Question

A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of water. This process of replacing 4 liters of the liquid in the container with an equal volume of water is continued repeatedly. The smallest number of times of doing this process, after which the volume of milk in the container becomes less than that of water, is

• A. 5
• B. 7
• C. 4
• D. None of Above

Answer 1

Answer: (B)

7

Answer 2

Each time $10\%$ of the mixture is substituted with pure adulterant (water), the concentration of the mixture decreases to $90\%$ of its initial concentration.
Likewise, when a proportion $p$ of a mixture is substituted with pure adulterant, the concentration of the resulting mixture becomes $(1 - p)$ times the previous concentration.
Where, $p = \frac {4}{40} = 0.1$
Initially, the mixture contains pure milk, resulting in a concentration or strength of $100\%$ or $1$ (in terms of proportion). After undergoing the substitution process $n$ times, the concentration of milk in the mixture will be represented by the expression $1 \times (0.9)^n$.
To ensure that the volume of milk is lower than the volume of water, p should be less than $0.5$.
Consequently, we aim to find the smallest n that satisfies this condition.
$1 \times (0.9)^n<0.5$
This is possible only when the value of $n = 7$.

So, the correct option is (B): $7$