On a single diagram, sketch the graphs of $y = |2x - p|$ and $y = qx$ where the following conditions are satisfied, indicating the axial intercepts.

• $p$ and $q$ are constants, $p > 1$ and $q > 0$, and
• the graphs have only one point of intersection.

1. Determine the condition for exactly one intersection by analyzing the gradients.

The graph of $y = |2x - p|$ is V-shaped. Its right branch is defined by the line $y = 2x - p$ (for $x \ge \frac{p}{2}$) and has a gradient of $2$.
Since the line $y = qx$ passes through the origin $(0,0)$ and $q > 0$, it will naturally intersect the left branch $y = -2x + p$ exactly once.
To ensure there is only one point of intersection in total, $y = qx$ must not catch up to and intersect the right branch. This requires the line $y = qx$ to be parallel to or steeper than the right branch:
$q \ge 2$
Therefore, the least possible value for $q$ is $2$.

1. Find the $x$-coordinate of the single point of intersection.

Based on the graphs sketched earlier, the line $y = qx$ starts at $(0,0)$ and must intersect the left branch of the $V$-shaped modulus graph, which has the equation $y = -(2x - p) = p - 2x$.
Since it is given that there is only one point of intersection, we solve for $x$ on this branch:
$p - 2x = qx$
$qx + 2x = p$
$x(q + 2) = p$
$x = \frac{p}{q + 2}$

2. Determine the solution to the inequality using the intersection point.

The inequality $|2x - p| > qx$ is satisfied wherever the graph of $y = |2x - p|$ lies strictly above the line $y = qx$.
At $x = 0$, the modulus graph is at $y = p$ (where $p > 1$), while the line is at $y = 0$. Thus, the modulus graph starts out above the line.
It remains above the line until they meet at the single intersection point. Since there are no other intersections, the modulus graph is only above the line to the left of this point.
Therefore, the solution is:
$x < \frac{p}{q + 2}$