This being a leap year none of the next 3 year is a leap year. So, the day of the week will be 3 days beyond Monday i.e., it will be Thursday.
1600 years of 0 odd days and 30 years have 1 odd day. 49 years contain 12 leap years and 37 ordinary years and therefore (24 + 37) odd days i.e., 5 odd days i.e., 1949 years contain (0 + 1 + 5) = 6 odd days. 26 day of January contain 5 days.Total odd days = (6 + 5) = 11 or 4 days. So, the days was Thursday
1992 being a leap year, it has 2 odd days. So, the first day of the year 1993 will be two days beyond Wednesday i.e., it will be Friday.
The year 1979 being an ordinary year, it has 1 odd day. So, the day on 12th January, 1979
But January 12, 1980 being Saturday
? January 12, 1979 was Friday.
Counting the number of days after 3rd November, 1987 we have Nov., Dec., Jan., Feb., March., April.
So days = 27 + 31 + 31 + 29 + 31 + 4 = 153 days containing 6 odd days on 4th April, 1988.
So, the day was Tuesday.
As per the given above question ,we can say that
Odd number of days from September 6, 1970 to September 6, 1981 = 14 days
Hence, the Sunday will be on September 6, 1981.
Clock will make right angle at (5n + 15) x 12/11 min past n.
Given that, n = 3
? Clock will make right angle at ( 5 x 3 + 15 ) x 12/11 min past 3 . =30 x 12 / 11 min past 3.
= 328/11 min
So required time = 3 H 328/11 min
Angle traced by hour hand in 12 h = 360°
Angle traced by hour hand in 25/4 h = (360/12) x (25/4)° = 187.5°
Angle traced by minute hand in 60 min = 360°
Angle traced by it in 15 min = (360/60) x 15° = 90°
? Required angle = (187.5° - 90°) = 97.5°
Since, third Thursday is on 16th. So, for calculation of last day of the month, we should calculate the odd number of days, here total days are 15 till 31.
? Odd days = 15/7 = 1
? Day on last date of month = friday
That will be 5th Friday.
Angle of 360° is covered by hour hand in 12 h .
So, angle of 135° is covered by hour hand in 12/360° x 135° = 4.5 h
Thus, required time = (3 + 4.5)h = 7.5 h = 7 : 30
There are 11 intervals when the clock strikes 12.
Time taken for 11 intervals = 48 s
? Time taken for 1 interval = (48/11)s
In order to strike 3, there are 2 intervals, for which the time taken = (48/11) x 2) = 88/11 s
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