Boolean Algebra MCQ Quiz in मल्याळम - Objective Question with Answer for Boolean Algebra - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 16, 2025

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

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:

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 2 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 3:

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 3 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 4:

Consider the Boolean function z(a, b, c).

F1 R.S Madhu 09.04.20 D 4

Which one of the following minterm lists represents the circuit given above?

  1. z = ∑ (0, 1, 3, 7)
  2. z = ∑ (1, 4, 5, 6, 7)
  3. z = ∑ (2, 4, 5, 6, 7)
  4. z = ∑ (2, 3, 5)

Answer (Detailed Solution Below)

Option 2 : z = ∑ (1, 4, 5, 6, 7)

Boolean Algebra Question 4 Detailed Solution

The given circuit gives the output:

Z(a, b, c) = \(a + \;\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftharpoonup$}} \over b} \;c\) 

Expanding it into canonical form to obtain the minterms

Z(a, b, c) = \(a\left( {b + \;\bar b} \right)\left( {c + \bar c} \right) + \left( {a + \bar a} \right)\bar b\;c\) 

\(abc + ab\;\bar c + a\bar b\;c + a\bar b\bar c + \bar a\bar b\;c\)

After rearranging the canonical terms, this corresponds to min-terms: ∑ (1,4, 5, 6, 7)

Alternate solution:

The output of the circuit is \(a + \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftharpoonup$}} \over b} \;c\) 

K Map for this Boolean expression

F1 R.S Madhu 20.04.20  D1

The above K Map corresponds to min-terms: ∑ (1,4, 5, 6, 7)

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?

\(f\left( {X,Y,Z,W} \right)\; = X\bar YZ\bar W + \bar X\bar Y\bar Z\bar W + X\bar Y\bar Z\bar W + \bar X\bar YZ\bar W + \;\bar XY\bar ZW + \bar XYZW + \;XY\bar ZW + XYZW\)

Answer (Detailed Solution Below) 5

Boolean Algebra Question 6 Detailed Solution

\(f\left( {X,Y,Z,W} \right)\; = X\bar YZ\bar W + \bar X\bar Y\bar Z\bar W + X\bar Y\bar Z\bar W + \bar X\bar YZ\bar W + \;\bar XY\bar ZW + \bar XYZW + \;XY\bar ZW + XYZW\)

\(f\left( {X,Y,Z,W} \right)\; = \bar X\bar Y\bar Z + \bar X\bar YZ\bar W + \;\bar XY\bar ZW + \bar XYZW + X\bar Y\bar Z\bar W + X\bar YZ\bar W + \;XY\bar ZW + XYZW\)

\(f\left( {X,Y,Z,W} \right)\; = \bar X\bar Y\bar W\;\left( {Z + \bar Z} \right) + \;\bar XYW\left( {Z + \bar Z} \right) + X\bar Y\bar W\left( {Z + \bar Z} \right) + \;XYW\left( {Z + \bar Z} \right)\)

\(f\left( {X,Y,Z,W} \right)\; = \bar X\bar Y\bar W\; + X\bar Y\bar W + \;\bar XYW + \;XYW\)

\(f\left( {X,Y,Z,W} \right)\; = \left( {\bar X + X} \right)\left( {\bar Y\bar W\;} \right) + \;\left( {\bar X + X} \right)\left( {YW} \right)\)

\(f\left( {X,Y,Z,W} \right)\; = \bar Y\bar W + \;YW\;\;\)

\(f\left( {X,Y,Z,W} \right)\; = Y \odot W\)

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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