Boolean Algebra MCQ Quiz - Objective Question with Answer for Boolean Algebra - Download Free PDF
Last updated on Jun 6, 2025
Latest Boolean Algebra MCQ Objective Questions
Boolean Algebra Question 1:
What is the result of applying the complement law A + A' in Boolean algebra?
Answer (Detailed Solution Below)
Boolean Algebra Question 1 Detailed Solution
Explanation:
To understand why the complement law holds, we need to analyze the truth table for the expression A + A'. A truth table lists all possible values of the variables involved and the resulting value of the expression for each combination of variable values. In this case, we are dealing with a single variable A, which can either be 0 or 1.
A | A' | A + A' |
---|---|---|
0 | 1 | 1 |
1 | 0 | 1 |
Name |
AND Form |
OR Form |
Identity law |
1.A=A |
0+A=A |
Null Law |
0.A=0 |
1+A=1 |
Idempotent Law |
A.A=A |
A+A=A |
Inverse Law |
AA’=0 |
A+A’=1 |
Commutative Law |
AB=BA |
A+B=B+A |
Associative Law |
(AB)C |
(A+B)+C = A+(B+C) |
Distributive Law |
A+BC=(A+B)(A+C) |
A(B+C)=AB+AC |
Absorption Law |
A(A+B)=A |
A+AB=A |
De Morgan’s Law |
(AB)’=A’+B’ |
(A+B)’=A’B’ |
Boolean Algebra Question 2:
Which of the following is the complement of the boolean function.
A'B + CD' + A'B + CD'
Answer (Detailed Solution Below)
Boolean Algebra Question 2 Detailed Solution
The correct answer is Option 3) (A + B')(C' + D).
Key Points
Given Boolean function: A'B + CD' + A'B + CD'
- First simplify the given expression:
- A'B + A'B = A'B (redundant term)
- CD' + CD' = CD' (redundant term)
- So, the simplified function = A'B + CD'
Now, we are asked to find the complement of this function:
Let F = A'B + CD'
Then the complement is: F' = (A'B + CD')'
Apply De Morgan’s Law:
- (A'B + CD')' = (A'B)' · (CD')'
- (A'B)' = A + B'
- (CD')' = C' + D
Therefore, F' = (A + B')(C' + D)
Hence, the correct answer is: Option 3) (A + B')(C' + D)
Boolean Algebra Question 3:
Let A' represents complement of A. Which of the following boolean expressions is/are true?
(A). A + AB = A
(B). (A + B)' = A'B'
(C). (A')' = A
(D). (AB)' = A' + B'
Choose the correct answer from the options given below:
Answer (Detailed Solution Below)
Boolean Algebra Question 3 Detailed Solution
The correct answer is: Option 3.
Important Points
Let's evaluate each Boolean expression to determine which ones are true:
(A). A + AB = A This expression simplifies using the Distributive Law and the Absorption Law: A + AB = A (1 + B) A (1 + B) = A (Since 1 + B = 1) Therefore, this expression is true.
(B).(A + B)' = A'B' This expression is based on De Morgan's Law: (A + B)' = A'B' Therefore, this expression is true.
(C).(A')' = A This expression is based on the Double Negation Law: (A')' = A Therefore, this expression is true.
(D).(AB)' = A' + B' This expression is also based on De Morgan's Law: (AB)' = A' + B' Therefore, this expression is true.
All the expressions (A), (B), (C), and (D) are true, So the correct answer is option 3) (A), (B), (C) and (D).
Boolean Algebra Question 4:
De Morgan’s law states
Answer (Detailed Solution Below)
Boolean Algebra Question 4 Detailed Solution
The correct answer is Both.
Key Points
- De Morgan's laws are two transformation rules that are both valid rules of inference.
- The first law states that the negation of a conjunction is the disjunction of the negations:
- The second law states that the negation of a disjunction is the conjunction of the negations:
- Both these laws are essential in the field of Boolean algebra and digital logic design.
Additional Information
Commutative Law
- Commutative law states that changing the sequence of the variables does not have any effect on the output of a logic circuit.
- A + B = B + A
Associative Law
- It states that the order in which the logic operations are performed is irrelevant as their effect is the same.
