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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
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