How would you go about solving this problem?
A 2000 L tank containing 550 L of water is being filled with water at the rate of 75 L per minute from a full 1600 L tank. How long will it be before the two tanks have the same amount of water?
550 + x = 1600 -x
2x = 1050
so
525 L moves
time = 525 L / 25 L/min
I think Professor Damon means
time=525/75 min
Thank you!
To solve this problem, we need to determine the time it takes for the two tanks to have the same amount of water. Here's how we can approach it:
Step 1: Determine the rate at which the water level in the 2000 L tank is increasing.
Since the tank is being filled at a rate of 75 L per minute, the water level in the 2000 L tank increases by 75 L every minute.
Step 2: Determine the rate at which the water level in the 1600 L tank is decreasing.
Since the 1600 L tank is being emptied into the 2000 L tank, the water level in the 1600 L tank decreases at the same rate as the water level in the 2000 L tank increases, which is 75 L per minute.
Step 3: Calculate the difference in water levels between the two tanks.
At the start, the 2000 L tank contains 550 L of water, so the initial difference in water levels between the two tanks is 2000 L - 550 L = 1450 L.
Step 4: Calculate how long it takes for the two tanks to have the same amount of water.
Since the water level in both tanks changes at the same rate of 75 L per minute, we can divide the initial difference in water levels (1450 L) by the rate of change (75 L/min) to get the time in minutes it takes for them to be equal.
Time = Difference in water levels / Rate of change
Time = 1450 L / 75 L/min
Time = 19.33 minutes (approximately)
Therefore, it will take approximately 19.33 minutes for the two tanks to have the same amount of water.