Calendar 2 - Free MCQ Practice Test with solutions, UPSC Prelims Paper

=
= 134/7
=1
= Sunday

20^th June, 1837 means 1836 complete years + first 5 months of the year 1837 + 20 days of June
1600 years give no odd days
200 years give 3 odd days
36 years give (36 + 9 ) or 3 odd days
Thus 1836 years give 6 odd days
From 1^st January to 20^th June there are 3 odd days
Odd days: January 3, February 0, March 3, April 2, May 3, June 6 = 17
Thus the total number of odd days = 6 + 3 or 2 odd days
This means that the 20th of June fell on 2nd day commencing from Monday.
The required day was Tuesday.

Number of days from 26-Jan-1996 to 15-May-1996 (both days included)
= 6(Jan) + 29(Feb) + 31 (Mar) + 30(Apr)+ 15(May)
= 111

94 days = (13 × 7) + 3 = 3 odd days.
The required day is 3 days beyond Thursday i.e., Sunday

Counting the years 1600 + 300 + 74
In 1600 years, there are zero odd days
In 300 years, there is one odd day
In 74 years, there are 18 leap years and 56 normal years, so the odd days are:
18(2) + 56(1) = 36 + 56 = 92,
Which is 13 weeks and 1 odd day
In 25 days of January, 1975, there are 3 weeks and 4 odd days
Total odd days = 0 + 1 + 1 + 4
six odd days, so it was a Saturday.

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

1 - May - 2007

=17/7
=3
=Tuesday
= May 1^st → Tuesday + 5 days = Saturday = 5^th may
5^th may + 7 days = Saturday = 12^th may
12^th may + 7 days = Saturday = 19^th may
19^th may + 7 days = Saturday = 26^th may
= Answer = 26^th may

Given that seventh day of a month is three days earlier than Friday
⇒ Seventh day is Tuesday
⇒ 14^th is Tuesday
⇒ 19^th is Sunday

This is a leap year. So, none of the next 3 years will be leap years. Each year will give one odd day so the day of the week will be 3 odd days beyond Monday i.e. it will be Thursday.

Given that 1st October is Sunday
Number of days in October = 31
31 days = 3 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)
Hence 1^st November = (Sunday + 3 odd days) = Wednesday

Odd days ⇒ The number of days more than complete number of weeks in the given period are odd days.
123 = 7 × 19 + 4 ⇒ 4 odd days.

After every 400 years, the same day occurs. (Because a period of 400 years has 0 odd days)
Thus, 18^th April 1603 is Thursday, After 400 years i.e., on 18^th April 2003 has to be Thursday.

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

Given that 25th August = Thursday
Hence 29th August = Monday
So 22nd,15th and 8th and 1st of August also will be Mondays
Number of Mondays in August = 5

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

Mentioned month begins on a Saturday and has 30 days
Sundays = 2^nd, 9^th, 16^th, 23^rd, 30^th
⇒ 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

Total number of days = 30 × 365 + 8 days from leap years = 10958
Thus number of weeks = 1565
Hence December 9, 1971 must have been Thursday.

1 Jan 1901 = (1900 years + 1st Jan 1901)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601 - 1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
1st Jan 1901 = 1 odd day
Total number of odd days = (0 + 1 + 1) = 2
2 odd days = Tuesday
Hence 1 January 1901 is Tuesday.

This is a leap year.
So, none of the next 3 years will be leap years.
Each ordinary year has one odd day, so there are 3 odd days in next 3 years.
So the day of the week will be 3 odd days beyond Wednesday i.e. it will be Saturday

Given that January 1, 2004 was Thursday.
Odd days in 2004 = 2 (because 2004 is a leap year)
(Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004, that is the whole year 2004)
Hence January 1, 2005 = (Thursday + 2 odd days) = Saturday