Adders MCQ Quiz - Objective Question with Answer for Adders - Download Free PDF
Last updated on Jun 20, 2025
Latest Adders MCQ Objective Questions
Adders Question 1:
The minimum number of 2-input NAND gates required to realise sum and carry of a half adder
Answer (Detailed Solution Below)
Adders Question 1 Detailed Solution
Sum expression of a half adder is given by:
Sum = A XOR B = \(\bar A\;B + A\;\bar B\)
The implementation of half adder circuit is shown below:
We can conclude that to generate the sum bit, we only need 4 NAND gates.
Logic Gates |
Min. number of NOR Gate |
Min. number of NAND Gate |
NOT |
1 |
1 |
AND |
3 |
2 |
OR |
2 |
3 |
EX-OR |
5 |
4 |
EXNOR |
4 |
5 |
NAND |
4 |
1 |
NOR |
1 |
4 |
Half-Adder |
5 |
5 |
Half-Subtractor |
5 |
5 |
Full-Adder |
9 |
9 |
Full-Subtractor |
9 |
9 |
Adders Question 2:
A 4-bit parallel adder uses four full adders. What is the maximum number of bits in the sum output, including the carry-out bit?
Answer (Detailed Solution Below)
Adders Question 2 Detailed Solution
Explanation:
4-Bit Parallel Adder and Its Output:
A 4-bit parallel adder is a digital circuit used for adding binary numbers. It consists of four full adders connected in series, where each full adder is responsible for adding corresponding bits from two binary numbers along with a carry from the previous stage. This configuration allows for the addition of two 4-bit binary numbers.
Understanding the Sum Output:
In a parallel adder, the number of bits in the sum output depends on two factors:
- The number of bits in the numbers being added.
- The carry-out bit from the most significant bit (MSB).
When two 4-bit binary numbers are added, the sum can theoretically require 5 bits to represent the result, because:
- A 4-bit binary number has a maximum value of 1111 (in binary), which equals 15 in decimal.
- The sum of two maximum 4-bit binary numbers (15 + 15) equals 30 in decimal, which is represented as 11110 in binary—a 5-bit result.
- The fifth bit arises from the carry-out of the MSB addition.
Detailed Analysis:
The operation of the 4-bit parallel adder can be broken down as follows:
- Input: Two 4-bit binary numbers (A and B) and an initial carry-in (Cin, typically 0).
- Output: A 4-bit sum (S3, S2, S1, S0) and a carry-out (Cout) from the MSB addition.
The sum bits (S0, S1, S2, S3) represent the 4-bit result of the addition, while the carry-out (Cout) represents the fifth bit, indicating an overflow from the MSB. Thus, the total number of bits in the sum output, including the carry-out, is 5.
Why Option 1 is Correct:
Option 1 states that the maximum number of bits in the sum output, including the carry-out bit, is five. This is correct because:
- The addition of two 4-bit numbers can produce a result requiring 5 bits to represent.
- The fifth bit arises from the carry-out generated during the addition of the MSB.
- Therefore, the sum output consists of 4 bits from the addition itself and an additional carry-out bit, making a total of 5 bits.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 2: Three bits.
This option is incorrect because a 4-bit parallel adder must produce at least a 4-bit result for the sum of two 4-bit binary numbers. Including the carry-out bit, the total number of bits in the sum output is 5, not 3.
Option 3: Eight bits.
This option is incorrect because a 4-bit parallel adder is designed for adding two 4-bit numbers, not two 8-bit numbers. The addition of two 4-bit numbers results in a sum output requiring a maximum of 5 bits (4 bits for the sum and 1 bit for the carry-out). An 8-bit sum output would require an entirely different circuit, such as an 8-bit adder.
Option 4: Four bits.
This option is partially correct but does not account for the carry-out bit. While the sum of two 4-bit numbers produces a 4-bit result, the carry-out from the MSB adds an additional bit, making the total number of bits in the sum output 5.
Conclusion:
In a 4-bit parallel adder, the maximum number of bits in the sum output, including the carry-out bit, is five. This result accounts for the 4-bit sum and the additional carry-out bit from the MSB addition. Understanding the operation of parallel adders is essential for analyzing binary addition circuits and their outputs.
Adders Question 3:
What is the Boolean expression for the Carry-out (C) output of a Half Adder?
Answer (Detailed Solution Below)
Adders Question 3 Detailed Solution
A half adder circuit is made up of an AND gate with an XOR gate as shown below:
- A half adder is also known as XOR gate because XOR is applied to both inputs to produce the sum
- Half adder can add only two bits (A and B) and has nothing to do with the carry
- If the input to a half adder has a carry, then it will neglect it and adds only the A and B bits
- That means the binary addition process is not complete and that's why it is called a half adder
Sum (S) = A⊕B, Carry = A.B
INPUTS |
OUTPUTS |
||
A |
B |
Sum |
CARRY |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
Adders Question 4:
Number of NAND gates required to realise a half adder circuit is _________.
Answer (Detailed Solution Below)
Adders Question 4 Detailed Solution
The correct option is 1
Half Adder using NAND Gates:
The half adder can also be designed with the help of NAND gates. NAND gate is considered as a universal gate. A universal gate can be used for designing any digital circuitry. It is always simple and efficient to use the minimum number of gates in the designing process of our circuit. The minimum number of NAND gates required to design a half adder is 5.
- The first NAND gate takes the inputs which are the two 1-bit numbers.
- The resultant NAND-operated inputs will be again given as input to 3- NAND gates along with the original input.
- Out of these 3 NAND gates, 2-NAND gates will generate the output which will be given as input to the NAND gate connected at the end.
- The gate connected at the end will generate the sum bit. Out of the 3 considered NAND gates, the third NAND gate will generate the carry bit.
Adders Question 5:
A full adder adds:
Answer (Detailed Solution Below)
Adders Question 5 Detailed Solution
Concept:
A binary adder is a logic circuit in the form of a half adder and full adder which is used to add the binary digits.
Explanation:
Half adder: it is a logic circuit that performs addition on two binary digits. It produces a sum and carry.
Full adder: It is a logic circuit that takes three inputs to perform addition. Two binary input and one carry-in input of the previous stage is used. It generates sum and carry (C-out). C-in is a carry from a less significant digit and c-out is a carry from the most significant bit.
If we want to add two n-bit binary adders then it requires 1 half adder and n-1 full adder to complete the circuit. So, in the given question to add 4- bit binary numbers requires 1 half adder and 3 full adders.
Top Adders MCQ Objective Questions
A ________ arithmetic circuit adds two binary digits, giving a sum bit and a carry bit.
Answer (Detailed Solution Below)
Adders Question 6 Detailed Solution
Download Solution PDFHalf adder circuit have two inputs and two outputs (sum and carry).
A half adder circuit is made up of an AND gate with an XOR gate as shown below:
- A half adder is also known as XOR gate because XOR is applied to both inputs to produce the sum
- Half adder can add only two bits (A and B) and has nothing to do with the carry
- If the input to a half adder has a carry, then it will neglect it and adds only the A and B bits
- That means the binary addition process is not complete and that's why it is called a half adder
Sum (S) = A⊕B, Carry = A.B
INPUTS |
OUTPUTS |
||
A |
B |
Sum |
CARRY |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
How many number of 2-input NAND gates are required to realise a half adder circuit?
Answer (Detailed Solution Below)
Adders Question 7 Detailed Solution
Download Solution PDFHalf Adder using NAND Gates:
The half adder can also be designed with the help of NAND gates. NAND gate is considered as a universal gate. A universal gate can be used for designing any digital circuitry. It is always simple and efficient to use the minimum number of gates in the designing process of our circuit. The minimum number of NAND gates required to design a half adder is 5.
- The first NAND gate takes the inputs which are the two 1-bit numbers.
- The resultant NAND-operated inputs will be again given as input to 3- NAND gates along with the original input.
- Out of these 3 NAND gates, 2-NAND gates will generate the output which will be given as input to the NAND gate connected at the end.
- The gate connected at the end will generate the sum bit. Out of the 3 considered NAND gates, the third NAND gate will generate the carry bit.
Answer (Detailed Solution Below)
Adders Question 8 Detailed Solution
Download Solution PDFA half adder circuit is made up of an AND gate with an XOR gate as shown below:
- A half adder is also known as XOR gate because XOR is applied to both inputs to produce the sum
- Half adder can add only two bits (A and B) and has nothing to do with the carry
- If the input to a half adder has a carry, then it will neglect it and adds only the A and B bits
- That means the binary addition process is not complete and that's why it is called a half adder
Sum (S) = A⊕B, Carry = A.B
INPUTS |
OUTPUTS |
||
A |
B |
Sum |
CARRY |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
In a half adder, the carry output is high if the inputs are:
Answer (Detailed Solution Below)
Adders Question 9 Detailed Solution
Download Solution PDFThe truth table for half adder is given below:
A |
B |
Sum |
Carry |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
Therefore, the carry is given by AB.
Hence the carry output is high if the both inputs are high i.e. 1,1
If the inputs are P, Q and R, then in the full adder, find the output expression of the sum.
Answer (Detailed Solution Below)
Adders Question 10 Detailed Solution
Download Solution PDFFull Adder:
Full Adder is the adder that adds three inputs and produces two outputs. The first two inputs are P and Q and the third input R is an input carry.
It is the advantage of full adder over half adder that it can take three inputs together and third input can be an input carry.
Full Adder Circuit:
Full Adder Truth Table:
Inputs | Outputs | |||
A | B | C-N | Sum | C - out |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 |
Half adder is a logic circuit that accepts ________ single bit inputs.
Answer (Detailed Solution Below)
Adders Question 11 Detailed Solution
Download Solution PDFHalf adder circuit:
Half adder circuit has 2 inputs and 2 outputs.
A half adder circuit is basically made up of an AND gate and an XOR gate as shown below:
Sum (S) = A⊕B
Carry = A.B
The truth table is as shown:
INPUTS |
OUTPUTS |
||
A |
B |
Sum |
CARRY |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
A full adder circuit has three binary digit inputs (two input bits and one carry input bit) and two binary digit outputs, Sum bit and carry output bit.
A Full adder can be realized using two half adders as shown:
The truth table of a full adder logic is:
A |
B |
C |
Cin |
S |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
The Sum output bit of a full adder is given by:
S = A ⊕ B ⊕ C
The carry output bit of a full adder is given by:
X1 = AB + BC + AC
In full adder, there are
Answer (Detailed Solution Below)
Adders Question 12 Detailed Solution
Download Solution PDFA full adder circuit has three binary digit inputs (two input bits and one carry input bit) and two binary digit outputs, Sum bit and carry output bit.
A Full adder can be realized using two half adders as shown:
A full adder can be implemented using 2 XOR, 2 AND, 1 OR as shown in figure:
The truth table of a full adder logic is:
A |
B |
C |
Cin |
S |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
The Sum output bit of a full adder is given by:
S = A ⊕ B ⊕ C
The carry output bit of a full adder is given by:
X1 = AB + BC + AC
A full adder circuit has
Answer (Detailed Solution Below)
Adders Question 13 Detailed Solution
Download Solution PDFThe basic block diagram for a Full Adder is as shown:
A Full adder can be realized using two half adders as shown:
We, therefore, conclude that a full adder combinational circuit has 3 inputs and 2 outputs.
A full adder can be implemented using 2 XOR, 2 AND, 1 OR as shown in figure:
The truth table of a full adder logic is:
A |
B |
C |
Cin |
S |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
S = A ⊕ B ⊕ C
The Sum output bit of a full adder is given by:
The carry output bit of a full adder is given by:
X1 = AB + BC + AC
In half adder, the total number of inputs and outputs are:
Answer (Detailed Solution Below)
Adders Question 14 Detailed Solution
Download Solution PDFConcept:
A half adder circuit is basically made up of and a AND gate with XOR gate as shown below.
- A half adder is also known as the XOR gate because XOR is applied to both inputs to produce the sum.
- Half adder can add only two bits (A and B) and has nothing to do with the carry.
- Half adder has two inputs (A and B) and two outputs (S and C).
- If the input to a half adder has a carry, then it will neglect it and adds only the A and B bits.
- That means the binary addition process is not complete and that's why it is called a half adder.
AND gate EXOR gate combination is ____
Answer (Detailed Solution Below)
Adders Question 15 Detailed Solution
Download Solution PDFA half adder circuit is basically made up of and a AND gate with an XOR gate as shown below.
- A half adder is also known as XOR gate because XOR is applied to both inputs to produce the sum
- Half adder can add only two bits (A and B) and has nothing to do with the carry
- If the input to a half adder has a carry, then it will neglect it and adds only the A and B bits
- That means the binary addition process is not complete and that's why it is called a half adder
Sum (S) = A⊕B, Carry = A.B
INPUTS |
OUTPUTS |
||
A |
B |
Sum |
CARRY |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
A Full adder can be realized using two half adders as shown:
A full adder can be implemented using 2 XOR, 2 AND, 1 OR as shown in the figure: