1964 AHSME Problems/Problem 40
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
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