1. Use the information about reactant A

We are told that for reactant $A$:

• time to become $\dfrac{1}{2}$ of initial concentration $= t_{1/2}$
• time to become $\dfrac{1}{4}$ of initial concentration $= 2t_{1/2}$

Now, if a reaction is first order in $A$, then: $[A]=[A]_0 e^{-kt}$ so the half-life is constant: $t_{1/2}=\frac{\ln 2}{k}$

Then going from $[A]_0$ to $\dfrac{[A]_0}{4}$ means two half-lives, so $t_{1/4}=2t_{1/2}$ which matches the given condition.

Hence, the reaction is first order with respect to $A$.

So, order in $A = 1$.

2. Use the information about reactant B

It is given that when concentration of $B$ is plotted against time, the graph is a straight line with negative slope and positive intercept.

A straight-line decrease of concentration with time means: $[B]=[B]_0-kt$ This is the integrated rate law for a zero-order dependence.

Hence, the reaction is zero order with respect to $B$.

So, order in $B = 0$.

3. Find the overall order

Overall order $=$ sum of individual orders: $n = 1 + 0 = 1$

4. Final answer

The overall order of the reaction is: $\boxed{1}$

5. Comparison with stored correct answer

Stored correct answer = $1$

Our derived answer also = $1$

So, the answer agrees with the stored correct answer.