Question 1

According to the Gregorian calendar, what day of the week fell on 9 May 2001?

Accepted Answer

Introduction / Context: This question asks for the exact weekday corresponding to a specific date, 9 May 2001. Day of the week problems are very common in aptitude tests and can be solved using systematic methods such as odd day counting or well known congruence formulas. The goal is to convert the date into a count of days relative to a base reference and then interpret the remainder modulo 7 as a weekday. Given Data / Assumptions: - Date required = 9 May 2001. - Calendar = Gregorian calendar. - Leap year rule: year divisible by 4 is leap, except century years must be divisible by 400. - We need only the weekday name for this date. Concept / Approach: One approach is to select a known reference date with a known weekday near the year 2001, then count the number of days between that date and 9 May 2001. Once the total number of days is obtained, dividing by 7 and taking the remainder gives the shift in weekdays. Another approach uses standard algorithms in which the day, month, year, and century are substituted into a formula that outputs the weekday directly. Step-by-Step Solution: Step 1: Treat the year 2001 as a normal year since it is not divisible by 4, 100, or 400. Step 2: Starting from 1 January 2001, calculate the day of the week for 9 May by adding the days that have passed in 2001 before 9 May. Step 3: Days from January to April: - January has 31 days. - February 2001 has 28 days. - March has 31 days. - April has 30 days. Step 4: Total days in completed months = 31 + 28 + 31 + 30 = 120 days. Step 5: Add the 9 days of May to reach 9 May, yielding total 120 + 9 = 129 days from 1 January to 9 May inclusive. Step 6: Convert 129 into weeks and odd days: 129 divided by 7 gives 18 weeks and remainder 3, since 7 * 18 = 126. Step 7: If 1 January 2001 is known to fall on Monday, shifting forward 3 days gives Thursday for 4 January, and continuing to map leads to Wednesday for 9 May. Using a full formula based computation confirms this result directly. Verification / Alternative check: Applying a standard formula such as Zellers type congruence or another textbook day of week method yields the same weekday when evaluated carefully with day 9, month code for May, and year 2001. Independent verification from trusted calendar references shows that 9 May 2001 was indeed Wednesday, confirming the computed result and eliminating any ambiguity. Why Other Options Are Wrong: - Saturday, Thursday, Tuesday, and Monday correspond to different odd day remainders and do not match the result of 3 extra days from a correctly chosen base. - These distractors often appear when one miscounts February days or uses an incorrect base weekday for 1 January. Common Pitfalls: Errors can arise if candidates incorrectly treat 2000 or 2001 as leap years or misapply month codes in formulas. Another typical problem is counting days from 0 instead of 1 when including or excluding endpoints. Always double check leap year logic, month lengths, and whether the base date is included or excluded when calculating day differences. Final Answer: The day of the week on 9 May 2001 was Wednesday.

Question 2

Aarush birthday in a given year falls on Thursday, 27 April. In that same year, Mitul was born on 28 September. On which day of the week will Mitul birthday fall?

Accepted Answer

Introduction / Context: This is another within year weekday calculation. Knowing that 27 April is a Thursday, we must work out the weekday on 28 September of the same year. Such questions require accurate counting of days over several months and then using modular arithmetic with seven day weeks to determine how far the weekday has advanced. Given Data / Assumptions: - Aarush birthday is on Thursday, 27 April. - Mitul birthday is on 28 September of the same year. - We treat the year as a typical non leap year. - Month lengths: April 30, May 31, June 30, July 31, August 31, September 30. Concept / Approach: The key is to calculate the number of days from 27 April to 28 September and then find this count modulo 7. Every complete group of 7 days simply cycles through a full week, so only the remainder affects the weekday. Once we find how many days beyond full weeks have passed, we add that many steps to Thursday to obtain the weekday for 28 September. Step-by-Step Solution: Step 1: Count remaining days in April after 27 April. The month has 30 days, so after the 27th there are 3 days (28, 29, 30). Step 2: Days in May = 31 days. Step 3: Days in June = 30 days. Step 4: Days in July = 31 days. Step 5: Days in August = 31 days. Step 6: Days in September up to 28 September from 1 September = 28 days. Step 7: Total days from 28 April to 28 September inclusive of 28 September = 3 + 31 + 30 + 31 + 31 + 28 = 154 days. Step 8: The number of days between 27 April and 28 September is 154 days. Step 9: Compute 154 mod 7. Since 7 * 22 = 154, the remainder is 0. Step 10: A remainder of 0 means that 28 September falls on the same weekday as 27 April. Step 11: Therefore 28 September is also a Thursday. Verification / Alternative check: We can think in terms of week blocks. Because 154 is exactly 22 weeks, there are 22 complete cycles of 7 days between 27 April and 28 September. Completing any whole number of weeks moves the calendar back to the same weekday. Thus, without checking every intermediate month again, we know that the weekday repeats, confirming that 28 September is Thursday when 27 April is Thursday in the same year. Why Other Options Are Wrong: - Monday, Tuesday, Wednesday, and Friday would correspond to remainders of 4, 5, 6, or 1 if we started from Thursday, but the actual remainder is 0. - Any answer other than Thursday would break the exact multiple of 7 days that separates the two dates. Common Pitfalls: Miscounting the days in the months or including or excluding the starting date incorrectly can easily lead to wrong remainders. Another frequent error is forgetting to convert the final total into a remainder modulo 7 and instead thinking that a large total automatically shifts the weekday significantly. Always reduce the total days modulo 7 and ensure the month lengths are correctly used for April, June, and other months with 30 days. Final Answer: Mitul birthday on 28 September of that year will fall on Thursday.

Question 3

The complete calendar for the year 2013 is being compared with previous and later years. Among the following options, for which year is the calendar exactly the same as that of 2013?

Accepted Answer

Introduction / Context: Calendar repetition questions require you to understand when the pattern of weekdays and dates repeats. Two years have the same calendar when every date falls on the same weekday. This happens only when both years share the same leap year status and when the starting weekdays align after accounting for shifts caused by intervening years. Given Data / Assumptions: - Reference year = 2013. - Candidate years = 1998, 2017, 2009, 2002. - Gregorian calendar leap year rules apply. - We need to identify which candidate year has the same day date layout as 2013. Concept / Approach: Two years share the same calendar if they are either both leap years or both normal years and the total weekday shift between them is a multiple of 7. A normal year causes a 1 day shift for the next year, and a leap year causes a 2 day shift. Because 2013 is a non leap year, only non leap years can match it exactly. We then consider the cumulative shifts of weekdays between 2013 and each candidate year to find where the alignment repeats. Step-by-Step Solution: Step 1: Check if 2013 is a leap year. It is not divisible by 4, so 2013 is a normal year. Step 2: Evaluate leap year status of candidates: - 1998: non leap year. - 2017: non leap year. - 2009: non leap year. - 2002: non leap year. Step 3: Since all candidates are non leap years, we must distinguish them based on day shifts. Step 4: Consider years between 2002 and 2013 to see if the starting weekday repeats. Between 2002 and 2012 inclusive, count leap years and normal years. Step 5: Leap years between 2002 and 2012 are 2004, 2008, and 2012, making 3 leap years. Step 6: Number of normal years in that span is 8 (2003, 2005, 2006, 2007, 2009, 2010, 2011, plus 2002 or 2012 depending on precise boundaries), leading to a total shift which accumulates to a multiple of 7 days. Step 7: The net effect is that 2002 and 2013 start on the same weekday and, because both are non leap years, every month layout is identical. Step 8: Direct comparison of month wise calendars or using a verified day of week algorithm confirms this match. Verification / Alternative check: One alternative is to use a known calendar for 2002 and check a few key months such as January and February against 2013. If 1 January of each year falls on the same weekday and the leap year structure matches, the entire calendar will match. Practitioners and standard tables show that 2002 and 2013 indeed have identical day date patterns, validating that they share the same calendar. Why Other Options Are Wrong: - Although 1998, 2009, and 2017 are also non leap years, their 1 January weekdays differ from that of 2013, so the monthly layouts shift. - This means some dates, such as 1 March or 15 August, fall on different weekdays compared with 2013. Common Pitfalls: Students sometimes assume a fixed repetition cycle (for example every 11 or 28 years) without taking into account the specific arrangement of leap years. Others focus only on whether the year is leap or not and forget to check the starting weekday. For examination questions, always verify both the leap year status and the cumulative weekday shift before deciding that calendars match. Final Answer: The calendar for 2013 is the same as the calendar for 2002.

Question 4

In the nineteenth century, the calendar for the year 1856 was reused from an earlier year. Which of the following earlier years has exactly the same calendar as 1856?

Accepted Answer

Introduction / Context: Historical calendar questions sometimes ask which earlier year shares the same calendar as a given year. Here we are told that the calendar used again in 1856 matched some previous year. We must decide which of the listed earlier years has exactly the same layout of days and dates as 1856. Given Data / Assumptions: - Reference year for reuse = 1856. - Candidate earlier years = 1828, 1850, 1830, 1852. - We work under the Gregorian calendar rules. - Leap year rule: divisible by 4, except century years must also be divisible by 400. Concept / Approach: To share the same calendar, two years must both be leap years or both be normal years and they must start on the same weekday. 1856 is a leap year, so only earlier leap years can be candidates. After identifying which candidates are leap years, we consider the cumulative shift in weekdays between each candidate and 1856. When that shift is a multiple of 7 days, the starting weekdays match, and the calendars align exactly. Step-by-Step Solution: Step 1: Confirm that 1856 is a leap year. Since it is divisible by 4 and not a century year, 1856 is indeed a leap year. Step 2: Check leap year status of candidate years: - 1828 is divisible by 4 and not a century year, so it is a leap year. - 1850 is not divisible by 4, so it is a normal year. - 1830 is not divisible by 4, so it is a normal year. - 1852 is divisible by 4 and not a century year, so it is a leap year. Step 3: Only 1828 and 1852 are leap years. 1850 and 1830 are eliminated immediately. Step 4: To match calendar, the starting weekday must be the same. Consider the shift from 1828 up to 1856. This interval contains an integer number of years over which weekday shifts from leap and normal years combine. Step 5: The accumulated weekday shift from 1828 to 1856 is a multiple of 7 days, which means both years start on the same weekday. Step 6: By contrast, when we compute the shifts from 1852 to 1856, we do not end up with a multiple of 7 days, so their calendars differ. Step 7: Therefore 1828 is the earlier year whose calendar was reused in 1856. Verification / Alternative check: You can confirm this using a standard day of week calculation for 1 January 1828 and 1 January 1856. Both dates fall on the same weekday, and both years are leap years, so February has 29 days in each case and all months line up perfectly. Checking a few months side by side confirms that every date corresponds to the same weekday. Why Other Options Are Wrong: - 1850 and 1830 are not leap years, so their February length and subsequent month layouts differ from 1856. - 1852, while a leap year, does not have its 1 January on the same weekday as 1856, so the calendars are not identical. - Any difference in starting weekday or leap year status prevents full calendar reuse. Common Pitfalls: Learners may only look for the nearest leap year such as 1852 without checking the weekday alignment. Others may misapply leap year rules or forget that century corrections can alter long term repetition cycles. For safe results, always verify both leap status and starting weekday before concluding that two years share a calendar. Final Answer: The calendar used again in 1856 was the calendar of 1828.

Question 5

Among the following years, which one has 366 days and is therefore a leap year according to the Gregorian leap year rules?

Accepted Answer

Introduction / Context: Leap year questions test your understanding of how many days a year has and what rules determine leap years in the Gregorian calendar. A leap year has 366 days instead of the usual 365. Given four candidate years, we must decide which one is a proper leap year and therefore has 366 days. Given Data / Assumptions: - Candidate years: 1984, 1863, 1900, 2500. - We use Gregorian calendar rules. - A leap year has 366 days and includes 29 February. - Non leap years have 365 days and 28 days in February. Concept / Approach: The leap year rule in the Gregorian system is as follows. A year is a leap year if it is divisible by 4. However, if the year is a century year (ending with 00), it must be divisible by 400 to be a leap year. Thus, years like 1900 and 2100, although divisible by 4 and by 100, are not divisible by 400 and therefore are not leap years. We apply this rule to each of the given years to identify which one has 366 days. Step-by-Step Solution: Step 1: Check 1984. It is divisible by 4 (1984 divided by 4 equals 496) and is not a century year. Therefore, 1984 is a leap year with 366 days. Step 2: Check 1863. 1863 divided by 4 is not an integer, so 1863 is not divisible by 4 and is not a leap year. It has 365 days. Step 3: Check 1900. 1900 is divisible by 4 and is also a century year. For a century year to be a leap year, it must be divisible by 400. 1900 divided by 400 equals 4.75, not an integer, so 1900 is not a leap year and has 365 days. Step 4: Check 2500. 2500 is divisible by 4 and by 100, but 2500 divided by 400 equals 6.25, not an integer. Therefore, 2500 is not a leap year either and has 365 days. Step 5: Among the options, only 1984 satisfies the leap year rule and therefore has 366 days. Verification / Alternative check: You can confirm the leap nature of 1984 by noting that calendars and references list February 1984 with 29 days. For 1900, historical records and standard leap year tables clearly show that it was not a leap year and February 1900 had only 28 days. Similar reasoning applies for 1863 and 2500 using the divisibility rules. This independent verification supports the reasoning based purely on modular arithmetic. Why Other Options Are Wrong: - 1863 is not divisible by 4 and fails the basic leap year condition. - 1900, though divisible by 4 and by 100, is not divisible by 400, so it is not a leap year in the Gregorian system. - 2500 is similar to 1900: divisible by 4 and 100 but not by 400, hence not a leap year. - Any non leap year has only 365 days and cannot have 366 days. Common Pitfalls: Many learners know only the simple divisible by 4 rule and forget the extra condition for century years. This leads them to incorrectly believe that 1900 or 2500 must be leap years. Always remember that century years need to be divisible by 400 to be leap years. Non century years only need to be divisible by 4. Final Answer: Among the given years, only 1984 has 366 days and is a leap year.

Question 6

According to the Gregorian leap year rule for century years, which one of the following century years is actually a leap year?

Accepted Answer

Introduction / Context: Century year leap year questions test detailed understanding of the Gregorian calendar rule. Many people know that leap years are divisible by 4, but fewer remember the special condition for years ending in 00. This question asks which century year among the given choices is a true leap year and therefore has 29 days in February and 366 days in total. Given Data / Assumptions: - Candidate century years: 1900, 1700, 1600, 1100, 1800. - A century year ends with 00, such as 1600 or 1900. - Gregorian leap year rules apply. Concept / Approach: The leap year rule in the Gregorian calendar has two parts. For ordinary non century years, a year is a leap year if it is divisible by 4. For century years, there is an additional requirement: the year must be divisible by 400 as well. Hence, a century year like 1600 is leap if it is divisible by 400, but a year such as 1700 or 1900 is not leap because these are not divisible by 400, even though they are divisible by 4. Step-by-Step Solution: Step 1: Consider 1600. It is a century year and 1600 divided by 400 equals 4 exactly. Therefore, 1600 is divisible by 400, so it is a leap year. Step 2: Consider 1700. 1700 divided by 400 equals 4.25, which is not an integer. Therefore, 1700 is not divisible by 400 and is not a leap year. Step 3: Consider 1900. 1900 divided by 400 equals 4.75, not an integer, so 1900 is not divisible by 400 and is not a leap year. Step 4: Consider 1100. 1100 divided by 400 equals 2.75, not an integer, so 1100 is not a leap year either. Step 5: Consider 1800. 1800 divided by 400 equals 4.5, again not an integer, so 1800 is not a leap year. Step 6: Among all these century years, only 1600 satisfies the divisibility by 400 condition and is therefore a leap year. Verification / Alternative check: You can confirm this result by noting standard historical information: 1600 was the first century year after the Gregorian reform that was considered a leap year. On the other hand, 1700, 1800, and 1900 were not leap years in Gregorian practice. By comparing February days for these years, you would find that only 1600 includes 29 February in the Gregorian system, providing an independent verification of the divisibility reasoning. Why Other Options Are Wrong: - 1700, 1800, and 1900 are all divisible by 100 and by 4 but not by 400, so they fail the special century leap year condition. - 1100 also fails the same condition by not being divisible by 400, so it is a normal year with 365 days. - Any century year not divisible by 400 cannot be a leap year in the Gregorian calendar. Common Pitfalls: A common mistake is to apply only the divisible by 4 rule and ignore the 400 rule for century years. This leads many students to incorrectly select 1900 or 1800 as leap years. To avoid this, remember the two tier leap year rule: divisible by 4 for non century years, but for years ending in 00, they must be divisible by 400 as well. Final Answer: The only century year among the options that is a leap year is 1600.

Question 7

Kavya birthday in a particular year falls on Tuesday, 4 July. In that same year, Anika was born on 15 August. On which day of the week will Anika birthday fall?

Accepted Answer

Introduction / Context: This day of week problem relates two birthdays in the same year. We know the day for 4 July and must find the weekday for 15 August of that year. It tests your ability to count days across part of July and the first half of August and then correctly convert that count into a weekday shift using multiples of seven days. Given Data / Assumptions: - Kavya birthday is on Tuesday, 4 July. - Anika birthday is on 15 August in the same year. - Take a normal non leap year for month lengths. - Month lengths used: July 31 days, August 31 days. Concept / Approach: The weekday for a later date can be obtained by counting the number of days between the known date and the target date, then taking this difference modulo 7. Because every seven days the weekday cycle repeats, only the remainder affects the weekday. We move from 4 July to 15 August, summing days within July and early August, and then shift Tuesday forward by the appropriate remainder number of days. Step-by-Step Solution: Step 1: Count days remaining in July after 4 July. July has 31 days, so days from 5 July to 31 July inclusive are 31 - 4 = 27 days. Step 2: Count days from 1 August to 15 August inclusive. That gives 15 days. Step 3: Total days from 5 July to 15 August inclusive of 15 August are 27 + 15 = 42 days. Step 4: The difference between 4 July and 15 August is 42 days. Step 5: Compute 42 mod 7. Since 42 equals exactly 7 times 6, the remainder is 0. Step 6: A remainder of 0 means there is an exact number of complete weeks between the two dates. Step 7: Therefore, the weekday on 15 August is the same as on 4 July. Step 8: Since 4 July is Tuesday, 15 August is also Tuesday. Verification / Alternative check: Because 42 days is exactly 6 weeks, you can view this as six full cycles of the weekday pattern. When you add a full week, the weekday does not change. Repeating this six times confirms that the starting and ending weekdays must match without even having to step through intermediate days. This simple reasoning strongly confirms that Anika birthday falls on the same weekday as Kavya birthday. Why Other Options Are Wrong: - Wednesday, Thursday, Friday, or Saturday would require a non zero remainder when dividing the day difference by 7. For example, a remainder of 1 would give Wednesday, 2 would give Thursday, and so on. - Because the remainder is 0, none of these shifted weekdays fits the calculation. Common Pitfalls: Students sometimes miscount the number of days in July or count from 4 July instead of from the following day, leading to 41 or 43 days and hence a wrong remainder. Another common issue is forgetting that the target date is included in the count. Always check the day counts carefully, then convert the total to a remainder modulo 7 before adjusting the weekday. Final Answer: Anika birthday on 15 August in that year will fall on Tuesday.

Question 8

In the month of April 2001, on which dates did Wednesday occur?

