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31. Find the number of odd days in 123 days
123 = 7 × 17 + 4 ⇒ 4 odd days.
Solution:
Odd days ⇒ The number of days more than complete number of weeks in the given period are odd days.123 = 7 × 17 + 4 ⇒ 4 odd days.
32. Given that on 18th April 1603 is Thursday. What was the day on 18th April 2003?
Thus, 18th April 1603 is Thursday, After 400 years i.e., on 18th April 2003 has to be Thursday.
Solution:
After every 400 years, the same day occurs. (Because a period of 400 years has 0 odd days)Thus, 18th April 1603 is Thursday, After 400 years i.e., on 18th April 2003 has to be Thursday.
33. Today is Friday, after 126 days, it will be:
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
Solution:
= 0Each day of the week is repeated after 7 days.
So, after 126 days, it will be Friday.
After 126 days, it will be Friday
34. If 25th of August in a year is Thursday, the number of Mondays in that month is
Hence 29th August = Monday
So 22nd,15th and 8th and 1st of August also will be Mondays
Number of Mondays in August = 5
Solution:
Given that 25th August = ThursdayHence 29th August = Monday
So 22nd,15th and 8th and 1st of August also will be Mondays
Number of Mondays in August = 5
35. How many times does the 29th day of the month occur in 400 consecutive years?
Thus the 29th day of the month occurs
= 4400 + 97
= 4497 times.
Solution:
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.
36. Second Saturday and every Sunday is a holiday. How many working days will be there in a month of 30 days beginning on a Saturday?
Sundays = 2nd, 9th, 16th, 23rd, 30th
⇒ 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
Solution:
Mentioned month begins on a Saturday and has 30 daysSundays = 2nd, 9th, 16th, 23rd, 30th
⇒ 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
37. December 9, 2001 is Sunday. What was the day on December 9, 1971?
We know leap year has 2 odd days
Here, Number of Normal years= 22 and number of Leap years =8
So odd days =22+16 =38 i.e 3 odd days (remainder when 38 is divided by 7,i.e. 3)
Hence, it was a Thursday.
Solution:
We know every year has 1 odd daysWe know leap year has 2 odd days
Here, Number of Normal years= 22 and number of Leap years =8
So odd days =22+16 =38 i.e 3 odd days (remainder when 38 is divided by 7,i.e. 3)
Hence, it was a Thursday.
38. What day of the week was 1st January 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.
Solution:
1st 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.
39. Today is 5th August. The day of the week is Wednesday. This is a leap year. What will be the day of the week on this date after 3 years?
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
Solution:
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
40. January 1, 2004 was a Thursday, what day of the week lies on January 1, 2005.
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
Solution:
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