Given year 2003, when divided by 4 leaves a remainder of 3.
NOTE: When remainder is 3, 11 is added to the given year to get the result.
So, 2003 + 11 = 2014
Given that 2.12.91 is the first Tuesday
Hence we can assume that 4.12.91 is the first Thursday
If we add 7 days to 4.12.91, we will get second Thursday
If we add 14 days to 4.12.91, we will get third Thursday
If we add 21 days to 4.12.91, we will get fourth Thursday
=> fourth Thursday = (4.12.91 + 21 days) = 25.12.91.
We cannot find out the answer because the number of days of the current month is not given.
NOTE :
Repetition of leap year ===> Add +28 to the Given Year.
Repetition of non leap year
Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.
Step 2: Add +6 to the Given Year.
Solution :
Given Year is 1897, Which is a non leap year.
Step 1 : Add +11 to the given year (i.e 1897 + 11) = 1908, Which is a leap year.
Step 2 : Add +6 to the given year (i.e 1897 + 6) = 1903
Therfore, The calendar of the year 1897 can be used again in the year 1903
As month begins on Saturday, so 2nd, 9th, 16th, 23rd, 30th days will be Sundays. While 8th and 22nd days are second Saturdays. Thus, there are 7 holidays in all.
Hence, no. of working days = 30 ? 7 =23
From March 6th to November 11th,
there are 250 days and
Thus 250 x 24 hours
So from Mar 6th 9AM to Nov 11th 8PM, there will be 250 x 24 + 11 hours = 6011 hrs.
Days in 4 years =>
Let the first year is Normal year i.e, its not Leap year. A Leap Years occurs once for every 4 years.
4 years => 365 + 365 + 365 + 366(Leap year)
4 years => 730 + 731 = 1461
Therefore, Number of Days in 4 Years = 1461 Days.
Given Today is 20th January 2017, Sunday
In january, we have 31 days
February - 28 days (Non leap year)
March - 31 days
April - 30 days
=> Remaining days => 31 - 20 = 11 in Jan
+ 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days
More 20 days to complete 90 days => upto 20th April
Therefore, after 90 days from today i.e, 20th Jan 2017 is 20th Apr 2017.
Now, the day of the week will be
90/7 => Remainder '6'
As the day starts with '0' on sunday
6 => Saturday.
Required day is 20th April, Saturday.
NOTE: When a century year leaves a remiander 0, when divided by 400 then it is a leap year (366 days).
So, 1200 has 366 days.
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 24th Nov, 2007 will be 1 day beyond the day on 24th Nov, 2006.
But, 24th Nov, 2007 is Thursday.
24 - 10 = 14 days.
Therefore, 2 weeks ago it is same day.
Thus, 10th Nov, 2006 is one day before 10th Nov, 2007 i.e. it is Wednesday.
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