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(D) The given circuit consists of an $OR$ gate $(G_1)$ followed by a $NAND$ gate $(G_2)$.Let the output of the $OR$ gate be $Y$. Then $Y = A + B$.The

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$1, 1$

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(D) The given circuit consists of an $OR$ gate $(G_1)$ followed by a $NAND$ gate $(G_2)$.Let the output of the $OR$ gate be $Y$. Then $Y = A + B$.The final output $D$ is the output of the $NAND$ gate with inputs $Y$ and $C$, so $D = \overline{Y \cdot C} = \overline{(A + B) \cdot C}$.Case $1$: $A = 0, B = 0, C = 0$.$Y = A + B = 0 + 0 = 0$.$D = \overline{Y \cdot C} = \overline{0 \cdot 0} = \overline{0} = 1$.Case $2$: $A = 1, B = 1, C = 0$.$Y = A + B = 1 + 1 = 1$.$D = \overline{Y \cdot C} = \overline{1 \cdot 0} = \overline{0} = 1$.Thus, the output states are $1, 1$.

$A$	$B$	$C$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$0$

For the given combination of logic gates, if the logic states of inputs $A, B, C$ are as follows: $A = B = C = 0$ and $A = B = 1, C = 0$, then the logic states of output $D$ are:

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$A$ $B$ $C$
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