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.
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