A 100 gallon tank is filled with a salt solution containing 10 lbs of salt. Pure water is pumped in at a rate of 5 gallons per minute and pumped out at the same rate. How many minutes will it take the amount of salt to drop to 8 lbs? Round to one decimal place.

see the related questions below. They should help you set up and solve your differential equation.

If you get stuck, come back with what you have done so far.

I keep getting an negative answer. I know the answer is 4.5 only because the teacher tells us what it I just cannot figure out how to set it up. I got e^t/20 +1/20Ae^t/20 = 5e^t/20, which is way wrong

check to see how much salt is coming in and going out. Since pure water contains no salt, and the 5 gallons leaving contain 5/100 of the salt present, we have

ds/dt = 5/100 * 0 - 5/100 s
ds/dt = -1/20 s
ds/s = -1/20 dt
s = c*e^(-t/20)
Since s(0) = 10,
s = 10*e^-(t/20)

so, let's find t when s=8

10 e^(-t/20) = 8
e^(-t/20) = 0.8
-t/20 = ln0.8
t = -20 ln0.8 = 4.46

Looks like you need to review the logic that led to your original DE.

Thanks (:

I knew it had t/20 in the solution somehow. I see what I did wrong now.

To solve this problem, we need to figure out the rate at which the amount of salt in the tank is changing.

Let's break down the problem step by step:

Step 1: Find the initial concentration of salt in the tank.
The tank initially contains 10 lbs of salt in 100 gallons of solution. Thus, the initial concentration of salt in the tank is 10 lbs / 100 gallons = 0.1 lbs/gallon.

Step 2: Determine how the concentration of salt changes over time.
Since pure water is being pumped in and out at the same rate, the total volume of the solution in the tank remains constant at 100 gallons. Therefore, the concentration of salt in the tank will decrease linearly over time.

Step 3: Find the rate of change of salt concentration.
The rate at which the salt concentration changes can be calculated by finding the difference between the initial concentration and the desired concentration, and then dividing it by the time it takes to reach the desired concentration.

Difference in concentration = 0.1 lbs/gallon - 8 lbs / 100 gallons = -7.9 lbs / 100 gallons
Rate of change = Difference in concentration / time

Step 4: Calculate the time required to reach the desired concentration.
Using the given rate of change, the equation for the rate of change becomes:

Rate of change = -7.9 lbs / 100 gallons / time

By rearranging this equation, we can solve for time:

Time = -7.9 lbs / 100 gallons / (-5 gallons/minute)

Calculating the above expression gives us:
Time = 7.9 lbs / (100 gallons * 5 gallons/minute)

Simplifying further:
Time = 7.9 lbs / (500 gallons * minute/gallon)

Therefore, Time = 0.0158 minutes, or rounded to one decimal place, Time = 0.02 minutes.

Thus, it will take approximately 0.02 minutes for the amount of salt to drop to 8 lbs.