SOLUTION: With 2 inflowing pipes open, a water tank can be filled in four hours. if the larger pipe can fill the tank alone in 7 hours, how long would the smaller pipe take to fill the tank?

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Question 1190160: With 2 inflowing pipes open, a water tank can be filled in four hours. if the larger pipe can fill the tank alone in 7 hours, how long would the smaller pipe take to fill the tank?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.
With 2 inflowing pipes open, a water tank can be filled in four hours.
if the larger pipe can fill the tank alone in 7 hours, how long would
the smaller pipe take to fill the tank?
~~~~~~~~~~~~~~~ 

Two pipes, working together, fill    of the tank volume per hour.


One larger pipe, working alone, fills    of the tank volume per hour.


Hence, one smaller pipe, working alone, fills   =  =  of the tank volume per hour.


It means that one smaller pipe, working alone, will fill the tank in  hours = 9  hours = 9 hours and 20 minutes.

Solved.

---------------

It is a standard and typical joint work problem.

There is a wide variety of similar solved joint-work problems with detailed explanations in this site.  See the lessons
    - Using Fractions to solve word problems on joint work
    - Solving more complicated word problems on joint work
    - Selected joint-work word problems from the archive


Read them and get be trained in solving joint-work problems.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic
"Rate of work and joint work problems"  of the section  "Word problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



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


The other tutor showed a solution using the standard algebraic method. Here is an alternative method.

Consider the least common multiple of the two times, which is 28 hours.

In 28 hours, the two pipes together could fill 28/4=7 of those tanks; in 28 hours the large pipe alone could fill 28/4=7 of those tanks.

That means in 28 hours the smaller pipe could fill 7-4=3 of those tanks; and that in turn means it would take the smaller pipe 28/3 hours, or 9 1/3 hours, to fill the one tank by itself.

ANSWER: 9 1/3 hours
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