Two pumps are being filled by two different pumps. The first vat has 888 gallons is being filled at a rate of 161616 gallons per second. The second vat has 383838 gallons and is being filled at a rate of 141414 gallons per second. After how many seconds will both vats have the same amount of liquid?

To determine after how many seconds both vats will have the same amount of liquid, we can denote the time in seconds as \( t \).

1. **First vat**:
- Initial amount: 888 gallons
- Filling rate: 161616 gallons per second
- Amount after \( t \) seconds: \( 888 + 161616t \)

2. **Second vat**:
- Initial amount: 383838 gallons
- Filling rate: 141414 gallons per second
- Amount after \( t \) seconds: \( 383838 + 141414t \)

To find when both vats have the same amount of liquid, we set the two expressions equal to each other:

\[
888 + 161616t = 383838 + 141414t
\]

Next, we rearrange the equation to isolate \( t \):

Subtract \( 141414t \) from both sides:

\[
888 + 161616t - 141414t = 383838
\]

Combine like terms:

\[
888 + (161616 - 141414)t = 383838
\]
\[
888 + 20102t = 383838
\]

Now, subtract 888 from both sides:

\[
20102t = 383838 - 888
\]
\[
20102t = 382950
\]

Next, solve for \( t \) by dividing both sides by 20102:

\[
t = \frac{382950}{20102}
\]

Perform the division:

\[
t \approx 19.04
\]

Since time is typically expressed in whole seconds, we can either round to 19 seconds or check both 19 seconds and 20 seconds for precision.

**Calculate amounts at \( t = 19 \) and \( t = 20 \)**:

1. For \( t = 19 \):

- First vat:
\[
888 + 161616 \times 19 = 888 + 3060704 = 3061592
\]

- Second vat:
\[
383838 + 141414 \times 19 = 383838 + 2686846 = 3060684
\]

**They are not equal.**

2. For \( t = 20 \):

- First vat:
\[
888 + 161616 \times 20 = 888 + 3232320 = 3233208
\]

- Second vat:
\[
383838 + 141414 \times 20 = 383838 + 2828280 = 3212118
\]

**Again they are not equal.**

Finding the exact second where they match is challenging without decimals, but the calculation indicates they will meet between 19 and 20 seconds.

Thus, to conclude:
\[
\text{They will have the same amount of liquid at approximately } 19.04 \text{ seconds.}
\]