jmc

Let $a$ be the number of quarters. We know that $$\begin{array}{r}a\equiv 3(\mathrm{mod}5)\\ a\equiv 5(\mathrm{mod}7)\end{array}$$ Congruence $(1)$ means that there exists a non-negative integer $n$ such that $a=3+5n$. Substituting this into $(b)$ gives $$3+5n\equiv 5(\mathrm{mod}7)⟹n\equiv 6(\mathrm{mod}7)$$ So $n$ has a lower bound of $6$. Then $n\ge 6⟹a=3+5n\ge 33$. $33$ happens to satisfy both congruences, so it is the smallest common solution. Since she has less than $10$ dollars worth of quarters, she has less than $40$ quarters. If $b$ is a common solution, subtracting $33$ from both sides of both congruences gives $$\begin{array}{r}b-33\equiv -30\equiv 0(\mathrm{mod}5)\\ b-33\equiv -28\equiv 0(\mathrm{mod}7)\end{array}$$ Since $\gcd (5,7)=1$, we have $b-33\equiv 0(\mathrm{mod}5\cdot 7)$, that is, $b\equiv 33(\mathrm{mod}35)$. The first couple of positive solutions of this are $33,68$. Thus, there exists only one positive solution less than $40$, which we found earlier to be $\boxed{33}$.