Calendar Questions and Answers

Answer: A

The correct statement is that 2nd February of that year is a Tuesday.

February can have a fifth Monday only in a leap year in which 1 February is a Monday. The Mondays then fall on the 1st, 8th, 15th, 22nd and 29th, and the fifth one is 29 February, the person's birthday. So 2 February is a Tuesday.

The other statements fail. With 1 February a Monday, the month starts on Monday, not Sunday, and its Sundays are the 7th, 14th, 21st and 28th, only four. The Mondays have dates 1, 8, 15, 22 and 29, mostly odd, so they do not all have even dates.

Answer: D

The 3rd of July falls on a Sunday.

Count the days from 2 June to 3 July. June has 30 days, so from 2 June to 2 July is 30 days, and one more day reaches 3 July: 31 days in all.

\[ 31 = 4 \times 7 + 3 \]

Four complete weeks bring the weekday back to Thursday, and the 3 extra days move it forward three places: Friday, Saturday, Sunday.

Thursday would need a gap that is a whole number of weeks (28 or 35 days). Friday and Saturday are what a count of 29 or 30 days gives, for instance by stopping at 2 July instead of 3 July.

Answer: D

September 11, 1919 is a Thursday.

Count the odd days (days beyond whole weeks) from May 5, 1911.

May 5, 1911 to May 5, 1919 is 8 years, including the leap years 1912 and 1916, so 8 + 2 = 10 odd days, which is 3 after removing a week. From May 5 to September 11, 1919: 26 days left in May, then 30, 31, 31 and 11:

\[ \begin{aligned} 26 + 30 + 31 + 31 + 11 &= 129 \\ 129 &= 18 \times 7 + 3 \end{aligned} \]

Total odd days: 3 + 3 = 6. Friday plus 6 days is Thursday. Forgetting the two leap days would give Tuesday, which is not offered.

Answer: A

The key gives Monday.

Days repeat every 7. Reading the question, as the key does, as the day that comes \(7^{9}\) days after today: \(7^{9}\) is an exact multiple of 7, so that many days is a whole number of weeks and the weekday is unchanged.

\[ 7^{9} \equiv 0 \pmod 7 \]

The point of the question is that any power of 7 leaves remainder 0 on division by 7, so no counting is needed. Wednesday, Saturday and Tuesday would each need a remainder of 2, 5 or 1 days.

Answer: C

The father's birthday falls on Thursday.

Both birthdays (September and December) come after 29 February, so they shift by exactly the same number of weekdays from year to year. An ordinary year moves a date on by 1 day and a leap year by 2.

\[ \begin{aligned} \text{2017 to 2020} &= 1 + 1 + 2 \\ &= 4\ \text{days} \end{aligned} \]

Sunday moved on 4 days is Thursday, which agrees with Atul's birthday. The father's birthday, also a Sunday in 2017, moves by the same 4 days, so it too falls on a Thursday.

Wednesday comes from counting 2020 as an ordinary year (a shift of 3), and Friday from adding an extra day. The months matter only to show both dates fall after February.

Answer: A

The next Tuesday anniversary falls in 2020.

A date moves forward one weekday each year, and two when a 29 February falls in between. Starting from Tuesday, 15 September 2015:

• 2016: +2 (leap year), Thursday
• 2017: +1, Friday
• 2018: +1, Saturday
• 2019: +1, Sunday
• 2020: +2 (leap year), Tuesday

The total shift is 7 days, a whole week, so 15 September 2020 is again a Tuesday. 2021 is a year too many, and 2019 falls on a Sunday.

Answer: A

The Sunday is 20 March 2050.

From 15 March 2025 to 15 March 2050 is 25 years. Each ordinary year moves the weekday on by 1, and each 29 February in between adds 1 more. The leap days fall in 2028, 2032, 2036, 2040, 2044 and 2048, which is 6.

\[ \begin{aligned} \text{shift} &= 25 + 6 = 31 \\ 31 &\equiv 3 \pmod 7 \end{aligned} \]

So 15 March 2050 falls 3 days after Saturday, on a Tuesday. The following Sunday is 15 + 5 = 20 March.

Forgetting the leap days gives a shift of 25, which leaves 4, making 15 March a Wednesday and pointing wrongly to the 19th.

Answer: B

1 Oct, 1994 = (1993 years + Period from 1.1.1993 to 1.10.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

93 years = (70 ordinary years + 23 leap year) = (70 x 1 + 23 x 2)= 4 odd days

Jan  Feb  March  April  May  June  July  Aug  sept  Oct

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 1 ) = 274 days

274 days = (39 weeks + 1 day) =1 odd day.

Total number of odd days = (0 + 1 + 4 + 1) = 6 odd days.

Given day is Saturday.

Answer: A

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 = 4 odd days.

Jan. Feb. Mar.  Apr.  May.  Jun.  Jul.  Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days = (32 weeks + 3 days) = 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 =  0 odd days.

Given day is Sunday

Answer: C

The year 1985 is an ordinary year. So, it has 1 odd day.

So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984.

But, 19th June,1984 is Monday

S0, 19th June, 1985 is Tuesday.