- (A + B) + C = A + (B + C)
Distributive Law
- A + BC = (A + B)(A + C)
Boolean Algebra Question 5:
Match the following:
List I | List II |
A. Identity Law | I. A (A + B) = A, A + A ∙ B = A |
B. Distributive Law | II. 1 ∙ A = A, 0 + A = A |
C. Absorption Law | III. (A ∙ B)’ = A’ + B’, (A + B)’ = A’ ∙ B’ |
D. De Morgan Law | IV. (A ∙ B) C = A (B ∙ C), (A + B) + C = A + (B + C) |
Answer (Detailed Solution Below)
Boolean Algebra Question 5 Detailed Solution
The correct answer is A - II, B - IV, C - I, D - III
Key Points
Name of Law |
AND Law |
OR Law |
Identity Law |
1 ∙ A = A |
0 + A = A |
Null Law |
0 ∙ A = 0 |
1 + A = 1 |
Inverse Law |
A ∙ A = A |
A + A = A |
Idempotent Law |
A ∙ A’ = 0 |
A + A’ = 1 |
Associative Law |
A ∙ B = B ∙ A |
A + B = B + A |
Distributive Law |
(A ∙ B) C = A (B ∙ C) |
(A + B) + C = A + (B + C) |
Absorption Law |
A (A + B) = A |
A + A ∙ B = A |
De Morgan Law |
(A ∙ B)’ = A’ + B’ |
(A + B)’ = A’ ∙ B’ |
Top Boolean Algebra MCQ Objective Questions
Boolean algebra obeys
Answer (Detailed Solution Below)
Boolean Algebra Question 6 Detailed Solution
Download Solution PDF
Name |
AND Form |
OR Form |
Identity law |
1.A = A |
0 + A = A |
Null Law |
0.A = 0 |
1 + A = 1 |
Idempotent Law |
A.A = A |
A + A = A |
Inverse Law |
AA’ = 0 |
A + A’ = 1 |
Commutative Law |
AB = BA |
A + B = B + A |
Associative Law |
(AB)C |
(A + B) + C = A + (B + C) |
Distributive Law |
A + BC = (A + B)(A + C) |
A(B + C) = AB + AC |
Absorption Law |
A(A + B) = A |
A + AB = A |
De Morgan’s Law |
(AB)’ = A’ + B’ |
(A + B)’ = A’B’ |
Considering X and Y as binary variables, the Boolean expression X + Y + 1 is equivalent to
Answer (Detailed Solution Below)
Boolean Algebra Question 7 Detailed Solution
Download Solution PDFFROM laws of Boolean algebra
1 + any variable = 1
Name |
AND Form |
OR Form |
Identity law |
1.A=A |
0+A=A |
Null Law |
0.A=0 |
1+A=1 |
Idempotent Law |
A.A=A |
A+A=A |
Inverse Law |
AA’=0 |
A+A’=1 |
Commutative Law |
AB=BA |
A+B=B+A |
Associative Law |
(AB)C |
(A+B)+C = A+(B+C) |
Distributive Law |
A+BC=(A+B)(A+C) |
A(B+C)=AB+AC |
Absorption Law |
A(A+B)=A |
A+AB=A |
De Morgan’s Law |
(AB)’=A’+B’ |
(A+B)’=A’B’ |
Determine the minimised expression of Boolean function
F = X̅ Z̅ + Y̅ Z̅ + Y Z̅ + XYZ
Answer (Detailed Solution Below)
Boolean Algebra Question 8 Detailed Solution
Download Solution PDFConcept:
All Boolean algebra laws are shown below
Name |
AND Form |
OR Form |
Identity law |
1.A = A |
0 + A = A |
Null Law |
0.A = 0 |
1 + A = 1 |
Idempotent Law |
A. A = A |
A + A = A |
Inverse Law |
AA’ = 0 |
A + A’ = 1 |
Commutative Law |
AB = BA |
A + B = B + A |
Associative Law |
(AB)C |
(A + B) + C = A + (B + C) |
Distributive Law |
A + BC = (A + B) (A + C) |
A (B + C) = AB + AC |
Absorption Law |
A (A + B) = A |
A + AB = A |
De Morgan’s Law |
(AB)’ = A’ + B’ |
(A + B)’ = A’B’ |
Calculation:
F = X̅ Z̅ + Y̅ Z̅ + Y Z̅ + XYZ
= X̅ Z̅ + Z̅ (Y̅ + Y) + XYZ
= X̅ Z̅ + Z̅ + XYZ
= Z̅ (1 + X̅) + XYZ
= Z̅ + XYZ
Now using Distributive Law
= (Z̅ + Z)(Z̅ + XY)
= Z̅ + XY
A ’literal’ in Boolean Algebra means
Answer (Detailed Solution Below)
Boolean Algebra Question 9 Detailed Solution
Download Solution PDF- A variable is a symbol that may take on the value 0 or 1.
