jmc

Let $m$ represent the total number of muffins in the original batch. Hence, each tray contains $\frac{m}{6}$ muffins and $5$ croissants, for a total of $\frac{m}{6}+5$ baked goods. We know that this value is at least $20$, hence we have the inequality $$\frac{m}{6}+5\ge 20.$$To solve, we first subtract $5$ from both sides of the inequality, yielding $\frac{m}{6}\ge 15$, and then multiply both sides by $6$ to get $m\ge 90.$

Therefore the original batch contained at least $\boxed{90}$ muffins.