Asked by ujjval
Q:20
A clock was 4
minutes slow at noon on Friday and 4 minutes 36 seconds fast on
the following Friday at 4 P.M. If the clock gains uniformly, when
does it show the correct time?
A:8 P.M. on Monday (after
second Friday)
B:8 P.M. on Monday
(between two Fridays)
C:10 P.M. on Monday
(after second Friday)
D:8 A.M. on Monday
(between two Fridays)
A clock was 4
minutes slow at noon on Friday and 4 minutes 36 seconds fast on
the following Friday at 4 P.M. If the clock gains uniformly, when
does it show the correct time?
A:8 P.M. on Monday (after
second Friday)
B:8 P.M. on Monday
(between two Fridays)
C:10 P.M. on Monday
(after second Friday)
D:8 A.M. on Monday
(between two Fridays)
Answers
Answered by
Steve
time elapsed was 7 days + 4 hours = 619200 seconds
during that time the amount the clock was off went from -240 to +276, so it gained 516 seconds.
So, if it was right on after x seconds, then
-240 + 516n/619200 = 0
x = 288000 seconds = 3 days 8:00:00
so, (B)
during that time the amount the clock was off went from -240 to +276, so it gained 516 seconds.
So, if it was right on after x seconds, then
-240 + 516n/619200 = 0
x = 288000 seconds = 3 days 8:00:00
so, (B)
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