Question 1

Today is given as Sunday, 20th January 2017. Exactly 90 days from today, what will be the calendar date and day of the week?

Accepted Answer

Introduction / Context: This question combines counting days across months with basic weekday arithmetic. You are told that today is Sunday, 20 January 2017, and asked what the date and day of the week will be exactly 90 days later. Even if the given weekday for 20 January 2017 does not match the real-world calendar, you must treat the information in the question as correct and work consistently from it. Given Data / Assumptions: Today: 20 January 2017 (assumed to be a Sunday for the purpose of this problem). We must move forward 90 days. Year 2017 is a common year (February has 28 days). Month lengths: January 31, February 28, March 31, April 30. We calculate both the resulting date and the weekday, given the assumed starting weekday. Concept / Approach: First, find the calendar date that is 90 days after 20 January by counting days month by month. Then, we determine the weekday by taking 90 modulo 7, because the weekday pattern repeats every 7 days. A positive remainder tells us how many days ahead the weekday will be relative to Sunday. Step-by-Step Solution: Step 1: Count days remaining in January after 20 January. January has 31 days, so from 21 January to 31 January there are 11 days. We need a total of 90 days, so remaining after January = 90 - 11 = 79 days. Step 2: Subtract full months. February (28 days): remaining = 79 - 28 = 51. March (31 days): remaining = 51 - 31 = 20. So, 20 days into April after 31 March gives 20 April. Step 3: Determine the weekday after 90 days. 90 divided by 7 gives 12 full weeks and a remainder of 6 days (since 90 mod 7 = 6). Starting from Sunday, 6 days ahead: Monday (1), Tuesday (2), Wednesday (3), Thursday (4), Friday (5), Saturday (6). Therefore, 90 days after a Sunday is a Saturday. Verification / Alternative check: You can verify by using the ordinal day-of-year method: convert 20 January to its day-of-year number, add 90, and see that it lands on 20 April. The modulo 7 approach for weekdays gives a 6-day shift, which always moves Sunday to Saturday. This matches the option “20th April, Saturday”. Why Other Options Are Wrong: 18th April and 19th April are 2 and 1 days earlier than the correct date, and 21st April is 1 day later, so they do not match the counted 90 days. Even if some of them have a Saturday or Sunday attached, the date plus weekday combination must both be correct, and only 20th April, Saturday satisfies that. Common Pitfalls: Some learners count the starting date (20 January) as day 1 instead of counting forward from the next day, which leads to an off-by-one error. Others mis-handle the month lengths and make a mistake when February or March is passed. Always subtract month by month carefully, and then use modulo 7 for the weekday shift. Final Answer: Exactly 90 days from Sunday, 20 January 2017, the date and day will be 20th April, Saturday.

Question 2

According to the leap-year rules of the Gregorian calendar, which of the following years has 366 days?

Accepted Answer

Introduction / Context: Leap-year questions test your understanding of the rules that determine which years have 366 days (a leap year) rather than the usual 365 days. In the Gregorian calendar, the leap-year rule for centuries is slightly different from the simple “divisible by 4” rule used for non-century years. Given Data / Assumptions: We are given the years: 1900, 1200, 2500, and 1700. We must determine which of these has 366 days (i.e., is a leap year). Gregorian leap-year rules: - Every year divisible by 4 is a leap year, - But if the year is divisible by 100, it is not a leap year, - Unless it is also divisible by 400, in which case it is a leap year. Concept / Approach: We apply the leap-year rule step by step to each option. Since all given years are century years (ending in 00), we pay special attention to the “divisible by 400” condition. Only century years divisible by 400 are leap years with 366 days. Step-by-Step Solution: Step 1: Examine 1900. 1900 is divisible by 100, but 1900 / 400 = 4.75, not an integer. Therefore, 1900 is not divisible by 400 and is not a leap year. Step 2: Examine 1200. 1200 is divisible by 100, and 1200 / 400 = 3, an integer. So 1200 is divisible by 400 and is a leap year with 366 days. Step 3: Examine 2500. 2500 is divisible by 100, but 2500 / 400 = 6.25, not an integer. Therefore, 2500 is not a leap year. Step 4: Examine 1700. 1700 is divisible by 100, but 1700 / 400 = 4.25, not an integer. Therefore, 1700 is not a leap year. Verification / Alternative check: You can summarise the rule: a century year is a leap year only if it is divisible by 400. Testing each option quickly with this rule immediately identifies 1200 as the only year satisfying that condition. All others fail the divisibility-by-400 test and hence must have only 365 days. Why Other Options Are Wrong: 1900, 2500, and 1700 are all divisible by 100 but not by 400, so they do not qualify as leap years under the Gregorian rules. Treating them as leap years would incorrectly assign them 366 days, which contradicts the well-established calendar system. Common Pitfalls: Many learners remember only the “divisible by 4” rule and forget the additional conditions for century years. This leads to incorrectly marking years like 1900 as leap years. Always remember the full rule: years divisible by 100 require an extra check for divisibility by 400 before being called leap years. Final Answer: Among the given options, the year that has 366 days is 1200.

Question 3

The calendar for the year 1989 is exactly the same (same weekdays on the same dates) as the calendar for which of the following previous years?

Accepted Answer

Introduction / Context: Calendar repetition questions ask you to find another year whose calendar matches exactly with a given year. For two years to have the same calendar, the years must both be leap or both be common (non-leap) years, and the total shift in weekdays between them must be a multiple of 7 days so that all dates fall on the same weekdays. Given Data / Assumptions: Reference year: 1989 (a non-leap year). Candidate years: 1978, 1970, 1980, 1985. We use the Gregorian leap-year rules: - Years divisible by 4 are leap years, except century years not divisible by 400. We need to compare weekday shifts from the candidate years up to 1989. Concept / Approach: We calculate the total day shift from each candidate year to 1989. Each common year between them adds a shift of +1 day to the next New Year’s Day, and each leap year adds +2 days. If the total shift from a candidate year to 1989 is a multiple of 7 and both years are common years, then their calendars match. Step-by-Step Solution: Step 1: Note that 1989 is not divisible by 4, so it is a common year. Step 2: Check 1978. Years between 1978 and 1989: 1978 to 1988 inclusive (11 years). Leap years in this range: 1980, 1984, 1988 (3 leap years). Common years = 11 - 3 = 8. Total shift = 8 * 1 + 3 * 2 = 8 + 6 = 14 days. 14 mod 7 = 0, so 1 January 1978 and 1 January 1989 fall on the same weekday. Also, 1978 is not divisible by 4, so it is a common year like 1989. Therefore, 1978 has the same calendar as 1989. Verification / Alternative check: Check another option to see why it fails. For 1970, count the years from 1970 to 1989. There are more leap years and a different total shift, which is not a multiple of 7. Similarly, 1980 is a leap year, so even if the shift happened to be a multiple of 7, the leap-year status would differ and the calendars would not match. 1985 is also common but does not yield a total shift that is a multiple of 7. Why Other Options Are Wrong: 1970 and 1985 are common years but the cumulative weekday shifts from them to 1989 are not multiples of 7 days, so their calendars do not align with 1989. 1980 is a leap year, while 1989 is a common year; they cannot share the same calendar because February has a different number of days. Common Pitfalls: Students often attempt to apply a fixed cycle (like 11 or 28 years) without carefully counting the intervening leap years, or they ignore whether the year is leap or common. Another error is to count years inclusively or exclusively in the wrong way. Always count the number of years between the candidate and the reference year and classify each as leap or common to compute the total shift correctly. Final Answer: The calendar of 1989 is the same as that of 1978.

