
Question 6
A tank of blood which is 50 gallons in volume is full of this red delicious substance. The man who owns this tanks want to replace the blood with ethanol. Since the tank is heavy he cannot pour all the water out. Rather he has a 1 gallon pail. Each time he pours the ethanol in the mixture displaces 1 gallon.
For example,.....
He had 50 gallons of blood.
He pours 1 gallon of ethanol and the blood displaces 1 gallon.
Now he has 49 gallons of blood and 1 gallon of ethanol.
He pours 1 gallon of ethanol now 1 gallon of the mixture displaces (not the blood, otherwise this be too easy).
And so on.....
1)Show that the man can never fully (ideally) clean the tank to ehtanol only.
2)But he would be satisfied with if the concetration of blood in the tank is at least .01%
How many pouring are required?
=WARNING=
This problem seems awesomely similar to your differencial equations rate in/rate out mixture equation. But it is not. Because this is not a continous time time. The amout of concetration is not based on the time and continuity of it going in and out but rather on the number of pouring which is a natural number and hence a discrete model.

With each pouring, 2% of the mixture is displaced.
Hence, 98% of the blood remains.
After n pourings, there are: B = 50(0.98)^n gallons of blood in the tank.
We can see that B will never equal 0 (in a finite number of pourings).
If B < 0.01%, we have: 50(0.98)^n < 0.0001 . . . at most 0.01%
Then: n = ln(0.000002) ÷ ln(0.98) = 649.5348951
Therefore, it will take 650 pourings.

Trick question!
This was the blood of a murdered Soviet whose blood was full of ethanol.

I will solve this problem in a general sense.
Let $\displaystyle V$ be the volume in whatever units.
Let $\displaystyle p$ be the volume of the pail.
Define a sequence,
$\displaystyle a_n$ as the amount of ethanol in the tank.
The initial condition is,
$\displaystyle a_0=0$.
To solve this "differencial equation lookalike" we will define a recurrence relation. To define $\displaystyle a_{n+1}$ through $\displaystyle a_n$.
$\displaystyle a_{n+1}=\mbox{Pour In}\mbox{Pour Out}$ (because of displacement)
$\displaystyle \mbox{Pour In}=p$ because you pour in a constant amount of ethanol.
$\displaystyle \mbox{Pour Out}=\frac{a_n}{V}$ because you have $\displaystyle a_n$ amount of ethanol and because it is a homogenous solution everything is divided evenly. Thus the amount displace is the one shown above.
$\displaystyle a_{n+1}=p\frac{a_n}{V}$
Thus,
$\displaystyle a_0=0$
$\displaystyle a_1=p\frac{0}{V}=p$
$\displaystyle a_2=p\frac{a_1}{V}=p\frac{p}{V}$
$\displaystyle a_3=p\frac{a_2}{V}=p\frac{p}{V}+\frac{p}{V^2}$
$\displaystyle a_4=p\frac{a_3}{V}=p\frac{p}{V}+\frac{p}{V^2}\frac{p}{V^3}$
The pattern continues...
$\displaystyle a_n=\sum_{k=0}^{n1}\frac{(1)^kp}{V^k}$ for $\displaystyle n\geq 1$
It takes a trained eye to recognize the geometric series.
Thus, we have,
$\displaystyle a_n=p\frac{1(V)^n}{1+V}$ for $\displaystyle n\geq 1$
The rest is trivial all you need to solve $\displaystyle a_n\geq K$ where $\displaystyle K$ is the amount specified in the problem.
The problem is not mine while the solution is. I took this problem from the Talmud which deals with rainwater. I did not check Soroban's solution but it is probably correct and shorter than mine.