jmc

Let the number of quarters in the first, second, third, and fourth piles be $a$, $b$, $c$, and $d$, respectively. We have the equations $$\begin{array}{rl}a+b+c+d& =20\\ a& =b-3\\ b& =c+2\\ d& =2b\end{array}\text{\hspace{1em}(1)(2)(3)(4)}$$ We want to find the value of $d$. We will express each of $a$, $b$, and $c$ in terms of $d$ and then substitute these equations into Equation (1) to find the value of $d$. From Equation (4), we have $b=d/2$. From Equation (3), we have $c=b-2$. Since $b=d/2$, we can rewrite Equation (3) as $c=d/2-2$. We can substitute $b=d/2$ into Equation (2) to get $a=d/2-3$. Substituting $b=d/2$, $c=d/2-2$, and $a=d/2-3$ into Equation (1) to eliminate $a$, $b$, and $c$, we get $(d/2-3)+d/2+(d/2-2)+d=20$, so $d=10$. Thus, there are $\boxed{10}$ quarters in the fourth pile.