Laws of Boolean Algebra MCQ Quiz in తెలుగు - Objective Question with Answer for Laws of Boolean Algebra - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

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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 = ________.

  1. Bubbled AND
  2. Bubbled NOR
  3. Bubbled XOR
  4. Bubbled OR

Answer (Detailed Solution Below)

Option 4 : Bubbled OR

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.

\(\overline {A.B} = \bar A + \bar B\)

F1 U.B Madhu 27.03.20 D1

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.

\(\overline {A + B} = \bar A.\bar B\)

F1 U.B Madhu 27.03.20 D2

Laws of Boolean Algebra Question 2:

{A + A̅·B} can also be represented as:

  1. A
  2. B
  3. A + B
  4. A̅ + B

Answer (Detailed Solution Below)

Option 3 : A + B

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)

  1. C
  2. B
  3. CD
  4. D

Answer (Detailed Solution Below)

Option 2 : B

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 _________.

  1. Z
  2. Y + Z
  3. X ⊕ Y
  4. (X + Y)Z

Answer (Detailed Solution Below)

Option 3 : X ⊕ Y

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 _______

  1. Involution law 
  2. Absorption law 
  3. Complementation law  
  4. DeMorgan’s law  

Answer (Detailed Solution Below)

Option 4 : DeMorgan’s law  

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:

\(\overline {\left( {{A} + {B} + {C}} \right)} = \left( {\overline {{A}} \;.\;\overline {{B}} \;.\overline {{C}} } \right)\)

26 June 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’

Laws of Boolean Algebra Question 6:

The output Y of the circuit shown in the figure is

F1 Shubham Madhu 27.07.21 D11

  1. \(\overline{A + B + C} + \overline{DE}\)
  2. \(\overline{ABC} + D + E\)
  3. \(ABC + \overline{D + E}\)
  4. ABC + D + E

Answer (Detailed Solution Below)

Option 4 : ABC + D + E

Laws of Boolean Algebra Question 6 Detailed Solution

The correct answer is 'option 4'

Solution:

From the figure,

F1 Shubham Madhu 27.07.21 D11

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

  1. \(B + C \overline D\)
  2. A
  3. \(AC \overline D\)    
  4. AB

Answer (Detailed Solution Below)

Option 2 : A

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.

  1. AND
  2. OR
  3. XOR
  4. NAND

Answer (Detailed Solution Below)

Option 4 : NAND

Laws of Boolean Algebra Question 8 Detailed Solution

\(\begin{array}{l} \left( {\bar A + \bar B} \right)\left[ {\overline {A\left( {B + C} \right)} } \right] + A\left( {\bar B + \bar C} \right)\\ = \left( {\bar A + \bar B} \right)\left[ {\bar A + \left( {\overline {B + C} } \right)} \right] + A\left( {\bar B + \bar C} \right) \end{array}\)

\(\begin{array}{l} = \left( {\bar A + \bar B} \right)\left[ {\bar A + \bar B\bar C} \right] + A\bar B + A\bar C\\ = \bar A + \bar A\bar B\bar C + \bar A\bar B + \bar B\bar C + A\bar B + A\bar C \end{array}\)

\(\begin{array}{l} = \bar A\left( {1 + \bar B\bar C + \bar A\bar B} \right) + \bar B\bar C + A\bar B + A\bar C\\ = \bar A + \bar B\bar C + A\bar B + A\bar C \end{array}\)

=+ A.B̅ + B̅.C̅ + A.C̅

= A̅ + B̅   + B̅.C̅ + A.C̅

= A̅ + B̅ (1 + C̅) + A.C̅

+ 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) 

  1. AB
  2. ABCD
  3. A
  4. ABC

Answer (Detailed Solution Below)

Option 1 : AB

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

  1. A+B+C
  2. ABC
  3. 1+ABC
  4. 1

Answer (Detailed Solution Below)

Option 4 : 1

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.

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