NIMCET 2027 Boolean Algebra Previous Year Questions (PYQs) – Page 3 of 4
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Showing 36 questions
Which of the following Boolean expression represents the shaded portion of the Venn diagram?
Note: Here "." represents an AND operation and "+" denotes an OR operation.
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Solution
The shaded region is the part common to all three sets X, Y, Z (the triple overlap).
That means the region represents:
X ∙ Y ∙ Z
Now check options:
Option (2):
Z.(X + Y) = ZX + ZY
This includes regions where Z overlaps with X OR Z overlaps with Y — this is larger than triple intersection → NOT correct.
Option (3):
(Z.X′) + Y
This includes all of Y plus some part of Z, totally incorrect.
Option (1):
Z′ + (X.Y)
This includes all outside Z plus XY, wrong.
Option (4):
Z′.(X + Y)
This is the region outside Z but inside X or Y, wrong.
The expression P+QR is the reduced
form of _____
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Solution
Boolean Simplification
Given Expression: P + QR
We ask: This is the simplified (reduced) form of which expression?
Try expanding: (P + Q)(P + R)
Using distributive law:
(P + Q)(P + R) = P(P + R) + Q(P + R) = P + PR + PQ + QR = P + PQ + PR + QR = P + QR (since P absorbs PQ and PR)
✅ Final Answer: (P + Q)(P + R)
The
Boolean expression for the following truth table is _________
| x | y | z | f |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
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Solution
The Boolean function $a+(\overline{a} \,b)$ is equivalent to
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Solution
$a+(\overline{a} \,b)$
=$ (a+\bar{a})(a+b)$
=$1.(a+b)=(a+b)$
A tautology is a Boolean formula that is always true. Which of the following is a tautology?
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Solution
$x+\overline{y}+\overline{x}$
=$\bar{y}+x+\bar{x}$
=$\bar{y}+1$
=$1$
The minimum number of NAND gates required for implementing the Boolean expression $AB+A\, \overline{B}C+A\, \overline{B}\, \overline{C}$
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Solution
$AB+A\, \overline{B}C+A\, \overline{B}\, \overline{C}$
=$AB+AB'C+AB'C'$
=$AB+AB'(C+C')$
=$AB+AB'$
=$A(B+B')$
=A
Which of the following is equivalent to the Boolean expression:
$(X+Y).(X+\overline{Y}).(\overline{X}+Y)$
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Solution
$(X+Y).(X+\overline{Y}).(\overline{X}+Y)$
=$(X+Y)(X+Y')(X'+Y)$
=$(XX+XY+YX+YY')(X'+Y)$
=$(X+XY)(X'+Y)$
=$X(1+Y)(X'+Y)$
=$X(X'+Y)$
=$XX'+XY$
=$XY$
Write the simplified form of the Boolean expression
(A+C)(AD+AD')+AC+C
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Solution
(A+C)(AD+AD')+AC+C
=(A+C)A(D+D')+C(A+1)
=(A+C)+C
=A+C
The Boolean expression AB+ AB' + A'C + AC is unaffected by the value of the Boolean variable
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Solution
$\begin{array}{ll}{A(B+\overline{B})+C(A+\overline{A})} \\ {A+C}\, \end{array}$
NIMCET 2027
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and More.
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