If $r>0$, then for all $p$ and $q$ such that $pq\ne 0$ and $pr>qr$, we have
$\text{(A) } -p>-q\quad \text{(B) } -p>q\quad \text{(C) } 1>-q/p\quad \text{(D) } 1<q/p\quad \text{(E)  None of these}$
If $r>0$ and $pr>qr$, we can divide by the positive number $r$ and not change the inequality direction to get $p>q$.  Multiplying by $-1$ (and flipping the inequality sign because we're multiplying by a negative number) leads to $-p q$ by $p$ to get $1 > \frac{q}{p}$.  This contradicts $D$.  Thus, $D$ is sometimes false, which is bad enough to be eliminated.
If $(p, q, r) = (3, 2, 1)$, then the condition that $pr > qr$ is satisfied.  However, $-p = -3$ and $q = 2$, so $-p > q$ is false for at least this case, eliminating $B$.
If $(p, q, r) = (2, -3, 1)$, then $pr > qr$ is also satisfied.  However, $-\frac{q}{p} = 1.5$, so $1 > -\frac{q}{p}$ is false, eliminating $C$.
All four options do not follow from the premises, leading to $\fbox{E}$ as the correct answer.