pipe a that can fill a tank in an hour and pipe b that can fill the tank in 15 are opened simultaneously when the tank is empty . pipe b is shut 15 minutes before the tank overflows . when will the tank overflow ?
the last 15 minutes only pipe a was open . since it needs 1 hour to fill the tank , then in 15 minutes it fills 1 / 4 th of the tank , thus 3 / 4 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 4 = 5 tanks / hour , therefore to fill 3 / 4 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 3 / 4 ) / 5 = 3 / 20 hours = 15 minutes . total time = 15 + 15 = 30 minutes . answer : c
add(multiply(divide(divide(const_3, const_4), const_3), const_60), 15)