First pipe fill the tank in 1 hour = 112part of tankSecond pipe fill the tank in 1 hour = 116part of tankThird pipe empty the tank in 1 hour = 130part of tankWhen all three pipes are opened simultaneously,part of the tank filled in 1 hour=112+116−130LCM of 12, 16, and 30 = 240 = 20+15−8240=27240∴Required time taken by all the three pipes = 24027=809=889Hours
Rate of flow of water = x cm/minutes∴Valume of water that flowed in the in1 minutes = 5×4 ×x = 20xcu.cm∴Valume of water that flowed in the in6 hours 18 minutes i.e. (6×60+18)=378min= 2x×378 cu.cmAccording to question, 20x×378=700×400×450⇒x=(700×400×45020×378)cm /minutes⇒x=(700×400×450×60100000×20×378)km/hours⇒x = 10 km/hours
(A + B)'s 1 hour work = (112+115)=960=320(A + C)'s 1 hour work = (112+120)=860=215Part filled in 2 hrs = (320+215)=1760,Part filled in 6 hrs = (3×1760)=1720Remaining part = (1−1720)=320Now it is the turn of A and B and320 part is filled by A and B in 1 hour.∴Total time taken to fill tank = (6+1)hrs = 7 hrs
Part filled by (A + B) in 1 hour = (120+130)=112So, A and B together can fill thetank in 12 hrs, 13 part is filled by(A + B) in (13×12) = 4 hrsSince the leak empties one - third water,so time taken to fill the tank = Time taken by (A + B) to fill the whole tank + Time taken by (A + B) to fill one - third tank = (12 + 4) = 16 hrs.