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:
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
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