28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006) Odd days in 1600 years = 0 Odd days in 400 years = 0 5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) ≡ 6 odd days Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days ∴ 148 days = (21 weeks + 1 day) ≡ 1 odd day. Total number of odd days = (0 + 0 + 6 + 1) = 7 ≡ 0 odd day. Given day is Sunday.
2.
Second Saturday and every Sunday is a holiday. How many working days will be there in a month of 30 days beginning on a Saturday?
(A) 21
(B) 22
(C) 23
(D) 24
Solution:
Mentioned month begins on a Saturday and has 30 days Sundays = 2nd, 9th, 16th, 23rd, 30th ⇒ Total Sundays = 5 Every second Saturday is holiday. 1 second Saturday in every month Total days in the month = 30 Total working days = 30 - (5 + 1) = 24
3.
It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
(A) Sunday
(B) Saturday
(C) Friday
(D) Wednesday
Solution:
On 31st December, 2005 it was Saturday. Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days. ∴ On 31st December 2009, it was Thursday. Thus, on 1st Jan, 2010 it is Friday.
4.
Arun went for a movie nine days ago. He goes to watch movies only on Thursdays. What day of the week is today?
(A) Wednesday
(B) Thursday
(C) Friday
(D) Saturday
Solution:
Clearly it can be understood from the question that 9 days ago was a Thursday Number of odd days in 9 days = 2 (As 9 - 7 = 2, reduced perfect multiple of 7 from total days) Hence today = (Thursday + 2 odd days) = Saturday
5.
On 2007, What was the date of last Saturday in May month?
(A) 22th May
(B) 24th May
(C) 26th May
(D) 28th May
Solution:
1 - May - 2007 $$\eqalign{ & = \frac{{1 + 2 + 7 + 1 + 6}}{7} \cr & = \frac{{17}}{7} \cr & = 3 \cr & = {\text{Tuesday}} \cr} $$ = May 1st → Tuesday + 5 days = Saturday = 5th may 5th may + 7 days = Saturday = 12th may 12th may + 7 days = Saturday = 19th may 19th may + 7 days = Saturday = 26th may = Answer = 26th may
6.
On 17th March, 1997 Monday falls. What day of the week was it on 17th March, 1996?
(A) Monday
(B) Tuesday
(C) Saturday
(D) Sunday
Solution:
The year 1996 is a leap year. So, it has 2 odd days. But 17th March comes after 29th February. So, the day on 17th March, 1997 will be 1 day beyond the day on 17th March,1996. Here 17th March, 1997 is Monday. So, 17th March, 1996 is Sunday.
7.
How many seconds in 10 years?
(A) 31523500 sec
(B) 315360000 sec
(C) 315423000 sec
(D) 315354000 sec
Solution:
We know that, 1 year = 365 days 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds. Then, 1 year = 365 x 24 x 60 x 60 seconds. = 8760 x 3600 1 year = 31536000 seconds. Hence, 10 years = 31536000 x 10 = 315360000 seconds.
8.
Today is Friday, after 126 days, it will be:
(A) Thursday
(B) Friday
(C) Saturday
(D) Sunday
Solution:
$$\frac{{126}}{7}$$ = 0 Each day of the week is repeated after 7 days. So, after 126 days, it will be Friday. After 126 days, it will be Friday
9.
What was the day of the week on 16th August, 1947?
(A) Sunday
(B) Monday
(C) Saturday
(D) Thursday
Solution:
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'.
10.
It was Thursday on 12th January 2006. What day of the week it will be on January 12th 2007 ?
(A) Wednesday
(B) Thursday
(C) Friday
(D) Saturday
Solution:
There is exactly 1 year, (365 days) between two dates. 2006 is an ordinary year. It has one odd day. The day of the week on January 12th 2007 is one day beyond Thursday ⇒ Friday