A vessel contains 60 litres of milk. 6 litres of milk is taken out and 6 litres of water is added to the vessel. Again 6 litres of milk fro…

Solution
If we are taking out 6 litres out of a 60 litre solution and replacing it with water,
=> We are replacing $\frac{1}{10}$th of the solution with water
=> Fraction of milk will become $\frac{9}{10}$th of original.
If the process is repeated 'n' times, fraction of milk will become $(\frac{9}{10})^n$ of the original.
Here, the process is done twice.
=> Final quantity of milk = $(\frac{9}{10})^2 \times 60$ = 48.6 litres
and Final quantity of water = 60 - 48.6 = 11.4 litres
$\therefore$ Required ratio = $\frac{48.6}{11.4}$ = 81 : 19

Solution
If we are taking out 6 litres out of a 60 litre solution and replacing it with water,
=> We are replacing $\frac{1}{10}$th of the solution with water
=> Fraction of milk will become $\frac{9}{10}$th of original.
If the process is repeated 'n' times, fraction of milk will become $(\frac{9}{10})^n$ of the original.
Here, the process is done twice.
=> Final quantity of milk = $(\frac{9}{10})^2 \times 60$ = 48.6 litres
and Final quantity of water = 60 - 48.6 = 11.4 litres
$\therefore$ Required ratio = $\frac{48.6}{11.4}$ = 81 : 19