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(D) For a reaction,if the time taken for the concentration to become $1/4$ of its initial value $(t_{75\%})$ is twice the time taken to become $1/2$ o

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(D) For a reaction,if the time taken for the concentration to become $1/4$ of its initial value $(t_{75\%})$ is twice the time taken to become $1/2$ of its initial value $(t_{50\%})$,it indicates a first-order reaction with respect to $A$.Mathematically,for first-order: $t_{75\%} = 2 	imes t_{50\%}$.Regarding $B$,the graph of concentration $[B]$ versus time $t$ is a straight line with a negative slope and a positive intercept. This represents a zero-order reaction with respect to $B$,as $[B]_t = [B]_0 - kt$.Therefore,the overall order of the reaction $= 1 + 0 = 1$.

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