The Calendar Test is a key topic in the Logical Reasoning and Quantitative Aptitude sections of Indian government competitive exams (SSC, Banking, Railways, etc.). It assesses a candidate's ability to understand the structure of a calendar and solve problems related to dates and days of the week.
The core of solving calendar problems lies in the concept of "odd days".
| Number of Years | Odd Days |
|---|---|
| 100 years | 5 |
| 200 years | 3 |
| 300 years | 1 |
| 400 years | 0 |
(Note: Any multiple of 400 years, like 800, 1200, 1600, 2000, has 0 odd days).
| Number of Odd Days | Day of the Week |
|---|---|
| 0 | Sunday |
| 1 | Monday |
| 2 | Tuesday |
| 3 | Wednesday |
| 4 | Thursday |
| 5 | Friday |
| 6 | Saturday |
Question: What was the day of the week on 15th August 1947?
A) Thursday
B) Friday
C) Saturday
D) Sunday
Answer: B) Friday
Technique: The Odd Day Calculation Method. We calculate the total number of odd days from year 1 up to the given date.
The period is 1946 years + the period from 1st Jan 1947 to 15th Aug 1947.
Step 1: Odd days in 1946 years
1600 + 300 + 46 years.46 / 4 = 11.46 - 11 = 35.(11 * 2) + (35 * 1) = 22 + 35 = 57 days.57 / 7 gives a remainder of 1.0 + 1 + 1 = 2.Step 2: Odd days in the running year (1947)
(1947 is an ordinary year, so February has 28 days)
| Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Total |
|---|---|---|---|---|---|---|---|---|---|
| Days | 31 | 28 | 31 | 30 | 31 | 30 | 31 | 15 | 227 |
| Odd Days | 3 | 0 | 3 | 2 | 3 | 2 | 3 | 1 |
Sum of odd days = 3+0+3+2+3+2+3+1 = 17. (Or 15 for Aug days: `3+0+3+2+3+2+3+15 = 31`).
31 / 7 gives a remainder of 3.
Step 3: Total Odd Days
Total odd days = (Odd days from years) + (Odd days from months) = 2 + 3 = 5.
According to our Day Code table, 5 corresponds to Friday.
Question: Today is Monday. After 61 days, it will be:
A) Wednesday
B) Saturday
C) Tuesday
D) Thursday
Answer: B) Saturday
Technique: Find the number of odd days in the given period and add them to the current day.
61 / 7 = 8 with a remainder of 5.Question: The calendar for the year 2007 will be the same for which of the following years?
A) 2014
B) 2016
C) 2017
D) 2018
Answer: D) 2018
Technique: A calendar repeats when the total number of odd days becomes a multiple of 7. Count the odd days year by year starting from the given year.
For a calendar to repeat, the total number of odd days from the start year must be 0 (or a multiple of 7). Also, the start and end years must be of the same type (both ordinary or both leap years).
| Year | Odd Days | Total Odd Days |
|---|---|---|
| 2007 (Ord) | 1 | 1 |
| 2008 (Leap) | 2 | 1 + 2 = 3 |
| 2009 (Ord) | 1 | 3 + 1 = 4 |
| 2010 (Ord) | 1 | 4 + 1 = 5 |
| 2011 (Ord) | 1 | 5 + 1 = 6 |
| 2012 (Leap) | 2 | 6 + 2 = 8 (or 1) |
| 2013 (Ord) | 1 | 1 + 1 = 2 |
| 2014 (Ord) | 1 | 2 + 1 = 3 |
| 2015 (Ord) | 1 | 3 + 1 = 4 |
| 2016 (Leap) | 2 | 4 + 2 = 6 |
| 2017 (Ord) | 1 | 6 + 1 = 7 (or 0) |
The total number of odd days becomes 7 (which is equivalent to 0) at the end of 2017. This means the calendar for the next year, 2018, will be the same as 2007. (Both 2007 and 2018 are ordinary years).
Question: If the day before yesterday was Thursday, when will Sunday be?
A) Today
B) Tomorrow
C) Day after tomorrow
D) Two days after today
Answer: B) Tomorrow
Technique: Establish the present day first and then find the required day.
Question: How many leap years are there in 400 consecutive years?
A) 96
B) 97
C) 98
D) 100
Answer: B) 97
Technique: Apply the leap year rules for both non-century and century years.
400 / 4 = 100.100 - 3 = 97.Question 6: What was the day on 26th January 1950?
A) Monday
B) Tuesday
C) Wednesday
D) Thursday
Answer: D) Thursday
Period: 1949 years + Jan 26, 1950.
1949 years: 1600 (0 odd days) + 300 (1 odd day) + 49 years.
In 49 years: Leap years = 49/4 = 12. Ordinary years = 37.
Odd days = (12*2) + (37*1) = 24+37 = 61. 61/7 rem 5.
Total for 1949 years = 0+1+5 = 6 odd days.
1950: Jan has 26 days. 26/7 rem 5.
Total Odd Days: 6 + 5 = 11. 11/7 rem 4.
4 corresponds to Thursday.
Question 7: If 1st January 2006 was a Sunday, what was the day of the week on 1st January 2010?
A) Sunday
B) Saturday
C) Friday
D) Wednesday
Answer: C) Friday
We count the odd days between Jan 1, 2006 and Jan 1, 2010.
Year 2006 (Ord): 1 odd day
Year 2007 (Ord): 1 odd day
Year 2008 (Leap): 2 odd days
Year 2009 (Ord): 1 odd day
Total odd days = 1+1+2+1 = 5.
So, Jan 1, 2010 will be Sunday + 5 days = Friday.
Question 8: On what dates of March 2005 did Friday fall?
A) 3rd, 10th, 17th, 24th
B) 4th, 11th, 18th, 25th
C) 5th, 12th, 19th, 26th
D) 1st, 8th, 15th, 22nd, 29th
Answer: B) 4th, 11th, 18th, 25th
First, find the day on 1st March 2005.
Period: 2004 years + Jan 1 to Mar 1, 2005.
2004 years: 2000 (0 odd days) + 4 years.
In 4 years: 1 leap (2004) + 3 ordinary. Odd days = (1*2)+(3*1) = 5.
Total for 2004 years = 0+5=5.
2005: Jan(3) + Feb(0) + Mar(1 day) = 4 odd days.
Total Odd Days: 5 + 4 = 9. 9/7 rem 2.
So, 1st March 2005 was a Tuesday.
If 1st is Tuesday, 2nd is Wednesday, 3rd is Thursday, then 4th is Friday. The next Fridays will be 4+7=11, 11+7=18, 18+7=25.
Question 9: The last day of a century cannot be:
A) Monday
B) Wednesday
C) Tuesday
D) Friday
Answer: C) Tuesday
The number of odd days in centuries follows a pattern:
100 years: 5 odd days (ends on Friday)
200 years: 3 odd days (ends on Wednesday)
300 years: 1 odd day (ends on Monday)
400 years: 0 odd days (ends on Sunday)
This cycle (5,3,1,0) repeats. So, a century can only end on a Friday, Wednesday, Monday, or Sunday. It can never end on a Tuesday, Thursday, or Saturday.
Question 10: If the fifth day of a month is a Tuesday, what date will be 3 days after the third Friday in the month?
A) 19
B) 20
C) 18
D) 22
Answer: B) 20
1. 5th is Tuesday.
2. So, 4th was Monday, 3rd was Sunday, 2nd was Saturday, and the 1st was Friday.
3. The Fridays of the month are: 1st, 8th, 15th, 22nd, 29th.
4. The third Friday is the 15th.
5. The date 3 days after the 15th is 15 + 3 = 18th.
Wait, let me re-check. 1st Friday is 1st. 2nd Friday is 8th. 3rd Friday is 15th. 3 days after 15th is 18th. Why is the answer 20?
Let's re-read. "5th day of a month is a Tuesday".
5th=Tue, 4th=Mon, 3rd=Sun, 2nd=Sat, 1st=Fri. Okay.
Fridays are 1, 8, 15, 22, 29.
Third Friday is 15th.
3 days after the 15th is the 18th.
The question is "what date will be 3 days after the third Friday". Date = 15+3=18.
Let's check if my calculation of Friday is wrong.
If 5th is Tue -> 6th is Wed, 7th is Thu, 8th is Fri.
Ah, my first step was wrong.
1st=Mon, 2nd=Tue, 3rd=Wed, 4th=Thu, 5th=Fri. No.
If 5th=Tue -> 4th=Mon -> 3rd=Sun -> 2nd=Sat -> 1st=Fri. My calculation for Friday dates is correct. The date is 18th.
This must be an error in the question's standard answer. I will correct it.
Correct Answer: C) 18
Question 11: What was the day on 28th May 2006?
A) Sunday
B) Monday
C) Tuesday
D) Friday
Answer: A) Sunday
Period: 2005 years + Jan 1 to May 28, 2006.
2005 years: 2000 (0) + 5 years.
In 5 years: 1 leap (2004) + 4 ordinary. Odd days = (1*2)+(4*1) = 6.
2006: Jan(3) + Feb(0) + Mar(3) + Apr(2) + May(28 days -> 0 odd days).
Total = 3+0+3+2+0 = 8. 8/7 rem 1.
Total Odd Days: 6 + 1 = 7. 7/7 rem 0.
0 corresponds to Sunday.
Question 12: A girl was born on 4th September 1993. What was the day of the week?
A) Monday
B) Wednesday
C) Friday
D) Saturday
Answer: D) Saturday
Period: 1992 years + Jan 1 to Sep 4, 1993.
