Question

There are two different vats. The first vat has 20 gallons is being filled at a rate of 3 gallons per minute. The second vat has 108 gallons and is leaking fluid at the rate of 2 gallons per minute. After how many seconds will both vats have the same amount of liquid?

Answer 1

To determine when both vats will have the same amount of liquid, we can create equations for the amount of liquid in each vat over time.

### First Vat:
- **Initial Amount**: 20 gallons
- **Filling Rate**: 3 gallons per minute

Let \( t \) be the time in minutes. The amount of liquid in the first vat after \( t \) minutes can be expressed as:
\[
V_1(t) = 20 + 3t
\]

### Second Vat:
- **Initial Amount**: 108 gallons
- **Leaking Rate**: 2 gallons per minute

The amount of liquid in the second vat after \( t \) minutes is:
\[
V_2(t) = 108 - 2t
\]

### Setting the Amounts Equal:
To find when the amounts are equal, we set \( V_1(t) \) equal to \( V_2(t) \):
\[
20 + 3t = 108 - 2t
\]

### Solving for \( t \):
1. Rearranging the equation:
\[
3t + 2t = 108 - 20
\]
\[
5t = 88
\]
\[
t = \frac{88}{5} = 17.6 \text{ minutes}
\]

### Converting Minutes to Seconds:
Since we need the answer in seconds, we convert \( 17.6 \) minutes to seconds:
\[
17.6 \text{ minutes} = 17.6 \times 60 = 1056 \text{ seconds}
\]

Thus, the time at which both vats will have the same amount of liquid is:
\[
\boxed{1056} \text{ seconds}
\]