Two pumps are being filled by two different pumps. The first vat has 16 gallons is being filled at a rate of 12 gallons per second. The second vat has 23 gallons and is being filled at a rate of 10 gallons per second. After how many seconds will both vats have the same amount of liquid?

Let \( t \) be the number of seconds after the filling begins.

For the first vat:
- Initial amount = 16 gallons
- Rate of filling = 12 gallons/second

The amount of liquid in the first vat after \( t \) seconds will be:
\[
\text{Amount in first vat} = 16 + 12t
\]

For the second vat:
- Initial amount = 23 gallons
- Rate of filling = 10 gallons/second

The amount of liquid in the second vat after \( t \) seconds will be:
\[
\text{Amount in second vat} = 23 + 10t
\]

We are looking for the time \( t \) when both vats have the same amount of liquid:
\[
16 + 12t = 23 + 10t
\]

Now, let's solve for \( t \).
1. Subtract \( 10t \) from both sides:
\[
16 + 2t = 23
\]
2. Subtract 16 from both sides:
\[
2t = 23 - 16
\]
\[
2t = 7
\]
3. Divide by 2:
\[
t = \frac{7}{2} = 3.5
\]

Thus, the time \( t \) when both vats have the same amount of liquid is:
\[
\boxed{3.5} \text{ seconds}
\]