Page 48 - ComputerScience_Class_11

Converting 110 000 011 101.101 100 to octal
110 000 011 101 . 101 100 4(2 ) 2(2 ) 1(1 ) Octal
0
1
2
6 0 3 5 . 5 4 1 1 0 1 × 4 + 1 × 2 + 0 × 1 = 6
0 0 0 0 × 4 + 0 × 2 + 0 × 1 = 0
0 1 1 0 × 4 + 1 × 2 + 1 × 1 = 3
1 0 1 1 × 4 + 0 × 2 + 1 × 1 = 5
1 0 0 1 × 4 + 0 × 2 + 0 × 1 = 4
(6035.54) 8
13. Perform the following Binary addition:
(1001.1) + (110110.11) + (10.1) 2
2
2
Ans. +1 +1 +1 +1 +1 +1 +1
1 0 0 1 . 1
1 1 0 1 1 0 . 1 1
+ 1 0 . 1

1 0 0 0 0 1 0 . 1 1
(1000010.11) 2
14. Perform binary subtraction (100110.01)2 - (10111.1)2 using:
a. Borrow method
b. 1’s Complement method
c. 2’s Complement method

Ans. a. +1 +1 +1 +1 +1 +1
1 0 0 1 1 0 -1 . 0 1
- 1 0 1 1 1 . 1
0 0 1 1 1 0 . 1 1
Discarding leading 0’s
(1110.11) 2

b. 1’s complement of 010111.10 is 101000.01 +1 +1
Adding 100110.01 and 101000.01, we get 1 0 0 1 1 0 . 0 1
Discarding MSB 1 and leading 0’s + 1 0 1 0 0 0 . 0 1
1 0 0 1 1 1 0 . 1 0
1
(1110.11) 2 0 0 1 1 1 0 . 1 1

c. 1’s complement of 010111.10 is 101000.01 +1
2’s complement of 101000.10 is 101000.10 1 0 0 1 1 0 . 0 1
Adding 100110.01 and 101000.10, we get + 1 0 1 0 0 0 . 1 0
Discarding MSB 1 and ignoring leading 0’s
(1110.11) 1 0 0 1 1 1 0 . 1 1
2
15. Perform binary subtraction (1000.01) - (110111.1) using:
2
2
a. 1’s complement method
b. 2’s complement method +1 +1
Ans. a. 1’s complement of 110111.10 is 001000.01 1 0 0 0 . 0 1
Adding 1000.01 and 001000.01, we get + 0 0 1 0 0 0 . 0 1
0 1 0 0 0 0 . 1 0
(-101111.01) 2 1’s complement of 010000.10 is 101111.01

46 Touchpad Computer Science (Ver. 3.0)-XI