Reasoning Aptitude
CALENDAR MCQs
Calender
Total Questions : 123
| Page 5 of 13 pages
Answer: Option A. -> 22 hrs 20 min
Answer: (a)
Duration of the journey = (Duration form 12.25 noon to midnight) + (Duration from 12.00 midnight to 10.45 a.m.)
= 11 hrs 35 min + 10 hrs 45 min = 22 hrs 20 min.
Answer: (a)
Duration of the journey = (Duration form 12.25 noon to midnight) + (Duration from 12.00 midnight to 10.45 a.m.)
= 11 hrs 35 min + 10 hrs 45 min = 22 hrs 20 min.
Answer: Option A. -> 2800
Answer: (a)
The century year which is completely divisible by 400, is a leap year.
Thus, the year 2800 is a leap year.
Answer: (a)
The century year which is completely divisible by 400, is a leap year.
Thus, the year 2800 is a leap year.
Answer: Option D. -> 2018
Answer: (d)
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd days.
Sum = 14 odd days = 0 odd day.
Therefore, Calendar for the year 2018 will be the same as for the year 2007.
Answer: (d)
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd days.
Sum = 14 odd days = 0 odd day.
Therefore, Calendar for the year 2018 will be the same as for the year 2007.
Answer: Option A. -> 5th Friday
Answer: (a)
Number of days left in the month after 16th = 31 - 16 = 15
Number of odd days = 15/7 = 2 weeks + 1 odd day
Therefore, Required day = Thursday + 1 odd day = Friday
As, 16th of the month is third Thursday, the day which is two weeks after this day is fifth Thursday.
So, one day after 5th Thursday is 5th Friday.
Answer: (a)
Number of days left in the month after 16th = 31 - 16 = 15
Number of odd days = 15/7 = 2 weeks + 1 odd day
Therefore, Required day = Thursday + 1 odd day = Friday
As, 16th of the month is third Thursday, the day which is two weeks after this day is fifth Thursday.
So, one day after 5th Thursday is 5th Friday.
Answer: Option B. -> 2
Answer: (b)
1977 is an ordinary year.
We know that the calendar of an ordinary year repeats after 6 yrs or 11 yrs.
Let us check for the number of odd days in 6th and 11th years.
Year
Number of odd days
1978
1
1979
1
1980
2
1981
1
1982
1
1983
1
1984
2
1985
1
1986
1
1987
1
1988
2
From the above table, Number of odd days from 18.09.1977 to 18.09.1983 = 7,
i.e., 0 odd days
It means that in 1983, 18th September would fall on Sunday.
From the above table, a number of odd days from 18.09.1977 to 18.09.1988 = 14
i.e., 0 odd days.
Now, it is clear that 2 marriage anniversaries would fall on Sunday in the next 15 yr.
Answer: (b)
1977 is an ordinary year.
We know that the calendar of an ordinary year repeats after 6 yrs or 11 yrs.
Let us check for the number of odd days in 6th and 11th years.
Year
Number of odd days
1978
1
1979
1
1980
2
1981
1
1982
1
1983
1
1984
2
1985
1
1986
1
1987
1
1988
2
From the above table, Number of odd days from 18.09.1977 to 18.09.1983 = 7,
i.e., 0 odd days
It means that in 1983, 18th September would fall on Sunday.
From the above table, a number of odd days from 18.09.1977 to 18.09.1988 = 14
i.e., 0 odd days.
Now, it is clear that 2 marriage anniversaries would fall on Sunday in the next 15 yr.
Answer: Option D. -> Cannot be specified
Answer: (d)
There are months of 30, 31 and 28 days and last day of the month are Wednesday.
So, using 28 and 30 days, there are 4 Mondays.
Using 31 days, there are 5 Mondays
So, it cannot be specified.
Answer: (d)
There are months of 30, 31 and 28 days and last day of the month are Wednesday.
So, using 28 and 30 days, there are 4 Mondays.
Using 31 days, there are 5 Mondays
So, it cannot be specified.
Answer: Option B. -> 8x
Answer: (b)
Answer: (b)
Answer: Option C. -> Saturday
Answer: (c)
The dates of three of the Sundays are even number is 2, 9, 16, 23, 30.
So, on 16th of that month = Sunday,
15th of that month falls on a Saturday.
Answer: (c)
The dates of three of the Sundays are even number is 2, 9, 16, 23, 30.
So, on 16th of that month = Sunday,
15th of that month falls on a Saturday.
Answer: Option A. -> 700
Answer: (a)
Answer: (a)
Answer: Option C. -> Tuesday
Answer: (c)
Answer: (c)