Aptitude - Clock - Discussion

I can't get that 55 min. are gained in 60 min

In 60 minutes, the minute hand gains 55 minutes on the hour hand.

@Sharanya

Can you explain this briefly please ?

theta=30h - 11/2 m
theta=0;
30h=11/2 m
m=180/11;

55 min. are gained in 60 min, what it mean pleas advise ?

In the duration of 1 hour an hour needle rotates only by duration of 5 min e.g from 1 o'clock to 2 o'clock and in the same time min hand will rotate by 60 min so the gain is (60 - 5) = 55 min.

Relative speed of hour and minute hand in 1 rotation is=60-5=55.

Does this take into account that the hour hand moves while the minute hand moves? So, that after 15 minutes, the minute hand has moved 15/60 (=1/4) of the way around, and similarly, the hour hand has moved proportionally the same. In other words, it has moved 1/4 of the way to the next hour (4 o'clock).

So, the minute hand is moving 12 times as fast as the hour hand. Conversely, the hour hand is moving 1/12 the speed of the minute hand. I have only had a moment to think about this, but if you consider both hands to be moving, but at different rates, it would seem that the minute hand would pass the hour hand (between the 1 o'clock and 2 o'clock hours) , between and 1:05 and 25 seconds and 1:05 and 50 seconds. From 2 to 3, it would pass over the hour hand somewhere between 2:11 and 15 seconds and 2:11 and 40 seconds. Roughly.

However, I haven't given thought to how to relate the two positions of the two hands via a single relationship for the precise points of intersection.

Both the needles coincide after 3 O' clock. Number 3 is on clock is situated after 15 minutes. So answer must have 15 minutes or more in options, We have only one option that is D 16(4/11).

I have formula for clocks coincide.

= (5(H)+0)*(12/11) where H = 3.