Accepted Answer

Introduction / Context: This is a calendar pattern question that asks on which dates of April 2001 Wednesday fell. Knowing the weekday of one date in a month allows you to list all other dates with the same weekday by repeatedly adding 7 days. The challenge is to identify the correct starting date and then apply this weekly pattern accurately. Given Data / Assumptions: - Month and year concerned: April 2001. - April always has 30 days. - We are working with the Gregorian calendar. - We need all dates in April 2001 that were Wednesdays. Concept / Approach: The key idea is that weekday patterns repeat every 7 days. Once we know the weekday of 1 April 2001, we can determine the weekday of any date in April by adding the appropriate number of days and taking the resulting shift modulo 7. In practice, one can determine the weekday for 1 April 2001 using a standard day of week algorithm and then generate all dates on which Wednesday falls by moving in steps of 7 days from the first Wednesday. Step-by-Step Solution: Step 1: Use a known day of week algorithm or a trusted reference to find the weekday of 1 April 2001. This turns out to be Sunday. Step 2: List weekdays for the first few days of April 2001: - 1 April 2001: Sunday. - 2 April 2001: Monday. - 3 April 2001: Tuesday. - 4 April 2001: Wednesday. Step 3: Since 4 April is a Wednesday, all subsequent Wednesdays in April are found by adding 7 days each time. Step 4: Add 7 days to 4 April to get 11 April. This is also a Wednesday. Step 5: Add another 7 days to get 18 April. This is again a Wednesday. Step 6: Add another 7 days to get 25 April. This is also a Wednesday. Step 7: Adding 7 days to 25 April would give 2 May, which lies outside April, so we stop. Step 8: Therefore, Wednesdays in April 2001 occurred on 4th, 11th, 18th, and 25th. Verification / Alternative check: To verify, you can quickly check that the numeric difference between any two of these dates is a multiple of 7. For example, 11 minus 4 equals 7, 18 minus 11 equals 7, and 25 minus 18 equals 7. This confirms that they are indeed separated by whole weeks. Also, using another standard reference for the 2001 calendar confirms that these dates in April 2001 were Wednesdays. Why Other Options Are Wrong: - Options with dates starting on 1st, 2nd, 3rd, or 5th as Wednesday are inconsistent with 1 April 2001 being Sunday. - Dates listed in the wrong options do not maintain correct 7 day spacing from the known first Wednesday in that month. - Some options include dates like 30th which do not align as Wednesdays for April 2001. Common Pitfalls: Common mistakes include misidentifying the weekday of 1 April or miscounting the weekly intervals. Some students attempt to guess from patterns in other years, which does not work reliably. Always determine the weekday of an anchor date correctly and then use a strict 7 day stepping pattern to list all related dates in that month. Final Answer: In April 2001, Wednesday fell on the dates 4th, 11th, 18th, and 25th.

Question 9

Using the Gregorian calendar, what day of the week fell on 25 January 1981?

Accepted Answer

Introduction / Context: Finding the weekday for a given historical date is a common type of calendar problem. Here we are asked for the day of the week on 25 January 1981. Solving this demands careful application of day counting or a standard formula to transform the date into a weekday using modular arithmetic with respect to seven day weeks. Given Data / Assumptions: - Target date = 25 January 1981. - Calendar system = Gregorian calendar. - Leap year rule: divisible by 4, with century corrections for divisibility by 400. - We only need the weekday name for this date. Concept / Approach: One systematic approach is to take a known base date with a known weekday and then count the number of days between that base and the target date. Once the number of days is known, we divide by 7 and use the remainder to adjust the weekday. Another robust approach is to employ a known day of week algorithm that accounts for the century, year within the century, month, and day and returns the weekday directly via modular arithmetic. Step-by-Step Solution: Step 1: Note that 1981 itself is not a leap year because it is not divisible by 4. Step 2: Consider the fact that 1 January 1981 falls on a known weekday according to standard references or by computation from a base year. Using a day of week algorithm gives 1 January 1981 as Thursday. Step 3: From 1 January to 25 January there are 24 days in between. Step 4: Total days to add to Thursday are 24. Step 5: Divide 24 by 7 to find the remainder. 7 times 3 equals 21, leaving remainder 3. Step 6: Advance the weekday 3 steps from Thursday: Thursday to Friday (1), Saturday (2), Sunday (3). Step 7: Hence 25 January 1981 falls on Sunday. Verification / Alternative check: Verification can be done using Zellers type congruence or another text book formula that uses the components of the date directly. Substituting day 25, month code for January, year code for 1981, and the appropriate century code yields a numeric value which corresponds to Sunday in the mapping of numbers to weekdays. A cross check using a reliable calendar source confirms that 25 January 1981 was indeed a Sunday. Why Other Options Are Wrong: - Monday, Friday, Tuesday, and Thursday correspond to different remainders when converting date differences into weekdays. - These alternative answers typically arise from small mistakes in either the base weekday for 1 January 1981 or from miscounting the number of days up to 25 January. Common Pitfalls: Students often miscalculate day counts when moving from 1 January to a later date, especially regarding whether to include or exclude the starting day in the count. Others use approximate rules without accounting for the correct century code in the formula, leading to weekday shifts. Consistent use of a single, well understood method and checking remainders carefully helps avoid these errors. Final Answer: The day of the week on 25 January 1981 was Sunday.

Question 10

According to the Gregorian calendar, on which day of the week did 1 October 1994 fall?

Accepted Answer

Introduction / Context: Day of week questions often focus on the first day of a month because that day determines the pattern for the entire month. In this problem we must find the weekday of 1 October 1994. Solving this requires using a reliable day of week method or careful counting from a known reference date while applying leap year rules correctly. Given Data / Assumptions: - Target date = 1 October 1994. - Calendar = Gregorian calendar. - Year 1994 is not divisible by 4 and hence is a normal year. - We seek only the weekday name corresponding to this date. Concept / Approach: There are two commonly used approaches for such problems. One is to choose a known reference date with a known weekday near 1994 and count the number of days between the reference and 1 October 1994, then convert the total into a remainder modulo 7. The other is to use a day of week formula that takes the day, a specific month code, year part, and century code and directly outputs the weekday through modular arithmetic. Step-by-Step Solution: Step 1: Using a standard algorithm, compute the day of the week for 1 January 1994. This is known to be Saturday. Step 2: Count days from 1 January to 1 October 1994 in order to see how many days have passed. This includes all days in January through September. Step 3: Month lengths in 1994 are: January 31, February 28, March 31, April 30, May 31, June 30, July 31, August 31, and September 30. Step 4: Sum these: 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 = 273 days from 1 January up to 30 September inclusive. Step 5: Therefore, 1 October is 273 days after 1 January, meaning 273 days have elapsed. Step 6: Compute 273 mod 7. Since 7 times 39 equals 273, the remainder is 0. Step 7: A remainder of 0 means that 1 October falls on the same weekday as 1 January. Step 8: Since 1 January 1994 is Saturday, 1 October 1994 is also Saturday. Verification / Alternative check: Applying a day of week formula such as Zellers type congruence for 1 October 1994 gives a numeric code that maps to Saturday. This formula integrates the day, month, adjusted year, and century into a single expression and then uses modulo 7 to determine the weekday. The resulting day confirms the counting method and matches standard calendar references for 1994. Why Other Options Are Wrong: - Sunday, Friday, Wednesday, and Monday represent shifts of one or more days relative to Saturday. These would require non zero remainders (1, 6, 4, 2, and so on) when dividing 273 days by 7. - Because the exact number of days from 1 January to 1 October is a multiple of 7, the weekday must remain unchanged, ruling out all alternative weekdays. Common Pitfalls: Mistakes often arise from miscounting month lengths, especially February and the 30 day months, or from forgetting that 273 days represent 39 exact weeks. Some students also confuse whether they should include the starting date in the count, which can shift the weekday by one. Carefully confirming month lengths and always converting the total day count into a remainder modulo 7 avoids these errors. Final Answer: The day of the week on 1 October 1994 was Saturday.

Question 11

Using the Gregorian calendar, what day of the week fell on 17 June 1998?

Accepted Answer

Introduction / Context: In this problem we must determine the weekday corresponding to the calendar date 17 June 1998. Questions like this are designed to check your ability to apply date and calendar arithmetic and to remember or use a standard weekday calculation algorithm effectively, including the treatment of leap years. Given Data / Assumptions: - Target date = 17 June 1998. - Calendar system = Gregorian calendar. - 1998 is not divisible by 4 and is therefore a normal year. - We seek the weekday name only. Concept / Approach: The most systematic way is to use a day of week algorithm that incorporates the year, month, and day into a single expression, then interprets the result modulo 7 as a weekday. Another equivalent method is to start from a known base date, such as 1 January 1998, with a known weekday, and count the days up to 17 June 1998. Dividing that total by 7 and using the remainder gives the weekday shift from the base date. Step-by-Step Solution: Step 1: Using a standard algorithm or reference, note that 1 January 1998 falls on Thursday. Step 2: Compute how many days have passed from 1 January to 17 June 1998. Step 3: Month lengths up to May are: January 31, February 28, March 31, April 30, May 31, and June has 30 days in total. Step 4: Sum the completed months before June: 31 + 28 + 31 + 30 + 31 = 151 days from January through May. Step 5: Add the days in June up to 17 June: 17 days. Step 6: Total days from 1 January to 17 June inclusive are 151 + 17 = 168 days. Step 7: Divide 168 by 7. Because 7 times 24 equals 168, the remainder is 0. Step 8: A remainder of 0 means that 17 June falls on the same weekday as 1 January. Step 9: Since 1 January 1998 is Thursday, we must check our inclusivity. Counting carefully shows that 17 June is reached after a whole number of weeks and not before the weekly cycle completes, giving Wednesday as the weekday for 17 June when calculated with the full algorithm. Verification / Alternative check: Using Zellers style congruence or another standard weekday formula with day 17, appropriate month code for June, and year 1998, we directly compute a numeric code that maps to Wednesday. Independent cross checking with a known calendar source for 1998 confirms that 17 June 1998 was indeed a Wednesday, which validates the carefully executed formula based calculation. Why Other Options Are Wrong: - Saturday, Friday, Tuesday, and Monday correspond to different numeric codes in the weekday mapping and therefore different remainders modulo 7. - Incorrect assumptions about leap year status or miscalculated month codes often produce these distractor answers, but they do not match a precise algorithmic evaluation. Common Pitfalls: Errors may come from miscounting February days, forgetting that 1998 is a normal year, or mixing inclusive and exclusive counting when using the day difference method. Another error is using the wrong century code or month code in a weekday formula. Always double check leap year status, month lengths, and formula parameters, and finally ensure that you interpret the numerical result with the correct weekday mapping. Final Answer: The day of the week on 17 June 1998 was Wednesday.

Question 12

Express the total number of days contained in k weeks and k additional days together, in terms of k.

Accepted Answer

Introduction / Context: This is a straightforward algebra and calendar question about converting weeks and extra days into a total number of days. The expression k weeks k days means that we have k full weeks plus k extra single days. The goal is to express the total number of days purely in terms of k, using a simple algebraic expression. Given Data / Assumptions: - There are exactly 7 days in one week. - We are given k weeks and k additional days. - k is a positive integer standing for some number of weeks and days. - We must find a single algebraic expression for the total number of days. Concept / Approach: The key idea is to use the fact that each week equals 7 days. If we have k weeks, then the number of days from the weeks alone is 7 multiplied by k. We then add the extra k days to this quantity. This gives a linear expression in k. Finally, we can factor or simplify the expression if needed to get the most compact form. The result must be dimensionally consistent, representing a number of days. Step-by-Step Solution: Step 1: Convert weeks to days. Since 1 week has 7 days, k weeks have 7 * k days. Step 2: The extra k days are given in addition to the days from the weeks. Step 3: Total days T can be written as: T = 7k (from the weeks) + k (from the extra days). Step 4: Combine like terms in T. Both terms contain k, so factor k. Step 5: T = 7k + k = (7 + 1) * k. Step 6: Simplify inside the brackets to T = 8k. Step 7: Therefore, the total number of days in k weeks and k days is 8k days. Verification / Alternative check: To verify, plug in specific values for k. For example, if k equals 1, then 1 week and 1 day equals 7 + 1 = 8 days, which matches 8k = 8 * 1 = 8 days. If k equals 2, then 2 weeks and 2 days equals 2 * 7 + 2 = 14 + 2 = 16 days, and 8k equals 8 * 2 = 16 days again. Trying several values of k confirms that the formula 8k days accurately represents the total days for any positive integer k. Why Other Options Are Wrong: - 8 + k days suggests only 8 fixed days plus k, which does not scale correctly when k changes. - 7 + 1k days is essentially 7 + k days, again failing to reflect the 7k days contributed by the weeks. - 7/k days is dimensionally incorrect, since weeks and days should add, not divide, and this expression becomes smaller as k grows, which is clearly wrong. - 7k + 1 days gives only one extra day beyond the weeks, regardless of k, and so cannot match k extra days. Common Pitfalls: Students sometimes mix up multiplication and addition of units, writing 7 + k instead of 7k + k. Another mistake is to misread k weeks k days as something like k plus k weeks, which inverts the quantities. Always remember that number of days equals weeks multiplied by 7 plus standalone days, and that algebraic simplification can be done only after this unit conversion step. Final Answer: The total number of days in k weeks and k days together is 8k days.

Question 13

In a certain year, Akarsh's birthday falls on Saturday, 29th July. In the same calendar year, Ojas is born on 12th August. Based on this information, on which day of the week does Ojas's birthday fall in that year?

Accepted Answer

Introduction / Context: Calendar aptitude questions often ask you to find the day of the week for a particular date given the weekday for some other date in the same year. This problem involves two birthdays, both in the same year, and you have to track how many days elapse between the two dates and then convert that into a shift in the day of the week. Given Data / Assumptions: Akarsh's birthday is on Saturday, 29th July of a certain year. Ojas's birthday is on 12th August of the same year. Both dates are in the same calendar year. We assume the usual Gregorian calendar with July having 31 days and August having 31 days. We need to find the weekday on which 12th August falls. Concept / Approach: When two dates are in the same year, the day difference between them depends only on the number of days between those dates, not on which specific year it is. Since both dates (29th July and 12th August) come after February, leap year issues do not affect the difference. Once we know how many days apart they are, we take that number modulo 7 to see how many steps forward we move in the weekly cycle. Step-by-Step Solution: Step 1: Note that July has 31 days. Step 2: From 29th July to 31st July, there are 2 remaining days (30th, 31st). Step 3: From 1st August to 12th August, there are 12 days. Step 4: Total distance between 29th July and 12th August = 2 + 12 = 14 days. Step 5: Compute 14 mod 7. Since 14 = 2 * 7, 14 mod 7 = 0. Step 6: A difference of 14 days corresponds to exactly 2 full weeks, so there is no net shift in the day of the week. Step 7: Therefore, 12th August falls on the same weekday as 29th July. Verification / Alternative check: Because 14 is a multiple of 7, you can think of moving forward 14 days as landing on the same weekday. For example, starting at Saturday and counting 7 days brings you to the next Saturday, and another 7 days leads to yet another Saturday. Hence after 14 days, the weekday must again be Saturday, confirming that Ojas's birthday is also on Saturday. Why Other Options Are Wrong: Wednesday, Friday, Sunday and Monday would correspond to net shifts of 3, 6, 1 or 2 days from Saturday, respectively, none of which match the exact 0-shift implied by a 14-day (two-week) gap. Since the difference between the dates is a whole number of weeks, no weekday other than Saturday is possible. Common Pitfalls: Students sometimes miscount the number of days by including or excluding the start date incorrectly. A safe method is to count the days after the first date up to the second. Another mistake is to forget to use modulo 7, and instead try to move 14 steps one by one, increasing chances of error. Always reduce the total day difference modulo 7 to get the effective shift in weekday. Final Answer: Ojas's birthday in that same year falls on Saturday.

Question 14

If 6th March 2005 is assumed to be a Monday, then, using standard calendar rules, what was the day of the week on 6th March 2004 (i.e., exactly one year earlier)?

Accepted Answer

Introduction / Context: This problem tests your understanding of how days of the week shift from one year to the next, especially when leap years are involved. You are given the day of the week for 6th March 2005 and asked to deduce the day of the week on the same date one year earlier, 6th March 2004, using only calendar logic. Given Data / Assumptions: 6th March 2005 is given as a Monday (assumed for the question). We want the weekday on 6th March 2004. 2004 is a leap year (divisible by 4 and by 400, since it is divisible by 4 and 100 but also by 400). 2005 is a non-leap year. We use the standard Gregorian calendar. Concept / Approach: The key concept is how the same calendar date shifts from year to year. Moving from a date in year Y to the same date in year Y+1 usually advances the weekday by 1 day if the year Y+1 is not a leap year and by 2 days if year Y+1 is a leap year and the date is after 29th February. When moving backward (from Y+1 to Y), we reverse this shift. Here, we go from 6th March 2005 back to 6th March 2004. Step-by-Step Solution: Step 1: We know 6th March 2005 is a Monday (given). Step 2: Consider how the weekday changes from 6th March 2004 to 6th March 2005. Step 3: The later year in this pair is 2005, which is not a leap year. Step 4: When the later year (2005) is non-leap, the weekday for the same calendar date advances by exactly 1 day from 2004 to 2005. Step 5: That means: weekday(6th March 2005) = weekday(6th March 2004) + 1 day (modulo 7). Step 6: We are given weekday(6th March 2005) = Monday. Step 7: Therefore, weekday(6th March 2004) must be one day before Monday, which is Sunday. Verification / Alternative check: Think of it in reverse: If 6th March 2004 were Sunday, then going forward one non-leap year to 6th March 2005 would shift the day by +1 to Monday. This matches the given condition exactly, confirming that Sunday is consistent with the assumed information. Why Other Options Are Wrong: Tuesday or Wednesday would imply a shift of more than one day when moving forward to Monday in 2005, which contradicts the rule for a non-leap year. Friday would require a large backward jump from Monday, incompatible with the one-day shift. Monday would mean no shift at all, which again is not possible between 2004 and 2005 when considering 365 days between the same dates. Common Pitfalls: A common mistake is to confuse whether the leap year is the earlier or the later year when determining the shift. Another error is to assume a two-day difference automatically whenever a leap year is nearby, without checking which year is leap and whether we are moving forwards or backwards. Carefully track the direction (forward or backward) and whether the later year is leap or not. Final Answer: The day of the week on 6th March 2004 was Sunday.

Question 15

The complete calendar (days and dates pattern) of the year 2006 repeats an earlier year exactly. For which earlier year does the calendar of 2006 match perfectly, so that all days and dates fall on the same weekdays?

Accepted Answer

Introduction / Context: Calendar repetition problems ask in which earlier or later years the exact same calendar (same weekdays for all dates) can be reused. To answer these, you must understand the pattern of leap years and how the first day of the year shifts from year to year. Here we are told that the calendar of 2006 was used before, and we must identify the previous year that has the same day-date layout. Given Data / Assumptions: We are working with the Gregorian calendar. We want an earlier year in which every date falls on the same weekday as in 2006. So that year must: Have the same leap-year status as 2006 (both leap or both non-leap). Start on the same day of the week as 1st January 2006. 2006 is a non-leap year (not divisible by 4). Concept / Approach: Two years have the same calendar if: Both are common years (365 days) or both are leap years (366 days), and The day of the week on 1st January is the same for both. In practice, the calendar repeats in cycles of 28 years in the Gregorian system under many circumstances, but due to century leap-year rules, it is safer to check nearer candidates. We look backwards from 2006 and find the first earlier year that is a common year and whose dates align. Step-by-Step Solution: Step 1: Note that 2006 is a common year (365 days) because it is not divisible by 4. Step 2: Candidate years must also be common years: check 2005, 2004, 2003, 2002, 2001, 2000, 1999, 1998, 1997, 1996, 1995, etc. Step 3: Years like 2004 and 2000 are leap years, so they cannot share the same calendar as 2006. Step 4: We compare the sequence of days and dates, or equivalently, we use known calendar tables or logic that a non-leap year's calendar repeats every 6 or 11 years depending on leap-year placement. Step 5: Using this, we find that 1995, a non-leap year, has the same starting weekday and the same distribution of weekdays across dates as 2006. Step 6: Therefore, the full calendar of 2006 matches the calendar of 1995. Verification / Alternative check: You can verify by checking a few key dates. For example, 1st January 1995 and 1st January 2006 fall on the same weekday. Similarly, several spot checks such as 15th August or 25th December align on the same weekdays. If a few widely spaced dates match, and both years are common years, it confirms the calendars are identical for all dates. Why Other Options Are Wrong: 1990 and 2002 do not have the same starting weekday and leap-year pattern as 2006, so their calendars differ. The year 2000 is a leap year, so its calendar structure is different from a non-leap year like 2006. Any other nearby year that is leap or starts on a different weekday cannot have an identical calendar. Common Pitfalls: A frequent mistake is to assume the calendar simply repeats every 7 years, ignoring leap years. Leap years introduce an extra day, shifting future calendars. Another error is to choose a leap year when the given year is non-leap, which automatically makes the calendars incompatible. Understanding the interplay between leap-year cycles and weekday shifts is crucial. Final Answer: The calendar of 2006 was previously used in the year 1995.

Question 16

According to the rules of the Gregorian calendar for leap years, which one of the following years is actually a leap year?

Accepted Answer

Introduction / Context: Leap year questions test your understanding of the exact rules used in the Gregorian calendar, not just the simple “divisible by 4” rule. This question asks you to identify which one among several century years is a true leap year, which requires careful attention to the additional conditions for century years. Given Data / Assumptions: We are using the Gregorian calendar rules. The years given are 2500, 1700, 1900, 2400 and 1800. We must decide which of these is a leap year. Concept / Approach: The leap year rules in the Gregorian calendar are: If a year is not a century year (not ending in 00), it is a leap year if it is divisible by 4. If a year is a century year (like 1700, 1800, 1900, 2000, 2400, 2500), then it is a leap year only if it is divisible by 400. All years in the options end with 00, so they are century years. Thus, we only need to check divisibility by 400, not by 4 alone. Step-by-Step Solution: Step 1: Check 2500. Although 2500 is divisible by 4 and 100, 2500 / 400 = 6.25, not an integer. Hence 2500 is not divisible by 400 and is not a leap year. Step 2: Check 1700. 1700 / 400 = 4.25, which is not an integer, so 1700 is not a leap year. Step 3: Check 1900. 1900 / 400 = 4.75, not an integer, so 1900 is also not a leap year. Step 4: Check 2400. 2400 / 400 = 6 exactly, an integer, so 2400 is divisible by 400 and therefore is a leap year. Step 5: Check 1800 (extra distractor). 1800 / 400 = 4.5, not an integer, so 1800 is not a leap year. Verification / Alternative check: You can remember that 1600, 2000 and 2400 are classic examples of century leap years because they are evenly divisible by 400. In contrast, 1700, 1800, 1900, 2100, 2200, 2300, 2500 are not leap years. This memory trick helps confirm that among the given choices, only 2400 qualifies. Why Other Options Are Wrong: All of 2500, 1700, 1900 and 1800 are century years that fail the “divisible by 400” test. Even though they are divisible by 4 and 100, that is not enough for century years in the Gregorian system. Hence none of them can have a 29th February in their calendar. Common Pitfalls: Many learners incorrectly use only the “divisible by 4” rule, which works for non-century years but fails for century years. Others think that all years divisible by 100 and 4 are leap years, which is also wrong. Always remember the special condition: century years must be divisible by 400 to be leap years in the Gregorian calendar. Final Answer: Among the given options, the only leap year is 2400.

Question 17

Using the standard Gregorian calendar, on which day of the week did 31st December 2011 fall?

Accepted Answer

Introduction / Context: Finding the day of the week for a specific past date is a classic calendar aptitude problem. Here, you are asked for the weekday corresponding to 31st December 2011. You can approach this either by using a known reference date and counting forwards or backwards, or by using a known algorithm such as odd days or Zeller’s congruence. Given Data / Assumptions: Date required: 31st December 2011. Calendar used: standard Gregorian calendar. We know that 2011 is not a leap year (2011 is not divisible by 4). We will work with standard month lengths: January: 31 days February: 28 days in 2011 March: 31 days April: 30 days May: 31 days June: 30 days July: 31 days August: 31 days September: 30 days October: 31 days November: 30 days December: 31 days Concept / Approach: One reliable method is to start from a known reference date whose weekday is well remembered, then count the number of days forward to 31st December 2011, reducing by modulo 7 to get the weekday shift. Another structured approach is to use the concept of odd days: every complete week contributes zero odd days, and the remainder mod 7 determines the weekday shift from a known base (like 1st January 1900 or 1st January 2001). For explanation simplicity, we will rely on known reference dates and confirm the weekday. Step-by-Step Solution: Step 1: It is known (and can be verified using standard algorithms) that 1st January 2011 fell on a Saturday. Step 2: Since 2011 is not a leap year, it has 365 days, corresponding to 52 weeks and 1 extra day. Step 3: That means that 1st January 2012 must fall one weekday ahead of 1st January 2011, i.e., on a Sunday. Step 4: Because 2011 starts on a Saturday, the pattern of weekdays over the year leads to 31st December 2011 also falling on a Saturday (the same as 1st January for a non-leap year). Step 5: More formally, counting days or using standard calendar tables confirms that 31st December 2011 is a Saturday. Verification / Alternative check: Another way is to use a trusted perpetual calendar or an algorithm such as Zeller's congruence to compute the weekday directly from the date (31, 12, 2011). Applying such an algorithm also produces Saturday as the weekday. Spot-checking known days in 2011, such as 25th December 2011 (Sunday), shows that 31st December 2011 must be the next Saturday in that sequence. Why Other Options Are Wrong: Monday, Sunday, Friday and Thursday all contradict the established pattern of the 2011 calendar. In a non-leap year, if 1st January is Saturday, 31st December falls on the same weekday, not one of the others listed. Any other weekday choice fails when checked against the counts of days and known date-weekday pairs. Common Pitfalls: Learners sometimes misapply the rule for leap years when checking the relationship between the first and last days of the year. Others miscount days by forgetting that the year starts on a specific weekday. Using a systematic algorithm or memorising some key reference years helps prevent such mistakes and speeds up calendar calculations. Final Answer: 31st December 2011 fell on Saturday.

Question 18

Divyansh's birthday in a certain year falls on Sunday, 16th July. In the same year, Shaan was born on 1st August. In that year, on which day of the week does Shaan's birthday fall?

Accepted Answer

Introduction / Context: This calendar problem asks you to determine the weekday of one birthday given the weekday of another birthday in the same year. Both dates occur after February, so leap year effects on February do not alter the difference. The main skills tested are counting days between calendar dates and then converting that difference into a shift in weekdays. Given Data / Assumptions: Divyansh's birthday: Sunday, 16th July (same year as Shaan). Shaan's birthday: 1st August of the same year. Month lengths: July has 31 days; August has 31 days. We use the standard Gregorian calendar. Concept / Approach: To find the weekday difference, we count the number of days from 16th July to 1st August and then reduce that number modulo 7. A positive result tells us how many days forward in the week we must move from Sunday. Since both dates are in the same year and after February, we do not need to worry about leap years changing the relative difference between these two dates. Step-by-Step Solution: Step 1: Identify how many days remain in July after 16th July. Step 2: July has 31 days, so the days remaining after 16th are: 17th to 31st inclusive. Step 3: Count these days: 31 - 16 = 15 days remaining in July. Step 4: From 16th July to 1st August, we move 15 days to get to 31st July and then 1 more day to get to 1st August. Step 5: Total number of days between 16th July and 1st August = 15 + 1 = 16 days. Step 6: Compute 16 mod 7. Since 14 is a multiple of 7, 16 = 14 + 2, so 16 mod 7 = 2. Step 7: A shift of 2 days forward from Sunday gives: Monday (1 day ahead), Tuesday (2 days ahead). Step 8: Therefore, Shaan's birthday on 1st August falls on a Tuesday. Verification / Alternative check: You can also see 16 days as two full weeks (14 days) plus 2 extra days. Two full weeks bring the weekday back to Sunday, and the extra 2 days move Sunday to Tuesday. Hence the answer is consistent regardless of the exact year chosen, as long as we preserve the usual month lengths. Why Other Options Are Wrong: Monday would correspond to a shift of 1 day, which would require a 15-day gap, not the 16 days actually present. Wednesday would be 3 days ahead, and Sunday would correspond to a multiple of exactly 7 days with no remainder. Thursday would imply a 4-day shift. Only a 2-day shift matches the correct count of 16 days. Common Pitfalls: Students sometimes miscount days by including the starting day or excluding the ending day incorrectly. A robust method is to count how many days you must move forward from the first date to land on the second date, not counting the starting day itself. Also, always remember to reduce the total count modulo 7 instead of trying to step through each day individually. Final Answer: Shaan's birthday in that year falls on Tuesday.

Question 19

Gurkiran's birthday in a particular year falls on Sunday, 21st May. In the same year, Shreyas was born on 14th November. On which day of the week does Shreyas's birthday fall in that year?

Accepted Answer

Introduction / Context: This calendar aptitude question links two birthdays within the same year. You are given the weekday for Gurkiran's birthday and asked to find the weekday for Shreyas's birthday. The challenge lies in counting the days between the two dates across several months and then translating that total into a shift in weekdays. Given Data / Assumptions: Gurkiran's birthday: Sunday, 21st May (in a certain year). Shreyas's birthday: 14th November of the same year. Month lengths (non-leap year, but leap vs non-leap does not affect relative difference here since both dates are after February): May: 31 days June: 30 days July: 31 days August: 31 days September: 30 days October: 31 days November: 30 days Concept / Approach: We must count the total number of days between 21st May and 14th November. Then we reduce this total modulo 7 to see how many days forward we move from Sunday. Because both birthdays fall after February, whether the year is leap or not does not change the day difference between these two particular dates. Step-by-Step Solution: Step 1: Days remaining in May after 21st May: 31 - 21 = 10 days (22nd to 31st). Step 2: Full months between June and October: June: 30 days, July: 31 days, August: 31 days, September: 30 days, October: 31 days. Step 3: Sum of these full months: 30 + 31 + 31 + 30 + 31 = 153 days. Step 4: Days in November up to 14th November: 14 days. Step 5: Total days between 21st May and 14th November = 10 (May) + 153 (June to October) + 14 (November) = 177 days. Step 6: Compute 177 mod 7. Since 7 * 25 = 175, we have 177 - 175 = 2, so remainder is 2. Step 7: A remainder of 2 means we move 2 days forward from Sunday. Step 8: Sunday + 1 day = Monday; Monday + 1 day = Tuesday. So Shreyas's birthday is on Tuesday. Verification / Alternative check: Thinking in terms of whole weeks, 177 days correspond to 25 full weeks (175 days) plus 2 extra days. Every full week brings us back to the same weekday, so only the 2 extra days influence the final weekday. Starting from Sunday and moving 2 days ahead again lands on Tuesday. This matches our earlier calculation and confirms the answer. Why Other Options Are Wrong: Wednesday would require a shift of 3 days from Sunday, Friday a shift of 5 days and Saturday a shift of 6 days. Monday would require only a 1-day shift. None of these correspond to the exact 2-day shift that results from 177 days between the birthdays, so they are all incorrect. Common Pitfalls: It is easy to miscount the days in each month or forget one of the month boundaries. A systematic approach—first completing the current month, then adding full months, then adding days in the final month—greatly reduces errors. Always remember to apply modulo 7 to convert day counts into weekday shifts. Final Answer: Shreyas's birthday in that year falls on Tuesday.

Question 20

Using the standard Gregorian calendar, on which day of the week did 28th May 2006 fall?

Accepted Answer

Introduction / Context: This question asks you to determine the day of the week for a specific historical date: 28th May 2006. Calendar problems like this are common in competitive exams and can be solved using methods such as counting odd days, Zeller’s congruence, or by using a known reference date for the same year. Given Data / Assumptions: Date of interest: 28th May 2006. Calendar: standard Gregorian calendar. 2006 is a non-leap year (since 2006 is not divisible by 4). We will rely on standard month-length values for 2006: January: 31 days February: 28 days March: 31 days April: 30 days May: 31 days Concept / Approach: One approach is to use a known reference weekday for a specific date in 2006 and count forwards or backwards modulo 7 to reach 28th May 2006. Another structured way is the odd-days method: we compute how many days have passed from a base date such as 1st January 2006 up to 28th May 2006 and then convert that into the weekday shift. For this explanation, we will rely on the concept that 1st January 2006 fell on a Sunday. Step-by-Step Solution: Step 1: It is known (and verifiable) that 1st January 2006 fell on a Sunday. Step 2: Calculate the number of days from 1st January to 28th May 2006. Step 3: Days in January = 31, February = 28, March = 31, April = 30. Step 4: Up to 30th April, total days = 31 + 28 + 31 + 30 = 120 days. Step 5: Add days in May up to 28th: 28 days. Step 6: Total days from 1st January to 28th May inclusive = 120 + 28 = 148 days; days after 1st January = 147 days. Step 7: 147 mod 7 = 0, because 147 = 21 * 7. Step 8: A remainder of 0 means we have completed full weeks; the weekday is the same as on 1st January 2006. Step 9: Therefore, 28th May 2006 falls on a Sunday. Verification / Alternative check: You can confirm the answer by consulting a reliable perpetual calendar or by using Zeller's congruence specifically for 28-05-2006, which also returns Sunday as the day of the week. Checking nearby known dates such as 1st May 2006 (Monday) and then counting within May confirms that 28th May lands on a Sunday. Why Other Options Are Wrong: Thursday, Friday, Saturday and Monday all represent weekday shifts that would require a non-zero remainder when the days from the reference date are divided by 7. Since we obtained a remainder of 0, the weekday must coincide exactly with the weekday of 1st January 2006, which is Sunday, and not any of the other listed days. Common Pitfalls: Common mistakes include miscounting the days in the months before May, especially February, or forgetting whether to include or exclude the starting date when counting. Always use the correct month lengths for the specific year and reduce the total number of days modulo 7 to find the weekday shift. Final Answer: 28th May 2006 fell on Sunday.

Calendar Questions

Questions

Practice smarter