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 2 of 3.
Subhra Sar said:
8 years ago
Answer is B.
X+YZ is the solved result of the equation so 1 AND & 1 OR. Total 2 Gates are required. @Sukriti is right. Thank you.
X+YZ is the solved result of the equation so 1 AND & 1 OR. Total 2 Gates are required. @Sukriti is right. Thank you.
Somavardhan said:
7 years ago
For XY 1 AND gate x+z 1OR gate x (x+z) 1 and gate, why (x+z) 1and gate xy+x (x+z) +y (x+z) 1 3 input or gate.
Total 5 gates.
Total 5 gates.
Ram93 said:
1 decade ago
P:xy - 2 input AND.
Q:x+z - 2 input OR.
R:xQ - 2 input AND.
S:yQ - 2 input AND.
P+R+S - 3 input OR.
Total logic gates 5.
Q:x+z - 2 input OR.
R:xQ - 2 input AND.
S:yQ - 2 input AND.
P+R+S - 3 input OR.
Total logic gates 5.
Thabiso said:
7 years ago
Answer is E(5).
XY=1 AND gate,
(X+Z)= OR gate,
X.(X+Z)= AND gate,
Y.(X+Z)= AND gate,
XY+X(X+Z)+Y(X+Z)= the last OR gate.
XY=1 AND gate,
(X+Z)= OR gate,
X.(X+Z)= AND gate,
Y.(X+Z)= AND gate,
XY+X(X+Z)+Y(X+Z)= the last OR gate.
McallenNJB said:
6 years ago
Answer is 2.
XY+ X (X+Z) + Y (X+Z),
XY + X + XZ + XY + YZ,
(XY + X + XZ + XY) + YZ,
X (Y + 1 + Z + Y) + YZ,
X + YZ.
XY+ X (X+Z) + Y (X+Z),
XY + X + XZ + XY + YZ,
(XY + X + XZ + XY) + YZ,
X (Y + 1 + Z + Y) + YZ,
X + YZ.
(2)
SUKHPREET SINGH said:
9 years ago
X.Y = AND GATE
X+Z = OR GATE
X(X+Z) = AND GATE
Y(X+Z) = AND GATE
X.Y+X(X+Z)+Y(X+Z) = OR GATE
5 GATES REQUIRED
X+Z = OR GATE
X(X+Z) = AND GATE
Y(X+Z) = AND GATE
X.Y+X(X+Z)+Y(X+Z) = OR GATE
5 GATES REQUIRED
Vinay gupta said:
8 years ago
Answer B is right. Because of after solving an expression, we got X+YZ so 2 gates required.
@Sukriti is right.
@Sukriti is right.
(1)
ABHISHEK MANKAR said:
9 years ago
XY + X + XZ + YX + YZ.
X(1 + Y + Z + Y) + YZ.
X.1 + YZ.
So, one AND gate & one OR gate are required.
X(1 + Y + Z + Y) + YZ.
X.1 + YZ.
So, one AND gate & one OR gate are required.
Mona said:
6 years ago
Assume that three input OR gate is used for ORing each operation so total 5 gates are required.
BONDs said:
1 decade ago
Shouldn't it be 6.
Because,
1. X+Z (let's say Q).
2. XY.
3. XQ.
4. YQ.
5 and 6. 2+3+4.
Because,
1. X+Z (let's say Q).
2. XY.
3. XQ.
4. YQ.
5 and 6. 2+3+4.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers