This page contains the NCERT Computer Science class 11 chapter 2 Encoding Schemes and Number System. You can find the solutions for the chapter 2 of NCERT class 11 Computer Science Exercise. So is the case if you are looking for NCERT class 11 Computer Science related topic Encoding Schemes and Number System questions and answers for the Exercise
Exercise
1. Write base values of binary, octal and hexadecimal number system.
•
Binary: base 2
•
Octal: base 8
•
Hexadecimal: base 16
2. Give full form of ASCII and ISCII.
ASCII:
American Standard Code for Information Interchange
ISCII:
Indian Script Code for Information Interchange
3. Try the following conversions.
(i)
(514)₈
= (?)₁₀
(iv)
(4D9)₁₆
= (?)₁₀
(ii)
(220)₈
= (?)₂
(v)
(11001010)₂
= (?)₁₀
(iii)
(76F)₁₆
= (?)₁₀
(vi)
(1010111)₂
= (?)₁₀
(i) (514)₈ = (?)₁₀
(514)₈
= (5 × 8²) + (1 × 8¹) + (4 × 8⁰)
= (5 × 64) + (1 × 8) + (4 × 1)
= 320 + 8 + 4
= (332)₁₀
(ii) (220)₈ = (?)₂
Convert each octal digit in (220)₈ to its equivalent 3-bit binary:
2 → 010 2 → 010 0 → 000
(220)₈
= (010010000)₂
= (10010000)₂
(iii) (76F)₁₆ = (?)₁₀
(76F)₁₆
= (7 × 16²) + (6 × 16¹) + (F × 16⁰)
= (7 × 256) + (6 × 16) + (15 × 1)
= 1792 + 96 + 15
= (1903)₁₀
(iv) (4D9)₁₆ = (?)₁₀
(4D9)₁₆
= (4 × 16²) + (D × 16¹) + (9 × 16⁰)
= (4 × 256) + (13 × 16) + (9 × 1)
= 1024 + 208 + 9
= (1241)₁₀
(v) (11001010)₂ = (?)₁₀
(11001010)₂
= (1 × 2⁷) + (1 × 2⁶) + (0 × 2⁵) + (0 × 2⁴) + (1 × 2³) + (0 × 2²) + (1 × 2¹) + (0 × 2⁰)
= 128 + 64 + 0 + 0 + 8 + 0 + 2 + 0
= (202)₁₀
(vi) (1010111)₂ = (?)₁₀
(1010111)₂
= (1 × 2⁶) + (0 × 2⁵) + (1 × 2⁴) + (0 × 2³) + (1 × 2²) + (1 × 2¹) + (1 × 2⁰)
= 64 + 0 + 16 + 0 + 4 + 2 + 1
= (87)₁₀
4. Do the following conversions from decimal number to other number systems.
(i)
(54)₁₀
= (?)₂
(iv)
(889)₁₀
= (?)₈
(ii)
(120)₁₀
= (?)₂
(v)
(789)₁₀
= (?)₁₆
(iii)
(76)₁₀
= (?)₈
(vi)
(108)₁₀
= (?)₁₆
(i) (54)₁₀ = (?)₂
Division
Quotient
Remainder
54 ÷ 2
27
0
27 ÷ 2
13
1
13 ÷ 2
6
1
6 ÷ 2
3
0
3 ÷ 2
1
1
1 ÷ 2
0
1
2 54 → 0 2 27 → 1 2 13 → 1 2 6 → 0 2 3 → 1 2 1 → 1 2 0 → 1
Reading remainders from bottom to top: (54)₁₀ = (110110)₂
(ii) (120)₁₀ = (?)₂
Division
Quotient
Remainder
120 ÷ 2
60
0
60 ÷ 2
30
0
30 ÷ 2
15
0
15 ÷ 2
7
1
7 ÷ 2
3
1
3 ÷ 2
1
1
1 ÷ 2
0
1
2 120 → 0 2 60 → 0 2 30 → 0 2 15 → 0 2 7 → 1 2 3 → 1 2 1 → 1 2 0 → 1
Reading remainders from bottom to top: (120)₁₀ = (1111000)₂
(iii) (76)₁₀ = (?)₈
Division
Quotient
Remainder
76 ÷ 8
9
4
9 ÷ 8
1
1
1 ÷ 8
0
1
8 76 → 4 8 9 → 1 8 1 → 1 8 0 → -
Reading remainders from bottom to top: (76)₁₀ = (114)₈
(iv) (889)₁₀ = (?)₈
Division
Quotient
Remainder
889 ÷ 8
111
1
111 ÷ 8
13
7
13 ÷ 8
1
5
1 ÷ 8
0
1
8 889 → 1 8 111 → 7 8 13 → 5 8 1 → 1 8 0 → -
Reading remainders from bottom to top: (889)₁₀ = (1571)₈
(v) (789)₁₀ = (?)₁₆
Division
Quotient
Remainder
789 ÷ 16
49
5
49 ÷ 16
3
1
3 ÷ 16
0
3
16 789 → 5 16 49 → 1 16 3 → 3 16 0 → -
Reading remainders from bottom to top: (789)₁₀ = (315)₁₆
(vi) (108)₁₀ = (?)₁₆
Division
Quotient
Remainder
108 ÷ 16
6
12 (C)
6 ÷ 16
0
6
16 108 → C 16 6 → 6 16 0 → -
Reading remainders from bottom to top: (108)₁₀ = (6C)₁₆
5. Express the following octal numbers into their equivalent decimal numbers.
(i)
145
(ii)
6760
(iii)
455
(iv)
10.75
(i) 145
(145)₈
= (1 × 8²) + (4 × 8¹) + (5 × 8⁰)
= (1 × 64) + (4 × 8) + (5 × 1)
= 64 + 32 + 5
= (101)₁₀
(ii) 6760
(6760)₈
= (6 × 8³) + (7 × 8²) + (6 × 8¹) + (0 × 8⁰)
= (6 × 512) + (7 × 64) + (6 × 8) + (0 × 1)
= 3072 + 448 + 48 + 0
= (3568)₁₀
(iii) 455
(455)₈
= (4 × 8²) + (5 × 8¹) + (5 × 8⁰)
= (4 × 64) + (5 × 8) + (5 × 1)
= 256 + 40 + 5
= (301)₁₀
(iv) 10.75
(10.75)₈
= (1 × 8¹) + (0 × 8⁰) + (7 × 8-1) + (5 × 8-2)
{= 8 + 0 + \dfrac{7}{8} + \dfrac{5}{64}}
= 8 + 0.875 + 0.078125
= (8.953125)₁₀
6. Express the following decimal numbers into hexadecimal numbers.
(i)
548
(ii)
4052
(iii)
58
(iv)
100.25
(i) (548)₁₀ = (?)₁₆
Division
Quotient
Remainder
548 ÷ 16
34
4
34 ÷ 16
2
2
2 ÷ 16
0
2
16 548 → 4 16 34 → 2 16 2 → 2 16 0 → -
Reading remainders from bottom to top: (548)₁₀ = (224)₁₆
(ii) (4052)₁₀ = (?)₁₆
Division
Quotient
Remainder
4052 ÷ 16
253
4
253 ÷ 16
15
13 (D)
15 ÷ 16
0
15 (F)
16 4052 → 4 16 253 → D 16 15 → F 16 0 → -
Reading remainders from bottom to top: (4052)₁₀ = (FD4)₁₆
(iii) (58)₁₀ = (?)₁₆
Division
Quotient
Remainder
58 ÷ 16
3
10 (A)
3 ÷ 16
0
3
16 58 → A 16 3 → 3 16 0 → -
Reading remainders from bottom to top: (58)₁₀ = (3A)₁₆
(iv) (100.25)₁₀ = (?)₁₆
Integer part:
