Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

GATE | GATE MOCK 2017 | Question 16

  • Last Updated : 19 Nov, 2018

The Boolean function f implemented in the figure shown below, using two input multiplexers is:

mock_1

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

(A) AB’C + ABC’



(B) A’B’C + A’BC’

(C) A’BC + A’B’C’

(D) ABC + AB’C’


Answer: (D)

Explanation: The output of a multiplexer (2X1) can be expressed in the form of:
mock_16

f= S’. L1 + S. L2

For the above diagram the output from the first multiplexer would be B’C + BC’ which acts a select line for the second multiplexer. Hence,
f= (B’C + BC’)’A + (B’C + BC’).0

= ((B’C)’ . (BC’)’)A + 0
= (B+C’)(B’ + C)A

= (BC + B’C’) A
= ABC + AB’C’



Quiz of this Question

My Personal Notes arrow_drop_up
Recommended Articles
Page :