(X) Ch-6 Number System

1. Answer the following questions:
  1. What is number system?
    2. Answer :
    The system concerned with the number and represented by sequence of digits is called number system. It plays vital role in computing and electronics. Number system also refers to the digits, its arrangement, positional value and base of number system.
  2. What do you mean by decimal number system?
    3. Answer:
    The number system having base 10 and consists of digits: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 is called Decimal (Denary) number system.
  3. What is Binary number system? Write its uses in computer system
    4. Answer :
    The number system having base two(2) and consists of digits: 0 and 1 is called Binary number system.
    It has two bits 0 and 1. An electronic circuit has two states either ON & OFF state. The bit 1 represents high voltage (ON state) and the bit 0 represents the low voltage (OFF state) of an electronic circuit.
  4. What do you meant by base or radix of number system?
    5. Answer :
    The total number of digits used by the particular number system is called base or radix of that number system. For Example: Base of Binary number system is 2 because it uses two digits 0 and 1 only.
  5. What are the numbers and characters sets of hexadecimal number system?
    6. Answer :
    Numbers of hexadecimal are 0 to 9 (0,1,2,3,4,5,6,7,8,9) and characters of hexadecimal are A - F (A,B,C,D,E,F).
  6. What do you mean by BCD and ASCII?
    7. Answer :
    BCD(Binary Coded Decimal)
    The process that represents the digits and alphabets with the help of binary digits is called BCD. It is 4 bit code, which can represent a maximum of 64 different characters.
    ASCII(American Standard Code for Information Interchange)
    An 8 bit code and is exclusively used to represent the data internally in the micro computers is called ASCII.
  7. Write the use of octal and hexadecimal number system in computer system?
    8. Answer :
    The octal and hexadecimal number systems are introduced to interpret large binary numbers in computer languages.
2. Convert the following binary number to decimal numbers:
a. (1101101)2 = (?)10
Answer:
Face Value 1 1 0 1 1 0 1
Place Value 26 25 24 23 22 21 20
=1×26+1×25+0×24+1×23+1×22+0×21+1×20
=64+32+0+8+4+0+1
=(109)10
∴ (1101101)2 = (109)10
b. (101010)2 = (?)10
Answer:
Face Value 1 0 1 0 1 0
Place Value 25 24 23 22 21 20
=1×25+0×24+1×23+0×22+1×21+0×20
=32+0+8+0+2+0
=(42)10
∴ (101010)2 = (42)10
c. (1100110)2 = (?)10
Answer:
Face Value 1 1 0 0 1 1 0
Place Value 26 25 24 23 22 21 20
=1×26+1×25+0×24+0×23+1×22+1×21+0×20
=64+32+0+0+4+2+0
=(102)10
∴ (1100110)2 = (102)10
d. (1100011)2 = (?)10
Answer:
Face Value 1 1 0 0 0 1 1
Place Value 26 25 24 23 22 21 20
=1×26+1×25+0×24+0×23+0×22+1×21+1×20
=64+32+0+0+0+2+1
=(99)10
∴ (1100011)2 = (99)10
e. (111011)2 = (?)10
Answer:
Face Value 1 1 1 0 1 1
Place Value 25 24 23 22 21 20
=1×25+1×24+1×23+0×22+1×21+1×20
=32+16+8+0+2+1
=(59)10
∴ (1101101)2 = (59)10
f. (1010101)2 = (?)10
Answer:
Face Value 1 0 1 0 1 0 1
Place Value 26 25 24 23 22 21 20
=1×26+0×25+1×24+0×23+1×22+0×21+1×20
=64+0+16+0+4+0+1
=(85)10
∴ (1010101)2 = (85)10
3. Convert the following decimal numbers to binary numbers:
