pipe a that can fill a tank in an hour and pipe b that can fill the tank in half an hour are opened simultaneously when the tank is empty . pipe b is shut 10 minutes before the tank overflows . when will the tank overflow ?
the last 10 minutes only pipe a was open . since it needs 1 hour to fill the tank , then in 10 minutes it fills 1 / 5 th of the tank , thus 5 / 6 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 5 / 6 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 5 / 6 ) / 3 = 5 / 18 hours = 27 minutes . total time = 10 + 27 = 37 minutes . answer : a .
subtract(add(divide(multiply(add(divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), divide(multiply(const_1, const_60), const_2)), divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), multiply(const_1, const_60))), 10), divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), multiply(const_1, const_60))), 10), const_3)