Recently I've been going through an exercise of Chemical Kinetics and Nuclear Chemistry and right now I am stuck on one question that was previously asked in JEE(Advanced)-2013. The question says,
In the reaction,
$\ce{P + Q -> R + S}$
The time taken for $75\%$ reaction of $\ce{P}$ is twice that time taken for $50\%$ reaction of $\ce{P}$. The concentration of $\ce{Q}$ varies with reaction time as shown in the figure. What is the overall order of the reaction?
And here was my approach to the question,
In the question different data have been given for different reactants (i.e. $\ce{P}$ and $\ce{Q}$). For $\ce{P}$ it has been given that it take $\ce{P}$ double the time to react $75\%$ than that from $50\%$. So assuming the time taken for $50\%$ reaction to be $t$ and that of $75\%$ to be $2t$, we can easily find that order of reaction w.r.t P is $1$. As,
$$\frac{t_{\frac12}}{t_{\frac34}}=\frac{\log 2}{\log 4}=\frac12$$
Further w.r.t $Q$ there is a graph given which is something like in form $\ce{[Q]=[Q_o]-mt}$. So comparing with the equation of zero order reaction which is like $\ce{[A]=[A_o]-kt}$ (where $\ce{[A]}$ is the concentration of reactant and $k$ is rate constant), we can conclude that order of reaction w.r.t to $Q$ is $0$.
So I concluded that overall order of reaction will be
order of reaction of $P +$ order of reaction of $Q$
$\implies 1+0=1$
So was this the right approach from my side or there is any kind of error in any step?