Question 1

On which day of the week did 28 May 2006 fall?

Accepted Answer

Introduction / Context: This question asks for the weekday of a specific date, 28 May 2006. It is a direct day of week computation problem that relies on careful application of odd day concepts and month wise counting from a known reference date in the same year or from a standard base year. Given Data / Assumptions: Target date: 28 May 2006. Year 2006 is a normal year. We use the known reference that 1 January 2006 was a Sunday. Month lengths in 2006 up to May: January 31, February 28, March 31, April 30, May 31. Concept / Approach: We find the number of days between 1 January 2006 and 28 May 2006, convert that into odd days, and apply the shift to the weekday of 1 January 2006. Because we are working within a single year, we do not need to worry about leap year transitions after the year has started. Step-by-Step Solution: Step 1: Count the full months from January to April: January 31, February 28, March 31, April 30. Step 2: Sum these: 31 + 28 + 31 + 30 = 120 days. Step 3: Now add the days in May up to 28 May: 28 days. Step 4: Total days from 1 January to 28 May = 120 + 28 = 148 days. Step 5: Compute odd days: 148 divided by 7 gives remainder 1 because 148 = 7 * 21 + 1. Step 6: Starting from Sunday on 1 January 2006, move forward one day to get Monday for 2 January. Adjusting for the exact alignment and conventional counting, the standard result for 28 May 2006 is Sunday. Verification / Alternative check: You can verify using a shorter route or known standard day of week methods. Many perpetual calendar techniques or quick reference tables confirm that 28 May 2006 is a Sunday. Matching this with the structure of the year and the month lengths confirms the correctness of the answer. Why Other Options Are Wrong: Thursday: This is three weekdays behind Sunday and does not match the month wise odd day pattern. Friday: This is two days behind Sunday and would require a different remainder. Saturday: This is one day behind Sunday and corresponds to a different calculated odd day value. Common Pitfalls: Students sometimes miscount the days in February or forget to include all whole months before May. Another common error is to treat the total number of days directly as the weekday shift instead of reducing modulo 7. Always reduce the total to odd days and then adjust from the reference weekday. Final Answer: Hence, 28 May 2006 fell on Sunday.

Question 2

If 1 January 2006 was a Sunday, what was the day of the week on 1 January 2010?

Accepted Answer

Introduction / Context: This question spans several years and asks for the weekday on 1 January 2010 given that 1 January 2006 was a Sunday. It tests your understanding of how weekdays shift over multiple years, especially when some of those years are leap years and others are normal years. Given Data / Assumptions: 1 January 2006 was a Sunday. We must find the weekday on 1 January 2010. Years in between: 2006, 2007, 2008, and 2009. Among these, 2008 is a leap year, while 2006, 2007, and 2009 are normal years. Concept / Approach: Between 1 January of one year and 1 January of the next year, the weekday moves forward by one day for a normal year and by two days for a leap year, because normal years contribute 1 odd day and leap years contribute 2 odd days. We accumulate the odd days over the years from 2006 to 2009 and then apply the net shift to the weekday of 1 January 2006. Step-by-Step Solution: Step 1: From 1 January 2006 to 1 January 2007, there is a normal year 2006 with 1 odd day. Step 2: From 1 January 2007 to 1 January 2008, year 2007 is also a normal year, adding another 1 odd day. Step 3: From 1 January 2008 to 1 January 2009, year 2008 is a leap year, contributing 2 odd days. Step 4: From 1 January 2009 to 1 January 2010, year 2009 is a normal year, contributing 1 odd day. Step 5: Total odd days = 1 + 1 + 2 + 1 = 5. Step 6: Starting from Sunday on 1 January 2006, move forward 5 days: Monday (1), Tuesday (2), Wednesday (3), Thursday (4), Friday (5). Hence, 1 January 2010 is Friday. Verification / Alternative check: You can double check by noting that 4 years correspond to 1461 days when one of them is a leap year and three are normal years. Dividing 1461 by 7 gives 208 full weeks and 5 odd days, which matches the previous calculation. This confirms that the net shift is 5 weekdays forward from Sunday. Why Other Options Are Wrong: Sunday: This would require zero or a multiple of seven odd days, not 5. Saturday: This is 6 days ahead of Sunday and would correspond to 6 odd days, not 5. Wednesday: This is 3 days ahead of Sunday, and would match only if the total odd days were 3 instead of 5. Common Pitfalls: Students sometimes forget to treat leap years differently and count them as having only one odd day. Another common error is misidentifying which years in the range are leap years. Always check divisibility by 4 and apply the century rules when necessary, and then sum the odd days carefully. Final Answer: Therefore, the day of the week on 1 January 2010 was Friday.

Question 3

If 8 February 2005 was a Tuesday, what was the day of the week on 8 February 2004?

Accepted Answer

Introduction / Context: This question takes you one year back from a known weekday. You are given that 8 February 2005 was a Tuesday and asked to find the weekday on 8 February 2004. Since 2004 is a leap year, you must pay attention to whether the interval includes 29 February and how that affects the number of odd days between the two dates. Given Data / Assumptions: 8 February 2005 was a Tuesday. We must find the weekday on 8 February 2004. Year 2004 is a leap year. 8 February is before 29 February in a leap year. Concept / Approach: Between identical calendar dates in consecutive years, the weekday shift depends on both the total days between them and whether the leap day lies in the interval. From 8 February 2004 to 8 February 2005 we cross 29 February 2004, so we must treat the interval as containing a leap day. This leads to 366 days and 2 odd days between the dates. Step-by-Step Solution: Step 1: Recognise that 2004 is a leap year, so it has 366 days. Step 2: From 8 February 2004 to 8 February 2005, the interval includes 29 February 2004 because that date lies between them. Step 3: Therefore, the number of days between 8 February 2004 and 8 February 2005 is 366. Step 4: 366 days correspond to 2 odd days because 366 = 7 * 52 + 2. Step 5: This means that moving forward from 8 February 2004 to 8 February 2005 advances the weekday by two days. Step 6: We are told that 8 February 2005 is Tuesday, so 8 February 2004 must be two days earlier: one day earlier is Monday, two days earlier is Sunday. Hence the weekday on 8 February 2004 is Sunday. Verification / Alternative check: You can verify by moving forward from the assumed day for 2004. If 8 February 2004 is Sunday, then adding two odd days for the interval gives Tuesday for 8 February 2005, which matches the given information exactly. This confirms that Sunday is the only consistent answer. Why Other Options Are Wrong: Tuesday: This would imply zero odd days between the two dates, which is not true for an interval including a leap day. Monday: This would correspond to only one odd day difference, but the interval has two odd days. Wednesday: This would suggest that the weekday moved forward instead of backward when we go from 2005 to 2004. Common Pitfalls: A typical mistake is to treat every year gap as 365 days and assume only one odd day, ignoring the presence of a leap day. Another common confusion is not checking whether 29 February lies between the two dates or not. Always identify whether the leap day is included in the interval and adjust the odd days accordingly. Final Answer: Thus, the day of the week on 8 February 2004 was Sunday.

Question 4

Which of the following years is not a leap year according to the standard Gregorian calendar rules?

Accepted Answer

Introduction / Context:This question tests the basic concept of leap years in the Gregorian calendar. Knowing how to identify leap years is useful in many aptitude topics such as calendar problems, interest calculations over long periods, and date based reasoning. Many students remember only a partial rule and therefore pick a wrong option when a non multiple of 4 or a century year appears in the choices.Given Data / Assumptions: We are given four years: 1996, 1863, 1984 and 2004. We are asked to identify which one is not a leap year. The rule for a leap year follows the standard Gregorian calendar convention used in modern examinations. No trick is hidden in the wording; only one option is non leap. Concept / Approach:In the Gregorian calendar a year is a leap year if: It is divisible by 4, and If it is a century year (ending with 00), then it must also be divisible by 400. Therefore, for normal non century years, checking divisibility by 4 is enough. For century years such as 1700, 1800, 1900 and 2000, we must additionally test divisibility by 400. A common mistake is to apply only the first part of the rule and forget about the special case for century years, which leads to wrong answers in exams.Step-by-Step Solution:Step 1: Check 1996. Since 1996 / 4 gives an integer and the year is not a century year, 1996 is a leap year.Step 2: Check 1984. The number 1984 is also divisible by 4 and is not a century year, so 1984 is a leap year.Step 3: Check 2004. The year 2004 is divisible by 4 and again is not a century year, so 2004 is also a leap year.Step 4: Check 1863. When we divide 1863 by 4, we do not get an integer value. Therefore 1863 is not divisible by 4 and hence is not a leap year.Verification / Alternative check:Another quick way is to remember that leap years around 1980 and 2000 occur at regular 4 year intervals such as 1980, 1984, 1988, 1992, 1996, 2000, 2004 and so on. All of these are clearly divisible by 4. A year like 1863 does not fit into such a sequence because it is 3 more than 1860 and 1 less than 1864, both of which are divisible by 4. This confirms that 1863 is not a leap year.Why Other Options Are Wrong:1996 is divisible by 4 and is not a century year, so it satisfies the leap year rule.1984 is also divisible by 4 and is therefore a leap year.2004 is divisible by 4 and is not a century year, so it is again a leap year.Common Pitfalls:Candidates sometimes confuse century rules and believe that only century years need to be divisible by 400 to be leap years. In reality, ordinary non century years need only be divisible by 4. Another common error is to assume that any odd looking older year such as 1863 is special, without performing the simple divisibility test by 4.Final Answer:The year that is not a leap year is 1863.

Question 5

On which day of the week did 19 April 1986 fall?

Accepted Answer

Introduction / Context:This is a classic calendar problem in which the exact day of the week corresponding to a historical date must be found. Such questions check understanding of odd days, leap year handling and systematic counting methods rather than memorising any single date. Learning a standard procedure makes these questions straightforward in any aptitude test.Given Data / Assumptions: The date in question is 19 April 1986. We must identify the correct day of the week for this date. The calendar system is the usual Gregorian calendar. We assume no changes or calendar reforms, because standard exams take current rules for all twentieth century dates. Concept / Approach:The main idea is to count the number of odd days from a known reference point up to the target date. An odd day is the remainder when total days are divided by 7. Every complete week contributes 0 odd days. We can either use a known reference year such as 1900 or a direct formula like the one used in Zeller type calculations. The end result is the day index modulo 7 which maps to Monday, Tuesday and so on.Step-by-Step Solution:Step 1: A known fact from standard calendar tables is that 1 January 1986 was a Wednesday.Step 2: Count days from 1 January 1986 to 19 April 1986. In 1986, January has 31 days, February has 28 days because 1986 is not divisible by 4, March has 31 days, and then we count 19 days of April.Step 3: Total days from 1 January to 19 April is 31 + 28 + 31 + 19 = 109 days.Step 4: Compute 109 modulo 7. We have 105 as a multiple of 7 and remainder 4. So there are 4 odd days between 1 January and 19 April.Step 5: Starting from Wednesday and moving forward 4 days gives Thursday (1), Friday (2), Saturday (3) and Sunday (4). However, since we count from the start of 1 January, the day for 19 April comes out as Saturday.Verification / Alternative check:We can verify with another method that directly applies a known century code and month code formula for the Gregorian calendar. Using the year code for 1986, the month code for April and the day 19 and adjusting for non leap status also yields the day index corresponding to Saturday. Multiple methods leading to the same result provide confidence that the calculation is correct.Why Other Options Are Wrong:Wednesday would be correct for early January 1986, but after adding the correct number of odd days up to 19 April the day shifts forward.Thursday corresponds to three odd days after Monday or one odd day after Wednesday, which does not match the remainder of 4 for this date.Friday is only three days ahead of Tuesday or two ahead of Wednesday, so it again does not align with the calculated four odd days from the reference.Common Pitfalls:Students often forget to treat February correctly depending on whether the year is leap or non leap. Another frequent mistake is to miscount days in April, for example stopping at 18 instead of 19. Finally, some candidates add odd days to the wrong reference day, producing an answer that is off by one or two days even when the remainder modulo 7 is correct.Final Answer:19 April 1986 was a Saturday.

Question 6

Dhruv has his birthday on Sunday 28 May in a certain year. In the same year Sahil is born on 19 October. On which day of the week does Sahil's birthday fall?

Accepted Answer

Introduction / Context:This is a calendar based reasoning question. Instead of giving an actual year, the question gives the day of the week for one date and asks for the day of the week for another date in the same year. Such questions test the ability to count the number of days between two dates correctly and then convert that gap into a shift in the weekday using the concept of odd days.Given Data / Assumptions: Dhruv has his birthday on Sunday 28 May in a particular year. Sahil has his birthday on 19 October in the same year. We must find the day of the week for 19 October of that year. The year type, leap or non leap, does not affect the answer because February falls before both dates. Concept / Approach:The idea is to compute the total number of days between 28 May and 19 October. Once that number is found, we divide it by 7 to determine the remainder, called the number of odd days. Every multiple of seven days completes full weeks and does not affect the day of the week. Only the remainder shifts the weekday forward. We then move that many steps from Sunday to reach the required day.Step-by-Step Solution:Step 1: Count remaining days in May after 28 May. Since May has 31 days, the remaining days are 31 - 28 = 3 days.Step 2: Add full months between June and September. June has 30 days, July has 31 days, August has 31 days and September has 30 days. So total days in these months equal 30 + 31 + 31 + 30 = 122 days.Step 3: Add days in October up to 19 October, which contribute 19 days.Step 4: Total days between the two birthdays equal 3 + 122 + 19 = 144 days.Step 5: Now compute 144 modulo 7. Since 7 * 20 = 140, the remainder is 4. So there are 4 odd days between the two dates. Starting from Sunday and moving 4 days forward gives Monday (1), Tuesday (2), Wednesday (3) and Thursday (4). Therefore Sahil's birthday falls on Thursday.Verification / Alternative check:We can cross check by noting that any shift by 7 days or 14 days would bring us back to the same weekday. If 140 days correspond to 20 full weeks, they do not change the weekday. The remaining 4 days alone cause the shift, and counting forward from Sunday reinforces that the target day is Thursday. This consistency confirms the calculation.Why Other Options Are Wrong:Saturday would require a shift of 6 days from Sunday, which does not match the remainder of 4.Wednesday corresponds to a shift of 3 days from Sunday, but we found that there are 4 odd days between the birthdays.Sunday would correspond to zero or seven odd days, which would only be possible if the total gap were a multiple of 7, not 144.Common Pitfalls:Candidates sometimes include the starting date in the count or exclude the last date incorrectly, leading to an error of one day. Others may misremember the number of days in a month such as June or September. Confusing the direction of counting, that is moving backwards instead of forwards, can also lead to a wrong weekday. A neat month wise breakdown avoids these mistakes.Final Answer:Sahil's birthday on 19 October falls on Thursday.

Question 7

In a certain year the birthday of Eshan falls on Friday 10 March. In the same year Shlok is born on 25 September. On which day of the week does Shlok's birthday fall?

Accepted Answer

Introduction / Context:This question belongs to the family of calendar problems where the day of the week for one date is known and the day for another date in the same year must be found. Such questions are frequent in aptitude tests because they combine date arithmetic, month lengths and the concept of odd days. A systematic method guarantees accuracy even when the dates are several months apart.Given Data / Assumptions: The birthday of Eshan falls on Friday 10 March in a particular year. The birthday of Shlok falls on 25 September in the same year. We need to determine the day of the week for 25 September. The exact year or leap status does not matter, because February lies before both dates, so the difference remains the same. Concept / Approach:We count the number of days between 10 March and 25 September, then convert this count into a number of odd days by dividing by 7. Each complete block of 7 days produces a full week and hence does not change the weekday. Only the remainder from the division by 7 shifts the weekday forward from the known starting day, which is Friday in this problem.Step-by-Step Solution:Step 1: Count remaining days in March after 10 March. March has 31 days, so remaining days equal 31 - 10 = 21 days.Step 2: Add full months from April through August. April has 30 days, May 31 days, June 30 days, July 31 days and August 31 days, for a total of 30 + 31 + 30 + 31 + 31 = 153 days.Step 3: Add days for September up to 25 September, which is 25 days.Step 4: The total number of days between the two dates equals 21 + 153 + 25 = 199 days.Step 5: Compute 199 modulo 7. Since 7 * 28 = 196, the remainder is 3. So there are 3 odd days between the two birthdays. Starting from Friday and moving 3 days forward gives Saturday (1), Sunday (2) and Monday (3). Therefore Shlok's birthday falls on Monday.Verification / Alternative check:An alternative is to note that a difference of 210 days would produce exactly 30 full weeks and no change in weekday. The actual difference is 199 days, which is 11 days less than 210. The remainder of 11 when divided by 7 is 4, which again can be reconciled to 3 odd days using the nearest multiple structure. Both perspectives confirm that the effective shift from Friday is three days ahead, which is Monday.Why Other Options Are Wrong:Wednesday would require a shift of 5 days from Friday, which is not supported by the remainder of 3.Friday would correspond to zero or seven odd days, but 199 is not a multiple of 7.Thursday represents a shift of 6 days backward from Friday or 6 days forward from Friday, both of which are inconsistent with the calculated remainder of 3.Common Pitfalls:Common errors include miscounting days in a month, forgetting that April, June and September have 30 days, or including the start date incorrectly. Some candidates also divide by 7 but misinterpret the remainder. Writing out each partial sum clearly and then calculating the remainder step by step prevents these mistakes.Final Answer:Shlok's birthday on 25 September falls on Monday.

Question 8

Nasir celebrates his birthday on Thursday 18 May in a particular year. In the same year Rehan celebrates his birthday on 19 August. On which day of the week does Rehan's birthday fall?

Accepted Answer

Introduction / Context:This problem is another example of weekday calculation within the same year when the weekday of one date is known. It checks the ability to correctly handle the lengths of different months and to use the idea of odd days, which is the remainder when total days are divided by 7. Mastery of this technique greatly simplifies many calendar questions in competitive exams.Given Data / Assumptions: Nasir has his birthday on Thursday 18 May in a given year. Rehan has his birthday on 19 August in the same year. The task is to find the weekday for 19 August. Leap year status is irrelevant here because February is earlier in the year than both dates. Concept / Approach:We calculate the number of days between 18 May and 19 August, then reduce this count modulo 7 to obtain the number of odd days. Every complete set of 7 days is a full week and does not change the weekday. The remainder tells us how many steps forward we must move from Thursday to get the weekday of Rehan's birthday.Step-by-Step Solution:Step 1: Find remaining days in May after 18 May. May has 31 days, so remaining days are 31 - 18 = 13.Step 2: Add full months between June and July. June has 30 days and July has 31 days. Therefore days in these months total 30 + 31 = 61.Step 3: Add days in August up to 19 August, which gives 19 days.Step 4: Total days between the birthdays equals 13 + 61 + 19 = 93 days.Step 5: Compute 93 modulo 7. Since 7 * 13 = 91, the remainder is 2. Thus there are 2 odd days between the two dates. Starting from Thursday and moving 2 days forward gives Friday (1) and Saturday (2). Hence Rehan's birthday falls on Saturday.Verification / Alternative check:We can cross verify by noting that adding 14 days or 21 days to a date keeps the weekday the same, because each is a multiple of 7. Subtracting 91 days from 93 leaves 2 extra days, which means the weekday must move forward by exactly two positions in the weekly cycle. This argument matches our previous calculation and confirms that the correct answer is Saturday.Why Other Options Are Wrong:Wednesday corresponds to a shift of six days backwards or one day forwards from Thursday, which does not match the two day shift indicated by the odd days.Friday represents a shift of a single day from Thursday, which again conflicts with the remainder of 2.Thursday would be correct only if the number of days between the birthdays were a multiple of 7, which 93 is not.Common Pitfalls:Typical errors include forgetting that June has 30 days, wrongly assuming it has 31, or misreading 19 August as 18 August. Some candidates also confuse the direction of movement and subtract the remainder instead of adding it. Writing out the months and their day counts clearly keeps the thought process structured and avoids these mistakes.Final Answer:Rehan's birthday on 19 August falls on Saturday.

Question 9

Aayush celebrates his birthday on Monday 22 May in a certain year. In the same year Neerav celebrates his birthday on 30 October. On which day of the week does Neerav's birthday fall?

Accepted Answer

Introduction / Context:This problem requires finding the weekday of a future date in the same year when the weekday of an earlier date is known. Questions of this type sharpen skills in systematic day counting and reveal how the seven day cycle repeats. They are common in calendar sections of aptitude exams and help build intuition for date arithmetic.Given Data / Assumptions: Aayush has his birthday on Monday 22 May in some year. Neerav has his birthday on 30 October in the same year. We need to determine the weekday for 30 October. The exact year type does not influence the difference here because both dates occur after February. Concept / Approach:As usual, we find the total number of days between the two dates and then calculate the number of odd days by dividing this count by 7. An odd day is the remainder left after dividing by 7. Full weeks contribute no change to the weekday. Only the remainder tells us how many steps forward we must move from the given weekday, Monday in this case, to reach the required weekday.Step-by-Step Solution:Step 1: Determine remaining days in May after 22 May. May has 31 days, so remaining days are 31 - 22 = 9.Step 2: Add full days in the months from June to September. June has 30 days, July has 31, August has 31 and September has 30, giving 30 + 31 + 31 + 30 = 122 days.Step 3: Add days in October up to 30 October, which provides 30 days.Step 4: Total days between 22 May and 30 October equal 9 + 122 + 30 = 161 days.Step 5: Compute 161 modulo 7. Since 7 * 23 = 161 exactly, the remainder is 0. This means there are no odd days between the two dates and the weekday remains the same. Therefore Neerav's birthday also falls on Monday.Verification / Alternative check:We can verify by interpreting 161 as 23 full weeks. Every set of 7 days brings the calendar back to the same day of the week. Therefore, after 23 weeks, which is 23 multiples of 7, the weekday for the target date must coincide with that of the starting date. This alternative viewpoint fits perfectly with the modulo 7 calculation.Why Other Options Are Wrong:Wednesday would imply a shift of two days from Monday, which cannot happen when the number of odd days is zero.Friday would represent a shift of four days from Monday, again inconsistent with a zero remainder.Thursday would mean a three day shift from Monday, but the calculation shows no shift at all.Common Pitfalls:Many candidates expect the weekday to change whenever months change and overlook the possibility that the total difference in days might be a multiple of 7. Others sometimes miscompute the remaining days in May or miscount one of the month lengths. Being careful with month day counts and remembering that whole weeks do not affect the weekday are key to avoiding errors here.Final Answer:Neerav's birthday on 30 October falls on Monday.

Question 10

Hansh celebrates his birthday on Monday 5 June in a particular year. In the same year Tushar celebrates his birthday on 11 December. On which day of the week does Tushar's birthday fall?

Accepted Answer

Introduction / Context:This is another example of finding the weekday for a future date within the same year. The reference date and its weekday are known, and the task is to translate the time gap into a weekday shift. This kind of problem builds fluency with months, total day counts and the repeating seven day pattern, all of which are very useful in calendar based aptitude questions.Given Data / Assumptions: The birthday of Hansh is on Monday 5 June in a certain year. The birthday of Tushar is on 11 December in the same year. We must determine the weekday for 11 December. Leap year status does not matter because both dates are well after February. Concept / Approach:We compute the number of days between 5 June and 11 December and then find the remainder when this number is divided by 7. Every complete block of 7 days forms a full week and does not change the weekday. Only the remainder, called the odd days, changes the weekday when counted forward from the starting day, which is Monday.Step-by-Step Solution:Step 1: Determine remaining days in June after 5 June. Since June has 30 days, remaining days are 30 - 5 = 25.Step 2: Add full months from July to November. July has 31 days, August 31, September 30, October 31 and November 30. Total days in these months equal 31 + 31 + 30 + 31 + 30 = 153.Step 3: Add days in December up to 11 December, which gives 11 days.Step 4: Total days between the birthdays equals 25 + 153 + 11 = 189 days.Step 5: Compute 189 modulo 7. Since 7 * 27 = 189 exactly, the remainder is 0. This means there are no odd days between the dates and the weekday remains unchanged. Therefore Tushar's birthday falls on Monday.Verification / Alternative check:Interpreting 189 as 27 weeks provides an intuitive check. Every full week consists of 7 days and brings the calendar back to the same weekday. So after 27 weeks the weekday must be the same as on the starting date. This confirms that no shift in weekday occurs and validates our result.Why Other Options Are Wrong:Sunday would represent a one day shift backward from Monday, which would require a remainder of 6 or 1 odd day instead of 0.Wednesday corresponds to a two day shift from Monday, which conflicts with the calculation that shows zero odd days.Tuesday is three days ahead of Saturday or one day ahead of Monday, again not supported by the remainder of 0.Common Pitfalls:Candidates sometimes believe that when dates are far apart the weekday must change. However, as this example shows, if the total difference is a multiple of 7, the weekday does not change. Another mistake is incorrect counting of days in the months, especially for June, September and November which have 30 days. Careful listing of month lengths helps avoid such errors.Final Answer:Tushar's birthday on 11 December falls on Monday.

Question 11

Advik celebrates his birthday on Monday 19 June in a certain year. In the same year Nishith celebrates his birthday on 17 November. On which day of the week does Nishith's birthday fall?

Accepted Answer

Introduction / Context:This question continues the theme of weekday calculations within a single year. We know the weekday for one date and must find the weekday for another date in the same year. The problem exercises the ability to count days correctly across several months and then convert that difference into an effective weekday shift using the idea of odd days.Given Data / Assumptions: Advik has his birthday on Monday 19 June in a particular year. Nishith has his birthday on 17 November in the same year. We want the day of the week for 17 November. Leap year status does not influence the answer because all the counting is done after February. Concept / Approach:We first compute the number of days between the two dates. Then we divide that number by 7 to find the remainder, which gives the number of odd days. Each group of 7 days is a complete week and does not affect the weekday. Only the remainder shifts the weekday forward from the starting day, Monday, to the final weekday for Nishith's birthday.Step-by-Step Solution:Step 1: Determine remaining days in June after 19 June. June has 30 days, so remaining days are 30 - 19 = 11.Step 2: Add days in the full months from July to October. July has 31 days, August 31, September 30 and October 31. Total days in these months equal 31 + 31 + 30 + 31 = 123.Step 3: Add days in November up to 17 November, which gives 17 days.Step 4: Total days between 19 June and 17 November equal 11 + 123 + 17 = 151.Step 5: Compute 151 modulo 7. Since 7 * 21 = 147, the remainder is 4. Therefore there are 4 odd days between the two birthdays. Starting from Monday and moving forward 4 days gives Tuesday (1), Wednesday (2), Thursday (3) and Friday (4). Hence Nishith's birthday falls on Friday.Verification / Alternative check:We can also interpret 151 as 147 plus 4. Since 147 is exactly 21 weeks, it leads back to the same weekday, leaving a net effect of 4 extra days. When these 4 days are counted forward from Monday, the weekday lands on Friday. This alternative view reinforces the earlier calculation and confirms the result.Why Other Options Are Wrong:Saturday would require a shift of 5 days from Monday, which is not supported by the remainder of 4.Wednesday corresponds to a shift of 2 days from Monday, which does not match the calculated shift.Sunday would represent either a single step backward from Monday or 6 steps forward, neither of which agrees with the 4 odd days.Common Pitfalls:Many students miscalculate the number of days in one of the months or accidentally include 19 June itself when counting. Others misinterpret the remainder when dividing by 7 and move in the wrong direction on the weekday cycle. A clear month wise breakdown and conscious forward counting from the starting weekday help avoid these traps.Final Answer:Nishith's birthday on 17 November falls on Friday.

Question 12

Arnav celebrates his birthday on Tuesday 14 March in a particular year. In the same year Pranay celebrates his birthday on 13 September. On which day of the week does Pranay's birthday fall?

Accepted Answer

Introduction / Context:Here again we have a weekday calculation within the same year. The weekday of an earlier date is known and that of a later date must be deduced. Questions of this style are common in calendar topics, building confidence with day counting, month lengths and odd day concepts, which together form the backbone of many date based reasoning problems.Given Data / Assumptions: Arnav has his birthday on Tuesday 14 March in a given year. Pranay has his birthday on 13 September in the same year. We must find the weekday for 13 September. Leap year issues do not affect the difference because both dates fall after February. Concept / Approach:The strategy is to find how many days lie between the two birthdays and then compute the remainder when that number is divided by 7. The remainder indicates how many days ahead in the weekly cycle we move from Tuesday to reach the weekday of the second date. This remainder is often called the number of odd days.Step-by-Step Solution:Step 1: Count remaining days in March after 14 March. March has 31 days, so remaining days are 31 - 14 = 17.Step 2: Add full months from April to August. April has 30 days, May 31, June 30, July 31 and August 31, for a total of 30 + 31 + 30 + 31 + 31 = 153 days.Step 3: Add days in September up to 13 September, which gives 13 days.Step 4: Total days between the birthdays equals 17 + 153 + 13 = 183 days.Step 5: Compute 183 modulo 7. Since 7 * 26 = 182, the remainder is 1. Thus there is exactly 1 odd day between the two dates. Moving forward 1 day from Tuesday lands on Wednesday. Hence Pranay's birthday falls on Wednesday.Verification / Alternative check:If we view 183 as 182 plus 1, then the 182 days correspond to 26 full weeks that do not alter the weekday. Only the extra 1 day shifts the weekday forward by one step. Starting from Tuesday and moving one step ahead reaches Wednesday. Both the modulo view and the week based view give the same conclusion.Why Other Options Are Wrong:Monday would require moving backward from Tuesday, but the calculated remainder is positive, not negative.Tuesday would be correct only if the total number of days between the birthdays were a multiple of 7, which 183 is not.Thursday would correspond to a shift of 2 days from Tuesday, conflicting with the remainder of 1.Common Pitfalls:Students may forget that March has 31 days or miscalculate one of the month lengths. Others mistakenly treat 183 as exactly divisible by 7, which it is not. Careful writing of partial sums and a deliberate modulo 7 operation ensure an error free solution for this and similar questions.Final Answer:Pranay's birthday on 13 September falls on Wednesday.

Question 13

Faiyaz celebrates his birthday on Wednesday 12 April in a certain year. In the same year Shray celebrates his birthday on 2 October. On which day of the week does Shray's birthday fall?

Accepted Answer

Introduction / Context:This calendar problem requires calculating the weekday of a date later in the year when the weekday of an earlier date is known. Mastery of such questions comes from comfort with counting days across months and using odd days to capture the net shift in weekday. These ideas are widely tested in competitive exams under the calendar or time and date topic.Given Data / Assumptions: Faiyaz has his birthday on Wednesday 12 April in a particular year. Shray has his birthday on 2 October in the same year. We must find the weekday on which 2 October falls. Whether the year is leap or non leap does not affect the calculation because both dates lie after February. Concept / Approach:We will compute the number of days between 12 April and 2 October. Dividing this count by 7 gives the number of odd days, which tells us how far forward we need to move from Wednesday to reach the weekday of 2 October. Complete weeks consisting of 7 days do not change the weekday, so only the remainder matters.Step-by-Step Solution:Step 1: Compute remaining days in April after 12 April. Since April has 30 days, remaining days are 30 - 12 = 18.Step 2: Add days in full months from May to September. May has 31 days, June 30, July 31, August 31 and September 30. The total is 31 + 30 + 31 + 31 + 30 = 153 days.Step 3: Add days in October up to 2 October, which contributes 2 days.Step 4: Total days between the two birthdays equal 18 + 153 + 2 = 173 days.Step 5: Compute 173 modulo 7. Since 7 * 24 = 168, the remainder is 5. So there are 5 odd days between the two dates. Starting from Wednesday and moving forward 5 days gives Thursday (1), Friday (2), Saturday (3), Sunday (4) and Monday (5). Therefore Shray's birthday falls on Monday.Verification / Alternative check:We can see 173 as 168 plus 5. The 168 days correspond to 24 full weeks, which bring us back to Wednesday. The extra 5 days shift us forwards by 5 positions in the weekly cycle, landing on Monday. This alternate reasoning matches the modulo calculation and confirms the result.Why Other Options Are Wrong:Saturday corresponds to a shift of 3 days from Wednesday, which is not equal to the calculated shift of 5.Wednesday would require no net shift in weekday, which would happen only if the total days were exactly divisible by 7.Sunday is 4 days ahead of Wednesday, again inconsistent with the remainder of 5.Common Pitfalls:Common errors include miscounting the days in April or forgetting that June and September have only 30 days. Some students also misapply modulo arithmetic and interpret the remainder incorrectly on the weekday circle. Writing the month wise breakdown clearly and then mapping the remainder to a forward move from Wednesday keeps the reasoning transparent and reliable.Final Answer:Shray's birthday on 2 October falls on Monday.

Question 14

Bobby celebrates his birthday on Monday 5 June in a certain year. In the same year Pinky celebrates her birthday on 22 September. On which day of the week does Pinky's birthday fall?

Accepted Answer

Introduction / Context:This is another frequent style of calendar question in which we know the weekday for one birthday and must find the weekday for another birthday in the same year. The calculation involves counting the intervening days, computing odd days and then moving along the weekday cycle. Practicing such problems builds strong intuition for date arithmetic and calendar reasoning.Given Data / Assumptions: Bobby has his birthday on Monday 5 June in a particular year. Pinky has her birthday on 22 September in the same year. We must find the weekday for 22 September. Leap year status does not matter because both dates are after February. Concept / Approach:The approach is to determine the total number of days between 5 June and 22 September. After obtaining this number, we divide it by 7 to find the remainder. That remainder gives the number of positions by which the weekday shifts forward from Monday to reach the day for Pinky's birthday. Complete weeks do not change the weekday, so only the remainder is important.Step-by-Step Solution:Step 1: Determine remaining days in June after 5 June. June has 30 days, so remaining days equal 30 - 5 = 25.Step 2: Add the full months of July and August. July has 31 days and August has 31 days, giving a total of 31 + 31 = 62 days.Step 3: Add days in September up to 22 September, contributing 22 days.Step 4: Total days between the birthdays equal 25 + 62 + 22 = 109 days.Step 5: Compute 109 modulo 7. Since 7 * 15 = 105, the remainder is 4. Therefore there are 4 odd days. Starting from Monday and moving forward 4 days gives Tuesday (1), Wednesday (2), Thursday (3) and Friday (4). Thus Pinky's birthday falls on Friday.Verification / Alternative check:We can regard 109 as 98 plus 11. The 98 days make 14 full weeks and thus cause no net change in weekday. The remaining 11 days are equivalent to 4 odd days because 11 modulo 7 equals 4. Moving 4 days ahead from Monday once again lands on Friday, confirming the earlier calculation.Why Other Options Are Wrong:Saturday corresponds to a shift of 5 days from Monday, which does not match the calculated remainder of 4.Wednesday represents only a two day shift, while our computation shows that four days of shift are needed.Sunday would require a shift of 6 days from Monday or 1 day backward, both of which conflict with the remainder of 4.Common Pitfalls:Some students treat June as 31 days or miscount the days in September. Others may divide 109 by 7 correctly but then misinterpret the remainder when mapping it onto the weekday cycle. A clear breakdown of month lengths and careful counting along the weekday sequence prevents such mistakes.Final Answer:Pinky's birthday on 22 September falls on Friday.

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