Boolean Algebra MCQ Quiz in বাংলা - Objective Question with Answer for Boolean Algebra - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Apr 18, 2025

পাওয়া Boolean Algebra उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Boolean Algebra MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest Boolean Algebra MCQ Objective Questions

Top Boolean Algebra MCQ Objective Questions

Boolean Algebra Question 1:

Let, x1 ⊕ x2 ⊕ x3 ⊕ x4 = 0 where x1, x2, x3, x4 are Boolean variables, and ⊕ is the XOR operator.

Which one of the following must always is TRUE?

  1. x1x2x3x4 = 0
  2. x1x3 + x2 = 0
  3. 1 ⊕ x̅3 = x̅2 ⊕ x̅4
  4. x1 + x2 + x3 + x4 = 0

Answer (Detailed Solution Below)

Option 3 : x̅1 ⊕ x̅3 = x̅2 ⊕ x̅4

Boolean Algebra Question 1 Detailed Solution

Concept:

XOR gate is a gate that gives a true output when the number of true inputs is odd.

Explanation:

Given, x1 ⊕ x2 ⊕ x3 ⊕ x4 = 0

Where, x1, x2, x3, x4 are Boolean variables, and ⊕ is the XOR operator

Consider x1 = 1, x2 =1, x3 =1 and x4= 1

1 ⊕ 1 ⊕ 1 ⊕ 1 = 0

Now, consider all the options one by one.

1) x1x2x3x4 = 0 [Incorrect]

Here, put the value of x1,x2, x3, x4 as 1

So, 1.1.1.1 = 1

2) x1x3 + x2 = 0 [Incorrect]

1.1 + 1 =1

3) x̅1 ⊕ x̅3 = x̅2 ⊕ x̅4 [Correct]

Here, x̅1 = x̅3 = x̅2 = x̅4 = 0,

So, 0 ⊕ 0 = 0 ⊕ 0,

0 = 0

4) x1 + x2 + x3 + x4 = 0 [Incorrect]

As, 1+1+1+1 = 1

Boolean Algebra Question 2:

Which one of the following is NOT a valid identity?

  1. (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)
  2. (x + y) ⊕ z = x ⊕ (y + z)
  3. x ⊕ y = x + y, if xy = 0
  4. x ⊕ y = (xy + x'y')'

Answer (Detailed Solution Below)

Option 2 : (x + y) ⊕ z = x ⊕ (y + z)

Boolean Algebra Question 2 Detailed Solution

1. (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) 

x

y

z

(x ⊕ y) ⊕ z

x ⊕ (y ⊕ z)

0

0

0

0

0

0

O

1

1

1

0

1

0

0

1

0

1

1

0

0

1

0

0

1

1

1

0

1

0

0

1

1

0

0

0

1

1

1

1

1

 

Therefore exclusive OR is associative and hence (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) 

2.

x

y

z

(x + y) ⊕ z

x ⊕ (y + z)

0

0

0

0

0

0

O

1

1

1

0

1

0

0

1

0

1

1

0

1

1

0

0

1

1

1

0

1

0

0

1

1

0

1

0

1

1

1

0

0

 

(x + y) ⊕ z ≠ x ⊕ (y + z) ∴ is it not a valid identity

3.

xy = 0

Checking validity

X

y

x + y

x ⊕ y

0

0

0

0

0

1

1

1

1

0

1

1

x + y = x ⊕ y / if xy = 0

4.

(xy + x'y')'

= (x’ + y’).(x+y) / Demorgan’s Law

= x’y +xy’

= x ⊕ y

Boolean Algebra Question 3:

If function f(A, B) = ∑ m(0, 1, 2, 3) is implemented using SOP form, the resultant Boolean function would be:

  1. A + B
  2. A + B̅
  3. AB
  4. 1

Answer (Detailed Solution Below)

Option 4 : 1

Boolean Algebra Question 3 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(A, B) = ∑ m(0, 1, 2, 3)

= A̅ B̅ + A̅ B + A B̅ + AB

= A̅ ( B + B̅) + A (B̅ + B)

= A̅ + A = 1

Boolean Algebra Question 4:

Find the Boolean function for the shaded region of the following diagram represented.

F1 R.S-D.K 13.09.2019 D5

  1. (A + C’) (B’ + C)
  2. (A + C) (B’ + C)
  3. (A + C’) (B’ + C’)
  4. (A + C) (A + B’)

Answer (Detailed Solution Below)

Option 2 : (A + C) (B’ + C)

Boolean Algebra Question 4 Detailed Solution

F(A, B, C) = C + AB’C’

F(A, B, C) = (C + C’)(C + AB’)

F(A, B, C) = (A + C)(B’ + C)

Boolean Algebra Question 5:

X

Y

F(X,Y)

0

0

0

0

1

0

1

0

1

1

1

1

The truth table represents the Boolean function:

  1. X
  2. X - Y
  3. X + Y
  4. Y

Answer (Detailed Solution Below)

Option 1 : X

Boolean Algebra Question 5 Detailed Solution

From truth table:

