1
GATE ECE 1988
MCQ (Single Correct Answer)
+1
-0.3
For the identity AB+$$\overline A $$ C + BC= AB + $$\overline A $$ C, The dual form is
A
(a+b)($$\overline A + c$$)(b+c)=(a+b)($$\overline A $$+C)
B
($$\overline A + \overline B )(A + \overline C )(\overline B + \overline C ) = \overline {(A} \, + \overline B )(A + \overline C )$$
C
$$(A + B)(\overline A + C)(B + C) = (\overline A + \overline {B)} (A + \overline C )$$
D
$$\overline A \,\overline B + A\,\overline C + \overline B \,\overline C = \,\overline A \overline B + A\,\overline C $$
2
GATE ECE 1988
MCQ (Single Correct Answer)
+1
-0.3
The Boolean function A+BC is a reduced form of
A
AB+BC
B
(A+B).(A+C)
C
$$\overline A $$B+A$$\overline B $$ C
D
(A+C).B
3
GATE ECE 1988
MCQ (Single Correct Answer)
+1
-0.3
The minimum number of 2-input NAND gates required to implement the Boolean function Z=A$$\overline {B\,} $$C, assuming that A, B and C are available, is
A
two
B
three
C
five
D
six
4
GATE ECE 1988
MCQ (Single Correct Answer)
+1
-0.3
For the circuit shown below the output Fis given by GATE ECE 1988 Digital Circuits - Logic Gates Question 31 English
A
F=1
B
F=0
C
F=X
D
F= $$\overline X $$
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