Difference between revisions of "1964 AHSME Problems/Problem 40"
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Revision as of 22:21, 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
| 1983 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
