Skip to content
Related Articles

Related Articles

GATE | GATE-CS-2014-(Set-1) | Question 65

Improve Article
Save Article
  • Difficulty Level : Basic
  • Last Updated : 29 Sep, 2021
Improve Article
Save Article

Consider a 4-to-1 multiplexer with two select lines S1 and S0, given below

GATECS2014Q55

The minimal sum-of-products form of the Boolean expression for the output F of the multiplexer is
(A) P’Q + QR’ + PQ’R
(B) P’Q + P’QR’ + PQR’ + PQ’R
(C) P’QR + P’QR’ + QR’ + PQ’R
(D) PQR’


Answer: (A)

Explanation: For 4 to 1 mux

=p’q’(0)+p’q(1)+pq’r+pqr’

=p’q+pq’r+pqr’

=q(p’+pr’)+pq’r

=q(p’+r’)+pq’r

=p’q+qr’+pq’r

Ans (a)



Quiz of this Question

My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!