Boolean Algebra and K-Map : MCQs & PYQs with Explanation

Option D

In Boolean algebra, a function can be written either in Sum of Minterms (SOP): Z=Σ_m(m₀ ,m₁ ,m₅ ,m₇) or in Product of Maxterms (POS): Z=Π_M(…)
For a function of three variables, the indexes run from 0 to 7.

• Given Minterms (Where Z = 1) : 0, 1, 5, 7
• Remaining Indices (Where Z = 0) : 2, 3, 4, 6

Thus, the POS form is : Z=Π_M(2,3,4,6)

Option A

Given Boolean Expression:
=x(x+y)
=xx+xy
=x+xy
=x(1+y)
=x.1
=x

Option A

Given Boolean expression:
=(A+C)(AD+AD’)+AC+C
=(A+C)(AD+AD’)+C(A+1)
=(A+C)(AD+AD’)+C.1
=(A+C)(AD+AD’)+C
=AAD+AAD’+ACD+ACD’+C
=AD+AD’+ACD+ACD’+C
=AD+AD’+C(AD+AD’+1)
=AD+AD’+C.1
=AD+AD’+C
=A(D+D’)+C
=A.1+C
=A+C
Therefore, (A+C)(AD+AD’)+AC+C = A+C

Option A

The given Boolean expression:
= x’y’+xy+x’y
= x’y’+x’y+xy
=x’(y’+y)+xy
=x’.1+xy
=x’+xy
=(x’+x)(x’+y)
=1.(x’+y)
=x’+y
Therefore, x’y’+xy+x’y = x’+y

Option C

Truth Table Confirmation:

Option C

The number of minterms in an n-variable truth table is 2ⁿ. This is because each Boolean variable can have two possible values, 0 or 1. When there are n variables, the total number of possible input combinations is 2ⁿ. Each unique input combination corresponds to exactly one minterm. Therefore, an n-variable truth table contains 2ⁿ minterms.

Option C

Boolean algebra is used to simplify logical expressions by applying Boolean laws and theorems. In the given expression, AB+A’C+BC, the term BC is redundant and can be removed using the Consensus Theorem.
The Consensus Theorem states : XY+X′Z+YZ = XY+X′Z
So, AB+A′C+BC = AB+A′C
Therefore, the simplified Boolean expression is : AB+A′C

Option C

• A.A’ = 0
• A+AB = A(1+B) = A.1 = A
• A+A’B = (A+A’)(A+B) = 1.(A+B) = A+B
• A(A+B) = A.A+A.B = A+A.B = A(1+B) = A.1 = A

Option B

Option A

Given Expression:
F=A’C+A’B+AB’C+BC
F=A’C+A’B+C(B+AB’)
F=A’C+A’B+C((A+B)(B+B’))
F=A’C+A’B+C(A+B)
F=A’C+A’B+AC+BC
F=A’C+AC+A’B+BC
F=C(A’+A)+A’B+BC
F=C+BC+A’B
F=C(1+B)+A’B
F=C+A’B

Option B

In the given K-map, two quads are formed. The first quad includes minterms 0, 2 and 8 and the don’t care terms 10, while the second quad includes minterms 0, 2 and 6 and the don’t care terms 4.
Quad 1 : a’b’c’d’+ab’c’d’+a’b’cd’+ab’cd’ : b’d’
Quad 2 : a’b’c’d’+a’b’c’d+ab’c’d’+ab’c’d : b’c’
Therefore, the minimal form is b’d’+b’c’.

Option B

Given boolean expression:
=X.(X+Y)
=X.X+X.Y
=X+X.Y
=X(1+Y)
=X.1
=X
So, X.(X+Y) = X

Option B

In digital electronics, a Karnaugh Map (K-Map) is used to simplify Boolean expressions by grouping adjacent 1’s (minterms). For 3 variables, the K-Map has 8 cells (0 to 7), and we place 1’s according to given minterms.
Given:
F(x,y,z) = ∑(0, 2, 3, 4, 6)
Now place 1’s at positions: 0, 2, 3, 4, 6. Then make the largest possible groups.

• Group 1: (0, 2, 4, 6) : Here z=0 (z’) is common.
• Group 2: (2, 3) : Here x=0 and y=1 (x’y) is common.

So the simplified expression becomes: F = z’+x’y

Option D

Given Boolean expression,
=((X+Y)’+Z’)’
=(X+Y)’’.Z’’
=(X+Y).Z
So, ((X+Y)’+Z’)’ = (X+Y).Z

Option A

In digital logic, a “don’t care” condition indicates that the value of a particular input does not affect the final output of the circuit. These conditions are used during simplification of Boolean expressions or Karnaugh maps (K-maps). 
Because the output does not depend on certain inputs, we are free to treat those inputs as either 0 or 1, whichever helps produce the simplest possible expression.
“Don’t care” conditions do not prioritize inputs nor ensure completeness of truth tables, instead, they are a tool for simplification.