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

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<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>
  
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==Solution==
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From March 15 <math>1</math> P.M. on the watch to March 21 <math>9</math> A.M. on the watch, the watch passed <math>20 + 5 \times 24 = 140</math> hours.
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Since <math>1</math> watch hour equals <math>\frac{24}{23 + \frac{57.5}{60}} = \frac{576}{575}</math> real hour, the difference between the watch time and the actual time passed is <math>140 \times \left( \frac{576}{575} - 1 \right) = \frac{28}{115}</math> hour <math>=14\frac{14}{23}</math> minutes.
  
 
==See Also==
 
==See Also==

Latest revision as of 20:25, 21 October 2021

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:

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

Solution

From March 15 $1$ P.M. on the watch to March 21 $9$ A.M. on the watch, the watch passed $20 + 5 \times 24 = 140$ hours.

Since $1$ watch hour equals $\frac{24}{23 + \frac{57.5}{60}} = \frac{576}{575}$ real hour, the difference between the watch time and the actual time passed is $140 \times \left( \frac{576}{575} - 1 \right) = \frac{28}{115}$ hour $=14\frac{14}{23}$ minutes.

See Also

1964 AHSME (ProblemsAnswer KeyResources)
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