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.