ANS:C - Explicit

No answer description is available. 

ANS:B - False

No answer description is available. Let's discuss.

ANS:C -

See the output = 1.

0 0 1 = 1 : A B C.

1 1 0 = 1 : A B C.

Answer: C.

ANS:A - 10111100

Option A = 10 11 11 00

A XOR B = 1 XOR 0 => 1
C XOR D = 1 XOR 1 => 0

E XOR F = 1 XOR 1 => 0
G XOR H = 0 XOR 0 => 0

(A XOR B) XOR (C XOR D) = 1 XOR 0 => 1 .....(1)
(E XOR F) XOR (G XOR H) = 0 XOR 0 => 0 .....(2)

(1) XOR (2) = 1 XOR 0 => 1.

Hence the Option A is correct.

ANS:A - (A)

First of all, "read" the question. They asks using NAND gate. So option c having OR gate is canceled out, option D having AND gate also cancel out, now there are A and B. We are to find first AB we see option A has AB. So, accurately answer is A.

ANS:D - 5

AB = 1.
C' = 1.
C'D = 1.
EF = 1.
C'D + EF=1.
AB(C'D+EF) = 5.

Totally 5 gates are used.

ANS:B - 2-input OR gate

Since it is pulses it corresponds to 0's and 1's.
Input A= 101010101
Input B= 111110111

Ouput x shows= 111110111, which corresponds to a characteristic of an OR Gate.

ANS:A - (A)

According to De-Morgan's Theorem.
(A'.B')' = (A+B).
So, ((A'.B') + (C+D+E))' ==> (A+B) + (C+D+E)'.

ANS:C -

1st AND gate input 00 and second AND gate input is 11 and the outputs are 0, 1 this is given to or gate, the output is 1 and is given to invert and output is 0. so answer C is correct.

ANS:C - Exclusive-OR gate

Exclusive OR means either of the Inputs(A or B) is high then the output (X) will be high otherwise X will be low. So waveform satisfies this condition. If both i/p high or both i/p are low then o/p will be low. So, one i/p should be low and other should be high or vice versa for being o/p high.