Logical Sequences Test - 6

10 p.m. – 5 × (40 minutes) = 6:40 p.m.

Day before yesterday was Thursday,
Yesterday – Friday
Today- Saturday
So, Tomorrow will be Sunday.

4^th + 21 = 25^th ⇒ 4^th (Tuesday)
25^th + 2 = 24^th + 3 = 27^th (Thursday)

Data is inadequate because we don't know the number of days in the present month.

As given, Saturday falls on 26th January and we have to find the day on 14th February. Clearly, 2nd, 9th and 16th February each is a Saturday. Thus, 14th February was a Thursday.

Mentioned month begins on a Saturday and has 30 days
Sundays = 2^nd, 9^th, 16^th, 23^rd, 30^th
⇒ 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

In the given month, Mondays would fall on 1^st, 8^th, 15^th, 22^nd and 29^th of August.
Therefore, the month has 5 Mondays.

As 7^th day next to the given date has the same day, so a Sunday would fall on
1 + 7 = 8^th
8 + 7 = 15^th
15 + 7 = 22^nd
22 + 7 = 29^th

As October is the month with 31 days, so 30^th is Monday, 31^st is Tuesday and 1^st November is Wednesday.

As 2004 is a leap year, so the number of days in each month from 26^th January, 2004 to 15^th May, 2004 is given as follows.
January = 6 days
Feb = 29 days
March = 31 days
April = 30 days
May = 15 days

Therefore, total number of days = 111 days

Only April and July are the two months which have the same calendar in a year.

This can be checked by calculating the period between the months, which is divisible by 7.

From June to October = June + July + August + September
= 30 + 31 + 31 + 30 = 122 (which is not divisible by 7)

From April to November = April + May + June + July + August + September + October
= 30 + 31 + 30 + 31 + 31 + 30 + 31 = 214 (which is not divisible by 7)

From April to July = April + May + June
= 30 + 31 + 30 = 91 (which is divisible by 7)

From October to December = October + November
= 31 + 30 = 61 (which is not divisible by 7)

So, April and July will have the same calendar.