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> | ||
| + | ==Solution== | ||
| + | |||
| + | 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. | ||
| + | |||
| + | 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 19:25, 21 October 2021
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:
Solution
From March 15 P.M. on the watch to March 21 A.M. on the watch, the watch passed 
hours.
Since watch hour equals 
real hour, the difference between the watch time and the actual time passed is 
hour 
minutes.
See Also
| 1964 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 39 |
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| All AHSME Problems and Solutions | ||
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