Let's denote the amount of liquid in the first vat as \( V_1 \) and the amount in the second vat as \( V_2 \).
Initially, we have:
- \( V_1 = 121212 \) gallons (currently filling)
- \( V_2 = 100100100 \) gallons (currently leaking)
The first vat is being filled at a rate of \( 333 \) gallons per minute, meaning that after \( t \) minutes, the volume of liquid in the first vat will be: \[ V_1(t) = 121212 + 333t \]
The second vat is leaking at a rate of \( 777 \) gallons per minute, so after \( t \) minutes, the volume of liquid in the second vat will be: \[ V_2(t) = 100100100 - 777t \]
We want to find the time \( t \) when both vats have the same amount of liquid, i.e., when \( V_1(t) = V_2(t) \): \[ 121212 + 333t = 100100100 - 777t \]
Now, we'll solve this equation for \( t \):
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Rearranging the equation: \[ 333t + 777t = 100100100 - 121212 \]
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Combine the terms: \[ 1110t = 100100100 - 121212 \]
Calculating the right-hand side: \[ 100100100 - 121212 = 100100100 - 121212 = 100100100 - 121212 = 100100100 - 121212 = 99978888 \]
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Now, substitute back into the equation: \[ 1110t = 99978888 \]
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Solve for \( t \): \[ t = \frac{99978888}{1110} \]
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Performing the division: \[ t \approx 90099.036036 \]
This is in minutes. To convert that to seconds, we multiply by \( 60 \): \[ t_{seconds} \approx 90099.036036 \times 60 \approx 5405942.1616 \]
So, rounding to the nearest whole number: \[ t_{seconds} \approx 5405942 \]
Thus, both vats will have the same amount of liquid after approximately 5405942 seconds.