### In a quadrilateral $PQRS,$ if $PQ \parallel RS, \angle S = 2 \angle Q, PS = q$ and $RS = p.$ Find the length of the side $PQ.$

$p + q$

Step by Step Explanation:
1. Let us first draw the quadrilateral $PQRS$.
Let $\angle Q = x$
So, $\angle S = 2x$
Let us join $R$ to a point $E$ on the side $PQ$ such that $PERS$ is a parallelogram.
2. We know that opposite sides of a parallelogram are equal.
So, $\angle PSR = \angle PER = 2x \ldots (i)$
3. Also, \begin{align} & \angle PER + \angle REQ = 180^\circ &&[ \text{ Angles on a straight line } ] \\ \implies & \angle REQ = 180^ \circ - \angle PER \\ \implies & \angle REQ = 180^ \circ - 2x && [\text{From } (i)] \\ \end{align}
4. The sum of angles of a triangle is $180^ \circ.$
In $\triangle ERQ$ \begin{align} & \angle REQ + \angle EQR + \angle QRE = 180^ \circ \\ \implies & 180^\circ - 2x + x + \angle QRE = 180^\circ \\ \implies & \angle QRE = x \end{align}
5. In $\triangle ERQ,$ \begin{align} & \angle QRE = \angle EQR \\ \implies & ER = EQ \ldots (ii) && [\text{ Sides opposite to equal angles are equal. }] \\ \end{align}
6. We are given that $RS = p$ and $PS = q.$
As, opposite sides of a parallelogram are equal,
$RS = PE = p$
and $PS = ER = q$
$\implies EQ = q \ldots [\text{From } (ii)]$
7. We can see that $PQ = PE + EQ = p + q$
8. Thus, $PQ = p + q$