There are two different vats. The first vat has 12 gallons is being filled at a rate of 3 gallons per minute. The second vat has 108 gallons and is leaking fluid at a rate of 7 gallons per minute. After how many seconds will both vats have the same amount of liquid?

Answer 1

Let's call the number of seconds it takes for both vats to have the same amount of liquid "t".

First, let's convert the rate per minute into the rate per second. There are 60 seconds in a minute:
The first vat is being filled at a rate of 3 gallons per minute, which is 3/60 = 1/20 gallons per second.
The second vat is leaking fluid at a rate of 7 gallons per minute, which is 7/60 = 7/60 gallons per second.

After t seconds, the first vat will have 12 + (1/20)t gallons of liquid.
After t seconds, the second vat will have 108 - (7/60)t gallons of liquid.

We want to find the value of t when the first vat has the same amount of liquid as the second vat:
12 + (1/20)t = 108 - (7/60)t

Let's multiply through by 60 to get rid of the fractions:
720 + 3t = 6480 - 7t

Combining like terms:
10t = 5760

Dividing both sides of the equation by 10:
t = 576

So, after 576 seconds, both vats will have the same amount of liquid.