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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| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1964|num-b=31|num-a=Last Problem}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
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| + | {{MAA Notice}} | ||
Revision as of 22:20, 24 July 2019
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:
See Also
| 1964 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 31 |
Followed by Problem 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
