Page 264 - Services Selection Board (SSB) Interviews

260 Calender

Concept Cracker

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1. (d) Lets find out 1 April from Zeller’s Formula: Number of days in a period of 100 years is 365 ×
D
C
13×  −m 1     100 + 23 or 365 × 100 + 24.
2 C
f = +   + D +   +   −× .
k
 5   4   4  If century year is not a leap year then number of days
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In this case k = 1 (since 1 April) = 365 × 100 + 23, and number of weeks is 5217
and 4 odd days and for leap year it will be 5217
Month m = 2 (As march = 1, April = 2) weeks and 5 odd days, hence number of Sundays is
D is the last two digit of year here D = 01 (As year is either 5217 or 5218.
2001) 3. (d) From Zeller’s formula lets find the 1 day of 1940
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C is 1 two digit of century here C = 20 (As year is 13×  −m 1    
C
D
k
2 C
2001) f = +   5  + D +   +   −× .
4 
4 



× −
 13 2 1  01  20
f = +   + 01+   +   −× In this case k = 1 (since 1 January)
2 20
1
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 5   4   4 
Month m = 11(As march = 1, April = 2, and hence
 25
f = 1 +   + 01 + [0.5 + ] [ ] 5 − 40. January = 11)
 5  D is the last two digit of year here D = 39 (As year
f = 1 + 5 + 01 + 0 + 5 – 40 = – 28 1940 will start from March)
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This – ve value of f can be made positive by adding C is 1 two digit of century here C = 19 (As year is
multiple of 7 1940)
So f = – 28 + 28 = 0 f = 1 + 13 ×  11 1 − + 39 +  39 +  19 − 2 19.
×
So number of odd days is 0,   5     4     4  
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So 1 April 2001 is Sunday,  142
th
st
th
So 1 Friday is on 6 April, so next Fridays is 13 , f = 1 +   5   + 39 + [9.75 ] [4.75+ ] − 38.
th
th
20 , 27 April.
2. (c) In a period of 100 years there are 23 or 24 leap Or f = 1 + 28 + 39 + 9 + 4 – 38 = 43 when
divided by 7 gives remainder 1 hence it has one odd
years (as for century year it might be or might not be day or 1 January 1940 was Monday
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a leap year, as 1900 was not a leap year)
Then from next year we can make table.
Year 1940 1941 1942 1943 1944 1945 1946 1947 1948 1949
Odd Day 2 1 1 1 2 1 1 1 2 1
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Day of 1 Jan Monday Wed Thu Fri Sat Mon Tue Wed Thu Sat
So in the range from 1940 to 1949 not a single year hence calendar of these two years is not similar.
started with Sunday. 5. (c) Number of odd days between 2013 and 2015 is
4. (d) Calendar of 2 years is similar if number of odd 2, so we have to find the year which will have 2 odd
days between these two years is zero. Consider days between 1977 and required year.
options one by one Consider options one by one-
(a) Between 1940 to 1946 we have two leap years (a) Number of odd days between 1977 and 1981 is
1940 and 1944 so number of odd days is 2 + 1 + 1 1 + 1 + 1 + 2 = 5 odd days
+ 1 + 2 + 1 = 8 or 1 hence calendar of these two (b) Number of odd days between 1977 and 1985 is
years is not similar.
1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 = 10 odd days or
(b) Between 1977 to 1982 we have one leap years 3 odd days
1980 so number of odd days is 1 + 1 + 1 + 2 + 1 (c) Number of odd days between 1977 and 1990 is
= 6 calendar of these two years is not similar. 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2
(c) Between 1912 to 1916 we have one leap years + 1 = 16 odd days or 2 odd days, hence calendar
1912 so number of odd days is 2 + 1 + 1 + 1 = 5 of 1990 is the answer.