**Digital Circuits Questions and Answers – 1’s, 2’s, 9’s & 10’s Complements – 1**

This set of Digital Electronics/Circuits Multiple Choice Questions & Answers (MCQs) focuses on “1’s, 2’s, 9’s & 10’s Complements – 1”.

**1. 1’s complement of 1011101 is ____________**

a) 0101110

b) 1001101

c) 0100010

d) 1100101

**Answer: c**

**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

**Answer: c**

**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

**Answer: d**

**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.

1 1 1 Minuend - 1 1 1 1 0 1’s complement of subtrahend - 1 0 1 0 1 ____________ Carry over - 1 1 0 0 1 1 1 _____________ 1 0 1 0 0

**4. On subtracting (010110)2 from (1011001)2 using 2’s complement, we get ____________**

a) 0111001

b) 1100101

c) 0110110

d) 1000011

**Answer: d**

**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.

1’s complement of subtrahend - 1 1 0 1 0 0 1 _________________ 1 1 1 Minuend - 1 0 1 1 0 0 1 2’s complement of subtrahend - 1 1 0 1 0 1 0 _________________ Carry over - 1 1 0 0 0 0 1 1 Answer: 1000011

**5. On subtracting (001100)2 from (101001)2 using 2’s complement, we get ____________**

a) 1101100

b) 011101

c) 11010101

d) 11010111

**Answer: b**

**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.

1’s complement of subtrahend - 1 1 0 0 1 1 _________________ Minuend - 1 0 1 0 0 1 2’s complement of subtrahend - 1 1 0 1 0 0 _________________ Carry over - 1 0 1 1 1 0 1 Answer: 011101

**6. On addition of 28 and 18 using 2’s complement, we get ____________**

a) 00101110

b) 0101110

c) 00101111

d) 1001111

**Answer: b**

**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.

Augend - 0 0 1 1 1 0 0 Addend - 0 0 1 0 0 1 0 _________________ 0 1 0 1 1 1 0 Answer: 0 1 0 1 1 1 0

**7. On addition of +38 and -20 using 2’s complement, we get ____________**

a) 11110001

b) 100001110

c) 010010

d) 110101011

**Answer: c**

**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.

Augend - 0 1 0 0 1 1 0 2’s Complement of Subtrahend: 1 1 0 1 1 0 0 _________________ 1 0 0 1 0 0 1 0 Answer: 0 1 0 0 1 0

**8. On addition of -46 and +28 using 2’s complement, we get ____________**

a) -10010

b) -00101

c) 01011

d) 0100101

**Answer: a**

**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.

Augend - 0 0 1 1 1 0 0 .....(a) 2’s Complement of Subtrahend: 1 0 1 0 0 1 0 .....(b) _________________ Addiing (a) and (b): 1 1 0 1 1 1 0 Since, there is no carry, so answer will be negative and 2's complement of the above result is determined. 0 0 1 0 0 0 1 + 1 _________________ 0 0 1 0 0 1 0 Answer: - 1 0 0 1 0

**9. On addition of -33 and -40 using 2’s complement, we get ____________**

a) 1001110

b) -110101

c) 0110001

d) -1001001

**Answer: d**

**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.

Augend - 1 0 1 0 0 0 0 1 .....(a) 2’s Complement of Subtrahend: 1 1 0 1 1 0 0 1 .....(b) ______________________ Addiing (a) and (b): 1 0 1 0 0 1 0 0 0 Since, there is no carry, so answer will be negative and 2's complement of the above result is determined. 1 0 0 1 0 0 0 + 1 _________________ 1 0 0 1 0 0 1 Answer: -1001001

**10. On subtracting +28 from +29 using 2’s complement, we get ____________**

a) 11111010

b) 111111001

c) 100001

d) 1

**Answer: d**

**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.

1’s complement of subtrahend - 1 0 0 0 1 1 Minuend - 0 1 1 1 0 1 2’s complement of subtrahend - 1 0 0 1 0 0 ____________________ Carry over - 1 0 0 0 0 0 1 Answer: 000001 = 1

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