# Difference between revisions of "1964 AHSME Problems/Problem 40"

(Created page with "==Problem== A watch loses <math>2\frac{1}{2}</math> minutes per day. It is set right at <math>1</math> P.M. on March 15. Let <math>n</math> be the positive correction, in minute...") |
m (Fixed a typo) |
||

Line 1: | Line 1: | ||

==Problem== | ==Problem== | ||

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

<math>\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}</math> | <math>\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}</math> |

## Revision as of 18:07, 6 March 2014

## Problem

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