Division
Quotient
Remainder
100 ÷ 16
6
4
6 ÷ 16
0
6
16 100 → 4 16 6 → 6 16 0 → -
Integer part: (64)₁₆
Fractional part: 0.25 × 16 = 4.0
Fractional part: (0.4)₁₆
Therefore, (100.25)₁₀ = (64.4)₁₆
7. Express the following hexadecimal numbers into equivalent decimal numbers.
(i)
4A2
(ii)
9E1A
(iii)
6BD
(iv)
6C.34
(i) (4A2)₁₆ = (?)₁₀
(4A2)₁₆
= (4 × 16²) + (A × 16¹) + (2 × 16⁰)
= (4 × 256) + (10 × 16) + (2 × 1)
= 1024 + 160 + 2
= (1186)₁₀
(ii) (9E1A)₁₆ = (?)₁₀
(9E1A)₁₆
= (9 × 16³) + (E × 16²) + (1 × 16¹) + (A × 16⁰)
= (9 × 4096) + (14 × 256) + (1 × 16) + (10 × 1)
= 36864 + 3584 + 16 + 10
= (40474)₁₀
(iii) (6BD)₁₆ = (?)₁₀
(6BD)₁₆
= (6 × 16²) + (B × 16¹) + (D × 16⁰)
= (6 × 256) + (11 × 16) + (13 × 1)
= 1536 + 176 + 13
= (1725)₁₀
(iv) (6C.34)₁₆ = (?)₁₀
Integer part:
(6C)₁₆
= (6 × 16¹) + (C × 16⁰)
= (6 × 16) + (12 × 1)
= 96 + 12 = 108
Fractional part:
(0.34)₁₆
(= 3 × 16-1) + (4 × 16-2)
{= \dfrac{3}{16} + \dfrac{4}{256}}
= 0.1875 + 0.015625 = 0.203125
∴ (6C.34)₁₆ = (108.203125)₁₀
8. Binary to octal and hexadecimal
(i)
1110001000
(ii)
110110101
(iii)
1010100
(iv)
1010.1001
(i) (1110001000)₂
Octal (group 3 bits)
Make perfect group of 3 bits (from right to left):
•
1110001000₂ = 001 110 001 000₂
Convert each 3-bit group to octal digit:
001 → 1 110 → 6 001 → 1 000 → 0
So, 1110001000₂ = 1610₈
Hexadecimal (group 4 bits)
Make perfect group of 4 bits (from right to left):
•
1110001000₂ = 0011 1000 1000₂
Convert each 4-bit group to hex digit:
0011 → 3 1000 → 8 1000 → 8
So, 1110001000₂ = 388₁₆
✅ Final: 1110001000₂ = 1610₈ = 388₁₆
(ii) (110110101)₂
Octal (group 3 bits)
Make perfect group of 3 bits (from right to left):
•
110110101₂ = 110 110 101₂
Convert each 3-bit group to octal digit:
110 → 6 110 → 6 101 → 5
So, 110110101₂ = 665₈
Hexadecimal (group 4 bits)
Make perfect group of 4 bits (from right to left):
•
110110101₂ = 0001 1011 0101₂
Convert each 4-bit group to hex digit:
0001 → 1 1011 → B 0101 → 5
So, 110110101₂ = 1B5₁₆
✅ Final: 110110101₂ = 665₈ = 1B5₁₆
(iii) (1010100)₂
Octal (group 3 bits)
Make perfect group of 3 bits (from right to left):
•
1010100₂ = 001 010 100₂
Convert each 3-bit group to octal digit:
001 → 1 010 → 2 100 → 4
So, 1010100₂ = 124₈
Hexadecimal (group 4 bits)
Make perfect group of 4 bits (from right to left):
•
1010100₂ = 0101 0100₂
Convert each 4-bit group to hex digit:
0101 → 5 0100 → 4
So, 1010100₂ = 54₁₆
✅ Final: 1010100₂ = 124₈ = 54₁₆
(iv) (1010.1001)₂
Octal (group 3 bits)
Integer part (grouping right → left):
•
1010₂ = 001 010₂
Convert integer groups:
001 → 1 010 → 2
Integer part = 12₈
Fractional part (grouping left → right; pad zeros on right):
•
1001₂ = 100 100₂
Convert fractional groups:
100 → 4 100 → 4
Fractional part = .44₈
So, 1010.1001₂ = 12.44₈
Hexadecimal (group 4 bits)
•
1010.1001₂ = 1010 . 1001₂
Convert each 4-bit group:
