basic calendar problems Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (a)

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

Thus, the year 2800 is a leap year.

Answer: (b)

1977 is an ordinary year.

We know that the calendar of an ordinary year repeats after 6 yrs or 11 yrs.

Let us check for the number of odd days in 6th and 11th years.

From the above table, Number of odd days from 18.09.1977 to 18.09.1983 = 7,

i.e., 0 odd days

It means that in 1983, 18th September would fall on Sunday.

From the above table, a number of odd days from 18.09.1977 to 18.09.1988 = 14

i.e., 0 odd days.

Now, it is clear that 2 marriage anniversaries would fall on Sunday in the next 15 yr.

Answer: (d)

There are months of 30, 31 and 28 days and last day of the month are Wednesday.

So, using 28 and 30 days, there are 4 Mondays.

Using 31 days, there are 5 Mondays

So, it cannot be specified.

Answer: (a)

Number of days left in the month after 16th = 31 - 16 = 15

Number of odd days = 15/7 = 2 weeks + 1 odd day

Therefore, Required day = Thursday + 1 odd day = Friday

As, 16th of the month is third Thursday, the day which is two weeks after this day is fifth Thursday.

So, one day after 5th Thursday is 5th Friday.

Answer: (d)

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd days.

Sum = 14 odd days = 0 odd day.

Therefore, Calendar for the year 2018 will be the same as for the year 2007.