BPSC 70th CCE Re-exam Prelims General Studies · Question 105 of 142

Correct — D, Thursday. Count the days between the two dates, not the dates themselves. From 1 January 2008 to 1 January 2009 the calendar passes through the whole of 2008, and 2008 is a leap year: it is divisible by 4 and is not a century year, so February had 29 days and the year ran to 366 days. Divide by seven and the remainder is what matters — 366 = 52 × 7 + 2, so the year contributes 2 odd days. The weekday therefore advances by two: Tuesday + 2 = Thursday. Written out, 1 January 2008 was a Tuesday, so 30 December 2008 (day 365) was a Tuesday again and 1 January 2009 fell two weekdays later, on a Thursday — which is what the calendar actually shows. The structure worth carrying away is that any date repeats one weekday later after an ordinary year (365 = 52 weeks + 1 odd day) and two weekdays later after a span containing a 29 February. Notice also what the question does not require: you never need to know that 1 January 2008 was in fact a Tuesday, because the stem supplies it. Everything after that is division by seven, and a candidate who starts hunting for a remembered calendar has already spent more time than the mark is worth.

• (a)Wednesday — The answer you get by adding only one odd day — that is, by treating 2008 as an ordinary 365-day year. It is the single most common error on this question type, and it is caught by one check: does the span from the first date to the second contain a 29 February? Here it does, so the shift is two days, not one.
• (b)Monday — Tuesday minus one — the result of counting backwards instead of forwards. Going from 1 January 2008 to 1 January 2009 moves forward in time, so the weekday must advance; Monday would be the answer to the reverse question, what day was 1 January 2007, and even then only if 2007 were a leap year, which it is not.
• (c)Sunday — Tuesday minus two — the same backwards error as (b), with the leap-year shift applied in the wrong direction. Anyone who subtracts the two odd days rather than adding them lands here, which is why the option is on the page at all.

Calendar questions are solved by odd days — the remainder left when a number of days is divided by seven. An ordinary year has 365 days, and 365 = 52 × 7 + 1, so it carries 1 odd day; a leap year has 366 days and carries 2. The leap-year rule is: a year is a leap year if it is divisible by 4, except that a century year must be divisible by 400. So 1996, 2004, 2008, 2012 and 2000 are leap years, while 1900, 2100 and 2200 are not. That century exception has real consequences — 1 January 2100 falls on a Friday, not on the Saturday a naive four-year rule would give. Building upward, a century contains 24 leap years and 76 ordinary years, giving 5 odd days; 200 years give 3, 300 give 1, and 400 years give 0, which is why the Gregorian calendar repeats exactly every 400 years. Month codes matter for shorter spans: January has 31 days and so contributes 3 odd days, February 0 in an ordinary year and 1 in a leap year, April 2, and so on.

The reasoning route is mechanical if it is done in the right order. First, identify the span — here exactly one year, so the only question is what kind of year lies inside it. Second, test for leap: 2008 ÷ 4 = 502 with no remainder and 2008 is not a century year, so it is a leap year. Third, convert the span to odd days: 366 mod 7 = 2. Fourth, add, never subtract, when moving forward: Tuesday + 2 = Thursday. The single fact that discriminates between the two serious options, Wednesday and Thursday, is whether 29 February 2008 lies inside the interval — and it does, because the interval starts on 1 January 2008 and ends after February. Change the anchor date and the answer changes with it: 1 March 2008 to 1 March 2009 is only 365 days, because the leap day falls before the start of that span, so that shift would be one day, not two. Examiners set exactly this trap by choosing dates on either side of February, so read the two dates before reaching for the leap-year rule.

• An ordinary year of 365 days carries 1 odd day (365 = 52 × 7 + 1); a leap year of 366 days carries 2 odd days
• Leap-year rule: divisible by 4, except century years, which must be divisible by 400 — so 2000 was a leap year but 1900 and 2100 are not
• 2008 was a leap year, so 1 January 2009 fell two weekdays after 1 January 2008 — Tuesday to Thursday
• The leap day only counts if 29 February falls inside the span: 1 January 2008 to 1 January 2009 is 366 days, but 1 March 2008 to 1 March 2009 is 365 days and shifts by one day only
• Odd days accumulate as 5 in 100 years, 3 in 200, 1 in 300 and 0 in 400, which is why the Gregorian calendar repeats every 400 years

• Adding one odd day instead of two by forgetting that the span contains a 29 February
• Subtracting instead of adding — moving forward in time always advances the weekday
• Applying the leap rule mechanically to century years; 1900 and 2100 are not leap years because they are not divisible by 400

BPSC puts two or three pure reasoning items of this kind in every paper and keeps them one step deep — one span, one leap-year test, one addition — so they are the cheapest marks on the paper if the odd-days method is automatic. UPSC does not ask calendars in the General Studies prelims at all; the same reasoning appears in the CSAT paper, where it is usually embedded in a longer data or arrangement problem rather than asked as a bare date.

• practice — not a real PYQ

  If 1 March 2008 was a Saturday, what day was 1 March 2009 ?

  • (a)Saturday
  • (b)Sunday
  • (c)Monday
  • (d)Friday

  Answer(b) Sunday — the span 1 March 2008 to 1 March 2009 contains no 29 February (the 2008 leap day falls before it), so it is 365 days = 1 odd day, and Saturday + 1 = Sunday.

• practice — not a real PYQ

  Which of the following years is NOT a leap year ?

  • (a)1600
  • (b)1900
  • (c)2000
  • (d)2400

  Answer(b) 1900 — a century year is a leap year only if it is divisible by 400. 1600, 2000 and 2400 satisfy that test; 1900 does not.