Question

How many two-input AND and OR gates are required to realize Y = CD+EF+G?

(a) 2, 2

(b) 2, 3

(c) 3, 3

(d) 3, 2

This question was addressed to me in an international level competition.

This interesting question is from Logic Gates and Networks in division Boolean Algebra and Minimization Techniques and Logic Gates of Digital Circuits

Answer 1

Right option is (a) 2, 2

To elaborate: Y = CD + EF + G

             The number of two input AND gate = 2

             The number of two input OR gate = 2.