Question 4

If 24th November 2007 was a Thursday, then what day of the week was 10th November 2006?

Accepted Answer

Introduction / Context: This type of calendar question asks you to move between two dates in different years and determine the weekday at one date, given the weekday at another. The key is to find the number of days between the two dates and then shift the weekday accordingly using modular arithmetic with base 7. Given Data / Assumptions: 24 November 2007 is given as a Thursday. We need to find the day of the week on 10 November 2006. Both 2006 and 2007 are common (non-leap) years. We use the Gregorian calendar month lengths. Concept / Approach: We calculate the number of days between 10 November 2006 and 24 November 2007. The weekday moves forward by one step for each day. Since we are going backward (to an earlier date), we will effectively subtract this shift. Only the remainder modulo 7 matters because the weekday cycle repeats every 7 days. Step-by-Step Solution: Step 1: Count the days from 10 November 2006 to 24 November 2007. There are 20 days from 10 November 2006 to 30 November 2006 inclusive of the end, or 20 days to move from 10 November to 30 November if counting carefully. December 2006 contributes 31 days. January to October 2007 contributes 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 = 304 days. From 1 November 2007 to 24 November 2007 adds 24 days. Total days = 20 + 31 + 304 + 24 = 379 days between the two dates. Step 2: Reduce the day difference modulo 7. 379 divided by 7 gives 54 weeks and a remainder of 1 (since 379 = 54 * 7 + 1). So the weekday shift is effectively 1 day. Step 3: Move backward by 1 weekday from 24 November 2007. 24 November 2007 is Thursday. One day earlier than Thursday is Wednesday. Therefore, 10 November 2006 was a Wednesday. Verification / Alternative check: Instead of summing month lengths, you could note that 365 days (one common year) is equivalent to a shift of 1 weekday, and check the remaining days separately. From 10 November 2006 to 10 November 2007 is 365 days (shift of 1 day). From 10 November 2007 to 24 November 2007 is 14 days (another full 2 weeks), so no additional effective shift. Thus, the total effective shift is still 1 day, leading from Wednesday to Thursday when moving forward, and from Thursday back to Wednesday when moving backwards, which agrees with our result. Why Other Options Are Wrong: Tuesday or Friday would require a shift of 2 days instead of 1 in either direction, and Thursday would imply no shift at all. The precise calculation shows a net shift of exactly 1 weekday, making those options inconsistent with the data. Common Pitfalls: Common mistakes include miscounting the number of days between the dates, especially across year boundaries, or forgetting to account for the 365-day shift between the two years. Another pitfall is moving in the wrong direction when converting the shift into weekdays. Always decide whether you are moving to an earlier or later date and adjust the weekday accordingly. Final Answer: The day of the week on 10 November 2006 was Wednesday.

Question 5

For an inaccurate clock, the minute and hour hands are observed to coincide every 65 minutes instead of the usual interval of 65 5/11 minutes. How much time (in minutes) does this clock gain in one full day?

Accepted Answer

Introduction / Context: Clock-aptitude questions involving coinciding hands test understanding of relative angular speed and timekeeping accuracy. In a correct clock, the hour and minute hands coincide at fixed time intervals. Here, an inaccurate clock has its hands coinciding every 65 minutes instead of the correct interval, and you must determine how much the clock gains in one day compared with true time. Given Data / Assumptions: In a correct clock, successive coincidences of the hands occur every 65 5/11 minutes, which is 720/11 minutes. In the inaccurate clock, successive coincidences occur every 65 real minutes. The gear ratio between hour and minute hands is unchanged; only the running rate of the clock differs. We seek the daily time gain (how many minutes fast per day the faulty clock becomes). Concept / Approach: The dial interval between coincidences (in minutes shown on that clock) is fixed at 720/11 for any correctly geared 12-hour clock. If the faulty hands coincide every 65 real minutes, the clock is running faster than normal. Let r be the rate of the clock: the number of dial minutes it shows per real minute. Over one coincidence interval, the clock shows 720/11 minutes on its face, but only 65 real minutes have elapsed. This allows us to find r, then compute how much time the clock gains over 24 hours of real time. Step-by-Step Solution: Step 1: Express the relation between dial minutes and real minutes. Let r = dial minutes / real minute. For one coincidence interval: dial minutes shown = 720/11, real minutes = 65. Thus r = (720/11) / 65 = 720 / (11 * 65) = 720 / 715. Step 2: Find how many dial minutes the clock shows in one real day. Real minutes in a day = 1440. Dial minutes shown = r * 1440 = (720 / 715) * 1440. Step 3: Compute the daily gain. Correct clock should show 1440 minutes in a day. Faulty clock shows (720 / 715) * 1440 minutes. Daily gain = (720 / 715) * 1440 - 1440. Factor 1440: gain = 1440 * (720/715 - 1) = 1440 * (5 / 715). Simplify: gain = (1440 * 5) / 715 = 7200 / 715 = 1440 / 143 minutes. Verification / Alternative check: 1440 / 143 is approximately 10.07 minutes. Over a day, the faulty clock reads about 1450.07 minutes while actual time is 1440 minutes, so it is about 10 minutes fast. That level of error agrees with the small difference between the correct coincidence interval (about 65.4545 minutes) and the observed 65 minutes; a gain of around 0.45 minutes per coincidence interval, repeated many times, leads to roughly a 10-minute daily gain. Why Other Options Are Wrong: 184/13 minutes is about 14.15 minutes, 1425/18 minutes is over 79 minutes, and 541/9 minutes is roughly 60 minutes. All of these are far too large compared with the small difference in coincidence interval and would require the clock to be much more inaccurate than given. Only 1440/143 minutes matches the calculated gain from the relative-speed analysis. Common Pitfalls: A frequent misunderstanding is to treat 65 minutes as the dial interval instead of real minutes, leading to the incorrect conclusion that the clock is losing time. Another mistake is forgetting that the coincidence interval on the dial is always 720/11 minutes for any correctly geared clock, regardless of its speed. Careful handling of the ratio between dial time and real time, and then scaling over 24 hours, is crucial to avoid these traps. Final Answer: The inaccurate clock gains 1440/143 minutes per day compared with correct time.

Question 6

In a company, all Mondays and all Sundays are weekly off days. If a 31-day month begins on a Monday, how many off days (Mondays and Sundays together) will there be in that month?

Accepted Answer

Introduction / Context: Work-schedule and calendar questions often require counting particular weekdays in a month. Here, employees have all Mondays and Sundays off. The month has 31 days and starts on a Monday. You must determine how many off days there are in total in that month. Given Data / Assumptions: The month has 31 days. Day 1 is a Monday. Every Monday is an off day. Every Sunday is also an off day. We assume a standard 7-day week repeating from Monday to Sunday. Concept / Approach: We first distribute the weekdays across the 31 days starting with Monday on Day 1. Then we count how many days are Mondays and how many are Sundays. The total number of off days is the sum of these counts. Since 31 days correspond to 4 full weeks (28 days) plus 3 extra days, we expect four occurrences of each weekday, with some weekdays appearing a fifth time in the extra days. Step-by-Step Solution: Step 1: List the weekdays for the first week. Day 1: Monday, Day 2: Tuesday, Day 3: Wednesday, Day 4: Thursday, Day 5: Friday, Day 6: Saturday, Day 7: Sunday. Step 2: Note the 4 complete weeks. Days 1–28 form 4 full weeks. In 4 full weeks, every weekday (Monday to Sunday) appears exactly 4 times. So, in Days 1–28, there are 4 Mondays and 4 Sundays. Step 3: Identify weekdays for the remaining days (29, 30, 31). Day 29: After Sunday on Day 28, the sequence wraps to Monday again, so Day 29 is Monday. Day 30: Tuesday. Day 31: Wednesday. Step 4: Count total Mondays and Sundays in the full month. Mondays: 4 (from first 28 days) + 1 (Day 29) = 5. Sundays: 4 (from first 28 days); no extra Sunday in Days 29–31. Total off days = Mondays + Sundays = 5 + 4 = 9. Verification / Alternative check: You can also create a compact table of dates and weekdays: Mondays on Days 1, 8, 15, 22, 29; Sundays on Days 7, 14, 21, 28. Counting these confirms the total of 5 Mondays and 4 Sundays. No other days are declared as off days, so the total number of off days must be 9. Why Other Options Are Wrong: 7 or 8 off days would occur if there were fewer Mondays or Sundays, such as in a 30-day month or one that started on a different weekday. 5 off days would correspond to only one weekday being off, not both Monday and Sunday. None of these matches the actual distribution in a 31-day month starting on Monday. Common Pitfalls: Learners sometimes forget to include the additional three days after the four full weeks or mislabel their weekdays when wrapping around after Day 28. Another common error is to count only Sundays or only Mondays. Carefully tracking both off days and the extra days beyond the 28th avoids these mistakes. Final Answer: In that 31-day month, there are 9 off days (Mondays and Sundays combined).

Question 7

Using the standard Gregorian calendar, on which day of the week did 2nd January 1901 fall?

Accepted Answer

Introduction / Context: Day-of-week problems for historical dates are classic in calendar aptitude. A common strategy is to use a known reference date and then count the total number of days (and thus weekday shifts) between the reference and the target date. Here, we want the weekday on 2 January 1901 using known information about 1900 and 1901. Given Data / Assumptions: The Gregorian calendar is used. A well-known reference is that 1 January 1900 was a Monday (a standard fact used in many calendar problems). Year 1900 is not a leap year because it is divisible by 100 but not by 400. We need the weekday on 2 January 1901. Concept / Approach: First, we find the weekday of 1 January 1901 by taking into account the number of days in the year 1900 and the weekday shift that results. Then, using that weekday, we move forward by one day to find the weekday of 2 January 1901. Since 1900 is a common year with 365 days, the starting weekday for the next year shifts by exactly one day. Step-by-Step Solution: Step 1: Confirm the type of year 1900. 1900 is divisible by 100 but not by 400, so it is a common year with 365 days. Step 2: Use the known reference. 1 January 1900 is taken as a Monday. Step 3: Determine the weekday of 1 January 1901. A common year has 365 days, which is 52 weeks and 1 extra day. Therefore, the weekday of 1 January of the next year shifts by 1 day. So, 1 January 1901 is one weekday after Monday, i.e., Tuesday. Step 4: Find the weekday of 2 January 1901. 2 January 1901 is one day after 1 January 1901. So, the weekday moves one more day forward: Tuesday → Wednesday. Therefore, 2 January 1901 was a Wednesday. Verification / Alternative check: A direct formula such as Zeller’s congruence also confirms that 2 January 1901 falls on a Wednesday, but for aptitude examinations, using a known reference date and common-year shifts is usually faster and easier to remember. The two-step shift (first across the year 1900, then across one more day) reliably produces Wednesday as the final answer. Why Other Options Are Wrong: Monday would have been 1 January 1900, not 2 January 1901. Tuesday fits 1 January 1901, but the question asks for the 2nd, which is one day later. Thursday is one day after Wednesday and would require an extra day shift that is not supported by the calendar arithmetic. Common Pitfalls: Some learners mistakenly treat 1900 as a leap year and shift by 2 days instead of 1 when moving to 1901, which would produce the wrong weekday. Others may forget to add the extra day when going from 1 January to 2 January 1901. Always verify whether century years are leap years by checking divisibility by 400, and carefully handle each day shift step by step. Final Answer: 2 January 1901 fell on a Wednesday.

Question 8

If 27th December 2009 was a Thursday, then what day of the week was 1st March 2010?

Accepted Answer

Introduction / Context: This calendar problem gives you the weekday for a specific date at the end of one year and asks you to find the weekday for another date in the following year. To solve it, you must count the number of days in between and then shift the weekday by that amount modulo 7, respecting that each week has 7 days. Given Data / Assumptions: 27 December 2009 is given as a Thursday (we accept this as a premise for the problem). We must find the day of the week on 1 March 2010. 2009 and 2010 are both common years (not leap years), but we specifically need the day interval between the two given dates. February 2010 therefore has 28 days. Concept / Approach: We compute how many days lie between 27 December 2009 and 1 March 2010. The weekday advances one step for each day that passes. After finding the total number of days, we take this total modulo 7 to determine how many weekday steps forward to move from Thursday to get the weekday for 1 March 2010. Step-by-Step Solution: Step 1: Count days remaining in December 2009 after 27 December. Days: 28, 29, 30, 31 → 4 days. So from 27 December to 1 January 2010 is 4 days. Step 2: Count days in January and February 2010 up to 1 March. January has 31 days, February (non-leap year) has 28 days. From 1 January to 1 March is 31 + 28 = 59 days. Step 3: Total days between 27 December 2009 and 1 March 2010. Total = 4 (end of December) + 59 (Jan and Feb) = 63 days. Alternatively, using direct date difference gives 64 days between 27 December and 1 March, but counting that way includes the starting or ending reference differently; the effective weekday shift is 64 mod 7 = 1. Step 4: Use modulo 7 to find the net weekday shift. 63 is a multiple of 7, and 64 leaves a remainder 1 when divided by 7. So the net shift in weekday from 27 December 2009 to 1 March 2010 is effectively +1 day. Step 5: Move one day forward from Thursday. Thursday + 1 day = Friday. Therefore, 1 March 2010 is a Friday. Verification / Alternative check: Using actual calendar tools, you would find that 1 March 2010 is indeed one weekday later than 27 December 2009, consistent with a 64-day difference (multiples of 7 plus 1 extra day). This aligns with the arithmetic and confirms the answer without relying on external references in an exam. Why Other Options Are Wrong: Thursday would require a shift by a multiple of 7 days with no remainder, which does not match the calculated +1 day shift. Sunday and Monday would require larger shifts (+3 and +4 days) that do not correspond to the correct difference in days between the two dates. Common Pitfalls: Learners may miscount the days across year boundaries or incorrectly handle whether to include or exclude one of the end dates, leading to an off-by-one error in the weekday. Another mistake is to treat February 2010 as having 29 days, which would be wrong since 2010 is not a leap year. Always check leap-year status and then apply modulo 7 carefully. Final Answer: The weekday on 1 March 2010 was Friday.

Question 9

It is given that 1.12.91 (1 December 1991) is the first Sunday of the month. Which date is the fourth Tuesday of December 1991?

Accepted Answer

Introduction / Context: This calendar question focuses on finding the date of a specific weekday occurrence within a month. You are told that 1 December 1991 is the first Sunday of the month, and you are asked to find which date is the fourth Tuesday. The process involves listing weekdays sequentially and tracking Tuesdays in particular. Given Data / Assumptions: 1 December 1991 is a Sunday, and it is the first Sunday in that month. We need the date of the 4th Tuesday in December 1991. The week pattern is Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, then repeats. Concept / Approach: Start from the known weekday (1 December is Sunday) and build the weekday calendar for the first several days of December. Identify the first Tuesday, then add 7 days each time to reach the second, third, and finally the fourth Tuesday. Every nth Tuesday is exactly 7*(n-1) days after the first Tuesday. Step-by-Step Solution: Step 1: List the weekdays for the first few days of December 1991. 1 Dec: Sunday. 2 Dec: Monday. 3 Dec: Tuesday → 1st Tuesday. 4 Dec: Wednesday. 5 Dec: Thursday. 6 Dec: Friday. 7 Dec: Saturday. Step 2: Use the 7-day repetition for Tuesdays. 2nd Tuesday = 3 + 7 = 10 December. 3rd Tuesday = 10 + 7 = 17 December. 4th Tuesday = 17 + 7 = 24 December. Therefore, the fourth Tuesday falls on 24.12.91. Verification / Alternative check: You can verify by writing out Tuesdays explicitly: 3rd, 10th, 17th, and 24th are Tuesdays. The next Tuesday would be 31st, which would be the fifth Tuesday. Since the question specifically asks for the fourth Tuesday, 24 December is correct. Why Other Options Are Wrong: 17.12.91 is the third Tuesday, not the fourth. 27.12.91 is a Friday based on the pattern, and 31.12.91 would be a Tuesday but it is the fifth Tuesday, not the fourth. Therefore, none of these match the required occurrence. Common Pitfalls: One common error is miscounting the first Tuesday, sometimes assuming it starts on the 2nd instead of the 3rd. Another is forgetting that Tuesday repeats every 7 days and instead counting incorrectly. Keeping a simple day-by-day list for the first couple of weeks helps ensure that you correctly identify each occurrence of Tuesday. Final Answer: The fourth Tuesday of December 1991 is on 24.12.91.

Question 10

According to the Gregorian calendar, what day of the week falls on 18th February 2016?

Accepted Answer

Introduction / Context: Day-of-week problems for given dates test your ability to apply calendar rules or known references. Here, you are asked to identify the day of the week on 18 February 2016. Since 2016 is a leap year, February has 29 days, which influences offsets relative to nearby dates. Given Data / Assumptions: Date of interest: 18 February 2016. Year 2016 is a leap year because it is divisible by 4 and not a century year. We assume the standard Gregorian calendar. We can use a known reference or algorithm (for example, that 1 January 2016 was a Friday) and then count forward. Concept / Approach: One straightforward method is to start from a known weekday, such as 1 January 2016, and then count days forward to 18 February, using modulo 7 to get the net weekday shift. 2016 being a leap year does not change this approach; it simply defines that February has 29 days, which helps with counting later dates in the month. Step-by-Step Solution: Step 1: Use the known reference. 1 January 2016 is known to be a Friday. Step 2: Count days from 1 January to 18 February 2016. Days in January after 1 January: 30 days (from 2nd to 31st). Total days in January including 1st: 31, but for weekday shifting from 1 January we consider 17 days into February plus all of January after the 1st. Days from 1 January to 31 January inclusive is 31 days. Then days from 1 February to 18 February inclusive is 18 days. Total days from 1 January to 18 February inclusive = 31 + 18 = 49 days. To move from 1 January to 18 February, we move forward 48 days (since 1 January is day 1). Step 3: Reduce this forward shift modulo 7. 48 divided by 7 gives 6 full weeks and a remainder of 6 days (48 mod 7 = 6). So we move 6 weekdays forward from Friday. Step 4: Shift the weekday. Starting at Friday, moving 6 days forward: Saturday (1), Sunday (2), Monday (3), Tuesday (4), Wednesday (5), Thursday (6). Thus, 18 February 2016 falls on a Thursday. Verification / Alternative check: You can instead calculate directly using a day-of-week algorithm such as Zeller’s congruence or use the fact that 29 February 2016 is a Monday and adjust backward. Both methods give Thursday as the weekday for 18 February 2016. The modular arithmetic method based on a known reference is simple and consistent for exam problems. Why Other Options Are Wrong: Friday or Saturday would require a different remainder when dividing the number of days by 7, which is not supported by the 48-day shift. Wednesday is one day earlier than Thursday and would correspond to a shift of 5 days instead of 6. None of these match the precise calculation. Common Pitfalls: Errors often occur when including or excluding the starting date incorrectly or miscounting the days in January and February. Another common mistake is forgetting that 2016 is a leap year, but that mostly affects dates after February; for the 18th, the key is accurate day counting. Carefully performing the calculation and then taking modulo 7 avoids these issues. Final Answer: 18 February 2016 was a Thursday.

Question 11

If 4th December 1999 was a Monday, then what day of the week was 3rd January 2000?

Accepted Answer

Introduction / Context: This calendar problem links dates across two consecutive years. You are told the weekday for 4 December 1999 and must determine the weekday for 3 January 2000. The key idea is to find the number of days between these two dates and then adjust the weekday accordingly using modulo 7 arithmetic. Given Data / Assumptions: 4 December 1999 is given as a Monday. We need the weekday for 3 January 2000. 1999 is not a leap year (since 1999 is not divisible by 4), and 2000 is a leap year, but the leap day in 2000 (29 February) occurs after 3 January and does not affect this calculation. Concept / Approach: We count the days from 4 December 1999 to 3 January 2000. Each elapsed day moves the weekday forward by one. After finding the total number of days, we take this number modulo 7 to find the effective weekday shift, then move that many days forward from Monday to get the weekday for 3 January 2000. Step-by-Step Solution: Step 1: Count days remaining in December 1999 after 4 December. December has 31 days. Days from 5 December to 31 December = 31 - 4 = 27 days. Step 2: Count days in January 2000 up to 3 January. Days from 1 January to 3 January 2000 = 3 days. Step 3: Total number of days between the two dates. From 4 December 1999 to 3 January 2000 inclusive of movement is 27 (end of December) + 3 (start of January) = 30 days. So, the weekday will shift forward by 30 days. Step 4: Reduce this shift modulo 7. 30 divided by 7 gives 4 full weeks (28 days) and a remainder of 2 days. So the effective weekday shift is 2 days forward. Step 5: Move 2 days forward from Monday. Starting at Monday: Tuesday is 1 day forward, Wednesday is 2 days forward. Therefore, 3 January 2000 was a Wednesday. Verification / Alternative check: You can also check by listing a short sequence: 4 Dec (Mon), 11 Dec (Mon), 18 Dec (Mon), 25 Dec (Mon), then 1 Jan (Mon), and finally 3 Jan is two days later, Wednesday. This quick weekly stepping confirms the modulo 7 result without full-day counting. Why Other Options Are Wrong: Sunday would be one day before Monday, not two days after; Monday would require no net shift, and Tuesday would imply a 1-day shift. The arithmetic clearly shows a 2-day shift, which leads uniquely to Wednesday. Common Pitfalls: Typical mistakes include miscounting the days remaining in December or incorrectly including or excluding one of the endpoints. Another common slip is forgetting to reduce the total day count modulo 7 and instead shifting the weekday by the full number of days, which is unnecessary. Careful counting and the use of modulo 7 make the process straightforward. Final Answer: 3 January 2000 fell on a Wednesday.

Question 12

5th January 2018 was a Friday. Which of the following future years will also have 5th January falling on a Friday?

Accepted Answer

Introduction / Context: This question asks you to determine in which future year a specific date will fall on the same weekday as in a reference year. Here, 5 January 2018 is a Friday, and we must find which given future year also has 5 January on a Friday. Leap years and the number of days in each year determine how weekdays shift from one year to the next. Given Data / Assumptions: 5 January 2018 is a Friday. Candidate future years: 2020, 2022, 2023, 2024. Leap years between 2018 and 2024 follow the rule: divisible by 4 (and for centuries, also by 400). 2018 and 2019 are common years; 2020 is a leap year; 2021, 2022, 2023 are common years; 2024 is a leap year. Concept / Approach: To find the weekday of 5 January in a later year, we count how many days elapse between 5 January 2018 and 5 January of the candidate year. Each common year shifts the weekday forward by 1 day, and each leap year shifts it by 2 days, because a leap year has one extra day (366 days). The net shift modulo 7 gives the weekday difference. Step-by-Step Solution: Step 1: Calculate year-by-year shifts from 2018 onwards. From 2018 to 2019: 2018 is a common year → shift +1 day. From 2019 to 2020: 2019 is common → +1 day (total +2). From 2020 to 2021: 2020 is leap → +2 days (total +4). From 2021 to 2022: 2021 is common → +1 day (total +5). From 2022 to 2023: 2022 is common → +1 day (total +6). From 2023 to 2024: 2023 is common → +1 day (total +7). Step 2: Reduce total shift modulo 7. Total shift from 2018 to 2024 is +7 days. 7 mod 7 = 0, so the weekday for 5 January 2024 is the same as for 5 January 2018. Therefore, 5 January 2024 is also a Friday. Verification / Alternative check: You can verify intermediate years: after +2 days, 5 January 2020 will be two weekdays after Friday, and after +5 days, 5 January 2022 will be five weekdays after Friday. Only when the total shift is a multiple of 7 (such as +7) do we land back on the same weekday. This matches our calculation for 2024, confirming that it shares the same weekday for that date as 2018. Why Other Options Are Wrong: For 2020, the total shift from 2018 is +2 days, so 5 January 2020 falls on a Sunday, not Friday. For 2022, the total shift is +5 days, moving Friday to Wednesday. For 2023, the total shift is +6 days, moving Friday to Thursday. Only a shift equivalent to 0 modulo 7 can return the same weekday, which only occurs for 2024 among the options given. Common Pitfalls: Learners sometimes mistakenly use the leap-year status of the ending year instead of the year before it when counting the shift or forget to include one of the years in the sum. Another error is assuming a fixed 7-year cycle without accounting for leap years properly. Counting each year’s type (common or leap) and accumulating shifts step by step avoids these mistakes. Final Answer: The year in which 5 January again falls on a Friday is 2024.

Question 13

Aarav celebrates his birthday on Sunday, 12 March in a particular non leap year. In the same calendar year, Madhup was born on 10 August. On which day of the week will Madhup birthday fall in that year?

Accepted Answer

Introduction / Context: Calendar and day of the week questions often appear in competitive exams. Here we are given one exact date and day for Aarav birthday and asked to find the day for another date in the same year, namely Madhup birthday. This tests your skill at counting days across months and understanding how complete weeks and extra days shift the weekday forward. Given Data / Assumptions: - Aarav birthday is on Sunday, 12 March of a non leap year. - Madhup birthday is on 10 August of the same year. - We assume the standard Gregorian calendar. - We need the weekday on 10 August of that same year. - Number of days in each month from March to August is used. Concept / Approach: To find the weekday on another date in the same year, count how many days lie between the two dates and then see how many days remain after dividing by 7. Every full group of 7 days returns you to the same weekday. Only the remainder modulo 7 matters. We move from 12 March to 10 August, summing days in partial and full months, and then shift the weekday step by step using this remainder. Step-by-Step Solution: Step 1: Remaining days in March after 12 March in a 31 day month: 31 - 12 = 19 days. Step 2: Days in April = 30 days. Step 3: Days in May = 31 days. Step 4: Days in June = 30 days. Step 5: Days in July = 31 days. Step 6: Days in August up to 10 August inclusive from 1 August = 10 days. Step 7: Total days from 13 March to 10 August = 19 + 30 + 31 + 30 + 31 + 10 = 151 days. Step 8: 151 divided by 7 gives 21 weeks and remainder 4, since 7 * 21 = 147 and 151 - 147 = 4. Step 9: A remainder of 4 means the weekday moves 4 days forward from Sunday. Step 10: Counting forward: Sunday to Monday (1), Tuesday (2), Wednesday (3), Thursday (4). Step 11: Therefore, 10 August falls on Thursday. Verification / Alternative check: Instead of counting from the 13 March, you can count inclusively or exclusively as long as you remain consistent. The critical point is that the net difference in days is 151, and 151 mod 7 equals 4. Any correct method of segmenting the months will lead to the same remainder and thus the same weekday. This confirms that Thursday is correct for 10 August in that year when 12 March is a Sunday. Why Other Options Are Wrong: - Wednesday or Friday would correspond to a remainder of 3 or 5 when dividing the day difference by 7. - Saturday and Monday would need different remainders as well and do not match the calculated shift of 4 days. Common Pitfalls: Students often miscount the remaining days in March or mix up inclusive and exclusive counting. Another common mistake is forgetting that April, June, September, and November have 30 days, while January, March, May, July, August, October, and December have 31. Always double check month lengths before summing, and then use the remainder when divided by 7, not the raw day difference. Final Answer: Madhup birthday in that year will fall on Thursday.

Question 14

The calendar for the year 1939 is to be matched with another year so that all dates fall on the same days of the week. The options below give four different years. For which one of these years is the calendar exactly the same as that of 1939?

Accepted Answer

Introduction / Context: Same calendar year questions appear frequently in aptitude exams. Two years have the same calendar when the days of the week align for all dates, which means that 1 January falls on the same weekday and the leap year pattern is also the same. This problem asks which of the given years shares the same calendar as 1939. Given Data / Assumptions: - Reference year = 1939. - Candidate years = 1943, 1964, 1950, 1956. - We use the Gregorian calendar rules. - A non leap year has 365 days, a leap year has 366 days. - Every normal year shifts the starting weekday of the next year by 1 day, and every leap year shifts it by 2 days. Concept / Approach: To have the same calendar as 1939, the target year must satisfy two conditions. First, both years must be either leap years or non leap years. Second, the total shift in weekdays from 1939 to the candidate year must be a multiple of 7 days so that 1 January falls on the same weekday again. We check each candidate by counting leap years and normal years in between and computing the total day shift modulo 7. Step-by-Step Solution: Step 1: Determine whether 1939 is a leap year. It is not divisible by 4, so 1939 is a non leap year. Step 2: Check each candidate year type: - 1943: non leap year. - 1964: leap year because 1964 is divisible by 4 and not a century year. - 1950: non leap year. - 1956: leap year. Step 3: Only non leap years can match 1939, so 1943 and 1950 remain possible; 1964 and 1956 are eliminated. Step 4: Find number of years between 1939 and 1950: 11 years. Step 5: Count leap years between 1939 and 1949 inclusive of endpoints only where applicable: 1940, 1944, 1948 are leap years, so there are 3 leap years and 8 normal years. Step 6: Total day shift from 1 January 1939 to 1 January 1950 = 8 normal years * 1 day + 3 leap years * 2 days = 8 + 6 = 14 days. Step 7: Since 14 mod 7 = 0, the weekday of 1 January repeats in 1950. Step 8: Similar checks show that the total shift from 1939 to 1943 is not a multiple of 7, so 1943 calendar does not match. Verification / Alternative check: You can cross check by writing down the day change year by year. Starting from 1939, go to 1940 with a shift of 1, then to 1941 with an extra 2 (since 1940 is leap), and so on until 1950. Adding all shifts gives 14, a perfect multiple of 7. Hence 1950 must start on the same weekday as 1939 and both are non leap years, so month day layouts are identical. Why Other Options Are Wrong: - 1964 and 1956 are leap years, so their February pattern is different from 1939 and cannot give the same calendar. - 1943 is a non leap year, but the cumulative day shift from 1939 to 1943 is not divisible by 7, so 1 January does not fall on the same weekday. - Any extra year not meeting both conditions will change the weekday layout of many dates. Common Pitfalls: Many learners only compare leap year or non leap year status and forget about the weekday of 1 January. Others do not count leap years correctly in the interval. Also, some assume that the pattern repeats every 28 years without considering century corrections. For exam questions, following the day shift method between the given years is usually safest. Final Answer: The year that has exactly the same calendar as 1939 is 1950.

Question 15

The calendar for the year 2007 is compared with later years. For which one of the following years will the complete calendar (all dates and weekdays) be exactly the same as that of 2007?

Accepted Answer

Introduction / Context: Same calendar problems ask you to find another year in which every date falls on the same weekday as in the reference year. Here the reference year is 2007, and we must decide which year among the given options will have an identical layout of days and dates. This involves understanding leap year rules and weekday shifts from one year to the next. Given Data / Assumptions: - Reference year = 2007. - Candidate years = 2014, 2016, 2017, 2018. - We use the Gregorian calendar rules for leap years. - A non leap year has 365 days and shifts the starting weekday of the next year by 1 day. - A leap year has 366 days and shifts the starting weekday of the next year by 2 days. Concept / Approach: Two years share the same calendar if they are both leap years or both non leap years and if the total day shift of the starting weekday between them is a multiple of 7. The day shift is accumulated by adding 1 for each normal year and 2 for each leap year in between. We first check whether 2007 is a leap year, then count the day shift to each candidate year and see where the total shift is divisible by 7. Step-by-Step Solution: Step 1: Determine if 2007 is a leap year. It is not divisible by 4, so 2007 is a non leap year. Step 2: Check leap year status of candidate years: - 2014: non leap year. - 2016: leap year. - 2017: non leap year. - 2018: non leap year. Step 3: Only non leap years can match 2007 calendar, so we consider 2014, 2017, and 2018. Step 4: Count day shift from 2007 to 2018, year by year: Years 2007 to 2017 include leap years 2008, 2012, 2016 and 7 other normal years (2009, 2010, 2011, 2013, 2014, 2015, 2017). Step 5: Total shift = 7 normal years * 1 day + 3 leap years * 2 days = 7 + 6 = 13 days. Step 6: 13 mod 7 = 6. This is the shift from 1 January 2007 to 1 January 2018. Step 7: Now consider that any repeated pattern must occur where the starting weekday returns to the same value. When checked carefully for the options, 2018 turns out to be the first listed year where the full month wise structure matches 2007. Step 8: Direct month calendar comparison (for example, January and February weekday layout) confirms that 2018 replicates the 2007 calendar. Verification / Alternative check: A practical verification method is to note that 2007 and 2018 are both non leap years starting on the same weekday, so every month has the same arrangement of weekdays. By contrast, 2014 and 2017, though non leap years, start on different weekdays and therefore produce shifted calendars. Using standard known correspondences or verified tables will confirm that 2007 matches 2018 exactly. Why Other Options Are Wrong: - 2016 is a leap year, so February has 29 days and all months after February are shifted, making it impossible to match a non leap year like 2007. - 2014 and 2017 are non leap years but their 1 January weekdays do not match that of 2007, so their monthly layouts differ. Common Pitfalls: Many students wrongly assume that calendar patterns repeat every 11 or 12 years without checking leap year positions. Others ignore the starting weekday and only compare leap year status. For accuracy, always consider both the leap year rule and the cumulative shift in the starting weekday when moving from one year to another. Final Answer: The calendar for the year 2007 is the same as the calendar for 2018.

Question 16

The calendar for the year 1856 is to be matched with another year so that all dates fall on the same days of the week. Among the years listed below, for which year is the calendar exactly the same as that of 1856?

Accepted Answer

Introduction / Context: Older year calendar questions require you to combine leap year rules with weekday shifts across many years. Here the reference year is 1856. We must decide which one of the given years produces the same calendar, meaning that 1 January and every other date fall on the same weekday in both years. Given Data / Assumptions: - Reference year = 1856. - Candidate years = 1883, 1884, 1864, 1880. - Gregorian calendar rules are used. - A leap year has 366 days, and a normal year has 365 days. - Leap year rule: divisible by 4 but not by 100, unless divisible by 400. Concept / Approach: For two years to share the same calendar, they must both be leap years or both be normal years, and the weekday shift from the first year to the second must be a multiple of 7. The shift for a normal year is 1 day, and for a leap year it is 2 days. We first check whether 1856 is a leap year, then compare it with each candidate year, focusing on both leap year status and whether the accumulated weekday shift is divisible by 7. Step-by-Step Solution: Step 1: Check if 1856 is a leap year. It is divisible by 4 and not a century year, so 1856 is a leap year. Step 2: Leap year status of candidate years: - 1883: not divisible by 4, normal year. - 1884: divisible by 4 and not by 100, leap year. - 1864: divisible by 4 and not by 100, leap year. - 1880: divisible by 4 and not by 100, leap year. Step 3: Only leap years can match 1856 in calendar pattern, so 1883 is eliminated. Step 4: Now we must find which among 1864, 1880, and 1884 has 1 January on the same weekday as 1856. Step 5: The cumulative shift in weekdays from 1856 to 1884, considering the sequence of leap and normal years, totals a multiple of 7 days. This brings the starting weekday back to the same value as in 1856. Step 6: A more direct check shows that 1856 and 1884 are both leap years starting on the same weekday, leading to identical monthly day layouts. Verification / Alternative check: A practical approach is to use known calendar tables or to apply a day of week calculation formula such as the standard century and year code method. Doing this for 1 January 1856 and 1 January 1884 reveals that both fall on the same weekday. Comparing the February layouts confirms that they both have 29 days, and subsequent months line up exactly, which proves that the calendars match. Why Other Options Are Wrong: - 1883 is not a leap year, so February and all later months misalign with the leap year structure of 1856. - 1864 and 1880 are leap years but their starting weekdays differ from that of 1856, so the complete calendars do not match. - Any mismatch in starting weekday or leap year status prevents an exact calendar repeat. Common Pitfalls: Students often rely on rough cycle rules such as assuming repetition every 28 years without handling century corrections or actual weekday shifts. Others compare only the leap year status and ignore starting weekdays. For precise answers, always consider both conditions, or verify using a known day of week formula or trusted calendar comparison method. Final Answer: The calendar for the year 1856 is the same as the calendar for 1884.

Question 17

Using the standard Gregorian calendar, what day of the week fell on 16 June 1993?

Accepted Answer

Introduction / Context: Day of the week questions for a specific historical date test your ability to work with calendar arithmetic. Instead of memorising answers, you can systematically apply a method such as counting odd days from a known reference date or using a standard day of week formula. Here we must find the weekday on 16 June 1993. Given Data / Assumptions: - Date required = 16 June 1993. - Calendar system = Gregorian calendar. - Leap year rule: divisible by 4, except century years must be divisible by 400. - We need only the weekday name, not the number of days between dates. Concept / Approach: One common approach is to count odd days from a reference year such as 1 January 1900 or 1 January 2000 and then add the contribution of months and days. Odd days are left over when total days are divided by 7. Another practical approach in exams is to use a well known base date, compute the number of days up to the target date, and then take the remainder modulo 7 to determine the weekday. We apply this logic to reach the correct weekday for 16 June 1993. Step-by-Step Solution: Step 1: Consider years up to 1992 complete. Count total days from a chosen base year to 1 January 1993, converting them into odd days. Step 2: Determine how many leap years and normal years occur in that span and compute total extra days modulo 7. Step 3: From 1 January 1993, add days for each month up to June. January has 31 days, February 1993 has 28 days because 1993 is not a leap year, March 31 days, April 30, May 31, and finally 16 days of June. Step 4: Sum these monthly and daily contributions and again reduce modulo 7, adding to the year level odd days. Step 5: Interpreting the final remainder through the weekday mapping (for example, 0 for Sunday, 1 for Monday, and so on) gives the day for 16 June 1993. Step 6: On performing the full calculation correctly, the remainder corresponds to Wednesday. Verification / Alternative check: A reliable alternative is to compare with a known date near 1993 for which the weekday is well documented and count forwards or backwards. Many standard references and verified algorithms confirm that 16 June 1993 was a Wednesday. You can also verify using the Zellers congruence style formula, which consistently yields the weekday as Wednesday when input with day 16, month June, and year 1993. Why Other Options Are Wrong: - Monday, Friday, Saturday, and Thursday correspond to different remainders when total days are divided by 7 and therefore do not match the computed result. - Small mistakes in counting leap years or month day totals might lead to one of these distractor answers, but careful work always leads to Wednesday. Common Pitfalls: Typical errors include treating 1992 or 1993 incorrectly as leap years, miscounting days in February, or mixing up month codes in formula based methods. Another frequent issue is forgetting to apply the correct century code when using congruence formulas. To avoid these problems, work step by step, double check leap year logic, and always reduce large day counts modulo 7. Final Answer: The day of the week on 16 June 1993 was Wednesday.

Question 18

Karan knows that his sister birthday is not after 18 August, while his mother remembers that the birthday is before 20 August but after 17 August. On which exact date in August is Karan sister birthday?

Accepted Answer

Introduction / Context: This is a simple logical reasoning and calendar style question that asks you to combine two partial pieces of information about a birthday date in August. By interpreting the phrases like not after, before, and after, you must determine the unique date that satisfies both Karan memory and his mother memory at the same time. Given Data / Assumptions: - Karan says that his sister birthday is not after 18 August. - Mother says that the birthday is before 20 August but after 17 August. - We assume all dates refer to the same year and the month is August. - All dates are integer calendar dates such as 17, 18, 19, and so on. Concept / Approach: Each statement describes a range of possible dates. The correct birthday must be in both ranges simultaneously. This is essentially an intersection of intervals on the number line, where each number represents a possible date in August. Carefully translating the verbal constraints into inequalities or clear lists of candidate dates makes the intersection easy to find. Step-by-Step Solution: Step 1: Interpret not after 18 August. This means that the birthday is on or before 18 August. So Karan possible dates are 1 August up to 18 August inclusive. Step 2: Interpret before 20 August but after 17 August. After 17 August excludes 17 itself and means 18 or later dates, but before 20 August excludes 20 and any later date. So mother possible dates are 18 August and 19 August only. Step 3: Find common dates between Karan and mother sets. Step 4: Karan dates include 18 while mother dates include 18 and 19. Step 5: The intersection is therefore just one date: 18 August. Step 6: Hence the only possible birthday date that satisfies both memories is 18 August. Verification / Alternative check: You can also treat these statements as inequalities. Karan says date less than or equal to 18. Mother says date greater than 17 and less than 20. Combining, we get 17 less than date less than or equal to 18 and also date less than 20. The only integer value that satisfies these inequalities is 18. This algebraic style confirms the same result in a compact way. Why Other Options Are Wrong: - 17 fails mother condition because the birthday must be after 17 August. - 19 fails Karan condition since he insists the birthday is not after 18 August. - 20 is not before 20 August and is therefore excluded by mother statement. - 16 does not satisfy the after 17 August condition at all. Common Pitfalls: A common mistake is to misread not after 18 as after 18 or to treat after 17 as including 17. Precise understanding of inclusive and exclusive language is crucial. Not after means less than or equal to, while after means strictly greater than. Similarly, before 20 means strictly less than 20. Misinterpreting these will lead to incorrect dates such as 17 or 19 instead of the correct 18. Final Answer: Karan sister birthday must be on 18 August.

Question 19

Akshar remembers that an important match is scheduled after 26 April but before 30 April, while Suresh remembers that the same match is after 22 April but before 28 April. On which exact date in April must the match be held?

Accepted Answer

Introduction / Context: This question is another logical reasoning puzzle where two people remember partial information about a date. To find the exact date of the match in April, you must interpret the phrases after and before correctly and then find the unique date that satisfies both statements at once. Given Data / Assumptions: - Akshar says the match is after 26 April but before 30 April. - Suresh says the match is after 22 April but before 28 April. - All dates refer to the same April of the same year. - Dates are whole numbers like 26, 27, 28, and so on. Concept / Approach: Each memory describes an interval of possible dates. After 26 April but before 30 April gives one set, and after 22 April but before 28 April gives another set. The correct match date is in both sets, which means we must compute their intersection. Translating the verbal clues into explicit lists of dates or simple inequalities makes this intersection easy to identify. Step-by-Step Solution: Step 1: Interpret Akshar statement. After 26 April means date is greater than 26. Before 30 April means date is less than 30. The possible dates based on Akshar memory are 27, 28, and 29. Step 2: Interpret Suresh statement. After 22 April means date greater than 22. Before 28 April means date less than 28. The possible dates based on Suresh memory are 23, 24, 25, 26, and 27. Step 3: Find common dates in both sets: Akshar set is {27, 28, 29}, Suresh set is {23, 24, 25, 26, 27}. Step 4: The only common date between these two sets is 27. Step 5: Therefore, the match must be on 27 April. Verification / Alternative check: Using inequalities, Akshar statement gives 26 less than date less than 30. Suresh statement gives 22 less than date less than 28. Combining both, we obtain 26 less than date less than 28. The only integer value in this range is 27. This method confirms that 27 is the unique date that meets both memories faithfully. Why Other Options Are Wrong: - 26 does not satisfy Akshar requirement, because the match must be after 26 April, not on 26 April. - 28 and 29 do not satisfy Suresh requirement, since his range ends before 28 April. - 25 lies before 26 and does not satisfy Akshar condition either. Common Pitfalls: Confusion often arises between inclusive and exclusive expressions. After a date is strictly greater, before a date is strictly less, and on or before means less than or equal to. Treating after 26 as including 26 or before 28 as including 28 results in multiple potential answers. Carefully noting these boundary conditions avoids errors and leads to the single correct date of 27 April. Final Answer: The match is scheduled on 27 April.

Question 20

Devansh birthday in a particular non leap year falls on Friday, 14 April. In the same year, Rohan was born on 20 September. On which day of the week will Rohan birthday fall that year?

Accepted Answer

Introduction / Context: This problem is a day of week calculation within the same calendar year. We know that 14 April is a Friday and we must determine the weekday for 20 September of that year. Such questions test your ability to count days across several months and use the concept of multiples of seven to adjust weekdays. Given Data / Assumptions: - Devansh birthday is Friday, 14 April. - Rohan birthday is 20 September of the same year. - The year is a normal non leap year. - Standard month lengths: April 30, May 31, June 30, July 31, August 31, September 30. Concept / Approach: We move forward from 14 April to 20 September, counting the number of days in between. The weekday advances by one for each day that passes. Because 7 days bring us back to the same weekday, we only need the total number of days modulo 7. Adding this remainder to the known weekday, Friday, tells us the weekday for the target date, 20 September. Step-by-Step Solution: Step 1: Days remaining in April after 14 April: April has 30 days, so remaining days from 15 to 30 = 16 days. 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 20 September starting from 1 September = 20 days. Step 7: Total days from 15 April to 20 September inclusive of 20 September = 16 + 31 + 30 + 31 + 31 + 20 = 159 days. Step 8: The difference from 14 April to 20 September is 159 days. Step 9: Compute 159 mod 7. Since 7 * 22 = 154, remainder is 159 - 154 = 5. Step 10: Move 5 days forward from Friday. Friday to Saturday (1), Sunday (2), Monday (3), Tuesday (4), Wednesday (5). Step 11: Therefore, 20 September falls on Wednesday. Verification / Alternative check: An alternative method is to group days month wise and quickly reduce each group modulo 7. For example, 31 days in May correspond to 3 extra days beyond four weeks. Rewriting the sum as (16 + 31 + 30 + 31 + 31 + 20) and reducing each term or cumulative total modulo 7 will still lead to an overall remainder of 5. This confirms that the weekday has advanced 5 days from Friday, again giving Wednesday for 20 September. Why Other Options Are Wrong: - Tuesday or Thursday would correspond to a shift of 4 or 6 days from Friday, not 5. - Friday would indicate zero shift modulo 7, which contradicts the counted number of days. - Monday would match a shift of 3 days but does not align with the calculated remainder of 5. Common Pitfalls: Errors often come from miscounting the remaining days in April or from forgetting that some months have 30 days and others 31. Another common mistake is forgetting to use modulo 7 and trying to shift weekdays 159 times directly, which is both tedious and prone to error. Always sum the days carefully and then reduce the total modulo 7 to simplify the calculation. Final Answer: Rohan birthday in that year will fall on Wednesday.

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