SOLUTION: A tank can be filled by a pipe in 5.0 h and emptied by another pipe in 6.0 h. How much time will be required to fill an empty tank if both are running?

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Question 1098065: A tank can be filled by a pipe in 5.0 h and emptied by another pipe in 6.0 h. How much time will be required to fill an empty tank if both are running?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
tank can be filled in 5 hours using one pipe.
same tank can be emptied in 6 hours using another pipe.

rate * time = quantity.

quantity equals 1 full tank.

rate * time = 1

when filing the tank, time = 5 hours, therefore rate = 1/5 of the tank in one hour.

when emptying the tank, time = 6 hours, therefore rate = 1/6 of the tank in one hour.

the filling pipe fills the tank at the rate of 1/5 of the tank in one hour.

the emptying pipe empties the tank at the rate of 1/6 of the tank in one hour.

when they are both runjning, they oppose each other, therefore the combined rate is fill rate minus the empty rate.

(1/5 - 1/6) * T = 1 is the combined equation when they are both running.

simplify by placing the fractions under the saem denominator to get:

(6/30 - 5/30) * T = 1

combined rate becomes 1/30, therefore 1/30 * T = 1

solve for T to get T = 30 hours.

it would take 30 hours to fill the tank if both the filling pipe and the emptying pipe are tunning at the same time.

in 30 hours, the amount of water poured into the tank is equal to 1/5 * 30 = 6 times the capacity of the tank.

in 30 hours, the amount of water empties out of the tank is equal to 1/6 * 30 = 5 times the capacity of the tank.

you are left with the capacity of the tank which is 1 full tank.

to place this into gallons of water, assume 1 full tank is equal to 600 gallons.

the tank is filled in 5 hours, therefore rate * time = quantity becomes rate * 5 = 600 and you get a fill rate of 120 gallons per hour.

the tank is emptied in 6 hours, therefore rate * time = quantity becomes rate * 6 = 600 and you get an empty rate of 100 gallons per hour.

when both pipes are turned on at the same time, you get a fill rate of 120 gallons per hour and an empty rate of 100 gallons per hour which give you a net fill rate of 20 gallons per hour.

at 20 gallons per hour net fill rate, it will take 30 hours to fill the tank to 600 gallon capacity.

you get the same answer except we're using 600 gallons as the quantity rather than 1 full tank as quantity.


Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
The filling pipe adds 1%2F5 of the tank volume per hour.


The drainage pipe takes out 1%2F6 of the tank volume per hour.


When both pipes work together, the tank receives 1%2F5 - 1%2F6 of the volume per hour.


1%2F5 - 1%2F6 = 6%2F30-5%2F30 = 1%2F30.


So, the tank receives 1%2F30 of its volume per hour.


Hence, it will require 30 hour to fill the tank under the given condition.


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It is a typical joint work problem.

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


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Also,  you have this free of charge online textbook in ALGEBRA-I in this site
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The referred lessons are the part of this textbook under the topic
"Rate of work and joint work problems"  of the section  "Word problems".


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