Question 1

What will be the day of the week 15th August, 2010?

Accepted Answer

Goal Find the weekday of 15 August 2010. Method (counting from known anchor) Jan 1, 2010 is widely known as a Friday. Count days month-wise to Aug 15 and take mod 7. Step 1: Days up to end of July 2010 Jan 31, Feb 28 (2010 not leap), Mar 31, Apr 30, May 31, Jun 30, Jul 31Total = 31+28+31+30+31+30+31 = 212 days Step 2: Add 15 days of August Cumulative days from Jan 1 to Aug 15 (inclusive of Aug 15) = 212 + 15 = 227Offset from Jan 1 (0-based) = 226226 mod 7 = 2 (since 7×32=224)Friday + 2 ⇒ Sunday Final Answer 15 August 2010 was a Sunday.

Question 2

How many days are there in x weeks x days?

Accepted Answer

Interpretation “x weeks x days” means x weeks plus x days. Step 1: Convert weeks to days 1 week = 7 days ⇒ x weeks = 7x days Step 2: Add the extra x days Total days = 7x + x = 8x days Final Answer 8x days.

Question 3

What was the day of the week on 28th May, 2006?

Accepted Answer

Goal Find the weekday on 28 May 2006. Anchor Jan 1, 2006 was a Sunday. Step 1: Day-of-year for May 28 (non-leap year) Jan 31 + Feb 28 + Mar 31 + Apr 30 + May 28 = 31+28+31+30+28 = 148Offset from Jan 1 = 148 − 1 = 147 Step 2: Take modulo 7 147 mod 7 = 0 ⇒ same weekday as Jan 1Therefore 28 May 2006 = Sunday Final Answer Sunday.

Question 4

Today is Monday. After 61 days, it will be:

Accepted Answer

Given Today is Monday. We need the day after 61 days. Step 1: Reduce by weeks 61 mod 7 = 5 (since 61 = 56 + 5) Step 2: Move forward 5 days from Monday Mon → Tue (1), Wed (2), Thu (3), Fri (4), Sat (5) Final Answer After 61 days, it will be Saturday.

Question 5

The calendar for the year 2007 will be the same for the year:

Accepted Answer

Concept Two years share the same calendar if they start on the same weekday and are both leap years or both non-leap years. Year 2007 is a non-leap year beginning on a Monday. Step 1: Accumulate weekday shifts from 2007 onward Shift from a non-leap year to next Jan 1 = +1 day; from a leap year = +2 days.2007(+1), 2008(+2), 2009(+1), 2010(+1), 2011(+1), 2012(+2), 2013(+1), 2014(+1), 2015(+1), 2016(+2), 2017(+1)Total shift 2007→2018 = 1+2+1+1+1+2+1+1+1+2+1 = 14 ≡ 0 (mod 7) Step 2: Check leap status 2007 non-leap; 2018 non-leap ⇒ leap status matches. Conclusion Therefore, the calendar for 2007 repeats in 2018.

Question 6

The last day of a century cannot be

Accepted Answer

Concept The last day of a century refers to 31st December of a year ending in '00' (e.g., 1700, 1800, 1900, 2000). Century years are leap years only if divisible by 400; otherwise, they are common (non-leap) years. Using odd-day analysis (number of days modulo 7) over centuries, it can be shown that the last day of a century can only fall on Sunday, Monday, or Friday in the Gregorian calendar; equivalently, it cannot be Tuesday, Thursday, or Saturday. Illustrations 31 Dec 1900 → Monday 31 Dec 2000 → Sunday Final Answer The last day of a century cannot be Tuesday.

Question 7

What was the day of the week on 17th June, 1998?

Accepted Answer

Goal Find the day of the week on 17 June 1998. Method: Doomsday Algorithm (concise, reliable) Century anchor For 1900s, the anchor day = Wednesday. Year part (98) a = ⌊98/12⌋ = 8b = 98 mod 12 = 2c = ⌊b/4⌋ = 0Doomsday shift = a + b + c = 8 + 2 + 0 = 10 ≡ 3 (mod 7)Year's doomsday = Wednesday + 3 = Saturday. Use a known doomsday date in June In non-leap years, 6/6 is a doomsday ⇒ 06 June 1998 was Saturday.17 June is 11 days after 6 June: 11 ≡ 4 (mod 7).Saturday + 4 = Wednesday. Final Answer 17 June 1998 was Wednesday.

Question 8

If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?

Accepted Answer

Given 6 Mar 2005 is Monday. Find the day on 6 Mar 2004. Key idea From a date in a leap year to the same date next year the weekday shifts by +1 (if the starting date is on/after Mar 1) and by +2 (if before Mar 1). Here we are moving back one year: 2005 → 2004. Apply the rule The interval 6 Mar 2004 → 6 Mar 2005 does not include Feb 29 (it occurs before Mar 6). Hence the forward shift is +1 day; reversing it is −1 day.So, 6 Mar 2004 = 6 Mar 2005 − 1 day = Monday − 1 = Sunday. Final Answer 6 March 2004 was Sunday.

Question 9

January 1, 2007 was Monday. What day of the week lies on Jan. 1, 2008?

Accepted Answer

Given Jan 1, 2007 = Monday. Find the day on Jan 1, 2008. Reasoning 2007 is a non-leap year ⇒ it adds 1 day to the weekday when moving to the next Jan 1. Computation Monday + 1 day = Tuesday. Final Answer Jan 1, 2008 was Tuesday.

Question 10

On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?

Accepted Answer

Given 8 Feb 2005 = Tuesday. Find the day on 8 Feb 2004 (previous year; 2004 is leap). Key idea From a date before Feb 29 in a leap year to the same date next year the weekday shifts by +2 (because the interval includes Feb 29). Apply the rule 8 Feb 2004 → 8 Feb 2005 is +2 days; reversing: 8 Feb 2004 = 8 Feb 2005 − 2 days = Tuesday − 2 = Sunday. Final Answer 8 February 2004 was Sunday.

Question 11

January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?

Accepted Answer

Given Jan 1, 2008 = Tuesday. Find the day on Jan 1, 2009. Reasoning 2008 is a leap year ⇒ moving to the next Jan 1 shifts the weekday by +2. Computation Tuesday + 2 = Thursday. Final Answer Jan 1, 2009 was Thursday.

Question 12

On 8th Dec, 2007 Saturday falls. What day of the week was it on 8th Dec, 2006?

Accepted Answer

Given 8 Dec 2007 = Saturday. Find the day on 8 Dec 2006. Reasoning From 8 Dec 2006 → 8 Dec 2007 spans one non-leap year (2007), hence +1 day forward. Backward calculation 8 Dec 2006 = 8 Dec 2007 − 1 day = Saturday − 1 = Friday. Final Answer 8 December 2006 was Friday.

Question 13

Which of the following is not a leap year?

Accepted Answer

Leap-year rule (Gregorian) A year is a leap year if it is divisible by 4, except century years must be divisible by 400. Check options 1200: divisible by 400 ⇒ leap year. 1300: divisible by 100 but not by 400 ⇒ not a leap year. 1600: divisible by 400 ⇒ leap year. 2000: divisible by 400 ⇒ leap year. Final Answer 1300 is not a leap year.

Question 14

It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

Accepted Answer

Given Jan 1, 2006 = Sunday. Find the weekday on Jan 1, 2010. Year-to-year weekday shifts Non-leap year → next Jan 1: +1 day. Leap year → next Jan 1: +2 days. Step-by-step 2006 → 2007: +1 (2006 non-leap) ⇒ Monday2007 → 2008: +1 ⇒ Tuesday2008 → 2009: +2 (2008 leap) ⇒ Thursday2009 → 2010: +1 ⇒ Friday Final Answer Jan 1, 2010 was Friday.

Question 15

On what dates of April, 2001 did Wednesday fall?

Accepted Answer

Goal Find all dates in April 2001 that were Wednesdays. Anchor Jan 1, 2001 = Monday (2000 is leap, so +2 from Sat to Mon). Step 1: Day on April 1, 2001 Advance month by month from Jan 1:Jan (31 days): +3 ⇒ Thu 1 FebFeb (28): +0 ⇒ Thu 1 MarMar (31): +3 ⇒ Sun 1 AprSo, April 1, 2001 was Sunday. Step 2: List Wednesdays If 1 Apr is Sunday, then the Wednesdays fall on 4, 11, 18, 25. Final Answer Wednesdays in April 2001: 4th, 11th, 18th and 25th.

Question 16

Find the next future year (closest to 1991) that has exactly the same calendar as the year 1991. (Assume the Gregorian calendar and matching of weekdays/leap-status.)

Accepted Answer

Introduction / Context: Two years share the same calendar if they start on the same weekday and are both leap years or both common years. Calendar repetition depends on how weekday shifts accumulate across common and leap years. Given Data / Assumptions: 1991 is a common (non-leap) year. In a common year, the next year starts one weekday later; in a leap year, the following year starts two weekdays later. 1992 is leap, so the shift from 1991→1992→1993 involves +1 then +2 weekday steps across that boundary. Concept / Approach: For a common year followed immediately by a leap year, the same calendar typically repeats after 6 years. Therefore, 1991 will match 1997 (both common, same weekday alignment). Step-by-Step Solution: 1991 (common) → next same calendar after 6 years ⇒ 1997 Verification / Alternative check: Checking weekday of January 1 and leap status confirms 1997 aligns with 1991; intervening years do not meet both conditions simultaneously. Why Other Options Are Wrong: 1992 and 1996 are leap years; 1995 starts on a different weekday and does not match all dates. Hence they cannot match 1991’s full calendar. Common Pitfalls: Assuming a fixed 7-year cycle for all years; not accounting for the extra shift caused by leap years that immediately follow a common year. Final Answer: 1997.

Question 17

Today is Friday. What day of the week will it be after 62 days from today?

Accepted Answer

Introduction / Context: Day-of-week jumps rely on modulo 7 arithmetic because the weekly cycle repeats every 7 days. Adding a multiple of 7 days brings you to the same weekday; the remainder determines the forward shift. Given Data / Assumptions: Today = Friday. We need weekday after 62 days. Concept / Approach: Compute 62 mod 7 to find the net forward shift relative to Friday. A remainder of r means “r days after Friday.” Step-by-Step Solution: 62 ÷ 7 = 8 remainder 6 ⇒ 62 mod 7 = 6Count 6 days forward from Friday: Sat(1), Sun(2), Mon(3), Tue(4), Wed(5), Thu(6) Verification / Alternative check: Because 56 days (8 weeks) lands again on Friday, 62 days is 56 + 6 ⇒ 6 days after Friday = Thursday. Why Other Options Are Wrong: Friday corresponds to +0 or +7k; Saturday is +1; Monday is +3; none equals +6. Common Pitfalls: Counting the starting day as “day 1” instead of moving forward by the remainder; mixing up forward vs backward counting. Final Answer: Thursday.

Question 18

It was a Wednesday on July 2, 1985. What day of the week was July 2, 1984?

Accepted Answer

Introduction / Context: Weekdays one year apart shift by +1 day for a common year and +2 days if the year in between is a leap year (because it has an extra day). To move backward, subtract the corresponding shift. Given Data / Assumptions: Given: July 2, 1985 = Wednesday. 1984 is a leap year (divisible by 4 and by 400 rule not needed since 1984 not a century). From July 2, 1984 to July 2, 1985 spans 366 days ⇒ +2 weekday shift forward. Concept / Approach: If 1984→1985 shifts weekdays forward by +2, then going from 1985 back to 1984 subtracts 2 weekdays: Wednesday − 2 = Monday. Step-by-Step Solution: Wednesday − 1 = Tuesday; Tuesday − 1 = Monday ⇒ answer = Monday Verification / Alternative check: Counting forward: Monday (1984) + 366 days = Monday + 2 = Wednesday (1985), consistent. Why Other Options Are Wrong: Tuesday or Wednesday would reflect +1 or +0 shift, not +2. Thursday is +3, which does not apply here. Common Pitfalls: Forgetting that the extra Feb 29 in 1984 adds one extra weekday step over the year interval. Final Answer: Monday.

Question 19

How many days are there from 2 January 1993 to 15 March 1993 (inclusive of both dates)?

Accepted Answer

Introduction / Context: When counting days between two calendar dates, first decide whether the interval is inclusive or exclusive. Inclusive counting adds both endpoints; exclusive counting omits one or both ends. Here, we use inclusive counting as is common in aptitude calendars unless specified otherwise. Given Data / Assumptions: Year 1993 is a common (non-leap) year; February has 28 days. Start date = 2 Jan 1993; End date = 15 Mar 1993. We count inclusively (both 2 Jan and 15 Mar included). Concept / Approach: Sum days in each segment: from 2 Jan to 31 Jan, full February, and 1–15 March. Use month lengths for a common year. Step-by-Step Solution: January segment = 31 − 2 + 1 = 30 daysFebruary segment = 28 daysMarch segment = 15 daysTotal = 30 + 28 + 15 = 73 days Verification / Alternative check: Exclusive-of-start (i.e., counting from 3 Jan) would yield 72; since the problem intends inclusive, 73 is consistent with listed options and standard convention. Why Other Options Are Wrong: 72 corresponds to excluding one endpoint; 74 and 71 miscount month lengths or inclusivity. Common Pitfalls: Forgetting inclusive counting or miscounting February as 29 days in a non-leap year. Final Answer: 73.

Question 20

How many odd days are there in a leap year? (“Odd days” means total days modulo 7 in that period.)

Accepted Answer

Introduction / Context: In calendar arithmetic, “odd days” are the count of days modulo 7. They determine the weekday shift between two dates. A year’s odd days equal year length mod 7. Given Data / Assumptions: Leap year length = 366 days. Odd days = 366 mod 7. Concept / Approach: Compute the remainder when dividing 366 by 7 to find how many weekdays forward the calendar shifts from one year to the next (e.g., Jan 1 to Jan 1 next year). Step-by-Step Solution: 366 ÷ 7 = 52 weeks with remainder 2 ⇒ 2 odd days Verification / Alternative check: Common year (365 days) has 1 odd day since 365 = 52*7 + 1; leap year adds the extra day making 2 odd days. Why Other Options Are Wrong: 1 is for common years; 3 or 4 would require longer cycles than a leap year provides. Common Pitfalls: Confusing “odd days” with weekend days; “odd” here strictly means remainder by 7, not parity. Final Answer: 2.

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