a. (44)10=(?)2
2 44 Rem
2 22 0
2 11 0
2 5 1
2 2 1
2 1 0
0 1
∴ (44)10 = (101100)2
b. (404)10=(?)2
2 404 Rem
2 202 0
2 101 0
2 50 1
2 25 0
2 12 1
2 6 0
2 3 0
2 1 1
0 1
∴ (404)10 = (110010100)2
c. (100)10=(?)2
2 100 Rem
2 50 0
2 25 0
2 12 1
2 6 0
2 3 0
2 1 1
0 1
∴ (100)10 = (1100100)2
d. (145)10=(?)2
2 145 Rem
2 72 1
2 36 0
2 18 0
2 9 0
2 4 1
2 2 0
2 1 0
0 1
∴ (145)10 = (10010001)2
e. (325)10=(?)2
2 325 Rem
2 162 1
2 81 0
2 40 1
2 20 0
2 10 0
2 5 0
2 2 1
2 1 0
0 1
∴ (325)10 = (101000101)2
f. (505)10=(?)2
2 505 Rem
2 252 1
2 126 0
2 63 0
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
0 1
∴ (505)10 = (111111001)2
4. Convert the following decimal numbers to octal numbers:
a. (44)10=(?)8
8 44 Rem
8 5 4
0 5
∴ (44)10 = (54)8
b. (104)10=(?)8
8 104 Rem
8 13 0
8 1 5
0 1
∴ (104)10 = (150)8
c. (540)10=(?)8
8 540 Rem
8 67 4
8 8 3
8 1 0
0 1
∴ (540)10 = (1034)8
d. (230)10=(?)8
8 230 Rem
8 28 6
8 3 4
0 3
∴ (230)10 = (346)8
e. (190)10=(?)8
8 190 Rem
8 23 6
8 2 7
0 2
∴ (190)10 = (276)8
f. (980)10=(?)8
8 980 Rem
8 122 4
8 15 2
8 1 7
0 1
∴ (980)10 = (1724)8
5. Convert the following octal numbers to decimal numbers.
a. (740)8 = (?)10
Answer:
Face Value 7 4 0
Place Value 82 81 80
=7×82+4×81+0×80
=7x64+4x8+0x1
=448+32+0
=(480)10
∴ (740)8 = (480)10
b. (570)8 = (?)10
Answer:
Face Value 5 7 0
Place Value 82 81 80
=5×82+7×81+0×80
=5x64+7x8+0x1
=320+56+0
=(376)10
∴ (570)8 = (376)10
c. (250)8 = (?)10
Answer:
Face Value 2 5 0
Place Value 82 81 80
=2×82+5×81+0×80
=2x64+5x8+0x1
=128+40+0
=(168)10
∴ (250)8 = (168)10
d. (1230)8 = (?)10
Answer:
Face Value 1 2 3 0
Place Value 83 82 81 80
=1×83+2×82+3×81+0×80
=1x512+2x64+3x8+0x1
=512+128+24+0
=(664)10
∴ (1230)8 = (664)10
e. (330)8 = (?)10
Answer:
Face Value 3 3 0
Place Value 82 81 80
=3×82+3×81+0×80
=3x64+3x8+0x1
=192+24+0
=(216)10
∴ (330)8 = (216)10
f. (2020)8 = (?)10
Answer:
Face Value 2 0 2 0
Place Value 83 82 81 80
=2×83+0×82+2×81+0×80
=2x512+0x64+2x8+0x1
=1024+0+16+0
=(1040)10
∴ (2020)8 = (1040)10
6. Convert the following decimal numbers to hexadecimal numbers.
a. (25)10=(?)16
16 25 Rem
16 1 9
0 1
∴ (25)10 = (19)16
b. (40)10=(?)16
16 40 Rem
16 2 8
0 2
∴ (40)10 = (28)16
c. (45)10=(?)16
16 45 Rem
16 2 13 (D)
0 2
∴ (45)10 = (2D)16
d. (46)10=(?)16
16 46 Rem
16 2 14 (E)
0 2
∴ (25)10 = (2E)16
e. (145)10=(?)16
16 145 Rem
16 9 1
0 9
∴ (145)10 = (91)16
f. (115)10=(?)16
16 115 Rem
16 7 3
0 7
∴ (115)10 = (73)16
7. Convert the following hexa-decimal numbers to decimal numbers.
a. (130)16 = (?)10
Answer:
Face Value 1 3 0
Place Value 162 161 160
=1×162+3×161+0×160
=1x256+3x16+0x1
=256+48+0
=(304)10
∴ (130)16 = (304)10
b. (1A7)16 = (?)10
Answer:
Face Value 1 A 7
Place Value 162 161 160
=1×162+A×161+7×160
=1x256+10x16+7x1
=256+160+7
=(423)10
∴ (1A7)16 = (423)10
c. (240)16 = (?)10
Answer:
Face Value 2 4 0
Place Value 162 161 160
=2×162+4×161+0×160
=2x256+4x16+0x1
=512+64+0
=(576)10
∴ (240)16 = (576)10
d. (2AB)16 = (?)10
Answer:
Face Value 2 A B
Place Value 162 161 160
=2×162+A×161+B×160
=2x256+10x16+11x1
=512+160+11
=(683)10
∴ (2AB)16 = (683)10
e. (4FD)16 = (?)10
Answer:
Face Value 4 F D
Place Value 162 161 160
=4×162+F×161+D×160
=4x256+15x16+13x1
=1024+240+13
=(1277)10
∴ (4FD)16 = (1277)10
f. (530F)16 = (?)10
Answer:
