Page 263 - Services Selection Board (SSB) Interviews

Calender 259
f = 2 + 10 + 10 + 2 + 5 – 40 = – 11  103
This – ve value of f can be made positive by adding f = 18 +   5   + 50 + [12.5 ] [ ] 5+ − 40.
multiple of 7
So f = – 11 + 14 = 3 f = 18 + 20 + 50 + 12 + 5 – 40 = 65.
When divided by 7 we will get remainder 3, hence When divided by 7 we will get remainder 2, hence
number of odd days is 2,
number of odd days is 3, th
nd
So 2 june 2010 is 3 days more than Monday, i.e So 18 October 2050 is 2 days more than Sunday,
i.e Tuesday.
Wednesday.
3. (b) From Zeller’s Formula: 5. (b) Here we have to find the number of odd days
th
th
between, 5 march and 5 November,
13×  −m 1    
C
D
k
f = +   + D +   +   −× . Number of days in March is 26 or 5 odd days
2 C
 5   4   4  (Here we have not included 5 march)
th
th
In this case k = 15 (since 15 August) Number of days in April is 30 or 2 odd days
Month m = 6 (As march = 1, April = 2, May = 3, Number of days in May is 31 or 3 odd days
August = 6) Number of days in June is 30 or 2 odd days
D is the last two digit of year here D = 47 (As year is Number of days in July is 31 or 3 odd days
1947) Number of days in August is 31 or 3 odd days
st
C is 1 two digit of century here C = 19 (As year is Number of days in September is 30 or 2 odd days
1947)
Number of days in October is 31 or 3 odd days
×−
 13 6 1  47  19
2 19.
f = 15 +   + 47 +   +   −× Number of days in November is 5 or 5 odd days
 5   4   4  (Here 5 November is included)
th
 77 So total number of odd days = 5 + 2 + 3 + 2 +
f = 15 +   + 47 + [11.75 ] [4.75+ ] − 38.
 5  3 + 3 + 2 + 3 + 5 = 28 when divided by 7 gives
th
remainder 0 hence 5 November will be same as
f = 15 + 15 + 47 + 11 + 4 - 38 = 54. that of 5 march.
th
When divided by 7 we will get remainder 5, hence 6. (b) From Zeller’s Formula we can find that 1 March
st
number of odd days is 3, 2009 is Sunday so we will have 5 Saturdays and 5
A remainder of 0 corresponds to Sunday, 1 means Sundays in total 10 weekends.
Monday, 7 (c) January and August or October depends on leap
th
So 15 August 1947 is 5 days more than Sunday, i.e year or non leap year. But if we find the number of
Friday. odd days between March and November we will get
4. (a) From Zeller’s Formula number of odd days is 0 hence they will have same
C
13×  −m 1     calendar.
D
2 C
f = +   5   + D +   +   −× . 8. (c) The year 2008 is a leap year. It has 2 odd days.
k

4 

4 
th
th
In this case k = 18 (since 18 October) The day on 14 Feb, 2008 is 2 days before the day
th
on 14 Feb, 2009. Hence, this day is Thursday.
Month m = 8 (As march = 1, April = 2, May = 3, 9. (a) Number of days in K weeks is 7K hence total
October = 8)
number of days is 7K + K = 8k or number of days
D is the last two digit of year here D = 50 (As year is must be a multiple of 8.
2050) 10. (a) Number of days in K weeks is 7K hence total
st
C is 1 two digit of century here C = 20 (As year is number of days is 7K + K = 8k
2050)
Similarly number of days in 2Kth day of the 2Kth
week is 2k × 7 + 2k = 16k
13 ×  8 − 1  50  20
f = 18 +   + 50 +   +   − 2 × 20. Required number of days is 16K – 8K = 8k or
 5   4   4  number of days must be a multiple of 8.