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?
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.
To find out when both vats will have the same amount of liquid, we can set up an equation based on the rates at which each vat is being filled or emptied.
Let's assume t is the number of minutes it takes for both vats to have the same amount of liquid.
For the first vat, the rate of filling is 3 gallons per minute, so the amount of liquid in the first vat after t minutes will be 3t gallons.
For the second vat, the rate of leaking is 7 gallons per minute, so the amount of liquid in the second vat after t minutes will be 108 - 7t gallons.
To find when both vats have the same amount of liquid, we can set up the equation:
3t = 108 - 7t
Now, let's solve for t:
3t + 7t = 108
10t = 108
t = 10.8
Therefore, it will take approximately 10.8 minutes for both vats to have the same amount of liquid.
To convert minutes to seconds, we multiply by 60:
10.8 * 60 = 648
So, it will take approximately 648 seconds for both vats to have the same amount of liquid.
To find out when both vats will have the same amount of liquid, we can set up an equation based on the rate of change of the liquid levels in each vat.
Let's assume that after time t (in minutes), the first vat will have a total of x gallons of liquid, and the second vat will have a total of y gallons of liquid.
For the first vat being filled at a rate of 3 gallons per minute, we can say that x = 12 + 3t.
For the second vat leaking fluid at a rate of 7 gallons per minute, we can say that y = 108 - 7t.
To find the time when both vats will have the same amount of liquid, we need to solve for t when x = y.
So, we set up the equation 12 + 3t = 108 - 7t.
Simplifying the equation, we get 10t = 96.
Dividing both sides by 10, we find t = 9.6.
Since we want the answer in seconds, we multiply t by 60:
t_in_seconds = 9.6 * 60
Therefore, both vats will have the same amount of liquid after 576 seconds.