A vessel contains 79.99 liters mixture of milk and water in which milk is 50% more than water. 20.03 liters of mixture is taken out and 59.99 liters mixture of milk and water mixed in the remaining mixture into the vessel. If in the resultant mixture the milk becomes two times of the water, then find approximate quantity (in liters) of water added.

The correct answer is 16 (or Option 16).

Step-by-step Explanation:

1. Initial Mixture Analysis:

Total volume of initial mixture ≈ 80 liters.

It is given that milk is 50% more than water. Therefore, the ratio of milk to water is:

$\text{Ratio of Milk : Water}=150:100=3:2$

2. Mixture Removed:

Volume of mixture removed ≈ 20 liters.

Remaining mixture volume = 80 - 20 = 60 liters.

In this remaining 60 liters, milk and water remain in the ratio 3 : 2.

Quantity of milk in remaining mixture =

$\frac{3}{5}\times 60=36\text{ liters}$

Quantity of water in remaining mixture =

$\frac{2}{5}\times 60=24\text{ liters}$

3. Addition of New Mixture:

Now, 60 liters (59.99 liters ≈ 60 liters) of a new mixture containing milk and water is added.

Let the volume of water added in this new mixture be w liters.

Then, the volume of milk added in the new mixture is (60 - w) liters.

4. Final Resultant Mixture Equation:

Total Milk in final mixture = 36 + (60 - w) = 96 - w liters.

Total Water in final mixture = 24 + w liters.

According to the question, in the resultant mixture, milk is two times water:

$\text{Total Milk}=2\times \text{Total Water}$

$96-w=2(24+w)$

$96-w=48+2w$

$3w=48$

$w=16\text{ liters}$

Thus, the approximate quantity of water added is 16 liters.