GATE CSE 2008 | Question: 2

Question

If $P, Q, R$ are subsets of the universal set U, then $$(P\cap Q\cap R) \cup (P^c \cap Q \cap R) \cup Q^c \cup R^c$$ is

• A. $Q^c \cup R^c$
• B. $P \cup Q^c \cup R^c$
• C. $P^c \cup Q^c \cup R^c$
• D. U

Answer 1

Treating as a boolean expression like suggested in an answer here:

PQR + P'QR + Q' + R'

= (P+P') QR + Q' + R'

= QR + Q' + R'

= QR + Q'R' + Q'R + R'

= R(Q+Q') + R'(Q'+1)

= R + R'

= 1

Also R+R' means RUR' so its equal to U.

Answer 2

If someone is good in digital Logic part or in vein Diagram part, than this question is easy for them :)

Just convert Union into + and intersection in . and try to solve it.

else Vein diagram becomes very easy for understanding.

If Understood UpVoted :)

Answer 3

$(p \cap q \cap r) \cup (p' \cap q \cap r) \cup q' \cup r’$

$(p \cap q \cap r) \cup ((p' \cap q \cap r)' \cap q \cap r)'$                          {demorgan’s law}

$(p \cap q \cap r) \cup (p \cap q \cap r)'$                                  {after solving the second part}

$\bigcup$                                                                                            {$A \cup A’$ = $U$}

Answer 4

The correct option is D 

The given set theory expression can be convert in to equivalent boolean algebra expression follows:

(p ∩ q ∩ r) ∪ (pc ∩ q ∩ r) ∪ qc ∪ rc
= pqr + p'qr + q' + r'
= qr + (p + P') + q' + r'
= qr + q' + r'
=(q+q').(r+q')+r'
= r + q' + r'
= r + r' + q'
= 1 + q'
= 1