smc

Since the arithmetic mean of the $50$ numbers is $38$, their sum must be $50\ast 38=1900$. After $45$ and $55$ are discarded, the sum decreases by $45+55=100$, so it must become $1900-100=1800$. But this means that the new mean of the remaining $50-2=48$ numbers must be $\frac{1800}{48}=37.5$. Thusly, our answer is $\boxed{37.5}$, and we are done.