To find out after how many seconds both vats will have the same amount of liquid, we can set up equations for the amount of liquid in each vat over time.
Let's denote \( t \) as the time in minutes.
For the first vat:
It starts with 0 gallons and is being filled at a rate of 1.5 gallons per minute.
The amount of liquid in the first vat after \( t \) minutes is:
\[
V_1(t) = 1.5t
\]
For the second vat:
It starts with 120 gallons and is leaking at a rate of 2.5 gallons per minute.
The amount of liquid in the second vat after \( t \) minutes is:
\[
V_2(t) = 120 - 2.5t
\]
We want to find the time \( t \) when both vats have the same amount of liquid, so we set the two equations equal to each other: \[ 1.5t = 120 - 2.5t \]
Now, let's solve for \( t \): \[ 1.5t + 2.5t = 120 \] \[ 4t = 120 \] \[ t = \frac{120}{4} = 30 \text{ minutes} \]
Now, we want to express this in seconds. Since there are 60 seconds in a minute: \[ 30 \text{ minutes} = 30 \times 60 = 1800 \text{ seconds} \]
Thus, the answer is:
30 minutes
1800 seconds