There are two different vats. The first vat has 121212 gallons is being filled at a rate of 333 gallons per minute. The second vat has 100100100 gallons and is leaking fluid at a rate of 777 gallons per minute. After how many seconds will both vats have the same amount of liquid?

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 \( t \):
1. Rearranging the equation:
\[
333t + 777t = 100100100 - 121212
\]

2. Combine the terms:
\[
1110t = 100100100 - 121212
\]

Calculating the right-hand side:
\[
100100100 - 121212 = 100100100 - 121212 = 100100100 - 121212 = 100100100 - 121212 = 99978888
\]

3. Now, substitute back into the equation:
\[
1110t = 99978888
\]

4. Solve for \( t \):
\[
t = \frac{99978888}{1110}
\]

5. 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**.

minutes

Let’s go through the calculations again, focusing on finding the time in minutes.

We already set up the equations for the volumes in the first and second vats:
- Volume in the first vat after \( t \) minutes:
\[
V_1(t) = 121212 + 333t
\]
- Volume in the second vat after \( t \) minutes:
\[
V_2(t) = 100100100 - 777t
\]

We want to find the time \( t \) when \( V_1(t) = V_2(t) \):
\[
121212 + 333t = 100100100 - 777t
\]

Rearranging the equation:
\[
333t + 777t = 100100100 - 121212
\]
\[
1110t = 100100100 - 121212
\]

Calculating the right-hand side:
\[
100100100 - 121212 = 99978888
\]

Now we have:
\[
1110t = 99978888
\]
To solve for \( t \):
\[
t = \frac{99978888}{1110}
\]

Calculating this gives:
\[
t \approx 90099.036036
\]

Now, converting to the nearest whole number:
\[
t \approx 90099 \text{ minutes}
\]

Therefore, both vats will have the same amount of liquid after approximately **90099 minutes**.