Laws of Boolean Algebra MCQ Quiz in मल्याळम - Objective Question with Answer for Laws of Boolean Algebra - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 11, 2025
Latest Laws of Boolean Algebra MCQ Objective Questions
Top Laws of Boolean Algebra MCQ Objective Questions
Laws of Boolean Algebra Question 1:
According to De-Morgan's theorem: NAND = ________.
Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 1 Detailed Solution
De Morgan’s First Theorem:
- According to De Morgan’s first theorem, a NAND gate is equivalent to a Bubbled OR gate.
- The Boolean expressions for the bubbled OR gate can be expressed by the equation shown below.
\(\overline {A.B} = \bar A + \bar B\)
De Morgan’s second theorem:
- According to De Morgan’s first theorem, a NOR gate is equivalent to a Bubbled AND gate.
- The Boolean expressions for the bubbled AND gate can be expressed by the equation shown below.
\(\overline {A + B} = \bar A.\bar B\)
Laws of Boolean Algebra Question 2:
{A + A̅·B} can also be represented as:
Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 2 Detailed Solution
{A + A̅ ·B} = (A + A̅) (A + B)
= 1 . (A + B)
= (A + B)
Laws of Boolean Algebra Question 3:
If the following Boolean expression is reduced, what will be the reduced result?
(B + BC) (B + B̅C) (B + D)
Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 3 Detailed Solution
Concept:
Some of the important boolean identities are as follows:
- 1 + A = 1
- \(A \space +\space \overline{A}\space = \space 1 \)
- 0 + A = A
- \(A \space\space \overline{A}\space = \space 0 \)
Calculation:
Y = (B + BC) (B + B̅C) (B + D)
Y = { B (1 + C) } { (B + B̅) (B + C) } (B + D)
Y = { B } { (B + C) } (B + D)
Y = BBB + BBD + BBC + BCD
Y = B + BD + BC + BCD
Y = B (1 + D) + BC (1 + D)
Y = B + BC
Y = B (1 + C)
Y = B
Laws of Boolean Algebra Question 4:
If function f(X, Y, Z) = ∑ m(2, 3, 4, 5) is implemented using SOP form, the resultant Boolean function would be _________.
Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 4 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(X, Y, Z) = ∑ m(2, 3, 4, 5)
= X̅YZ̅ + X̅YZ + XY̅Z̅ + XY̅Z
= X̅Y(Z + Z̅) + XY̅(Z + Z̅)
= X̅Y + XY̅
Laws of Boolean Algebra Question 5:
The equality (A + B + C)I = AI.BI.CI is better known as _______
Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 5 Detailed Solution
The correct option is 4
Concept:
De Morgan’s law states that:
\(\overline {\left( {{A_1}.{A_2} \ldots {A_n}} \right)} = \left( {\overline {{A_1}} + \overline {{A_2}} + \ldots + \overline {{A_n}} } \right)\)
\(\overline {\left( {{A_1} + {A_2} + \ldots + {A_n}} \right)} = \left( {\overline {{A_1}} \;.\;\overline {{A_2}} \;.\;..\;\overline {{A_n}} } \right)\)
∴ De-Morgan's expression will be:
\(\overline {\left( {{A} + {B} + {C}} \right)} = \left( {\overline {{A}} \;.\;\overline {{B}} \;.\overline {{C}} } \right)\)
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’ |
Laws of Boolean Algebra Question 6:
The output Y of the circuit shown in the figure is
Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 6 Detailed Solution
The correct answer is 'option 4'
Solution:
From the figure,
At 1 the combination is AB
At2 the combination is C
At 3 combination is \(\overline {(AB).C}\)
At 4 combination is \(\overline{(D+E)}\)
\(\implies Y=\overline{\overline{(ABC)}.\overline{(D+E)}}\)
Using De Morgan's law,
\(\implies Y=ABC+D+E\)
Laws of Boolean Algebra Question 7:
The simplified form of the expression \(f = A \overline B + AB + AC \overline D\) is
Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 7 Detailed Solution
Concept:
Some of the important boolean expressions are:
1.) \(A + \bar{A} = 1\)
2.) \(A \bar{A} = 0\)
3.) \(1+A=1\)
4.) \(0+A= A\)
Calculation:
Given, \(f = A \overline B + AB + AC \overline D\)
\(f = A (\overline B + B) + AC \overline D\)
\(f = A + AC \overline D\)
\(f = A ( 1+ C \overline D)\) : Use Domination Law 1 + A = 1
\(f = A \)
Laws of Boolean Algebra Question 8:
Consider the following Boolean expression:
\(\left( {\bar A + \bar B} \right)\left[ {\overline {A\left( {B + C} \right)} } \right] + A\left( {\bar B + \bar C} \right)\)
It can be represented by a single three-input logic gate. Identify the gate.Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 8 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
\(\begin{array}{l} = \bar A + \bar B + \bar C\\ = \overline {ABC} \end{array}\)
The given Boolean expression represents NAND gate.Laws of Boolean Algebra Question 9:
Using Boolean algebra, if the following Boolean expression is minimised, then the result will be:
AB + ABC + AB (D + E)
Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 9 Detailed Solution
Concept:
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’ |
Calculation:
Given,
F = AB + ABC + AB (D + E)
or, F = AB (1 + C) + AB (D + E) [∵ 1 + C = 1]
or, F = AB + AB (D + E)
or, F = AB (1 + D + E) [∵ 1 + D+ E = 1]
or, F = AB
Laws of Boolean Algebra Question 10:
According to Boolean Algebra (1+A+B+C) can be simplified as
Answer (Detailed Solution Below)
Laws of Boolean Algebra Question 10 Detailed Solution
Concept:
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’ |
Analysis:
Given:
Y = (1+A+B+C)
By using Null Law;
Y = 1
Hence, option 4 is correct.