Minimization of Boolean Expression MCQ Quiz in मल्याळम - Objective Question with Answer for Minimization of Boolean Expression - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 10, 2025
Latest Minimization of Boolean Expression MCQ Objective Questions
Top Minimization of Boolean Expression MCQ Objective Questions
Minimization of Boolean Expression Question 1:
For the function F, the Karnaugh map is shown in the figure below. The minimal representation of F is
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 1 Detailed Solution
Minimization of Boolean Expression Question 2:
Which Boolean law is represented below 2
P + Q·R = (P+Q)·(P+R)
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 2 Detailed Solution
The correct answer is "option 4".
EXPLANATION:
Option 1: FALSE
De Morgan's law equations are :
(P.Q)' → P' + Q'
(P+Q)' → P' . Q'
Option 2: FALSE
Associative law equations are :
P + (Q + R) → (P + Q) + R
P . (Q . R) → (P . Q) . R
Option 3: FALSE
Commutative law equations are :
P + Q → Q + P
P . Q → Q . P
Option 4: TRUE
Distributive law equation is :
P + Q . R → (P+Q) . (P+R)
Hence, the correct answer is "option 4".
Minimization of Boolean Expression Question 3:
Which of the following is a NOT a correct identity in Boolean Algebra ?
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 3 Detailed Solution
Concept:
Laws of Boolean algebra:
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 |
Minimization of Boolean Expression Question 4:
For an n-variable Boolean function, the maximum number of prime implicants is
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 4 Detailed Solution
In a n variable Boolean function, the maximum number of prime implicant is given by:
\({{\rm{P}}_{{\rm{max}}}} = \frac{{{2^{\rm{n}}}}}{2} = {2^{{\rm{n}} - 1}} \)
Example with n = 4:
Maximum number of prime applicants = 2n-1 = 24 -1 = 8
Minimization of Boolean Expression 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)
Minimization of Boolean Expression 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’ |
Minimization of Boolean Expression Question 6:
What is the dual of the Boolean function x + yz?
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 6 Detailed Solution
Concept:-
Dual of Boolean function:- It is the expression one obtains by interchanging addition and multiplication and interchanging 0's and 1's. The dual of the function F is denoted as Fd.
Explanation:-
To get a dual of any Boolean Expression, replace-
OR with AND i.e. + with .
AND with OR i.e. .with +
Consider the function: F(x, y, z) = x + yz
The dual of this function is-
Fd (x,y,z) = x(y + z).
Additional InformationSelf-Dual Function:- When a function is equal to its dual, it is called a Self-Dual Function.
Minimization of Boolean Expression Question 7:
Solve the following Boolean expression:
\(Y = A\left( {\bar A + C} \right)\left( {\bar AB + C} \right)\left( {\bar ABC + \bar C} \right)\)
Select the correct option.
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 7 Detailed Solution
Given 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’ |
Minimization of Boolean Expression Question 8:
Find the boolean expression for the logic circuit shown below:
(1-NAND gate, 2-NOR gate, 3-NOR gate)
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 8 Detailed Solution
De Morgan’s’ Law
\(\overline {X + Y} = \bar X.\bar Y\)
\(\overline {XY} = \bar X + \bar Y\)
Let P be the output of 1- NAND gate
\({\rm{P}} = \overline {{\rm{A}}.{\rm{B}}}\)
Let Q be the output of 1- NAND gate
\({\rm{Q}} = \overline {\bar A + B} \)
\({\rm{Y}} = \overline {{\rm{P}} + {\rm{\;Q}}} = \bar P\bar Qa\)
\({\rm{Y}} = {\rm{\;}}\overline {\overline {{\rm{A}}.{\rm{B}}} \;} .\;\overline {\overline {\bar A + B} } \;\;\)
Y = (A.B). (A̅ + B)
Y = A.B.A̅ + A.B.B
∵ A.A̅ = 0 and B.B = B
Y = AB
Minimization of Boolean Expression Question 9:
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)
Minimization of Boolean Expression Question 9 Detailed Solution
\(\left( {P \ne Q} \right) \ne R = \overline {\left( {\bar P + \bar Q} \right)} + \bar R = PQ + \bar R\)
\(\\ P \ne \left( {Q \ne R} \right) = \bar P + \overline {\left( {\bar Q + \bar R} \right)} = \bar P + Q\bar R \)
∴ S1 is FALSE
But
\(Q \ne R = \bar Q + \bar R = \bar R + \bar Q = R \ne Q\)
∴ S2 is TRUE
Important Points
𝑋 \(\ne\) 𝑌 = 𝑋′ + 𝑌′ = (𝑋.𝑌)'
It is a NAND operation and NAND is commutative but not associative
Minimization of Boolean Expression Question 10:
Consider the following Boolean expression:
\(F(A, B, C) =\left( {\bar A + \bar B} \right)\left[ {\overline {A\left( {B + C} \right)} } \right] + A\left( {\bar B + \bar C} \right)\)
Which of the following is/are true about F(A, B, C)?
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 10 Detailed Solution
\(\begin{array}{l} \left( {\bar A + \bar B} \right)\left[ {\overline {A\left( {B + C} \right)} } \right] + A\left( {\bar B + \bar C} \right)\\ = \left( {\bar A + \bar B} \right)\left[ {\bar A + \left( {\overline {B + C} } \right)} \right] + A\left( {\bar B + \bar C} \right) \end{array}\)
\(\begin{array}{l} = \left( {\bar A + \bar B} \right)\left[ {\bar A + \bar B\bar C} \right] + A\bar B + A\bar C\\ = \bar A + \bar A\bar B\bar C + \bar A\bar B + \bar B\bar C + A\bar B + A\bar C \end{array}\)
\(\begin{array}{l} = \bar A\left( {1 + \bar B\bar C + \bar A\bar B} \right) + \bar B\bar C + A\bar B + A\bar C\\ = \bar A + \bar B\bar C + A\bar B + A\bar C \end{array}\)
= A̅ + A.B̅ + B̅.C̅ + A.C̅
= A̅ + B̅ + B̅.C̅ + A.C̅
= A̅ + B̅ (1 + C̅) + A.C̅
= A̅ + B̅ + A.C̅
=A̅ + C̅ + B̅
\(\begin{array}{l} = \bar A + \bar B + \bar C\\ = \overline {ABC} \end{array}\)
Step 1:
Step 2:
Step 3:
It can be represented by:
- one 3-input NAND gate
- one 2-input NAND gate and one 2-input AND gate
- three 2-input NAND gate