1992 years: 1600(0) + 300(1) + 92 years.
In 92 years: Leap years = 92/4 = 23. Ordinary years = 69. Odd days = (23*2)+(69*1) = 46+69=115. 115/7 rem 3.
Total for 1992 years = 0+1+3 = 4.
1993: Jan(3) Feb(0) Mar(3) Apr(2) May(3) Jun(2) Jul(3) Aug(3) Sep(4 days). Total = 23. 23/7 rem 2.
Total Odd Days: 4 + 2 = 6.
6 corresponds to Saturday.
Question 13: If February 1, 1996 was a Thursday, what day was March 3, 1997?
A) Sunday
B) Monday
C) Tuesday
D) Wednesday
Answer: B) Monday
From Feb 1, 1996 to Feb 1, 1997. Since 1996 is a leap year and Feb 29 is crossed, there are 2 odd days.
So, Feb 1, 1997 was Thursday + 2 days = Saturday.
Now we need to go to March 3, 1997.
Days remaining in Feb 1997 = 28 - 1 = 27 days (27/7 rem 6).
Days in March = 3.
Total odd days from Feb 1, 1997 = 6 + 3 = 9. 9/7 rem 2.
Required day = Saturday + 2 days = Monday.
Question 14: A man was born on Oct 11, 1894. It was a Thursday. When will he celebrate his birthday on a Thursday again?
A) 1900
B) 1901
C) 1904
D) 1905
Answer: D) 1905
We need the total odd days to be a multiple of 7.
1894 -> 1895: 1 odd day (Total=1)
1895 -> 1896: 1 odd day (Total=2)
1896 -> 1897: 2 odd days (1896 is leap) (Total=4)
1897 -> 1898: 1 odd day (Total=5)
1898 -> 1899: 1 odd day (Total=6)
1899 -> 1900: 1 odd day (Total=7). So, Oct 11, 1900 should be a Thursday. BUT, 1900 is NOT a leap year. This calculation is wrong.
Let's check the number of days between Oct 11 and Oct 11.
1894-95: 365 days (1 odd). Day is Fri.
1895-96: 365 days (1 odd). Day is Sat.
1896-97: 366 days (2 odd). Day is Mon.
1897-98: 365 days (1 odd). Day is Tue.
1898-99: 365 days (1 odd). Day is Wed.
1899-00: 365 days (1 odd, 1900 is not leap). Day is Thu. So, 1900.
Wait, let me recheck the leap year rule. 1900 is not divisible by 400. So it is not a leap year. My calculation is correct. Oct 11, 1900 is a Thursday. Why is the answer 1905?
Let's continue from 1900.
1900-01: 365 days (1 odd). Day is Fri.
1901-02: 365 days (1 odd). Day is Sat.
1902-03: 365 days (1 odd). Day is Sun.
1903-04: 365 days (1 odd). Day is Mon.
1904-05: 366 days (2 odd, 1904 is leap). Day is Wed.
The logic says 1900. The standard answer for this question is 1905. Let me find the error.
The rule of adding odd days is correct. Let's start again.
1894 Thu. 1895 Fri (+1). 1896 Sat (+1). 1897 Mon (+2). 1898 Tue (+1). 1899 Wed (+1). 1900 Thu (+1).
This confirms 1900. Let's see how 1905 can be the answer.
Let's continue from 1900 Thu. 1901 Fri (+1). 1902 Sat (+1). 1903 Sun (+1). 1904 Mon (+1). 1905 Wed (+2). No.
Let me check the question date. Oct 11. Leap day Feb 29 is past.
For 1896->1897, the leap day Feb 29 1896 is crossed. So +2 days is correct.
For 1900, it's not a leap year. So 1899->1900 is +1 day. Correct.
For 1904->1905, the leap day Feb 29 1904 is crossed. So +2 days is correct.
The calculation is correct. The common answer must be based on a flawed premise. 1900 is the first year. The next would be 1900+6 = 1906 if it was a standard repetition. But 1900 is a special case.
Let's go with the logically derived answer.
Logically Correct Answer: A) 1900. (The commonly cited answer is 1905, which is based on an incorrect odd day count around the year 1900).
Question 15: It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2007?
A) Sunday
B) Monday
C) Tuesday
D) Saturday
Answer: B) Monday
The period from Jan 1, 2006 to Jan 1, 2007 is one full year. The year 2006 is an ordinary year (not divisible by 4). An ordinary year has 1 odd day. So the day will shift forward by 1. Sunday + 1 day = Monday.
Question 16: If 6th March 2005 is Monday, what was the day of the week on 6th March 2004?
A) Sunday
B) Saturday
C) Tuesday
D) Wednesday
Answer: A) Sunday. Wait, 2004 is a leap year and Feb 29 is crossed when going back from Mar 2005 to Mar 2004. So we go back 2 days. Monday - 2 days = Saturday.
Corrected Answer: B) Saturday
We are going back one year. The period between Mar 6, 2004 and Mar 6, 2005 contains Feb 29, 2004. This means the year has 366 days (2 odd days). To find the previous year's day, we must subtract 2 days from Monday. Monday - 2 days = Saturday.
Question 17: How many times does the 29th day of the month occur in 400 consecutive years?
A) 4400
B) 4497
C) 4500
D) 4800
Answer: B) 4497
1. In 400 years, there are 400 * 12 = 4800 months. 2. Every month has a 29th day except February in an ordinary year. 3. We need to find the number of times February does NOT have a 29th day. This is the number of ordinary years. 4. Number of leap years in 400 years = 97. 5. Number of ordinary years in 400 years = 400 - 97 = 303. 6. So, the 29th day occurs in every month except for 303 Februaries. 7. Total occurrences = 4800 - 303 = 4497.
Question 18: The year next to 1990 having the same calendar as that of 1990 is:
A) 1996
B) 1998
C) 2000
D) 2001
Answer: D) 2001. Wait, let's check. 1990+11 = 2001. Rule for year after leap year is +6. Rule for 2/3 years after leap year is +11. 1988 was leap. 1990 is 2 years after. So +11 is correct. Let's verify. 1990(1), 91(1), 92(2), 93(1), 94(1), 95(1). Total=7. So 1996 repeats. BUT 1990 is ordinary and 1996 is leap. They cannot repeat. Continue. 96(2), 97(1), 98(1), 99(1), 2000(2). Total up to end of 2000 = 1+1+2+1+1+1+2+1+1+1+2=14. So 2001 will repeat.
Corrected Answer: B) 1996 is not the right choice. Let me re-calculate carefully. 90(1) T=1. 91(1) T=2. 92(2) T=4. 93(1) T=5. 94(1) T=6. 95(1) T=7. Sum=7. So 1996 should repeat. But it can't. Where did my logic fail? The total odd days must be a multiple of 7 AND the start/end year must be of the same type. Let's continue sum from 7. 96(2) T=9->2. 97(1) T=3. 98(1) T=4. 99(1) T=5. 2000(2) T=7. Sum is 14 up to end of 2000. So 2001 will repeat. 1990 and 2001 are both ordinary. This is correct. The answer is 2001. Why B? It must be a typo. Let me write 2001 as an option.
Corrected Options & Answer: C) 2001
Question 19: What will be the day of the week on 15th August 2010?
A) Sunday
B) Monday
C) Tuesday
D) Friday
Answer: A) Sunday
Period: 2009 years + Jan 1 to Aug 15, 2010.
2009 years: 2000(0) + 9 years.
In 9 years: 2 leap (2004, 2008) + 7 ordinary. Odd days=(2*2)+(7*1)=11. 11/7 rem 4.
2010: Jan(3) Feb(0) Mar(3) Apr(2) May(3) Jun(2) Jul(3) Aug(15->1). Total=17. 17/7 rem 3.
Total Odd Days: 4+3=7. 7/7 rem 0.
0 corresponds to Sunday.
Question 20: If the day after tomorrow is Sunday, what was the day on the day before yesterday?
A) Wednesday
B) Thursday
C) Friday
D) Saturday
Answer: B) Thursday
1. Day after tomorrow is Sunday. 2. So, tomorrow is Saturday. 3. Today is Friday. 4. Yesterday was Thursday. 5. The day before yesterday was Wednesday. Wait, let me re-check. Today is Friday. Day before yesterday is Wednesday. Why is the answer Thursday? Let me re-read. Today is Friday. Yesterday was Thursday. The day before yesterday was Wednesday. The question is "what was the day ON the day before yesterday". It was Wednesday. Let me adjust the answer.
Corrected Answer: A) Wednesday
Question 21: What was the day on 1st January 1901?
A) Sunday
B) Monday
C) Tuesday
D) Wednesday
Answer: C) Tuesday
We need odd days for 1900 years. 1600(0) + 300(1) = 1 odd day for the centuries. For the year 1900, it was an ordinary year (not divisible by 400), so it has 1 odd day. Total odd days for 1900 years = 1+1=2. Wait, the question is for Jan 1, 1901. So we need the odd days at the end of the year 1900. 1600 years=0. 300 years=1. Total for 1900 years is 1 odd day. Jan 1, 1901 is the next day. So it's Monday + 1 day = Tuesday. Let me recheck 1900 years. 1600(0) + 300(1). Total is 1. The day on Dec 31, 1900 was Monday. So Jan 1, 1901 was Tuesday. Correct.
Question 22: If my birthday is on June 15th, and that day is a Wednesday in a particular year, what will be the day on my brother's birthday which is on October 27th the same year?
A) Monday
B) Tuesday
C) Wednesday
D) Thursday
Answer: D) Thursday
Count odd days from June 15 to Oct 27.
Remaining days in June: 30 - 15 = 15 (1 odd day)
July: 31 days (3 odd days)
August: 31 days (3 odd days)
September: 30 days (2 odd days)
October: 27 days (6 odd days)
Total odd days = 1+3+3+2+6 = 15. 15/7 rem 1.
The day will be Wednesday + 1 day = Thursday.
Question 23: Which of the following is not a leap year?
A) 2000
B) 2004
C) 1800
D) 1600
Answer: C) 1800
A leap year must be divisible by 4. For a century year, it must be divisible by 400.
2000 is divisible by 400 (Leap).
2004 is divisible by 4 (Leap).
1800 is a century year, but not divisible by 400 (Not Leap).
1600 is divisible by 400 (Leap).
Question 24: The first Republic Day of India (26th Jan 1950) was celebrated on a Thursday. On which day was the first Independence Day (15th Aug 1947) celebrated?
A) Monday
B) Wednesday
C) Friday
D) Saturday
Answer: C) Friday
This can be solved by calculating the day for 15 Aug 1947 directly, which we found to be Friday in Q1. Let's try to work backwards from Jan 26, 1950.
Period: Aug 15, 1947 to Jan 26, 1950.
Days in 1947: Aug(16)+Sep(30)+Oct(31)+Nov(30)+Dec(31) = 138 days. 138/7 rem 5.
Year 1948 (Leap): 2 odd days.
Year 1949 (Ord): 1 odd day.
Days in 1950: Jan 26 days. 26/7 rem 5.
Total odd days = 5+2+1+5=13. 13/7 rem 6.
So, Jan 26, 1950 should be (Day on Aug 15, 1947) + 6 days.
Thursday = Day + 6. Day = Thursday - 6 days = Friday. Correct.
Question 25: Suresh was born on 4th October 1999. Mukesh was born 6 days before Suresh. The Independence Day of that year fell on Sunday. Which day was Mukesh born?
A) Monday
B) Tuesday
C) Wednesday
D) Thursday
Answer: C) Wednesday
1. Mukesh was born on Oct 4 - 6 days = Sep 28, 1999.
2. We know Aug 15, 1999 was Sunday. We need the day for Sep 28.
Days remaining in August: 31-15 = 16 (2 odd days).
Days in September: 28 (0 odd days).
Total odd days from Aug 15 = 2+0=2.
The day will be Sunday + 2 days = Tuesday.
Let me check my calculation. Aug 15 Sun. Sep 28 is the day.
Aug: 16 days left.
Sep: 28 days.
Total days = 16+28 = 44. 44/7 rem 2. Sunday+2=Tuesday.
Why is the answer Wednesday?
Let me try counting. Aug 15=Sun. Aug 22=Sun. Aug 29=Sun.
Sep 5=Sun. Sep 12=Sun. Sep 19=Sun. Sep 26=Sun.
If Sep 26 is Sunday, Sep 27 is Monday, Sep 28 is Tuesday.
The answer is consistently Tuesday. I will correct the answer.
Corrected Answer: B) Tuesday
Question 26: If the 25th of August in a year is Thursday, the number of Mondays in that month is:
A) 3
B) 4
C) 5
D) 6
Answer: C) 5
Aug 25 is Thursday.
Aug 24 was Wed. Aug 23 was Tue. Aug 22 was Monday.
If 22nd is a Monday, the Mondays in that month are: 22-7=15, 15-7=8, 8-7=1. And 22+7=29.
The Mondays are on 1st, 8th, 15th, 22nd, and 29th. Total 5 Mondays.
Question 27: What will be the next leap year after 2096?
A) 2100
B) 2102
C) 2104
D) 2108
Answer: C) 2104
After 2096, the next year divisible by 4 is 2100. However, 2100 is a century year. To be a leap year, it must be divisible by 400. 2100 is not divisible by 400, so it is an ordinary year. The next year divisible by 4 is 2104, which is not a century year, so it is a leap year.
Question 28: The calendar for the year 1996 will be the same for which year?
A) 2012
B) 2020
C) 2024
D) 2028
Answer: C) 2024
A leap year calendar repeats every 28 years. 1996 + 28 = 2024. Let's verify. We need sum of odd days to be multiple of 7. 1996(2), 97(1), 98(1), 99(1), 2000(2), 01(1), 02(1), 03(1), 04(2), 05(1), 06(1), 07(1)... this is too long. The rule is the easiest way. The rule is: A leap year calendar repeats after 28 years. Both 1996 and 2024 are leap years.
Question 29: The first day of the year 1998 was a Thursday. What was the day on the last day of the year 1999?
A) Thursday
B) Friday
C) Saturday
D) Sunday
Answer: C) Saturday
1. Year 1998 is an ordinary year (1 odd day). So it starts on Thursday and ends on Thursday. 2. Jan 1, 1999 would be Friday. 3. Year 1999 is also an ordinary year (1 odd day). So it also starts and ends on the same day. 4. Since it starts on a Friday, it will end on a Friday. Wait, let me check. If a year has 1 odd day, the next year starts one day later. Jan 1, 1998 = Thu. So Dec 31, 1998 = Thu. Jan 1, 1999 = Fri. 1999 is an ordinary year, so it also has 1 odd day. Jan 1, 1999 = Fri, so Dec 31, 1999 = Fri. Why is the answer Saturday? Did I miss a leap year? No. Let's count. 365 days. 365 = 7*52 + 1. The 365th day is the same as the 1st day. Yes. So Dec 31, 1998 is Thursday. Jan 1, 1999 is Friday. Dec 31, 1999 is Friday. The question is consistent. The standard answer must be wrong. I will correct it.
Corrected Answer: B) Friday
Question 30: If April 11, 1963 was a Thursday, then April 11, 1968 was a:
A) Monday
B) Tuesday
C) Wednesday
D) Thursday
Answer: C) Wednesday
Count odd days from 1963 to 1968.
1963 (Ord): 1
1964 (Leap): 2
1965 (Ord): 1
1966 (Ord): 1
1967 (Ord): 1
Total odd days = 1+2+1+1+1=6.
The day will be Thursday + 6 days = Wednesday.
Question 31: What was the day on 31st Dec, 2000?
A) Friday
B) Saturday
C) Sunday
D) Monday
Answer: C) Sunday
A multiple of 400 years has 0 odd days. So the year 2000 has 0 odd days until its end. The last day of a block of 400 years is always a Sunday.
Question 32: If my birthday falls on the day after the day before yesterday, and today is Friday, when is my birthday?
A) Wednesday
B) Thursday
C) Saturday
D) Sunday
Answer: B) Thursday
Today is Friday. Yesterday was Thursday. The day before yesterday was Wednesday. The day AFTER the day before yesterday is Thursday.
Question 33: Martin's birthday is on the fourth Saturday of the month. If the month starts on a Wednesday, on what date is Martin's birthday?
A) 22nd
B) 23rd
C) 24th
D) 25th
Answer: D) 25th
1st is Wed. 2nd is Thu. 3rd is Fri. 4th is the first Saturday. The Saturdays are: 4th, 11th, 18th, 25th. The fourth Saturday is the 25th.
Question 34: What day would it be on 1 September 2020?
A) Sunday
B) Monday
C) Tuesday
D) Wednesday
Answer: C) Tuesday
Period: 2019 years + Jan 1 to Sep 1, 2020.
2019 years: 2000(0) + 19 years.
In 19 years: 4 leap (04,08,12,16) + 15 ordinary. Odd days=(4*2)+(15*1)=8+15=23. 23/7 rem 2.
2020 (Leap): Jan(3) Feb(1) Mar(3) Apr(2) May(3) Jun(2) Jul(3) Aug(3) Sep(1 day). Total=21. 21/7 rem 0.
Total Odd Days: 2+0=2.
2 corresponds to Tuesday.
Question 35: If the national day of a country was celebrated on the 4th Saturday of a month, find the date of celebration, given that the first day of that month was a Tuesday.
A) 24th
B) 25th
C) 26th
D) 27th
Answer: C) 26th
1st is Tue. 2nd Wed. 3rd Thu. 4th Fri. 5th is the first Saturday. The Saturdays are: 5th, 12th, 19th, 26th. The fourth Saturday is the 26th.
Question 36: January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?
A) Monday
B) Wednesday
C) Thursday
D) Friday
Answer: C) Thursday
The year 2008 is a leap year (divisible by 4). A leap year has 2 odd days. So the day will shift forward by 2. Tuesday + 2 days = Thursday.
Question 37: The calendar for the year 2005 is the same as for the year?
A) 2010
B) 2011
C) 2012
D) 2013
Answer: B) 2011
Count odd days from 2005. 2004 was a leap year. 2005 is 1 year after a leap year. The rule is to add 6 years. 2005+6=2011. Let's verify: 2005(1), 06(1), 07(1), 08(2), 09(1), 10(1). Total=7. So 2011 repeats.
Question 38: A century will always begin on which of the following days, assuming the previous century ended on a Sunday?
A) Sunday
B) Monday
C) Tuesday
D) Saturday
Answer: B) Monday
If a century (e.g., Dec 31, 1900) ends on a Sunday, the next century (Jan 1, 1901) will begin on the very next day, which is Monday.
Question 39: An event happens every 4 years on a fixed date. If the first event happened in 1997 on a Monday, when will the 5th event happen and on what day?
A) 2013, Monday
B) 2013, Tuesday
C) 2017, Monday
D) 2017, Tuesday
Answer: B) 2013, Tuesday
1st event: 1997 (Mon). 2nd event: 2001 (1997+4). Leap years crossed: 2000. Odd days: 1+1+1+2=5. Day=Mon+5=Sat. 3rd event: 2005 (2001+4). Leap years: 2004. Odd days: 1+1+1+2=5. Day=Sat+5=Thu. 4th event: 2009 (2005+4). Leap years: 2008. Odd days: 1+1+1+2=5. Day=Thu+5=Tue. 5th event: 2013 (2009+4). Leap years: 2012. Odd days: 1+1+1+2=5. Day=Tue+5=Sun. My calculation gives 2013, Sunday. Let me check the question logic again. The years are 1997, 2001, 2005, 2009, 2013. Day shifts: 1997->2001 (+5 days), 2001->2005 (+5 days), 2005->2009 (+5 days), 2009->2013 (+5 days). Total shift = 5+5+5+5 = 20 days. 20/7 rem 6. Final day = Monday + 6 days = Sunday. The standard answer is B. Let me see how. Maybe the question is simpler. 1st 1997. 5th event is 4*4=16 years later. 1997+16 = 2013. Total odd days in 16 years. Leap years: 2000, 2004, 2008, 2012 (4 leap years). Ordinary years = 12. Odd days = 4*2+12*1=8+12=20. 20/7 rem 6. Monday+6=Sunday. Calculation is consistently Sunday. I will correct the answer.
Logically Correct Answer: 2013, Sunday
Question 40: If the 1st of the month is a Saturday, and the month has 31 days, how many Wednesdays are in that month?
A) 3
B) 4
C) 5
D) 6
Answer: C) 5
1st=Sat. 2nd=Sun. 3rd=Mon. 4th=Tue. 5th=Wednesday. The Wednesdays are on: 5th, 12th, 19th, 26th. And 26+7=33, which is outside the month. Wait, this is 4 Wednesdays. Let me check again. 1st=Sat. -> 7th=Fri, 6th=Thu, 5th=Wed. Correct. Wednesdays are 5, 12, 19, 26. This gives 4. How can it be 5? If the 1st, 2nd, or 3rd day is a Wednesday, there will be 5. Let me re-read. 1st is Saturday. Yes. Month has 31 days. Calculation for Wed: 1(Sat), 2(Sun), 3(Mon), 4(Tue), 5(Wed). Correct. Dates: 5, 12, 19, 26. This is 4 Wednesdays. This is another flawed question. I will correct it. Let's make the first day a Thursday. Then 1=Thu, 2=Fri, 3=Sat, 4=Sun, 5=Mon, 6=Tue, 7=Wed. Then Wed are 7,14,21,28. Still 4. Let's make the first day a Wednesday. Then Wed are 1,8,15,22,29. 5 Wednesdays.
Revised Question 40: If the 1st of the month is a Wednesday, and the month has 31 days, how many Wednesdays are in that month? Answer is 5.
Question 41: What was the day on 31 Oct 1984 (Indira Gandhi's assassination)?
A) Monday
B) Tuesday
C) Wednesday
D) Thursday
Answer: C) Wednesday
Period: 1983 years + days in 1984. 1984 is a leap year.
1983 years: 1600(0)+300(1)+83 years. In 83 years: 20 leap, 63 ord. Odd=(20*2)+(63*1)=40+63=103. 103/7 rem 5. Total=1+5=6.
1984: J(3)F(1)M(3)A(2)M(3)J(2)J(3)A(3)S(2)O(31->3). Total=25. 25/7 rem 4.
Total Odd Days: 6+4=10. 10/7 rem 3. Wednesday.
Question 42: The day on 5th March of a year is the same as the day on 5th November of the same year. This statement is true for which year?
A) Any year
B) Any leap year
C) Any ordinary year
D) Never
Answer: A) Any year
For the days to be same, odd days between the dates must be a multiple of 7.
Mar(31-5=26->5) Apr(2) May(3) Jun(2) Jul(3) Aug(3) Sep(2) Oct(3).
Total = 5+2+3+2+3+3+2+3 = 23. This is not a multiple of 7.
Let me check the standard mnemonic for same day months. "March and November" are a pair for ordinary years.
Let's count odd days for full months between Mar and Nov.
Mar(3), Apr(2), May(3), Jun(2), Jul(3), Aug(3), Sep(2), Oct(3) = 21. 21/7 rem 0.
This applies to full months. So 1st March and 1st November are the same day. Therefore, 5th March and 5th November will also be the same. The calculation doesn't depend on Feb, so it's true for any year.
Question 43: Anju remembers that her sister's birthday is after the 18th but before the 21st of a month, while her father remembers it's after the 19th but before the 22nd. On which date is the birthday?
A) 19th
B) 20th
C) 21st
D) Cannot be determined
Answer: B) 20th
Anju's memory: Dates can be 19 or 20. Father's memory: Dates can be 20 or 21. The only date common to both statements is the 20th.
Question 44: On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?
A) Tuesday
B) Monday
C) Sunday
D) Wednesday
Answer: C) Sunday
We are going back one year. The period from Feb 8, 2004 to Feb 8, 2005 contains Feb 29, 2004 (since 2004 is a leap year). This means there are 2 odd days in this period. We must go back 2 days from Tuesday. Tuesday - 2 days = Sunday.
Question 45: If 15th June falls 3 days after tomorrow, which is a Friday, on what day will the last day of the month fall?
A) Monday
B) Tuesday
C) Wednesday
D) Thursday
Answer: C) Wednesday
1. "Tomorrow is a Friday". So Today is Thursday. 2. 3 days after tomorrow (Fri) is Mon. So June 15th is a Monday. Wait, let me re-read. "3 days after tomorrow, WHICH IS a Friday". So Tomorrow is Friday. Today is Thursday. 3 days after Friday is Monday. So June 15 = Monday. June has 30 days. We need the day on June 30th. Days from 15th to 30th = 15 days. 15/7 rem 1. Day on June 30th = Monday + 1 day = Tuesday. Let me check the answer. C) Wednesday. Let me try another interpretation. "15th June falls on the day which is 3 days after tomorrow. Tomorrow is Friday". Today is Thursday. Tomorrow is Friday. The day 3 days after Friday is Monday. So June 15 is Monday. Same result. Let me try this: "3 days after tomorrow" - Tomorrow is Friday. Day after is Saturday. Day after that is Sunday. So June 15th is Sunday. If June 15 is Sunday. Days left = 15. 15/7 rem 1. Day on June 30 = Sunday+1 = Monday. Still not Wednesday. Let's try "tomorrow, which is Friday" as a separate clause. 1. 15th June is 3 days after tomorrow. Let today be X. Tomorrow is X+1. 3 days after is X+4. So June 15 is X+4. 2. Tomorrow is Friday. So X+1 = Fri. X=Thursday. 3. Day on June 15 = X+4 = Thursday+4 = Monday. All interpretations give Monday or Tuesday. The question must be flawed. Let's assume June 15 = Wednesday. Last day of month is 30th. Days remaining = 15. 1 odd day. Last day = Wed+1 = Thursday. I will create a clean question.
Revised Question 45: If June 10th in a year was a Wednesday, on what day will the last day of that month fall?
Answer: B) Tuesday
June has 30 days. We know June 10th is a Wednesday. We need the day on June 30th. Days between them = 30-10=20. 20/7 rem 6. The day will be Wednesday + 6 days = Tuesday.
Question 46: How many odd days are there in 10 years?
A) 10
B) 11
C) 12
D) 13
Answer: C) 12. Wait, 10 years (e.g., 2001-2010) has 2 leap years (04,08). 8 ordinary. Odd days = 2*2+8*1=12. This is the total odd days, not mod 7. If the question asks for the remainder, 12/7 rem 5. Let's assume the question asks for the total odd days count. But if the 10 years are 1997-2006, leap years are 2000, 2004 (2 leap years). 12 odd days. If 10 years are 2000-2009, leap years are 2000, 2004, 2008 (3 leap years). 7 ordinary. Odd days = 3*2+7*1=13. The question is ambiguous. Let's assume a standard 10 year period not spanning a special century. It will have 2 leap years.
Answer (assuming 2 leap years): 12 odd days. Answer (assuming 3 leap years): 13 odd days.
Question 47: The year 2024 started on a Monday. On what day will it end?
A) Monday
B) Tuesday
C) Wednesday
D) Sunday
Answer: B) Tuesday
2024 is divisible by 4, so it is a leap year. A leap year has 366 days (2 odd days). It starts on a Monday. The day shifts by 2-1=1 for the last day. So it ends one day after it started. Monday + 1 = Tuesday.
Question 48: If day before yesterday was Saturday, what day will be the day after tomorrow?
A) Tuesday
B) Wednesday
C) Thursday
D) Friday
Answer: B) Wednesday
Day before yesterday = Sat. Yesterday = Sun. Today = Monday. Tomorrow = Tuesday. Day after tomorrow = Wednesday.
Question 49: If the 3rd of a month is a Monday, which of the following is the 5th day after the 21st of that month?
A) Monday
B) Tuesday
C) Wednesday
D) Thursday
Answer: C) Wednesday
We need the day on the 21+5 = 26th of the month. 3rd is Monday. Days between 3rd and 26th = 23 days. 23/7 rem 2. The required day is Monday + 2 days = Wednesday.
Question 50: What was the day on 1st Jan 0001?
A) Sunday
B) Monday
C) Saturday
D) Friday
Answer: B) Monday
The Gregorian calendar system is based on the assumption that the first day, Jan 1, 0001, was a Monday. Our entire odd day calculation system is built on this premise. 0 years have passed, so 0 odd days. The first day is day 1. 1 corresponds to Monday.