$\begin{align}  & \frac{\tfrac{1}{10}\times \tfrac{2}{3}+\tfrac{1}{4}}{\tfrac{1}{2}\div \tfrac{3}{5}-\tfrac{1}{4}}=\frac{(\tfrac{1}{10}\times \tfrac{2}{3})+\tfrac{1}{4}}{(\tfrac{1}{2}\times \tfrac{5}{3})-\tfrac{1}{4}}=\frac{\tfrac{1}{15}+\tfrac{1}{4}}{\tfrac{5}{6}-\tfrac{1}{4}} \\ & =\frac{\tfrac{4+15}{60}}{\tfrac{10-3}{12}}=\frac{\tfrac{19}{60}}{\tfrac{7}{12}}=\frac{19}{60}\times \frac{12}{7}=\frac{19}{35} \\\end{align}$