Question 1

The calendar for the year 2003 will be repeated in which one of the following years?

Accepted Answer

Introduction / Context:This question is another example of calendar repetition. You are asked to find the next year after 2003 that has an identical calendar, meaning all dates fall on the same weekdays. This requires you to consider both leap-year patterns and weekday shifts over a span of years. Given Data / Assumptions: Reference year: 2003. We must find a later year with exactly the same calendar. Options include: 2024, 2014, 2009, 2020, 2010. We use Gregorian leap year rules. Concept / Approach:Two conditions must be satisfied for two years to share the same calendar: Both years must be of the same type: leap years or non-leap years. The weekday on which 1 January falls must be the same in both years. A non-leap year typically repeats its calendar after 6 or 11 years, depending on how leap years are positioned in between, and fully stabilises in a 28-year cycle. Step-by-Step Solution:Step 1: Identify the type of 2003. It is not divisible by 4, so it is a non-leap year.Step 2: A matching year must also be a non-leap year.Step 3: The next candidate years in the list are 2009, 2010, 2014, 2020 and 2024. Among these, 2009 and 2010 are non-leap; 2014 is also non-leap; 2020 and 2024 are leap years (divisible by 4).Step 4: We also need the same starting weekday. For non-leap years, after a non-leap year the weekday of 1 January shifts by +1, and after a leap year it shifts by +2.Step 5: If you track the weekday shifts or consult a year–calendar table, you find that 1 January 2014 falls on the same weekday as 1 January 2003 and that both years are non-leap years.Step 6: Therefore, 2014 is the first year after 2003 that has an identical calendar to 2003. Verification / Alternative check:A more concrete verification can be done by counting shifts: from 2003 to 2014 you have a specific combination of leap and non-leap years that results in a net shift that is a multiple of 7 days, restoring the weekday alignment. Checking key dates (for example, whether 1 March or 1 October fall on the same weekdays in both years) confirms that the calendars indeed match. Why Other Options Are Wrong:2024 and 2020 are leap years and cannot share a calendar with the non-leap year 2003. 2009 and 2010 are non-leap years but do not align in weekday pattern with 2003. Only 2014 satisfies both the non-leap condition and the correct weekday alignment. Common Pitfalls: Assuming that any year that is a multiple of 7 years ahead must have the same calendar, which ignores leap-year effects. Matching only leap vs. non-leap status and forgetting about the starting weekday. Trying to perform year-by-year counting of days without a systematic or tabular approach. Final Answer:The calendar for 2003 is repeated in the year 2014.

Question 2

If the third day of a month is a Tuesday, which day of the week will be the 25th day of that month?

Accepted Answer

Introduction / Context:This question is a straightforward example of working with days of the month and their corresponding weekdays. It tests your ability to move forward by a specific number of days using the 7-day cycle of the week. Given Data / Assumptions: The 3rd day of the month is a Tuesday. We are asked for the weekday on the 25th day of the same month. The month length is not needed as long as it has at least 25 days. Concept / Approach:The basic idea is to calculate the difference in days between the 3rd and 25th dates, then use the remainder modulo 7 to shift forward from Tuesday. Because the week repeats every 7 days, only the remainder after dividing the difference by 7 affects the weekday. Step-by-Step Solution:Step 1: Compute the difference in dates between the 3rd and the 25th: 25 − 3 = 22 days.Step 2: Find 22 mod 7. Since 7 * 3 = 21, 22 = 7 * 3 + 1.Step 3: The remainder is 1, which means the 25th day falls 1 weekday after the 3rd.Step 4: The 3rd is a Tuesday (given).Step 5: Moving one day forward from Tuesday gives Wednesday.Step 6: Therefore, the 25th day of the month is a Wednesday. Verification / Alternative check:To verify, you can list a shorter segment: 3rd is Tuesday, 10th (7 days later) is Tuesday, 17th is Tuesday, 24th is Tuesday, and the 25th is one day after that, which is Wednesday. This matches the modular arithmetic approach and confirms the result. Why Other Options Are Wrong:Tuesday would be correct if the difference in days were a multiple of 7 with remainder 0, not 1. Monday or Sunday would require moving backward in time or having a different remainder. Friday is several days ahead in the cycle and does not correspond to a remainder of 1. Common Pitfalls: Miscomputing the difference in dates (e.g., using 23 instead of 22). Dividing by 7 but confusing the quotient and remainder. Moving in the wrong direction on the weekday cycle (backward instead of forward). Final Answer:The 25th day of the month will be a Wednesday.

Question 3

In a given year, Holi is celebrated on 6 March at 9:00 AM and Diwali is celebrated on 11 November at 8:00 PM. What is the total time interval between these two events, in hours?

Accepted Answer

Introduction / Context:This is a detailed calendar and time-interval question, combining date differences with hour-level precision. You are asked to find the total number of hours between two events in the same year: Holi and Diwali, given specific dates and times. Given Data / Assumptions: Holi: 6 March at 9:00 AM. Diwali: 11 November at 8:00 PM. Both festivals occur in the same (non-leap) year. We must find the total time interval in hours. Concept / Approach:We break the problem into two parts: First, find the number of full days between 6 March and 11 November. Second, adjust for the difference in time on the start and end days (from 9:00 AM to 8:00 PM). Once the total number of days and the additional hours are found, we convert everything into hours. Step-by-Step Solution:Step 1: Assume a non-leap year. The months from March to October have the following days: March 31, April 30, May 31, June 30, July 31, August 31, September 30, October 31, November 30.Step 2: From 6 March 9:00 AM to the end of 6 March 9:00 AM next day is 24 hours, but we will count days from calendar dates first and handle times afterward.Step 3: Compute days from 6 March to 11 November. From 7 March to 31 March there are 25 days. Then add full months: April (30), May (31), June (30), July (31), August (31), September (30), October (31).Step 4: Up to 31 October, total days after 6 March = 25 + 30 + 31 + 30 + 31 + 31 + 30 + 31 = 239 days.Step 5: From 1 November to 11 November is 11 days. So the total from 6 March to 11 November (date-to-date) is 239 + 11 = 250 days.Step 6: Now account for the time difference on these dates: From 9:00 AM on 6 March to 9:00 AM on 11 November is 250 full days. From 9:00 AM to 8:00 PM on 11 November adds 11 extra hours.Step 7: Convert 250 days into hours: 250 * 24 = 6000 hours.Step 8: Add the extra 11 hours: 6000 + 11 = 6011 hours. Verification / Alternative check:Another way to check is to carefully sum month-by-month from 6 March to 11 November ensuring no month is missed and then independently confirm that 250 days * 24 + 11 hours = 6011 hours. Because the difference in time of day is less than 24 hours, it is safe to treat the 250 days as an integer and then adjust by +11 hours. Why Other Options Are Wrong:2101, 4125 or 5124 hours are far too small and correspond to much shorter intervals than eight months. 5890 hours is close but still undercounts the true interval, indicating a missed month or incorrect day count. Only 6011 hours matches the exact calculation based on 250 days plus 11 additional hours. Common Pitfalls: Forgetting to include all intermediate months or miscounting the days from March to November. Ignoring the time-of-day difference and calculating only whole days. Miscomputing the total number of days or incorrectly multiplying by 24 when converting to hours. Final Answer:The total time interval between Holi and Diwali is 6011 hours.

Question 4

How many total days are there in four consecutive years in the Gregorian calendar, assuming that exactly one of the years is a leap year with 366 days and the remaining three years are ordinary years with 365 days each?

Accepted Answer

Introduction / Context: This problem is a basic calendar aptitude question that checks whether you understand how many days occur in ordinary years and leap years, and how these combine when we look at a block of consecutive years. Such questions are common in competitive exams because they build the foundation for more advanced calendar and day of week problems. Given Data / Assumptions: We are considering four consecutive years in the Gregorian calendar. An ordinary year has 365 days. A leap year has 366 days. In a normal cycle of four consecutive years there is exactly one leap year and three ordinary years, ignoring rare century exceptions for this level of question. Concept / Approach: The idea is to split the total days into contributions from ordinary years and the leap year. Once we know how many of each type occur in the four year block, we simply add their respective day counts. This uses only simple multiplication and addition, but it is important not to miscount the number of leap years in the given range. Step-by-Step Solution: Step 1: In four consecutive years, assume there are three ordinary years and one leap year. Step 2: Each ordinary year has 365 days, so days in ordinary years = 3 * 365. Step 3: Calculate 3 * 365 = 1095 days. Step 4: A leap year has 366 days, so days in the leap year = 366. Step 5: Add them to get the total days in four years = 1095 + 366. Step 6: Compute 1095 + 366 = 1461 days. Verification / Alternative check: Another way is to use average length of a year. Over a long period, average days per year in the Gregorian calendar is close to 365.25. If we multiply 365.25 * 4, we get 1461, which confirms the same result. This matches the direct calculation using three ordinary years and one leap year. Why Other Options Are Wrong: 1460 is too small because it assumes exactly 365 days for every year, ignoring the leap year. 1462 is too large and would correspond to two leap years in four consecutive years, which is not typical. 1459 is even smaller and clearly does not fit any reasonable combination of ordinary and leap years in four consecutive years. Common Pitfalls: A frequent mistake is to assume all four years are ordinary and simply do 4 * 365. Another mistake is to think there are two leap years within every set of four consecutive years, which is not correct in the usual non century situation. Students may also forget the exact number of days in a leap year and mix up 365 and 366, so always recall that February has 29 days in a leap year. Final Answer: Therefore, the total number of days in four consecutive years, with three ordinary years and one leap year, is 1461 days.

Question 5

Starting from 20 January 2017, which is given as a Sunday, what will be the exact calendar date and day of the week after 90 days from that day?

Accepted Answer

Introduction / Context: This question tests your understanding of how to move forward through the calendar by a fixed number of days, and how to track both the date and the day of the week. Problems that say something like “today is” followed by a specific date and weekday are very common in calendar aptitude topics and help build confidence with date arithmetic. Given Data / Assumptions: Today is 20 January 2017. The day on 20 January 2017 is stated to be Sunday. We need to find the date and weekday that is exactly 90 days after this given day. Months have their usual lengths: January 31 days, February 28 days in a non leap year, and March 31 days. We treat 2017 as a non leap year since February 2017 has 28 days. Concept / Approach: The approach is to calculate how many days remain in January after the twentieth, then add full months, and finally add any remaining days to reach a total of 90. In parallel, we use the fact that every block of 7 days brings us back to the same weekday, so we only need the remainder when 90 is divided by 7 to shift the weekday correctly from the starting Sunday. Step-by-Step Solution: Step 1: Remaining days in January after 20 January are 31 minus 20 = 11 days. Step 2: Days in February 2017 are 28 days, and days in March are 31 days. Step 3: Add these to get 11 + 28 + 31 = 70 days from 21 January to 31 March inclusive of 31 March. Step 4: We need 90 days total, so remaining days after 31 March are 90 − 70 = 20 days. Step 5: Counting 20 more days into April leads us to 20 April as the target date. Step 6: To find the weekday, note that 90 divided by 7 gives a remainder of 6, so the day moves forward by 6 days from Sunday. Step 7: Starting at Sunday and moving forward 6 days gives Saturday as the weekday on 20 April. Verification / Alternative check: You can also group 90 days as 84 days plus 6 days. Exactly 84 days correspond to 12 full weeks, so they cycle the weekday back to Sunday. Then only the extra 6 days matter. Moving 6 days ahead from Sunday goes to Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, which again confirms that the weekday is Saturday. This double check supports our calculation of both the date and the day. Why Other Options Are Wrong: 18 April, Friday is two days earlier than the correct date and also one weekday off. 21 April, Sunday is one day ahead of the correct date and does not match the weekday shift. 19 April, Saturday is one date too early compared to the proper count of 90 days. Only 20 April, Saturday matches both the number of days and the weekday shift from the starting Sunday. Common Pitfalls: A common mistake is to forget whether to include or exclude the starting day when counting. In standard aptitude questions, when we say “90 days from today” we usually count the days after the current day, not including it. Another error is miscounting the days in February or March, which can shift the result by one or more days. Finally, when handling weekdays, some candidates incorrectly add full multiples of 7 to the remainder, which is unnecessary since multiples of 7 do not change the weekday. Final Answer: Therefore, 90 days from 20 January 2017, given as a Sunday, falls on 20th April, Saturday.

Question 6

In the context of the Gregorian leap year rule, which of the following given years has exactly 366 days and therefore qualifies as a leap year?

Accepted Answer

Introduction / Context: This calendar question focuses on identifying leap years, that is, years with 366 days. Leap year identification is crucial in many aptitude questions, because it affects the total number of days and therefore the day of the week for given dates. Understanding the century rules in the Gregorian calendar avoids common traps in exam questions. Given Data / Assumptions: We are given four candidate years: 1900, 1200, 2500, and 1700. We need to decide which of these years has 366 days. The Gregorian leap year rule applies to these years. Concept / Approach: The Gregorian calendar defines a leap year as a year divisible by 4, except for century years. For century years, there is an additional condition: the year must be divisible by 400 to be a leap year. If a century year is divisible by 100 but not by 400, it remains an ordinary year of 365 days. Applying this rule carefully to each option allows us to determine which year actually has 366 days. Step-by-Step Solution: Step 1: Check 1900. It is divisible by 100 but 1900 / 400 = 4 remainder 300, so it is not divisible by 400. Step 2: Therefore 1900 is not a leap year and has 365 days. Step 3: Check 1200. It is divisible by 100, and 1200 / 400 = 3 exactly, so it is divisible by 400. Step 4: A century year divisible by 400 is a leap year, so 1200 has 366 days. Step 5: Check 2500. It is divisible by 100 but 2500 / 400 leaves a remainder, so it is not divisible by 400. Hence it is not a leap year. Step 6: Check 1700. It is divisible by 100 but 1700 / 400 also leaves a remainder, so it is not divisible by 400, and therefore not a leap year. Verification / Alternative check: We can verify the rule by recalling well known leap years such as 2000 and 1600 which are divisible by 400. Both are recognised as leap years. In contrast, years like 1900 and 2100 are commonly cited examples of century years that are not leap years because they do not satisfy the divisibility by 400 condition. When we apply the same logic to 1200, it clearly behaves like 1600 or 2000 and thus is a genuine leap year with 366 days. Why Other Options Are Wrong: 1900 is a century year that is divisible by 4 and 100 but not by 400, so it has only 365 days. The same reasoning applies to 2500 and 1700, which are also divisible by 100 but not by 400. Since none of these meet the divisibility by 400 test, they remain ordinary years with 365 days each. Only 1200 passes the 400 divisibility test and so qualifies as a leap year with 366 days. Common Pitfalls: Many learners incorrectly assume that every year divisible by 4 is a leap year, forgetting the special rule for century years. This leads to mistakes in questions involving years like 1900 and 2100. Another common error is mixing up the direction of the rule and thinking that divisibility by 400 is optional. In reality, for century years the divisibility by 400 requirement is essential. Always separate the cases into general years and century years, and then apply the appropriate conditions. Final Answer: Therefore, among the given options, the only year with 366 days is 1200.

Question 7

On 24 November 2007 it is given that Thursday falls on this date; based on this information, what day of the week was it on 10 November 2006?

Accepted Answer

Introduction / Context: This problem involves working backward and forward with days of the week over a span of more than one year. You are given the weekday for a particular date in 2007 and asked to determine the weekday for an earlier date in 2006. Such questions test your ability to manage long gaps of days using the idea that every 7 days the calendar repeats its weekday pattern. Given Data / Assumptions: 24 November 2007 is given as a Thursday. We need to find the day of the week on 10 November 2006. We assume the Gregorian calendar with February 2007 as a non leap month of 28 days. The day of the week repeats every 7 days. Concept / Approach: The key idea is to calculate how many days lie between 10 November 2006 and 24 November 2007, then use the remainder when that number is divided by 7 to shift weekdays backward. Since the target date is earlier than the reference date, we will move backwards in time from the known Thursday by the appropriate number of days modulo 7 to determine the weekday for 10 November 2006. Step-by-Step Solution: Step 1: Count the number of days between 10 November 2006 and 24 November 2007. Step 2: The exact difference is 379 days between these two dates. Step 3: Since weekdays repeat every 7 days, find 379 mod 7, which is 1. Step 4: A remainder of 1 means that 10 November 2006 falls one day earlier in the week compared to 24 November 2007. Step 5: The given day on 24 November 2007 is Thursday. Step 6: One day before Thursday is Wednesday. Step 7: Therefore, 10 November 2006 was a Wednesday. Verification / Alternative check: Instead of computing the full 379 days directly, you can break the interval into manageable pieces. For example, count days from 10 November 2006 to 10 November 2007, then from 10 November 2007 to 24 November 2007. The one year span from 10 November 2006 to 10 November 2007 contains 365 days, because 2007 is not a leap year affecting February. Then from 10 November 2007 to 24 November 2007 there are 14 days. Adding 365 and 14 again gives 379 days, and 379 mod 7 equals 1. This reconfirms that the weekday shifts by one day when moving between the two dates. Why Other Options Are Wrong: Tuesday would be two days earlier than Thursday, but we only require a shift of one day. Friday is one day after Thursday, which corresponds to moving forward instead of backward. Another Thursday would correspond to a shift of a multiple of 7 days, but our remainder is 1, not 0, so that cannot be correct. Therefore only Wednesday matches the computed backward shift of one day from the known Thursday. Common Pitfalls: A common source of error is confusing whether you should move forward or backward in time when the target date is earlier than the reference date. Another frequent mistake is ignoring the modulus operation and attempting to track the full number of days, which is unnecessary and can cause arithmetic slips. Students may also miscount the number of days between the two dates by forgetting that year 2007 is not a leap year or by making month length errors. Always remember that only the remainder upon division by 7 matters for the weekday. Final Answer: Hence, the day of the week on 10 November 2006 was Wednesday.

Question 8

For which of the following years does the calendar of the year 1989 repeat exactly, meaning that dates and days of the week fall in the same pattern for all twelve months?

Accepted Answer

Introduction / Context: This question belongs to the advanced part of calendar aptitude, where you are asked to determine another year whose calendar is identical to a given year. Two years have the same calendar if they are of the same type, both leap or both ordinary, and if the weekdays align correctly for all dates. Knowing how and when calendars repeat is useful for quickly answering day of week questions without recomputing from scratch. Given Data / Assumptions: The reference year is 1989. Candidate years are 1978, 1970, 1980, and 1985. We are using the Gregorian calendar and standard leap year rules. The calendar repeats when the year type and weekday alignment for 1 January match. Concept / Approach: To have the same calendar, two years must satisfy two conditions. First, both years must be either leap years or ordinary years. Second, the day of the week on 1 January must be the same for both. Calendar repetition generally follows a cycle related to 28 years, but century and leap year effects can shift this pattern. A practical exam approach is to reason about total day shifts between 1 January of the given year and 1 January of a candidate year, and then use the modulo 7 idea. Step-by-Step Solution: Step 1: Determine whether 1989 is a leap year. It is not divisible by 4, so it is an ordinary year with 365 days. Step 2: Any matching year must also be an ordinary year, not a leap year. Step 3: Check candidate years. 1978, 1970, 1980, and 1985 are all not divisible by 4, so each is an ordinary year. Step 4: Next, we consider the shifts in total days between 1 January 1978 and 1 January 1989, accounting for leap years in between, and reduce this shift modulo 7. Step 5: When this calculation is carried out correctly, 1978 turns out to have the same weekday on 1 January as 1989, while the other candidate years do not. Step 6: Therefore the full layout of weekdays versus dates in each month matches between 1978 and 1989. Verification / Alternative check: Another way is to recall that calendars often repeat in cycles where a year can match another year 11 years earlier or later in some cases, depending on leap years in between. For 1989, the year 1978 is known to be a matching year in many calendar tables. If you construct miniature monthly calendars for January and February of both years and see that the weekdays line up exactly for all dates, you will observe that 1978 and 1989 align, while 1970, 1980, and 1985 do not align in the same way. This comparative check confirms the choice of 1978 as the correct answer. Why Other Options Are Wrong: 1970 is an ordinary year but the weekday on 1 January 1970 does not match the weekday on 1 January 1989, so the entire calendar shifts. Similarly, 1980 is an ordinary year but its starting weekday and internal distribution of weekdays across months does not coincide with 1989. The same mismatch happens for 1985. Thus, none of these years are calendar twins of 1989. Common Pitfalls: A common pitfall is to assume that any year which differs by 7, 11, or 28 years automatically has the same calendar. In reality, you must also check how many leap years lie between the two years, because leap years add an extra day of shift. Another mistake is ignoring the difference between leap years and ordinary years and matching them incorrectly. Always confirm both the year type and the cumulative day shift modulo 7 when identifying a matching calendar year. Final Answer: Hence, the year whose calendar is the same as that of 1989 is 1978.

Question 9

If 5 March 2005 is given to be a Monday, then what day of the week would 5 March 2004 have fallen on in the same calendar system?

Accepted Answer

Introduction / Context: This question is a classic example of moving from one year to a neighbouring year to determine the day of the week on the same date. The information about the weekday for 5 March 2005 is provided, and we are asked to infer the weekday for 5 March 2004. These questions highlight how leap years and ordinary years change the weekday when you move forward or backward by full years. Given Data / Assumptions: 5 March 2005 is given as a Monday. We need the weekday on 5 March 2004. Year 2004 is a leap year, but the date 5 March comes after 29 February. We are using the Gregorian calendar and standard leap year definitions. Concept / Approach: When you move from a given date in one year to the exact same date in the next year, the weekday normally shifts by 1 day if the year in between contributes 365 days, and by 2 days if the year in between contributes 366 days that fall after the considered date. However, here we must be careful: the interval from 5 March 2004 to 5 March 2005 does not include the leap day of 2004, because 29 February 2004 occurred before 5 March 2004. Hence the effective day shift is only 365 days, not 366 days. Step-by-Step Solution: Step 1: Consider the interval from 5 March 2004 to 5 March 2005. Step 2: Count the number of days between these two dates. Step 3: Although 2004 is a leap year, 5 March is after 29 February, so the extra day of the leap year lies before the start of our interval. Step 4: Therefore, the number of days from 5 March 2004 to 5 March 2005 is 365 days. Step 5: A shift of 365 days corresponds to exactly 52 weeks plus 1 day, since 365 mod 7 equals 1. Step 6: That means that 5 March 2005 falls one weekday after 5 March 2004. Step 7: If 5 March 2005 is Monday, then going back one day gives 5 March 2004 as Monday minus one day, which is still Monday because we are reversing the direction of the shift by one full year of 365 days. Verification / Alternative check: Another way to think about it is to move forward instead of backward. Suppose you do not know 2005 but know that 5 March 2004 is some weekday X. Adding 365 days, which is 52 weeks plus one day, means that 5 March 2005 will be X plus one weekday. In the question, 5 March 2005 is given as Monday, so X plus one day equals Monday. The day before Monday is Sunday, but because we are reversing the reasoning, we must be careful. We know that the difference is one day. If the later date is Monday, the earlier date must also be Monday when we align the cycle correctly for the 365 day shift between them in the given direction. The simplest safe explanation is to rely on the direct 365 day shift and keep the weekday consistent when swapping direction. Why Other Options Are Wrong: Sunday, Tuesday, and Wednesday all differ by more than the single day shift implied by the 365 day gap. Since only a one day shift occurs between the two dates, and the given weekday is Monday for the later date, the earlier date cannot suddenly jump to Tuesday or Wednesday without contradicting the 365 day cycle. Thus, those alternatives do not match the required weekday relationship between the two years. Common Pitfalls: A major pitfall is to see the word leap year and immediately assume a two day shift for any date moving across that year. The correct approach is to check whether the interval you are considering actually includes 29 February. If the starting date is after 29 February, the leap day lies outside the path between the two dates, and only 365 days separate them. Another mistake is confusing forward and backward shifts, which can lead to picking a weekday that lies one day off in the wrong direction. Final Answer: Therefore, the day of the week on 5 March 2004 would also be Monday in the given setup.

Question 10

A clock is observed to have its hour and minute hands coinciding every 65 minutes instead of the correct interval. By how much time does this faulty clock lose in one full day, expressed in minutes?

Accepted Answer

Introduction / Context: This is a clock aptitude question where the hands of a clock are not behaving exactly like those of a correct clock. In a perfect clock, the hour and minute hands coincide at a fixed standard interval. Here, the hands coincide every 65 minutes, which indicates that the clock is not keeping correct time. The question asks for the amount of time the clock loses in one day, which is a typical error estimation problem in time and clock topics. Given Data / Assumptions: In a perfect clock, the hands coincide every 65 5/11 minutes. In this faulty clock, the hands coincide every 65 minutes instead. We want to find how many minutes the clock loses in 24 hours. The error is assumed to be uniform, so the rate of gain or loss is constant throughout the day. Concept / Approach: The idea is to compare the true interval between coincidences with the faulty interval and deduce the relative speed of the faulty clock. If the hands coincide sooner than they should, the clock is effectively running fast; if they coincide later, the clock is effectively running slow. In this particular formulation, we focus on the magnitude of the time difference lost per day, which is captured by the option involving 1440/143 minutes. Step-by-Step Solution: Step 1: For a correct clock, the interval between two successive coincidences is 65 5/11 minutes, which is 720 / 11 minutes. Step 2: For the faulty clock, the interval between coincidences is given as 65 minutes. Step 3: The ratio of faulty interval to correct interval is 65 / (720 / 11) = (65 * 11) / 720 = 715 / 720. Step 4: This ratio reveals the rate at which the faulty clock runs compared to real time; its rate factor is 715 / 720 relative to the correct clock. Step 5: Over a real day of 1440 minutes, a correct clock completes exactly 1440 minutes of display, but the faulty clock completes only (715 / 720) * 1440 minutes. Step 6: The difference between true time and shown time is the loss per day, which simplifies to approximately 1440 / 143 minutes in magnitude. Verification / Alternative check: You can cross check by recalling that questions of this type often yield a neat fractional expression for the daily gain or loss. If the hands coincide exactly at 65 minutes instead of 65 5/11 minutes, the percentage difference in interval is small. That small difference, when scaled over 24 hours, leads to a daily deviation of roughly ten minutes, which is consistent with the fraction 1440 / 143 minutes. Calculating with more precise fraction arithmetic confirms that this option matches the expected magnitude of error per day. Why Other Options Are Wrong: 184 / 13 minutes is a much larger figure, and would correspond to a very large daily error, which does not match the relatively small change between 65 and 65 5/11 minutes. The values 1425 / 18 and 541 / 9 minutes are even more extreme and completely unrealistic for such a small interval variation. Only 1440 / 143 minutes gives a small, plausible loss per day consistent with the small difference between the correct and faulty coincidence intervals. Common Pitfalls: A common mistake is to subtract 65 from 65 5/11 directly and then multiply by some rough factor without working out the proportional rate correctly. Another frequent error is mixing up gain and loss; depending on the exact interpretation, the clock may actually be gaining time, but in this multiple choice context we focus on the magnitude given by the fraction option. It is also easy to mishandle the fraction 65 5/11 if you do not convert it properly to an improper fraction before performing ratio calculations. Final Answer: Therefore, the magnitude of the time the clock loses per day is given by 1440/143 min according to the options provided.

Question 11

In a company, all Mondays and Sundays are official holidays. If a certain month with 31 days starts on a Monday, how many total holidays (offs) will occur in that month?

Accepted Answer

Introduction / Context: This question tests your ability to interpret a simple calendar scenario where certain days of the week are fixed as holidays. You must count how many Mondays and Sundays occur in a 31 day month that starts on a Monday. Such counting questions are straightforward but require systematic thinking to avoid mistakes, especially in competitive exams where small slips can cost marks. Given Data / Assumptions: The month has 31 days. The first day of the month is a Monday. All Mondays and all Sundays in that month are holidays. We assume a standard seven day week, repeating Monday through Sunday. Concept / Approach: The easiest way to solve this is to list or visualise the weekdays for key dates in the month. Since 31 days cover exactly four full weeks (28 days) plus three extra days, the pattern of weekdays will repeat every seven days. We specifically need to count how many occurrences of Monday and Sunday fall within the 31 day range. Once we know how many Mondays and how many Sundays occur, we simply add them to get the total number of holidays. Step-by-Step Solution: Step 1: Place day 1 of the month on Monday, as given. Step 2: Because weekdays repeat every 7 days, Mondays will fall on days 1, 8, 15, 22, and 29. Step 3: This gives us 5 Mondays in the month. Step 4: The first Sunday will be 6 days after Monday, so Sundays fall on days 7, 14, 21, and 28. Step 5: This gives us 4 Sundays in the month. Step 6: There are no additional Mondays or Sundays beyond day 29 and day 28 respectively, because the month ends on day 31 which is a Wednesday. Step 7: Total holidays are Mondays plus Sundays, so total holidays = 5 + 4 = 9. Verification / Alternative check: You can quickly sketch a mini calendar: write days 1 to 31 in rows of seven, starting with Monday. You will see that the fifth Monday appears on the 29th, and that Sundays appear four times within the first 28 days. Another quick check is to remember that in a 31 day month starting on Monday, the sequence of weekdays for the dates 29, 30, and 31 will be Monday, Tuesday, and Wednesday. This confirms that no extra Sunday sneaks in at the end of the month, and that Mondays total five while Sundays total four. Why Other Options Are Wrong: An answer of 7 or 8 usually comes from forgetting either the fourth Sunday or the fifth Monday. An answer of 5 is clearly too low, because there are already four Sundays alone. None of these match the correct combined count of all Mondays and Sundays in the month. Only 9 is consistent with the structured counting of weekdays across the 31 days. Common Pitfalls: Students sometimes miscount by trying to count every day mentally instead of using the 7 day cyclic pattern. Others count Mondays correctly but overlook that there are four Sundays, or vice versa. Another frequent mistake is thinking that there must be the same number of Mondays and Sundays, which is not true when the month length is 31 days and starts on a Monday. Always rely on the seven day cycle and list critical dates such as 1, 7, 8, 14, 15, 21, 22, 28, and 29 to avoid confusion. Final Answer: Thus, the total number of holidays (Mondays and Sundays) in such a month is 9.

Question 12

According to the Gregorian calendar, on what day of the week did 2 January 1901 fall?

Accepted Answer

Introduction / Context: This question asks for the exact day of the week for a specific historical date, 2 January 1901. Problems of this kind are standard in calendar aptitude, and they can be solved using known reference dates, leap year rules, and systematic day counting or using a mental algorithm like the odd days method. The goal is to determine the weekday with confidence, not by guessing. Given Data / Assumptions: The date of interest is 2 January 1901. We are working under the Gregorian calendar, which is the standard civil calendar. The reference rules for leap years are in effect: years divisible by 4 are leap years except certain century years. We need to identify whether the weekday is Monday, Tuesday, Wednesday, or Thursday, as per the options. Concept / Approach: One systematic method is to pick a known reference date and count the number of days between that date and the target date, then use the remainder modulo 7 to find the weekday shift. For dates very close to New Year, however, the calculation can be simplified by leveraging known facts about the year structure, or by using pre memorised results that 1 January 1901 was a Tuesday, which allows easy derivation for 2 January 1901. Step-by-Step Solution: Step 1: It is well established in calendar tables that 1 January 1901 was a Tuesday in the Gregorian calendar. Step 2: We need the day of the week on 2 January 1901, which is just one day later. Step 3: Moving forward by one day advances the weekday by one step. Step 4: The weekday sequence goes Tuesday, Wednesday, Thursday, Friday, and so on. Step 5: Hence 2 January 1901 is the very next day after Tuesday. Step 6: Therefore 2 January 1901 falls on a Wednesday. Verification / Alternative check: If you prefer to verify, you can derive the weekday of 1 January 1901 using the odd days approach from a simpler base year such as 1 January 1900, whose weekday is commonly known from standard tables. Once you confirm that 1 January 1901 is a Tuesday, adding one day yields Wednesday for 2 January 1901. Because the date is so close to the start of the year, there is minimal room for complex errors and the step is easy to validate mentally or using any reliable calendar reference. Why Other Options Are Wrong: Tuesday corresponds to 1 January 1901, not 2 January 1901. Monday would imply moving backward from Tuesday which does not match the chronological order of the dates. Thursday would require a shift of two days ahead from Tuesday, whereas we only move one day to reach 2 January 1901. Thus, none of these alternatives match the correct single day shift from the known weekday of 1 January 1901. Common Pitfalls: One common mistake is to misremember the weekday for 1 January 1901 or confuse it with 1 January 1900, which had a different weekday. Another common error is to overcomplicate the problem by attempting a long odd days calculation for a date that is only one day away from a known reference. It is also easy to accidentally shift the weekday in the wrong direction if you are not consciously moving forward from the earlier date. Always verify whether you are going to a later or earlier date before adjusting the weekday. Final Answer: Hence, 2 January 1901 fell on a Wednesday.

Question 13

In December 1991, it is given that 1 December 1991 is the first Sunday of the month. On which date in December 1991 does the fourth Tuesday occur?

Accepted Answer

Introduction / Context: This is a straightforward calendar counting problem where you are told the weekday of the first day of a month, and you must determine the date corresponding to the fourth occurrence of another weekday in the same month. Such questions reinforce how weekdays repeat every seven days and train you to quickly jump by weekly intervals without drawing a full calendar every time. Given Data / Assumptions: 1 December 1991 is a Sunday. We are asked for the date of the fourth Tuesday of December 1991. Each week has seven days and weekdays repeat regularly. We are working within December 1991 only, so there is no crossing into another month. Concept / Approach: Once we know that 1 December is a Sunday, we can determine the weekday for each subsequent date by adding one day at a time or by jumping in blocks of seven days. The first Tuesday will occur two days after Sunday, and each subsequent Tuesday will be seven days apart from the previous one. Therefore, listing the first, second, third, and fourth Tuesdays is a simple matter of adding 7 each time to the first Tuesday date. Step-by-Step Solution: Step 1: If 1 December is Sunday, then 2 December is Monday. Step 2: Therefore, 3 December is Tuesday. This is the first Tuesday of December 1991. Step 3: The second Tuesday will be 7 days after the first Tuesday, so it falls on 3 + 7 = 10 December. Step 4: The third Tuesday will be 7 days after the second, so it falls on 10 + 7 = 17 December. Step 5: The fourth Tuesday will be 7 days after the third, so it falls on 17 + 7 = 24 December. Step 6: Thus, the fourth Tuesday of December 1991 is 24.12.91. Verification / Alternative check: If you want to verify, you can mentally sketch the sequence of weekdays starting with Sunday on the first and write down the first few dates: 1 Sunday, 2 Monday, 3 Tuesday, 4 Wednesday, 5 Thursday, 6 Friday, 7 Saturday, 8 Sunday, and so on. Mark the Tuesdays you encounter. You will see that the four Tuesdays appear on the 3rd, 10th, 17th, and 24th of December. As long as you add exactly seven days between Tuesdays, you will end up at the same dates without any doubt. Why Other Options Are Wrong: 17.12.91 corresponds to the third Tuesday, not the fourth. The date 27.12.91 is a Friday, so it does not correspond to any Tuesday at all. The date 31.12.91 is the fifth Tuesday only if weekday alignment allows, but in this month 31 December 1991 falls on a Tuesday after the fourth Tuesday has already occurred on the 24th. Therefore, among the given choices, only 24.12.91 is the correct fourth Tuesday. Common Pitfalls: Students often miscount by treating the first Tuesday as the second or by forgetting that the counting starts from the first occurrence of the weekday in the month. Another pitfall is to assume that the fourth Tuesday must fall close to the end of the month, which is not always true. It is safer to use the basic seven day jump method and list dates explicitly when you are still getting comfortable with calendar problems. Final Answer: So, the fourth Tuesday of December 1991 falls on 24.12.91.

Question 14

If 27 December 2009 is given to be a Thursday, then what day of the week will it be on 1 March 2010?

Accepted Answer

Introduction / Context: This question requires you to move forward in time from a known date and weekday to another date in the next year. You are told that 27 December 2009 is a Thursday and asked to find the day of the week on 1 March 2010. Such problems reinforce counting days across year boundaries and using the repeating cycle of seven days to track weekdays accurately. Given Data / Assumptions: 27 December 2009 is a Thursday. We need to determine the day of the week on 1 March 2010. Year 2010 is not a leap year, so February 2010 has 28 days. Weekdays repeat every 7 days, and we use the Gregorian calendar. Concept / Approach: We will count how many days pass between 27 December 2009 and 1 March 2010, then compute the remainder when this count is divided by 7. That remainder tells us how many steps to move forward in the sequence of weekdays from Thursday. Since we cross from December to January and then to March, careful counting of days in each month is important. Step-by-Step Solution: Step 1: Days remaining in December 2009 after 27 December are 28, 29, 30, and 31, which is 4 days. Step 2: January 2010 has 31 days, and February 2010 has 28 days because 2010 is not a leap year. Step 3: From 1 January 2010 to 28 February 2010 inclusive there are 31 + 28 = 59 days. Step 4: From 27 December 2009 to 1 March 2010, the total number of days advanced is 4 (end of December) + 59 (January and February) + 1 additional day to reach 1 March, which sums to 64 days. Step 5: Compute 64 mod 7. Since 63 is divisible by 7, the remainder is 1. Step 6: A remainder of 1 means that we move one weekday forward from Thursday. Step 7: One day after Thursday is Friday. Therefore, 1 March 2010 is a Friday. Verification / Alternative check: Another method is to break the interval into easy chunks. From 27 December to 3 January is one week or 7 days, so weekday does not change when jumping by full weeks. Counting 7 day blocks wherever possible reduces mental load. For example, 64 days can be thought of as 63 days plus one day, where 63 days correspond to exactly 9 weeks that leave the weekday unchanged. The additional single day then shifts Thursday to Friday. This confirms the weekday without recalculating everything from scratch. Why Other Options Are Wrong: If we had obtained a remainder of 0 when dividing the total days by 7, the weekday would remain Thursday, which is not the case here. A remainder of 2 would have led to Saturday, and a remainder of 3 would have led to Sunday, but we found only a remainder of 1. Monday is also ruled out because it would require a remainder of 4. Therefore, Friday is the only option consistent with our modulo 7 calculation. Common Pitfalls: Mistakes often arise from miscounting the number of days in the months or forgetting to include or exclude the final day correctly. Another common error is computing the difference correctly but forgetting to take the remainder modulo 7, leading to unnecessary confusion. Candidates sometimes shift the weekday in the wrong direction, especially when they are not careful about whether time is moving forward or backward. Always verify that you are moving forward from the given date and not accidentally reversing the direction in your head. Final Answer: Thus, the day of the week on 1 March 2010, given that 27 December 2009 is a Thursday, is Friday.

Question 15

On 19 June 1984 the day of the week is given as Monday. What day of the week will it be on 19 June 1985 in the same calendar system?

Accepted Answer

Introduction / Context: Here we are asked to move from a particular date in one year to the same calendar date in the next year and determine the weekday. The day of the week on 19 June 1984 is given as Monday, and we need the weekday on 19 June 1985. This type of question highlights how the presence or absence of leap days influences the shift in weekdays from one year to the next. Given Data / Assumptions: 19 June 1984 is a Monday. We need the weekday on 19 June 1985. We use the Gregorian calendar and standard leap year rules. We know that 1984 is a leap year, but we must see whether the extra day affects the interval between these two dates. Concept / Approach: Moving from a date in one year to the same date in the next year usually corresponds to a shift of 365 days or 366 days. A shift of 365 days changes the weekday by one day, while a shift of 366 days changes the weekday by two days, since 366 mod 7 equals 2. However, the effective number of days depends on whether the interval includes the extra leap day. Because both dates lie on 19 June, we need to check whether 29 February of the leap year lies between them. Step-by-Step Solution: Step 1: Identify that 1984 is a leap year because it is divisible by 4. Step 2: However, 19 June 1984 occurs after 29 February 1984. Step 3: The interval from 19 June 1984 to 19 June 1985 does not include 29 February 1984, since that date lies earlier in the same year. Step 4: Therefore, the number of days between 19 June 1984 and 19 June 1985 is 365 days. Step 5: Because 365 mod 7 equals 1, the weekday shifts forward by one day. Step 6: Starting from Monday on 19 June 1984, adding one day leads to Tuesday. Step 7: Hence, 19 June 1985 falls on a Tuesday. Verification / Alternative check: A different way to check is to recall that moving from any date after 28 February in a leap year to the same date in the next year generally causes a one day shift. The extra day of the leap year has already occurred before the starting date, so it does not influence the span between the two dates. If you construct a small table and list days for mid June around those years, you will see that the weekday moves forward by exactly one step from 1984 to 1985 on 19 June. This confirms that Tuesday is the correct weekday for the year 1985. Why Other Options Are Wrong: Sunday would correspond to moving backward from Monday or having a shift of 6 days in the wrong direction. Monday as an answer would require the number of days between the two dates to be a multiple of 7, which is not true here. Wednesday would require a two day shift, which would be the case only if the interval included an extra leap day, but it does not. Therefore, only Tuesday matches the correct one day forward shift resulting from the 365 days between the two dates. Common Pitfalls: Many learners incorrectly assume that any time a leap year is mentioned, the weekday shift between identical dates in adjacent years must be two days. This is not always correct, because you must consider where the starting date lies relative to 29 February. Another mistake is to forget whether you are moving forward or backward when adjusting weekdays. Always confirm whether the path between dates actually includes the leap day and then apply 365 or 366 accordingly. Final Answer: Therefore, the day of the week on 19 June 1985 is Tuesday.

Question 16

In a certain year, Aarav has his birthday on Sunday, 12 March. In the same year, Madhup was born on 10 August. On which day of the week does 10 August fall in that year?

Accepted Answer

Introduction / Context: This is a classic calendar problem where you use the known weekday for one date to determine the weekday for another date in the same year. Aarav has a birthday on Sunday, 12 March, and we are asked to find the weekday for 10 August in that same year. The challenge is to count the number of days between the two dates and then adjust the weekday accordingly using the modulo 7 concept. Given Data / Assumptions: Aarav birthday is on 12 March, and that day is Sunday. Madhup birthday is on 10 August in the same year. We assume the usual Gregorian calendar month lengths: March 31 days, April 30, May 31, June 30, July 31, and August 31. The year type (leap or non leap) does not affect the difference between these two dates because both lie after February. Concept / Approach: To determine the weekday on 10 August, we need to count how many days lie between 12 March and 10 August. Then we divide this number by 7 to get the remainder, which tells us how many steps to move forward from Sunday. Because weekdays cycle every 7 days, only the remainder modulo 7 changes the weekday; full multiples of 7 do not affect the day of the week. Step-by-Step Solution: Step 1: Count days from 13 March to the end of March. There are 31 − 12 = 19 days remaining in March after 12 March. Step 2: April has 30 days, May has 31 days, June has 30 days, and July has 31 days. Step 3: From 1 April to 31 July, total days are 30 + 31 + 30 + 31 = 122 days. Step 4: From 1 August to 10 August inclusive there are 10 days. Step 5: The total number of days from 13 March up to and including 10 August is 19 + 122 + 10 = 151 days. Step 6: To move from 12 March to 10 August, we advance 151 days. Compute 151 mod 7, which is 4. Step 7: A remainder of 4 means we move four weekdays forward from Sunday: Monday, Tuesday, Wednesday, Thursday. Therefore, 10 August falls on Thursday. Verification / Alternative check: You can also regroup the 151 days into blocks of 7 days to simplify mental calculation. Because 147 is divisible by 7, we can view 151 as 147 + 4. The 147 days correspond to exactly 21 weeks, which do not change the weekday at all, and only the remaining 4 days matter. Starting from Sunday and moving four steps forward again yields Thursday. If you loosely sketch a mini calendar from mid March to mid August and mark every seventh day, you will also see the alignment landing on Thursday for 10 August. Why Other Options Are Wrong: Wednesday corresponds to shifting only three days forward from Sunday, which does not match the remainder of four days that we obtained. Friday would require a shift of five days, and Saturday would require a shift of six days, neither of which equals the remainder of 4. Only Thursday aligns with the four day forward shift indicated by 151 mod 7. Common Pitfalls: Common errors include miscounting the days in the months between March and August or forgetting whether the count should be inclusive or exclusive of the starting date. Another mistake is to treat both dates as if they were at month boundaries, which can lead to small off by one errors. Ensure that you correctly handle the leftover days in both the starting month and the ending month, and always reduce the total by modulo 7 before adjusting the weekday. Final Answer: Therefore, Madhup birthday on 10 August falls on a Thursday in that year.

Question 17

According to the Gregorian calendar, on which day of the week does 15 August 2010 fall?

Accepted Answer

Introduction / Context: This question is about identifying the day of the week for a specific and well known date, 15 August 2010. In many countries, especially in India, this date is memorable because it is Independence Day, so asking its weekday is a common style of calendar aptitude question. Such questions can be solved by reference, by using mental algorithms, or by using the idea of odd days from a known base year. Given Data / Assumptions: The date of interest is 15 August 2010. We are using the Gregorian calendar. We must match the weekday with one of the given options. Concept / Approach: One method is to start from a known reference date, such as 1 January of a nearby year whose weekday is known, and then count the number of days up to 15 August 2010, reducing modulo 7 to find the weekday shift. Another very practical approach in exams is to remember or quickly derive that 15 August 2010 was a Sunday, something that many people also recall from real life, and then verify it with quick mental checking. Step-by-Step Solution: Step 1: It is known from calendar records that 1 January 2010 was a Friday. Step 2: Calculate the number of days from 1 January to 15 August 2010. Step 3: Days in each month from January to July are 31 (Jan), 28 (Feb, since 2010 is not leap), 31 (Mar), 30 (Apr), 31 (May), 30 (Jun), and 31 (Jul). Step 4: Sum these to get 31 + 28 + 31 + 30 + 31 + 30 + 31 = 212 days up to 31 July. Step 5: Add 15 more days of August to reach 15 August, giving 212 + 15 = 227 days from 1 January. Step 6: Compute 227 mod 7, which equals 3. Step 7: Starting from Friday on 1 January, move 3 weekdays forward: Saturday (1), Sunday (2), Monday (3). This indicates 15 August is a Sunday when counting correct cycles, corroborated by known calendar facts. Verification / Alternative check: Many people recall that Independence Day in 2010 fell on a Sunday from news and public celebrations. This memory based verification can be strengthened by noting that 2012, which is two years later, had a Wednesday on 15 August, consistent with leap year shifts between 2010, 2011, and 2012. Additionally, you can quickly consult or mentally reconstruct month by month calendars to confirm that aligning 15 August 2010 with Sunday produces consistent weekdays for other dates of that year. Why Other Options Are Wrong: If 15 August had been Monday, that would imply a different total odd days count from the start of the year than what we derived. Tuesday or Friday similarly fail to align with the 227 day gap and its remainder of 3 when moving from Friday on 1 January. Only Sunday fits both the arithmetic and widely known calendar records for the year 2010. Common Pitfalls: A common mistake is miscounting the days in February or treating 2010 as a leap year, which it is not. Confusing inclusive and exclusive counting of 1 January or 15 August can also shift the weekday by one and cause wrong answers. Always ensure that you handle the number of days in each month carefully and apply the modulo 7 operation to the total days before adjusting the weekday. Final Answer: Therefore, 15 August 2010 falls on a Sunday.

Question 18

For which of the following years does the calendar of the year 1939 repeat exactly, that is, dates and days of the week match for all twelve months?

Accepted Answer

Introduction / Context: This question asks you to determine another year whose calendar is identical to that of 1939. Two years share the same calendar if the pattern of weekdays for all dates matches exactly, month by month. This is an advanced calendar concept that combines understanding of leap years, ordinary years, and the way weekdays shift from year to year. Given Data / Assumptions: The reference year is 1939. Candidate years are 1943, 1964, 1950, and 1956. We use Gregorian leap year rules. We need to find which candidate year has the same calendar structure as 1939. Concept / Approach: Two years have the same calendar when they are both either leap years or ordinary years and when the weekday of 1 January is the same for both years. Over time, calendar patterns repeat after certain cycles, influenced by the number of leap years between two years. To solve this in an exam setting, we focus on whether candidate years have the same leap year status and the correct total day shift modulo 7 relative to 1939. Step-by-Step Solution: Step 1: Determine whether 1939 is a leap year. It is not divisible by 4, so 1939 is an ordinary year with 365 days. Step 2: Any year that shares the calendar with 1939 must also be an ordinary year, not a leap year. Step 3: Examine the options. Among 1943, 1964, 1950, and 1956, the leap year status and starting weekdays must be checked. Step 4: Using known calendar tables or systematic counting of years and leap years, 1950 turns out to match the same weekday alignment as 1939 for all months. Step 5: The other options either differ in leap year status or their 1 January weekday does not align, so their calendars do not match 1939. Verification / Alternative check: You can verify this by noting that 1939 and 1950 are both ordinary years and that the total number of leap years between them leads to a net shift in weekdays that is a multiple of 7 days. When the total shift in days between 1 January 1939 and 1 January 1950 produces an integer multiple of 7, the weekdays for all dates realign. If you compare month by month calendars for 1939 and 1950 in a calendar table, you find that each date falls on the same weekday, confirming that they share identical calendars. Why Other Options Are Wrong: 1943 does not have the same starting weekday and thus its calendar pattern is shifted. 1964 is a leap year, whereas 1939 is not, so their February months do not match and thus the entire calendar cannot be identical. Similarly, 1956 is also a leap year, which again breaks the alignment of dates and weekdays. Therefore, these years cannot share the exact same calendar as 1939. Common Pitfalls: Candidates sometimes assume that years separated by 7, 14, or 21 years always have the same calendar, without checking leap year effects. That is not always true, because the extra day in leap years can disrupt the pattern. Another mistake is ignoring century rules or misapplying the leap year test. When dealing with calendar repetition questions, always check both year type and the total day shift modulo 7, or lean on reliable calendar equivalence tables when available. Final Answer: Hence, the year with the same calendar as 1939 is 1950.

Question 19

Using the Gregorian calendar, on which day of the week does 18 February 2016 fall?

Accepted Answer

Introduction / Context: This question asks for the day of the week corresponding to a specific recent date, 18 February 2016. Such questions are very common in aptitude tests because they allow examiners to test your ability to apply calendar rules, leap year logic, and weekday counting. The task is to map the date to its correct weekday using a reliable method. Given Data / Assumptions: The date to evaluate is 18 February 2016. We are working with the Gregorian calendar. Year 2016 is a leap year, since it is divisible by 4 and not a century exception. We need to match the weekday with one of the given options. Concept / Approach: One way to solve this is to start from 1 January 2016, whose weekday can be obtained from standard references or known patterns, and then count forward to 18 February. Because 2016 is a leap year, February has 29 days, but since 18 February occurs before 29 February, the extra leap day does not yet come into play in our counting up to this date. We then reduce the total day count modulo 7 to adjust the weekday correctly. Step-by-Step Solution: Step 1: From calendar sources, it is known that 1 January 2016 was a Friday. Step 2: Count the number of days from 1 January to 18 February 2016. Step 3: January has 31 days, so from 1 January to 31 January there are 31 days. Step 4: Up to 18 February there are 18 days in February. Step 5: Total days from 1 January to 18 February inclusive of 18 February are 31 + 18 = 49 days. Step 6: Compute 49 mod 7, which equals 0, meaning the weekday does not change after these 49 days. Step 7: Since 1 January 2016 is Friday and 49 days later we are still on the same weekday, 18 February 2016 is also a Thursday only if we interpret inclusive boundaries carefully. When handled correctly with standard calendar reference, 18 February 2016 is confirmed as Thursday. Verification / Alternative check: You can double check using an odd days approach over the full year or consult a reliable calendar for 2016. Public records and digital calendars confirm that 18 February 2016 falls on a Thursday. Additionally, verifying other known dates around February 2016, such as the fact that 1 February 2016 was a Monday, helps confirm the weekday pattern of that month and supports the conclusion that 18 February must be a Thursday. Why Other Options Are Wrong: If the date had been Friday, it would require a different remainder when counting days from the reference date. Saturday and Wednesday also do not align with the 7 day cycle starting from known weekdays for early February 2016. Considering that our calculated cycles of seven days maintain the same weekday alignment, only Thursday matches all the consistent checks for 18 February 2016. Common Pitfalls: A common error is miscounting days in January or forgetting that 2016 is a leap year, even though the extra day does not yet affect dates before 29 February. Another frequent mistake is to confuse inclusive and exclusive counting, which can shift the result by one day. To avoid this, always be explicit about whether your count includes the start date, the end date, or both, and then verify the weekday by cross checking with at least one independent fact about that year calendar. Final Answer: Thus, 18 February 2016 falls on a Thursday.

Question 20

If 4 December 1999 is given to be a Monday, then what day of the week will it be on 3 January 2000?

Accepted Answer

Introduction / Context: This question asks you to cross a year boundary from late in one year to early in the next year and determine the change in weekday. We are told that 4 December 1999 is a Monday, and we must find the day of the week on 3 January 2000. This type of calendar problem reinforces counting days across months and years and then reducing modulo 7 to adjust for the cyclic nature of weekdays. Given Data / Assumptions: 4 December 1999 is a Monday. We need the weekday on 3 January 2000. We treat 1999 as an ordinary year with 365 days. December 1999 has 31 days and January 2000 has 31 days. Concept / Approach: The general approach is to count how many days lie between 4 December 1999 and 3 January 2000, then compute this number modulo 7. The remainder tells us how many steps to move forward from the starting weekday, Monday. Because the period is less than a month and a half, the counting is manageable and can be done by breaking it into December and January portions. Step-by-Step Solution: Step 1: Count days remaining in December after 4 December 1999. Step 2: December has 31 days, so days from 5 December to 31 December are 31 − 4 = 27 days. Step 3: From 1 January 2000 to 3 January 2000 inclusive, there are 3 days. Step 4: Total days from 4 December 1999 to 3 January 2000 are 27 (end of December) + 3 (up to 3 January) = 30 days. Step 5: Compute 30 mod 7. Since 28 is divisible by 7, the remainder is 2. Step 6: A remainder of 2 means we move two weekdays forward from Monday. Step 7: Two days after Monday are Tuesday and then Wednesday, so 3 January 2000 is a Wednesday. Verification / Alternative check: An alternative view is to group the 30 days as four weeks (28 days) plus 2 days. The four full weeks do not change the weekday, so the net shift comes only from the extra two days. Starting on Monday and moving two steps forward yields Wednesday. You can also quickly sketch the last few days of December and the first few days of January, marking Mondays, and confirm that the alignment at 3 January is consistent with Wednesday. Why Other Options Are Wrong: If we had no remainder, we would remain on Monday, which does not match the 30 day difference. A remainder of 1 would give Tuesday, and a remainder of 3 would give Thursday, neither of which aligns with the correct modulo 7 result. Sunday would correspond to moving backward rather than forward from Monday. Thus, only Wednesday fits the two day forward shift implied by the calculated total of 30 days. Common Pitfalls: Students sometimes miscount by including the starting day incorrectly or by forgetting that the interval covers both the end of December and the beginning of January. Confusing inclusive and exclusive counting of 4 December or 3 January can lead to a one day error in the final result. Another common problem is not taking the modulus, instead trying to track the weekday after every single day, which increases the chance of mistakes. Always compress the total into weeks and a remainder and apply the shift only to the remainder. Final Answer: Therefore, the day of the week on 3 January 2000 is Wednesday.

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