Bilal maqbool assignment no.01 computing

1. Submitted.To.
Prepared By. Bilal Maqbool.
Department SE-I
Roll no. 10
University University of Azad Jammu
and Kashmir.
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2. Q No1. Convert Hexadecimal no (68BE)16 to Binary and from Binary to Octal?
6 8 B E
0110 1000 1011 1110
(110100010111110)2
By taking the three pair numbers
110 100 010 111
1×2²+1×2¹+0×20 1×2²+0×2¹+0×20
0×2²+1×2¹+0×20
1×2²+1×2¹+1×20
4+2+0 =6 4+0+0 =4 0+2+0=2 4+2+1=7
110
1×2²+1×2¹+0×20
4+2+0=6
(64276)8
Q No2. Convert Decimal no (345)10 to Binary into two ways?
(a) Directly to Binary
(101011001)2
(b) Decimal---Hexadecimal---Binary ( 345)10
(159)16
Converting each number separately into binary in three pair numbers.
1 5 9
0001 0101 1001
(101011001)2
Q No3.Hexadecimal to Binary conversion. Direct and indirect methods.
(a) (0A2B)16
Direct method:
Converting in binary in four pair
0 A 2 B
0000 1010 0010 1011
(101000101011)2
Indirect method:
Converting to Decimal
2 345
2 172-1
2 86-0
2 43-0
2 21-1
2 10-1
2 5-0
2 2-1
1-0
16 345
16 21-9
1-5
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3. ( 0A2B)16
0×16³+10×16²+2×16¹+11×160
0+2560+32+11
(2603)10
(101000101011)2
(b) (1E3)16
Direct method:
1 E 3
0001 1110 0011
(111100011)2
Indirect method:
Converting into Decimal
(1E3)16
1×16²+E×16¹+3×160
256+224+3
(483)10
(483)10 to Binary
(111100011)2
Q no4. Binary to Hexadecimal conversion (Direct and Indirect method)
(a) (111001)2
Direct Method:
The four pair number of (111001)2
0011 1001
2 2603
2 1301-1
2 650-1
2 325-0
2 162-1
2 81-0
2 40-1
2 20-0
2 10-0
2 5-0
2 2-1
1-0
2 483
2 241-1
2 120-1
2 60-0
2 30-0
2 15-0
2 7-1
2 3-1
1-1
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4. 0×2³+0×2²+1×2¹+1×20 1×2³+0×2²+0×2¹+1×20
0+0+2+1 8+0+0+1
3 9
(39)16
Indirect method:
Converting to Decimal
(111001)2
1 ×25
+1 ×24
+1×2³+0×2²+0×2¹+1×20
32+16+8+0+0+1
57
Converting (57)10 to Hexadecimal
Hexadecimal to Decimal:
(39)16
(b) (1110)2
Direct method:
Binary to hexadecimal
1110
1×2³+1×2²+1×2¹+0×20
8+4+2+0
14
=(E)16
Indirect method:
Converting to Decimal
(1110)2
1×2³+1×2²+1×2¹+0×20
8+4+2+0
(14)16
Q No5. Convert the following Binary number to Decimal, Hexadecimal and Octal then convert Octal and
Hexadecimal back to the Decimal number.
(1001001)2
Binary to Decimal:
(1001001)2
1×26
+0 ×25
+0 ×24+1×2³+0×2²+0×2¹+1×20
16 57
3-9
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5. 64+0+0+8+0+0+1
73
(73)10
Binary to Hexadecimal:
(1001001)2
Four pair number of (1001001)2
0100 1001
0×2³+1×2²+0×2¹+0×20
1×2³+0×2²+0×2¹+1×20
0+4+0+0 8+0+0+1
4 9
By combining (49)16
Binary to Octal:
(1001001)2
Three number pair
001 001 001
0×2²+0×2¹+1×20
0×2²+0×2¹+1×20
0×2²+0×2¹+1×20
1 1 1
= (111)8
Hexadecimal to Decimal:
= (49)16
= 1×8²+1×8¹+1×80
=64+8+1
=(73)10
Octal to Decimal:
(111)8
=4×16¹+9×16
=64+9
=(73)10
Q No6.Subtraction problem using their respective compliment’s method.
(a) (5434)8 – (3556)8
(b) 7’s compliment.
7777
- 3556
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6. 4221
8’scompliment
4221
+ 1
4222
5434
+4222
(1656)8
(b) (100100)2 – (11101)2
1’s compliment
111111
-011101
100010
2’s compliment.
100010
+ 1
100011
100100
+ 100011
1000111
(111)2
(c) (86A1)16 – (8B12)16
15’s compliment
15 15 15 15
- 8 B 1 2
7 4 E D
16’s complement
7 4 E D
+ 1
7 4 E E
8 6 A 1
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7. +7 4 E E
1 5 B 8 F
(5B8F)16