Page 265 - Services Selection Board (SSB) Interviews

Calender 261
6. (b) In a year we have 53 Sundays only when year 9. (c) Counting the number of days after 3rd November,
start with Sunday (For non leap year) and either 1994 we have:
with Saturday or Sunday (For leap year). Number of Days in November = 27 or 6 odd days.
From zeller’s formula 1st January 2001 was Monday Number of Days in December = 31 or 3 odd days.
st
From number of odd days 1 January 2006 will start Number of Days in January = 31 or 3 odd days
with Sunday so it had 53 Sunday Number of Days in February = 28 or 0 odd days
st
Similarly 1 January 2009 will start with Thursday.
Number of Days in March = 20 or 6 odd days.
7. (b) Let us take two cases So total number of odd days = 6 + 3 + 3 + 0 + 6
Case (i): When year start with Sunday then next = 18 when divided by 7 gives remainder 4.
4 years will always have 52 Sundays hence total Number of odd days = 4.
number of Sundays are 53 + 3 × 52 = 209 Sundays
Case (ii): When year start with Saturday and then The day on 3rd November, 1994 is (7 - 4) days
beyond the day on 20th March, 1995.
we have 53 Sundays that means year is a leap year
then next 4 years will always have 52 Sundays hence So, the required day is Thursday.
total number of Sundays are 53 + 3 × 52 = 209 10. (b) 94 when divided by 7 gives remainder 3 hence
Sundays. today it must be Thursday, after 3 more days we will
8. (c) If month ends with Thursday then next month will get a Sunday, next Sunday will be after 3 + 7 = 10
start with Friday and it may have 5 Friday otherwise days and so on so we will get Sunday after a day 7K
it may have 4 Fridays. + 3 days.

Concept Deviator
rd
\ Last day of 3 century is Monday.
1. (b) 100 years contain 5 odd days.
st
\ Last day of 1 century is Friday. 400 years contain 0 odd day.
th
\ Last day of 4 century is Sunday.
200 years contain (5 × 2) = 3 odd days.
nd
\ Last day of 2 century is Wednesday. This cycle is repeated that means last day of century
is Friday, Wednesday, Monday, or Sunday and it
300 years contain (5 × 3) = 15 = 1 odd day. repeats the cycle.
2. (c) Count the number of odd days from the year 2011 onwards. Calendar will repeat when we will have number
of odd days = 0.

Year 2011 2012 2013 2014 2015 2016 2017 2018 2019 2020 2021
Odd day 1 2 1 1 1 2 1 1 1 2 1
Total odd day 1 3 4 5 6 8 ( = 1) 2 3 4 6 7

So after ending of 2021 the new year 2022 will start Number of days in April: 30 or number of odd
and that will have same calendar as that of 2011. days: 2
3. (c) If June month has 5 Mondays then dates must Number of days in May: 31 or number of odd
th
nd
th
st
th
be 1 , 8 , 15 , 22 , and 29 or second group of days: 3
rd
th
th
nd
dates must be 2 , 9 , 16 , 23 , and 30 th Number of days in June: 1 or number of odd
Now lets find out the number of days or odd days days: 1
st
st
between 1 January till 1 June. So total number of odd days are 2 + 0 + 3 + 2 + 3
Number of days in January: 30 or number of odd + 1 = 11 when divided by 7 gives remainder 4,
st
days are 2 (here we have not included 1 Jan so Hence 1 January must be 4 days back from
st
number of days is 31 – 1 = 30) Monday i.e Thursday. And if 2 January
nd
st
Number of days in February: 28 or number of is Monday then 1 January must be Friday.
odd days: 0 (Assuming non leap Year) Now consider a leap year then number of odd days
st
st
Number of days in March: 31 or number of odd between 1 January and 1 June is 5 days hence
days 3