find a day without reference day Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

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.

Answer: (c)

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.

Answer: (a)

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.

Answer: (a)

18^th September 1991 means,

1990 complete years + 8 months of 1991 + 18 days of September

Number of odd days in 1600 yrs = 0

Number of odd days in 300 yrs = 1

Number of odd days in 90 yrs (22 leap year + 68 ordinary years )

= 22 × 2 + 68 × 1

= 44 + 68 = 112

⇒ 0 odd days

Number of odd days in 1991,

= 3 + 0 + 3 + 2 +3 + 2 + 3 + 3 + 4

= 23 = 7 × 3 + 2

= 2 odd days

Total number of odd days till 18th September, 1991

= 0 + 1 + 0 + 2 = 3

So, the required day was Wednesday.

Answer: (d)

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.