find the time (to the nearest second) between 10:00 and 11:00 when the hours and the minutes hands of the clock are symmetric about the axis passing through 12:00 and 6:00
i forgot the options given!
1divide the clock into 3600*12=43200 divisions.
it is clear the seconds hand is between 1:00 and 2:00
the hour hand moves 1 division per second
the minute hand moves 720 divisions per 60 seconds ie 12 divisions per second
let the hour hand move x divisions starting from 11:00
by then the minute hand would have moved 12x divisions starting from 12:00
so it is clear x=7200-12x
or x=553.846 ie 9 minutes 23 seconds
so the hands would be symmetric at 11:09:23
1sorry calculation mistake,
x=553.846 ie 9 min 13 seconds and 84.6 milliseconds :)
hands would be symmetric at 11:09:13
49it must be 11:09:14 dear Fibonacci. Let us see how......
let at 10:00 be the initial angular position of the hour and minute clock....
now, total clock=360°,
so, 60 small divisions=360°
so, 1 small div=6°
So, angular dist of hour hand from 12=(2*5*6)°=60°
& angular dist of mins hand from 12=0°
now, let the minute hand travel an angle θ before the required situation of symmetry is reached........
for 1 round(60 small div) of min hand the hr hand moves by 5 divs.
so, angle moved by hr hand for θangle of min hand=θ/12
by the problem,
60°-θ/12=θ,
or, 720-θ=12θ
or, 720=13θ,
or, θ=720/13=55.384°
now, for minute hand,
1 minute=6°
1 second=0.1°
thus 55.384°=54°+1.384°=9min13.84sec
=9min14sec(correct to nearest second)
for the previous solution it was wrong to write 13sec84.6millisecond
instead it should have been 84.6 centi second.
1Did everyone miss this one : find the nearest time BETWEEN 10:00 and 11:00 and all are giving the answer 11:09:14 ????? Is it possible ?
1nice observation Aveek :) the time should be 10:09:14 as correctly pointed out by subhomoy