Problem: A watch loses 221 minutes per day. It is set right at 1 P.M. on March 15. When the watch shows 9 A.M. on March 21, the positive correction to be added to the time shown by the watch, in minutes, equals:
Answer Choices:
A. 142314
B. 14141
C. 13115101
D. 1311583
E. 132313
Solution:
First we note that the number of minutes in a day is 24×60=1440, while the number of minutes as recorded by the watch in a day is 221 less or 143721, so that the correction factor for this watch is 1440/143721 or, more simply, 576/575.
Designate a minute recorded according to the watch as a "watch-minute" and a day recorded by the watch as a "watch-day." Since the time interval given is 565 watch-days, or 565×24×60 watch-minutes, we have 563×24×60+n=565×24×60×575576∴n=140⋅60(575576−1)=142314.