Calendar Practice Problems: Level 02

Sol : Option D
The year 2004 was a leap year. So, it had 2 odd days.
The day on Feb 8, 2004 must be 2 days before the day on Feb 8, 2005.
Hence, this day was Sunday

Sol : Option B
28 May, 2006 = (2005 years + time period from 1.1.2006 to 28.5.2006)
1600 years » 0 odd days
400 years » 0 odd days
5 years = (1 leap year + 4 ordinary years) = (1 x 2 + 4 x 1) ≡ 6 odd days
Jan. Feb. March April May
(31 + 28 + 31 + 30 + 28 ) = 148 days
  ∴ 148 days = (1 day + 21 weeks) ≡ 1 odd day. Total number of odd days = (0 + 0 + 6 + 1) = 7 ≡ 0 odd day. So, the given day is Sunday.

Sol : Option C
June 17, 1998 = (1997 years + time period from 1.1.1998 to 17.6.1998)
1600 years » 0 odd days
300 years » (5 x 3) » 1 odd day
97 years has 73 ordinary years + 24 leap years.
Number of odd days in 97 years (73 + 24 x 2) = 121 days = 2 odd days.
Jan. Feb. March April May June
(31 + 28 + 31 + 30 + 31 + 17) = 168 days
∴168 days = 24 weeks = 0 odd days
Total number of odd days = (0 + 1 + 2 + 0) = 3
So, the given day is Wednesday.

Sol : Option C
We know that in 1600 years, there will be 0 odd days. And in the next 300 years, there will be 1 odd day. From 1901 to 1978 we have 19 leap years and 59 non-leap years.
So, the total number of odd days up to 31st Dec. 1978 is 19 x 2 + 59 = 97. On dividing 97 by 7 we get 6 as the remainder, which is the total number of odd days in these years.
So, till 31st Dec. 1978 we have 1 + 6 = 7 odd days, which forms one complete week. Now, in 1979, we have 3 odd days in January, and 2 odd days in the month of February (up to 9th Feb). So, the total odd days are 3 + 2 = 5. Hence, 9th February 1979 was a Friday.

Sol : Option D
In this question, the reference point is May 10, 1997 and we need to find the number of odd days from May 10, 1997 up to Oct 10, 2001.
Now, from May 11, 1997 - May 10, 1998 = 1 odd day
May 11, 1998 - May 10, 1999 = 1 odd day
May 11, 1999 - May 10, 2000 = 2 odd days (2000 was leap year)
May 11, 2000 - May 10, 2001 = 1 odd day
Thus, the total number of odd days up to May 10, 2001 = 5.
The remaining 21 days of May will give 0 odd days. In June, we have 2 odd days; in July, 3 odd days; in August, 3 odd days; in September,2 odd days and up to 10th October, we have 3 odd days. Hence, total number of odd days = 18 i.e. 4 odd days.
Since, May 10, 1997 was a Monday, and then 4 days after Monday would be Friday. So, Oct 10, 2001 was Friday.

Sol : Option A
Firstly in terms of years, the year 1911 to 1912 would give us 2 odd days and 1913, 1914, 1915 would give 1, 1 and 1 odd day respectively.
Now shift the focus on months. If you move one month ahead i.e. from 11th April to 11th May, the month ending in between is April, which gives you 2 days. Now after that the month of May, June, July, and August gives you 3, 2, 3, and 3 odd days respectively.
With this you reach on 11th September 1915. After this, there are 6 more September days (from 11th to 17th September).
The total number of odd days is 2 + 1 + 1 + 1 + 2 + 3 + 2 + 3 + 3 + 6 = 24.
Subtracting 21 (3 full weeks) from this the odd number of days left is 3.
Adding three days to the day given i.e. Tuesday, the answer becomes Friday.

Sol : Option D
We have to find the day on 1st April, 2002.
1st April, 2002 = (2000 years + time period from 1.1.2002 to 1.4.2002)
1600 years » 0 odd days
400 years » 0 odd days
Jan. Feb. March April
(31 + 28 + 31 + 1)     = 91 days ≡ 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On April 1, 2002 it was Monday.
So in April 2002, Tuesday falls on 2nd, 9th, 16th and 23rd & 30th.

Sol : Option A
First, we count the number of odd days for the left over days in the given period.
Here, given period is from 15.8.2012 to 11.6.2013
Aug Sept Oct Nov Dec Jan Feb Mar Apr May Jun
16   30   31  30   31   31   28  31   30   31   11(left days)
Therefore, 2 + 2 + 3 + 2 + 3 + 3 + 0 + 3 + 2 + 3 + 4 (odd days) = 6 odd days
So, the given day Thursday + 6 = Wednesday.

Sol : Option B
29th Feb, 2016 = Monday => 28th Feb, 2012 = Sunday
28th Feb, 2017 = Tuesday (because 2016 is a leap year, there will be 2 odd days)
Therefore» Feb 28th 2018 (Wednesday), Feb 28th 2019 (Thursday), Feb 28th 2020 (Friday), Feb 29th 2020 (Saturday)
Or, Feb 29th to Feb 29th after 4 years, we have 5 odd days.
So, every subsequent birthday, would come after 5 odd days.
2020 birthday – 5 odd days
2024 birthday – 10 odd days = 3 odd days
2028 birthday – 8 odd days = 1 odd day
2032 birthday – 6 odd days
2036 birthday – 11 odd days = 4 odd days
2040 birthday – 9 odd days = 2 odd days
2044 birthday – 7 odd days = 0 odd days. So, after 28 years, his birthday would fall on Monday.
The next birthday on Monday would be in year 2072 (further 28 years later), the one after that would be in year 2100. But we are told that she lives upto year 2099.
So, there are 2 occurrences of his birthday falling on Monday – 2044 & 2072.

Sol : Option C
June, 1999 = (1998 years + time period from 1.1.1999 to 16.6.1998)
1600 years » 0 odd days
300 years » 1 odd days
98 years has 74 ordinary years + 24 leap years.
Number of odd days in 98 years (24 × 2 + 74) = 122 days = 3 odd days.
Jan. Feb. March. April. May. June.
(31 + 28 + 31 + 30 + 31 + 16) = 167 days
167 days = 23 weeks & 6 days = 6 odd days.
Total number of odd days = (0 + 1 + 3 + 6) = 3
So, the given day is Wednesday.