1. 1’s complement of 1011101 is ____________
a) 0101110
b) 1001101
c) 0100010
d) 1100101
Explanation: 1’s complement of a binary number is obtained by reversing the binary bits. All the 1’s to 0’s and 0’s to 1’s. Thus, 1’s complement of 1011101 = 0100010.
2. 2’s complement of 11001011 is ____________
a) 01010111
b) 11010100
c) 00110101
d) 11100010
Explanation: 2’s complement of a binary number is obtained by finding the 1’s complement of the number and then adding 1 to it.
2’s complement of 11001011 = 00110100 + 1 = 00110101.
3. On subtracting (01010)2 from (11110)2 using 1’s complement, we get ____________
a) 01001
b) 11010
c) 10101
d) 10100
Explanation: Steps For Subtraction using 1’s complement are:
-> 1’s complement of the subtrahend is determined and added to the minuend.
-> If the result has a carry, then it is dropped and 1 is added to the last bit of the result.
-> Else, if there is no carry, then 1’s complement of the result is found out and a ‘-’ sign preceeds the result.
4. On subtracting (010110)2 from (1011001)2 using 2’s complement, we get ____________
a) 0111001
b) 1100101
c) 0110110
d) 1000011
Explanation: Steps For Subtraction using 2’s complement are:
-> 2’s complement of the subtrahend is determined and added to the minuend.
-> If the result has a carry, then it is dropped and the result is positive.
-> Else, if there is no carry, then 2’s complement of the result is found out and a ‘-’ sign preceeds the result.
5. On subtracting (001100)2 from (101001)2 using 2’s complement, we get ____________
a) 1101100
b) 011101
c) 11010101
d) 11010111
Explanation: Steps For Subtraction using 2’s complement are:
-> 2’s complement of the subtrahend is determined and added to the minuend.
-> If the result has a carry, then it is dropped and the result is positive.
-> Else, if there is no carry, then 2’s complement of the result is found out and a ‘-’ sign preceeds the result.
6. On addition of 28 and 18 using 2’s complement, we get ____________
a) 00101110
b) 0101110
c) 00101111
d) 1001111
Explanation: Steps for Binary Addition Using 2’s complement:
-> The binary equivalent of the two numbers are obtained and added using the rules of binary addition.
7. On addition of +38 and -20 using 2’s complement, we get ____________
a) 11110001
b) 100001110
c) 010010
d) 110101011
Explanation: Steps for Binary Addition Using 2’s complement:
-> The 2’s complement of the addend is found out and added to the first number.
-> The result is the 2’s complement of the sum obtained.
8. On addition of -46 and +28 using 2’s complement, we get ____________
a) -10010
b) -00101
c) 01011
d) 0100101
Explanation: The BCD form is written of the two given numbers, in their signed form. After which, normal binary addition is performed.
Augend is 28 and Subtrahend is -46.
9. On addition of -33 and -40 using 2’s complement, we get ____________
a) 1001110
b) -110101
c) 0110001
d) -1001001
Explanation: The BCD form is written of the two given numbers, in their signed form. After which, normal binary addition is performed.
Augend is -40 and Subtrahend is -33.
10. On subtracting +28 from +29 using 2’s complement, we get ____________
a) 11111010
b) 111111001
c) 100001
d) 1
Explanation: Steps For Subtraction using 2’s complement are:
-> 2’s complement of the subtrahend is determined and added to the minuend.
-> If the result has a carry, then it is dropped and the result is positive.
-> Else, if there is no carry, then 2’s complement of the result is found out and a ‘-’ sign preceeds the result.