Calendar PYQs for UPSC (2011-2024) | Solved Questions & PDF Download

Master Calendar-Based Reasoning for UPSC with solved Previous Year Questions (PYQs) from 2011 to 2024. This page provides detailed solutions, practice questions, and a free downloadable PDF to help you ace your UPSC preparation.

Why Calendar Questions Matter in UPSC?

Calendar-based questions are a crucial part of the UPSC CSAT (Civil Services Aptitude Test). They test your ability to:

➡️ Calculate days, dates, and years.
➡️ Understand leap years and odd days.
➡️ Solve complex date-based puzzles.

This page breaks down the concepts and provides year-wise solved questions to help you master this topic.

Year-Wise Calendar PYQs (2011-2024)

Q1. If the 3^rd day of a month is Monday, which one of the following will be the fifth day from 21^st of this month? [CSAT 2014]

(a) Monday

(b) Tuesday

(d) Friday

Solution:

Given that,

3rd day of a month is Monday

Now,

fifth day from 21^ST of this month = 21 + 5 = 26th day of the month

3rd day of a month is Monday so left days = 26 - 3 = 23 days

23/7 = 3 weeks and 2 odd days left

Thus Monday + 2 more odd days = Wednesday

Fifth day from 21^ST of this month = Wednesday

Hence option (c) is correct

Q2. If second and fourth Saturdays and all the Sundays are taken as only holidays for an office, what would be the minimum number of possible working days of any month of any year? [CSAT 2017]

(a) 23

(b) 22

(d) 20

Solution:

Given that,

Second and fourth Saturdays and all the Sundays are taken as only holidays for an office.

So, the minimum number of possible days would be in the month of February not having a leap year

Now,

Total number of days = 28 days/7 = 4 weeks

Thus, the month of Feb and Non-leap year will consists of 4 Saturdays and 4 Sundays

Number of working days = 28 - 2 (Saturdays) - 4 (Sundays) = 22 days

The minimum number of possible working days = 22 days

Hence option (b) is correct

Q3. Consider the sequence given below: 4/12/95, 1/1/96, 29/1/96, 26/2/96, what is the next term of the series? [CSAT 2018]

(a) 24/3/96

(b) 25/3/96

(d) 27/3/96

Solution:

Given that,

4/12/95, 1/1/96, 29/1/96, 26/2/96, .....

The given sequence consists of 28 days difference

4/12/95 ......+^28days........1/1/96

1/1/96......+^28days........ 29/1/96

29/1/96......+^28days........ 26/2/96

26/2/96......+^28days........ 25/3/96 (as the year 1996 is a leap year consisting of 29 days)

Hence option (b) is correct

Q4. Which year has the same calendar as that of 2009? [CSAT 2019]

(a) 2018

(b) 2017

(d) 2015

Solution:

After every 7 odd days the years repeats or the calendar is same

Now,

Number of odd days in a leap year = 366/7 = 2 odd days

Number of odd days in a non- leap year = 365/7 = 1 odd days

Total odd days should be 7 to repeat the calendar

2010 = 1 odd day, 2011 = 1 odd day, 2012 = 2 odd days, 2013 = 1 odd day, 2014 = 1 odd day and 2015 = 1 odd day

Total after 6 years = 7 odd days

2009 + 6 = 2015

Hence option (d) is correct

Q5. Mr 'X' has three children. The birthday of the first child falls on the 5th Monday of April, that of the second one falls on the 5th Thursday of November. On which day is the birthday of his third child, which falls on 20th December? [CSAT 2019]

(a) Monday

(b) Thursday

(d) Sunday

Solution:

Given that,

The birthday of the first child falls on the 5th Monday of April, that of the second one falls on the 5th Thursday of November.

1st child 5th of Monday have 2 possible days = 29th or 30th of April

Possibility 1: 29th April is Monday and 30th April as Tuesday

No. of days between 30th April to 1st November = (1 + 31 + 30 + 31 + 31 + 30 + 31 + 1) = 186 days

Number of weeks and days = 186/7 = 26 weeks and 4 odd days

So 1st November = Tuesday + 4 odd days = Friday

29th of November = Friday and 30th November = Saturday

Hence this possibility cannot satisfy the given condition of 5th Thursday of November

Possibility 2: 30th April is Monday

No. of days between 1st May to 1st November = ( 31 + 30 + 31 + 31 + 30 + 31 + 1) = 185 days

Number of weeks and days = 185/7 = 26 weeks and 3 odd days

So 1st November = Tuesday + 3 odd days = Thursday

29th of November = 5th Thursday

Thursdays in December = 6, 13, 20 and 27th

Third child = 20th of December is Thursday

Hence option (b) is correct

Q6. If in a particular year 12th January is a Sunday, then which one of the following is correct? [CSAT 2020]

(a) 15th July is a Sunday if the year is a leap year.

(b) 15th July is a Sunday is the year is not a leap year.

(d) 12th July is not a Sunday if the year is a leap year.

Solution:

Given that,

12th January is a Sunday

There are two possibilities if the year is a leap year or not a leap year

Possibility 1: Leap year

Total Number of days up to 12th July = (19 + 29 + 31 + 30 + 31 + 30 + 12) = 182/7 = 26 weeks and 0 odd days

So, 12th July is Sunday

15th July = Wednesday

Possibility 2: Not a Leap year

Total Number of days up to 12th July = (19 + 28 + 31 + 30 + 31 + 30 + 12) = 181/7 = 25 weeks and 6 odd days

So, 12th July is Sunday + 6 odd days = Saturday

15th July = Tuesday

Hence option (c) is correct

Q7. Which day is 10th October, 2027? [CSAT 2021]

(a) Sunday

(b) Monday

(d) Saturday

Solution:

Given that,

Date = 10th October , and Year = 2027 (Which is not a leap year)

Now,

Odd days are 0 after every 400 years, so after 2000 years = 0 odd days

Number of odd days from 2000 to 2026 = 6 leap years + 20 non leap years = 2 x 6 + 20 = 32 odd days

32/7 = 4 weeks + 4 odd days

Number of odd days from 1st Jan to 10th October = (31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 10) = 283 days/7 = 40 weeks and 3 odd days

Thus total odd days = 4 + 3 = 7

Therefore odd days = 0

Hence 10th of October 2027 is Sunday as Sunday consists of 0 odd days

Q8. Consider two Statements and a Question: [CSAT 2021]

Statement-1: The last day of the month is a Wednesday.

Statement-2: The third Saturday of the month was the seventeenth day.

Question: What day is the fourteenth of the given month?

(a) Statement-1 alone is sufficient to answer the Question.

(b) Statement-2 alone is sufficient to answer the Question.

(d) Neither Statemetn-1 alone nor Statement-2 alone is sufficient to answer the Question.

Solution:

Given that,

From Statement-1: The last day of the month is a Wednesday.

A Month can have 28, 29, 30 or 31 days thus to determine the 14th day of the month is not possible.

Hence statement 1 alone is not sufficient

From Statement-2: The third Saturday of the month was the seventeenth day.

17th is Saturday so 14th will be Wednesday

Hence statement 2 alone is sufficient.

Q9. Joseph visits the club on every 5th day; Harsh visits on every 24th day, while Sumit visits on every 9th day. If all three of them met at the club on a Sunday, then on which day will all three of them meet again? [CSAT 2021]

(a) Monday

(b) Wednesday

(d) Sunday

Solution:

Given that,

Joseph visits the club on every 5th day.

Harsh visits on every 24th day

Sumit visits on every 9th day.

They all meet on Sunday

Now,

Taking LCM = 5, 9, 24 = 360

Thus all meet on 360^th day

1st meeting is on Sunday

So, second meeting = 360/7 = 51 weeks and 3 odd days

Thus second meeting is on Sunday + 3 odd days = Wednesday

Hence option (b) is correct

Q10. Which date of June 2099 among the following is Sunday? [CSAT 2022]

(a) 4

(b) 5

(d) 7

Solution:

Given that,

Year = 2099 (not a leap year)

Now,

2000 = 0 odd days (after every 400 years 0 odd days)

From 2000 to 2098 = 98 years

Total odd days in 98 years = 24 leap years and 74 ordinary years = 24 x 2 + 74 = 122/7 = 3 odd days

Total odd days in month from 1st Jan to 31st May of 2099 = (31 + 28 + 31 + 30 + 31) = 151/7 = 4 odd days

Total odd days = 3 + 4 = 7 = 0 odd days

So, 31^st may is Sunday, then 7^th June 2099 would be Sunday.

Hence option (d) is correct

Q11. If today is Sunday, then which day is it exactly on 10¹⁰ th day? [CSAT 2023]

(a) Wednesday

(b) Thursday

(d) Saturday

Solution:

Given that,

Today is Sunday

Now on 10¹⁰th day = 10¹⁰/7 (dividing 10 by 7 the remainder is 3) = 3¹⁰/7 (the remainder becomes the base) = (3²)⁵/7 = 9⁵/7 = 2⁵/7 (the remainder becomes the base) = 32/7 = 4 odd days

So Sunday + 4 odd days = Wednesday as (Sunday is included as 1^st odd day)

Hence option (a) is correct

Q12.The calendar for the year 2025 is same for: [CSAT 2024]

(a) 2029

(b) 2030

(d) 2033

Solution:

Given that,

Calendar for the year = 2025

Now,

Year	Repetition after years
Leap year	28 years
Leap year + 1	6 years
Leap year + 2	11 years
Leap year + 3	11 years

Thus for 2025

2024 is a leap year

So, 2024 leap year + 1 = 2025 = Addition of 6 year in 2025

= 2025 + 6 = 2031

Hence option (c) is correct

ANSWER KEY

1. C

2. B

3. B

4. D

5. B

6. C

7. A

8. B

9. B

10. D

11. A

12. C

Download Calendar PYQs PDF

➡️ Download the complete set of Calendar PYQs (2011-2024) with solutions in PDF format.
➡️ Click Here to Download PDF

FAQs on Calendar Reasoning for UPSC

1. Is Calendar Reasoning important for UPSC?

Yes, Calendar Reasoning is a crucial topic for the UPSC CSAT paper. It tests your ability to calculate dates, days, and years accurately.

2. How to prepare for Calendar Reasoning in UPSC?

➡️ Understand basic concepts like odd days, leap years, and day calculations.
➡️ Practice PYQs regularly.
➡️ Use reliable resources like NCERT books and online platforms like iassetu.com.

3. What are the most frequently asked Calendar Reasoning questions in UPSC?

➡️ Odd days and century-based questions
➡️ Leap year-based problems
➡️ Day and date calculations

Conclusion

Practicing Calendar PYQs is essential for cracking the UPSC CSAT paper. This page provides year-wise solved questions, practice problems, and a free PDF download to help you prepare effectively. Bookmark this page and revisit it regularly for updates.

For more UPSC preparation resources, explore iassetu.com. Happy learning!