An analog watch gains 3 minutes every hour. If it is set right at 11 a.m. on February 21st, 2012 when will the hour hand of this defective watch and a correct watch be at the same position ?
My attempt :-
My understanding regarding the gaining of time is as follows :-
Correct watch----Defective watch
11 am--------------11 am
12 pm--------------12:03 pm
1 pm--------------1:06 pm
2 pm--------------2:09 pm
and so on
I assumed that t hrs after 11 a.m. in the normal clock , both the clocks show the same time
so t hrs in normal clock will be equivalent to $t+\frac{3t}{60}$ hrs
I am stuck at this point on what further equation to make to solve for t
Alternative method :- Why cant we do as follows as well :- for every 60 minutes elapsed on normal clock, time elapsed on defective clock= 63 minutes, so after LCM(60,63) minutes = 1260 minutes , both should show the same time ? why is this not acceptable ?
Edit :- if possible please explain me using the equations only from what I have made, how shall I find "t" from there ?
I have no clue what the official solution has done by simply calculating :-
$\frac{12*60 min}{3 min/hr}$=240 hours= 10 days
The actual position of the hour hand is running away from the correct position at a rate of $3$ minutes per hour. The actual position needs to do a complete "lap" of $12\cdot 60$ minutes on the face of the watch, so that's the total "distance" it needs to cover, where distance is interpreted as minutes on the watch relative to the correct hour hand. So now the official solution is just using the formula $$\text{time}=\frac{\text{distance}}{\text{rate}} = \frac{12\cdot 60\text{ minutes on clock}}{3\text{ (minutes on clock)/hour}} = 240 \text{ hours}\text{.}$$