1964 AHSME Problems/Problem 40
Revision as of 18:44, 5 March 2014 by Bobthesmartypants (talk | contribs) (Created page with "==Problem== A watch loses <math>2\frac{1}{2}</math> minutes per day. It is set right at <math>1</math> P.M. on March 15. Let <math>n</math> be the positive correction, in minute...")
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 te time shown by the watch at a given time. When the watch shows
A.M. on March 21,
equals: