Practice Finding day without ref day - verbal reasoning Online Quiz (set-1) For All Competitive Exams

15^th August 1949 means,

1948 complete year + First 7 months of the year 1949 + 15 days of August

Number of odd days in 1600 yrs = 0

Number of odd days in 300 yrs = 1

Number of odd days in 48 yr (36 non - leap years + 12 leap years)

= 36 × 1 + 12 × 2

= 60 = 7 × 8 + 4 = 4

odd days From 1^st January 1949 to 15^th August 1949

Number of odd days in 1949,

Total number of odd days in 1949 = 3 + 0 + 3 + 2 + 3 + 2 + 3 + 1 = 17

= 7 × 2 + 3 = 3 odd day

Total odd days = 1 + 4 + 3 = 8

= 1 odd days

Since, 1 is the code for Monday.

Therefore, the required day was Monday.

1^st January 2001 means 2000 complete years + 1 day of January 2001

Number of odd days in 2000 yrs = 0

Number of odd days in January 2001 = 1

Total number of odd days = 0+1 = 1

So, the required day was Monday

28^th August 2009 means,

2008 complete years + First 7 months of the year 2009 + 28 days of August

Number of odd days in 2000 yrs = 0

Number of odd days from 2001 yrs to 2008 yrs

2001 2002 2003 2004 2005 2006 2007 2008  1 1 1 2 1 1 1 2

=1+1+1+ 2+ 1+1+1+2 = 10

= 7 x1+ 3 = 3 odd days

Number of odd days in 2009,

January February March April May July  August 3 0 3 2 3 2 3 0

= 3 + 0 + 3 + 2 + 3 + 2 + 3 + 0

= 16 = 7 x 2 + 2 = 2 odd days

Total number of odd days till 28^th August 2009

= 0 + 3 + 2 = 5

So, the required day is Friday.

Odd days in 1600 yrs = 0

Odd days in 300 yrs = 1

46 yrs = (11 leap year + 35 ordinary year)

= (11x2 + 35 x 1) = 1 odd day

∴ Odd days in 1946 yrs = (0+ 1+ 1) = 2

Total  17  17 ÷ 7= remainder 3 odd days

Total odd days =2 + 3 = 5

∴ Required day = Friday

First of all, we have to find the day on 1^st April, 2012 1^st April 2012 means

(2011 years 3 months and 1 day)

Now, 2000 years have 0 odd days 11 years have

(2 leap years and 9 ordinary years)

= ( 2 × 2 + 9 × 1 ) odd days

= (4 + 9) odd days = 13

= 6 odd days

3 months and 1 day

= 92 days = 1 odd day

Total number of odd days = ( 6 + 1 ) = 7

⇒ 0 odd day

Hence, it was Sunday on 1^st April 2012. (1^st Sunday).

Subsequently, Sundays of the month were on 1^st, 8^th, 15^nd, 22^nd, and 29^th.

1^st April 1901 means 1900 complete years + first 3 months of 1901 + 1 day of April

Number of odd days in 1600 yrs = 0

Number of odd days in 300 yrs = 1

Number of odd days in 1901 yrs

= 3 + 0 + 3 + 1 = 7

⇒ 0 odd days

Total number of odd days till 1^st April 1901 = 0 + 1 + 0 = 1

So, the required day was Monday.

30^th June 1980 means 1979 complete years + 6 months of 1980

Number of odd days in 1600 yrs = 0

Number of odd days in 300 yrs = 1

Number of odd days in 79 yrs = (19 leap yrs + 60 ordinary years)

= 19 × 2 + 60 × 1 = 38 + 60 = 98

⇒ o odd days

Number of odd days in 1980 = 3 + 1 + 3 + 2 + 3 + 2 = 14

⇒ 0 odd days

Total number of odd days till 30^th June, 1980 = 0 + 1 + 0 + 0 = 1

So, the required day was Monday.

19th August 1992 means,

1991 complete years + First 7 months of 1992 + 19 days of August

Number of odd days in 1600 years = 0

Number of odd days in 300 yrs = 1

Number of odd days in 91 yrs (22 leap year + 69 non-leap years)

= 22 × 2 + 69 × 1

= 44 + 69 113

= 7 × 16 + 1 = 1 odd day

From 1st January, 1992 to 19th August, 1992

Number of odd days in 1992,

= 3 + 1 + 3 + 2 + 3 + 2 + 3 + 5

= 22 = 7 × 3 + 1

= 1 odd day

∴ Number of odd days till 19th August, 1992 = 0 + 1 + 1 + 1 = 3

So, the required day was Wednesday.