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

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