1964 AHSME Problems/Problem 40: Difference between revisions

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Problem

A watch loses $2\frac{1}{2}$ 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:

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

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1964 AHSME Problems/Problem 40: Difference between revisions

Revision as of 22:22, 24 July 2019

Problem

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