Question
A watch which loses a half a minute ever hour was set to read the correct time at 0445hr on Monday. Determine in twelve hour system the time the watch will show on Friday at 1845hr the same week.
Answers
JAC OB254
1hr>30sec
from mon 0445hr to fri 1845hr is 4 days 14 hours
4×24+14=110 hours
1hr=30sec
110hrs= 110×30
=3300sec
3300/3600=11/12hr
11/12×60=55 min
1845hr-55min
=1750hrs
=5:50p.m
from mon 0445hr to fri 1845hr is 4 days 14 hours
4×24+14=110 hours
1hr=30sec
110hrs= 110×30
=3300sec
3300/3600=11/12hr
11/12×60=55 min
1845hr-55min
=1750hrs
=5:50p.m