pipe a can fill a tank in 60 min . there is a second pipe in the bottom of the cistern to empty it . if all the two pipes are simultaneously opened , then the cistern is full in 180 min . in how much time , the second pipe alone can empty the cistern ?
work done by the third pipe in 1 min = 1 / 180 - ( 1 / 60 ) = - 1 / 45 . [ - ve sign means emptying ] the third pipe alone can empty the cistern in 45 min . answer : b
inverse(add(divide(const_1, 180), divide(const_1, 60)))