Chapter 2 : Encoding Schemes and Number System | class 11th | Ncert solution for computer

NCERT Solutions Class 11 Computer Science Chapter 2 Encoding Schemes and Number System

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

Answer:

NUMBER SYSTEM	BASE VALUE
Binary number system	2
Octal number system	8
hexadecimal number system	16

Question 2:  Give full form of ASCII and ISCII.

Answer:

• The full form of ASCII is American Standard Code for Information
• The full form of ISCII is Indian Script Code for Information Interchange.

Question 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

Answer:

• (514)₈ = (?)₁₀

Digits	5	1	4
Position	2	1	0
Weight	82	81	80

Therefore,

                  Decimal number = 5×8² + 1×8¹ + 4×8⁰ 

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

                                      = 320 + 8 + 4

= (332)₁₀

• (220)₈ = (?)₂

Octal Digits	2	2	0
Binary value    (3 bits)	010	010	000

Therefore,

                  Binary number = (010010000)₂

• (76F)₁₆ = (?)₁₀

Digits	7	6	F(15)
Position	2	1	0
Weight	162	161	160

Therefore,

Decimal number = 7×16² + 6×16¹ + F×16⁰ 

                         = 7×256 + 6×16 + F× 1

                                      = 1792 + 96 + 15

                                         = (1903)₁₀

• (4D9)16 = (?)10

Digits	4	D	9
Position	2	1	0
Weight	162	161	160

Therefore,

Decimal number = 4×16² + 13×16¹ + 9×16⁰ 

                         = 4×256 + 13×16 + 9× 1

                                      = 1024 + 208 + 9

                                         = (1241)₁₀

• (11001010)2 = (?)10

Digits	1	1	0	0	1	0	1	0
Position	7	6	5	4	3	2	1	0
Weight	27	26	25	24	23	22	21	20

 Therefore,

Decimal number = 1×2⁷+ 1×2⁶+0×2⁵ +0×2⁴ +1×2³ +0×2² +1×2¹ +0×2⁰

                         = 128+64 +8 + 2

                                      = (202)₁₀

• (1010111)2 = (?)10

Digits	1	0	1	0	1	1	1
Position	6	5	4	3	2	1	0
Weight	26	25	24	23	22	21	20

Therefore,

Decimal number = 1×2⁶+ 0×2⁵+1×2⁴ +0×2³ +1×2² +1×2¹ +1×2⁰

                         = 64 +16 + 4+ 2 +1

                                      = (87)₁₀

Question 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

Answer:

Question 5: Express the following octal numbers into their equivalent decimal numbers.

(i) 145

(ii) 6760

(iii) 455

(iv) 10.75

Answer:

(i) 145

Digits	1	4	5
Position	2	1	0
Weight	82	81	80

Therefore,

 Decimal number = 1×8² +4×8¹ + 5×8⁰ 

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

                                      =64 + 32 + 5

= (101)₁₀

(ii) 6760

Digits	6	7	6	0
Position	3	2	1	0
Weight	83	82	81	80

Therefore,

Decimal number = 6×8³ +7×8² +6×8¹ + 0×8⁰ 

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

                                      =3072 + 448 + 48+0

= (3568)₁₀

(i) 455

Digits	4	5	5
Position	2	1	0
Weight	82	81	80

Therefore,

Decimal number = 4×8² +5×8¹ + 5×8⁰ 

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

                                      =256 + 40 + 5

= (301)₁₀

(iv) 10.75

Digits	1	0	7	5
Position	1	0	-1	-2
Weight	81	80	8-1	8-2

 Therefore,

Decimal number = 1×8¹+0×8⁰+7×8^-1+5×8^-2 

                                        =1×8+0×1+7×0.125+5×0.015625

                                        =8+0+0.875+0.078125

= (8.953125)₁₀

Question 6:  Express the following decimal numbers into hexadecimal numbers. (i) 548 (ii) 4052 (iii) 58 (iv) 100.25