Aptitude Chapter-4

Bodhak competitive bank exams calendars in this chapter we will discuss CALENDARS. We will see how o solve through equations and through shortcut methods.

The year is 365.2422 days approximately.

The common year consists of 365 days. Consequently, in every 4th year, there are 366 days.(leap year)

Leap years have an extra day. The extra day in the end of February.

 

 

competitive bank exams calendars

CALENDARS :

This module will have the following type of questions:

1: What was the day of the week on 15th August 1947?

2: The calendar of the year 2011 will be used again in the year?

3: The maximum gap between two successive leap years?

4: If 15th  August 2011 falls on Tuesday then 10th may 2012 falls on which day?

 

  • The time in which the earth travels around the sun is a solar year and is equal to 365 days 5 hrs. 48 minutes and 47 1/2   seconds
  • A  year is 365.2422 days approximately.
  • The common year consists of 365 days.
  • Consequently in every 4 th year there are 366 days.(leap year)
  • The years which have the extra day are called leap years. The day is inserted at the end of February.

One non-leap year contains one odd day

One leap year contains two odd days & 366 days

For finding the leap years we divide by 4

Every leap year contains 2 odd days.

24  x  2  = 48.  (odd days in leap years)

100  –  24  =76 (non leap years).

Every non-leap year contains 1 odd day.

76  x  1  =  76(odd days in non leap years).

48  +  76  =  124.

124  /  7  =  5(odd days in 100 years)

100  =  5  odd days

200  =  5  +  5  =10/7  =  3  odd days

300  =  5+5+5  =15/7  =  1 odd days

400  =  5+5+5+5=20+1=21/7  = 0 odd days

(400th years is a leap year hence 1 day extra is added)

Example :

What was the day of the week on 15th august 1947?

Step 1: find number of odd days up to 15th august 1947

Step 2: odds days are lies between 0  –  6

0=Sunday,1=Monday,2=Tuesday,3=wednesday,4=thuresday,5-Friday,6=Saturday

Step 3: first find up to 1946 =1900+46

1900  =  1600  +  300.

1600  =  0 odd days(400=0,800=0,1200=0,1600=0).

300  =  1

NOW 1900 = 0+1=1.

Step 4: out of 46 years 11 leap years are there (46/4=11).

11  x  2  =22.

46  –  11  =35  x  1=35.

22  +  35  =57

57  /  7  =1.

So 46 years =1.

1946 =  1  +  1  =2

Step 5: Now 1947

January 31/7 3
February 28/7 0
March 31/7 3
April 30/7 2
May 31/7 3
June 30/7 2
July 31/7 3
August 15/7 1
Total 17
17/7=3.

Step 6: 1946   =2

Up to august     =3.

i.e.  2  +  3  =5.

5  =  FRIDAY.  i.e 15th august 1947 is a FRIDAY