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 30 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 \( t \) represent the number of seconds after which both vats will have the same amount of liquid.

For the first vat:

• Initial amount = 8 gallons
• Filling rate = 16 gallons per second
• Amount in first vat after \( t \) seconds = \( 8 + 16t \)

For the second vat:

• Initial amount = 30 gallons
• Filling rate = 14 gallons per second
• Amount in second vat after \( t \) seconds = \( 30 + 14t \)

We are interested in finding \( t \) when the amounts in both vats are equal:

\[ 8 + 16t = 30 + 14t \]

Now we can solve for \( t \).

1. Subtract \( 14t \) from both sides: \[ 8 + 2t = 30 \]

2. Subtract 8 from both sides: \[ 2t = 22 \]

3. Divide by 2: \[ t = 11 \]

Thus, after \( \boxed{11} \) seconds, both vats will have the same amount of liquid.

To verify, we can find the amounts in each vat after 11 seconds:

• Amount in the first vat: \[ 8 + 16 \times 11 = 8 + 176 = 184 \text{ gallons} \]

• Amount in the second vat: \[ 30 + 14 \times 11 = 30 + 154 = 184 \text{ gallons} \]

Both vats indeed have 184 gallons after 11 seconds, confirming that our solution is correct.