There are two different vats the first that has 12 gallons and is being filled at the rate of 3 gallons per minute the second that has 100 gallons and is leaking fluid at the rate of 2 gallons per minute. How many seconds both bats have the same amount of liquid.

Answer 1

To find the time when both vats have the same amount of liquid, we can set up equations for the amount of liquid in each vat over time.

1. First vat:

   • Initial amount of liquid = 12 gallons
   • Filling rate = 3 gallons/minute
   • Amount of liquid after \( t \) minutes: \[ V_1(t) = 12 + 3t \]

2. Second vat:

   • Initial amount of liquid = 100 gallons
   • Leaking rate = 2 gallons/minute
   • Amount of liquid after \( t \) minutes: \[ V_2(t) = 100 - 2t \]

To find when both vats have the same amount of liquid, we set \( V_1(t) \) equal to \( V_2(t) \): \[ 12 + 3t = 100 - 2t \]

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

1. Combine like terms: \[ 3t + 2t = 100 - 12 \] \[ 5t = 88 \]
2. Solve for \( t \): \[ t = \frac{88}{5} = 17.6 \text{ minutes} \]

To convert minutes to seconds: \[ t = 17.6 \text{ minutes} \times 60 \text{ seconds/minute} = 1056 \text{ seconds} \]

Thus, both vats will have the same amount of liquid after 1056 seconds.