1010 → A 1001 → 9
So, 1010.1001₂ = A.9₁₆
✅ Final: 1010.1001₂ = 12.44₈ = A.9₁₆
9. Write binary equivalent of the following octal numbers.
(i)
2306
(ii)
5610
(iii)
742
(iv)
65.203
(i) (2306)₈ = (?)₂
Convert each octal digit to its 3-bit binary equivalent:
2 → 010 3 → 011 0 → 000 6 → 110
(2306)₈
= (010 011 000 110)₂
= (10011000110)₂
(ii) (5610)₈ = (?)₂
Convert each octal digit to its 3-bit binary equivalent:
5 → 101 6 → 110 1 → 001 0 → 000
(5610)₈
= (101 110 001 000)₂
= (101110001000)₂
(iii) (742)₈ = (?)₂
Convert each octal digit to its 3-bit binary equivalent:
7 → 111 4 → 100 2 → 010
(742)₈
= (111 100 010)₂
= (111100010)₂
(iv) (65.203)₈ = (?)₂
Integer part
Convert each octal digit (group from right → left):
6 → 110 5 → 101
(65)₈
= (110 101)₂
= (110101)₂
Fractional part
Convert each octal digit (left → right; pad to 3 bits):
2 → 010 0 → 000 3 → 011
(.203)₈
= (.010 000 011)₂
= (.010000011)₂
✅ Final: (65.203)₈ = (110101.010000011)₂
10. Write binary representation of the following hexadecimal numbers.
(i)
4026
(ii)
BCA1
(iii)
98E
(iv)
132.45
(i) (4026)₁₆ = (?)₂
Convert each hex digit to its 4-bit binary equivalent:
4 → 0100 0 → 0000 2 → 0010 6 → 0110
(4026)₁₆
= (0100 0000 0010 0110)₂
= (100000000100110)₂
(ii) (BCA1)₁₆ = (?)₂
Convert each hex digit to its 4-bit binary equivalent:
B → 1011 C → 1100 A → 1010 1 → 0001
(BCA1)₁₆
= (1011 1100 1010 0001)₂
= (1011110010100001)₂
(iii) (98E)₁₆ = (?)₂
Convert each hex digit to its 4-bit binary equivalent:
9 → 1001 8 → 1000 E → 1110
(98E)₁₆
= (1001 1000 1110)₂
= (100110001110)₂
(iv) (132.45)₁₆ = (?)₂
Integer part
Convert each hex digit to 4-bit binary:
1 → 0001 3 → 0011 2 → 0010
(132)₁₆
= (0001 0011 0010)₂
= (100110010)₂ (leading zeros dropped)
Fractional part
Convert each hex digit to 4-bit binary (left → right):
4 → 0100 5 → 0101
(.45)₁₆
= (.0100 0101)₂
= (.01000101)₂
✅ Final: (132.45)₁₆ = (100110010.01000101)₂
11. How does computer understand the following text?
(hint: 7-bit ASCII code)
(i)
HOTS
(ii)
Main
(iii)
CaSe
ASCII assigns each printable character a numeric code (0–127). Below we show the decimal code and the 7-bit binary for each character. Note: bytes typically store this in 8 bits with a leading 0.
(i) HOTS
Character
Decimal
7-bit binary
H
72
1001000
O
79
1001111
T
84
1010100
S
83
1010011
(ii) Main
Character
Decimal
7-bit binary
M
77
1001101
a
97
1100001
i
105
1101001
n
110
1101110
(iii) CaSe
Character
Decimal
7-bit binary
C
67
1000011
a
97
1100001
S
83
1010011
e
101
1100101
Note: ASCII is a 7-bit standard (0–127). Many systems store ASCII in 8-bit bytes (adding a leading 0). For characters beyond 127 (e.g., accented letters or other scripts) use extended ASCII or Unicode (UTF-8).
