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== | ||
− | {{AHSME | + | {{AHSME box|year=1964|num-b=39|after=Last Problem}} |
+ | |||
+ | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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Latest revision as of 20:25, 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 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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