1
GATE ECE 2004
+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
2
GATE ECE 2003
+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= $$\overline Z$$
3
GATE ECE 1999
+2
-0.6
For a binary half-subtractor having two inputs A and B, the correct set of Logical expressions for the output D(=Aminus B) and X(=Borrow) are
A
D = AB + $$\overline A B,X = \overline A B$$
B
$$D = \,\overline A B + A\overline B + A\overline B , X= A\overline B$$
C
$$D = \overline A B + A\overline B ,X = \overline A B$$
D
$$D = AB + \overline A \,\overline {\,B} ,X = A\overline B$$
4
GATE ECE 1999
+2
-0.6
The minimized form of the logical expression ($$\overline A \,\overline B \,\overline C + B\overline C + \overline A B\overline C + \overline A BC + AB\overline C )$$
A
$$\overline A \overline C + B\overline C + \overline A B$$
B
$$A\overline C + \overline B C + \overline A B$$
C
$$\overline A C + \overline B C + \overline A B$$
D
$$A\overline C + \overline B C + A\overline B$$
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