Calendar questions

In a normal year there are 365 days, so there will be 52 weeks and 1 day.

100 years contain 5 odd days.

Last day of 1st century is Friday.

200 years contain (5 x 2) 3 odd days.

Last day of 2nd century is Wednesday.

300 years contain (5 x 3) = 15 1 odd day.

Last day of 3rd century is Monday.

400 years contain 0 odd day.

Last day of 4th century is Sunday.

This cycle is repeated.

Last day of a century can be Friday, Wednesday, Monday and Sunday.

If today is tuesday, then after 7 days it will occur again. For further occurrence this day will occur also after 14, 21, 28…days.

So, according to the question:

\(\frac{50}{7}\)=7 (remainder=1)

Hence after 49 days tuesday will occur.

and on 50th day, it will be wednesday

Let us find the day on 1st July, 2004.

2000 years have 0 odd day. 3 ordinary years have 3 odd days.
Jan. Feb. March April May June July \(31 + 29 + 31 + 30 + 31 + 30 + 1 = 183 days \)
(26 weeks + 1 day) = 1 .
Total number of odd days = \((0 + 3 + 1)\) odd days = 4 odd days.
1st July 2004 was 'Thursday'.
Thus, 1st Monday in July 2004 as on 5th July.

Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th.

The two conditions that decide that a year is a leap year or not is:

• For a year to be a leap year, it should be divisible by 4.
• No century is a leap year unless it is divisible by 400.

Hence, the year 2100 is not a leap year as it is not divisible by 400.

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 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Number of days between 20th March and 20th September:

11 + 30 + 31 + 30 + 31 + 31 + 20 = 184

So, 184 ÷ 7 = 2 odd days

Friday + 2 = sunday

This can be illustrated with an example. Ex: 1896 is a leap year. The next leap year comes in 1904 (1900 is not a leap year).
For a century year to be a leap year it has to be divisible by 400.

The year 2008 is a leap year.

So, it has 2 odd days.

But, Feb 2008 not included because we are calculating from March 2008 to March 2009. So it has 1 odd day only.

Given that, 5th March, 2009 is Monday.

5th March, 2008 is Sunday (1 day before to 5th March, 2009).

On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday.

• Ans. (c)
• The last day of a century cannot be either Tuesday, Thursday or Saturday.

As the 2nd day is Sunday, so the first day is Saturday

So, total we have 30+31 = 61 days = 8 weeks + 5 odd days

So, Saturday+5 = Thursday.

On Dec 31, 2005, it was Saturday.
No. odd days for the years 2006 and 2007
= Remainder of \(\frac{365+365}{7}\) = 2 odd days
On Dec 31, 2007 it has to be Monday
Thus on Jan 1, 2008 it is Tuesday.

Dividing 105 by 7 we get 105/7 = 15
No remainder =>no odd days
After 105 days it will be again Wednesday.

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Year =2007,2008,2009,2010,2011,2012,2013,2014,2015,2016,2017
Odd day=1 ,2 ,1, 1, 1, 2, 1, 1, 1, 2 ,1
respectively

Sum = 14 odd days ≡ 0 odd days. ∴ Calendar for the year 2018 will be the same as for the year 2007.

282 = \( (40\times 7)+ 2 \)

\( 40\times7 \) = 280

So, 280th day is Friday .

Hence 282th day is Sunday.

Last day of a century cannot be Tuesday or Thursday or Saturday.

• Days from 16th August 1947 to 1st January 2017

\( 16 + 30 + 31 + 30 + 31 + 18 \) leap years + \( 51 \) years

\( 138 + 18\times366 + 51\times365 \) days

\( 138 + 6588 + 18615 \)days = \( 25341 \)days

\(\frac {25341}{7} = 3620 \) weeks + 1 day

To go in past Sunday - 1 = Saturday

16th August 1947 was a Saturday.

To repeat a year there should be no odd days

1993 has 365 days
means \((52\times7)+1\)means 1 odd day
Odd day = In a given period,the number of days more than the complete weeks are called odd days.

Which means if 1 January 1993 is Sunday,then 1 January 1994 will be Monday.

1993 - 1 odd day

1994 - 1 odd day

1995- 1 odd days

1996- 2 odd day

1997- 1 odd day

1998- 1 odd day

Now total 7 odd days mean 0 odd days .

So 1999 will show same Calder as 1993 .

We shall find the day on 1st April 2001.

1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April (31 + 28 + 31 + 1) = 91 days 0 odd days.

Total number of odd days = (0 + 0 + 0) = 0

On 1st April, 2001 it was Sunday.

In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.

100/4 = 25

100 , 200 ,etc will not be a leap year

so there are 24 leap years

100- 24= 76 years.

So In a century we have 24 leap years and 76 ordinary years.

On 31st December, 2005 it was Saturday. Number of odd days from the year 2006 to the year 2008 = (1 + 1 + 2 ) = 4 days.
On 31st December 2008, it was Wednesday. Thus, on 1st Jan, 2009 it is Thursday.

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 6th March, \( 2005 \) will be \( 1 \) day beyond the day on \( 6th \) March, \( 2004 \).

Given that, \( 6th \) March, \( 2005 \) is Monday.

\( 6th \) March, \( 2004 \) is Sunday (\( 1 \) day before to \( 6th \) March, \( 2005 \))

If the third day of a month is three days before a Sunday, the Sunday of that month will be on the 6th.

27th day = Sunday

Count the number of odd days from 2014 to get sum equal to 0 odd day

2014 - 1

2015 - 1

2016 -2

2017 - 1

2018 - 1

2019 - 1

2020 -2

2021 -1

2022 -1

2023 -1

2024 - 2

sum = 14 odd days = 0 odd day

so 2025 is same as 2014

Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday.

After 61 days, it will be Saturday.

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of odd days :

1600 years have 0 odd day.

100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years) = [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.

Jan Feb Mar Apr May Jun Jul

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days)

Total number of odd days = (0 + 2) = 2.

Required day was 'Tuesday'.

We will show that the number of odd days between last day of March and last day of June is zero. .

April May June

31 + 30 + 31

= 91 days = 13 weeks = 0 odd day. ,Number of odd days during this period = 0.

Thus, 1st April of an year will be the same day as 1st July of that year. Hence, the result follows

On 31st December, 2009 it was Saturday.
Number of odd days from the year 2010 to the year 2013 = (1 + 1 + 2 + 1) = 5 days.
On 31st December 2013, it was Thursday.
Thus, on 1st Jan, 2014 it is Friday.

If the period between the two months is divisible by 7, then that two months will have the same calendar.

(a) Oct + Nov \( = 31 + 30 = 61 \) (not divisible by 7)

(b). Apr + May + Jun + Jul + Aug + Sep + Oct \( = 30 + 31 + 30 + 31 + 31 + 30 + 31 = 214 \)(not divisible by 7)

(c) Jun + July + Aug + Sep \( = 30 + 31 + 31 + 30 = 122 \)(not divisible by 7)

(d) Apr + May + June \( = 30 + 31 + 30 = 91 \) (divisible by 7)

Hence, April and July months will have the same calendar.

In 100 years, leap year =24 years.

Therefore, in 800 years, leap years will be =8 × 24 = 192

Now, since the leap year occurs once after every four years, means the 400th year and 800th also leap years.

So, the total number of leap years in 800 years=192 + 2 = 194

Hence, option A is the correct answer.

We know that in 365 days we have 1 odd day and in 366 days we have 2 odd days.

13 March 2018 = Tuesday

13 March 2019 = Wednesday

13 March 2020 = Friday

13 March 2021 = Saturday.

We cannot find out the answer because the number of days of the current month is not given.

Odd days from 10th March to 8th August:
Remaining days in March = 21
Odd days in April = 2
Odd days in May = 3
Odd days in June = 2
Odd days in July = 3
Given day in August = 8
Total odd days = (21 + 2 + 3 + 2 + 3 + 8) ÷ 7
= 39 ÷ 7
= 4
So Monday + 4 =Friday

The fourth day of the month is Wednesday

so, the first day of the month = Sunday

The third day after the 19th day of the month = the 22nd day of the month

1st day of the month = 8th day of the month = 15th day of the month = 22nd day of the month = Sunday.

We know that in 365 days we have 1 odd day.

26 January 2017 = Thursday

26 January 2018 = Friday

26 January 2019 = Saturday

26 January 2020 = Sunday.

2020 is a leap year so February has 29 days.

26 January 2021 = Sunday + 2 = Tuesday (+2 because of leap year, In 366 days we have 2 odd days)

For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0.

<b>Take the year 1992 from the given choices.</b> Total odd days in the period 1990-1991= 2 normal years ≡ 2 x 1 = 2 odd days

<b>Take the year 1995 from the given choices.</b>

Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year ≡ 4 x 1 + 1 x 2 = 6 odd days

<b>Take the year 1996 from the given choices.</b>

Number of odd days in the period 1990-1995= 5 normal years + 1 leap year ≡ 5 x 1 + 1 x 2 = 7 odd days ≡ 0 odd days (As we can reduce multiples of 7 from odd days which will not change anything)

Though number of odd days in the period 1990-1995 is 0, there is a catch here. 1990 is not a leap year whereas 1996 is a leap year. Hence calendar for 1990 and 1996 will never be the same.

<b>Take the year 2001 from the given choices.</b>

Number of odd days in the period 1990-2000= 8 normal years + 3 leap years ≡ 8 x 1 + 3 x 2 = 14 odd days ≡ 0 odd days Also, both 1990 and 2001 are normal years. Hence 1990 will have the same calendar as that of 2001

X weeks X days =(7X+X) days = 8X days.

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 \(\times \) 2 + 7 \(\times \) 1) = 11 odd days 4 odd days.

Jan. Feb. March April May June July Aug. (31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days 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.

The day after the 11th of March is Thursday means on 12th of March is Thursday.

March has 31 days.

19th march is Thursday then 26th March is Thursday.

26 + 5 = 31st March.

Thursday + 5 = Tuesday.

Hence, option D is the correct answer.

15 Aug, 1947 = (1946 years + Period from 1.1.1947 to 15.8.1947)

Odd days in 1600 years = 0

Odd days in 300 years = 1

46 years = (35 ordinary years + 11 leap years) = (35 x 1 + 11 x 2)= 57 (8 weeks + 1 day) = 1 odd day

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 + 1 + 1 + 3) = 5 odd days. Hence, as the number of odd days = 5 , given day is Friday.

After 27 days

⇒ Friday + 27

27 days\( = \frac{27}{7} \)

= 3 weeks + 6 odd days (remainder)

Friday + 27 ⇒ Friday + 6

= Thursday

Answer = Thursday (Option c)

According to the question, 14th March is Tuesday
Number of days left in March = 31-14=17
Similarly, odd days from 14th March till 13th September = \((17+30+31+30+31+31+13)\% 7\) = \(3+2+3+2+3+3+6=22\)
Now, dividing 22 by 7, remainder = 1
Thus, day on 13th September = Tuesday (+1) = Wednesday

Today is Friday.

Each day gets repeated after 7 days.

so, on dividing 363 by 7, we get remainder = 6

1 Saturday

2 Sunday

3 Monday

4 Tuesday

5 Wednesday

6 Thursday

Hence, option C is the correct answer.

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 \(\times \) 3) 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 \(\times \) 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 17) = 168 days 168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday.