Jamia Millia Islamia MCA Boolean Algebra Previous Year Questions (PYQs)
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Showing 9 questions
Which of the following is false?
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Solution
$x·x' = 0$ (not 1) — this is false.
The Boolean expression $A + BC$ is the reduced form of —
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Solution
$$(A + B)(A + C) = A + BC$$ (by distributive law)
The Boolean expression $AB + AB' + A'C + AC$ is unaffected by the value of the Boolean variable:
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Solution
$AB + AB' = A$. So the expression becomes $A + A'C + AC = A + C$.
Variable $B$ disappears ⇒ expression doesn’t depend on $B$.
When you simplify algebraically the given expression to a minimum sum of products,
how many terms do you get?
(A + B' + C + E') (A + B' + D' + E) (B' + C' + D' + E')
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Solution
Solution:
Let’s analyze:
We can simplify using Boolean algebra rules.
After simplification (by K-map or expansion reduction), the minimum sum of products results in **4 terms**.
The simplified form of the given Boolean expression is:
A'CD'E + A'B'D' + ABCE + ABD
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Solution
Solution:
Simplify step-by-step:
$A'CD'E + A'B'D' + ABCE + ABD$
→ Combine using absorption and distributive laws.
$A'B'D' + ABD + ACD'E$
Hence, the final simplified expression is:
The time required for an electronic circuit to change its state is called
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Solution
Output delay between input and output change = propagation time.
$\boxed{\text{Propagation time}}$
Which of the following is true for $(p \land q) \to (p \lor q)$?
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Solution
$(p \land q) \to (p \lor q)$
This statement is always true, since whenever both $p$ and $q$ are true, $p \lor q$ is also true.
Hence, it represents a **Tautology**, not negation.
Simplify the Boolean expression (three variables):
$
F=\;A'BC\;+\;A'B'C\;+\;ABC'\;+\;A'B'C'\;+\;ABC\;+\;AB'C'
$
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Solution
Group terms:
- $ABC + ABC' = AB$
- $A'B'C' + AB'C' = B'C'$
- $A'BC + A'B'C = A'C$
Hence
$
F = AB \;+\; B'C' \;+\; A'C.
$
This is already minimal (no further absorption).
$\boxed{F = AB + A'C + B'C'}$
Minimize the following 3-variable function:
$F(A,B,C)=\sum(0,1,6,7)$
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Solution
Minterms 0(000) and 1(001) → pair ⇒ $A'B'$.
Minterms 6(110) and 7(111) → pair ⇒ $AB$.
So $F = A'B' + AB$ (XNOR, independent of $C$).
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