What will be the day on 18th Feb,2016?
Answer: B
18 Feb, 2016 = (2015 years + Period from 1.1.2016 to 18.2.2016)
Odd days in 1600 years = 0
Odd days in 400 years = 0
15 years = (12 ordinary years + 3 leap year) = (12 x 1 + 3 x 2)= 4 odd days
Jan Feb
(31 + 18 ) = 49 days
49 days = (7 weeks + 0 day)= 0 odd day.
Total number of odd days = (0 + 0 + 4 + 0) = 4 odd day.
Given day is Thursday
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On 17th March, 1997 Monday falls. What day of the week was it on 17th March, 1996?
Answer: C
Only the remainder on division by 7 matters - the odd days.
Step back one year from 17 March 1997 to 17 March 1996. Although 1996 is a leap year, 29 February 1996 falls before 17 March 1996, so it lies outside the interval.
\[ 365 \text{ days} \rightarrow 1 \text{ odd day} \]
\[ \text{Monday} - 1 = \textbf{Sunday} \]
Whether the leap day counts depends on where it falls relative to the two dates, not merely on whether the year is a leap year.
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What was the day on 15th august 1947 ?
Answer: A
Count the odd days up to 15 August 1947.
\[ 1600 \text{ years} \rightarrow 0 \text{ odd days} \]
\[ 300 \text{ years} \rightarrow 1 \text{ odd day} \]
\[ 46 \text{ years} = 35 \text{ ordinary} + 11 \text{ leap} \rightarrow 35 + 22 = 57 \rightarrow 1 \text{ odd day} \]
From 1 January to 15 August 1947 (not a leap year):
\[ 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 = 32 \times 7 + 3 \]
\[ \text{total odd days} = 0 + 1 + 1 + 3 = 5 \]
Five odd days corresponds to Friday - India's Independence Day fell on a Friday.
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In a year, Holi was enjoyed on 6th March 9 AM and Deepawali was celebrated on 11th November 8 PM. The total time interval (in hours) between the two was ?
Answer: D
Count the days from 6 March to 11 November, then add the part-day hours.
\[ \text{March (after the 6th)} = 25, \ \text{Apr } 30, \ \text{May } 31, \ \text{Jun } 30 \]
\[ \text{Jul } 31, \ \text{Aug } 31, \ \text{Sep } 30, \ \text{Oct } 31, \ \text{Nov } 11 \]
\[ \text{total} = 250 \text{ days} \]
From 9 a.m. to 9 a.m. is \( 250 \times 24 = 6000 \) hours, and from 9 a.m. to 8 p.m. is another 11:
\[ 6000 + 11 = 6011 \text{ hours} \]
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If 5th March, 2005 is Monday, what was the day of the week on 5th March, 2004 ?
Answer: A
The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
The day on 5th March, 2005 will be 1 day beyond the day on 5th March, 2004.
Given that, 5th March, 2005 is Monday.
5th March, 2004 is Sunday (1 day before to 5th March, 2005).
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April 19th, 1986 falls on which day?
Answer: A
19th April, 1986 = (1985 years + Period from 1.1.1986 to 19.4.1986)
Odd days in 1600 years = 0
Odd days in 300 years = 1
85 years has 21 leap years + 64 ordinary years.
Number of odd days in 85 years =( 21 x 2 + 64) = 1 odd day.
Jan. Feb. March April
(31 + 28 + 31 + 19) = 109 days
109 days = 15 weeks + 4 days= 4 odd day.
Total number of odd days = (1 + 0 + 1 + 4) = 6.
Given day is Saturday.
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What day of the week was 15th August 1947 ?
Answer: C
Correct answer: Friday
Count the odd days up to that date. 1600 years give 0, the next 300 give 1, and the 46 years from 1901 to 1946 give \(35 + 22 = 57\), that is 1 odd day.
From 1 January to 15 August 1947 is 227 days, and \(227 = 32 \times 7 + 3\), giving 3 more.
\[0 + 1 + 1 + 3 = 5 \text{ odd days}\]
Counting 0 as Sunday, 5 odd days lands on Friday.
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Prove that any date in March of a year is the same day of the week corresponding date in November that year.
Answer: A
We will show that the number of odd days between last day of February and last day of October is zero. .
March April May June July Aug. Sept. Oct.
31 + 30 + 31 + 30 + 31 + 31 + 30 + 31
= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.
Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows
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It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010 ?
Answer: B
Only the remainder on division by 7 matters - the odd days.
Count the odd days from 1 January 2006 to 1 January 2010.
\[ 2006:1, \quad 2007:1, \quad 2008:2 \text{ (leap)}, \quad 2009:1 \]
\[ \text{total} = 5 \text{ odd days} \]
\[ \text{Sunday} + 5 = \textbf{Friday} \]
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The calendar of year 1989 was same as which year?
Answer: A
Look backwards from 1989 until the odd days total a multiple of seven.
\[ 1988:2,\ 1987:1,\ 1986:1,\ 1985:1,\ 1984:2,\ 1983:1,\ 1982:1,\ 1981:1,\ 1980:2,\ 1979:1,\ 1978:1 \]
\[ \text{total} = 14 \quad\Longrightarrow\quad 0 \text{ odd days} \]
So 1989 repeats the calendar of 1978.
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