1. Today is 1st August and the day of the week is Monday. This is a leap year. The day of the week
on this day after 3 years will be
A. Wednesday
B. Friday
C. Thursday
D. Sunday
E. Tuesday
Answer - Option C
Explanation -
This being a leap year none of the next 3 years is a leap year. So, the day of the week will be 3 days
beyond Monday i.e., it will be Thursday
2. Monday falls on 4th April, 1988. What was the day on [latex]{3}^{rd}[/latex] November, 1987?
A. Tuesday
B. Sunday
C. Monday
D. Wednesday
E. Saturday
Answer - Option
Explanation -
Counting the number of days after 3rd November, 1987 we have :
Month:Nov. Dec. Jan. Feb. Mar. Apr.
Days: 27 + 31 + 31 + 29 + 31 + 4
= 153 days, containing 6 odd days.
i.e., (7 – 6) = 1 day beyond the day on 4th April, 1988. So, the day was Tuesday
3. The year next to 1990 having the same calendar as that of 1990 is ____
A. 1998
B. 2001
C. 2002
D. 2004
E. 2000
Answer - Option B
Explanation -
We go on counting the no. of odd days from 1990 onward till the sum is exactly divisible by 7.
The number of such days are 14 upto the year 2000. So, the calendar for 1990, was repeated in the year 2001
4. Which dates of April, 2012 will be Sunday?
A. 1, 8, 15, 22, 29
B. 3, 10, 17, 24, 31
C. 2, 9, 16, 23, 30
D. 5, 12, 19, 26
E. 4, 11, 18, 25
Answer - Option
Explanation -
[latex]{1}^{st}[/latex] April, 2012:
2000 + 11 + Number of days from 1st January 2012 to 1st April, 2012.
Number of odd days in 2000 years = 0
Number of odd days in 11 years = 13
Months : Jan Feb Mar April
Odd days: 3 + 1 + 3 + 1 = 8
Total number of odd days = 8 + 13 + 0 = 21
= 0 odd days.
Hence, [latex]{1}^{st}[/latex] April, 2012 is a Sunday.
[latex]{1}^{st}[/latex] ,[latex]{8}^{th}[/latex], [latex]{15}^{th}[/latex], [latex]{22}^{nd}[/latex] and [latex]{29}^{th}[/latex] of April, 2012 are Sunday’s.
5. If 30 January 1989 is Monday then what was the day of week on 26 Oct 2003.
A. Monday
B. Sunday
C. Tuesday
D. Saturday
D. Friday
Answer - Option A
Explanation -
Total number of odd days from 30 January 1989 to 30 January 2003
= odd days in 3 leap year + 11 ordinary years
= 3 odd days
Total odd days from 30 January 2003 to 26 October 2003 =
Jan |
Feb |
March |
April |
1 |
28 |
31 |
30 |
June |
July |
August |
Sep |
30 |
31 |
31 |
30 |
= 3 odd days
So, total odd days = 6 odd days
So, day on 26 October 2003 will be Sunday.