ×  8  =  8

        0  ×  2²  =  0  ×  4  =  0

        1  ×  2¹  =  1  ×  2  =  2

        0  ×  2⁰  =  0  ×  1  =  0

Step 3: Sum the results

        128  +  64  +  0  +  0  +  8  +  0  +  2  +  0  =  202

Thus, the decimal equivalent of the binary number 11001010 is 202.

(vi) (1010111)₂  =  (?)₁₀

Step 1: Identify the values of each digit

Each digit in a binary number represents a power of 2, starting from the rightmost digit (which is 2⁰).

Step 2: Calculate the decimal value

        1  ×  26  =  1  ×  64  =  64

        0  ×  25  =  0  ×  32  =  0

        1  ×  24  =  1  ×  16  =  16

        0  ×  23  =  0  ×  8  =  0

        1  ×  22  =  1  ×  4  =  4

        1  ×  21  =  1  ×  2  =  2

        1  ×  20  =  1  ×  1  =  1

Step 3: Sum the results

        64  +  0  +  16  +  0  +  4  +  2  +  1  =  87

Thus, the decimal equivalent of the binary number 1010111 is 87. 

4. Try the following conversions:

(i) (54)₁₀  =   (?)₂

Step-by-Step Conversion

54 ÷ 2 = 27, remainder 0

27 ÷ 2 = 13, remainder 1

13 ÷ 2 = 6, remainder 1

6 ÷ 2 = 3, remainder 0

3 ÷ 2 = 1, remainder 1

1 ÷ 2 = 0, remainder 1

Thus 54 in binary is: 110110

(ii)  (120)₁₀  =  (?)₂

Step-by-Step Conversion

120 ÷ 2 = 60, remainder 0

60 ÷ 2 = 30, remainder 0

30 ÷ 2 = 15, remainder 0

15 ÷ 2 = 7, remainder 1

7 ÷ 2 = 3, remainder 1

3 ÷ 2 = 1, remainder 1

1 ÷ 2 = 0, remainder 1

Thus 120 in binary is 1111000

(iii) (76)₁₀  =  (?)₈

Step-by-Step Conversion

76 ÷ 8 = 9, remainder 4

9 ÷ 8 = 1, remainder 1

1 ÷ 8 = 0, remainder 1

This 76 in Octal is 114

(iv) (889)₁₀  =  (?)₈

Step-by-Step Conversion

889 ÷ 8 = 111, remainder 1

111 ÷ 8 = 13, remainder 7

13 ÷ 8 = 1, remainder 5

1 ÷ 8 = 0, remainder 1

Thus 889 in Octal is 1571

(v)  (789)₁₀  =  (?)₁₆

Step-by-Step Conversion

789 ÷ 16 = 49, remainder 5

49 ÷ 16 = 3, remainder 1

3 ÷ 16 = 0, remainder 3

Thus 789 in hexadecimal is 315

(vi)  (108)₁₀  =  (?)₁₆

Step-by-Step Conversion

108 ÷ 16 = 6, remainder 12 (which is represented as C in hexadecimal)

6 ÷ 16 = 0, remainder 6

Thus 108 in Hexadecimal is 6C

5. Try the following conversions from Octal to Decimal:

(i) 145

Steps:

        ( 1 × 82 ) + (4 × 81 ) + ( 5 × 80 )

        = ( 1 × 64 ) + ( 4 × 8 ) + ( 5 × 1 )

        = 64 + 32 + 5

        = 101

Thus 145 in Octal is 101 in Decimal

(ii) 6760

Steps:

        ( 6 × 8³ ) + ( 7 × 8² ) + ( 6 × 8¹ ) + ( 0 × 8⁰ )

        = ( 6 × 512 ) + ( 7 × 64 ) + ( 6 × 8 ) + ( 0 × 1 )

        = 3072 + 448 + 48 + 0

        = 3568

Thus 6760 in Octal is 3568 in Decimal.

(iii) 455

Steps:

( 4 × 8² ) + ( 5 × 8¹ ) + ( 5 × 8⁰ )

= ( 4 × 64 ) + ( 5 × 8 ) + ( 5 × 1 )

= 256 + 40 + 5

= 301

Thus 455 in Octal is 301 in Decimal

(iv) 10.75

Steps:

Calculate the Decimal value:

( 1 × 8¹ ) + ( 0 × 8⁰ )

= ( 1 × 8 ) + ( 0 × 1 )

= 8 + 0

        = 8

Convert the fractional part (0.75)

( 7 × 8⁻¹ ) + ( 5 × 8⁻² )

= ( 7 × 1/8 ​) + ( 5 × 1/64​ )

= 7/8  +   5/64

= 61/64

Now add both parts:

8  +  61/64

= 8.953125

6. Express the following decimal numbers to hexadecimal numbers:

(i) 548

Steps:

548 ÷ 16 = 34, remainder 4

34 ÷ 16 = 2, remainder 2

2 ÷ 16 = 0, remainder 2

Thus 548 is 224 in Hexadecimal

(ii) 4052

Steps:

4052 ÷ 16 = 253, remainder 4

253 ÷ 16 = 15, remainder 13 (which is represented as D in hexadecimal)

15 ÷ 16 = 0, remainder 15 (which is represented as F in hexadecimal)

Thus 4052 is FD4 in Hexadecimal

(iii) 58

Steps:

58 ÷ 16 = 3, remainder 10 (which is represented as A in hexadecimal)

3 ÷ 16 = 0, remainder 3

Thus 58 is 3A in Hexadecimal

(iv) 100.25

Step 1: Convert the Integer Part (100)

100 ÷ 16 = 6, remainder 4

6 ÷ 16 = 0, remainder 6

Step 2: Convert the Fractional Part (0.25)

Multiply by 16:

0.25 × 16 = 4.0

Thus 100.25 in Hexadecimal is 64.4

7. Express the following hexadecimal numbers into equivalent decimal numbers.

(i) 4A2

Identify the positions and their values:

The hexadecimal number is read from right to left, where each digit’s position corresponds to a power of 16.

The digits in 4A2 are:

• 22 in the 16⁰ (ones) place
• AA (which is 10 in decimal) in the 16¹ (sixteens) place
• 4 in the 16² (two hundred fifty-sixes) place

Calculate the decimal value for each digit:

        2 × 16⁰ = 2 × 1 = 2

        10 × 16¹ = 10 × 16 = 16⁰

        4 × 16² = 4 × 256 = 1024

Sum all the values:

        1024 + 160 + 2 = 1186 

Thus 4A2 in Decimal is 1186

(ii) 9E1A

Calculate decimal value for each digit:

        9 × 16³ = 9 × 4096 = 36864

        14 × 16² = 14 × 256 = 3584

        1 × 16¹ = 1 × 16 = 16

        10 × 16⁰ = 10 × 1 = 10

Sum all the values:  40474

(iii)  6BD

Calculate decimal value for each digit:

        6 × 16² = 6 × 256 = 1536

        11 × 16¹ = 11 × 16 = 176

        13 × 16⁰ = 13 × 1 = 13

Sum all the values: 1725

(iv)  6C.34

Calculate decimal value for integer part (6C)

6 × 16¹ = 6 × 16 = 96

12 × 16⁰ = 12 × 1 = 12

Convert the fractional part (.34)

        3 × 16⁻¹   =   0.1875

        4 × 16⁻²  =  0.015625  

Sum all values and combine both parts to get the answer: 108.203125.

8. Convert the following binary numbers into octal and hexadecimal numbers.

(i) 1110001000

Convert to Decimal first: 904

Now, convert 904 to Octal: 

        904 ÷ 8 = 113 with a remainder of 0

        113 ÷ 8 = 14 with a remainder of 1

        14 ÷ 8 = 1 with a remainder of 6

        1 ÷ 8 = 0 with a remainder of 1

Octal value: 1610

Now, convert 904 to Hexadecimal:

        904 ÷ 16 = 56 with a remainder of 8

        56 ÷ 16 = 3 with a remainder of 8

        3 ÷ 16 = 0 with a remainder of 3

Hexadecimal value: 388

(ii)  110110101

Convert to Decimal first: 437

Now convert 437 to Octal:

        437 ÷ 8 = 54 with a remainder of 5

        54 ÷ 8 = 6 with a remainder of 6

        6 ÷ 8 = 0 with a remainder of 6

Octal value: 665

Now convert 437 to Hexadecimal:

        437 ÷ 16 = 27 with a remainder of 5

        27 ÷ 16 = 1 with a remainder of 11 (which is represented as B in hexadecimal)

        1 ÷ 16 = 0 with a remainder of 1

Hexadecimal value: 1B5

(iii)  1010100

Convert to Decimal first: 84

Now convert 84 to Octal: 

        84 ÷ 8 = 10 with a remainder of 4

        10 ÷ 8 = 1 with a remainder of 2

        1 ÷ 8 = 0 with a remainder of 1

Octal value: 124

Now convert 84 to Hexadecimal:

        84 ÷ 16 = 5 with a remainder of 4

        5 ÷ 16 = 0 with a remainder of 5

Hexadecimal value: 54

(iv)  1010.1001

Convert to Decimal first: 10.5625

Now convert 10.5625 to Octal:

Integer Part:

        10 ÷ 8 = 1 with a remainder of 2

        1 ÷ 8 = 0 with a remainder of 1

Fractional Part:

        0.5625 × 8 = 4.5 (Take the integer part, which is 4)

        0.5 × 8 = 4.0 (Take the integer part, which is 4)

Octal value: 12.44

Now convert 10.5625 to Hexadecimal

Integer Part:

        10 ÷ 16 = 0 with a remainder of 10 (which is represented as A in hexadecimal)

Fractional Part:

        0.5625 × 16 = 9.0 (Take the integer part, which is 9)

Hexadecimal value: A.9

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:

ह: Hex = 0939,      Binary = 0000100100111001

म: Hex = 092E,     Binary = 0000100100101110

स: Hex = 0938,     Binary = 0000100100111000

ब: Hex = 092C,     Binary = 0000100100101100

ए: Hex = 090F,      Binary = 0000100100001111

क: 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.
2. Multilingual Support: Allows seamless integration of multiple Indian languages on a single platform.
3. Consistent Representation: Maintains uniformity in text appearance, reducing misinterpretation when sharing content.
4. 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.