Two pipes A and B can fill a cistern in     minutes and 35 minutes respectively. Both pipes are opened. The cistern will be filled in just 25 minutes, if the pipe B is turned off after?

Solution

Let the pipe B be turned off after x minutes. Then, Part filled by (A + B) in x minutes + Part filled by A in (25 - x) minutes = 1 ∴ x (2/55- 1/35) + (25 - x) ×  2/55 =1 x/35 = 1 - 50/55  ∴ x = 35/11  or 3*2/11 minutes  Alternate method: Let total capacity of cistern = LCM(55/2, 35 ) = 385 L So A can fill in 1 min  =  385/27.5 = 14L & B can fill in 1 min  = 385/35 = 11L Now B is left in between , it means A is opened all time So in 25 min, A can fill  = 14×25 = 350 So remaining tank = 385 -350 = 35L So B can do it in = 35/11 = 3*2/11min So B is turned off  =  3*2/11min