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

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Discussion Overview

The discussion revolves around a mathematical problem involving the mixing of milk and water, specifically how many times a person can repeat the process of selling half a litre of milk and mixing the remainder with half a litre of water before the milk constitutes 50% of the mixture. The focus is on mathematical reasoning and calculations related to this scenario.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant introduces the problem and requests a detailed explanation of how to solve it.
  • Another participant defines \(M_n\) as the amount of milk in the mixture after the \(n\)th transaction and provides a formula for \(M_n\) based on previous amounts of milk.
  • A similar calculation is repeated by another participant, who arrives at the same formula for \(M_n\) and suggests a specific numerical answer for the number of repetitions.
  • Further elaboration on the characteristic root and closed form of the equation is provided, leading to a derived formula for determining when the milk content drops below 50%.
  • Participants calculate the number of repetitions needed to reach a mixture that is less than 50% milk, with one participant confirming the result with a numerical approximation.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical approach and calculations, but there is no explicit consensus on the interpretation of the results or the implications of the final answer.

Contextual Notes

Some assumptions regarding the process and the mathematical steps involved may not be fully articulated, and the dependency on specific definitions of variables is present.

WMDhamnekar
MHB
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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.
 
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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:

$$M_0=40$$

$$M_1=39.5$$

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

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

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

What is the root of the characteristic equation?
 
MarkFL said:
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:

$$M_0=40$$

$$M_1=39.5$$

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

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

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

What is the root of the characteristic equation?
So the answer to this question is $\frac{\ln{(0.5)}}{\ln{(0.9875)}}=55.1044742773$
 
The characteristic root is:

$$r=\frac{2M_0-1}{2M_0}$$

And so the closed form is:

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

Now, we know:

$$M_0=c_1$$

Hence:

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

To answer the question, we now want to solve:

$$M_n=\frac{1}{2}M_0$$

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

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

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

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

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

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

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