Face Value 5 3 0 F
Place Value 163 162 161 160
=5x163+3×162+0×161+F×160
=5x4096+3x256+0x16+15x1
=20480+768+0+15
=(21263)10
∴ (530F)16 = (21263)10
8. Convert the following octal numbers to binary numbers.
a. (77)8 = (?)2
Answer:
Octal Value 7 7
Binary value in 3 digits 111 111
∴ (77)8 = (111111)2
b. (76)8 = (?)2
Answer:
Octal Value 7 6
Binary value in 3 digits 111 110
∴ (76)8 = (111110)2
c. (47)8 = (?)2
Answer:
Octal Value 4 7
Binary value in 3 digits 100 111
∴ (47)8 = (100111)2
d. (17)8 = (?)2
Answer:
Octal Value 1 7
Binary value in 3 digits 001 111
∴ (17)8 = (1111)2
e. (65)8 = (?)2
Answer:
Octal Value 6 5
Binary value in 3 digits 110 101
∴ (65)8 = (1110101)2
f. (127)8 = (?)2
Answer:
Octal Value 1 2 7
Binary value in 3 digits 001 010 1111
∴ (127)8 = (1010111)2
9. Convert the following binary numbers to octal numbers.
a. (1010)2 = (?)8
Answer:
Binary numbers in group of 3 bits
(From Right to left)		 001 010
Octal Value 1 2
∴ (1010)2 = (12)8
b. (1110)2 = (?)8
Answer:
Binary numbers in group of 3 bits
(From Right to left)		 001 110
Octal Value 1 6
∴ (1010)2 = (16)8
c. (111)2 = (?)8
Answer:
Binary numbers in group of 3 bits
(From Right to left)		 111
Octal Value 7
∴ (111)2 = (7)8
d. (10111)2 = (?)8
Answer:
Binary numbers in group of 3 bits
(From Right to left)		 010 111
Octal Value 2 7
∴ (10111)2 = (27)8
e. (11101)2 = (?)8
Answer:
Binary numbers in group of 3 bits
(From Right to left)		 011 101
Octal Value 3 5
∴ (1101)2 = (35)8
f. (10111)2 = (?)8
Answer:
Binary numbers in group of 3 bits
(From Right to left)		 010 101
Octal Value 2 7
∴ (10111)2 = (27)8
10. Convert the following binary numbers to hexadecimal numbers.
a. (10101)2 = (?)16
Answer:
Binary numbers in group of 4 bits
(From Right to left)		 0001 0101
Hexadecimal Value 1 5
∴ (10101)2 = (15)16
b. (1111)2 = (?)16
Answer:
Binary numbers in group of 4 bits
(From Right to left)		 1111
Hexadecimal Value 15 (F)
∴ (1111)2 = (F)16
c. (11110)2 = (?)16
Answer:
Binary numbers in group of 4 bits
(From Right to left)		 0001 1110
Hexadecimal Value 1 14 (E)
∴ (10101)2 = (1E)16
d. (110)2 = (?)16
Answer:
Binary numbers in group of 4 bits
(From Right to left)		 0110
Hexadecimal Value 6
∴ (110)2 = (6)16
e. (1101)2 = (?)16
Answer:
Binary numbers in group of 4 bits
(From Right to left)		 1101
Hexadecimal Value 13 (D)
∴ (1101)2 = (D)16
f. (1111)2 = (?)16
Answer:
Binary numbers in group of 4 bits
(From Right to left)		 1111
Hexadecimal Value 15 (F)
