SOLUTION: A large pipe can fill a tank 4 times as rapidly as a small pipe. When both pipes are operating, they require 4 hours to fill the tank. How many hours would the small pipe requir

Algebra ->  Inverses -> SOLUTION: A large pipe can fill a tank 4 times as rapidly as a small pipe. When both pipes are operating, they require 4 hours to fill the tank. How many hours would the small pipe requir      Log On


   

Question 1163865: A large pipe can fill a tank 4 times as rapidly as a
small pipe. When both pipes are operating, they require
4 hours to fill the tank. How many hours would the small
pipe require to fill the tank operating alone?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.

When both pipes are operating, they are equivalent to 4+1 = 5 working small pipes.


Hence, 5 small pipes can fill the tank in 4 hours.


It means that one small pipe will do it in 4*5 = 20 hours.    ANSWER

Solved.



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Tutor @ikleyn showed an excellent informal common sense solution that gets you to the answer quickly.

Here is a variation of exactly the same method -- same logical reasoning, but different calculations.

The large pipe works 4 times as fast as the small pipe; so when the two together fill the pipe in 4 hours, the large pipe does 4/5 of the work and the small pipe does 1/5 of the work.

Since the small pipe does 1/5 of the work in 4 hours, the number of hours it would take the small pipe alone to do the job is 4*5 = 20 hours.