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Jamia Millia Islamia MCA Boolean Algebra PYQ

Jamia Millia Islamia MCA PYQ
Which of the following is false?





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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

$x·x' = 0$ (not 1) — this is false.
Jamia Millia Islamia MCA PYQ
The Boolean expression $A + BC$ is the reduced form of —





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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

$$(A + B)(A + C) = A + BC$$ (by distributive law)
Jamia Millia Islamia MCA PYQ
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 $B$.
Jamia Millia Islamia MCA PYQ
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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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2023 PYQ

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**.
Jamia Millia Islamia MCA PYQ
The simplified form of the given Boolean expression is: A'CD'E + A'B'D' + ABCE + ABD





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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2023 PYQ

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:
Jamia Millia Islamia MCA PYQ
The time required for an electronic circuit to change its state is called





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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

Output delay between input and output change = propagation time. $\boxed{\text{Propagation time}}$
Jamia Millia Islamia MCA PYQ
Which of the following is true for $(p \land q) \to (p \lor q)$?





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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2020 PYQ

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.
Jamia Millia Islamia MCA PYQ
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). $\boxed{F = AB + A'C + B'C'}$
Jamia Millia Islamia MCA PYQ
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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