A takes 25 days more than that taken by A and B together to do a work, while B takes 9 days more than that taken by A and B together to do the work. In how many days can A complete the work if he is assisted by B on every alternate day starting from the second day?

If A+B take T days, A takes T+25 and B takes T+9. 1/(T+25) + 1/(T+9) = 1/T. Solving gives T^2 = 225, so T = 15. A takes 40 days, B takes 24 days. Working alternate days: Day 1 (A): 1/40. Day 2 (A+B): 1/40 + 1/24 = 3/120 + 5/120 = 8/120 = 1/15. In 2 days, 1/40 + 1/15 = 11/120. In 20 days (10 cycles), 110/120 done. Remaining 10/120 = 1/12. Day 21 (A): 1/40. Remaining 1/12 - 1/40 = 7/120. Day 22 (A+B): 1/15. Remaining 7/120 - 8/120 (negative). Time is 21 + (7/120)/(8/120) = 21 + 7/8.