Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 14)
14.
How many gates would be required to implement the following Boolean expression before simplification? XY + X(X + Z) + Y(X + Z)
Discussion:
28 comments Page 3 of 3.
Ahlem said:
4 years ago
XY AND.
X+Z OR.
X (X+Z) AND.
Y (X+Z) AND.
XY+X (X+Z)+Y (X+Z) ONE OR.
The sum is 5.
X+Z OR.
X (X+Z) AND.
Y (X+Z) AND.
XY+X (X+Z)+Y (X+Z) ONE OR.
The sum is 5.
Shubhangi_n said:
1 decade ago
Answer is D as they want before simplification. Nidhya solved it after simplification.
Rajan said:
6 years ago
The sixth gate could be the OR gate for adding the individual terms of the expression.
Hema said:
10 years ago
5 gates only I didn't see properly its not x+z it was X(X+Z). So, 5 gates is answer.
Sasha said:
6 years ago
@All.
The Answer is 2 if you have simplified with boolean algebra down to x+ yz.
The Answer is 2 if you have simplified with boolean algebra down to x+ yz.
(5)
Ramso said:
1 decade ago
The answer is 5, first it become xy+(x+y)(y+z)
It need 2 and gate and 3 or gate.
It need 2 and gate and 3 or gate.
Abhishek said:
9 years ago
Without simplification also gates will do. x+z gate can be used twice.
Nandan said:
1 decade ago
Using K-map ,you get X+Y+Z, giving 2 gates. So answer is B
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers