Page 262 - Services Selection Board (SSB) Interviews

258 Calender

Answer with Solution

Concept Applicator

1. (b) Since in 1 year we have one odd day while in 6. (a) The century divisible by 400 is a leap year. 2100
one leap year we have two odd days. is not divisible by 400 hence is not a leap year.
Number of odd days are 1 (for 2006) 7. (d) The century divisible by 400 is a leap year. 2100
1 days more than Friday means Saturday. is not divisible by 400 hence is not a leap year so
2. (c) Since a day of the week repeat after every 7 days, 2100 is not a leap year, out of remaining 1128 is
divide 59 by 7, remainder is 3 so number of odd divisible by 4, so 1128 is a leap year.
st
days is 3, and 3 days after Monday is Thursday. 8. (d) On 31 December, 2005 it was Saturday.
3. (b) Since in 1 year we have one odd day while in Number of odd days from the year 2006 to the year
one leap year we have two odd days. 2009 = (1 + 1 + 2 + 1) = 5 days.
st
Number of odd days are: 1(for 2006) + 1(for 2007) \ On 31 December 2009, it was Thursday.
st
+ 2 (For leap year 2008) + 1(for year 2009) so total Thus, on 1 Jan, 2010 it is Friday.
number of odd days are 1 + 1 + 2 + 1 = 5 days. 5 9. (b) When 69 is divided by 7 we will get remainder
days more than Friday means Wednesday. 6 so the day will be 6 days more than Sunday i.e
4. (d) Number of days in June is 30 – 2 = 28, and Saturday.
number of days in July is 7, total number of days is 10. (d) Number of odd days between 6 march 2002 to
th
28 + 7 = 35, when we divide 35 by 7 remainder is 6 march 2003 is 1
th
0, or number of odd days is 0 hence 7th july must be th
the same day as that of 2nd June i.e Saturday in this Number of odd days between 6th march 2003 to 6
case. march 2004 is 2
th
5. (a) Number of years between 2001 to 2013 is 12 Number of odd days between 6th march 2004 to 6
years (2013 – 2001 = 12 years) march 2005 is 1
th
Number of leap years between 2001 to 2013 is 3, Hence total Number of odd days between 6 march
th
since a year has 1 odd days and a leap year has 2 2002 to 6 march 2005 is 1 + 2 + 1 = 4 odd
th
odd days, so total number of odd days is 12 + 3 = days so 6 march 2002 is 4 days before Sunday i.e
15 odd days. Wednesday.
When we divide 15 by 7 remainder is 1 or in total we
have 1 odd day so 1st jan 2013 is Tuesday.

Concept Builder

nd
1. (b) When we divide 79 by 7 we will get remainder In this case k = 2 (since 2 June)
2 so we have 2 odd days, so required day must be Month m = 4 (As march = 1, April = 2, May = 3,
2 days back from today (i.e Sunday) and that day June = 4)
should be Friday. D is the last two digit of year here D = 10 (As year is
nd
nd
2. (a) Consider from 2 June 2010 to 2 June 2013 2010)
we have total 2 non leap year and one leap year so C is 1st two digit of century here C = 20 (As year is
nd
number of odd days are 1 + 1 + 2 = 4 so 2 June
2010 must be 4 days back from Sunday and that 2010)
× 
day is Wednesday. f = 2 + 13 4 1 − + 10 +  10 +  20 −×
2 20.
From Zeller’s Formula:   5     4     4  
C
D
13×  −m 1      51
f = +   + D +   +   −× . f = 2 +   + 10 + [2.5 ] [ ] 5+ − 40.
2 C
k
 5   4   4   5 