Consider the following reaction
\( A + B \rightarrow C \quad \) [NCERT : PL-72|NV, April 8, 2024 (I)] The time taken for A to become \( 1 / 4^{\text {th }} \) of its initial concentration is twice the time taken to become \( 1 / 2 \) of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis.
The overall order of the reaction is \( \qquad \)
