Question 1205982
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There are a few approaches we could take. I'll discuss two methods.


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Method 1


Let's consider a tank that is 12 gallons. 
Feel free to pick any positive number you want since it turns out the capacity doesn't matter. 
I'm picking 12 because it is the LCM of 3 and 4.


Pump A fills the 12 gallon tank in 4 hours when working alone.
The unit rate is 12/4 = 3 gallons per hour.
Formula:  rate = (amount done)/time


Pump B fills the 12 gallon tank in 3 hours when working alone.
Unit rate = 12/3 = 4 gallons per hour.


When both pumps work together, neither pump hindering the other, their combined rate is 3+4 = 7 gallons per hour.


Then we'll use this formula
time = (amount done)/rate
to determine that <font color=red>12/7 hours</font> is the amount of time it takes when both pumps work together.


Side notes:<ul><li>12/7 = 1.71429 (approximate)</li><li>1.71429 hrs = 102.8574 minutes (Multiply by 60 to convert from hours to minutes)</li><li>102.8574 min = 60 min + 42.8574 min = 1 hr + 42.8574 min</li></ul>

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Method 2


Pump A does 1 job in 4 hours when working alone.
"1 job" is defined in this case as "filling the entire tank with oil".
It's unit rate is 1/4 of a job per hour.


Pump B has a unit rate of 1/3 of a job per hour through similar logic.


Combined rate = 1/4 + 1/3 = 3/12 + 4/12 = 7/12 of a job per hour


x = number of hours to do 1 job if both pumps work together
rate*time = amount done
(7/12 of a job per hour)*(x hours) = 1 job
(7/12)x = 1
x = <font color=red>12/7 hours</font>
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