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?
I suspect a typo. Since they start with the same amount, and the first tank fills more slowly, it will never have more than the second tank.
To solve this problem, we need to find the time it takes for the first tank to contain 7/3 times the amount of water in the second tank.
Let's assume that t is the time in minutes it takes for this to happen.
First, we need to determine the amount of water in each tank at time t.
For the first tank, the amount of water filled in t minutes is given by the equation:
Amount in first tank = 100 + (15 * t)
For the second tank, the amount of water filled in t minutes is given by the equation:
Amount in second tank = 100 + (20 * t)
Now, we can set up an equation based on the given condition:
Amount in first tank = (7/3) * Amount in second tank
Substituting the equations for the amounts in each tank, we get:
100 + (15 * t) = (7/3) * (100 + (20 * t))
To solve this equation, we can first simplify the right side:
100 + (15 * t) = (700/3) + (140/3 * t)
Next, we can multiply all terms by 3 to get rid of the fraction:
300 + 45t = 700 + 140t
Now, let's isolate the t term by subtracting 140t from both sides:
300 - 700 = 140t - 45t
-400 = 95t
Finally, we can solve for t by dividing both sides by 95:
t = -400 / 95
t ≈ -4.21
We cannot have negative time, so we discard this solution.
Therefore, there is no solution for when the first tank will contain 7/3 times the amount of water in the second tank.