# Tips to Solve Schedule Day& Date Question in Reasoning Section

SCHEDULE DAY& DATE

IMPORTANT POINTS TO KNOW:

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• In this type of questions a particular day/date in a particular year is given
• Either a day before or after the date is to found out
• Using this information we have to find out the day which falls on the given date in the question
• A normal year consists of 365 days
• Every fourth year consists of 366 days i.e., 1 day extra known as the leap year
• The extra day in the leap year is 29thof February
• When the number of days are converted into number of weeks we may get some remainder days known as odd days
• Normal year = 365 days = 52 weeks + 1 day

o  Here the remaining 1 day = odd day

• Similarly, Leap year = 366 days = 52 weeks + 2 days

o  Here there are 2 odd days in a leap year

• A month consists of 30, 31, 28 or 29 days depending on name of the month
• 31 days months are = January, March, May, July, August, October, December
• 30 days months are = April, June, September, November
• February consists of 28 days in a normal year and 29 days in a leap year
• A year can be determined as a leap year if it is divisible by 4

o  Leap year = [year/4]

• There are 7 days in a week
• The number of odd days are crucial to find the day of the date given in the question

TABLE1:IF THE DAY TO BE FOUND IS AFTER THE REFERENCE DATE GIVEN IN THE QUESTION:

 ODD DAY DAY OF THE WEEK 0 Previous Day 1 Given/Same day 2 Same day + 1 3 Same day + 2 4 Same day + 3 5 Same day + 4 6 Same day + 5 7 Same day + 6

TABLE 2:IF THE DAY TO BE FOUND IS BEFORE THE REFERENCE DATE GIVEN IN THE QUESTION:

 ODD DAY DAY OF THE WEEK 0 Next  Day 1 Given/Same day 2 Same day – 1 3 Same day – 2 4 Same day – 3 5 Same day – 4 6 Same day – 5 7 Same day – 6

QUESTION1:

The 1st day of the year 1998 was Thursday. If the birthday of Arun falls on 26th June, then on which day in 1998 was his birthday?

GIVEN:

1st January 1998 = Thursday

SOLUTION:

First we have to check whether 1998 is a normal or a leap year

1998 is not divisible by 4 i.e., it is a normal year with February = 28 days

Number of days to be calculated from 1stJanuary 1998 to 26th June 1998

January + February + March + April + May + June

= 31 + 28 + 31 + 30 + 31 + 26

= 177 days

Now to find the number of odd days convert the days into weeks

[177/7] = 25 weeks + 2 days

Therefore odd days = 2

Hence from table 1, 2 odd days = same day + 1

àThursday + 1 = Friday

Therefore,26th June 1998 was a Friday

QUESTION2:

The Independence Day was celebrated on Friday, 15th August in 1996. What was the 1stday of 1996?

GIVEN:

15th August 1996 = Friday

SOLUTION:

Here the day prior to the reference date is to be calculated

1st we have to find out whether 1996 is a normal/leap year

[1996/4] = 499

Hence, 1996 is a leap year = February = 29 days

Now the number of days from 1stJanuary 1996 to 15th August 1996 is to calculated

January + February + March + April + May + June + July + 15th August

= 31 + 29 + 31 + 30 + 31 + 30 + 31 + 15

= 228 days

To find the number of odd days

[228/7] = 32 weeks + 4 days

From Table 2,

4 odd days = Same day – 3 = Friday – 3 = Tuesday (3 days before Friday)

Therefore1stJanuary 1996 was a Tuesday