If a vessel contains \(x\) litres of wine and \(y\) litres are withdrawn and replaced with water, then \(y\) litres of the mixture is withdrawn and replaced with water, and this is continued to a total of \(n\) times, then 

\[\color{blue}{\frac{{{\rm{Final}}\;{\rm{quantity}}\;{\rm{of}}\;{\rm{wine}}\;{\rm{in}}\;{\rm{the}}\;{\rm{vessel}}}}{{{\rm{Initial}}\;{\rm{quantity}}\;{\rm{of}}\;{\rm{wine}}\;{\rm{in}}\,\,{\rm{the}}\,{\rm{vessel}}}} = {\left( {\frac{{x - y}}{x}} \right)^n}}\]

Some time this formula does not work. Suppose there is a vessel containing 100 litre pure milk, now 10 litres are withdrawn and replaced with water, again 20 litres are with drawn and replaced with water, again 30 litres are with drawn and replaced with water, now we have to calculate the final quantity of the milk in the vessel

Initially we have withdrawn 10 litres and replaced with water, this means we have withdrawn 10 % of the complete solution and replaced with water that means we have diluted the milk to 90%. Again we withdrawn 20 % of the solution and replaced with water, which means we have diluted the existing solution to 80% of the existing concentration. In the third case we have diluted the existing solution to 70%.

So the final concentration of the milk 

= (0.90)(0.80)(0.70) =0.504

Hence final quantity of milk = \(0.504 \times 100 = 50.4\) Litres 

To generalize this concept, suppose initial quantity of milk = \(a\) litres, if \({a_1},{a_2},{a_3} \ldots {a_n}\) litres of the solution are with drawn and replaced by water each time, then 

\[\color{blue}{\left[ {\frac{{{\rm{Final}}\,\,\,{\rm{quantity}}\,\,{\rm{of}}\,\,{\rm{milk}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{vessel}}}}{{{\rm{Intitial}}\,\,{\rm{quantity}}\,\,{\rm{of}}\,\,{\rm{milk}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{vessel}}}}} \right]}\]

\[\color{blue}{ = \left[ {\frac{{a - {a_1}}}{a}} \right]\,\left[ {\frac{{a - {a_2}}}{a}} \right]\,\left[ {\frac{{a - {a_3}}}{a}} \right].....\left[ {\frac{{a - {a_n}}}{a}} \right]}\]

Example 01: A container has 500 litre of pure alcohol. 100 litres of this alcohol are withdrawn and replaced by water. If this operation is repeated for another 3 times, Find the ratio of final quantities of alcohol and water in the mixture.

Solution: This is the case of repeated withdrawal and replacement,

\[\frac{{{\rm{Final}}\;{\rm{quantity}}\;{\rm{of}}\;{\rm{alcohol}}}}{{{\rm{Initial}}\;{\rm{quantity}}\;{\rm{of}}\;{\rm{alcohol}}}} = {\left( {\frac{{500 - 100}}{{500}}} \right)^4}=\frac{256}{625}\]

Hence ratio of alcohol and water = 256 : (625 – 256) = 256:369

Example 02: A container of capacity 100 liters is filled with milk. 10 liters milk is drawn out and replaced by 20 liters of water. Again 20 liters of solution is replaced by 30 liters of water and 30 liters of solution is replaced by 40 liters of water. Then what is amount of milk left in the container?    

Solution: Initially there are 100 liters of milk, 10 liters of milk is taken out and 20 liters of water is poured in 

Obviously, there will be 90 liters milk and 20 liters water. Now 20 liters of solution is taken out and 30 liters of water is poured in, then the quantity milk in the solution \(90 - \frac{{90}}{{110}} \times 20 = \frac{{90 \times 90}}{{110}}\)

Again 30 liters of solution is taken out and 40 liters of water is poured in. Thus, the volume of milk in the container is :

\(\frac{{90 \times 90}}{{110}} - \frac{{90 \times 90}}{{110}} \times \frac{{30}}{{120}} = \frac{{90 \times 90 \times 90}}{{110 \times 120}}\) = 55.23 litre

Example 03: From a container of wine, a thief has stolen 15 liters of wine and replaced it with same quantity of water. He again repeated the same process. Thus, in three attempts the ratio of wine and water becomes 216:127. Find the initial amount of wine in the container. 

Solution: Given that,

\[\frac{{{\rm{Wine}}\;{\rm{(left)}}}}{{{\rm{Water}}\;{\rm{(added)}}}} = \frac{{{\rm{216}}}}{{{\rm{127}}}}\]

\[\Rightarrow \frac{{{\rm{wine}}\,{\rm{(left)}}}}{{{\rm{wine(initial}}\,\;{\rm{amount)}}}} = \frac{{216}}{{343}}\]

Thus, \(216 x = 343 x {\left( {1 - \frac{{15}}{k}} \right)^3}\)

\(\frac{{216}}{{343}} = {\left( {\frac{6}{7}} \right)^3} = {\left( {1 - \frac{{15}}{k}} \right)^3}\)

\(\left( {1 - \frac{{15}}{k}} \right) = \frac{6}{7}\) \( \Rightarrow k = {\rm{ }}105\)

Thus, initial amount of wine was 105 litres. 

Example 04: A jar was full of alcohol. A person used to draw out 25% of the alcohol from the jar and replaced it with spirit solution. He has repeated the same process 4 times and thus there were only 567 gm of alcohol left in the jar, the rest part of the jar was filled with the solution. What was the initial amount of alcohol in the jar?