Aptitude - Clock - Discussion

How is come 5*5/11?

Please explain anyone.

Thanks for your solution @Shekhar.

Here, is some Equation:

Angle between hands = [(Hour Hand * 30) + (Minute hand * 1/2)] - Minute hand * 6.

(Hour Hand * 30) -> Because At 1 o'clock it will be 30 deg so at 7 it will be 150 deg.

Minute hand * 1/2 -> Hour hand moves 1/2 deg per min.

Minute hand * 6 -> Minute hand moves 6 deg per min.

180 = [7 * 30 + x/2] - 6x.

So, the answer is 5 5/1.

At 7o clock hour, the hand will be 210 degrees, at 5min minute hand will be 30 degrees, (same line).

So consider 5min 5/11 for hour hand and just 5/11min for minute hand.

For every 1min hour hand rotates by 0.5 degrees so for 60/11 min it rotates by 30/11 degree.

For every min minute hand rotates by 6 degrees for 5/11 min it rotates by 30/11 degree hence equal.

So at 7 past 5min 5/11 it will be 210+30/11 degree in hour hand and 30+30/11 for minute hand difference is 180 degree thus they are in same line.

Use formula.

(5A + or - 30) * 12/11min past A.

Where A = + whan A < 6 and - when a > 6.

According to the question A = 5.

Shortcut formula:

ANGLE = 30*H - 11/2M.

Together Means angle consider as zero.

Put into the formula.

0 = 30 * 7 - 11/2M.

11/2M = 210.

Imp together consider as zero.

Use this formula to find angle :

Angle=((11/2)*minutes )-30* hour.

Between 7 and 8 means that the time would be something around 7. So consider 7 hours and x minutes.
180 = (11* x )/2 - 30 * 7.
210 - 180 = 11*x /2.
30 = 11*x /2.
30*2 = 11x,
60 = 11x,
X = 5.5/11.
That is 7 hours and 5 5/11 min.

Trick is 60/11 *( time +or - 6).

So between 7 to 8. Time is smaller one is.7 now here 7 is big than 6 so formula become 60/11 * (7-6) which is the answer.

A=30H-5.5M.
Where A=180',
H=7,

Now,
180=30*7-5.5M,
30=5.5M,
M=300/55,
M=5*5/11.

Please solve how to find between 8 &9 o'clock as the same process?