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

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

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

27

On what dates of September 1972 did Friday fall?

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

1^{st} September 1972 falls on Friday

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

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

5^{th} August 2025 falls on Tuesday. Hence (d).

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

5

Number of odd days in 200 years à 3

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

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

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

So, 19

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).

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).

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).

After 237 days => 237/7 = 6 odd days

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

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

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

Given that, 25^{th} June, 2013 is Tuesday.

So, 25^{th}June, 2012 is Monday. Hence (b).

The day on 25

Given that, 25

So, 25

If the fifth day of a month is 2 days after the Wednesday, then 5^{th} day is Friday.

à 12^{th}, 19^{th} and 26^{th} day also falls on Friday

So, 27^{th} day of the month will be Saturday. Hence (c).

à 12

So, 27

A’s birth date = 3 May 1997

1 May 1997 àThursday

3 May 1997 à Saturday. Hence (d).

1 May 1997 àThursday

3 May 1997 à Saturday. Hence (d).

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

1^{st} August 2012 falls on Wednesday.

4^{th} August 2012 falls on Saturday

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

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

1

4

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

Find the third day of fourth year.

Let the five consecutive years be V, W, X, Y and Z. Here V and Z are leap years.

If 175^{th} day of first year is Thursday (Given), then 366^{th} 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).

If 175

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).

Which day of the week will fall on the middle day of the given period.

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

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

Then the middle day will be 914

How many Tuesdays are there in the month of December in last year?

335^{th} day of the last year is Monday(Given)

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

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

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

Total number of days from January to November in last year(Z) = 335 days. Then 1

8, 15, 22 and 29

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

Vini celebrated her 10^{th} 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).

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).

Madhan was born on 29^{th} 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 29^{th} February of 1996 to 29^{th} 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).

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 29

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).