Boolean Algebra MCQ Quiz in मल्याळम - Objective Question with Answer for Boolean Algebra - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 16, 2025
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?Answer (Detailed Solution Below)
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.
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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 ⊕ yBoolean Algebra Question 4:
Consider the Boolean function z(a, b, c).
Which one of the following minterm lists represents the circuit given above?
Answer (Detailed Solution Below)
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
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 |
Answer (Detailed Solution Below)
Boolean Algebra Question 5 Detailed Solution
From truth table:
F(X, Y) = XY̅ + XY
= X[Y̅ + Y]
F(x, y) = XBoolean 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:
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
Answer (Detailed Solution Below)
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 = BBoolean Algebra Question 8:
Which of the following is equivalent of the Boolean expression given below?
A + A̅.B + A.B̅Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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 = 1Boolean 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
Answer (Detailed Solution Below)
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