By which of the following a leap year must be divided

Here, n = 3 and (n + 1) = 4
? Two hands will coincide at 60n / 11 min past n
? 60 x 3 / 11 min past 3
? 180 /11 min past 3.
? 16⁴/Sub>11 min past 3

Total hours 6 : 14 am morning to evening 6 : 14 pm = 12 h
and total hours from 6 : 14 pm evening to 8 : 02 pm evening = 8 : 02 - 6 : 14 = 1 : 18 h i.e., 1 h and 48 min
? Total hours and minutes = 12 h + 1 h 48 min
= 13 h and 48 min

A leap year has (52 week + 2 days).
So, the number of odd days in a leap year is 2

Starting with 1988, we go on counting thye number of odd days till the sum is divisible by 7
Year 1988 odd days = 2
Year 1989 odd days = 1
Year 1990 odd days = 1
Year 1991 odd days = 1
Year 1992 odd days = 2

So total odd days = 7;
? Calendar for 1993 is the same as that of 1988.

No of days = Jan. + Feb. + March
30 + 28 + 15 = 73 days

Count the number of odd days from the year 2002 onwards to get the sum equals to 0 odd day.

Years 2002 2003 2004 2005 2006 2007
Odd days 1 1 2 1 1 1

Sum = 7 odd days = 0 odd day
Calender for the year 2008 is the same as that for the year 2002.

The century year which is completely divisible by 400, is a leap year. Thus, the year 2800 is a leap year.

Angle traced by hour hand in 12 h = 360°
Angle traced by hour hand in 5 h = (360/12) x 5° = 150°

2 h 20 min = 140 min
In every hour the minute hand move over 360°.
So, in every minute it will move 360°/60° = 6°
So, in 140 min the minute hand will move over = 6 x 140° = 840°

The actual time show is 8 h 15 min .