Practice Calendar problems - verbal reasoning Online Quiz (set-2) For All Competitive Exams
Q-1) The year next to 1990 will have the same calendar as that of the year 1990.
(a)
(b)
(c)
(d)
The year 1990 has 365 days. i.e. 1 odd day,
the year 1991 has 365 days, i.e. 1 odd day,
the year 1992 has 366 days i.e. 2 odd days.
The likewise year 1993, 1994, 1995 have 1 odd day each.
The sum of odd days, so calculated from years 1990 - 95
= (1 + 1 + 2 + 1 + 1 + 1) = 7 days = 0 odd day
Hence, the year 1996 will have the same calendar as that of the year 1990.
Q-2) If a day before yesterday was Thursday, then when will Sunday fall?
(a)
(b)
(c)
(d)
A day before yesterday = Thursday
Yesterday = Thursday + 1 = Friday
Today = Friday + 1 = Saturday
Therefore, Sunday = Saturday + 1 = Tomorrow
Q-3) If the day before yesterday was Wednesday, when will Sunday be?
(a)
(b)
(c)
(d)
If the day before yesterday was Wednesday, then today will be Friday and the day after tomorrow will be Sunday.
Q-4) On 8th Dec 2007, Saturday falls. What day of the week was it on 8th Dec. 2006?
(a)
(b)
(c)
(d)
The year 2006 is an ordinary year.
So, it has 1 odd day.
So, the day of 8th Dec 2007 will be 1 day beyond the day on 8th Dec 2006.
But, 8th Dec 2007 is Saturday.
Therefore, 8th Dec 2006 is Friday.
Q-5) January 1, 2007, was Monday. What day of the week lies on Jan. 1, 2008?
(a)
(b)
(c)
(d)
The year 2007 is an ordinary year.
So, it has 1 odd day.
1st day of the year 2007 was Monday.
1st day of the year 2008 will be 1 day beyond Monday.
Hence, it will be Tuesday.
Q-6) On 8th Feb 2005 it was Tuesday. What was the day of the week on 8th Feb 2004?
(a)
(b)
(c)
(d)
Q-7) What was the day of the week on 28th May 2006?
(a)
(b)
(c)
(d)
Q-8) What was the day of the week on 15th August 1947?
(a)
(b)
(c)
(d)
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
Month | Odd days |
January | 3 |
February |
0 (ordinary year) |
March | 3 |
April | 2 |
May | 3 |
June | 2 |
July | 3 |
August | 1 i.e, (15 ÷ 7) |
Total 17 17 ÷ 7= remainder 3 odd days
Total odd days =2 + 3 = 5
∴ Required day = Friday
Q-9) Subha went to watch the movie 9 days ago. She goes to watch movies only on Thursday, what is the day of the week today?
(a)
(b)
(c)
(d)
Day 9 days ago = Thursday
Therefore, Today = Thursday + 9
= Thursday + 7 + 2
= Thursday + 2 = Saturday
Q-10) What was the day of the week on 15 August 1947?
(a)
(b)
(c)
(d)
15 August 1947 means,
1946 complete years + first 7 months up to July 1947 + 15 days of August 1947
1600 years have 0 odd days.
300 years have 1 odd day
46 years have 11 leap years and
35 ordinary years = (11 × 2) + (35 × 1)
= 22 + 35 = 57 odd days
= 8 × 7 + 1 odd days = 8 weeks + 1 odd day
Up to 1946 there are 1 + 1 = 2 odd days
January 1947 = 3 odd days
February 1947 = 0 odd days (1947 is a normal year)
March 1947 = 3 odd days
April 1947 = 2 odd days
May 1947 = 3 odd days
June 1947 = 2 odd days
July 1947 = 3 odd days
Up to 15 August = 15 odd days
Total number od odd days up to 15 August 1947
= 2 + 3 + 0 + 3 + 2 + 3 + 2 + 3 + 15
= 33 odd days.
Hence, 15th August 1947 was Friday.
Q-11) If 1st January 2008 is Tuesday, then what day of the weeklies on 1st January 2009?
(a)
(b)
(c)
(d)
Since 2008 is a leap year.
In a leap year, last day = 1st day 1 + odd day
= Tuesday + 1 odd day
= Wednesday = 31st December
∴ 1st January, 2009 = Wednesday + 1 odd day = Thursday
Q-12) If 4th Saturday of a month was 22nd day, then what was the 13th day of that month?
(a)
(b)
(c)
(d)
4th Saturday = 22nd day
3rd Saturday = 22 - 7 = 15th day
∴ 13th day - Saturday - 2 = Thursday
Q-13) If 15th June falls on 3 days after tomorrow which is Friday, then what day of the week will fall on the last date of the month?
(a)
(b)
(c)
(d)
Tomorrow = Friday
3 days after tomorrow = 15th June = Friday + 3 odd days = Monday
Days from 15th to 30th June = 15
15 ÷ 7 = 7)15(2 =>remainder 1 odd day
30th June = Monday + 1 odd day = Tuesday
Q-14) If 27 March 1995 was a Monday, then what day of the week was 1 November 1994?
(a)
(b)
(c)
(d)
Here, 27th March 1995 was Monday.
Now, for calculating the total number of odd days. first, we calculate the total number of days till 1 November 1994.
∴ Number of days in March, 1995 = 27
Number of days in February 1995 = 28
Number of days in January 1995 = 31
Number of days in December 1994 = 31
Number of days in November 1994 = 29
Total number of days = 146
∴ Number of odd days = $146/7$ = 20$6/7$
So, 6 odd days
∴ On November 1, 1994 = Monday - 6
= Tuesday
Q-15) What was the day of the week on 1st April 1901?
(a)
(b)
(c)
(d)
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.
Q-16) 2 days before yesterday was Friday, then what day of the week will be the day after tomorrow?
(a)
(b)
(c)
(d)
2 days before yesterday = Friday
∴ Yesterday = Friday + 2 = Sunday
∴ Today = Sunday + 1 = Monday
∴ Day after tomorrow = Monday + 2 = Wednesday
Q-17) If day after tomorrow is Tuesday, then what day of the week will it be on 2 days after the day after tomorrow?
(a)
(b)
(c)
(d)
A day after tomorrow = Tuesday
∴ Two days after the day after tomorrow = Tuesday + 2
= Thursday
Q-18) If Thursday falls 2 days after tomorrow, then what day of the week was it on three days before yesterday?
(a)
(b)
(c)
(d)
Two days after tomorrow = Thursday
Tomorrow = Thursday - 2 = Tuesday
Today = Tuesday - 1 = Monday
Yesterday = Monday - 1 = Sunday
∴ 3 days before yesterday = Sunday - 3 = Thursday
Q-19) If a day before yesterday was Tuesday, then what day of the week will it be on a day after tomorrow?
(a)
(b)
(c)
(d)
A day before yesterday = Tuesday
Yesterday = Tuesday + 1 = Wednesday
Today = Wednesday + 1 = Thursday
Tomorrow = Thursday + 1= Friday
∴ Day after tomorrow = Friday + 1 = Saturday
Q-20) On which day of the week does 18th September 1991 fall?
(a)
(b)
(c)
(d)
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.