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41. What was the day of the week on 26-January-1950?
Solution:
Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
42. Today is Thursday. What day of the week it was 30 days?
The day is 2 days before Thursday i.e Tuesday
Solution:
30 days = 4 x 7 + 2 = 2 odd daysThe day is 2 days before Thursday i.e Tuesday
43. 01-Jan-2007 was Monday. What day of the week lies on 01-Jan-2008?
Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007)
Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday
Solution:
Given that January 1, 2007 was Monday.Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007)
Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday
44. 1.12.91 is the first Sunday. Which is the fourth Tuesday of December 91?
Hence we can assume that 3.12.91 is the first Tuesday
If we add 7 days to 3.12.91, we will get second Tuesday
If we add 14 days to 3.12.91, we will get third Tuesday
If we add 21 days to 3.12.91, we will get fourth Tuesday
⇒ Fourth Tuesday = (3.12.91 + 21 days) = 24.12.91
Solution:
Given that 1.12.91 is the first SundayHence we can assume that 3.12.91 is the first Tuesday
If we add 7 days to 3.12.91, we will get second Tuesday
If we add 14 days to 3.12.91, we will get third Tuesday
If we add 21 days to 3.12.91, we will get fourth Tuesday
⇒ Fourth Tuesday = (3.12.91 + 21 days) = 24.12.91
45. Today is Thursday. The day after 59 days will be?
Hence if today is Thursday, After 59 days, it will be
= (Thursday + 3 odd days)
= Sunday
Solution:
59 days = 8 weeks 3 days = 3 odd daysHence if today is Thursday, After 59 days, it will be
= (Thursday + 3 odd days)
= Sunday
46. What was day of the week on 21-September-1987?
Solution:
Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
47. How many odd days are there from 13th May, 2005 to 19th August 2005 (both inclusive)?
From 13th to 31st May = 19 days
In June = 30 days
In July = 31 days
From 1st to 19th April = 19 days
Total number of days = 19 + 30 + 31 + 19 = 99 days
The number of odd days are = 14 x 7 + 1 = 99
So there is 1 odd day in the given period
Solution:
Here we have to count the number days from 13th May, 2005 to 18rd August 2005 ( both inclusive)From 13th to 31st May = 19 days
In June = 30 days
In July = 31 days
From 1st to 19th April = 19 days
Total number of days = 19 + 30 + 31 + 19 = 99 days
The number of odd days are = 14 x 7 + 1 = 99
So there is 1 odd day in the given period
48. What is the year next to 1990 which will have the same calendar as that of the year 1990?
Total odd days in the period 1990-1991 = 2 normal years
⇒ 2 x 1 = 2 odd daysTake the year 1995 from the given choices.
Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
⇒ 4 x 1 + 1 x 2 = 6 odd daysTake the year 1996 from the given choices.
Number of odd days in the period 1990-1995 = 5 normal years + 1 leap year
⇒ 5 x 1 + 1 x 2 = 7 odd days = 0 odd days
(As we can reduce multiples of 7 from odd days which will not change anything) Though number of odd days in the period 1990-1995 is 0, there is a catch here.
1990 is not a leap year whereas 1996 is a leap year.
Hence calendar for 1990 and 1996 will never be the same.Take the year 2001 from the given choices.Number of odd days in the period 1990-2000 = 8 normal years + 3 leap years
⇒ 8 x 1 + 3 x 2 = 14 odd days = 0 odd days
Also, both 1990 and 2001 are normal years.
Hence 1990 will have the same calendar as that of 2001
Solution:
For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0. Take the year 1992 from the given choices.Total odd days in the period 1990-1991 = 2 normal years
⇒ 2 x 1 = 2 odd daysTake the year 1995 from the given choices.
Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
⇒ 4 x 1 + 1 x 2 = 6 odd daysTake the year 1996 from the given choices.
Number of odd days in the period 1990-1995 = 5 normal years + 1 leap year
⇒ 5 x 1 + 1 x 2 = 7 odd days = 0 odd days
(As we can reduce multiples of 7 from odd days which will not change anything) Though number of odd days in the period 1990-1995 is 0, there is a catch here.
1990 is not a leap year whereas 1996 is a leap year.
Hence calendar for 1990 and 1996 will never be the same.Take the year 2001 from the given choices.Number of odd days in the period 1990-2000 = 8 normal years + 3 leap years
⇒ 8 x 1 + 3 x 2 = 14 odd days = 0 odd days
Also, both 1990 and 2001 are normal years.
Hence 1990 will have the same calendar as that of 2001
49. Find the number of odd days in 126 years.
125 years = 100 years + 26 years
= 100 years + 6 leap years + 20 ordinary years
= 5 odd days + 12 odd days + 20 odd days
= 37 odd days = 5 x 7 +2 = 2 odd days
Solution:
A period of 100 years has 5 odd days . In 26 years , 4 are leap, remaining are ordinary years125 years = 100 years + 26 years
= 100 years + 6 leap years + 20 ordinary years
= 5 odd days + 12 odd days + 20 odd days
= 37 odd days = 5 x 7 +2 = 2 odd days
50. If the day before yesterday was Thursday, when will Sunday be?
⇒ Yesterday was a Friday
⇒ Today is a Saturday
⇒ Tomorrow is a Sunday
Solution:
Day before yesterday was Thursday⇒ Yesterday was a Friday
⇒ Today is a Saturday
⇒ Tomorrow is a Sunday