SOLUTION: "At noon, a pump is turned on to fill an empty pool. Normally, the pump would fill the pool in 12 hours, but at 12:00 P.M, a valve is accidentally opened that could drain a full po

Algebra ->  Rate-of-work-word-problems -> SOLUTION: "At noon, a pump is turned on to fill an empty pool. Normally, the pump would fill the pool in 12 hours, but at 12:00 P.M, a valve is accidentally opened that could drain a full po      Log On


   

Question 1131119: "At noon, a pump is turned on to fill an empty pool. Normally, the pump would fill the pool in 12 hours, but at 12:00 P.M, a valve is accidentally opened that could drain a full pool in 20 hours. If the valve remains open, at what time will the pool be full?"
This was a question I had on my Algebra 2 test and I got punched in the face with this question..., I wasn't able to fill in the chart at all, but if I'm right the chart should be a rate x time = distance (work) chart. I really have no idea how to fill in the chart...
Found 3 solutions by josgarithmetic, stanbon, ikleyn:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
Pump Filling Rate 1%2F12
valve FILLING Rate -1%2F20
Pump&Valve Together 1%2F12-1%2F20=%285%2F5%29%281%2F12%29-%283%2F3%29%281%2F20%29=2%2F60=1%2F30

The combined filling rate for pump and valve together is ONE pool in 30 hours.
Notice, the valve filling rate is NEGATIVE.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
"At noon, a pump is turned on to fill an empty pool. Normally, the pump would fill the pool in 12 hours, but at 12:00 P.M, a valve is accidentally opened that could drain a full pool in 20 hours. If the valve remains open, at what time will the pool be full?"
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1st pump data:
rate = 1/12 pool/hr
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2nd pump data:
rate = -1/20 pool/hr
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Rate at which the pool is filled if both are pumps are open:: 1/12 - 1/20
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Let the time to fill the pool be x-hours.
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x(1/12 - 1/20) = 1 pool
x((20-12)/(12*20)) = 1
x(8/240) = 1
x(1/30) = 1
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Ans:: Time to fill the pool = 30 hours.
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Cheers,
Stan H.
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Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.
The pump fills  1%2F12  of the pool volume per hour.  It is its rate of filling.


The valve drains  1%2F20  of the pool volume per hour.  It is its rate of draining.


If both the pump and the valve work together, then the net inflow rate into the pool is the difference


    1%2F12 - 1%2F20 = 5%2F60 - 3%2F60 = 2%2F60 = 1%2F30   of the pool volume per hour.


It means that the pool will be filled in 30 hours, if both the pump and the valve work simultaneously.

Solved.

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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.