2 water tanks are being set up.Each one starts with 100 liters of water.The first tank fills at the rate of 15 liters per min.The second tank fills at the rate of 20 liters per min.When will the first tank contain 7/3 as much as the second tank?
To find out when the first tank will contain 7/3 as much water as the second tank, we need to determine the time it takes for the first tank to reach that level.
Let's denote:
- V1 as the volume of water in the first tank
- V2 as the volume of water in the second tank
- t as the time in minutes
We know that both tanks start with 100 liters of water, so at time t = 0, we have:
V1(0) = V2(0) = 100 liters
The first tank fills at a rate of 15 liters per minute, so after time t, the volume in the first tank becomes:
V1(t) = V1(0) + 15t
Similarly, the second tank fills at a rate of 20 liters per minute, so after time t, the volume in the second tank becomes:
V2(t) = V2(0) + 20t
We are interested in finding the time when the first tank contains 7/3 times the volume of the second tank, so we can set up the equation:
V1(t) = (7/3)V2(t)
Substituting the expressions for V1(t) and V2(t) from above, we get:
100 + 15t = (7/3)(100 + 20t)
Now, let's solve this equation to find the value of t:
100 + 15t = (7/3)(100 + 20t)
Multiply both sides of the equation by 3 to get rid of the fraction:
300 + 45t = 7(100 + 20t)
300 + 45t = 700 + 140t
Collect like terms:
140t - 45t = 700 - 300
95t = 400
t = 400/95
Simplifying:
t = 4.21 (approximately)
Therefore, the first tank will contain 7/3 times the volume of the second tank after approximately 4.21 minutes.