Calendar

Answer: Option C
Solution:
To determine the day of the week for January 1, 2010, we can use the concept of leap years and the number of days between January 1, 2006, and January 1, 2010.

Steps:
1. Identify leap years: A leap year has 366 days (with an extra day in February).
– Leap years between 2006 and 2010: 2008 is a leap year.

2. Calculate the number of days from January 1, 2006, to January 1, 2010:
– From January 1, 2006, to January 1, 2007: 365 days (since 2006 is not a leap year).
– From January 1, 2007, to January 1, 2008: 365 days.
– From January 1, 2008, to January 1, 2009: 366 days (since 2008 is a leap year).
– From January 1, 2009, to January 1, 2010: 365 days.
3. Add the total days:
Total days = 365 + 365 + 366 + 365 = 1461 days.
4. Find the remainder when divided by 7 (since a week has 7 days) to determine the shift in the day of the week:
\[
1461 \div 7 = 208 \text{ weeks with a remainder of } 5
\] This means there is a shift of 5 days.
Since January 1, 2006, was a Sunday, we move forward by 5 days:
– Sunday + 5 days = Friday.
Thus, January 1, 2010, was a Friday.

Answer: Option D
Solution:
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) ≡ 6 odd days
Jan. Feb. March April May
(31 + 28 + 31 + 30 + 28 ) = 148 days
∴ 148 days = (21 weeks + 1 day) ≡ 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 ≡ 0 odd day.
Given day is Sunday.

Answer: Option C
Solution:
17^th 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 x 3) ≡ 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March April May June
(31 + 28 + 31 + 30 + 31 + 17) = 168 days

Answer: Option A
Solution:
To determine the day of the week for August 15, 2010, we can start with a known reference date and calculate the number of days between that reference date and August 15, 2010.

Known Reference:
We know that January 1, 2010, was a Friday.

Steps:
1. Calculate the number of days between January 1, 2010, and August 15, 2010:
– anuary 1 to January 31 = 31 days
– February 1 to February 28 = 28 days (2010 is not a leap year)
– March 1 to March 31 = 31 days
– April 1 to April 30 = 30 days
– May 1 to May 31 = 31 days
– June 1 to June 30 = 30 days
– July 1 to July 31 = 31 days
– August 1 to August 15 = 15 days

Total days = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days.

2. Find the remainder when the total days are divided by 7 (since a week has 7 days):
\[
227 \div 7 = 32 \text{ weeks with a remainder of } 3
\] This means there is a shift of 3 days forward.
Since January 1, 2010, was a Friday, moving forward by 3 days:
– Friday + 3 days = Monday.
Thus, August 15, 2010, was a Sunday.

Answer: Option B
Solution:
To determine the day of the week after 61 days from today (which is Monday), we can use the concept of the remainder when dividing the number of days by 7 (since there are 7 days in a week).

Steps:
1. Divide 61 by 7 to find the remainder:
\[
61 \div 7 = 8 \text{ weeks with a remainder of } 5
\] This means 61 days is equivalent to 8 full weeks and an additional 5 days.
2. Move forward by 5 days starting from Monday:
– Monday + 5 days = Saturday.
So, 61 days from today (Monday) will be a Saturday.

Answer: Option A
Solution:
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 6^th March, 2005 will be 1 day beyond the day on 6^th March, 2004.
Given that, 6^th March, 2005 is Monday.
∴ 6^th March, 2004 is Sunday (1 day before to 6^th March, 2005).

Answer: Option D
Solution:
To determine the dates of April 2001 on which Wednesday fell, we first need to know what day of the week April 1, 2001, was. From there, we can calculate the dates when Wednesday occurred.

Steps:
1. Find the day of the week for April 1, 2001:
We can use a reference date. For example, we know that **January 1, 2001, was a Monday.

2. Calculate the day of the week for April 1, 2001:
The number of days from January 1, 2001, to April 1, 2001, is:
– January has 31 days, so 31 – 1 = 30 days.
– February has 28 days (2001 is not a leap year), so 28 days.
– March has 31 days, so 31 days.

Total days = 30 + 28 + 31 = **89 days**.
3. Find the remainder when 89 is divided by 7 (since a week has 7 days):
\[
89 \div 7 = 12 \text{ weeks with a remainder of } 5
\] This means there is a shift of 5 days forward from Monday (since January 1, 2001, was a Monday).
So, April 1, 2001, was a Sunday.
4. Find the Wednesdays in April 2001:
If April 1, 2001, was a Sunday, then the Wednesdays in April 2001 will fall on the 4th, 11th, 18th, and 25th.
Thus, the Wednesdays in April 2001 were on April 4, 11, 18, and 25.

Answer: Option B
Solution:
To calculate the total number of days in x weeks and x days, we can use the following approach:

– 1 week = 7 days.
– Therefore, x weeks = \( 7x \) days.
– x days = \( x \) days.

So, the total number of days in \( x \) weeks and \( x \) days is:

\[
7x + x = 8x \text{ days}.
\] Thus, there are 8x days in x weeks and x days.

Answer: Option C
Solution:
100 years contain 5 odd days.
∴ Last day of 1^th century is Friday.
200 years contain (5 x 2) ≡ 3 odd days.
∴ Last day of 2^th century is Wednesday.
300 years contain (5 x 3) = 15 ≡ 1 odd day.
∴ Last day of 3^th century is Monday.
400 years contain 0 odd day.
∴ Last day of 4^th century is Sunday.
This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

Answer: Option C
Solution:
To determine the day of the week on February 8, 2004, given that February 8, 2005 was a Tuesday, we need to calculate the number of days between February 8, 2004, and February 8, 2005.

Since 2004 was a leap year, it has 366 days.

Steps:
1. Calculate the number of days between February 8, 2004, and February 8, 2005:
– From February 8, 2004, to February 8, 2005: Since 2004 is a leap year, there are 366 days.

2. Find the remainder when 366 is divided by 7 (since a week has 7 days):
\[
366 \div 7 = 52 \text{ weeks with a remainder of } 2
\] This means there is a shift of 2 days backward from Tuesday.

Since February 8, 2005, was a Tuesday, moving backward by 2 days:
– Tuesday – 2 days = Sunday.

Thus, February 8, 2004, was a Sunday.

Answer: Option D
Solution:
To find the year that has the same calendar as 2007, we need to check for a year that:

1. Has the same leap year status as 2007 (i.e., both should be either leap years or non-leap years).
2. Starts on the same day of the week.

Step-by-Step Calculation:
– 2007 was a normal year (not a leap year).
– To find the next year with the same calendar, we check when the total number of days passed is a multiple of 7 (since a week has 7 days).

Counting the extra days:
– 2008: Leap year → +2 days
– 2009: Normal year → +1 day
– 2010: Normal year → +1 day
– 2011: Normal year → +1 day
– 2012: Leap year → +2 days
– 2013: Normal year → +1 day
– 2014: Normal year → +1 day

Total extra days = (2+1+1+1+2+1+1) = 9 days ≡ 2 days (mod 7)

– 2015: Normal year → +1 day (total 3 extra days)
– 2016: Leap year → +2 days (total 5 extra days)
– 2017: Normal year → +1 day (total 6 extra days)
– 2018: Normal year → +1 day (total 7 extra days ≡ 0 mod 7 )

Answer:
The calendar for 2007 will be the same as 2018.

Answer: Option A
Solution:
The century divisible by 400 is a leap year.
∴ The year 700 is not a leap year.

Answer: Option D
Solution:
To determine the day of the week on 8th December 2006, given that **8th December 2007 was a Saturday, we need to count the number of leap years and regular years between these two dates.

Step 1: Understanding the year difference
– The year 2007 is a normal year (not a leap year).
– A normal year has 365 days, which is 65 mod 7 = 1 extra day.

Step 2: Moving backward
– Since 2007 contributes 1 extra day, the day of the week on 8th December 2006 would be one day before Saturday.

Step 3: Final Calculation
– 8th December 2007 → Saturday
– 8th December 2006 → (Saturday – 1 day) = Friday

Answer:
8th December 2006 was a Friday.

Answer: Option C
Solution:
To determine the day of the week on January 1, 2009, given that January 1, 2008, was a Tuesday, we follow these steps:

Step 1: Count the Extra Days
– 2008 is a leap year (divisible by 4), meaning it had 366 days.
– 366 days = 52 full weeks + 2 extra days (since 366 mod 7 = 2).

Step 2: Move Forward by Extra Days
– January 1, 2008 → Tuesday
– January 1, 2009 → (Tuesday + 2 days) = Thursday

Answer:
January 1, 2009, was a Thursday.

Answer: Option B
Solution:
To determine the day of the week on January 1, 2008, given that January 1, 2007, was a Monday, follow these steps:

Step 1: Count the Extra Days
– 2007 was a normal year (not a leap year), meaning it had 365 days.
– 365 days = 52 full weeks + 1 extra day (since 365 mod 7 = 1).

Step 2: Move Forward by Extra Days
– January 1, 2007 → Monday
– January 1, 2008 → (Monday + 1 day) = Tuesday
Answer:
January 1, 2008, was a Tuesday.

Answer: Option C
Solution:
To determine the day of the week on 12th January 2007, given that 12th January 2006 was a Thursday, follow these steps:

Step 1: Count the Extra Days
– 2006 was a normal year (not a leap year), meaning it had 365 days.
– 365 days = 52 full weeks + 1 extra day (since 365 mod 7 = 1 extra day).

Step 2: Move Forward by Extra Days
– 12th January 2006 → Thursday
– 12th January 2007 → (Thursday + 1 day) = Friday
Answer:
12th January 2007 was a Friday.

Answer: Option A
Solution:
22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22-Apr-2222)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)
Number of odd days in the period 2001-2200
= Number of odd days in 200 years
= 5 x 2 = 10 = 3
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
Number of odd days in the period 2201-2221
= 16 normal years + 5 leap years
= 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days
Number of days from 1-Jan-2222 to 22 Apr 2222
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112
112 days = 0 odd day
Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day
1 odd days = Monday
Hence 22 Apr 2222 is Monday

Answer: Option A
Solution:
To determine the day of the week 350 days from today (Monday), follow these steps:

1. A week has 7 days, so divide 350 by 7:
\[
350 \div 7 = 50 \text{ (exact, remainder 0)}
\] 2. Since there is no remainder, it means that exactly 50 full weeks will have passed.
3. Since a full week brings us back to the same day, the day remains Monday.
Thus, 350 days from today will still be Monday.

Answer: Option A
Solution:
To determine the day of the week on 24th July 2011, follow these steps using the day counting method:

Step 1: Choose a Reference Date
A commonly used reference date is 1st January 2000, which was a Saturday.

Step 2: Count the Number of Leap Years and Regular Years
We count the days from 1st January 2000 to 24th July 2011.

– From 2000 to 2010 (Full Years):
– Leap years: 2000, 2004, 2008 → 3 leap years × 2 extra days = 6 days
– Regular years: 2001, 2002, 2003, 2005, 2006, 2007, 2009, 2010 → 8 regular years × 1 extra day = 8 days
– Total extra days: 6 + 8 = 14 days → Equivalent to 2 weeks (0 extra days)**

– From 1st January 2011 to 24th July 2011:
– January: 31 days → 3 extra days
– February: 28 days → 0 extra days
– March: 31 days → 3 extra days
– April: 30 days → 2 extra days
– May: 31 days → 3 extra days
– June: 30 days → 2 extra days
– July (up to 24th): 24 days → 3 extra days
– Total extra days in 2011: 3 + 0 + 3 + 2 + 3 + 2 + 3 = 16 days → Equivalent to 2 weeks + 2 extra days

Answer: Option C
Solution:
Given that first day of a normal year was Friday
Odd days of the mentioned year = 1 (Since it is an ordinary year)
Hence First day of the next year = (Friday + 1 Odd day) = Saturday
Therefore, last day of the mentioned year = Friday

Answer: Option D
Solution:
To determine the day of the week on March 10, 1996, we can use a formula known as Zeller’s Congruence. Here’s how it works:

The formula is:
\[
h = (q + \left\lfloor \frac{13(m+1)}{5} \right\rfloor + K + \left\lfloor \frac{K}{4} \right\rfloor + \left\lfloor \frac{J}{4} \right\rfloor – 2J) \mod 7
\] Where:
– \(h\) is the day of the week (0 = Saturday, 1 = Sunday, 2 = Monday, …, 6 = Friday)
– \(q\) is the day of the month
– \(m\) is the month (3 = March, 4 = April, …, 12 = December; January and February are treated as months 13 and 14 of the previous year)
– \(K\) is the year of the century (i.e., year modulo 100)
– \(J\) is the century (i.e., year divided by 100)

For March 10, 1996:
– \(q = 10\)
– \(m = 3\)
– \(K = 96\) (since 1996 modulo 100)
– \(J = 19\) (since 1996 divided by 100)
Let’s plug these values into the formula to calculate the day of the week.
The day of the week on March 10, 1996, was a Sunday.

Answer: Option B
Solution:
Let’s use the same Zeller’s Congruence formula to find the day of the week for June 20, 1837.

For June 20, 1837:
– \( q = 20 \)
– \( m = 6 \)
– \( K = 37 \) (since 1837 modulo 100)
– \( J = 18 \) (since 1837 divided by 100)

I’ll calculate the day of the week using these values.
The day of the week on June 20, 1837, was a Tuesday.

Answer: Option B
Solution:
To calculate the number of days from January 26, 1996, to May 15, 1996 (both days included), we need to account for the number of days in each month between these two dates:

– January: From January 26 to January 31, there are 6 days.
– February: 1996 was a leap year, so February had 29 days.
– March: March has 31 days.
– April: April has 30 days.
– May: From May 1 to May 15, there are 15 days.

Now, let’s sum these values to get the total number of days.
There are a total of 111 days from January 26, 1996, to May 15, 1996, including both days.

Answer: Option A
Solution:
To find the day of the week after 94 days from Thursday, we divide 94 by 7 (since there are 7 days in a week) and find the remainder.

\[
94 \div 7 = 13 \text{ remainder } 3
\]

The remainder is 3, meaning that 94 days from Thursday will be 3 days later. Starting from Thursday:

– 1 day later: Friday
– 2 days later: Saturday
– 3 days later: Sunday
So, the day of the week after 94 days will be Sunday.

Answer: Option B
Solution:
Counting the years 1600 + 300 + 74
In 1600 years, there are zero odd days
In 300 years, there is one odd day
In 74 years, there are 18 leap years and 56 normal years, so the odd days are:
18(2) + 56(1) = 36 + 56 = 92,
Which is 13 weeks and 1 odd day
In 25 days of January, 1975, there are 3 weeks and 4 odd days.
Total odd days = 0 + 1 + 1 + 4
six odd days, so it was a Saturday.

Answer: Option D
Solution:
Clearly it can be understood from the question that 9 days ago was a Thursday
Number of odd days in 9 days = 2 (As 9 – 7 = 2, reduced perfect multiple of 7 from total days)
Hence today = (Thursday + 2 odd days) = Saturday

Answer: Option C
Solution:
To find the date of the last Saturday in May 2007, we can follow these steps:
1. Determine the day of the week for May 31, 2007.
2. Work backwards from May 31 to find the closest Saturday.
Let me calculate the day of the week for May 31, 2007, and then determine the last Saturday.
The last Saturday in May 2007 was on May 26th.

Answer: Option A
Solution:
If the seventh day of the month is three days earlier than Friday, we can first determine what day the 7th is:

– Three days earlier than Friday is Tuesday.

Now, we need to find out what day it will be on the 19th. Since the 7th is a Tuesday, we can count the days between the 7th and the 19th:

– From the 7th to the 19th, there are 12 days.
– Dividing 12 by 7 gives a remainder of 5, so we count 5 days forward from Tuesday:

– 1 day after Tuesday is Wednesday.
– 2 days after Tuesday is Thursday.
– 3 days after Tuesday is Friday.
– 4 days after Tuesday is Saturday.
– 5 days after Tuesday is Sunday.
Thus, the 19th day of the month will be a Sunday.

Answer: Option D
Solution:
This is a leap year. So, none of the next 3 years will be leap years. Each year will give one odd day so the day of the week will be 3 odd
days beyond Monday i.e. it will be Thursday.

Answer: Option C
Solution:
If October 1st is a Sunday, we can determine the day of the week for November 1st by first calculating how many days are in October.

Since October has 31 days, we need to find the remainder when 31 is divided by 7 (since there are 7 days in a week).

\[
31 \div 7 = 4 \text{ weeks with a remainder of } 3
\] This means that October 31st is 3 days after Sunday, which would be:
– Monday (1 day after Sunday)
– Tuesday (2 days after Sunday)
– Wednesday (3 days after Sunday)
So, November 1st will be Wednesday.

Answer: Option D
Solution:
To find the number of odd days in 123 days, we need to divide 123 by 7 (since there are 7 days in a week) and find the remainder.
\[
123 \div 7 = 17 \text{ weeks with a remainder of } 4
\] So, the number of odd days in 123 days is 4.

Answer: Option C
Solution:
After every 400 years, the same day occurs. (Because a period of 400 years has 0 odd days)
Thus, 18^th April 1603 is Thursday, After 400 years i.e., on 18^th April 2003 has to be Thursday.

Answer: Option B
Solution:
To find the day of the week after 126 days from Friday, we divide 126 by 7 (since there are 7 days in a week) and find the remainder.
\[
126 \div 7 = 18 \text{ with a remainder of } 0
\] Since the remainder is 0, this means that after 126 days, the day of the week will remain the same as today, which is Friday.

Answer: Option A
Solution:
Given that 25th August = Thursday
Hence 29th August = Monday
So 22nd,15th and 8th and 1st of August also will be Mondays
Number of Mondays in August = 5

Answer: Option C
Solution:
In 400 consecutive years there are 97 leap years. Hence, in 400 consecutive years February has the 29th day 97 times and the remaining
eleven months have the 29th day 400 × 11 or 4400 times.
Thus the 29th day of the month occurs
= 4400 + 97
= 4497 times.

Answer: Option D
Solution:
Let’s break this down step by step:

Given:
– The month has 30 days.
– The month starts on a Saturday.
– The second Saturday and every Sunday are holidays.

Step 1: Identify the holidays
– First Saturday: Since the month starts on a Saturday, the 1st is a Saturday, and the first Saturday will be on the 1st.
– Second Saturday: The second Saturday will be on the 8th.
– Sundays: Since the month starts on a Saturday, the Sundays fall on the 2nd, 9th, 16th, 23rd, and 30th.

Step 2: List the holidays
– Second Saturday: 8th
– Sundays: 2nd, 9th, 16th, 23rd, 30th

Step 3: Calculate the total number of holidays
– There are 1 second Saturday and 5 Sundays, so the total number of holidays is:
\[
1 + 5 = 6 \text{ holidays}.
\]

Step 4: Find the total number of working days
The total number of days in the month is 30, and there are 6 holidays. Therefore, the number of working days is:
\[
30 – 6 = 24 \text{ working days}.
\] Thus, there will be 24 working days in the month.

Answer: Option D
Solution:
To determine the day of the week on December 9, 1971, given that December 9, 2001, was a Sunday, we need to calculate the number of years between 1971 and 2001 and account for the leap years in this period.

Steps:
1.Number of years between 1971 and 2001:
This is 30 years (from December 9, 1971, to December 9, 2001).

2. Determine the number of leap years:
Leap years occur every 4 years, except for century years that are not divisible by 400. We need to count how many leap years occurred between 1971 and 2001.

– Leap years are: 1972, 1976, 1980, 1984, 1988, 1992, 1996, and 2000.
– There are 8 leap years between 1971 and 2001.

3. Total number of extra days:
Each regular year contributes 1 extra day (since 365 days mod 7 gives a remainder of 1), and each leap year contributes 2 extra days (since 366 days mod 7 gives a remainder of 2).

So, for 30 years:
– 22 regular years contribute 22 days.
– 8 leap years contribute 16 days.

Total extra days = \( 22 + 16 = 38 \) days.

4. Find the remainder when divided by 7:
Now, we calculate the number of odd days (days that don’t fit into full weeks) by dividing 38 by 7:
\[
38 \div 7 = 5 \text{ weeks with a remainder of } 3
\]

This means there are 3 extra days between December 9, 1971, and December 9, 2001.

5. Calculate the day of the week:
Since December 9, 2001, is a Sunday, we move 3 days back:
– 1 day back: Saturday
– 2 days back: Friday
– 3 days back: Thursday
Thus, December 9, 1971, was a Thursday.

Answer: Option B
Solution:
1 Jan 1901 = (1900 years + 1^th Jan 1901)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601 – 1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
1^th Jan 1901 = 1 odd day
Total number of odd days = (0 + 1 + 1) = 2
2 odd days = Tuesday
Hence 1 January 1901 is Tuesday.

Answer: Option D
Solution:
This is a leap year.
So, none of the next 3 years will be leap years.
Each ordinary year has one odd day, so there are 3 odd days in next 3 years.
So the day of the week will be 3 odd days beyond Wednesday i.e. it will be Saturday

Answer: Option D
Solution:
To determine the day of the week for January 1, 2005, given that January 1, 2004, was a Thursday, we need to account for whether 2004 was a leap year.
Steps:
1.2004 is a leap year, so it has 366 days. A leap year causes the day of the week to advance by 2 days (since 366 days divided by 7 leaves a remainder of 2).
2. Starting from Thursday (January 1, 2004):
– 1 day after Thursday is Friday.
– 2 days after Thursday is Saturday.
Therefore, January 1, 2005, will fall on a Saturday.

Answer: Option C
Solution:
Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
\( \frac{26 + 1 + 50 + 12 + 0}{7} \)
\( \frac{80}{7} \)
= 5
= Thursday

Answer: Option C
Solution:
To find the day of the week 30 days ago from today (Thursday), we divide 30 by 7 (since there are 7 days in a week) and find the remainder.

\[
30 \div 7 = 4 \text{ weeks with a remainder of } 2
\]

So, 30 days ago, the day of the week would be 2 days before Thursday.

– 1 day before Thursday is Wednesday.
– 2 days before Thursday is Tuesday.
Therefore, 30 days ago, it was Tuesday.

Answer: Option B
Solution:
To determine the day of the week for January 1, 2008, given that January 1, 2007, was a Monday, we need to consider whether 2007 is a leap year or not.

– 2007 is a regular (non-leap) year, which means it has 365 days.
– A regular year advances the day of the week by 1 day (because 365 days divided by 7 leaves a remainder of 1).

Since January 1, 2007, was a Monday, and 2007 is a regular year, we add 1 day:
– Monday + 1 day = Tuesday.
Therefore, January 1, 2008, was a Tuesday.

Answer: Option C
Solution:
To determine the fourth Tuesday of December 1991, given that December 1, 1991, is the first Sunday, we can break it down step by step:

Step 1: Identify the first Sunday of December 1991
– December 1, 1991 is a Sunday.
Step 2: Find the first Tuesday of December 1991
– Since December 1, 1991, is a Sunday, the next Tuesday will be December 3, 1991.
Step 3: Identify the subsequent Tuesdays
– The first Tuesday is December 3, 1991.
– The second Tuesday is December 10, 1991.
– The third Tuesday is December 17, 1991.
– The fourth Tuesday is December 24, 1991.
Thus, the fourth Tuesday of December 1991 is December 24, 1991.

Answer: Option A
Solution:
To find the day of the week after 59 days from today (Thursday), we divide 59 by 7 to find the remainder:

\[
59 \div 7 = 8 \text{ weeks with a remainder of } 3
\]

So, 59 days from Thursday will be 3 days ahead.

– 1 day after Thursday is Friday.
– 2 days after Thursday is Saturday.
– 3 days after Thursday is Sunday.

Therefore, the day 59 days from today will be Sunday.

Answer: Option C
Solution:
Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
\( \frac{21 + 6 + 87 + 21 + 0}{7} \)
\( \frac{135}{7} \)
= 2
= Monday

Answer: Option A
Solution:
To calculate the number of odd days from 13th May 2005 to 19th August 2005** (both inclusive), let’s break down the steps:

Step 1: Count the total number of days
We need to find the total number of days between 13th May 2005 and 19th August 2005. First, we’ll calculate the days in each of the months involved:

– May 2005: From 13th May to the end of May (31st May):
\[
31 – 13 + 1 = 19 \text{ days}
\]

– June 2005: June has 30 days.
\[
30 \text{ days}
\]

– July 2005: July has 31 days.
\[
31 \text{ days}
\]

– August 2005: From 1st August to 19th August:
\[
19 \text{ days}
\]

Step 2: Calculate the total number of days
Total days from 13th May to 19th August:
\[
19 + 30 + 31 + 19 = 99 \text{ days}
\]

Step 3: Calculate the number of odd days
To find the number of odd days, divide the total number of days by 7 and find the remainder.

\[
99 \div 7 = 14 \text{ weeks with a remainder of } 1
\] Therefore, there is 1 odd day between 13th May 2005 and 19th August 2005 (both inclusive).

Answer: Option B
Solution:
For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0. Take the year 1992 from the given choices.
Total odd days in the period 1990-1991 = 2 normal years
⇒ 2 x 1 = 2 odd daysTake the year 1995 from the given choices.
Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
⇒ 4 x 1 + 1 x 2 = 6 odd daysTake the year 1996 from the given choices.
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.Take the year 2001 from the given choices. 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

Answer: Option B
Solution:
To calculate the number of odd days in 126 years, we need to account for both leap years and non-leap years.

Step 1: Number of leap years in 126 years
A leap year occurs every 4 years, but years divisible by 100 are not leap years unless they are divisible by 400.

To find the number of leap years, we use the following formula:
– Leap years = (Number of years divisible by 4) – (Number of years divisible by 100) + (Number of years divisible by 400).

For 126 years:
– Leap years divisible by 4: \( \left\lfloor \frac{126}{4} \right\rfloor = 31 \)
– Leap years divisible by 100: \( \left\lfloor \frac{126}{100} \right\rfloor = 1 \)
– Leap years divisible by 400: \( \left\lfloor \frac{126}{400} \right\rfloor = 0 \)

Thus, the number of leap years in 126 years is:
\[
31 – 1 + 0 = 30
\]

Step 2: Number of non-leap years
Non-leap years = Total years – Leap years = \( 126 – 30 = 96 \)

Step 3: Calculate the number of odd days
– Each non-leap year contributes 1 odd day.
– Each leap year contributes 2 odd days.

Now, calculate the total number of odd days:
\[
\text{Total odd days} = (96 \times 1) + (30 \times 2) = 96 + 60 = 156
\]

Step 4: Find the remainder when divided by 7 (since a week has 7 days)
\[
156 \div 7 = 22 \text{ weeks with a remainder of } 2
\] Therefore, the number of odd days in 126 years is 2.

Answer: Option B
Solution:
If the day before yesterday was Thursday, then today is Saturday.

Now, to find out when Sunday will be:

– From Saturday, the next day is Sunday.

So, Sunday will be tomorrow.