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11. January 16, 1997 was a Thursday. What day of the week will it be on January 4, 2000?
  A.  Wednesday
  B.  Tuesday
  C.  Friday
  D.  Monday
     
   
View Answer

First we look for the leap years during this period.
1997, 1998, 1999 are not leap years.
1998 and 1999 together have net 2 odd days.
No. of days remaining in 1997 = 365 − 16 = 349 days = 49 weeks 6 odd days.
January 4, 2000 gives 4 odd days.
Total no. of odd days =2 + 6 + 4 = 12 days = 7 days (1 week) + 5 odd days
Hence, January 4, 2000 will be 5 days beyond Thursday ie it will be on Tuesday.

12. February 20, 1999 was Saturday. What day of the week was on December 30, 1997?
  A.  Monday
  B.  Friday
  C.  Tuesday
  D.  Wednesday
     
   
View Answer

The year during this interval was 1998 and it was not a leap year. Now, we calculate the no. of odd days in 1999 up to February 19:
19 February 1999 gives 3 odd days
1998, being ordinary year, gives 5 odd days
In 1997, December 30 and 31 give 1 odd days
In 1997, December 30 and 31 give 2 odd days
∴ total no. of odd days = 3 + 5 + 1 + 2 = 11 days = 4 odd days
Therefore, December 30, 1997 will fall 4 days before Saturday ie on Tuesday.

13. March 5, 1999 was on Friday, what day of the week will be on March 5, 2000?
  A.  Tuesday
  B.  Saturday
  C.  Monday
  D.  Sunday
     
   
View Answer

Year 2000 is a leap year.
No. of remaining days in 1999 = 365 − [31 days in January + 28 days in February + 5 days in March] = 301 days = 43 weeks ie 0 odd day.
No. of days passed in 2000 =
January (31 days) gives 3 odd days
February (29 days, being a leap year) gives 1 odd day
March (5 days) gives 5 odd days
∴ Total no of odd days = 0 + 3 + 1 + 5 = 9 days ie 2 odd days
Therefore, March 5. 2000 will be two days beyond Friday, ie on Sunday.

14. The first Republic Day of I(ndia was celebrated on 26th January 1950. What was the day of the week on that date?
  A.  Thursday
  B.  Tuesday
  C.  Monday
  D.  Friday
     
   
View Answer

Total number of odd days = 1600 years have 0 odd days + 300 years have 1 odd day + 49 years (12 leap + 37 ordinary) have 5 odd days + 26 days of January have 5 odd days = 0 + 1 + 5 + 5 = 4 odd days.
So, the day was Thursday.

15. Raju was born on 2 Oct 1869. The day of the week was
  A.  Friday
  B.  Sunday
  C.  Saturday
  D.  Thursday
     
   
View Answer

1600 years have 0 odd day.
200 years have 2 x 5 = 10 , i.e 3 odd day.
68 years contain 17 leap years and 51 ordinary years.
That is, 17 x 2 + 51 = 85 days, i.e 1 odd day.
In 1869, upto 2nd Oct, total number odd days = 31(Jan) +28(Feb) + 31(Mar) + 30(Apr) + 31(May) + 30(Jun) + 31(Jul)+ 31(Aug) + 30(Sep) + 2(Oct) = 275 days = 2 odd days.
∴ total odd days = 0 + 3 + 1 + 2 = 6 odd days.
∴ the day was Saturday.

Other method:

Whenever, the day of week for a year later than 1600 is asked, divide the years like following way:
2 Oct 1869 = 1600 + 200 + 68 + Jan 1 to 2 Oct of 1869
= 1600 + 200 + 68 + 365 days − 2 Oct to 31 Dec of 1869
= 1600 + 200 + 68 + (365 − 90) days
Number of odd days = 0 + 3 + 1 (for 17 leap years & 51 ordinary years) + 2 = 6 odd days.
∴ the day is Saturday.

16. What was the day of the week on 15th August 1947?
  A.  Thursday
  B.  sunday
  C.  Friday
  D.  Saturday
     
   
View Answer

15 Aug 1947 = (1600 + 300 + 46) years + 1 Jan to 15 Aug of 1947
= (1600 + 300 + 46)years + 365 − 16
Aug to 31 Dec 1947
= (1600 + 300 + 46)years + (365 − 138) days
Number of odd days = 0 + 1 + 1(from 11 leap years and 35 ordinary years) + 3 = 5 odd days.
∴ the day was Friday.
Remember the following table:
Months Odd days
Jan 3
Feb 0/1(ordinary/leap year)
Mar 3
Apr 2
May 3
June 2
Jul 3
Aug 3
Sep 2
Oct 3
Nov 2
Dec 3

17. Today is Friday. After 62 days it will be:
  A.  Monday
  B.  wednesday
  C.  Friday
  D.  Thursday
     
   
View Answer

Each day of the week is repeated after 7 days.
∴ After 63 days, it would be Friday.
So, after 62 days, it would be Thursday.

18. The year next to 1988 having the same calendar as that of 1988 is:
  A.  1989
  B.  1900
  C.  1993
  D.  1990
     
   
View Answer

Starting with 1988, we go on counting the num ber of odd days till the sum is divisible by 7
Years Odd days
1988 2
1989 1
1990 1
1991 1
1992 2

Total odd days = 7 i.e 0 odd days.
∴ Calendar for 1993 is the same as that of 1988

19. What day of the week 20th June, 1837?
  A.  Tuesday
  B.  Monday
  C.  Wednesday
  D.  Friday
     
   
View Answer

20th June, 1837 means "1836 complete years + first 5 months of the year 1837 + 20 days of June"
1600 years give 0 odd days
200 years give 3 odd days
36 years give 3 odd days

[36 years contain 9 leap years and 27 ordinary years and therefore, (27 + 18 =) 45 odd days = 3 odd days]
∴ 1836 years give (0 + 3 + 3) = 6 odd days
Now, from first January to 20th June We have:
Months Odd days
Jan 3
Feb 0
March 1
Apr 2
May 3
June 6

Total odd days = 17 i.e 3 odd days.
∴ Total number of odd days = 6 + 3 = 9 odd days i.e 2 odd days.
This means that the 20th June fell on the 2nd day commencing from Monday. Therefore the required day was Tuesday.

20. How many times does the 29th day of the month occur in 400 consecutive years?
  A.  4495 times
  B.  4497 times
  C.  4491 times
  D.  4493 times
     
   
View Answer

In 400 consecutive years there are 97 leap years.
Hence in 400 consecutive years February has the 29th day 97 times, and the remaining 11 months have the 29th day 400 x 11 or 4400 times.
∴ the 29th day of the month occurs (4400 + 97) = 4497 times.

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