Boolean-Expression
Question 1 |
Which one of the following is NOT a valid identity?
(x + y) ⊕ z = x ⊕ (y + z) | |
(x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) | |
x ⊕ y = x + y, if xy = 0 | |
x ⊕ y = (xy + x'y')' |
Question 1 Explanation:
Let x=1, y=1, z=0.
(x+y) ⊕ z = (1+1)⊕ 0 = 1 ⊕ 0 = 1
x ⊕ (y+z) = 1⊕(1+0) = 1 ⊕ 1 = 0
So,
(x+y) ⊕ z ≠ x ⊕ (y+z)
(x+y) ⊕ z = (1+1)⊕ 0 = 1 ⊕ 0 = 1
x ⊕ (y+z) = 1⊕(1+0) = 1 ⊕ 1 = 0
So,
(x+y) ⊕ z ≠ x ⊕ (y+z)
Question 2 |
The simplified SOP (Sum of Product) form the Boolean expression (P + Q’ + R’)(P + Q’ + R)(P + Q + R’)
(P’Q + R) | |
(P + Q’R’) | |
(P Q’ + R ) | |
(PQ + R) |
Question 2 Explanation:

Question 3 |
Evaluate (X XOR Y) XOR Y?
All 1’s | |
All 0’s | |
X | |
Y |
Question 3 Explanation:

Question 4 |
The complement of the Boolean expression AB ( B’C + AC ) is
( A’ + B’ ) + ( B + C’ )( A’ + C’ ) | |
( A’ + B’ ) + ( BC’ + A’C’ ) | |
( A’ + B’ )( B + C’) + ( A + C’ ) | |
( A + B )( B’ + C )( A + C ) |
Question 4 Explanation:
→ Finding complement boolean expression, we require Demorgan’s law.
→ DeMorgan’s law and after complementing:
⇒ ( AB ( B’C + AC ))’
⇒ (A + B)’ + (B’C + AC)’
⇒ (A’ + B’) + ( B’C)’ (AC)’
⇒ (A’ + B’) + ( B + C’)(A’ + C’)
→ DeMorgan’s law and after complementing:
⇒ ( AB ( B’C + AC ))’
⇒ (A + B)’ + (B’C + AC)’
⇒ (A’ + B’) + ( B’C)’ (AC)’
⇒ (A’ + B’) + ( B + C’)(A’ + C’)
Question 5 |
The boolean expression AB + AB’+ A’C + AC is independent of the boolean variable
A | |
B | |
C | |
None of these |
Question 5 Explanation:

→ AB + AB' + A'C + AC
= A(B + B') + C(A + A')
= A+C
As the expression is independent of 'B'
Question 6 |
The simplified form of the boolean expression:
(AB’(C+BD)+A’B’)C is:
B’C | |
A’B’C | |
AB’C | |
A’BC |
Question 6 Explanation:
(AB’(C+BD)+A’B’)C
(AB'C + AB'BD + A'B')C
(AB'C + A'B')C
AB'C + A'B'C
(A+A') B'C
B'C
(AB'C + AB'BD + A'B')C
(AB'C + A'B')C
AB'C + A'B'C
(A+A') B'C
B'C
Question 7 |
Consider the following sequence of instructions :
a=a⊕b, b=a⊕b, a=b⊕a
This Sequence
a=a⊕b, b=a⊕b, a=b⊕a
This Sequence
retains the value of the a and b | |
complements the value of a and b | |
swap a and b | |
negates values of a and b |
Question 7 Explanation:
⇒ a=a⊕b
⇒ b=a ⊕b
= a⊕b ⊕b (Substitute (1))
= 0 ⊕ b
= b
⇒ a = b⊕a
= b⊕a⊕b (Substitute (1))
= a⊕b⊕b
= a⊕0
= a
⇒ b=a ⊕b
= a⊕b ⊕b (Substitute (1))
= 0 ⊕ b
= b
⇒ a = b⊕a
= b⊕a⊕b (Substitute (1))
= a⊕b⊕b
= a⊕0
= a
Question 8 |
The logic expression x’yz’+x’yz+xyz’+xyz reduces to :
x’z | |
xyz | |
y | |
yz |
Question 8 Explanation:
We can use two methods to solve this problem.
Method-1 using K-Maps:

Method-2 using boolean simplification:
x’yz’+x’yz+xyz’+xyz
= x’y(z’+z)+xy(z’+z)
= x’y + xy
= y(x’+x)
= y
Method-1 using K-Maps:

Method-2 using boolean simplification:
x’yz’+x’yz+xyz’+xyz
= x’y(z’+z)+xy(z’+z)
= x’y + xy
= y(x’+x)
= y
There are 8 questions to complete.