- A literal is the use of a variable or its complement in an expression.
- A term is an expression formed by literals and operations at one level.
For example, the following function:
F1 = xy + xy'z + x'yz
Has 3 variables (x,y,z),
8 literals (x,y,x,y',z,x',y,z), and
4 terms (xy, xy'z, x'yz, and the OR term that combines the first level AND terms).
Find the boolean expression for the logic circuit shown below:
(1-NAND gate, 2-NOR gate, 3-NOR gate)
Answer (Detailed Solution Below)
Boolean Algebra Question 10 Detailed Solution
Download Solution PDFDe Morgan’s’ Law
Let P be the output of 1- NAND gate
Let Q be the output of 1- NAND gate
Y = (A.B). (A̅ + B)
Y = A.B.A̅ + A.B.B
∵ A.A̅ = 0 and B.B = B
Y = AB
The following expression may be simplified as (AB + C + DC)(AC + BC + D)
Answer (Detailed Solution Below)
Boolean Algebra Question 11 Detailed Solution
Download Solution PDF(AB + C + DC)(AC + BC + D)
= (AB + C[1+ D])(AC + BC + D)
= (AB + C) (AC + BC + D)
= ABC + ABC + ABD + AC + BC + CD
= ABC + ABD + AC + BC + CD
= AC (1 + B) + ABD + BC + CD
= AC + BC + CD + ABD
Solve the following Boolean expression:
Select the correct option.
Answer (Detailed Solution Below)
Boolean Algebra Question 12 Detailed Solution
Download Solution PDFGiven Y = A (A̅ + C) (A̅B + C) (A̅BC + C̅)
This can be written as:
Y = (A.A̅ + AC) (A̅B + C) (A̅BC + C̅)
Since A.A̅ = 0, the above expression can be written as:
Y = (AC) (A̅B + C) (A̅BC + C̅)
Y = (AC.A̅B + AC.C) (A̅BC + C̅)
With C.C = C, we can write:
Y = (AC) (A̅BC + C̅)
Y = AC.A̅BC + AC.C̅
Y = 0 + 0
Y = 0
Name |
AND Form |
OR Form |
Identity law |
1.A=A |
0+A=A |
Null Law |
0.A=0 |
1+A=1 |
Idempotent Law |
A.A=A |
A+A=A |
Inverse Law |
AA’=0 |
A+A’=1 |
Commutative Law |
AB=BA |
A+B=B+A |
Associative Law |
(AB)C |
(A+B)+C = A+(B+C) |
Distributive Law |
A+BC=(A+B)(A+C) |
A(B+C)=AB+AC |
Absorption Law |
A(A+B)=A |
A+AB=A |
De Morgan’s Law |
(AB)’=A’+B’ |
(A+B)’=A’B’ |
Let # be a binary operator defined as
X # Y = X’ + Y’ where X and Y are Boolean variables.
Consider the following two statements.
(S1) (P # Q) #R = P# (Q # R)
(S2) Q # R = R # Q
Which of the following is/are true for the Boolean variables P, Q and R?
Answer (Detailed Solution Below)
Boolean Algebra Question 13 Detailed Solution
Download Solution PDF∴ S1 is FALSE
But
∴ S2 is TRUE
Important Points
𝑋
It is a NAND operation and NAND is commutative but not associative
According to Boolean law:
Answer (Detailed Solution Below)
Boolean Algebra Question 14 Detailed Solution
Download Solution PDFConcept-
- The branch of algebra which deals with the values of the variables in the form of the truth values true and false is called Boolean algebra.
- True and false are usually denoted by 1 and 0 respectively.
Explanation-
If we take complement, we get negation of the variable but if we again take the complement of complemented variable, we get the same variable.
Consider the Boolean function z(a, b, c).
Which one of the following minterm lists represents the circuit given above?
Answer (Detailed Solution Below)
Boolean Algebra Question 15 Detailed Solution
Download Solution PDFThe given circuit gives the output:
Z(a, b, c) =
Expanding it into canonical form to obtain the minterms
Z(a, b, c) =
=
After rearranging the canonical terms, this corresponds to min-terms: ∑ (1,4, 5, 6, 7)
Alternate solution:
The output of the circuit is
K Map for this Boolean expression
The above K Map corresponds to min-terms: ∑ (1,4, 5, 6, 7)