Question

Two pumps are being filled by two different pumps. The first vat has 121212 gallons is being filled at a rate of 191919 gallons per second. The second vat has 404040 gallons and is being filled at a rate of 151515 gallons per second. After how many seconds will both vats have the same amount of liquid?

Answer 1

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

First Vat

The first vat starts with \( 121212 \) gallons and is being filled at a rate of \( 191919 \) gallons per second. Thus, the amount of liquid in the first vat after \( t \) seconds can be represented as:

\[ \text{Amount in first vat} = 121212 + 191919t \]

Second Vat

The second vat starts with \( 404040 \) gallons and is being filled at a rate of \( 151515 \) gallons per second. The amount of liquid in the second vat after \( t \) seconds can be represented as:

\[ \text{Amount in second vat} = 404040 + 151515t \]

Setting the Amounts Equal

We need to find \( t \) such that both vats contain the same amount of liquid:

\[ 121212 + 191919t = 404040 + 151515t \]

Solving for \( t \)

Rearranging the equation:

\[ 191919t - 151515t = 404040 - 121212 \]

This simplifies to:

\[ 40304t = 282828 \]

Now, we can solve for \( t \):

\[ t = \frac{282828}{40304} \]

Calculating:

\[ t = 7 \]

Final Result

Thus, both vats will have the same amount of liquid after \( \boxed{7} \) seconds.