Question

Two pumps are filling large vats of liquid. One vat is empty, and the pump is filling it at a rate of 6 gallons per minute. The other vat already contains 28 gallons and is continuing to fill at a rate of 4 gallons per minute. When will the two vats have the same amount of liquid?

Answer 1

Let \( t \) be the time in minutes after we start filling the vats.

For the first vat, which starts empty and fills at a rate of 6 gallons per minute, the amount of liquid in the vat after \( t \) minutes is given by:

\[
A(t) = 6t
\]

For the second vat, which already contains 28 gallons and fills at a rate of 4 gallons per minute, the amount of liquid in the vat after \( t \) minutes is given by:

\[
B(t) = 28 + 4t
\]

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

\[
6t = 28 + 4t
\]

Next, we will solve for \( t \):

1. Subtract \( 4t \) from both sides:

\[
6t - 4t = 28
\]

2. Simplify the left side:

\[
2t = 28
\]

3. Divide both sides by 2:

\[
t = 14
\]

Now, we need to verify that this indeed gives both vats the same amount of liquid. We can substitute \( t = 14 \) back into the equations for the amounts of liquid in both vats.

1. For the first vat:

\[
A(14) = 6 \times 14 = 84 \text{ gallons}
\]

2. For the second vat:

\[
B(14) = 28 + 4 \times 14 = 28 + 56 = 84 \text{ gallons}
\]

Since both vats contain 84 gallons when \( t = 14 \), we conclude that the solution is verified. Therefore, the time when the two vats will have the same amount of liquid is

\[
\boxed{14} \text{ minutes}.
\]