1964 AHSME Problem 40

A watch loses 2\tfrac{1}{2} minutes per day. It is set right at 1 P.M. on March 15. Let n be the positive correction, in minutes, to be added to the time shown by the watch at a given time. When the watch shows 9 A.M. on March 21, n equals:

\textbf{(A) }14\frac{14}{23}\qquad \textbf{(B) }14\frac{1}{14}\qquad \textbf{(C) }13\frac{101}{115}\qquad \textbf{(D) }13\frac{83}{115}\qquad \textbf{(E) }13\frac{13}{23}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.