Calendar Questions

On carefully inspecting this question, one can understand that there are two days which are important and these are:

A. My Birthday.

B. The day when I am making this statement.

If you think for a while, you will understand that such statements can be made only around the year’s end. So, if my birthday is on 31 December, then I will be making this statement on 1 January.

I will further explain using the following example:

1. Consider that today is 01 January 2017.

2. Then, the day before yesterday was 30 December 2016 and according to the question I was 25 then.

3. Yesterday was 31 December 2016, which happens to be my birthday too (Woohoo!), and my age increases by one to become 26.

4. I will turn 27 on my birthday this year (31 December 2017).

5. I will turn 28 on my birthday next year (31 December 2018).

Now, if you read the question again, it will make more sense:

The day before yesterday(30 December 2016), I was 25 years old and next year(31 December 2018) I will be 28.

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 2024, Which is a leap year.

So, Add +28 to the given year (i.e 2024 + 28) = 2052

Therfore, The calendar of the year 2024 can be used again in the year 2052.

We will show that the number of odd days between last day of February and last day of October is zero. .

March April May June July Aug. Sept. Oct.

31 + 30 + 31 + 30 + 31 + 31 + 30 + 31

= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.

Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )

Counting of odd days:

 1600 years have 0 odd day. 300 years have 1 odd day.

 47 years = (11 leap years + 36 ordinary years)= [(11 x 2) + (36 x 1) ]odd days = 58 odd days = 2 odd days.

Jan  Feb  Mar  Apr  May  Jun  Jul  Aug.

31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days = (32 weeks + 3 days) = 3,

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.

Hence, the required day was 'Saturday'.

First,we count the number of odd days for the left over days in the given period.Here,given period is 18.3.1994 to 25.2.1995

Therefore, No. of Odd Days = 6 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 3 + 4 = 36 = 1 odd day

So, given day Friday + 1 = Saturday is the required result.

We already know that the calendar after a leap year repeats again after 28 years.

Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.

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