SOLUTION: Container A has 200 L of water, and is being filled at a rate of 6 liters per minute. Container B has 500 L of water, and is being drained at 6 liters per minute. How many minutes

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Question 1205919: Container A has 200 L of water, and is being filled at a rate of 6 liters per minute. Container B has 500 L of water, and is being drained at 6 liters per minute. How many minutes will it take for the two containers to have the same amount of water?
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let the number of min be t
200%2B6t=500-6t
6t%2B6t=500-200
12t=300
t=300%2F12
t=25
it will take 25 minutes for the two containers to have the same amount of water


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


Informally....

Currently, the difference in the amounts of water in the two containers is 500-200 = 300L.

The amount of water in one container is increasing at the rate of 6L/min; the amount in the other is decreasing at the rate of 6L/min. The combined rate of change in the amounts of water in the two containers is 12L/min.

The number of minutes it takes for the two containers to have the same amount of water is 300/12 = 25.

ANSWER: 25

While a formal algebraic solution like the one provided by the other tutor was probably what the student wanted, it should be noted that solving the problem informally using logical reasoning and simple arithmetic is very good mental exercise.