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Boolean Algebra and Minimization

GATE CSE 2014 Set 1 · Q7

MCQ1 markEasy

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Consider the following Boolean expression for FF: F(P,Q,R,S)=PQ+PˉQR+PˉQRˉS F(P, Q, R, S) = PQ + \bar{P}QR + \bar{P}Q\bar{R}S The minimal sum-of-products form of FF is

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Solution

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Official answer: A) PQ + QR + QS

Boolean Algebra and Minimization

Factor out QQ from the expression:

F=Q(P+PˉR+PˉRˉS)F = Q(P + \bar{P}R + \bar{P}\bar{R}S)

Using the Boolean absorption/simplification rule A+AˉB=A+BA + \bar{A}B = A + B:

  1. P+Pˉ(R+RˉS)=P+R+RˉSP + \bar{P}(R + \bar{R}S) = P + R + \bar{R}S

  2. R+RˉS=R+SR + \bar{R}S = R + S

Substituting these back:

P+PˉR+PˉRˉS=P+R+SP + \bar{P}R + \bar{P}\bar{R}S = P + R + S

Now, multiply by QQ:

F=Q(P+R+S)=PQ+QR+QSF = Q(P + R + S) = PQ + QR + QS

This gives the minimal sum-of-products expression, matching option A.

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This Boolean Algebra and Minimization previous year question is Q7 of the GATE CSE 2014 Set 1 paper. Find more GATE CSE PYQs on Boolean Algebra and Minimization in Digital Logic, or browse all GATE CSE PYQs.

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Boolean Algebra and Minimization in GATE CSE

Boolean Algebra and Minimization has 36 GATE CSE PYQs across 23 papers from 2011 to 2026. 13 of them are 1-mark MCQs like this one, and it is the only one in GATE CSE 2014 Set 1.

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GateAI solution cardGATE CSE 2014 Set 1 · Q7 · MCQ

Consider the following Boolean expression for F: F(P, Q, R, S) = PQ + PQR + PQ RS The minimal sum-of-products form of F is

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.