Calendar Practice Questions

Over 3 years from 1 April there are normally 3 ordinary-year spans → 3 odd days (this assumes no 29 February falls in between, e.g. when the current year is itself the leap year). Wednesday + 3 = Saturday. (If a 29 February did fall within the 3 years, it would be 4 odd days → Sunday; review the key against your source.)

350 days = 50 weeks + 0 odd day(s). Weekdays repeat every 7 days, so only the 0 odd day(s) matter: Saturday + 0 = Saturday.

Count the odd days: from 10 Nov 2006 to 10 Nov 2007: 1 year(s), of which 0 contain 29 Feb → odd days = 1×1 + 0×2 = 1; then 10 Nov 2007 to 24 Nov 2007: 14 days → 0 odd day(s). Total odd days = 15 mod 7 = 1. Since we go backward in time, move 1 day(s) backward from Thursday → Wednesday. (Note: on the real calendar 24 Nov 2007 was a Saturday; the answer follows the day given in the question.)

Count the odd days: from 18 Apr 1603 to 18 Apr 2003: 400 year(s), of which 97 contain 29 Feb → odd days = 303×1 + 97×2 = 497. Total odd days = 497 mod 7 = 0. Since we go forward in time, move 0 day(s) forward from Thursday → Thursday. (Note: on the real calendar 18 Apr 1603 was a Friday; the answer follows the day given in the question.)

31-day month starting Friday: Saturdays = 2, 9, 16, 23, 30; Sundays = 3, 10, 17, 24, 31 (5 Sundays). Holidays = 2nd & 4th Saturdays (9th, 23rd) + 5 Sundays = 7. Working days = 31 − 7 = 24.