jmc

Let $r_1$ and $r_2$ be the roots of $x^2+px+m=0.$ Since the roots of $x^2+mx+n=0$ are $2r_1$ and $2r_2,$ we have the following relationships: $$m=r_1{r}_2,n=4r_1{r}_2,p=-(r_1+r_2),\text{and}m=-2(r_1+r_2).$$ So $$n=4m,p=\frac{1}{2}m,\text{and}\frac{n}{p}=\frac{4m}{\frac{1}{2}m}=\boxed{8}.$$ Alternatively, the roots of $${\left(\frac{x}{2}\right)}^2+p\left(\frac{x}{2}\right)+m=0$$ are twice those of $x^2+px+m=0.$ Since the first equation is equivalent to $x^2+2px+4m=0,$ we have $$m=2p\text{and}n=4m,\text{so}\frac{n}{p}=\boxed{8}.$$