Aptitude - Clock - Discussion

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.

Here, min space refers to the displacement of the minute hand, w.r.t y o' clock, in forms of mins(minute hand mins) where y = 3.

So min space = 15(5, 10, 15 - 1, 2, 3)

Visualize relative speed. There are two trains. One is goods train 15 km away travelling at 5/60 km per min.

And there is another train which is travelling at 1 km per minute.

Relative speed = (1-5/60)kmph.

= 55/60 = 11/12.

time = distance/speed = 15/(11/12)min = 180/11 min = 16 + 4/11 min.

55 min are gained in 60 min means;

Take an example minute hand and hour hand at 12 o clock i.e. both coincide.

Now when hour hand moves by 1 hour it has covered only 5 degrees from the start i.e.12 o clock.

However, the minute hand covers 60 degrees from the start i.e. 12o clock.

So in 60 minutes minute hand gains 55 minutes on hour hand (i.e. 60 - 5 = 55).

@Kajal.

When minute hand covers 60 minutes that is passed by hour hand only 5 minutes space which alternatively we can say 1 hour for hour hand. But here among space covered by minute hand i.e. 60 minutes. Hour hand expenses only 5 minute equals i.e. 1 hour. So gained by minute hand is 55 minutes.

No, The needles only move after every second. With time unit smaller than one second, all needles stay where they are. Since 16 (4/11) minutes = 16*60 (4/11) (in second) is not a natural number, So, it cannot be the answer.

Let M for a minute, T1 for the starting hour and Y for a nangle.

Then the formula be M= 2/11(T1*30- 0r + Y).
2/11(3*30+-0) because coincide means angle is zero.
2*90/11.
180/11.
16 4/11 answer.

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.

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).

Its very easy I am n. Vinodh kumar the formulae is x (x/11). So here x=3. We have to multiply with 5.
So 3*5=15.
Now x is 15 so as per formulae 15 (15/11) =16 (4/11).

In that formula,

M=2/11*(30*H +- degree) 2/11(3*30) =180/11 = 16 4/11 ans.

Note coincide means 0 degree.