Two pumps are being filled by two different pumps. The first vat has 8 gallons is being filled at a rate of 16 gallons per second. The second vat has 38 gallons and is being filled at a rate of 14 gallons per second. After how many seconds will both vats have the same amount of liquid?

Answer 1

Let's denote the amount of liquid in the first vat after \( t \) seconds as \( A(t) \) and the amount of liquid in the second vat as \( B(t) \).

For the first vat:

\[ A(t) = 8 + 16t \]

For the second vat:

\[ B(t) = 38 + 14t \]

We want to find the time \( t \) when both vats have the same amount of liquid, so we set \( A(t) \) equal to \( B(t) \):

\[ 8 + 16t = 38 + 14t \]

Now, we'll solve this equation for \( t \).

1. Subtract \( 14t \) from both sides:

\[ 8 + 2t = 38 \]

2. Subtract 8 from both sides:

\[ 2t = 30 \]

3. Divide both sides by 2:

\[ t = 15 \]

So, the two vats will have the same amount of liquid after \( \boxed{15} \) seconds.

To verify:

• For the first vat after 15 seconds:

\[ A(15) = 8 + 16 \times 15 = 8 + 240 = 248 \text{ gallons} \]

• For the second vat after 15 seconds:

\[ B(15) = 38 + 14 \times 15 = 38 + 210 = 248 \text{ gallons} \]

Both vats indeed have 248 gallons after 15 seconds, confirming our solution.