Calendar - aptitude test

ANSWER : Sunday

SOLUTION :

The year 1996 was a leap year. Hence, it adds 2 odd days. As 26th January, 1996 was a Friday, 26th January, 1997 was a Sunday.

ANSWER : 2007

SOLUTION :

The total number of odd days from 2001 onwards should be zero.
Now, as an ordinary year adds 1 odd day and a leap year adds 2 odd days, we have:
2001, 2002, 2003, 2005, 2006 – 1 odd day each
2004 – 2 odd days
Hence, at the end of 2006 total number of odd days = 7 or 0
Therefore, the calendar for the year 2001 is repeated in the year 2007.

ANSWER : Saturday

SOLUTION :

31st July, 1993 = (1992 years + period from 1st January, 1993 to 31st July, 1993)
1600 years have 0 odd days and 300 years have 1 odd day.
Now, the period from 1900 to 1992 have 69 ordinary years and 23 leap years
= (69*1 + 23*2) = 115 odd days = (16 weeks + 3 days) = 3 odd days.

January	February	March	April	May	June	July
31	28	31	30	31	30	31

= 212 days = (30 weeks + 2 days) = 2 odd days
Therefore, total number of odd days = 1 + 3 + 2 = 6 odd days.
Therefore, the required day was Saturday.

ANSWER : Saturday

SOLUTION :

100 years have 5 odd days. Hence the last day of 1st century is a Friday.
200 years have 10 odd days or 1 week + 3 odd days. Hence, the last day of the 2nd century is a Wednesday.
300 years have 15 odd days or 2 week + 1 odd day. Hence, the last day of the 3rd century is a Monday.
400 years have 0 odd days. Hence, the last day of the 4th century is a Sunday.

ANSWER : 2100

SOLUTION :

The two conditions that decide that a year is a leap year or not is:
• For a year to be a leap year, it should be divisible by 4.
• No century is a leap year unless it is divisible by 400.
Hence, the year 2100 is not a leap year as it is not divisible by 400.