Calendars

1) What was the day on 27 June 1999?

Show Solution
 Number of odd days in 1600 years à 0
 Number of odd days in 300 years à 1

Number of odd days for the remaining 98 years = (24 Leap years + 74 Ordinary years)

((24 * 2) + (74 * 1)) = 122

122/7 à 3

Number of odd days for the months January to May = (3 + 0 + 3 + 2 + 3) = 11/7 = 4 odd days

 Odd days in June = 27/7 à 6
   Total number of odd days = (0 + 1 + 3 + 4 + 6) = 14 à 0

 27th June 1999 falls on Sunday. Hence (a).

2)

On what dates of September 1972 did Friday fall?

Show Solution

1 September 1972
 Number of odd days in 1600 years à 0
 Number of odd days in 300 years à 1

Number of odd days for the remaining 71 years = (17 Leap years + 54 Ordinary years)

((17 * 2) + (54 * 1)) = 88

88/7 à 4

Number of odd days for the months January to August = (3 + 1 + 3 + 2 + 3 + 2 + 3 + 3) = 20/7 = 6 odd days

 Odd days in September = 1/7 à 1
   Total number of odd days = (0 + 1 + 4 + 6 + 1) = 12 à 5

1st September 1972 falls on Friday

In 1 September 1972, 1, 8, 15, 22 and 29 were Fridays.  Hence (b).

3) What will be the day on 5 August 2025?

Show Solution
 Number of odd days in 2000 years à 0

Number of odd days for the remaining 24 years = (6 Leap years + 18 Ordinary years)

((6 * 2) + (18 * 1)) = 30

30/7 à 2

Number of odd days for the months January to July = (3 + 0 + 3 + 2 + 3 + 2 + 3) = 16/7 = 2 odd days

 Odd days in August = 5/7 à 5
   Total number of odd days = (0 + 2 + 2 + 5) = 9 à 2
5th August 2025 falls on Tuesday. Hence (d).

4) If 19 October 1892 was Thursday, what would be the day on 19 October 2092?

Show Solution
Number of odd days in 200 years à 3

Difference in the number of the years between 1892 and 2092 = 200

So, 19th October 2092 would be Sunday. Hence (c).

5) The calendar for the year 1931 will be the same as which of the following years?

Show Solution
Number of odd days in year 1931 à 1

Number of odd days in year 1932 à 2

Number of odd days in year 1933 à 1

Number of odd days in year 1934 à 1

Number of odd days in year 1935 à 1

Number of odd days in year 1936 à 2

Number of odd days in year 1937 à 1

Number of odd days in year 1938 à 1

Number of odd days in year 1939 à 1

Number of odd days in year 1940 à 2

Number of odd days in year 1941 à 1

Total number of odd days till 1941 is 14.

Therefore the calendar will repeat in the year 1942. Hence (b).

6) Day before yesterday was Thursday. What day will be after 237 days from today?

Show Solution
 Two days before was Thursday. So, today is Saturday.

 After 237 days => 237/7 = 6 odd days

 The day after 237 days will be Friday. Hence (c).

7) If 25th June, 2013 is Tuesday, what was the day of the week on 25th June, 2012?

Show Solution
  The year 2013 is not a leap year. So, it has 1 odd day.

The day on 25th June, 2013 will be 1 day beyond the day on 25th June, 2012.

Given that, 25th June, 2013 is Tuesday.

So, 25thJune, 2012 is Monday. Hence (b).

8) If the fifth day of a month is 2 days after the Wednesday, what day will be the 27th day of the month?

Show Solution
 If the fifth day of a month is 2 days after the Wednesday, then 5th day is Friday.
 à 12th, 19th and 26th day also falls on Friday
So, 27th day of the month will be Saturday.  Hence (c).

9) A was born on May 23, 1997. B was born on 20 days before A. The May day of that year fell on Thursday. Which day was B born?

Show Solution
A’s birth date = 3 May 1997

1 May 1997 àThursday

3 May 1997 à SaturdayHence (d).

10) How many Saturdays are there in the month of August 2012?

Show Solution
 Number of odd days in 2000 years à 0

Number of odd days for the remaining 11 years = (2 Leap years + 9 Ordinary years)

((2 * 2) + (9 * 1)) = 13

13/7 à 6

   Number of odd days for the months January to July = (3 + 1 + 3 + 2 + 3 + 2 + 3) = 17/7 = 3 odd days

 Odd days in August = 1/7 à 1
   Total number of odd days = (0 + 6 + 3 + 1) = 10 à 3

1st August 2012 falls on Wednesday.

4th August 2012 falls on Saturday

So, 4, 11, 18 and 25 will also be Saturday. There are 4 Saturdays in August 2012. Hence (a).

11) Five consecutive years, out of which there are 2 leap years, are considered. 175th day of the first year is Thursday. 335th day of the last year is Monday.
Find the third day of fourth year.

Show Solution
 Let the five consecutive years be V, W, X, Y and Z. Here V and Z are leap years.
 If 175th day of first year is Thursday (Given), then 366th day of first year(V) is Saturday
 Hence first day of first year(V) should be Friday 
First day of the Second year(W)à Sunday
 First day of the Third year( X) à Monday
First day of the Fourth year(Y) à Tuesday
Third day of fourth year(Y) à Thursday Hence (d).

12) Five consecutive years, out of which there are 2 leap years, are considered. 175th day of the first year is Thursday. 335th day of the last year is Monday.
Which day of the week will fall on the middle day of the given period.

Show Solution
Total number of the days for five consecutive years = 366 + 365 + 365 + 365 + 366 à 1827 days.

 Then the middle day will be 914th day. If 175th day of first year is Thursday, then number of odd days for 914 days will have 4 odd days.910th day of the given period is Thursday.914th day will be after 4 days from Thursday. Then the middle day of the given period is Monday. Hence (c).

13) Five consecutive years, out of which there are 2 leap years, are considered. 175th day of the first year is Thursday. 335th day of the last year is Monday.
How many Tuesdays are there in the month of December in last year?

Show Solution
335th day of the last year is Monday(Given)

Total number of days from January to November in last year(Z) = 335 days. Then 1st December of Z will be Tuesday.

8, 15, 22 and 29th of December will also be Tuesdays.

Hence there are 5 Tuesdays in December. Hence (a).

14) Vini celebrated her 10th birthday on 15 May 2018. Vini’s sister is 1 year 15 days younger than her. At the time of her sister’s birth, their mother was 28 years old. On what date Vini’s mother born?

Show Solution
Vini celebrated her 10th birthday on 15 May 2018.

Vini was born on 15 May 2008.

Her Sister was 1 year 15 days younger than her. So her sister was born on 30 April 2009.

On that day, her mother completed 28 years old. So, Vini’s mother was born on 30 April 1971. Hence (b).

15) Madhan was born on 29th February of 1996 which happened to be Thursday. If he hopes to live up to the age of 115 years, how many birthdays will he celebrate on Thursday?

Show Solution
Madhan was born on 29th February of 1996 à Thursday

28 Feb 1997 à Friday, 28 Feb 1998 à Saturday, 28 Feb 1999 à Sunday, 28 Feb 2000 à Monday

29 February 2000 will be Tuesday.

Number of odd days from 29th February of 1996 to 29th February 2000 – 5 odd days.

Every subsequent birthday comes after 5 odd days

2000 Birthday à 5 odd days

2004 Birthday à 10 odd days à 3 odd days

2008 Birthday à 1 odd day

2012 Birthday à 6 odd days

2016 Birthday à 4 odd days

2020 Birthday à 2 odd days

2024 Birthday à 0 odd day

After 28 years he would have a birthday on Thursday

The next birthday on Thursday will be 2052, 2080, 2108.

Hence 4 times he will celebrate his birthday on Thursday.  Hence (d).