A works 25%faster than B,who is 33.33%faster than C. B takes 10 more days than A,to complete a piece of work. C works for 11 day and A works for 19 days then in how many days should B work to complete the remaining work????

18 days

let say C do 90 units/day 1/3 which 33.33% of 90 is 30 means B do 30+90 = 120 units/day and A do 1/4 which is 25% of 120 that is 30 means A do 30+120 = 150 units/day so, A = 150 units/day B = 120 units/day C = 90 units /day now , if A can complete the job in let say 40 days the he will do 6000 uints and same work will be done by B in 50 days , which satisfy first condition now , if C work for 11 days means 90*11 = 990 units A work for 19 days means 120*19 = 2850 units so out of 6000 , 3840 units is completed remaining 2160 work will be done by B so number of days = 2160/120 = 18

Try asking Bakugo....!! The answer is 18, as Paras explained. Else use the unitary method to solve. (Consider efficiencies , work done = efficiency * days)

18