F(X, Y) = XY̅ + XY

= X[Y̅ + Y]

F(x, y) = X

Boolean Algebra Question 6:

What is the minimum number of NAND gates needed for the below given?

f(X,Y,Z,W)=XY¯ZW¯+X¯Y¯Z¯W¯+XY¯Z¯W¯+X¯Y¯ZW¯+X¯YZ¯W+X¯YZW+XYZ¯W+XYZW

Answer (Detailed Solution Below) 5

Boolean Algebra Question 6 Detailed Solution

f(X,Y,Z,W)=XY¯ZW¯+X¯Y¯Z¯W¯+XY¯Z¯W¯+X¯Y¯ZW¯+X¯YZ¯W+X¯YZW+XYZ¯W+XYZW

f(X,Y,Z,W)=X¯Y¯Z¯+X¯Y¯ZW¯+X¯YZ¯W+X¯YZW+XY¯Z¯W¯+XY¯ZW¯+XYZ¯W+XYZW

f(X,Y,Z,W)=X¯Y¯W¯(Z+Z¯)+X¯YW(Z+Z¯)+XY¯W¯(Z+Z¯)+XYW(Z+Z¯)

f(X,Y,Z,W)=X¯Y¯W¯+XY¯W¯+X¯YW+XYW

f(X,Y,Z,W)=(X¯+X)(Y¯W¯)+(X¯+X)(YW)

f(X,Y,Z,W)=Y¯W¯+YW

f(X,Y,Z,W)=YW

XNOR gate with NAND gates:

F1 Raju Shraddha 05.05.2021. D6

Hence 5 NAND gates is needed

Tips and Tricks:

The above Boolean function can also be solved using K-map:

Boolean Algebra Question 7:

Minimize the Boolean expression x = ABC + A̅B + ABC̅

  1. B
  2. A
  3. D
  4. AB

Answer (Detailed Solution Below)

Option 1 : B

Boolean Algebra Question 7 Detailed Solution

X = ABC + A̅B + ABC̅

Rearranging:

X = ABC + ABC̅ + A̅B

X = AB(C + C̅) + A̅B

X = AB + A̅B   [C + C̅ = 1]

X = (A + A̅)B  

X = 1.B   [A + A̅ = 1]

X = B

Boolean Algebra Question 8:

Which of the following is equivalent of the Boolean expression given below?

A + A̅.B + A.B̅ 

  1. B̅ + A̅
  2. A + B̅
  3. B + A̅
  4. A + B

Answer (Detailed Solution Below)

Option 4 : A + B

Boolean Algebra Question 8 Detailed Solution

Concept-

  • The branch of algebra which deals with the values of the variables in the form of the truth values true and false is called Boolean algebra.
  • True and false are usually denoted by 1 and 0 respectively


Calculation:

Let F(A, B) = A + A̅ .B + A.B̅

F(A, B) = A (1 + B̅) + A̅.B

F(A, B) = A + A̅.B

F(A, B) = (A + A̅).(A + B)

F(A, B) = A + B 

Boolean Algebra Question 9:

Consider W, Y, A and B be the Boolean variable and $ operator defined as A $ B = A̅ + B where Y = W̅ $ A̅ and W = A + B̅. Find Y?

  1. A
  2. 0
  3. 1

Answer (Detailed Solution Below)

Option 4 : 1

Boolean Algebra Question 9 Detailed Solution

A $ B = A̅ + B (1)

W = A + B̅ (2)

Y = W̅ $ A̅

Y = W + A̅ (From 1)

Y = A + B̅ + A̅ (From 2)

Y = 1 + B̅ = 1

Boolean Algebra Question 10:

The truth table

X

Y

F(X, Y)

0

0

0

0

1

0

1

0

1

1

1

1

represents the Boolean function

  1. X
  2. X + Y
  3. X ⊕ Y
  4. Y

Answer (Detailed Solution Below)

Option 1 : X

Boolean Algebra Question 10 Detailed Solution

The correct answer is "option 1".

EXPLANATION:

Option 1: TRUE

The truth table X is:

X

Y

F

 X

0

0

0

0

0

1

0

0

1

0

1

1

1

1

1

1

 

F is equal to X.

Option 2: FALSE

The truth table X+Y is:

X

Y

F

             X+Y

0

0

0

0

0

1

0

1

1

0

1

1

1

1

1

1

F is not equal to X+Y.

Option 3: FALSE

The truth table X’Y + XY' is:

X

Y

F

 

X’ Y + XY'

0

0

0

0

0

1

0

1

1

0

1

1

1

1

1

0

F is not equal to X’.Y +X.Y'.

Option 4: FALSE

The truth table Y is :

X

Y

F

              Y

0

0

0

              0

0

1

0

              1

1

0

1

              0

1

1

1

              1

 F is not equal to Y.

Hence, the correct answer is "option 1".

From truth table, the sum of minterms

F(X, Y) = X.Y̅ + X.Y = X(Y̅ + Y) = X

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