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(C) The logic circuit consists of an $OR$ gate,a $NAND$ gate,and a $NOR$ gate. Let the inputs be $X$ and $Y$. $1$. The upper branch has an $OR$ gate w

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$X = 1, Y = 0$

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(C) The logic circuit consists of an $OR$ gate,a $NAND$ gate,and a $NOR$ gate. Let the inputs be $X$ and $Y$. $1$. The upper branch has an $OR$ gate with inputs $\overline{X}$ and $Y$,so $P = \overline{X} + Y$. $2$. The lower branch has a $NAND$ gate with inputs $X$ and $\overline{Y}$,so $Q = \overline{X \cdot \overline{Y}} = \overline{X} + Y$. $3$. The final output $R$ is from a $NOR$ gate with inputs $P$ and $Q$,so $R = \overline{P + Q}$. Truth Table:  $X$$Y$$P = \overline{X} + Y$$Q = \overline{X} + Y$$R = \overline{P + Q}$ $0$$0$$1$$1$$0$ $0$$1$$1$$1$$0$ $1$$0$$0$$1$$0$ $1$$1$$1$$1$$0$  Wait,re-evaluating the circuit: The lower $NAND$ gate has inputs $X$ and $\overline{Y}$,so $Q = \overline{X \cdot \overline{Y}} = \overline{X} + Y$. If $X=1, Y=0$: $P = \overline{1} + 0 = 0 + 0 = 0$. $Q = \overline{1 \cdot \overline{0}} = \overline{1 \cdot 1} = 0$. Then $R = \overline{P + Q} = \overline{0 + 0} = 1$. Thus,for $R=1$,we must have $X=1, Y=0$.

List-$I$	List-$II$
$A$. Butane $\to$ Isobutane	$a$. Cracking
$B$. Butane $\to$ Lower hydrocarbons	$b$. Isomerisation
$C$. $n$-Heptane $\to$ Toluene	$c$. Reed reaction
$D$. Propane $\to$ $CH_3CH_2CH_2SO_2Cl$	$d$. Aromatisation

The figure shows a system of logic gates. From the study of the truth table,it can be found that to produce a high output $(1)$ at $R$,we must have

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