Boolean Algebra MCQ Quiz in తెలుగు - Objective Question with Answer for Boolean Algebra - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Mar 14, 2025
Latest Boolean Algebra MCQ Objective Questions
Top Boolean Algebra MCQ Objective Questions
Boolean Algebra Question 1:
If r ∈ {p, q, ~p, ~q} such that the Boolean expression (~(p ∨ q)) ∨ (∼p ∧ q) is equivalent to r, then r is equal to
Answer (Detailed Solution Below)
Boolean Algebra Question 1 Detailed Solution
Concept:
~ ( p ∨ q) = ∼ p ∧ ∼q
Calculation:
Given:
The Boolean expression
~ ( p ∨ q) ∨ (∼ p ∧ q) = r
Taking LHS
= (∼ p ∧ ∼ q) ∨ (∼p ∧ q)
= (∼ p ∧∼ q) ∨ (∼p ∧ q)
Taking ∼p common
= (∼ p) ∧ (∼q ∨ q)
r = ∼ p
So r is equal to ∼p.
Boolean Algebra Question 2:
Consider following gates :
1. NAND
2. NOR
3. XOR
Out of these gates, the universal gates are
Answer (Detailed Solution Below)
Boolean Algebra Question 2 Detailed Solution
A Universal Gate is a gate by which every other gate can be realized.
AND, OR, NOT, etc. are basic gates.
NAND, NOR, etc. are the universal gate.
Example:
NOT, AND and OR gate realization using NAND gate is as shown:
Boolean Algebra Question 3:
The Boolean function a + (a̅ b) is equivalent to
Answer (Detailed Solution Below)
Boolean Algebra Question 3 Detailed Solution
Calculation:
y = a + a̅ b
⇒ y = (a + a̅) (a + b)
Because, we have,
y = aa + ab + a̅ a + a̅ b
y = a + ab + a̅ b
y = a (1 + b) + a̅ b
y = a + a̅ b
Therefore, now consider,
y = (a + a̅) (a + b)
y = (1) (a + b)
∴ y = a + b
Boolean Algebra Question 4:
Consider the Boolean function f = (a + bc)⋅(pq + r). Complement f' of function f is:
Answer (Detailed Solution Below)
Boolean Algebra Question 4 Detailed Solution
By using Demorgan's law:
(A + B)' = A'.B' and (A.B)' = A' + B'
Boolean function f = (a + bc)⋅(pq + r).
f = (a + bc)⋅(pq + r)
f = ((a + bc)⋅(pq + r))'
f = (a + bc)' + (pq + r)'
f = a'.(bc)' + (pq)'.r'
f = a'.(b' + c') + (p' + q').r'
(a + bc). (pq + r) = a'(b' + c') + (p' + q').r'
Therefore option 2 is correct.
Boolean Algebra Question 5:
If f(x) = \(\frac{{3{\rm{x\;}} + {\rm{\;}}2}}{{4{\rm{x\;}}-{\rm{\;}}1}}\) and g(x) = \(\frac{{{\rm{x\;}} + {\rm{\;}}2}}{{4{\rm{x}} - {\rm{\;}}3}}\) then
I. fog(x) = 2x
II. gof(x) = x
Answer (Detailed Solution Below)
Boolean Algebra Question 5 Detailed Solution
fog(x) = f(g(x)) = \(\frac{{3\frac{{x\; + \;2}}{{4x - 3}}{\rm{\;}} + {\rm{\;}}2}}{{4\frac{{x + \;2}}{{4x\; - 3}}{\rm{\;}}-{\rm{\;}}1}}\) = \(\frac{{11x}}{{11}}\) = x
gof(x) = g(f(x)) = \(\frac{{\frac{{3x\; + \;2}}{{4x - 1}}{\rm{\;}} + {\rm{\;}}2}}{{4\frac{{3x + 2}}{{4x\; - \;1}} - {\rm{\;}}3}}\) = \(\frac{{11x}}{{11}}\) = xBoolean Algebra Question 6:
Let ⊕ and ⊙ denote the Exclusive OR and Exclusive NOR operations, respectively.
Which one of the following is NOT CORRECT?
Answer (Detailed Solution Below)
Boolean Algebra Question 6 Detailed Solution
The correct answer is option 4
Calculation:
LHS = (P ⊕ P̅) ⊕ Q=1 ⊕ Q=Q̅
RHS = (P ⊙ P̅) ⊙ Q̅= 0 ⊙ Q̅ =Q
Hence LHS is not equal to RHS in option 4
Boolean Algebra Question 7:
Comprehension:
Next five questions are based on the following passage.
Consider a domain consisting of three Boolean variables Toothache, Cavity, and Catch. The full joint distribution is a 2 × 2 × 2 table as shown in the figure below.
| toothache | ￢toothache | |||
| catch | ￢catch | catch | ￢catch | |
| cavity | 0.108 | 0.012 | 0.072 | 0.008 |
| ￢cavity | 0.016 | 0.064 | 0.144 | 0.576 |
The marginal probability of cavity P(cavity) is ________.
Answer (Detailed Solution Below)
Boolean Algebra Question 7 Detailed Solution
The correct answer is option 1.
Solution :
P(cavity )= 0.108+0.012+0.072+0.08 = 0.200
Note : P(cavity) means the probability value where cavity is found irrespective of any other value.
Boolean Algebra Question 8:
Which of the following Boolean rules is correct?
Answer (Detailed Solution Below)
Boolean Algebra Question 8 Detailed Solution
Concept:
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’ |
Boolean Algebra Question 9:
Boolean expression y.z + z is equal to which of the following?
Answer (Detailed Solution Below)
Boolean Algebra Question 9 Detailed Solution
y.z + z = z.(y+1) =z.1 / using Annulment Law
So option 3 will be correct
Boolean Algebra Question 10:
Which of the following is not a valid expression of Boolean Algebra?
Answer (Detailed Solution Below)
Boolean Algebra Question 10 Detailed Solution
Option 1 : x + x.y = x
LHS = x + x.y = x(1 + y) = x.1 = x / Boolen Expression law
Option 2 : x'.y+x.y=y
LHS =x'.y+x.y = y(x'+x)=y(1)=y / Boolen Expression law
Option 3 :x'.y'+x.y=x
XNOR(x, y) = x ⊙ y = x.y + x'.y' / Exclusive NOR Rule
Option 4 : (x + y + z)' = x'.y'.z'
(x + y + z)' = x'.y'.z' / Folloe Demorgan's Law
So the correct Answer is Option 3