∴ (1111)2 = (F)16
11. Convert the following hexa-decimal numbers to binary numbers.
a. (56)16 = (?)2
Answer:
Hexadecimal Value 5 6
Decimal Value 5 6
Binary value 0101 0110
∴ (56)16 = (1010110)2
b. (A27)16 = (?)2
Answer:
Hexadecimal Value A 2 7
Decimal Value 10 2 7
Binary value 1010 0010 0111
∴ (A27)16 = (101000100111)2
c. (76)16 = (?)2
Answer:
Hexadecimal Value 7 6
Decimal Value 7 6
Binary value 0111 0110
∴ (76)16 = (1110110)2
d. (3B)16 = (?)2
Answer:
Hexadecimal Value 3 B
Decimal Value 3 11
Binary value 0011 1011
∴ (3B)16 = (111011)2
e. (A16)16 = (?)2
Answer:
Hexadecimal Value A 1 6
Decimal Value 10 1 6
Binary value 1010 0001 0110
∴ (A16)16 = (101000010110)2
f. (F45)16 = (?)2
Answer:
Hexadecimal Value F 4 5
Decimal Value 15 4 5
Binary value 1111 0100 0101
∴ (F45)16 = (111101000101)2
12. Convert the following octal numbers to hexadecimal numbers.
a. (45)8 = (?)16
Answer:
Octal number 4 5
3 bits binary number 100 101
∵ (45)8 = (0010 0101)2
Again,
Binary numbers in 4 bits 0010 0101
Hexadecimal number 2 5
∴(45)8=(25)16
b. (27)8 = (?)16
Answer:
Octal number 2 7
3 bits binary number 010 111
∵ (27)8 = (0001 0111)2
Again,
Binary numbers in 4 bits 0001 0111
Hexadecimal number 1 7
∴(27)8=(17)16
c. (136)8 = (?)16
Answer:
Octal number 1 3 6
3 bits binary number 001 011 110
∵ (136)8 = (0101 1110)2
Again,
Binary numbers in 4 bits 0101 1110
Hexadecimal number 5 14 (E)
∴(136)8=(5E)16
d. (100)8 = (?)16
Answer:
Octal number 1 0 0
3 bits binary number 001 000 000
∵ (100)8 = (0100 0000)2
Again,
Binary numbers in 4 bits 0100 0000
Hexadecimal number 4 0
∴(100)8=(40)16
e. (237)8 = (?)16
Answer:
Octal number 2 3 7
3 bits binary number 010 011 111
∵ (237)8 = (1001 1111)2
Again,
Binary numbers in 4 bits 1001 1111
Hexadecimal number 9 15 (F)
∴(237)8=(9F)16
f. (321)8 = (?)16
Answer:
Octal number 3 2 1
3 bits binary number 011 010 001
∵ (321)8 = (1101 0001)2
Again,
Binary numbers in 4 bits 1101 0001
Hexadecimal number 13 (D) 1
∴(321)8=(D1)16
13. Convert the following hexa-decimal numbers to octal numbers.
a. (45)16 = (?)8
Answer:
Hexadecimal number 4 5
Binary number in 4 bits 0100 0101
∵ (45)16 = (001 000 101)2
Again,
Binary numbers in 3 bits 001 000 101
Octal number 1 0 5
∴(45)16=(105)8
b. (245)16 = (?)8
Answer:
Hexadecimal number 2 4 5
Binary number in 4 bits 0010 0100 0101
∵ (245)16 = (001 001 000 101)2
Again,
Binary numbers in 3 bits 001 001 000 101
Octal number 1 1 0 5
∴(245)16=(1105)8
c. (1259)16 = (?)8
Answer:
Hexadecimal number 1 2 5 9
Binary number in 4 bits 0001 0010 0101 1001
∵ (1259)16 = (001 001 001 011 001)2
Again,
Binary numbers in 3 bits 001 001 001 011 001
Octal number 1 1 1 3 1
∴(1259)16=(11131)8
d. (A16)16 = (?)8
Answer:
Hexadecimal number A 1 6
Binary number in 4 bits 1010 0001 0110
∵ (A16)16 = (101 000 010 110)2
Again,
Binary numbers in 3 bits 101 000 010 110
Octal number 5 0 2 6
∴(A16)16=(5026)8
e. (5B)16 = (?)8
Answer:
Hexadecimal number 5 B
Binary number in 4 bits 0101 1011
∵ (5B)16 = (001 011 011)2
Again,
Binary numbers in 3 bits 001 011 011
Octal number 1 3 3
∴(5B)16=(133)8
f. (F20)16 = (?)8
Answer:
Hexadecimal number F 2 0
Binary number in 4 bits 1111 0010 0000
∵ (F20)16 = (111 100 100 000)2
Again,
Binary numbers in 3 bits 111 100 100 000
Octal number 7 4 4 0
∴(F20)16=(7440)8
End of Chapter-6