12) The hexadecimal number system uses 16 literals (0 – 9, A – F). Write down its base value.
Hexadecimal (hex) uses 16 distinct symbols: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. Because there are 16 symbols, the base (radix) is 16.
Hex digits:
0 1 2 3 4 5 6 7 8 9 A B C D E FDecimal values:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Example:
(2A3)₁₆
= (2 × 16²) + (10 × 16¹) + (3 × 16⁰)
(2A3)₁₆
= (2 × 256) + (10 × 16) + (3 × 1)
= 512 + 160 + 3
= 675₁₀
13. If a number system X has B distinct symbols, what is its base (radix)?
The base (radix) equals the number of distinct symbols, so the base = B.
Explanation: A positional number system with B symbols uses place values that are powers of B: the rightmost digit is multiplied by B⁰, the next by B¹, then B², and so on. Digits allowed are 0 to B−1.
Example (B = 7):
(321)₇
= (3 × 7²) + (2 × 7¹) + (1 × 7⁰)
= (3 × 49) + (2 × 7) + (1 × 1)
= 147 + 14 + 1
= 162₁₀
14. Write the equivalent hexadecimal and binary values for each character of the phrase given below.
‘‘हम सब एक”
This uses UNICODE (each character has a unique hexadecimal code).
(Shown in the Devanagari Unicode table: ए=090F, क=0915, ब=092C, म=092E, स=0938, ह=0939, etc.)
Character
Code point (hex)
UTF-16 (16-bit binary)
UTF-8 (hex bytes)
ह
U+0939
0000100100111001
E0 A4 B9
म
U+092E
0000100100101110
E0 A4 AE
(space)
U+0020
0000000000100000
20
स
U+0938
0000100100111000
E0 A4 B8
ब
U+092C
0000100100101100
E0 A4 AC
(space)
U+0020
0000000000100000
20
ए
U+090F
0000100100001111
E0 A4 8F
क
U+0915
0000100100010101
E0 A4 95
•
(Example: ‘अ’ → U+0905 → UTF-16 0000100100000101 → UTF-8 bytes E0 A4 85.)
15. What is the advantage of preparing digital content in Indian languages using a Unicode font?
Using Unicode for Indian-language content provides several practical benefits:
•
Compatibility: The same encoded text works across operating systems, browsers and applications without corruption.
•
Correct rendering: Unicode-aware fonts and shaping engines render complex Indic scripts (conjuncts, matras) properly.
•
Search & interoperability: Text can be searched, copied, indexed, and transmitted reliably between systems.
•
Single standard for many scripts: Devanagari, Tamil, Bengali, Telugu, etc., are supported under one encoding scheme.
•
Web & storage friendly: UTF-8 (a Unicode encoding) is compact, widely supported on the web and recommended for storage and interchange.
•
Backward compatible: ASCII (0–127) maps directly into Unicode (U+0000–U+007F), preserving legacy data.
Example: the character ‘अ’ is U+0905; encoded in UTF-8 it becomes the byte sequence E0 A4 85, which will display correctly on systems with a Unicode font.
16. Explore and list the steps required to type in an Indian language using UNICODE.
•
Add the language/keyboard in your system settings (Hindi, Tamil, etc.).
•
Turn on the keyboard/input method (language bar / input selector).
•
Open any app (Word, browser, editor) and select a Unicode font that supports that script.
•
Switch to the chosen language keyboard and start typing.
•
Save/share normally—because Unicode keeps text compatible across systems.
17. Encode the word ‘COMPUTER’ using ASCII and convert the encode value into binary values.
Character
ASCII
Binary Value
C
67
1000011
O
79
1001111
M
77
1001101
P
80
1010000
U
85
1010101
T
84
1010100
E
69
1000101
R
82
1010010
∴ The binary value of the word ‘COMPUTER’ is
1000011 1001111 1001101 1010000 1010101 1010100 1000101 1010010