To find the amount of water in Tank A after \( t \) minutes, we start with the initial amount of water and subtract the amount lost over time due to the leak.
Tank A starts with 280 liters of water and loses 2 liters per minute. Therefore, the expression for the amount of water in Tank A after \( t \) minutes is:
\[ A(t) = 280 - 2t \]
where \( A(t) \) represents the amount of water in Tank A after \( t \) minutes.
Next, we can write a similar expression for the amount of water in Tank B after \( t \) minutes. Tank B starts with 390 liters of water and loses 12 liters per minute. The expression for the amount of water in Tank B after \( t \) minutes is:
\[ B(t) = 390 - 12t \]
where \( B(t) \) represents the amount of water in Tank B after \( t \) minutes.
To find when the two tanks have the same amount of water, we set the two expressions equal to each other:
\[ 280 - 2t = 390 - 12t \]
Now, we can solve for \( t \):
- First, isolate the terms involving \( t \):
\[ 12t - 2t = 390 - 280 \]
- Simplify:
\[ 10t = 110 \]
- Divide both sides by 10:
\[ t = 11 \]
Now that we have \( t = 11 \), we can find the amount of water in either tank at that time. We'll use Tank A's expression:
\[ A(11) = 280 - 2(11) = 280 - 22 = 258 \text{ liters} \]
We can also check Tank B:
\[ B(11) = 390 - 12(11) = 390 - 132 = 258 \text{ liters} \]
Thus, after 11 minutes, both tanks will have the same amount of water, which is 258 liters.
In summary:
- The two tanks will have the same amount of water after 11 minutes.
- At that time, each tank will contain 258 liters of water.
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