Aptitude - Clock - Discussion

Hour hand moves = 0.5 degrees/min.
Minute hand moves = 6 degrees/min.
The angle between the hour hand and the minute hand should be = 90 degrees.
Let the minutes at which hour and minute hand makes 90 degrees = x.

Therefore,
Minute hand angle - hour hand angle = 90 degrees.
6 degree/min * x - (5*60 mins + x) * 0.5 degree/min = 90 degrees.
6 degree/min * x - 150 degrees - x * 0.5 degree/min = 90 degrees.
(6 degree/min * x - x * 0.5 degree/min) = 90 degrees + 150 degrees.
(6 degree/min * x - x * 0.5 degree/min) = 90 degrees + 150 degrees.
5.5 degree/min * x = 240 degrees.
x = 240 mins / 5.5
x = 43.6363 mins.

Hence, at time 5 hour and 43.6363 mins hour hand and minute hand makes 90 degrees angle.

Since for 1 hour, the minute's arm move 360 degrees. And for each hour arm move 30 degrees.

Hence for every 1 degree move the minute's arm move 12 degrees.

When the time is 5 : 30 the angle between the min arm and hour arm is 15 degree.

Our target to find a time when the angle is 90 degree.

For this, we solve the equation :12x - x + 15 = 90.

From here we get x = 75/11.

30 degree = 1 hour.

75/11 degree = 1/30 * 75/11 = 5/22hours = 300/22min.

Hence final time is 30 + 300/22 = 480/22 min from 5 o'clock. (we add 30 because we have considered the arms move from 5:30).

@Bala.

Bala it is very simple.

For right angle we need 90 degree between hour hand and minute hand.

So here we are looking for that condition.

Now look at hour hand on 5 o'clock, you'll get that minute hand should be at 5. 40 to form right angle.

(90 degree between hour and minute hand).

Now move minute hand to 25 min, now minute hand and hour hand are together (forms 0 degree).

Now simply move 15 min more to form right angle.

Don't think why it is "25+15", think we need 40 to form right angle, consider on logic instead of calculation.

5’o clock, space between the hour and min 25 min.
1 hr hand =30 degree.

5:30 and 6 will hands (30 min) of the clock 15 degrees.
So now total 5’o + 5:30 = 25 + 15 = 40min.

Now we know that 55 min space is gained in 60 min.
Then 40 min space are gained =(55/60) * 40 min.
43(7/11) min,
So, the Required time 43(7/11) min .past 5.

There is a formula for finding time nd degree
ie angle=(11/2)*min-30*hr
in this problem the hr nd min hands should be in right angle that means angle=90 then we can find min
90=(11/2)*min-30*5
min=(90+150)*(2/11)
=43 7/11 ie 5hrs 43 7/11min

@Chandra Shekar.

At 12 pm both hour and second hands are in the same line. If it is 1 pm the hour hand is at 1 and minute hand at 12. So difference is 5 mins.

We can say, reach 1 from 12 (hour hand) needs 60 mins but space difference is 5 mins. So in 60 mins 55 mins space covered.

At 5.30, Angle between hands is 15 degree and 75' more required to make a right angle. Relative angle made by hands is 11/2 ' in a minute.

We can assume in x min they cover remaining 75.
So 11/2 * x = 75.
x = 150/11 min.
So at 30 + 150/11 min, they make a right angle.

If we consider the basic solution, and start at 5:30 then min. space between hands at that time is 5 spaces and we need 10 more spaces to get right angle.

So simply (60*10)/55 min past 5:30 i.e. 30 min.+(60*10)/55 min. past 5.

What is wrong in the solution?

A right angle is an angle that bisects the angle formed by two halves of a straight line is called right angle and the angle between the two halves is 90 degrees. In the given problem handles of clock are in right angle so we take 90 degrees angle.

If the hours are less than 6 or the angle less than 180 then use...theta =11/2(min)-30(hrs).

If the hours are greater than 6 or the angle greater than or equal to 180 then use...theta = 30(hrs)-11/2(min).

It's simple and helps a lot. Good luck.