Answer to Question #259698 in Calculus for Sha

Question

A tank can be filled by a pipe in 18 hours. Five hours after this pipe is opened, it is supplemented by a smaller pipe, which, by itself can fill the tank in 22 hours. Simultaneously, a drainpipe is opened, which, by itself, can completely empty the tank in 30 hours. Find the total time, measured from the opening of the larger pipe, to completely fill the tank.

Answer 1

Let $V=$ the volume of a tank. Then the first pipe filling speed is $V/18,$ the second pipe filling speed is $V/22,$ and the pipe emptying speed is$V/30.$

Let $t=$ the total time, measured from the opening of the larger pipe, to completely fill the tank.

Then

$$\dfrac{V}{18}t+\dfrac{V}{22}(t-5)-\dfrac{V}{30}(t-5)=V$$

$$55t+45(t-5)-33(t-5)=990$$

$$67t=1050$$

$$t=\dfrac{1050}{67}\ h$$

$$t=15h\ 40min\ 18sec$$