1
GATE ECE 2006
MCQ (Single Correct Answer)
+2
-0.6
The point p in the following figure is stuck- at-1. The output f will be
A
$$\overline {AB\overline {C\,} }$$
B
$$\overline A$$
C
$$AB\overline C$$
D
A
2
GATE ECE 2004
MCQ (Single Correct Answer)
+2
-0.6
The Boolean expression AC + B$$\overline C$$ is equivalent to
A
$$\overline A \,C + B\overline C + AC$$
B
$$\overline B C\, + AC + B\overline C + \overline A C\overline B$$
C
AC+$$B\overline C + \overline B C + ABC$$
D
$$\overline A \,B\,\overline C + A\,B\,\overline C + A\overline {\,B} \,C$$
3
GATE ECE 2004
MCQ (Single Correct Answer)
+2
-0.6
A Boolean function 'f' of two variables x and y is defined as follows: f(0,0)=f(0,1)=f(1,1)=1;f(1,0)=0 Assuming complements of x and y are not available, a minimum cost solution for realizing F using only 2-input NOR gates and 2-input OR gates (each having unit cost) would have a total cost
A
1 unit
B
4 unit
C
3 unit
D
2 unit
4
GATE ECE 2003
MCQ (Single Correct Answer)
+2
-0.6
If the functions W, X, Y and Z are as follows

W= R+$$\overline P Q + \overline R$$ S
X = $$X = PQ\overline R \,\overline S + \overline P \,\overline Q \,\overline R \,\overline S + P\overline Q \,\overline R \,\overline S$$
Y = $$RS + \overline {OR + P\overline Q + \overline {PQ} }$$
Z = $$R + S + \overline {PQ + \overline {PQR} + P\overline {QS} }$$

A
W=Z, X=$$\overline Z$$
B
W=$$\overline Z$$, X=Y
C
W=Z, X=Y
D
W=Y= $$\overline Z$$
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