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Updated 6 August 2026

NCERT Solutions

Encoding Schemes and Number System

CBSE · Class 11 · Computer Science

NCERT Solutions for Encoding Schemes and Number System — CBSE Class 11 Computer Science.

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EXERCISE

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

The base value of a number system is the number of unique symbols it uses. So binary has 2 digits, octal has 8 digits, and hexadecimal has 16 digits.

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2Give full form of ASCII and ISCII.Show solution

ASCII stands for American Standard Code for Information Interchange. ISCII stands for Indian Script Code for Information Interchange.

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3(i)

(514)_8 = (?)_{10}

Show solution

Convert

(514)_8

to decimal using positional values:

5 \times 8^2 + 1 \times 8^1 + 4 \times 8^0

= 5 \times 64 + 1 \times 8 + 4 \times 1

= 320 + 8 + 4 = 332

So,

(514)_8 = (332)_{10}

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3(ii)

(220)_8 = (?)_2

Show solution

Convert each octal digit to its 3-bit binary equivalent:

-

2 = 010

2 = 010

0 = 000

So,

(220)_8 = 010\ 010\ 000_2 = 100100000_2

after removing the leading zero.

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3(iii)

(76F)_{16} = (?)_{10}

Show solution

Convert

(76F)_{16}

to decimal:

7 \times 16^2 + 6 \times 16^1 + F \times 16^0

Here,

F = 15

= 7 \times 256 + 6 \times 16 + 15

= 1792 + 96 + 15 = 1903

So,

(76F)_{16} = (1903)_{10}

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3(iv)

(4D9)_{16} = (?)_{10}

Show solution

Convert

(4D9)_{16}

to decimal:

4 \times 16^2 + D \times 16^1 + 9 \times 16^0

Here,

D = 13

= 4 \times 256 + 13 \times 16 + 9

= 1024 + 208 + 9 = 1241

So the correct decimal value is 1241. If you computed 3481, that would not match the hexadecimal conversion. The book's answer from the given value is

(4D9)_{16} = (1241)_{10}

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3(v)

(11001010)_2 = (?)_{10}

Show solution

Convert

(11001010)_2

to decimal using place values:

1\times 2^7 + 1\times 2^6 + 0\times 2^5 + 0\times 2^4 + 1\times 2^3 + 0\times 2^2 + 1\times 2^1 + 0\times 2^0

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

So,

(11001010)_2 = (202)_{10}

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3(vi)

(1010111)_2 = (?)_{10}

Show solution

Convert

(1010111)_2

to decimal:

1\times 2^6 + 0\times 2^5 + 1\times 2^4 + 0\times 2^3 + 1\times 2^2 + 1\times 2^1 + 1\times 2^0

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

So,

(1010111)_2 = (87)_{10}

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4(i)

(54)_{10} = (?)_2

Show solution

Convert

54_{10}

to binary by dividing by 2:

54 \div 2 = 27

remainder

0

27 \div 2 = 13

remainder

1

13 \div 2 = 6

remainder

1

6 \div 2 = 3

remainder

0

3 \div 2 = 1

remainder

1

1 \div 2 = 0

remainder

1

Write remainders from bottom to top: 110110.

So,

(54)_{10} = (110110)_2

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4(ii)

(120)_{10} = (?)_2

Show solution

Convert

120_{10}

to binary:

120 \div 2 = 60

remainder

0

60 \div 2 = 30

remainder

0

30 \div 2 = 15

remainder

0

15 \div 2 = 7

remainder

1

7 \div 2 = 3

remainder

1

3 \div 2 = 1

remainder

1

1 \div 2 = 0

remainder

1

Reading remainders from bottom to top gives 1111000.

So,

(120)_{10} = (1111000)_2

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4(iii)

(76)_{10} = (?)_8

Show solution

Convert

(76)_{10}

to octal by dividing by 8:

76 \div 8 = 9

remainder

4

9 \div 8 = 1

remainder

1

1 \div 8 = 0

remainder

1

Write remainders from bottom to top: 114.

So,

(76)_{10} = (114)_8

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4(iv)

(889)_{10} = (?)_8

Show solution

Convert

889_{10}

to octal:

889 \div 8 = 111

remainder

1

111 \div 8 = 13

remainder

7

13 \div 8 = 1

remainder

5

1 \div 8 = 0

remainder

1

Write remainders from bottom to top: 1571.

So,

(889)_{10} = (1571)_8

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4(v)

(789)_{10} = (?)_{16}

Show solution

Convert

789_{10}

to hexadecimal by dividing by 16:

789 \div 16 = 49

remainder

5

49 \div 16 = 3

remainder

1

3 \div 16 = 0

remainder

3

Write remainders from bottom to top: 315.

So,

(789)_{10} = (315)_{16}

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4(vi)

(108)_{10} = (?)_{16}

Show solution

Convert

108_{10}

to hexadecimal:

108 \div 16 = 6

remainder

12

12

in hexadecimal is C.

Write remainders from bottom to top: 6C.

So,

(108)_{10} = (6C)_{16}

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5(i)145Show solution

Convert

145_{10}

to octal:

145 \div 8 = 18

remainder

1

18 \div 8 = 2

remainder

2

2 \div 8 = 0

remainder

2

Write remainders from bottom to top: 221.

So,

(145)_{10} = (221)_8

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5(ii)6760Show solution

Convert

6760_{10}

to octal:

6760 \div 8 = 845

remainder

0

845 \div 8 = 105

remainder

5

105 \div 8 = 13

remainder

1

13 \div 8 = 1

remainder

5

1 \div 8 = 0

remainder

1

Write remainders from bottom to top: 15150.

So,

(6760)_{10} = (15150)_8

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5(iii)455Show solution

Convert

455_{10}

to octal:

455 \div 8 = 56

remainder

7

56 \div 8 = 7

remainder

0

7 \div 8 = 0

remainder

7

Write remainders from bottom to top: 707.

So,

(455)_{10} = (707)_8

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5(iv)10.75Show solution

Convert

10.75_{10}

to octal.

Integer part:

10_{10} = 12_8

.

Fractional part:

0.75 \times 8 = 6.00

.
So the fractional part is

.6_8

.

Therefore,

(10.75)_{10} = (12.6)_8

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6(i)548Show solution

Convert

548_{10}

to hexadecimal first:

548 \div 16 = 34

remainder

4

34 \div 16 = 2

remainder

2

2 \div 16 = 0

remainder

2

So,

548_{10} = 224_{16}

.

Now convert each hex digit to 4-bit binary:

-

2 = 0010

2 = 0010

4 = 0100

Combine:

0010\ 0010\ 0100 = 1000100100

after removing the leading zeros.

So, the binary form is

(1000100100)_2

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6(ii)4052Show solution

Convert

4052_{10}

to binary by division by 2:

4052 \div 2 = 2026

0

2026 \div 2 = 1013

0

1013 \div 2 = 506

1

506 \div 2 = 253

0

253 \div 2 = 126

1

126 \div 2 = 63

0

63 \div 2 = 31

1

31 \div 2 = 15

1

15 \div 2 = 7

1

7 \div 2 = 3

1

3 \div 2 = 1

1

1 \div 2 = 0

1

Reading bottom to top gives 111111010100. So,

(4052)_{10} = (111111010100)_2

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6(iii)58Show solution

Convert

58_{10}

to hexadecimal:

58 \div 16 = 3

remainder

10

10

in hexadecimal is A.

So,

(58)_{10} = (3A)_{16}

.

To write its binary representation:
-

3 = 0011

A = 1010

So binary is

0011\ 1010 = 111010

after removing leading zeros.

Thus,

(58)_{10} = (3A)_{16}

and binary is

(111010)_2

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6(iv)100.25Show solution

Convert

100.25_{10}

to binary.

Integer part:

100_{10} = 1100100_2

.

Fractional part:

0.25 \times 2 = 0.5

gives

0

;

0.5 \times 2 = 1.0

gives

1

.
So fractional part is

.01_2

.

Therefore,

(100.25)_{10} = (1100100.01)_2

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7(i)4A2Show solution

Convert hexadecimal

(4A2)_{16}

to decimal:

4 \times 16^2 + A \times 16^1 + 2 \times 16^0

= 4 \times 256 + 10 \times 16 + 2

= 1024 + 160 + 2 = 1186

So, the decimal value is 1186. The hexadecimal number itself is already given as 4A2.

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7(ii)9E1AShow solution

Convert

(9E1A)_{16}

to decimal:

9 \times 16^3 + E \times 16^2 + 1 \times 16^1 + A \times 16^0

Here,

E=14

and

A=10

= 9\times 4096 + 14\times 256 + 1\times 16 + 10

= 36864 + 3584 + 16 + 10 = 40474

So,

(9E1A)_{16} = (40474)_{10}

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7(iii)6BD

7(iv)6C.34

8(i)1110001000

8(ii)110110101

8(iii)1010100

8(iv)1010.1001

9(i)2306

9(ii)5610

9(iii)742

9(iv)65.203

10(i)4026

10(ii)BCA1

10(iii)98E

10(iv)132.45

11(i)HOTS

11(ii)Main

11(iii)CaSe

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

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

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

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

16Explore and list the steps required to type in an Indian language using UNICODE.

17Encode the word ‘COMPUTER’ using ASCII and convert the encode value into binary values.

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Frequently Asked Questions

What are the important topics in Encoding Schemes and Number System for CBSE Class 11 Computer Science?

Encoding Schemes and Number System covers several key topics that are frequently asked in CBSE Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.

How to score full marks in Encoding Schemes and Number System — CBSE Class 11 Computer Science?

Understand the core concepts first, then work through the 86 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

Where can I get free NCERT Solutions for Encoding Schemes and Number System Class 11 Computer Science?

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