Minimization of Boolean Expression MCQ Quiz in मल्याळम - Objective Question with Answer for Minimization of Boolean Expression - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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നേടുക Minimization of Boolean Expression ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Minimization of Boolean Expression MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

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

F1 U.B Madhu 27.03.20 D6

  1. AB + C̅
  2. C + A̅B
  3. A + B + C
  4. A + B̅C

Answer (Detailed Solution Below)

Option 2 : C + A̅B

Minimization of Boolean Expression Question 1 Detailed Solution

F1 U.B Madhu 27.03.20 D7

F = C + A̅B

Minimization of Boolean Expression Question 2:

Which Boolean law is represented below 2

P + Q·R = (P+Q)·(P+R) 

  1. De Morgan's law
  2. Associative law
  3. Commutative law
  4. Distributive law

Answer (Detailed Solution Below)

Option 4 : Distributive law

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 ?

  1. A(A + B) = B
  2. AB + AB' = A
  3. A + AB = A
  4. (A + B)(A + B') = A

Answer (Detailed Solution Below)

Option 1 : A(A + B) = B

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

  1. 2(n - 1)
  2. n/2
  3. 2n
  4. 2(n-1)

Answer (Detailed Solution Below)

Option 4 : 2(n-1)

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:

F1 R.S Madhu 29.04.20 D8

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)

 

  1. A - III, B - IV, C - I, D - II
  2. A - II, B - IV, C - I, D - III
  3. A - I, B - IV, C - II, D - III
  4. A - II, B - I, C - IV, D - III

Answer (Detailed Solution Below)

Option 2 : A - II, B - IV, C - I, D - III

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?

  1. x + yz
  2. (x + y)(x + z)
  3. x(y + z)
  4. xyz

Answer (Detailed Solution Below)

Option 3 : x(y + z)

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.

  1. 0
  2. A + B
  3. A̅ + BC
  4. 1

Answer (Detailed Solution Below)

Option 1 : 0

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

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’

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)

F2 R.S. Nita 04.10.2019 D 15

  1. AB
  2. AB̅
  3. A̅ B̅
  4. A̅ B

Answer (Detailed Solution Below)

Option 1 : AB

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?

  1. Only S1 is true
  2. Only S2 is true
  3. Both S1 and S2 are true
  4. Neither S1 nor S2 are true

Answer (Detailed Solution Below)

Option 2 : Only S2 is true

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)?

  1. F(A, B, C) can be represented by one 3-input NAND gate
  2. F(A, B, C) can be represented by two 2-input NAND gate
  3. F(A, B, C) can be represented by three 2-input NAND gate
  4. F(A, B, C) can be represented by one 2-input NAND gate and one 2-input AND gate

Answer (Detailed Solution Below)

Option :

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:

F1 Raju Madhu 21.08.20 D1

Step 2:

F1 Raju Madhu 21.08.20 D2

Step 3:

F1 Raju Madhu 21.08.20 D3

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
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