Boolean Algebra MCQ Quiz in বাংলা - Objective Question with Answer for Boolean Algebra - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Apr 1, 2025
Latest Boolean Algebra MCQ Objective Questions
Top Boolean Algebra MCQ Objective Questions
Boolean Algebra Question 1:
Which one of the following is NOT a valid identity?
Answer (Detailed Solution Below)
Boolean Algebra Question 1 Detailed Solution
1. (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)
|
x |
y |
z |
(x ⊕ y) ⊕ z |
x ⊕ (y ⊕ z) |
|
0 |
0 |
0 |
0 |
0 |
|
0 |
O |
1 |
1 |
1 |
|
0 |
1 |
0 |
0 |
1 |
|
0 |
1 |
1 |
0 |
0 |
|
1 |
0 |
0 |
1 |
1 |
|
1 |
0 |
1 |
0 |
0 |
|
1 |
1 |
0 |
0 |
0 |
|
1 |
1 |
1 |
1 |
1 |
Therefore exclusive OR is associative and hence (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)
2.
|
x |
y |
z |
(x + y) ⊕ z |
x ⊕ (y + z) |
|
0 |
0 |
0 |
0 |
0 |
|
0 |
O |
1 |
1 |
1 |
|
0 |
1 |
0 |
0 |
1 |
|
0 |
1 |
1 |
0 |
1 |
|
1 |
0 |
0 |
1 |
1 |
|
1 |
0 |
1 |
0 |
0 |
|
1 |
1 |
0 |
1 |
0 |
|
1 |
1 |
1 |
0 |
0 |
(x + y) ⊕ z ≠ x ⊕ (y + z) ∴ is it not a valid identity
3.
|
xy = 0 |
Checking validity |
||
|
X |
y |
x + y |
x ⊕ y |
|
0 |
0 |
0 |
0 |
|
0 |
1 |
1 |
1 |
|
1 |
0 |
1 |
1 |
x + y = x ⊕ y / if xy = 0
4.
(xy + x'y')'
= (x’ + y’).(x+y) / Demorgan’s Law
= x’y +xy’
= x ⊕ y
Boolean Algebra Question 2:
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 2 Detailed Solution
The 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)
Boolean Algebra Question 3:
If function f(A, B) = ∑ m(0, 1, 2, 3) is implemented using SOP form, the resultant Boolean function would be:
Answer (Detailed Solution Below)
Boolean Algebra Question 3 Detailed Solution
Laws of Boolean Algebra:
|
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’ |
Application:
f(A, B) = ∑ m(0, 1, 2, 3)
= A̅ B̅ + A̅ B + A B̅ + AB
= A̅ ( B + B̅) + A (B̅ + B)
= A̅ + A = 1
Boolean Algebra Question 4:
Find the Boolean function for the shaded region of the following diagram represented.
Answer (Detailed Solution Below)
Boolean Algebra Question 4 Detailed Solution
F(A, B, C) = C + AB’C’
F(A, B, C) = (C + C’)(C + AB’)
F(A, B, C) = (A + C)(B’ + C)
Boolean Algebra Question 5:
|
X |
Y |
F(X,Y) |
|
0 |
0 |
0 |
|
0 |
1 |
0 |
|
1 |
0 |
1 |
|
1 |
1 |
1 |
Answer (Detailed Solution Below)
Boolean Algebra Question 5 Detailed Solution
From truth table:
F(X, Y) = XY̅ + XY
= X[Y̅ + Y]
F(x, y) = X
Boolean Algebra Question 6:
What is the minimum number of NAND gates needed for the below given?
Answer (Detailed Solution Below) 5
Boolean Algebra Question 6 Detailed Solution
XNOR gate with NAND gates:
Hence 5 NAND gates is needed
Tips and Tricks:
The above Boolean function can also be solved using K-map:
Boolean Algebra Question 7:
Minimize the Boolean expression x = ABC + A̅B + ABC̅
Answer (Detailed Solution Below)
Boolean Algebra Question 7 Detailed Solution
X = ABC + A̅B + ABC̅
Rearranging:
X = ABC + ABC̅ + A̅B
X = AB(C + C̅) + A̅B
X = AB + A̅B [C + C̅ = 1]
X = (A + A̅)B
X = 1.B [A + A̅ = 1]
X = B
Boolean Algebra Question 8:
Which of the following is equivalent of the Boolean expression given below?
A + A̅.B + A.B̅
Answer (Detailed Solution Below)
Boolean Algebra Question 8 Detailed Solution
Concept-
- 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
Calculation:
Let F(A, B) = A + A̅ .B + A.B̅
F(A, B) = A (1 + B̅) + A̅.B
F(A, B) = A + A̅.B
F(A, B) = (A + A̅).(A + B)
F(A, B) = A + B
Boolean Algebra Question 9:
Consider W, Y, A and B be the Boolean variable and $ operator defined as A $ B = A̅ + B where Y = W̅ $ A̅ and W = A + B̅. Find Y?
Answer (Detailed Solution Below)
Boolean Algebra Question 9 Detailed Solution
A $ B = A̅ + B (1)
W = A + B̅ (2)
Y = W̅ $ A̅
Y = W + A̅ (From 1)
Y = A + B̅ + A̅ (From 2)
Y = 1 + B̅ = 1
Boolean Algebra Question 10:
The truth table
|
X |
Y |
F(X, Y) |
|
0 |
0 |
0 |
|
0 |
1 |
0 |
|
1 |
0 |
1 |
|
1 |
1 |
1 |
represents the Boolean function
Answer (Detailed Solution Below)
Boolean Algebra Question 10 Detailed Solution
The correct answer is "option 1".
EXPLANATION:
Option 1: TRUE
The truth table X is:
|
X |
Y |
F |
X |
|
0 |
0 |
0 |
0 |
|
0 |
1 |
0 |
0 |
|
1 |
0 |
1 |
1 |
|
1 |
1 |
1 |
1 |
F is equal to X.
Option 2: FALSE
The truth table X+Y is:
|
X |
Y |
F |
X+Y |
|
0 |
0 |
0 |
0 |
|
0 |
1 |
0 |
1 |
|
1 |
0 |
1 |
1 |
|
1 |
1 |
1 |
1 |
F is not equal to X+Y.
Option 3: FALSE
The truth table X’Y + XY' is:
|
X |
Y |
F |
X’ Y + XY' |
|
0 |
0 |
0 |
0 |
|
0 |
1 |
0 |
1 |
|
1 |
0 |
1 |
1 |
|
1 |
1 |
1 |
0 |
F is not equal to X’.Y +X.Y'.
Option 4: FALSE
The truth table Y is :
|
X |
Y |
F |
Y |
|
0 |
0 |
0 |
0 |
|
0 |
1 |
0 |
1 |
|
1 |
0 |
1 |
0 |
|
1 |
1 |
1 |
1 |
F is not equal to Y.
Hence, the correct answer is "option 1".
From truth table, the sum of minterms
F(X, Y) = X.Y̅ + X.Y = X(Y̅ + Y) = X