two pipes a and b can separately fill a cistern in 60 min and 75 min respectively . there is a third pipe in the bottom of the cistern to empty it . if all the 3 pipes are simultaneously opened , then the cistern is full in 50 min . in how much time , the third pipe alone can empty the cistern ?
work done by the third pipe in 1 min = 1 / 50 - ( 1 / 60 + 1 / 75 ) = - 1 / 100 . [ - ve sign means emptying ] the third pipe alone can empty the cistern in 100 min . answer : b
inverse(subtract(add(divide(const_1, 60), divide(const_1, 75)), divide(const_1, 50)))