Question #259698

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.


1
Expert's answer
2021-11-02T11:37:31-0400

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

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

Then


V18t+V22(t5)V30(t5)=V\dfrac{V}{18}t+\dfrac{V}{22}(t-5)-\dfrac{V}{30}(t-5)=V

55t+45(t5)33(t5)=99055t+45(t-5)-33(t-5)=990

67t=105067t=1050

t=105067 ht=\dfrac{1050}{67}\ h

t=15h 40min 18sect=15h\ 40min\ 18sec






Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS