Find A Day Without Reference Day Practice Questions Answers Test With Solutions & More Shortcuts
CALENDAR PRACTICE TEST [4 - EXERCISES]
Basic Calendar Problems
Find A Day With Reference Day
Find A Day Without Reference Day
Finding A Week On Basis Of Another Week
Question : 1
What was the day of the week on 1st April 1901?
a) Sunday
b) Monday
c) Wednesday
d) Saturday
Answer »Answer: (b)
1st 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
January | 3 |
February | 0 |
March | 3 |
April | 1 |
= 3 + 0 + 3 + 1 = 7
⇒ 0 odd days
Total number of odd days till 1st April 1901 = 0 + 1 + 0 = 1
So, the required day was Monday.
Question : 2
What was the day of the week on 30th June 1980?
a) Friday
b) Wednesday
c) Monday
d) Saturday
Answer »Answer: (c)
30th 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
January | 3 |
February | 1 |
March | 3 |
April | 2 |
May | 3 |
June | 2 |
Number of odd days in 1980 = 3 + 1 + 3 + 2 + 3 + 2 = 14
⇒ 0 odd days
Total number of odd days till 30th June, 1980 = 0 + 1 + 0 + 0 = 1
So, the required day was Monday.
Question : 3
What day of the week was on 15th August 1949?
a) Monday
b) Tuesday
c) Thursday
d) Saturday
Answer »Answer: (a)
15th 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 1st January 1949 to 15th August 1949
Number of odd days in 1949,
January | 3 |
February | 0 |
March | 3 |
April | 2 |
May | 3 |
June | 2 |
July | 3 |
August | (15 ÷ 7) = 1 |
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.
Question : 4
On which day of the week does 18th September 1991 fall?
a) Wednesday
b) Tuesday
c) Friday
d) Saturday
Answer »Answer: (a)
18th 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,
January | 3 |
February | 0 |
March | 3 |
April | 2 |
May | 3 |
June | 2 |
July | 3 |
August | 3 |
September | 4 |
= 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.
Question : 5
Ashu was born on August 19, 1992, What day of the week was the born?
a) Sunday
b) Monday
c) Tuesday
d) Wednesday
Answer »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,
January | 3 |
February | 1 |
March | 3 |
April | 2 |
May | 3 |
June | 2 |
July | 3 |
August | 5 |
= 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.
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