smc

Let $x^2+px+m=0$ have roots $a$ and $b$. Then $$x^2+px+m=(x-a)(x-b)=x^2-(a+b)x+ab,$$ so $p=-(a+b)$ and $m=ab$. Also, $x^2+mx+n=0$ has roots $2a$ and $2b$, so $$x^2+mx+n=(x-2a)(x-2b)=x^2-2(a+b)x+4ab,$$ and $m=-2(a+b)$ and $n=4ab$. Thus $\frac{n}{p}=\frac{4ab}{-(a+b)}=\frac{4m}{\frac{m}{2}}=\boxed{8}$. Indeed, consider the quadratics $x^2+8x+16=0,x^2+16x+64=0$.