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[SOLVED] Sequences and series

Dhamnekar Winod

Active member
Nov 17, 2018
100
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

Administrator
Staff member
Feb 24, 2012
13,736
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?
 

Dhamnekar Winod

Active member
Nov 17, 2018
100
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

Administrator
Staff member
Feb 24, 2012
13,736
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.