There are two different vats. The first vat has 20 gallons is being filled at a rate of 3 gallons per minute. The second vat has 108 gallons and is leaking fluid at the rate of 2 gallons per minute. After how many seconds will both vats have the same amount of liquid?

Answer 1

To determine when both vats will have the same amount of liquid, we can create equations for the amount of liquid in each vat over time.

First Vat:

• Initial Amount: 20 gallons
• Filling Rate: 3 gallons per minute

Let \( t \) be the time in minutes. The amount of liquid in the first vat after \( t \) minutes can be expressed as: \[ V_1(t) = 20 + 3t \]

Second Vat:

• Initial Amount: 108 gallons
• Leaking Rate: 2 gallons per minute

The amount of liquid in the second vat after \( t \) minutes is: \[ V_2(t) = 108 - 2t \]

Setting the Amounts Equal:

To find when the amounts are equal, we set \( V_1(t) \) equal to \( V_2(t) \): \[ 20 + 3t = 108 - 2t \]

Solving for \( t \):

1. Rearranging the equation: \[ 3t + 2t = 108 - 20 \] \[ 5t = 88 \] \[ t = \frac{88}{5} = 17.6 \text{ minutes} \]

Converting Minutes to Seconds:

Since we need the answer in seconds, we convert \( 17.6 \) minutes to seconds: \[ 17.6 \text{ minutes} = 17.6 \times 60 = 1056 \text{ seconds} \]

Thus, the time at which both vats will have the same amount of liquid is: \[ \boxed{1056} \text{ seconds} \]