smc

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 $-p<-q$, which directly contradicts $A$. Thus, $A$ is always false. If $p<0$ (which is possible but not guaranteed), we can divide the true statement $p>q$ 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 $\boxed{\text{None of these}}$ as the correct answer.