Solve Math Problem: Mixing Milk & Water to Get 50% Milk

[WMDhamnekar]

Hi,

A person has 40 litres of milk. As soon as he sells half a litre, he mixes the remainder with half a litre of water. How often can he repeat the process, before the amount of milk in the mixture is 50% of the whole?
Detailed explanation is appreciated.:)
Solution:

I am working on this problem. Meanwhile if any member of math help boards knows the correct answer, may reply to this question with correct answer.

 

[MarkFL]

I would begin by letting \(M_n\) be the amount of milk in the mixture (in L) after the \(n\)th step/transaction. So, we have:

\(\displaystyle M_0=40\)

\(\displaystyle M_1=39.5\)

Now, during the second transaction, we don't have 0.5 L of milk leaving, we have:

\(\displaystyle \frac{M_1}{M_0}\cdot\frac{1}{2}\) liters leaving. This suggests to me that we may state:

\(\displaystyle M_n=M_{n-1}-\frac{1}{2}\cdot\frac{M_{n-1}}{M_{0}}=M_{n-1}\left(\frac{2M_0-1}{2M_0}\right)\)

What is the root of the characteristic equation?

 

[WMDhamnekar]

I would begin by letting \(M_n\) be the amount of milk in the mixture (in L) after the \(n\)th step/transaction. So, we have:

\(\displaystyle M_0=40\)

\(\displaystyle M_1=39.5\)

Now, during the second transaction, we don't have 0.5 L of milk leaving, we have:

\(\displaystyle \frac{M_1}{M_0}\cdot\frac{1}{2}\) liters leaving. This suggests to me that we may state:

\(\displaystyle M_n=M_{n-1}-\frac{1}{2}\cdot\frac{M_{n-1}}{M_{0}}=M_{n-1}\left(\frac{2M_0-1}{2M_0}\right)\)

What is the root of the characteristic equation?

So the answer to this question is $\frac{\ln{(0.5)}}{\ln{(0.9875)}}=55.1044742773$

 

[MarkFL]

The characteristic root is:

\(\displaystyle r=\frac{2M_0-1}{2M_0}\)

And so the closed form is:

\(\displaystyle M_n=c_1\left(\frac{2M_0-1}{2M_0}\right)^n\)

Now, we know:

\(\displaystyle M_0=c_1\)

Hence:

\(\displaystyle M_n=M_0\left(\frac{2M_0-1}{2M_0}\right)^n\)

To answer the question, we now want to solve:

\(\displaystyle M_n=\frac{1}{2}M_0\)

\(\displaystyle M_0\left(\frac{2M_0-1}{2M_0}\right)^n=\frac{1}{2}M_0\)

\(\displaystyle \left(\frac{2M_0-1}{2M_0}\right)^n=\frac{1}{2}\)

\(\displaystyle n=\frac{\ln(2)}{\ln\left(\dfrac{2M_0}{2M_0-1}\right)}\)

Using \(M_0=40\), we have:

\(\displaystyle n=\frac{\ln(2)}{\ln\left(\dfrac{80}{79}\right)}\approx55.10447\quad\checkmark\)

So, we find that on the 56th repetition of the process, the mixture will be less than 50% milk.