Kinetics of a simultaneous parallel radioactive decay Question:

A radioactive isotope, A undergoes simultaneous decay to different nuclei as:
\begin{array}{cc}
\ce{A-&gt;P}&amp;\,(t_{1/2}=9\ \mathrm h)\\
\ce{A-&gt;Q}&amp;\,(t_{1/2}=4.5\ \mathrm h)
\end{array}
Assuming that initially neither P nor Q was present, after how many hours, amount of Q will be just double to the amount of A remaining?

My solving:
I got the $t_{1/2}$ for $A$ as $T=\dfrac{T_1T_2}{T_1+T_2}\ =\pu{ 3 h}$
Now from Rutherford-soddy law the number of atoms left after $n$ half lives are,
$$N=N_0\left(\frac{1}{2}\right)^n$$
and number of half lives $n=\dfrac{\text{Total time of Decay}}{\text{effective half life}}$
For isotope $A$, $\dfrac{N_A}{N_{0_A}}=\left(\dfrac{1}{2}\right)^{n}=\left(\dfrac{1}{2}\right)^{T/3}$ where $T$ is the common time of decay for $\mathrm A$ and $\mathrm Q$
For $\mathrm Q$ I had to assume that $\mathrm Q$ is also undergoing radioactive decay to form a relation between $\mathrm A$ and $\mathrm Q$
Now using the same formula as above $\dfrac{N_\mathrm Q}{N_{0_\mathrm Q}}=\left(\dfrac{1}{2}\right)^{T/4.5}=\left (\dfrac{1}{2}\right )^{2T/9}$
Now applying the condition in the question, $$\left (\frac{1}{2}\right)^{2T/9}=2\cdot \left (\dfrac{1}{2}\right)^{T/3} $$
Raising by power $9$ on both the sides,
$$\left (\dfrac{1}{2}\right )^{2T}=2^9\cdot \left (\frac{1}{2}\right )^{3T}$$
which implies,
$$ 2^{-2T}=2^{9-3T}$$
Therefore,
$$9-3T=-2T$$
$$
{\bbox[10px, border:2px solid red]{
 T=9\ \mathrm h.
}}
$$
Although I have got an answer and it satisfies the condition I still think it's wrong because of the assumption made for $\mathrm Q$ and something else needed to be done to make the relation between $\mathrm A$ and $\mathrm Q$ Moreover, it's given in the question that initially neither $\mathrm P$ nor $\mathrm Q$ was present and I guess the assumption violates that.
The question has already been solved by Yashwini and the answer given is correct.$^2$ A more intuitive and specific to question explanation would follow here.
Now, the two reactions given are:
\begin{array}{cc}
\require{cancel}
\ce{A -&gt; P} &amp;(t_{1/2} = 9\,\mathrm h) \\
\ce{A -&gt; Q} &amp;(t_{1/2} = 4.5\,\mathrm h) \\
\end{array}
Now using the rate law, we get,
\begin{align}
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&amp;=k_\mathrm P [A] \tag{1} \\
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&amp;=k_\mathrm Q [A] \tag{2} \\
\end{align}
The rate constant for a first order reaction having a half life of $t_{1/2}$ is defined as:
$$k=\frac{\ln 2}{t_{1/2}} \tag{3}$$
Now, substituting the given values of $t_{1/2}$ into the equations, we get $2k_\mathrm P = k_\mathrm Q$ (since $k\, \alpha \frac{1}{t_{1/2}})$
Now, intuitively since both reactions take place together, it would mean that for every one mole of P formed, two moles of Q forms. Therefore, for every mole of P formed, three moles of A react (since one mole is required for each mole of P and Q).
Now, we add the rate laws ($1$) and $(2)$, since the reactions take place simultaneously, to get:
$$-\frac{\mathrm{d}[A]}{\mathrm{d}t}=(k_\mathrm P +k_\mathrm Q) [A] \tag{4} $$
Now, since using the relation between $k_\mathrm{P}$ and $k_\mathrm{Q}$, we get $k_\mathrm{P} + k_\mathrm{Q} = 3k_\mathrm{P}$
Therefore using the integrated rate law for a first order reaction on equation $(4)$, we get:
$$A=A_0e^{-3k_\mathrm Pt} $$
Now, the amount of $A$ used here would be $A_0 -A$, and we get that value to be:
$$A_\text{used}=A_0\left(1-e^{-3k_\mathrm Pt}\right)$$
Now, as we have previously noted, for every three moles of A used, two moles  Q are formed. This means that the amount of Q now in the mixture would be two thirds of $A_\text{used}$. Therefore the amount of Q would be:
$$Q=\frac{2A_0\left(1-e^{-3k_\mathrm Pt}\right)}{3}$$
Now, we are given the condition, $Q = 2A$, substituting values of $Q$ and $A$ into the given relation we get:
$$\begin{align}
\frac{\cancel{2A_0}\left(1-e^{-3k_\mathrm Pt}\right)}{3} &amp;= \cancel{2A_0}\left(e^{-3k_\mathrm Pt}\right) \\
\implies 1 -e^{-3k_\mathrm Pt} &amp;= 3e^{-3k_\mathrm Pt} \\
\implies 4e^{-3k_\mathrm Pt} &amp;= 1
\end{align}$$
Solving for $t$, we get:
\begin{align}
3k_\mathrm Pt&amp;=2\ln 2 \\ \\
t&amp;=\frac{2\ln 2}{3k_\mathrm P}\\
\end{align}
Now, using equation $(3)$, we get the rate constant $k_\mathrm P$ to be $\frac{\ln 2}{9}$. Substituting this value into the expression for time, we get:
$$t=\frac{2 \cancel{\ln 2}}{\cancel{3} \frac{\cancel{\ln 2}}{\cancelto{3}{9}}}$$
Therefore, time taken for this condition to happen is:
$$t=2\times 3 = 6\ \mathrm h$$