• # question_answer ) To get an output Y = 1 from the circuit shown in the figure, the input must be A) A-0      B-1      C-0B) A-0      B-0      C-1C) A-1      B-0      C-1D) A-1      B-0      C-0

Gate I is OR gate, $Y'=A+B$ Gate II is and gate, $Y=Y'.C$ $\therefore$ $A=1,B=0,C=1$ will give $Y=1$ Hence, the correction option is (c).